Access Physics - National Open University of Nigeria

Access Physics - National Open University of Nigeria
PHY 001
ACCESS PHYSICS
COURSE
GUIDE
PHY 001
ACCESS PHYSICS
Course Developers
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
and
Dr N Oyedun
Federal University of Technology
Minna
Course Editors
Prof C Utah
University of Jos, Jos
and
Dr J O Oludotun
Distance Learning Institute
University of Lagos, Lagos
Programme Leader
Dr M Oki
School of Science and Technology
National Open University of Nigeria
Lagos
Course Coordinator
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
NATIONAL OPEN UNIVERSITY OF NIGERIA
ii
PHY 001
ACCESS PHYSICS
National Open University of Nigeria
Headquarters
14/16 Ahmadu Bello Way
Victoria Island
Lagos
Abuja office
No. 5 Dar es Salaam Street
Off Aminu Kano Crescent
Wuse II, Abuja
Nigeria
e-mail: [email protected]
URL: www.nou.edu.ng
Published by
National Open University of Nigeria
Printed 2008
ISBN: 978-058-593-1
All Rights Reserved
iii
PHY 001
CONTENTS
ACCESS PHYSICS
PAGE
Introduction ………………………………………………..
What You will Learn in this Course …………………………
Course Aims ……………………………………………….
Course Objectives ………………………………………….
Working through the Course ……………………………….
Course Materials ……………………………………………
Study Units ………………………………………………….
Textbooks and References…………………………………..
Assignment File……………………………………………..
Assessment …………………………………………………
Tutor-Marked Assignment ……………...…………………..
Final Examination and Grading ……………….…………....
Summary…………………………………………………….
1
1
2
2
3
3
4
5
5
5
6
6
6
Introduction
PHY 001: Access Physics is a two-semester, 6 credit-hour foundation
(remedial) course in Physics. It is a remedial course offered to students
admitted to the School of Science and Technology who are deficient in
Physics at the Senior Secondary School level and for whom Physics is
compulsory to their programme of study. The course will be done in the
first semester simultaneously with other listed courses in the programme
on offer. Before registration into second semester is allowed, the student
must have remedied the Access Physics course. The examination grade
obtained in the remedial course does not count in the GPA of the student
at the end of the semester.
As a foundation course meant for students who are deficient in Physics
the communicative approach is used. The material has been developed
to stimulate and sustain interest in the study of Physics in Nigeria. This
is done in recognition that most students dread Physics; resulting in poor
performance and an increasing low enrolment in Physics in tertiary
institutions in Nigeria despite the desire for technological development.
No compulsory prerequisite is required for this course. It is, however,
essential that you revise the Physics topics of the JSS Integrated Science
Course to acquaint yourself with what to expect in the related concepts
and principles from the course.
The course guide tells you briefly what the course is all about, what
course materials you will be using and how you can work your way
through these materials. It gives you some guidance on your TutorMarked Assignments. You will have detailed information on your
Tutor-Marked Assignments in the separate Assignment file. There is/are
self-assessment exercise(s) within the body of a unit and/or at the end of
iv
PHY 001
ACCESS PHYSICS
each unit. The exercise(s) is/are an overview of the unit to help you
assess yourself at the end of every unit.
What You will Learn in this Course
This course contains several modules. These modules cover the scope
and depth of the Physics content level stipulated for the Senior
Secondary School programme in conformity with the JCC decision.
Mathematics is used to clarify the Physics concepts or principles under
consideration as much as possible.
Physics phenomena, facts, laws, definitions, concepts, theories and skills
appropriate to Physics permeate the entire modules. The modules dealt
with scientific vocabularies, terminologies and conventions including
symbols, quantities and units. They described broadly the use of
scientific apparatus, including techniques of operation and aspects of
safety. They further make you familiar with scientific and technological
applications with their social, economic and environmental applications.
In addition, they expose you to information handling, experimental and
problem -solving techniques.
You will at the end of this course not only have remedied your
deficiency in Physics at the Senior Secondary School level but also have
laid a solid foundation for further studies in Physics and Physics related
courses. I believe you will find the course interesting and motivating.
The Course PHY 001
Course Aims
The aim of this course is to enable you remedy your deficiency in
Physics at the
Senior Secondary School level. It will help you to lay a solid foundation
in Physics on which you will build in your programme of study in the
School of Science and Technology. Therefore, the aims of the
foundation course in Physics are to enable you:
(i)
(ii)
(iii)
(iv)
acquired proper understanding of the basic
principles and applications of Physics;
develop scientific skills and attitudes as prerequisites for further scientific activities;
recognize the usefulness and limitations of
scientific method and to appreciate its applicability
in other disciplines an in everyday life;
develop abilities, attitudes and skills that encourage
efficient and safe practices;
v
PHY 001
ACCESS PHYSICS
(v)
(vi)
develop attitudes relevant to science such as
concern for accuracy and
precision, objectivity, integrity, initiative and
inventiveness.
Course Objectives
In addition to the aims, the course sets overall objectives which must be
achieved. In addition to the course objectives, each of the units has it
own specific objectives. You are advised to read properly the specific
objectives for each unit at the beginning of that unit. This will help you
to ensure that you achieved the objectives. As you go through each unit,
you should from time to time go back to these objectives to ascertain the
level at which you have progressed.
By the time you have finish going through this course, you should have
satisfied these objectives of the Access Physics Course which are to:
•
•
•
•
provide you with basic literacy in Physics for functional living in
the society;
enable you acquire basic concepts and principles of Physics as a
presentation for further studies:
enable you acquire essential skills and attitude as a preparation
for the technological application of Physics;
stimulate you and enhance creativity.
Working through the Course
In order to get the maximum benefit from this Access Course, you
would be required to do all that has been stipulated in the Course. That
is, you are to STUDY the course units, read the recommended reference
textbooks and do all the unit(s) self-assessment exercise(s). At some
points in the course you are required to submit assignments for
assessment purposes. There is a final examination at the end of the
course. This is a 6-credit unit course. As such, it will take you about 40
weeks to complete. One unit requires a total of eight hours (8 hours)
study in a week. You will find has listed below all the components of
the course.
Course Materials
You will be provided with the following materials:
(i)
(ii)
(iii)
(iv)
vi
Course Guide
Study Units in four Modules
Assignment File
Presentation Schedule
PHY 001
ACCESS PHYSICS
In addition, the course comes with a list of recommended textbooks
which though are not compulsory for you to acquire, are necessary for
you to use as complements to the study units. It is advised that you
acquire some of these textbooks and read them to broaden your scope of
understanding and as means of gradual build up of your own library.
Study Units
The following are the units contained in this course:
Module 1
Motion
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Units and Dimensions, Scalars and Vectors.
Molecular Theory of Matter
Motion
Motion under Gravity
Simple Harmonic Motion (S. H. M)
Module 2
Forces
Unit 1
Unit 2
Unit 3
Unit 4
Forces, Centre of Gravity and Equilibrium
Friction in Solids and Liquids
Linear Momentum
Simple Machines
Module 3
Mechanical and Heat Energies
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Mechanical Energy
Elastic Properties of Solids
Effects of Heat
Measurement of Temperature
Transfer of Heat and Heat Capacities
Latent Heat and Evaporation
Expansion of Gases
Module 4
Light Energy
Unit 1
Reflection of Light
vii
PHY 001
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
ACCESS PHYSICS
Reflection at Curved (Spherical) Mirrors
Refraction of Light
Refraction at Curved Surfaces (Lenses)
Applications of Light Waves
Dispersion of Light and Colours
Textbooks and References
The following are the list of textbooks you will need to reference as the
course progresses.
Anyakoha, M. W. (2000). New School Physics for Senior Secondary
Schools. Onitsha: Africana FEP.
Awe, Olumuyiwa and Okunola, O. O. (1992). Comprehensive
Certificate Physics. SSCE Edition. Ibadan: University Press.
Nelkon, M. and Parker, P. (1995). Advanced Level Physics. 7th ed.
lbadan: Heinemann.
Nelkon, M. (2000). Principles of Physics for Senior Secondary Schools.
Ibadan: Longman.
Okpala, P. N. (1990). Physics. Certificate Year Series for Senior
Secondary School. Ibadan: NPS Educational.
Ravi, K; George, K. O. and Tay Chen Hui (1991). New School Physics.
Certificate Science Series. Singapore: FEP International.
Assignment File
The Assignment File will be posted to you in due course. In this file,
you will find all the details of the work you must submit to your tutor
for marking. The marks you obtain for these assignments will count
towards the final mark you obtain for this course. More information on
assignments will be found in the Assignment File itself and in the
section on assessment in this Course Guide.
viii
PHY 001
ACCESS PHYSICS
Assessment
There are three components of assessment for this course. These are:
•
•
•
The self-assessment questions (SAQs) or Exercises
The Tutor-Marked Assignments(TMAs)
The End of Course Examination.
The exercises within each unit of the modules are meant to probe your
understanding of the concepts in the unit. It is non-grading and as such
does not add up to your final grade in the course.
Tutor-Marked Assignment
There are some tutor-marked assignments. You must compulsorily
answer ALL OF THESE and submit them for grading. These
assignments are in set of five batteries. Each battery is scored over
20marks. Your best four batteries of assignments will be selected and
used for the final grading. The TMAs constitute 40% of your final grade
in this course.
Final Examination and Grading
At the end of the course you will now sit for the final examination which
will be of three hours duration. This examination will account for 60%
of the total course grade. The examination will consist of questions
which reflect the types of practice exercises (Self Assessment
Questions) and Tutor-Marked Assignments you have previously treated.
It will also reflect all the basic concepts you would have learnt through
the duration of the course. All areas of the course will be covered in the
assessment. This component completes the final part of your grade in
this course.
Summary
PHY 001 intends to help you remedy your deficiency in Physics at the
Senior Secondary School level and to build a solid foundation in
Physics. It is meant to be a scaffolding upon which you would build the
relevant related courses in the course of your programme of study in the
School of Science and Technology. Upon the completion of this course,
you are supposed to have laid that solid foundation that is expected of
you. Remember that your continuing in your programme in this School
depends on your successful performance in this course. That is you must
ix
PHY 001
ACCESS PHYSICS
have remedied your deficiency in this course. You are therefore
encouraged to use the time between finishing the last unit and sitting for
the examination to revise the whole course.
COURSE
GUIDE
PHY 001
ACCESS PHYSICS
Course Developers
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
and
Dr N Oyedun
Federal University of Technology
Minna
Course Editors
Prof C Utah
University of Jos, Jos
and
Dr J O Oludotun
Distance Learning Institute
University of Lagos, Lagos
Programme Leader
Dr M Oki
School of Science and Technology
National Open University of Nigeria
Lagos
Course Coordinator
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
x
PHY 001
ACCESS PHYSICS
NATIONAL OPEN UNIVERSITY OF NIGERIA
xi
PHY 001
National Open University of Nigeria
Headquarters
14/16 Ahmadu Bello Way
Victoria Island
Lagos
Abuja office
No. 5 Dar es Salaam Street
Off Aminu Kano Crescent
Wuse II, Abuja
Nigeria
e-mail: [email protected]
URL: www.nou.edu.ng
Published by
National Open University of Nigeria
Printed 2008
ISBN: 978-058-593-1
All Rights Reserved
xii
ACCESS PHYSICS
PHY 001
CONTENTS
ACCESS PHYSICS
PAGE
Introduction ………………………………………………..
What You will Learn in this Course …………………………
Course Aims ……………………………………………….
Course Objectives ………………………………………….
Working through the Course ……………………………….
Course Materials ……………………………………………
Study Units ………………………………………………….
Textbooks and References…………………………………..
Assignment File……………………………………………..
Assessment …………………………………………………
Tutor-Marked Assignment ……………...…………………..
Final Examination and Grading ……………….…………....
Summary…………………………………………………….
1
1
2
2
3
3
4
5
5
5
6
6
6
Introduction
PHY 001: Access Physics is a two-semester, 6 credit-hour foundation
(remedial) course in Physics. It is a remedial course offered to students
admitted to the School of Science and Technology who are deficient in
Physics at the Senior Secondary School level and for whom Physics is
compulsory to their programme of study. The course will be done in the
first semester simultaneously with other listed courses in the programme
on offer. Before registration into second semester is allowed, the student
must have remedied the Access Physics course. The examination grade
obtained in the remedial course does not count in the GPA of the student
at the end of the semester.
As a foundation course meant for students who are deficient in Physics
the communicative approach is used. The material has been developed
to stimulate and sustain interest in the study of Physics in Nigeria. This
is done in recognition that most students dread Physics; resulting in poor
performance and an increasing low enrolment in Physics in tertiary
institutions in Nigeria despite the desire for technological development.
No compulsory prerequisite is required for this course. It is, however,
essential that you revise the Physics topics of the JSS Integrated Science
Course to acquaint yourself with what to expect in the related concepts
and principles from the course.
The course guide tells you briefly what the course is all about, what
course materials you will be using and how you can work your way
through these materials. It gives you some guidance on your TutorMarked Assignments. You will have detailed information on your
Tutor-Marked Assignments in the separate Assignment file. There is/are
self-assessment exercise(s) within the body of a unit and/or at the end of
xiii
PHY 001
ACCESS PHYSICS
each unit. The exercise(s) is/are an overview of the unit to help you
assess yourself at the end of every unit.
What You will Learn in this Course
This course contains several modules. These modules cover the scope
and depth of the Physics content level stipulated for the Senior
Secondary School programme in conformity with the JCC decision.
Mathematics is used to clarify the Physics concepts or principles under
consideration as much as possible.
Physics phenomena, facts, laws, definitions, concepts, theories and skills
appropriate to Physics permeate the entire modules. The modules dealt
with scientific vocabularies, terminologies and conventions including
symbols, quantities and units. They described broadly the use of
scientific apparatus, including techniques of operation and aspects of
safety. They further make you familiar with scientific and technological
applications with their social, economic and environmental applications.
In addition, they expose you to information handling, experimental and
problem -solving techniques.
You will at the end of this course not only have remedied your
deficiency in Physics at the Senior Secondary School level but also have
laid a solid foundation for further studies in Physics and Physics related
courses. I believe you will find the course interesting and motivating.
The Course PHY 001
Course Aims
The aim of this course is to enable you remedy your deficiency in
Physics at the
Senior Secondary School level. It will help you to lay a solid foundation
in Physics on which you will build in your programme of study in the
School of Science and Technology. Therefore, the aims of the
foundation course in Physics are to enable you:
(vii)
acquired proper understanding of the basic
principles and applications of Physics;
(viii) develop scientific skills and attitudes as prerequisites for further scientific activities;
(ix)
recognize the usefulness and limitations of
scientific method and to appreciate its applicability
in other disciplines an in everyday life;
(x)
develop abilities, attitudes and skills that encourage
efficient and safe practices;
xiv
PHY 001
ACCESS PHYSICS
(xi)
(xii)
develop attitudes relevant to science such as
concern for accuracy and
precision, objectivity, integrity, initiative and
inventiveness.
Course Objectives
In addition to the aims, the course sets overall objectives which must be
achieved. In addition to the course objectives, each of the units has it
own specific objectives. You are advised to read properly the specific
objectives for each unit at the beginning of that unit. This will help you
to ensure that you achieved the objectives. As you go through each unit,
you should from time to time go back to these objectives to ascertain the
level at which you have progressed.
By the time you have finish going through this course, you should have
satisfied these objectives of the Access Physics Course which are to:
•
•
•
•
provide you with basic literacy in Physics for functional living in
the society;
enable you acquire basic concepts and principles of Physics as a
presentation for further studies:
enable you acquire essential skills and attitude as a preparation
for the technological application of Physics;
stimulate you and enhance creativity.
Working through the Course
In order to get the maximum benefit from this Access Course, you
would be required to do all that has been stipulated in the Course. That
is, you are to STUDY the course units, read the recommended reference
textbooks and do all the unit(s) self-assessment exercise(s). At some
points in the course you are required to submit assignments for
assessment purposes. There is a final examination at the end of the
course. This is a 6-credit unit course. As such, it will take you about 40
weeks to complete. One unit requires a total of eight hours (8 hours)
study in a week. You will find has listed below all the components of
the course.
Course Materials
You will be provided with the following materials:
(v)
Course Guide
(vi)
Study Units in four Modules
(vii) Assignment File
(viii) Presentation Schedule
xv
PHY 001
ACCESS PHYSICS
In addition, the course comes with a list of recommended textbooks
which though are not compulsory for you to acquire, are necessary for
you to use as complements to the study units. It is advised that you
acquire some of these textbooks and read them to broaden your scope of
understanding and as means of gradual build up of your own library.
Study Units
The following are the units contained in this course:
Module 1
Motion
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Units and Dimensions, Scalars and Vectors.
Molecular Theory of Matter
Motion
Motion under Gravity
Simple Harmonic Motion (S. H. M)
Module 2
Forces
Unit 1
Unit 2
Unit 3
Unit 4
Forces, Centre of Gravity and Equilibrium
Friction in Solids and Liquids
Linear Momentum
Simple Machines
Module 3
Mechanical and Heat Energies
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Mechanical Energy
Elastic Properties of Solids
Effects of Heat
Measurement of Temperature
Transfer of Heat and Heat Capacities
Latent Heat and Evaporation
Expansion of Gases
Module 4
Light Energy
Unit 1
Reflection of Light
xvi
PHY 001
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
ACCESS PHYSICS
Reflection at Curved (Spherical) Mirrors
Refraction of Light
Refraction at Curved Surfaces (Lenses)
Applications of Light Waves
Dispersion of Light and Colours
Textbooks and References
The following are the list of textbooks you will need to reference as the
course progresses.
Anyakoha, M. W. (2000). New School Physics for Senior Secondary
Schools. Onitsha: Africana FEP.
Awe, Olumuyiwa and Okunola, O. O. (1992). Comprehensive
Certificate Physics. SSCE Edition. Ibadan: University Press.
Nelkon, M. and Parker, P. (1995). Advanced Level Physics. 7th ed.
lbadan: Heinemann.
Nelkon, M. (2000). Principles of Physics for Senior Secondary Schools.
Ibadan: Longman.
Okpala, P. N. (1990). Physics. Certificate Year Series for Senior
Secondary School. Ibadan: NPS Educational.
Ravi, K; George, K. O. and Tay Chen Hui (1991). New School Physics.
Certificate Science Series. Singapore: FEP International.
Assignment File
The Assignment File will be posted to you in due course. In this file,
you will find all the details of the work you must submit to your tutor
for marking. The marks you obtain for these assignments will count
towards the final mark you obtain for this course. More information on
assignments will be found in the Assignment File itself and in the
section on assessment in this Course Guide.
xvii
PHY 001
ACCESS PHYSICS
Assessment
There are three components of assessment for this course. These are:
•
•
•
The self-assessment questions (SAQs) or Exercises
The Tutor-Marked Assignments(TMAs)
The End of Course Examination.
The exercises within each unit of the modules are meant to probe your
understanding of the concepts in the unit. It is non-grading and as such
does not add up to your final grade in the course.
Tutor-Marked Assignment
There are some tutor-marked assignments. You must compulsorily
answer ALL OF THESE and submit them for grading. These
assignments are in set of five batteries. Each battery is scored over
20marks. Your best four batteries of assignments will be selected and
used for the final grading. The TMAs constitute 40% of your final grade
in this course.
Final Examination and Grading
At the end of the course you will now sit for the final examination which
will be of three hours duration. This examination will account for 60%
of the total course grade. The examination will consist of questions
which reflect the types of practice exercises (Self Assessment
Questions) and Tutor-Marked Assignments you have previously treated.
It will also reflect all the basic concepts you would have learnt through
the duration of the course. All areas of the course will be covered in the
assessment. This component completes the final part of your grade in
this course.
Summary
PHY 001 intends to help you remedy your deficiency in Physics at the
Senior Secondary School level and to build a solid foundation in
Physics. It is meant to be a scaffolding upon which you would build the
relevant related courses in the course of your programme of study in the
School of Science and Technology. Upon the completion of this course,
you are supposed to have laid that solid foundation that is expected of
you. Remember that your continuing in your programme in this School
depends on your successful performance in this course. That is you must
xviii
PHY 001
ACCESS PHYSICS
have remedied your deficiency in this course. You are therefore
encouraged to use the time between finishing the last unit and sitting for
the examination to revise the whole course.
xix
PHY 001
ACCESS PHYSICS
Course Code
PHY 001
Course Title
Access Physics
Course Developers
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
and
Dr N Oyedun
Federal University of Technology
Minna
Course Editors
Prof C Utah
University of Jos, Jos
and
Dr J O Oludotun
Distance Learning Institute
University of Lagos, Lagos
Programme Leader
Dr M Oki
School of Science and Technology
National Open University of Nigeria
Lagos
Course Coordinator
Dr (Mrs) Charity Okonkwo
School of Science and Technology
National Open University of Nigeria
Lagos
.
NATIONAL OPEN UNIVERSITY OF NIGERIA
xx
PHY 001
ACCESS PHYSICS
National Open University of Nigeria
Headquarters
14/16 Ahmadu Bello Way
Victoria Island
Lagos
Abuja office
No. 5 Dar es Salaam Street
Off Aminu Kano Crescent
Wuse II, Abuja
Nigeria
e-mail: [email protected]
URL: www.nou.edu.ng
Published by
National Open University of Nigeria
Printed 2008
ISBN: 978-058-593-1
All Rights Reserved
xxi
PHY 001
ACCESS PHYSICS
CONTENTS
PAGE
Module 1
……………………………………………………
1
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Motion Units and Dimensions, Fundamental
and Derived Quantities and Units, Scalars
and Vectors ……………………………………….
Molecular Theory of Matter ………………………
Motion in a Straight Line …………………………
Motion under Gravity …………………………….
Simple Harmonic Motion (SHM) …………………
1
21
46
68
85
Module 2
…………………………………………………….
107
Unit 1
Unit 2
Unit 3
Unit 4
Force, Centre of Gravity and Equilibrium ……….
Friction in Solids and Liquids …………………….
Linear Momentum …………………………………
Simple Machines ………………………………….
107
147
165
187
Module 3
…………………………………………………….
209
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Mechanical Energy ……………………………….
Elastic Properties of Solids ………………………
Elastic Properties of Solids ……………………….
Measurement of Temperature …………………….
Transfer of Heat and Heat Capacities …………….
Latent Heat and Evaporation ……………………..
Expansion of Gases ………………………………
209
228
242
260
285
308
339
Module 4
…………………………………………………….
369
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Reflection of Light ……………………………….. 369
Reflection at Curved (Spherical) Mirrors …………. 395
Refraction of Light ……………………………….. 413
Refraction at Curved Surfaces (Lenses) ……………440
Applications of Light Waves ……………………….464
Dispersion of Light and Colours ……………………485
xxii
PHY 001
ACCESS PHYSICS
MODULE 1
Unit 1
Motion Units and Dimensions, Fundamental and Derived
Quantities and Units, Scalars and Vectors
Molecular Theory of Matter
Motion in a Straight Line
Motion under Gravity
Simple Harmonic Motion (SHM)
Unit 2
Unit 3
Unit 4
Unit 5
UNIT 1
MOTION
UNITS
AND
DIMENSIONS,
FUNDAMENTAL AND DERIVED QUANTITIES
AND UNITS, SCALARS AND VECTORS
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Units and Dimensions
3.1.1 Units
3.1.2 Dimensions
3.2
Fundamental and Derived Quantities and their Units
3.2.1 Fundamental Quantities and their Units
3.2.2 Derived Quantities and their Units
3.3
Scalars and Vectors
3.3.1 Concept of Scalars
3.3.2 Concept of Vectors
3.3.3 Distinction between Scalars and Vectors
3.3.4 Vector Representation
3.4
Addition of Scalars and Vectors
3.4.1 Addition of Scalars
3.4.2 Addition of Vectors
3.4.3 Addition of Vectors Inclined at an Angle
3.4.3.1Using Pythagoras Rule
3.4.3.2Using Scale Drawing
3.4.3.3 The Parallelogram Method
3.4.3.4 The Triangle Method
3.5
Resolution of Vectors
3.6
The Resultant of More than Two Vectors
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1
PHY 001
1.0
ACCESS PHYSICS
INTRODUCTION
This section is a foundation component of Physics and will introduce
you to the concepts of Units, Dimensions, Fundamental and Derived
Quantities and their Units as well as Scalars and Vectors. It permeates
the entire course and you will come across and in fact use the concepts
in subsequent sections.
2.0
OBJECTIVES
By the end of this unit, you should be able to:
•
•
•
•
•
discuss the significance of units in physical measurements
distinguish between fundamental and derived quantities
determine the dimension of at least 5 physical quantities
explain the concepts of scalars and vectors using specific
examples
resolve a vector in a given direction
resolve any number of vectors into two components at right
angles to each other
solve simple problems involving vectors using various methods.
3.0
MAIN CONTENT
3.1
Units and Dimensions
•
•
3.1.1 Units
A unit has to be defined before any kind of measurement can be made.
You are aware that different systems of units have been used in the past.
However, the system which has gained universal acceptance is the
System International d' Unites usually called S. I. Units (adopted in
1960). The S.1 Unit is based on the metre as the unit of length; the
kilogram (kg) as the Unit of Mass; the second (s) as the Unit of Time,
the ampere (A) the unit of current and the Kelion (K) the unit of
temperature. Other Units are given in Tables 1.1 and 1.2.
3.1.2 Dimensions
We often use the term dimension in Physics to describe the relationship
between a physical quantity and the fundamental quantities expressed in
terms of the symbols M, L, T of the fundamental quantities mass, length
and time respectively. You should note that physical quantities can
either be dimensional or dimensionless.
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•
Dimensionless Quantities do not depend on the system of unit in
which they are measured. That simply means that they have no
units. Examples are angles and their trigonometric ratios (ratio
of two length); relative density (or specific gravity), (ratio of two
densities); efficiency of a machine (ratio of two quantities of
work).
•
Dimensional Quantities on the other hand depend on the
magnitude of the fundamental units in which they are measured
and are different in different unit systems. If you now consider
that the fundamental quantities are denoted by the symbols M, L,
and T, you will be able to determine (derive) the dimensions of
Physical quantities by your careful study of these illustrations.
i)
ii)
Area (length x breadth) has dimension of L x L = L2
Volume (length x breath x depth) has dimensions of L x L x L =
L3
Velocity (distance ÷ time) has dimension of L ÷ T = LT-1
Density (mass ÷ volume) has dimensions of M ÷ L2 = ML-2
Acceleration (velocity ÷ time) has dimensions of
L ÷ T2 = LT-2
Momentum (mass x velocity) has dimensions of
(M x L) ÷ T = MLT-1
Pressure (force ÷ area) = (mass x acceleration) ÷ area has
dimension of MLT-2 ÷ L2 = MT-2 L-1
iii)
iv)
v)
vi)
vii)
SELF ASSESSMENT EXERCISE 1
Determine the dimensions of force, work, surface tension, (force per
unit length) and power (rate of doing work).
3.2
Fundamental and Derived Quantities and their Units
3.2.1 Fundamental Quantities and their Units
You may have observed that there are two types of physical quantities.
One group of such quantities keeps occurring again and again. They are
independent of others and cannot be defined in terms of other quantities.
You may also have noticed that most other quantities depend on them.
This group of quantities are called fundamental quantities.
•
•
Fundamental quantities are the basic quantities upon which other
quantities depend
Fundamental units are the basic units upon which other units
depend. They are the units of the fundamental quantities.
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You will now study Table 1.1 below showing the fundamental
quantities and their units.
Table 1.1:
Examples of Fundamental Quantities and their Units
Quantity
Length
Time
Mass
Electric Current
Temperature
Luminous Intensity
Amount of substance
Unit
Metre
Second
Kilogram
Ampere
Kelvin
Candela
Mole
Unit abbreviation
m
s
kg
A
K
cd
mol
The unit abbreviation is a capital letter when it refers to the name of a
person e.g. Ampere (A) or Kelvin (K).
•
Can you think of any other quantity that would belong to the
above? Give reasons for your answer.
3.2.2 Derived Quantities and their Units
At this point you may have realised that all the quantities with the
exception of the seven fundamental quantities belong to this group.
•
•
•
Derived quantities are those quantities derived by some simple
combination of the fundamental quantities.
Derived units are the units of the derived quantities.
Example of derived quantities, their derivation and their units are
summarised in Table 1.2.
Table 1.2:
4
Examples of Derived Quantities and their Units
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Derived Quantity
Area (A)
Volume (V)
Density
Velocity (V)
Acceleration (a)
Force (F)
Energy or Work (W)
Power (P)
Momentum
Pressure (P)
Frequency (f)
Electric charge
Derivation
Length x breadth
Length x breadth x height
mass
volume
displacement
time
change in velocity
time
Mass x acceleration
Force x distance
work
time
Mass x velocity
force
area
number of oscillation
time
∫ idt
Derived Unit
m2
m3
kgm-3
ms-1
MS -2
Newton; N
Joule, J (Nm)
Watt, W (J S-1)
kg. m. s -1, Ns
N m-2
Pascal, Pa
Per second or s -1
(Hertz, HZ)
Coulomb
(c)
Volt (V)
Electric
potential Work/charge
difference
Elector motive force
Work/charge
Volt (V)
Electric resistance
Electric
potential Ohm ( Ω )
difference/current
Electric capacitance
Charge/Volt
Farad (F)
•
Supplementary Units
You will notice that the Units of plane angle, the radian and of solid
angle, the steradian are not classified as fundamental or derived units.
They are sometimes called supplementary units.
•
For many measurements, you will realise that the units may be
too big or too small and so multiples and submultiples of the
basic units are used. These are formed by using the following
prefixes shown in Table 1.3.
Table 1.3:
Multiple and Submultiples of Units with their Prefixes
and Symbols
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Submultiples
Example
-1
0.1 or 10
decimetre = 10-1 m
-2
0.01 or 10
centimetre = 10-2 m
0.001 or 10-3
millimetre = 10-3 m
-6
0.000001 or 10
microfarad = 10-6 f
0.000000001 or 10-7
nanosecond
0.000000000001 or picoampere
Multiples
Example
Prefix
deci
centi
milli
micro
nano
pico
Prefix
Symbol
d
c
m
m
n
p
Symbol
101
102
1000 or 103
1,000,000 or 106
1,000,000,000 or 109
1,000,000,000,000
12
or10
deca
hector
kilo
mega
giga
Da
H
K
M
G
tera
T
decametre = 101 m
hectormetre= 102m
kilometre = 103 m
megawatts = 106W
gigahert tiga
SELF ASSESSMENT EXERCISE 2
Distinguish between:
a.
b.
Fundamental and derived quantities; and
Fundamental and derived units.
3.3
Scalars and Vectors
3.3.1 Concept of Scalars
Scalars quantities are those quantities which have only magnitude or
numerical value but no direction, examples are length or distance, mass,
volume, density, work, time, speed, temperature and energy.
3.3.2 Concept of Vectors
Though we know that many measurable physical, quantities have
magnitude or numerical value as well as direction, such quantities are
not completely described unless their magnitudes and directions are
specified. These groups of quantities are known as vector quantities.
Examples are: Weight, displacement, velocity, acceleration, force,
momentum, electric field etc.
Vector quantities are those quantities which have both magnitude (and
size) and direction.
3.3.3 Distinction between Scalars and Vectors
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You can distinguish between scalars and vectors by noting that:
•
•
•
whereas scalars have only magnitude or numerical value, vectors
have both magnitude and direction;
a scalar can be described completely by indicating only its
magnitude whereas a vector can only be specified completely
when both its magnitude and direction are included; and
Scalars add just like ordinary numbers but the addition of vectors
must be carried out with due regard to their direction.
3.3.4 Vector Representation
Here we consider the various ways of representing a vector.
•
In a diagram, a vector is represented by an arrow. The arrow is
drawn in the direction of the vector. The arrow head gives the
sense of direction. Using a scale, the length of the arrow is
chosen to be operational to the magnitude of the vector.
O
→
A
→
B
O
→
C
O
Fig. 1.1: Diagrammatic representation of vectors
•
In print, a vector is represented by a bold face symbol such as A and
V. The magnitude of the vector A is A or  A
• In writing, it is usual to put an arrow over the symbol to denote
a vector quantity e.g., → A, → B, → C, in Fig. 1.1
3.4
Addition and Subtraction of Scalars and Vectors
3.4.1 Addition and Subtraction of Scalars
You have been informed that the process of adding scalars is straight
forward. They add just like ordinary numbers by ordinary algebraic
methods.
Examples:
(i)
(ii)
5 cm3 + 7 cm3 = 12 cm3;
5 kg - 2 kg = 3 kg
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10km/h
5 km/h
5 km/h
10 km/h
15km/h
Fig. 1.2: Scalar addition
3.4.2 Addition and Subtraction of Vectors
Unlike the scalars, in adding or subtracting vectors, you have to consider
both sizes and directions of the vector quantities under consideration.
• Resultant vector
A single vector which has the same effect as two or more vectors acting
in the same direction is called the resultant vector. Addition of vectors
gives rise to a resultant vector.
Examples
Consider the following examples
i)
ii)
Vectors acting in a straight line
Vectors acting in the same direction
Suppose a man walks 2 km due East and his wife walks 3 km due East.
What is the resultant vector (velocity)?
Solution
Let us represent the magnitude by the scale I cm = l km; and direction
yan arrow head. Therefore, the man covering the distance of 2 km and
his wife 3 km both due East would be represented thus:
A
Man
2 cm
B
his wife
3 cm
The resultant vector (R) would take into consideration the magnitude
and directions of both journeys as shown below:
→
A
8
2cm
(2km)
+
→
C
→
B
3 cm
(3km)
=
5 cm
= 5km due East
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Fig. 1.3: Resultant vector (in one direction)
b)
Vectors acting in opposite directions
Suppose the man in our example above walks due East
while his wife walks due West, the resultant vector would
then be as shown in the diagram below:
→
A
+
2 cm
(2km)
→
B
3 cm
(3km)
→
R
1 cm
= 1 km due west
Fig. 1.4: Resultant vector (in opposite direction)
3.4.3 Vectors Inclined at an Angle
Let us consider a man starting a journey from a point A and walks 3 km
due North, then he turns and walks 5 km due East. What would be his
Resultant Vector (displacement)?
•
To find the resultant vector, you would either employ the
Pythagoras rule (because the vectors are at right angle to each
other) or by scale drawing.
Solution
(i)
Using Pythagoras rule
N
α
A
Scale l cm: l km
Fig. 1.5: Resultant vector using the Pythagoras rule
•
The magnitude of the resultant is given by:
→
→
→
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A E 2= A N2+NE2
∴ R = √ 32+52 = √ 9+25
=
•
34 = 5.83 km
To obtain the direction (angle ) we use the relation
opp
5
tan  =
= 1.667
Adj
3
 = tan -1 1.667 = 59.04o
•
The result of this calculation implies that the resultant vector is
5.83km, North 59.04° East.
(ii)
Using scale drawing
Here you should use a scale of say l cm to represent l km to find
resultant vector following the directives given below:
•
•
•
•
draw 3 cm due North to represent 3 km due North;
draw 5 cm due East from the head of 3 km due North to represent
the 5 km;
join the tail of 3 kmN to the head of 5 kmE;
using your ruler, measure the distance AC and convert the result
by the scale used; and find the direction by measuring the angle
∝ using your protractor.
α
R = 5.8 cm x l km = 5.8 km  = 59°±1°
Fig. 1.6: Resultant vector using scale drawing
•
10
the
Note that in general there are three methods of adding or
compounding vectors inclined at angles to each other for the
purpose of finding the resultant. These are:
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(i)
(ii)
(iii)
the parallelogram method;
the analytical method.
the triangle method.
(i)
The Parallelogram method
In this method you will determine the resultant of two vectors inclined
to each other at an angle from the diagonal of a parallelogram drawn
with the two vectors as adjacent sides.
These two vectors must be drawn from a common origin.
The parallelogram law of vectors states that if two vectors are
represented in magnitude and direction by the adjacent sides of a
parallelogram, the diagonal of the parallelogram drawn from the point of
intersection of the vectors represents the resultant vector in magnitude
and direction.
Worked example
Find the resultant of two vectors 2N and 5N acting at a point 0 at an
angle of 60° to each other.
Solution
•
By Scale drawing
-
-
using a scale (1 cm = l km or any other convenient scale)
draw a horizontal line;
using your ruler, measure (mark) out 5 cm to represent the
5N;
using protractor, measure (mark) out angle 60° to the
horizontal;
draw a line to correspond with the 60° mark;
use the ruler to measure (mark) out 2 cm on the line OP to
represent the vector 2N;
complete the parallelogram by drawing PC parallel and
equal to OQ;
QC parallel and equal to AC;
join OC ;
use the ruler to measure the magnitude OC and the
protractor to measure the direction (angle ∝).
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•
The length of OC gives the magnitude while angle ∝ gives the
direction of the resultant vector.
•
Your diagram should look like this.
P
2N
(2cm)
C
R
R=
 =
0
60

0
5N (5cm)
Q
Fig. 1.7: Parallelogram method of determining R
•
The Analytical method
You can also obtain the resultant vector by Cosine rule. In this case, you
only have to draw a sketch of the parallelogram and use the Cosine rule
and Sine rule to find the direction.
•
•
Cosine Rule: R2 = OP2+ 0Q2 + 2 OP. OQ Cos 60
Sine Rule: Sin∝ = Sin B
QC
OC
•
To obtain the magnitude using Cosine Rule
P
C
2N
R
0
600
β1200

5N
Q
Fig. 1.8: The Cosine rule of obtaining the resultant
OC 2 = OP 2 +OQ 2 + 2 OP x
x Cos60°
OP 2 + OQ 2 + 20P x OQ Cos60°
OC =
=
12
OQ
22 + 52 + 2 x 2 x 5Cos60o
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=
4 + 25 + 20 x 0.0.5
=
29 + 10
R = OC = 39 = 6.25N
•
To obtain the direction using Sine rule
Sin = Sin120° = 0.8660
QC
OC
OC
Sin 120° = Sin (180 - 120°) = Sin 60°
Sin  = QC x 0.8660
OC
=
(ii)
2 x 0.8660 = 1.732 = 0.2656
6.25
6.25
-1
 = Sin 0.2656
= 15.4°
The Triangle method
You can find the resultant of two vectors A and B inclined at an angle θ
to each other by using the triangle method. To do this, you have to
follow these steps:
•
•
•
•
Starting from a point 0, draw OP (to scale) to represent the vector
A
Next, draw the second vector B i.e. PQ (to scale) by placing its
tail at the head of the first vector A, ensuring that it is inclined to
A by the angle θ, and
Finally, the head of B is joined to the tail of A to make up the
third side of the triangle OPQ. OQ represents the resultant of the
two vectors A and B.
The arrows on OP, PQ and QO follow one another round the
triangle. The angle P0Q gives the direction of the resultant with
respect to the vector A (the above is known as the triangle law of
addition of vectors)
Worked example
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Two forces 3N and 2N are inclined to each other at 30°. Find the
resultant force by the triangle method.
Solution
Let us use a scale of l cm = 1N
•
•
•
•
first, we draw a horizontal line OA = 3 cm to represent the 3N
force;
from the tip (head) of A, we measure out angle 30°;
next, we draw a line 2 cm to represent the 2N along the 30° mark
(line of inclination to OA extended at 30° mark); and
finally, we join point 0 to point B.
The expected result is as shown
Scale: 2 cm = 1N
Fig. 1.9: The triangle rule
Measure R and  to obtain the magnitude and direction of the resultant
respectively.
SELF ASSESSMENT EXERCISE 3
1.
2.
Outline the differences between Scalars and vectors
Draw vector diagrams to represent the following vector quantities
(i)
(ii)
3.
A vector of 5 km due South
A vector of I0 km on a bearing of 135°.
Find the resultant of the following vectors
(i)
5N
7N
3N
3N
2N
5N
(ii)
(iii)
Fig. 1.10: Vectors in the same and opposite directions
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4.
5.
A captain directs a ship due North at 100 km h-1, while the river is
flowing due East at 30 km h-1. What is the resultant velocity of
the ship?
Find the resultant of two vectors A and B of 3N and 4N acting at
a point 0 at an angle of 450 to each other. Use:
(i)
(ii)
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Parallelogram method
Triangle method.
Resolution of Vectors
Here we consider more than two non-collinear vectors acting at a point.
You may obtain the resultant by first reducing the system to two
perpendicular vectors with the aid of resolution of vectors and then
compounding the two vectors.
•
This is a reverse process of finding two vectors which have the
same effect as a resultant vector. The resolution of a vector F
means that we can express this vector F as the sum of two other
vectors called the components of the original vector F.
•
Component of a vector in a given direction is its effective value
in that direction. Usually, a vector is resolved into components
along mutually perpendicular directions. That is, the horizontal or
x -direction and vertical or y - direction.
•
Suppose a vector F is inclined at an angle 0 to a horizontal
component of the vector F as
Fx = F Cos θ; and the vertical component of the vector F as
Fy = F Sin θ
Y
Fy
F
Fy =Fsin θ
θ
X
Fx = FCos θ
Fig.1.11: Resolution of a vector F
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Worked example
A plane flies with a velocity of 500 km /h in a direction N30°E. Find its
effective velocity in the easterly and northerly directions.
•
You will realise that the question, simply put means to find the
two perpendicular components of the velocity.
N
N
Vy 30o V
60o
E
Fig. 1.12: Components of V
•
To obtain the velocity in the easterly direction, you will find the
horizontal component of the force using the formula VX = V Cos
θ
i.e.
•
VX
Likewise, to obtain the velocity in the northerly direction, you
will find the vertical component of the force using the formula
i.e.
3.6
= 500 Cos 600
= 500 x 0.5
= 250 kmh-1
Vy = V Sin θ
Vy = 500 Sin 60°
= 500 x 0.8660
= 433km h-1
The Resultant of more than two Vectors
Whenever you are required to find the resultant of more than two
vectors, you have to resolve each of the vectors in two perpendicular
directions as illustrated in 3.5.
•
16
The right hand or easterly direction is taken as positive while the
left hand or westerly direction is taken as negative. Also the
southerly or downward direction is taken as negative while the
northerly or upward direction is taken as positive.
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After resolving all the forces you then add all the horizontal
components x and the vertical components y. You then obtain the
resultant of x and y by the use of the Pythagoras rule.
Worked example
Determine the resultant of three concurrent forces one of 100N acting
horizontally to the right, another of 140N acting upward at an angle of
60° to the horizontal force and the third of 240N acting vertically
upward.
Solution
The following diagram shows the direction of the three forces T 240N
Y
240N
140N
600
X
100N
Fig. 1.13: Resolution of three concurrent vectors
Using the resolution of vector method, we shall now resolve the forces
into the horizontal and the vertical components as shown in the Table
1.4.
Table 1.4: Resolution of a vector
Force
Inclination
to Horizontal
the horizontal 0 Component
Vertical Component
240N
140N
90°
60°
240 Cos 90° = 0
140 Cos 60° = 70 N
I OON
0
100 Cos 0 = 100N
240 Sin 90° = 240N
140 Sin 60° =
121.24N
100 Sin 0° = 0
X=170N
Y=361.24N
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Using Pythagoras rule
X2 +Y2
The resultant R =
=
1702 + 361.242
=
28900 + 130494.34
=
399.24N
The direction of R is the angle which the resultant R makes with the
horizontal and you will obtain this from
tan θ = RY = 361.24 = 2.125
Rx
170
θ = tan-' 2.125
= 64.8°
SELF ASSESSMENT EXERCISE 4
Three forces of magnitude 10N; 5N and 4N act on an object in the
directions North, West and East respectively. Find the magnitude and
direction of their resultant.
4.0
CONCLUSION
In this unit, you have learned about units of measurements, dimensions,
fundamental and derived quantities and their units, scalars and vectors.
You have also learned to solve numerical questions relating to the unit.
Furthermore, you have learned some definitions of the key concepts in
the unit using the language of Physics.
5.0
SUMMARY
• Fundamental units are the basic units upon which other units depend.
They are the units of the fundamental quantities
• Derived units are the units of the derived quantities
• Fundamental quantities are the basic quantities that are independent
of others and cannot be defined in terms of other quantities or
derived from them.
• Derived quantities are those obtained by some simple combination of
the fundamental quantities.
• Vector quantities, are those quantities which have both magnitude
and direction.
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• Scalar quantities are those which have only magnitude or numerical
value but no direction. Resultant vector is that single vector which
would have the same effect in magnitude and direction as the
original vectors acting together.
• Components (Resolved parts) of a vector. The component of a vector
in a given direction is its effective value in that direction.
• The parallelogram law of vectors states that if two vectors are
represented in magnitude and direction by the adjacent sides of a
parallelogram, the diagonal of the parallelogram drawn form the
point of intersection of the vectors represents the resultant vector in
magnitude and direction.
6.0
TUTOR-MARKED ASSIGNMENT
1.
Distinguish between fundamental and derived quantities. Give
five examples of fundamental quantities and their units
2.
Which of these quantities: Pressure, volume, velocity, density and
momentum are:
i)
ii)
iii)
Scalars; and
Vectors.
Determine their dimensions
3.
(a)
(b)
State the parallelogram law of forces
What is meant by the component of a vector in a given
direction?
4.
A body is under the action of two forces 7N and l0N. Find the
resultant of the two forces if:
(a)
(b)
the forces are parallel and act in the same direction.
The forces are parallel and act in opposite directions.
5. Four forces of magnitude 10N, 5N, 4N and 6N act on an object in
the directions North, West, East and South respectively. Find the
magnitude and direction of their resultant (use vector resolution
method).
19
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REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). New School Physics for Senior Secondary
Schools Onitsha:
Africana FEP Publishers.
Awe, O. and Okunola, O. O. (1998). Comprehensive Certificate Physics.
SSCE Edition. Ibadan: University Press PLC (1992 Edition)
Ndupu, B. L. N and Okeke, P. N. (2002). Round up Physics. A complete
Guide. Ikeja: Longman Nigeria PLC.
Ndupu, B. L. N, Okeke, P. N. and Ladipo, O. A. (1996). Senior
Secondary Physics 2 Ikeja: Longman Nigeria PLC.
Okpala, P. N. (1990) .Certificate Year Series for Senior Secondary
School. lbadan: NPS Educational Publishers Limited.
20
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MOLECULAR THEORY OF MATTER
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Concept of Matter
3.2
Structure of Matter
3.2.1 Evidence of the Particle Nature of Matter
3.2.2 Experimental Evidence of the Particle Nature of
Matter
3.2.3 Simple Atomic Structure
3.3
Molecules
3.3.1 Molecules Definition
3.3.2 Structure, Nature and Size of Molecules
3.3.3 Some Ideas about Molecular Size
3.3.4 Estimating the Size of a Molecule
3.4
States of Matter
3.5
The Kinetic Molecular Theory of Matter
3.5.1 Fundamental Assumptions of the Kinetic Molecular
Theory
3.5.2 Basic Assumptions of the Kinetic Theory of Gases
3.6
Explanation of Some Physical Properties and Processes
Using the Kinetic Molecular Theory
3.6.1 The Three Basic States of Matter
3.6.2 Pressure Exerted by a Gas
3.6.3 Diffusion
3.6.4 Change of State (Melting, Vaporisation)
3.6.5 Boiling
3.6.6 Evaporation
3.7
Properties of Fluids at Rest
3.7.1 Adhesion and Cohesion
3.7.2 Surface Tension
3.7.3 Capillarity
3.7.4 Osmosis
3.7.5 Viscosity
3.8
Crystalline and Amorphous Substance
3.8
Crystalline and Amorphous Substances
3.8.1 Crystals
3.8.2 Structure of Simple Crystals
3.8.3 Non - Crystalline and Amorphous Solids
3.8.4 Heating and Cooling of Crystals and Non- Crystals
Conclusion
Summary
Tutor- Marked Assignment
References/Further Readings
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INTRODUCTION
In this unit, the Kinetic molecular theory of matter is introduced to you.
The theory suggests that matter is made up of small particles called
molecules which are in constant motion. You will also learn about the
particle nature of matter, molecular and atomic structure and the nature
and size of a molecule. The three states of matter, assumptions of the
Kinetic molecular theory and use of the Kinetic theory to explain the
three states of matter and some common physical phenomena are also
considered. The unit ends with a brief discussion of the structure of
simple crystals.
2.0
OBJECTIVES
At the end of the unit, you should be able to:
•
•
•
•
•
•
•
•
•
state the molecular theory and explain its justification
explain how molecules of a substance move relative to other
molecules of the same substance
decide the atomic and molecular structures of matter.
state the constituents of the atom
estimate the size of a molecule
use the molecular theory to explain the three states of matter and
some physical phenomena such as change of state, pressure
exerted by a gas, osmosis diffusion and surface tension
state the fundamental assumptions of the kinetic molecular theory
of matter
distinguish between crystalline and non-crystalline or amorphous
substances
give at least two examples of a crystal and describe the structure
of simple crystals.
3.0
MAIN CONTENT
3.1
Concept of Matter
Matter is anything that has weight and occupies space. Every object or
substance is made up of matter. Many of the properties and behaviour of
substances can best be explained by assuming that all substances are
composed of small particles called molecules. The assumption that
matter is made up of tiny particles (molecules) which are in constant
motion is known as the "molecular theory of matter."
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3.2
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Structure of Matter
3.2.1 Evidence of the Particle Nature of Matter
(i)
Many substances in solid form can easily be crushed to powder
form e.g. piece of chalk, lump of clay, charcoal and piece of
stone.
(ii)
A dry stick or dry wood is easily broken into smaller bits
(iii)
Solubility - if you drop a cube of sugar into a cup of water and
turn the water, the sugar "disappears". That is, it dissolves in
water.
(iv)
If you scrape the surface of a piece of chalk, you will see
thousands of very tiny particles flake off and float through the air
(Note that what you see are not the smallest particles possible i.e.
molecules. There is a limit to the size of particles the eyes can
see)
(v)
If a beam of light (e.g. sunlight) is entering a dusty room through
a window, you will observe a chaotic motion of the dust particles
in the air.
3.2.2 Experimental Evidence of the Particle Nature of Matter
Experimental evidence of the atomic or molecular nature of matter is the
Brownian motion named after the botanist, Robert Brown, who
discovered the phenomenon in 1827. While he was observing tiny
pollen grains suspended in water under a microscope, Brown noticed
that the tiny pollen grains moved about in zigzag paths even though the
water appeared to be perfectly still. The here and there by the molecules
of water pollen grains were being knocked about which were vigorously
moving about.
Brownian motion can be demonstrated by the smoke-cell experiment.
The apparatus is shown in Fig. 2.1.
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(b
)
Fig. 2.1:
Molecular motion in gas: Brownian (random) motion
Collect some smoke from a smouldering piece of cloth or wood by
means of a syringe and introduce it into the cell. Replace the cover
quickly and adjust the focus of the microscope until the fine particles
come into view clearly. You will observe the smoke particles as black
dots which move about irregularly like a (drunkard) drunken man
staggering about. The particles dart from one place to another very
suddenly, some going out of focus, others coming into focus, but always
in motion.
Explanation of Brownian motion
The irregular movement of the smoke particles is due to the motion of
the invisible air molecules which bombard each particle from all sides.
The particle is very small and the number of molecules of air hitting one
side is not balanced by the number of molecules hitting the opposite side
at the same instant. Therefore, the particle moves in the direction of the
resultant force and when it moves to another place, the same thing
happens. Why can't the motion of the particles be due to convection? In
that case the particles would move upwards continually and not zigzag
from side to side.
Brownian motion in liquids can be demonstrated similarly as follows:
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Place on a clean microscope slide a few drops of diluted aquadag (fine
graphite particles suspended in water) or photopake (a similar
suspension used for blacking negatives) and cover the liquid with a
cover-slip. Project an image of the slide on a screen using a micro
projector so that the particles can be seen. The graphite particles are then
seen to be moving about in an irregular manner, thus showing Brownian
motion in a liquid. In this case, the irregular motion of the graphite
particles is due to their bombardment by the surrounding water
molecules which are constantly moving about in different directions.
Brownian motion is important for two reasons.
(i)
(ii)
It provides evidence for the existence of the tiny particles of
matter called molecules
It gives evidence that molecules are in a constant state of
random motion.
3.2.3 Simple Atomic Structure
Definition
An atom is the smallest indivisible particle of an element which can take
part in a chemical change.
Fig. 2.2: Structure of an atom
Fig.2.2 shows the simple structure of an atom. It consists of two parts:
the nucleus and the electrons. The nucleus is the heavy portion of the
atom and is made up of two types of particles called protons and
neutrons. The protons carry a positive charge while neutrons carry no
charge. The electrons carry a negative charge and circle in orbits around
the heavy nucleus. The numbers of orbits depend on the substance, for
example hydrogen has only one orbit while oxygen has two. An electron
is very light (about 1 /1840 of the mass of the proton). The negative charge
of an electron is equal to the positive charge of a proton and the number
of electrons in an atom is equal to the number of protons. The atom is
therefore electrically neutral.
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SELF ASSESSMENT EXERCISE 1
1.
2.
3.
4.
5.
3.3
Define matter and state the molecular theory of matter
Mention five simple evidences which you would use to justify the
molecular theory of matter
Describe an experiment to demonstrate Brownian motion. Why is
Brownian motion important?
With the aid of a labelled diagram describe the simple structure
of an atom
Name the constituents of an atom and briefly state the properties
of each one.
Molecules
3.3.1 Definition: A molecule is the smallest particle of a
substance which can have a separate existence and still
retain the properties of that substance.
3.3.2 Structure, Nature and Size of Molecules
(i)
Most substances cannot exist by themselves as individual atoms,
rather they combine their atoms with themselves or with other
atoms to form molecules. Thus a molecule may be made up of
similar atoms of the same element or different atoms of two or
more elements. For example a molecule of hydrogen is made up
of two atoms of hydrogen but a molecule of water consists of two
atoms of hydrogen and one atom of oxygen.
(ii)
The molecules of any pure substance are identical they have the
same structure, the same mass and the same mechanical
properties.
(iii)
Molecules are formed by atoms combining in simple proportions
(iv)
The simplest model of a molecule is that of a rigid sphere (like a
small billiard ball) which can move and collide with other
molecules or with a wall and exert attractive or repulsive forces
on neighbouring molecules. The molecular forces decrease as the
distance separating the molecules increase.
(v)
Molecules are in constant motion. The motion is random
(haphazard or zigzag) in liquids and gases but oscillatory or
vibrational in solids.
(vi)
The size of a molecule is extremely small - of the order of 10-9 10-10 m. as a result of the small size, molecules cannot be seen
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with the naked eyes or even with the aid of a microscope: Again
because of the small size, one gramme of an element contains
several millions of molecules. For example one gramme of
hydrogen contains about 1023 molecules.
3.3.3 Some Ideas about Molecular Size
It is difficult to imagine how tiny the diameter of a molecule actually is.
Even the numerical value estimated to be of the order of 10-10 m (10-9
cm) is also not easily and accurately conjectured. The following may
enable you to have a better idea of the size of a molecule:
(i)
If a fine hair is magnified until its thickness is that of a wide
street, a molecule in the hair would then look like a speck of
dust in the street
(ii)
The tip of a pin contains millions of molecules.
(iii)
Two grains of hydrogen contain 6 x 1023 molecules.
If the whole population of the world were to count such a huge number
of molecules individually at the rate of five molecules per second, it
would take 100 years to count all of them.
3.3.4 Estimating the Size of a Molecule
You can estimate the size or diameter of a molecule by performing the
oil film experiment. The principle of the oil film experiment was
discovered by Lord Raleigh in 1890. It was known that certain oils when
dropped on the surface of water, would spread to form a circular film
with the molecules standing up - right. Lord Raleigh argued that if a
drop of oil is placed on top of a water surface, the oil will spread out on
top of the water surface until the thickness of the oil film is one
molecule thick. He, therefore, used this reasoning to obtain the first
estimate of the diameter of a molecule. You can repeat Lord Raleigh's
experiment as follows:
Fig. 2.3: Oil film experiment
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(i)
(ii)
(iii)
(iv)
(v)
(vi)
ACCESS PHYSICS
Fill a shallow tray with water and allow it to stand until the
water is at rest.
Sprinkle some lycophodium powder lightly on the surface of
the water
Using a graduated pipette with a fine bore, take up a small
volume of olive oil and note the reading on the pipette scale
Drop a very small quantity of the oil on the water surface.
Note again the reading on the pipette and obtain the volume of
the oil dropped by subtracting the second from the first pipette
reading.
Allow the oil to spread, pushing the lycopoduim powder
outwards and forming a clear thin circular film of oil on the
water surface.
Measure the diameter of the oil film to the nearest centimeter
using a half millimetre scale. Calculate the thickness of the oil
film as follows:
Let diameter of oil film = dcm
Let volume of oil drop = Vcm3
Area of oil film
=π
d
2
2
cm2
.'. Thickness of oil film = Volume
Area
π
=
=
V
d
2
2
4V cm
πd2
If you have performed the experiment accurately you should get a value
of about 2 x 10-7 cm as the thickness of the oil film. Hence the size of an
oil molecule is taken as about 2 x 10-7 cm.
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SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.
3.4
Define molecules and distinguish between a molecule and an
atom of a substance.
Outline briefly the structure and nature of a molecule.
The actual size of a molecule is almost impossible to imagine.
Discuss.
Describe an experiment which you would use to estimate the
thickness of an oil molecule. State the principle employed in the
experiment you have described.
States of Matter
Matter exists in three main states, namely solid, liquid and gas. Solids
have fixed shape and volume. They cannot be poured. Liquids have
fixed or constant volumes but they assume the shape of the container.
They can be poured. Gases have no fixed shape, but always occupy the
shape of the container. They can be poured.
3.5
The Kinetic Molecular Theory of Matter
The theory states that matter is made up of tiny particles called
molecules which are in constant motion.
3.5.1 Fundamental Assumptions of the Kinetic Molecular
Theory
(i)
Matter exists either in solid, liquid or gaseous state.
ii)
All substances consist of molecules, the smallest particle which
can exist independently.
(iii)
In solids the
The forces
attractive or
in which the
lattices.
(i v)
In liquids the molecules move freely
in all directions. In
addition to vibrational energy, they have translational energy.
The Kinetic energy of the liquid molecules is greater than in
solids.
(v)
In gases the molecules are in constant motion and are further
apart than in solids and liquids. They move at high speeds and
have translational, vibrational and in addition rotational energy if
molecules vibrate about a mean or fixed position.
between the molecules are strong and may be
repulsive. All true solids have a crystalline structure
atoms are arranged in regular patterns called
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the molecules are made of two or more atoms. The attractive or
cohesive force is negligible, so, gases are perfectly free to expend
and completely fill the vessels containing them. Gas molecules
have the greatest Kinetic energy. Because the intermolecular
forces are small, the motion of molecules in the gaseous state is
linear until collision takes place either with other molecules or
with the walls of the container.
3.5.2 Basic Assumptions of the Kinetic Theory of Gases
The Kinetic theory of matter has been more completely developed for
gases than for solids and liquids. This is because the problems involved
are much simpler in the case of gases. The simplest substance to which
the theory has been applied is the ideal gas. The fundamental
assumptions of the theory are as follows:
(i)
Gases consist of many very small particles called molecules,
which are like perfectly elastic spheres and are usually in
constant random motion.
(ii)
Molecules exert no forces on one another except when they
collide. Therefore, between collisions with other molecules or
with the walls of the container, they move in straight lines.
(iii)
Collisions of molecules with one another or with the walls of the
container are perfectly elastic. This means that the total Kinetic
energy of two molecules before collision is the same as that after
collision, and that when a molecule collides with the wall its
Kinetic energy is unchanged.
(iv)
The duration of a collision is negligible compared with the time
between collisions
(v)
Molecules are separated by distances which are very large
compared with the size of the molecules (or the volume of the
molecules is negligible when compared with the volume of the
container); they are, however, distributed uniformly throughout
the container.
(vi)
Any finite volume of the gas contains a very large number of
molecules. This assumption is supported by experimental
evidence because under standard temperature and pressure (s.t.p),
there are about 3 x 1019 molecules per cm3 of any real gas.
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SELF ASSESSMENT EXERCISE 3
1.
2.
3.
3.6
What do you understand by the Kinetic molecular theory of
matter?
State (i) The assumptions of the Kinetic theory of matter, (ii) The
assumptions of the Kinetic theory of gases.
Distinguish between the Kinetic theory of matter and the Kinetic
theory of gases
Explanation of some Physical Properties and Processes
Using the Kinetic Molecular Theory
3.6.1 The Three Basic States of Matter
(a)
In solids, molecules vibrate about a mean position. The forces
between the neighbouring molecules are strong and the attractive
and repulsive forces between them balance. Solids therefore have
characteristic crystalline structures in which the atoms are
arranged in regular patterns called lattices.
(b)
In liquids, the forces between the molecules are less strongthan in
solids, but they are both cohesive forces. So, the molecules move
freely through the liquid, exchanging partners or neighbours as
they go. The liquid can thus take the shape of the container and
can be poured because the molecules are very free and are in
continuous random motion.
(c)
In gases, the molecules are much further apart than those of
solids and liquids. They move freely and at high speeds in the
space in which they exist. As in liquids they are in constant
random motion. The attractive or cohesive force is negligible, and
hence a gas can be poured and is perfectly free to expand and
completely fill the vessel containing it.
3.6.2 Pressure Exerted by a Gas
A gas consists of molecules in random motion. As the molecules collide
with one another and with the walls of the container, they change
direction of motion. Therefore, there is always change in momentum,
and rate of change of momentum exists at any given time. According to
Newton’s second law of motion, rate of change of momentum is force.
So, a force is exerted on the container as the molecules change
momentum. The force per unit area thus exerted is the pressure exerted
by the gas.
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3.6.3 Diffusion
If hydrogen sulphide gas is released at one end of the room, the rottenegg smell soon fills the whole room.
If a perfume bottle is opened at one corner of a room the perfume smell
soon spreads to every corner of the room.
In both cases the molecules of the gas, because of their incessant
random motion, have quickly travelled throughout the room and filled
the whole space with their smell. The process is known as diffusion.
Definition
Diffusion is the tendency of molecules to migrate and fill an-empty
space as a result of their random thermal motions.
Diffusion takes place in liquids and gases but more slowly in solids. Its
rate depends on the density and temperature of the gas. Light gases at
high temperatures diffuse faster than heavy gases at low temperatures.
3.6.4 Change of State (Melting, Vaporisation)
Melting - Heat lessens the cohesion which holds molecules together in
the crystal lattice of a solid. At the melting point heat breaks the lattice
completely and the molecules are free to move about randomly, though
still held together by the force of cohesion. During melting the
temperature is constant while heat is absorbed to break the lattice.
3.6.5 Boiling
At boiling point the temperature of the boiling liquid is again constant.
Heat absorbed is used to overcome the force of cohesion by breaking the
attractive forces holding the liquid molecules together. Thus the
molecules break away and become molecules of gas. The same amount
of heat absorbed for melting is given out for fusion (just like in the case
of vaporisation and condensation).
3.6.6 Evaporation
In a liquid the molecules move randomly with different speeds and they
have an average Kinetic energy (k.e). Some of the molecules near the
surface which are moving with speed greater than the average speed
succeed in escaping from the attraction of their neighbours and jump out
of the liquid. This is evaporation. The molecules would have left the
liquid to form the vapour. As they escape from the liquid with this extra
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energy the temperature of the liquid reduces. Thus we have the cooling
effect of evaporation.
SELF ASSESSMENT EXERCISE 4
Use the Kinetic molecular theory to explain the following:
a)
b)
c)
d)
e)
f)
The three states of matter
Pressure exerted by a gas
Diffusion
Change of state (melting,
condensation)
Boiling
Evaporation.
3.7
Properties of Fluids at Rest
solidification,
vaporisation,
3.7.1 Adhesion and Cohesion
Adhesion is the force of attraction between molecules of different
substances, e.g. molecules of water and glass.
Cohesion is the force of attraction between molecules of the same
substance, e.g. water molecules.
The different behaviours of water and mercury when spilled on a clean
glass surface is explained by adhesion and cohesion: The adhesion of
water to glass is greater than the cohesion between water molecules, so
water spreads out on clean glass and wets it. In the case of mercury, the
cohesion of mercury molecules is stronger than the adhesion of mercury
to glass. As a result mercury does not wet glass but rather forms into
globlets or spherical beads on the glass surface.
3.7.2 Surface Tension
The surface of a liquid behaves as if it were a stretched elastic skin with
the result that it can hold an object denser than the liquid gently placed
on it. This phenomenon is known as surface tension.
Definition
Surface tension is the force acting along the surface of a liquid, causing
the liquid surface to behave like a stretched elastic scene.
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Examples of the effect of surface tension
Fig. 2.4: Effects of surface tension
(i)
(ii)
(iii)
(iv)
If water is allowed to drip very slowly from a water tap, the shape
of each small drop is spherical (Fig. 2.4a)
If a dry paint brush is dipped in water and withdrawn, the hairs of
the brush cling to one another (Fig. 2.4b)
Droplets of mercury on a clean hard surface such as glass are
spherical in shape. Larger drops, because of their weight, are
flattened at the top with curved sides (Fig. 2.4c)
Soap bubbles are spherical in shape (Fig. 2.4d).
(You should be able to give some other examples of the effect of
surface tension)
Molecular explanation of surface tension
X
Fig. 2.5: Molecular forces in a liquid
Consider two molecules X and Y in a liquid as in Fig. 2.5. X is on the
surface while Y is well below the surface. Y will be attracted uniformly
in all directions and is not pulled in any one particular direction, i.e.
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there is no resultant force on it as a result of the attraction of the
molecules surrounding it. On the other hand X is only attracted by the
molecules within the body of the liquid. It therefore experiences a
resultant inward force (pull) at right angles to the liquid surface.
All molecules on or near the surface experience this resultant force
downwards with the result that the liquid surface will always tend to
shrink or contract and make the surface area as small as possible. The
surface is therefore in a state of strain or tension which is called surface
tension.
3.7.3 Capillarity
Definition
Capillarity (or capillary action) is the tendency of a liquid to rise or fall
in a narrow tube
Illustration of capillarity
Fig. 2.6: Demonstration of capillary action
(i)
Dip three capillary tubes with fine bores but with different
diameters into clean water as in Fig. 2.6 (a) above. You will
observe that the water rises in the tubes but the narrower the bore
the greater the height to which the water rises (Fig. 2.6 a).
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(ii)
Repeat the experiment with soap solution. You will observe a
similar situation except that the heights of soap solution are lower
than in the case of water (Fig. 2.6b)
(iii)
Repeat the experiment again with mercury. You will observe that
the level of mercury falls in the three tubes. The mercury level is
depressed below the level of mercury outside the tubes (i.e. in the
container). Again the narrower the tube the lower the level of
mercury (Fig 2.6c).
Explanation of Capillarity
(i)
The meniscus of water or soap solution is curved upwards
(concave) because the adhesion of water and soap solution to
glass is greater than the cohesion of water or soap. Therefore
water or soap solution wets the glass tube and so spreads a thin
film of water/soap solution on the inner surface of the tube. The
adhesive forces thus force the water (soap solution) to creep up
the inside of the tube. The water/soap solution is held up as it
creeps by surface tension forces acting around the circumference
of the meniscus. The water/soap solution thus keeps rising in the
tube until the weight of the column of water/soap solution
balances the surface tension acting at the top of the column.
(ii)
In the case of mercury the cohesion of mercury molecules which
is greater than the adhesion of mercury to glass causes the
mercury level to be depressed in the tube. The surface tension
forces acting around the circumference of the tube holds down
the mercury column as it is depressed by cohesive forces of the
mercury molecules. The depression continues until the weight of
the mercury column in the tube is equal to the surface tension.
You should try and identify other examples of capillarity e.g. absorption
of liquid by blotting paper, the rise of oil in wicks of lamps, etc.
3.7.4 Osmosis
Definition
Osmosis is the tendency of a solvent to pass from a dilute solution,
through a semi permeable membrane into a concentrated solution.
Semi-permeable membrane is a substance such as cellophane,
parchment or vegetable material which would allow some molecules of
liquid to diffuse through them but not others. Such a membrane may
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allow the molecules of a solvent to pass through it but not those of a
solute.
Demonstration of osmosis
Fig.
Demonstration of osmosis
(i)
(ii)
(iii)
2.7:
Tie a semi-permeable membrane across the mouth of a thistle
funnel and pour some concentrated sugar solution into the funnel.
Immerse the funnel in a beaker of water as shown in Fig. 2.7 and
note the level of sugar solution in the funnel.
Allow the set up to stand for a day, and check the level of sugar
solution in the funnel again. Test the water in the beaker for
sugar. You will observe that the level of the sugar solution in the
inverted thistle funnel has risen and that the water in the beaker is
free from sugar.
Explanation
The molecules of sugar (the solute) cannot pass through the semipermeable membrane, but molecules of water (the solvent) can do so.
Sugar and water molecules are bombarding the membrane on one side
and only water molecules on the other side. Thus, more water molecules
move up into the funnel than down out of it and so the level of liquid in
the thistle funnel rises.
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3.7.5 Viscosity
Brief mention should be made here of the force of viscosity which
occurs in fluids at rest or in motion.
Viscosity is "friction" in fluids (liquids and gases). It may be defined as
"the internal friction between the layers of a liquid or gas in
motion". Liquids that pour slowly are more viscous than those that
power faster. Thus very cold honey is more viscous than very cold
water.
The movement of one layer of fluid over a neighbouring layer is
opposed by forces of viscosity. Again, when a stone or a ball-bearing
falls through a viscous (think) liquid, the downward motion of the object
is opposed by the viscosity of the liquid.The greater the viscosity of the
liquid, the greater the opposition to the movement of the object and
hence the slower its motion.
A more detailed discussion of viscosity is done in Module 2, Unit 2.
SELF ASSESSMENT EXERCISE 5
1.
2.
3.
4.
5.
3.8
Distinguish between adhesion and cohesion.
Define surface tension and mention any three effects caused by
surface tension
Explain the meaning of
(i) Capillarity, (ii) Viscosity and describe an experiment to
illustrate each of them.
How does the Kinetic molecular theory explain surface
tension, capillarity and osmosis?
Why is it easier to pour out water from a container than to
pour out engine oil?
Crystalline and Amorphous Substances
Solids are usually classified into two groups:
(i)
(ii)
Crystals or crystalline solids;
Noncrystal or non-crystalline solids;
The difference between crystals and non-crystals is the arrangement of
atoms or molecules in the solid.
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3.8.1 Crystals
Definition
A crystal is a piece of solid matter in which the atoms, molecules or ions
are arranged in a highly regular repeating pattern called lattice.
Crystal Lattice
The particles in a crystal are arranged in regular 3-dimensional
framework or pattern called crystal lattice which repeats over and over
again in all directions. The high degree of regularity and order in the
arrangement of the molecules is the principal feature distinguishing
solids from liquids. Particles in a liquid are jumbled and highly
disorganised as they move about. They are even more disorganised in a
gas. Examples of common crystals are: sodium chloride, zinc
sulphide, chromium, iron and platinum salts.
3.8.2 Structure of Simple Crystals
A simple crystal is made up of a huge number of simple basic units or
building blocks called unit cells (Fig. 2.8a). If you stack these units up
and, down, side by side and in all directions, you can build the whole
lattice. Unit cells are of three types, giving rise to 3 types of lattice and
hence 3 types of crystals, simple cubic lattice, face- centred cubic lattice
and body-centred cubic lattice.
b.
Sodium chloride, NaCl
Fig. 2.8: Simple cubic crystal
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(a)
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Simple Cubic Crystal
Here the atoms or molecules or ions are placed at the corners of
imaginary cubes stacked side by side, up and down like building blocks
(Fig.2.8b). An example is the Sodium chloride (NaCl) crystal. In the
lattice, the atoms of Na and CI alternate positions in the cube in each of
the 3 directions. Each atom within solid, thus has six immediate
neighbours.
Zinc sulphide, ZnS
Fig. 2.9: Face-centred cubic crystal
(b)
Face-Centred Cubic Crystal
The unit cell has identical particles at each of the corners plus another
particle in the centre of each face as shown in Fig. 2.9. A typical facecentred cubic.
Crystal is Zinc sulphide (ZnS). Other crystals in this group include
crystals of common metals like copper, silver, aluminium, lead, etc.
Fig. 2.10: Body-centred cubic crystal
(c)
Body-centred Cubic Crystal
The unit cell has identical particles at each corner of the cube plus one
in the centre of the cell as illustrated in Fig. 2.10. Examples include;
chromium, iron and platinum salts.
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Fig.2.11: Chain-like molecules of amorphous substance
3.8.3 Non-Crystalline and Amorphous Solids
The atoms of non-crystals are not regularly arranged as in the case of
crystals. They are said to be "amorphous” that is having no definite
shape or form; not organised. In a number of ways amorphous
substances resemble liquids more than solids. Examples of amorphous
solids are glass and plastics. Amorphous solids never form crystals.
They are usually made up of long, chain-like molecules that are
intertwined in the liquid state just like strands of earthworms (Fig. 2.11)
3.8.4 Heating and Cooling of Crystals and Non-Crystals
When heated, crystalline solids, with time, melt at constant temperature.
When the melting process is completed, temperature rises until at
another constant temperature, the liquid begins to change to gas. Fig.
2.12 shows the heating curve for a crystalline solid.
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Temperature
Liquid changing to gas
Boiling
point
Solid changing
To liquid
Melting
point
solid
Time
Fig. 2.12: Heating curve for crystalline substances
When crystalline substances are cooled, they begin to solidify at some
constant temperature, forming solid crystal. Fig.2.13 shows a typical
cooling curve for a crystal cooling from the liquid phase. As soon as the
last bit of liquid has solidified, the temperature of the substance drops
sharply.
Temperature
Solid crystal formed
Melting point
Time
Fig. 2.13: Cooling curve for crystalline substance
Amorphous substances soften gradually when heated, unlike crystalline
substances which will melt suddenly. For this reason, glass or plastic
can be heated to soften it and then bend it. When cooled, amorphous
substances.
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Temperature
After heat is removed
Time
Supercooling with no
Fig. 2.14: Coolingcrystalline
curve forformation
amorphous substance
behave rather oddly. They lose heat steadily and very fast and solidify
without forming crystals. The cooling of the "liquid" continues until the
solid hardens. Crystals are never formed (and cannot be formed)
because the molecules of the amorphous substance cannot be tangled
before they are frozen. Fig.2.14 shows the cooling curve for a typical
amorphous substance. The process is called "supercooling". The
amorphous substance (glass, rubber or plastic) is "supercooled liquid”
and not a "solid" in the true sense of the word.
SELF ASSESSMENT EXERCISE 6
1.
2.
3.
Define and give three examples of a crystal
Distinguish between crystalline and amorphous substances. Give
two examples of each
Describe the structure of
(i)
(ii)
(ii)
4.
5.
Sodium chloride crystal,
Zinc sulphide crystal, and
Iron crystal
Name three types of crystal structure and give one example of
each.
Explain the following terms
(i)
(ii)
(iii)
Amorphous substance
Crystal lattice
Supercooled liquid
43
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4.0
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CONCLUSION
In this unit you learned about the molecular theory of matter;
specifically, you learned about the the concept, structure and nature of
matter. In addition, you learned about the molecule, states of matter, the
kinetic molecular theory and the assumptions of the theory.
Furthermore, you learned boiling, evaporation, surface tension,
capillarity, osmosis and viscosity. Finally, you learned crystal and
amorphous solids.
5.0
SUMMARY
•
Simple evidences of the molecular nature of matter exist in
everyday life, e.g. a number of substances in solid form is easily
broken into pieces or crushed into powder.
A molecule is the smallest particle of a substance which can have
a separate existence and still retain the properties of that
substance.
The size of a molecule is about 10-9 - 10-10 m.The molecular size
can be empirically estimated using the oil film experiment.
the molecular Kinetic theory of matter is based on a number of
fundamental assumptions which are used to explain the three
states of matter (solid, liquid and gas) as well as physical
properties such as pressure exerted by a gas, change of state,
boiling, evaporation, surface tension, cohesion, adhesion,
capillarity and osmosis.
Solids can be divided into two groups - crystalline solids and
non-crystalline or amorphous solids
A crystalline solid is one in which the atoms or molecules or ions
which make up the solid are arranged in a regularly repeating
pattern called lattice (crystal lattice). Crystals are very small
particles of a crystalline solid. They grow out from the solution of
the solid when the solution is cooled to freezing point. Crystalline
solids are usually soluble in water.
Non-crystalline solids do not form crystals. They are said to be
amorphous, i.e. without shape or form. When they lose heat, they
are supercooled and have no definite melting (or freezing) point.
They soften gradually when heated, e.g. glass, plastic, rubber.
They usually do not dissolve in water.
The structure of a crystalline solid can be any one of the
following:
•
•
•
•
•
•
•
(i)
(ii)
(iii)
44
Simple cubic,
face-centred cubic and
body-centred cubic.
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6.0
TUTOR-MARKED ASSIGNMENT
1.
Mention one experimental evidence and three evidences from
daily life experience which suggest that matter is made up of
small particles.
Using a labelled diagram only, illustrates the simple structure of
an atom. State two properties of each of any three constituents of
an atom which you have labelled.
Distinguish between the Kinetic molecular theory of matter and
the Kinetic theory of gases. State any three basic assumptions of
each theory.
Explain each of the following observations:
2.
3.
4.
(a)
(b)
(c)
(d)
5.
(a)
(b)
7.0
The smell of a dead rat located at a hidden corner of a
room, soon fills the entire room.
A thermometer suspended in boiling water does not
indicate any rise in temperature although heat is steadily
supplied to the vessel containing the water
Drops of water, which issue slowly from a tap, are
spherical in shape
A drop of water spreads out on a clean glass sheet and
wets it but a drop of mercury forms a spherical globlet on
the glass surface without wetting it.
Distinguish between crystalline and amorphous substances
and explain what is meant by "a true solid".
Draw sketch diagrams to illustrate the structure of each of
the following crystals: Chromium, Zinc sulphide and
Sodium chloride crystals.
REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). Physics for Senior Secondary Schools
Onitsha: Africana - FEP Publishers Ltd.
Awe, O. and Okunola, O. O. (1992). Comprehensive Certificate Physics.
SSCE Edition. lbadan: University Press Plc.
Ndupu, B. L. N and Okeke, P. N. (2000). Round up Physics for West
Africa Senior School Certificate Examination: A Complete
Guide. Lagos: Longman Nigeria Plc.
Nelkon, M. (1986). Principles of Physics for Senior Secondary Schools
(9th Edition). England: Longman Group Ltd.
45
PHY 001
UNIT 3
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MOTION IN A STRAIGHT LINE
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Types of Motion
3.1.1 Random Motion
3.1.2 Linear /Translational Motion
3.1.3 Rotational Motion
3.1.4 Oscillatory (or Vibrational) Motion
3.2
Relative Motion
3.3
Parameters used in Describing Motion
3.3.1 Concept of Speed
3.3.2 Concept of Velocity
3.3.3 Concept of Acceleration/Retardation
3.4
Distance/Displacement – Time graph
3.5
Speed Velocity – Time graph
3.6
Derivation of the Equations of Uniformly Accelerated
Linear Motion
3.7
Applications of the Three Equations of Motion in Solving
Numerical Problems
Conclusion
Summary
Tutor-Marked Assignment
References / Further Readings
1.0
INTRODUCTION
The term motion is usually associated with movement of objects when
forces act on them. You can use the term to describe any object that is
not at rest such as an aeroplane in flight or a man walking along the
road.
Motion can take place in one, two or three dimensions. When a body
moves along a straight line, we speak of one dimensional motion. Two
dimensional motion is motion in a plane. For example, the motion of
footballers on a field. Three dimensional motions take place in space.
Example is the flight of mosquito in a room.
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OBJECTIVES
At the end of this unit, you should be able to:
•
•
•
list the types of motion in a given environment
classify a list of familiar motion into random, rotational,
oscillatory and translational motion;
identify for a given motion the two bodies between which there is
a relative separation
distinguish between distance and displacement in a translational
motion
distinguish between speed and velocity
plot a distance-time graph and deduce the speed of motion from
the gradient or slope of the graph
determine speed and velocity when simple problems are set
involving distance or displacement and time
explain the concept of uniform motion, instantaneous
speed/velocity and uniform acceleration
determine acceleration from a velocity - time graph
deduce the three equations of motion from a (v - t) graph with
initial velocity and constant acceleration
explain the terms used in the equations of motion.
3.0
MAIN CONTENT
3.1
Types of Motion
•
•
•
•
•
•
•
•
There are four main types of motion:
(i)
(ii)
(iii)
(iv)
random motion
translational motion
rotational motion and
oscillatory motion.
3.1.1 Random motion
An object is said to undergo random motion when it moves in an
irregular manner with no preferred direction or orientation.
Examples
47
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•
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Movement of players in a football field;
Brownian motion - an irregular motion of particles suspended in
water or of smoke particles suspended in air.
Fig. 3.1:
3.1.2
Random motion of gas molecule
Linear/Translational Motion
When rigid objects move away from one point in space to another
without rotating, the motion is said to be translational. In translational
motion, you will observe that every point in the body remains relatively
fixed to one another. Each part of an object undergoing pure
translational motion follows the same path.
Examples
•
•
•
When you move from your home to the study centre;
A car travelling from Enugu to Abuja, along the road;
An insect crawling from one point to another.
3.1.3 Rotational Motion
An object undergoes rotational motion when its movement requires its
complete turning. We mean that all points in the body move
inconcentric circles such as the point P in a rotating wheel. The centre of
these circles all lie on a line called the axis of rotation.
Fig.
wheel
3.2:
Axis of rotation passes into the page
Examples
48
Rotational motion of a
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•
•
•
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Rotational motion of a record disc
The movement of the blades of an electric fan;
The earth spinning about its axis.
3.1.4 Oscillatory (or Vibrational ) Motion
You describe an object as undergoing oscillatory motion when its
movement is periodic. That is, the object undergoes a `to and fro'
movement about an axis, reversing the direction of its motion and
returning regularly to its original position.
String
(a) Simple pendulum
(b) Vibrating mass on a spring
Fig. 3.3: Oscillatory motion
bob
Example
•
•
The motion of a pendulum as it swings back and forth, Fig. 3.3.
(a)
The vertical motion of a disturbed mass hung from a spiral spring
or an elastic band Fig .3.3. (b)
You are aware that the actual motion of most objects is quite
complicated. The motion of an object is usually a combination of these
different types of motion.
You know that:
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•
The wheel of a car in motion combines rotational motion and
translational motion;
•
The earth moves round the sun (translational) as well as spins
about its north south axis (rotational);
•
A complex gas molecule with many atoms may undergo random,
translational, rotational and oscillatory motions.
3.2
Relative Motion
If you are sitting in a train travelling pass a railway platform on which
your friend is standing, there is said to be relative motion between you
and your friend.
A person sitting still at the back seat of a moving bus has the same speed
as that of the bus relative to the earth. But if he now walks towards the
driver of the bus, he has a speed relative to the earth which is more than
that of the bus.
Now, imagine the bus is moving at 30m/s and the person walks at 2m/s
towards the driver, his forward speed relative to the earth is 32m/s (30m/
s +2 m/s). But when he walks back from the driver to his back seat with
the speed of 2m/s, his speed relative to the earth is now 18m/s (30m/s 2m/s).
Relative to the earth, the speeds of the person are different in the two
occasions. In this illustration, the points of reference are the earth and
the bus.
When you travel in a bus look down on the floor of the bus. The bus will
appear to you to be stationary.
This is because you and the bus are moving with the same speed and
hence there is zero or no relative motion between two of you. But if you
look out of the window the trees outside the bus will appear to be
moving in opposite direction as the bus. Trees do not move. The illusion
of moving trees is due to the relative motion between you in the moving
bus and the stationary trees.
An object undergoes relative motion when there are two bodies between
which there exists a relative separation. Such a motion involves the
change of position of one object relative to another which may or may
not be fixed.
SELF ASSESSMENT EXERCISE 1
50
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1.
2.
3.
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5.
What is motion?
List and discuss the four main types of motion.
List five examples each of these main types of motion that you
come across in your environment (do not use examples already
given in the unit).
Give five examples of motions which involve more than one type
of motion
What do you understand by relative motion?
3.3
Parameters Used in Describing Motion
4.
The parameters used in describing motion are time, displacement
(distance in a specified direction), speed, velocity and acceleration.
Example: If a boy walks 30 m eastward and then 40 m northward the
distance he has walked is 70 m (30 + 40m). His displacement is 50 m
N38°E.(√[302 + 402]-àt an angle 38° East of North)
N
B
The displacement is 50 m N380 E
50 m
0
40 m
520
The displacement is 70 m
(i.e. 40 + 30m).
A
30 m
Fig. 3.4: Displacement and distance
Think of and give other good examples to show distance and
displacement.
3.3.1 Concept of Speed
In describing the motion of a body, we note both the distance and time it
takes to cover that distance. The rate at which a body covers a distance
is called the speed of the body.
Speed =
distance
time
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i.e.
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Speed = distance travelled
time taken
Symbolically, speed (v) = distance (s)
time (t)
i.e. v = s - ( 3.1)
t
You should note that:
•
•
•
Speed is a scalar quantity;
We talk of average speed because speed may not be constant all
the time. For instance when a driver slows down at road bends,
stops for school children to cross the road or may even increase
his speed;
Speed at a particular instance (instantaneous speed) is obtained
by timing a short section of the journey around that instance.
Speedometer is the instrument fixed in vehicle for this purpose.
Uniform Speed
When a body covers equal distances in equal time intervals, no matter
how small the time interval may be, the speed is said to be a uniform
speed or constant speed.
3.3.2 Concept of Velocity
You should, as a student of Physics, know that velocity and speed can
not be used interchangeable as is used in everyday language. In speed no
direction is specified (scalar) while in velocity the direction is specified
(vector).
Velocity is the rate of change of displacement with time or the rate of
change of distance with time or the rate of change of distance with time
in a given direction.
Velocity = displacement
time
i.e. v = s ---(3.2)
t
Note that:
52
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•
•
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When a car moves at constant speed in a straight line, its velocity
is constant;
When a car moves round a circular path with constant speed, the
velocity will be changing because of
the change in direction
at different points round the circle;
Velocity is a vector quantity and its unit is metre per second
(ms-1).
Non- Uniform Velocity
If a body moves round a circular path at constant speed, then it is said to
move with non - uniform velocity because its direction of motion is
constantly changing.
Uniform Velocity
When a body moves with equal displacement in equal time intervals no
matter how small the time intervals may be, the velocity is said to be
uniform or constant.
3.3.3 Concept of Acceleration/Retardation
An object may be moving with non-uniform velocity (changing
velocity). The change could be increase or decrease in velocity. When
the velocity of a moving object is increasing, the object is said to be
accelerating. When it is decreasing, the object is said to be retarding.
Acceleration (retardation) is defined as the rate of change of velocity
with time.
Specifically,
Acceleration (a) = increase in velocity (v)
time taken (t)
i.e. Acceleration (a)
=
velocity change
time taken to make the change
=> a
Retardation (- a)
=v
t
(3.3a)
=
velocity change
time taken to make the change
=> -a = v (3.3b)
t
S. 1. Unit = metre per second squared (ms-2)
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Uniform acceleration is defined as the motion of an object whose
velocity increases by equal amount in equal time interval however small
the time intervals may be.
•
Worked examples
l.
A car has a velocity of 72 km/h. How far does it travel in 1/2
min?
•
Solution
You must always use the same kind of units in calculations.
36 km/h = l0 m/s
.'. 72 km/h = 20m/s
1/2 min = 30s
velocity = displacement
time
=>
displacement = velocity x time
= 20m/s x 30s
= 600 m
2.
Suppose a train travels at 54 km/h and accelerates
uniformly to 144 km/h in 10s. Find the uniform
acceleration.
•
Solution
36 km/h = 10 m/s
54 km/h = 54 x 10 m/s = 15 m/s
36
114 km/h = 144 x 10 m/s = 40 m/s
36
acceleration = change in velocity
time taken
=> a = (40-15) m/s = 25m/s
10s
10s
= 2.5 m/s2
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SELF ASSESSMENT EXERCISE 2
1.
A car moves round a circular track with a uniform speed of
100 km/h. Show in a diagram, vectors which represent the
velocity of four consecutive positions A, B, C, D equally
spaced round the track. Is
(a)
(b)
2.
the velocity and
the speed the same at A, B, C and D?
Differentiate between:
(i)
(ii)
(iii)
average and instantaneous velocities
acceleration and retardation
distance and displacement.
3.
With an average velocity of 820 km/h an aeroplane in flight spent
1.25hrs to move from Lagos to Sokoto. How far is Lagos from
Sokoto by air?
4.
A car travels 120m in 5seconds. What is its average velocity?
5.
In an attempt to over take a luxury bus, a car driver increased his
velocity from 20 m/s to 60 m/s within a period of 15 seconds.
Determine the acceleration of the car.
3.4
Distance/Displacement - Time graph
You are familiar with the motion along a straight line in which the
actual distance covered is equal to the displacement and the speed is
equal to the magnitude of the velocity. Such motions can be represented
on distance/displacement - time graph as shown in Fig. 3.5 below.
distance/displacement
distance/displacement
S(m)
A
S(m)
A
S
C
t
S
C
B
B
So
time t(s)
0
time t(s)
0
Fig. 3.5(a)
Fig. 3.5(b)
In Fig. 3.5 (a), distance s = 0 when t = 0.
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That is, the timing is started when the body passes a reference point 0 on
the path.
In Fig..3.5b, S = So at t = 0.
This means that timing was started when the body was at a distance So
from the reference point on the path.
The straight line graph shows that the body covers equal distances in
equal time (speed is uniform).
The slope of the graph
=
AB = s represents the speed (v) of the body.
BC
t
The distances covered in the above diagrams are for:
Fig. 3.5a, S = vt i.e. slope x time
Fig. 3.5b, S = So + vt i.e. So + slope x time
For Fig.3.6 representing the motion of a body when the speed changes
with time, the instantaneous speed at any instance is the slope of the
graph at that instance.
For example, the instantaneous speed at P is represented by AB , the
slope of the target to the curve at P.
BC
distance/displacement
S(m)
A
P•
B
Time t(s)
Fig. 3.6: Motion of a body with varying speed
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3.5
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Speed/Velocity - Time graph
Speed/velocity
Speed/velocity
Speed/velocity
A
V1
C
t1
Time t(s)
B
Time t(s)
0
Time t(s)
(c)
(b)
(a)
Speed /velocity
Speed/velocity
S(m)
A
A
V1
V1
B t1
C
C
t1
B
Time t(s)
Time t(s)
(d)
(e)
Fig. 3.7: Speed/ velocity time graphs
The speed/velocity time graph of a moving object could be plotted as
shown in Fig. 3.7 above. The graph provides information about the
moving object. You can deduce the following from the graph.
(i)
(ii)
(iii)
(iv)
(v)
the slope of the graph gives the acceleration of the moving object;
the area under the graph is the distance covered;
a straight line graph parallel to X - axis shows that the object is
travelling with uniform velocity. That is, the object has zero
acceleration or retardation,
a straight line graph inclined to the X - axis shows that the object
is undergoing uniform acceleration or retardation;
a non-linear graph shows that the object is undergoing non
uniform acceleration or retardation. The acceleration or
retardation at any point is the slope of the tangent to the curve at
that point.
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Worked example
An object travelling along a straight path has the following velocities for
the first 12 seconds of the motion.
Velocities
(m/s)
Time (s)
0.0
5.0
10.0
15.0
15.0
7.5
0.0
0
2
4
6
8
10
12
Sketch the velocity - time graph for the motion.
Obtain from the graph:
(a)
(b)
(c)
(d)
the uniform acceleration during the first stage of the journey;
the retardation during the last stage of the journey;
for how long did the object travel altogether; and
what was the total distance travelled.
Solution
Fig. 3.8: Velocity time graph
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From the graph,
(a)
Acceleration OA = gradient of the line OA
AC = 5-0
OC
6-0
(b)
=
15mIs
6s
= 2.5m/s2
Retardation = gradient of the line BE
=
BD = 15-0 = 15m/s = 3.75m/s2
DE
12-8 4s
(c)
Total time of travel
= 6s + 2s + 4s = 12seconds.
(d)
The area under the velocity - time graph gives the total
distance travelled.
In time OC, distance covered,
= area of ∆OAC = 1/2 0C x AC
= ½ x 6 x 15 = 45m
In time CD, distance covered,
= area of rectangle CABD = CD x BC
= 2 x 15 = 30m
In time DE, distance covered
= area of ∆BDE ='/2 DE x BD
= '/2 x 4 x 15 = 30 m
Total distance travelled = (45 + 30 + 30) m
= 105 m
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SELF ASSESSMENT EXERCISE 3
An electric train moves from rest with a uniform acceleration of 1.5m/S2
for the first 10s and continues accelerating at 0.5m/s2 for a further 20s. It
continues at constant speed for 90s and fnally, takes 30s to decelerate
uniformly to rest.
a)
b)
c)
d)
e)
3.6
Draw a graph of speed against time for the journey;
From your graph or otherwise, deduce the total distance
travelled.
What is the average speed of the train for the whole journey?
What was the deceleration during the last stage of the motion?
For how long did the object travel altogether?
Derivation of the Equations of Uniformly Accelerated
Linear Motion
You can apply equations of uniformly accelerated motion when a body
moves under constant (uniform) acceleration along a straight line.
Here both the velocity and the acceleration are in the same direction
which is along the line of motion.
The speed - time graph of such a motion is shown in Fig. 3.9.
Distance s
(m)
Speed
(m/s)
v
Fig.
B
v-u
3.9: U Illustrating uniformly
accelerated
D
A
corresponding distance - time graph
If u = initial velocity
of t(s)
the body
Time
0
v = velocity at time t
a = the uniform acceleration
60
C
motion
with
Time t(s)
the
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You recall then that by definition,
a

change in velocity
=
time interval for the change
at = v-u
or v = u + at --- (3.4)
v-u
t
=
You already know that the distance S, travelled by the body during the
time interval is represented by the area of the trapezium OABC
That is, S = u + v x t
2
= average velocity x time interval
From equation 3.4; v = u + at
Therefore, substituting for v in the equation above we obtain
S = u + u + at x t
2
i.e
S = 2u + at x t = 2u + at x t
2
2
2
=> S = u + ½ at + t
S = u + ½ at2 ……. (3.5)
You can still make t the subject of the first equation v = u + at
i.e.
t=v–u
a
To obtain the third equation of motion, you will substitute for t in
equation 3.4 using the relation for t above.
i.e. S = ut +½ at2 becomes
S=u v-u
a
+½a v–u
a
2
Multiplying through by `2' gives
2S = 2u v- u
a
2S =
2uv – 2u
a
2
+
+
a v–u
2
a
1 v – 2uv + u2
a
2
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Multiplying through by `a'
2aS = 2UV - 2U2 + V2 - 2UV + U2
2aS = V2+U2 -2 U2
2aS = V2 - U2
If you make V2 the subject of the relation you obtain
V2 = U2 + 2aS--- (3.6)
The equations 3.4, 3.5 and 3.6 are generally referred to as the three
equations of uniformly accelerated linear motion.
You should note that if the body starts from rest, the initial velocity
U = 0 so that the
1st equation v = u + at becomes v = at
2nd equation S = ut +½at2 becomes S =½at2
3rd equation v2 = u2 + 2aS becomes V2 = 2aS
Similarly, if a body starts with a velocity u from a distance So, its
distance S, at time t, is given by
S = So + ut + ½ at2.
3.7
Applications of the Three Equations of Motion in Solving
Numerical Problems
•
Worked example
A train, starting from rest, accelerates uniformly so that it attains a speed
of 108 kmh-1 in 2 minutes. It travels at this speed for 10 minutes and is
then brought to rest with uniform retardation after 3 minutes. Calculate
the distance travelled by the train. Determine also the retardation.
The above problem can be solved by calculation and graphical methods.
We shall solve by calculation here while you repeat it by graphical
method later.
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Solution
First, you will identify the motion of the train which are in three parts:
(a)
(b)
(c)
under uniform acceleration;
at constant speed; and
under uniform retardation.
The sketch diagram is as shown in Fig. 3.9
Fig. 3.10: Sketch of the motion in our sample (Worked example)
In part (a), initial velocity, u = 0;
final velocity after 3 mins V1, = 108 kmh-l
=
108 x 1000 = 30m/s
3600
time interval t1, = 180s
•
distance travelled, S = ½ V1t1 (u = 0)
Sl = ½ x 30m/s x 180s = 2700 m
In part (b), constant velocity, u2 – v1 = 30 m/s
Time interval, t2 = 600s
•
distance travelled; S2 = u2 t2 (a = 0)
S2 = 30m/s x 600 = 18000 m
In part (c), initial velocity u3 = v1 = 30 m/s
final velocity, V3 =,0;
time interval , t3 = 120s
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distance travelled, S3 = ½ U3 t3
S3 = ½ x 30m/s x 120s = 1800 m
•
Total distance travelled by the train
S = S1 +S2+S3
= 2700m + 18000m + 1800 m
= 22, 500 m
= 22.5 km
The retardation took place at the ‘C’ part of the motion and is given by
Retardation = U3 - V3
t3
=
30-0
120s
= 0.25 ms -2
SELF ASSESSMENT EXERCISE 4
1.
A car is traveling with uniform velocity of 72 kmh-1. What
distance does it cover in 20s?
2.
Define uniform acceleration and average speed. A car starts from
rest with a uniform acceleration of 1.5m/s2 until it reaches a speed
of 24ms-1. It travels at this speed for 30s and then comes to rest
with uniform retardation in 10s. Draw on a graph paper a velocity
– time graph of the motion. Using calculation and graphical
methods determine:
(a)
(b)
(c)
the average speed of the car over the whole journey;
the distance travelled by the train; and
the retardation in the last 10 second.
3.
Use calculation method to solve all the problems treated earlier in
this unit.
4.0
CONCLUSION
In this unit you learned about the four types of motion namely random,
translational, rotational and oscillatory motion. The study highlighted
the fact that motion of an object is usually a combination of these
different types of motion. Also discussed was the idea of relative motion
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showing that motion involves the separation of an object relative to the
observer or point of reference. Furthermore, the parameters used in
describing motion namely time, position, distance, displacement, speed,
velocity and acceleration were examined. Finally, you studied
distance/displacement - time graph as well as speed/velocity - time
graph and how to solve problems relating to them.
5.0
SUMMARY
•
You can distinguish between four types of motion: -
(i)
random motion in which a body moves irregularly with no
preferred direction.
translational motion in which every point in a moving body
remains fixed in relation to other points,
rotational motion where every point in a body moves in
circles, and
oscillatory or vibratory motion in which a body moves to and fro,
reversing its direction, and returning regularly to its position.
(ii)
(iii)
(iv)
•
•
•
•
•
•
•
•
•
Relative motion of a body A with respect to another body B is the
motion A as observed from B. B is the frame of reference. In
general, all motion involves the separation of an object relative to
an observer (or point of reference).
Displacement is the distance travelled in a specified direction
Speed is the distance travelled in unit time.
Velocity is the rate of change of displacement with time
When a body moves with equal displacement in equal time
interval no matter how small the time interval may be, the
velocity is said to be a uniform or constant velocity.
The slope of a displacement/distance - time graph is the velocity/
speed.
Instantaneous velocity is the velocity at any instant in time
Acceleration is the rate of change of velocity with time.
The area under a velocity- time graph gives the total
displacement.
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6.0
TUTOR -MARKED ASSIGNMENT
1.
(a)
(b)
2.
Distinguish between
(a)
(b)
(c)
distance and displacement
speed and velocity
How can you find speed on a distance -time graph?
3.
A car starting from rest, accelerates uniformly so that it attains a
speed of 72 km/hr in 2 minutes. It travels at this speed for 5
minutes and is then brought to rest with uniform retardation after
3 minutes, Calculate the distance travelled by the train.
Determine also the retardation.
4.
The speedometer of a racing car reads the following values of
velocity v in the time t shown, starting from rest.
Time
(s)
Veloci
ty (m/
s)
List the four main types of motion
A bus is traveling along a straight road at 100kmh-1 and
the bus conductor walks at 8kmh-1 on the floor of the bus
and inthe same direction as the bus. Find the speed of the
conductor relative to the road in meters per second.
0 2
4
6
8
0 1
2
2
4
3
6
4
8
1
0
6
0
1
2
6
0
1
8
6
0
2
4
6
0
3
0
6
0
3
2
5
0
3
4
4
0
3
6
3
0
3
8
2
0
4
0
1
0
4
2
0
Sketch the velocity - time graph for the motion and obtain from the
graph:
(a)
(b)
(c)
66
the uniform acceleration during the first stage of the journey;
the retardation during the last stage of the journey; and
the total distance travelled.
PHY 001
7.0
ACCESS PHYSICS
REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). New School Physics for Senior Secondary
Schools Onitsha: Africana - FEP Publishers.
Awe, O. and Okunola, O. O. (1998). Comprehensive Certificate Physics.
SSCE Edition. Ibadan: University Press PLC (1992 Edition).
Nelkon, M. (2002). Principles of Physics for Senior Secondary Schools.
London: Longman Group Limited (1986 Edition).
Ndupu, B. L. N and Okeke, P. N. (2002). Round up Physics. A
Complete Guide. Ikeja: Longman Nigeria PLC.
Okpala, P. N. (1990). Certificate Year Series for Senior Secondary
School. Ibadan: NPS Educational Publishers Limited.
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UNIT 4
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MOTION UNDER GRAVITY
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Concept of Projectiles
3.2
Motion of Projectiles
3.3
Vertical and Horizontal Projections
3.3.1 Vertical Projection
3.3.2 Horizontal Projection
3.4
Resultant Velocity of a Projectile
3.5
Projection at an Angle to the Horizontal
3.6
Worked examples
3.7
Derivations of the Equations of Projectile Motion under
Gravity the Time of Flight (T)
3.7.1 The Maximum Height (H)
3.7.2 The Horizontal Range (R)
3.7.3 Applications of the Equations of Motion under
Gravity involving Numerical Problems
3.8
Applications of Projectiles
Conclusion
Summary
Tutor-Marked Assignments
References/Further Readings
1.0
INTRODUCTION
In unit 3 you studied motion in a straight line. Here you will learn how
fundamental laws of motion apply to motions along curved paths such
as the motion of projectiles along parabolic curves. A projectile motion
is a special case of motion in a plane. It has no motive power of its own
and will travel freely under the action of gravity and air resistance only
when it is launched into space.
2.0
OBJECTIVES
At the end of this unit, you will be able to:
•
•
•
•
68
identify a projectile motion
derive the time of flight, maximum height and range of a
projectile
solve simple problems involving the range, height and time of
flight of a projectile
note the applications of projectile motion in warfare and sport.
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3.0
MAIN CONTENT
3.1
Concept of Projectile
When you release a piece of stone from your catapult against a bird
perching on a tree branch, the stone travels in a parabolic path towards
the bird.
Also, if you throw a tennis ball against a wall, the path of the ball
towards the wall is a parabola. You will notice the same type of curve
when you project a ball horizontally from the top of a building. The
stone or ball you projected is
known as projectile, while the motion
of the stone or ball projected is known as projectile motion. The path
followed by a projectile is called its trajectory.
Examples of projectile motion are the motions of
•
•
•
•
•
•
•
3.2
kicked or thrown balls;
jumping animals;
objects dropped from windows;
bombs released from aircraft;
bullets fired from a rifle or gun
a stone released from a catapult;
an athlete doing the high jump.
Motion of Projectile
We will first consider the simple case of a stone thrown horizontally
with an initial velocity u from the top of a high wall or a cliff of height
h.
The stone is subject to two independent motions - a horizontal and a
vertical motion. Once the stone is free and before it hits the ground, no
horizontal force acts upon it. Its horizontal velocity will remain
unchanged until it hits the ground as shown in Fig. 4.1.
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U
U
U
V1
V2
U
V3
U
Fig. 4.1: Motion of a projectile
When you throw the stone forward initially, it has no vertical velocity.
The vertical pull of gravity accelerates it downwards at a rate of about
9.8 ms-2. The vertical or y - component of the velocity increases by
9.8ms-1 every second.
Hence, you will observe that any projectile in flight is doing two things
at once:
i.
ii.
it is flying horizontally with a constant speed.
It is moving up or down with an acceleration g.
Once you are able to recognise the above, solutions of projectile
problems
become easy for you. You simply split each problem into
two parts. One
involving horizontal motion at constant velocity the
other is exactly the same as the free fall due to gravity.
3.3
Vertical and Horizontal Projections
3.3.1 Vertical Projection
If you thrown a body such as a piece of stone vertically upwards, the
maximum height it will attain depends on the initial vertical velocity.
If H = maximum height reached by the object;
U = upward velocity of projection, i.e., initial velocity; and
g = acceleration due to gravity.
At the time t after projection,
the upward velocity v = u - gt; and the height
reached h = ut -½ gt2.
At the maximum height reached,
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V=0 and h=H
∴0= u-gt,
i.e. t1 =
u
g
Thus H = utl - ½ gt21
u2 – 1 u2
g 2 g
=
(substituting for tl i.e. tl = u/g)
Leading to H = u2 or 1 gt12
2g
2
(4.1)
Now, the body, starts from rest from the highest point (P) reached and
falls to earth under constant acceleration g (Fig. 4.2)
Speed
u
P
C
A
V
H
(a)
B
0
Fig. 4.2: Free fall:
vertical projection
Scope of
AB = -g
Slope of
BC = +g
Time
h
H
height
reached
(b)
Time
Fig. 4.3: Graphs of vertical projection
•
You should be able to show easily that the body strikes the
ground with a velocity which is equal in magnitude to the initial
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velocity of projection but opposite in direction. The time the body
takes to return to earth from the point P is also equal totI.
Therefore, total time of flight, T. of the body is 2t1
= 2u ;
g
⇒ T = 2u
(4.2)
g
Fig. 4.3: Illustrates the graphs of vertical projection
3.3.2 Horizontal Projection
When you roll a ball at fast speed along a horizontal table, after reaching
the edge of the table it continues with its horizontal motion. Now it also
experiences the pull of gravity. This downward pull deflects the ball.
From its original direction and it finally strikes the floor at a distance
which depends on the original horizontal speed and the height of the
table. This is illustrated in Fig. 4.4.
• • • •
•
•
•
•
•
•
•
Fig. 4.4: Flight of a ball with initial velocity horizontal
You can regard the motion of the ball as the combination of two
components: i.
a horizontal motion at constant velocity, and
ii.
a vertical motion under constant acceleration.
Suppose the total time of flight of the body from the instant it was
projected horizontally at A until it strikes the ground at B is T. Because
the horizontal velocity, u is constant, the horizontal range Ris given by
R = uT as illustrated in Fig. 4.5
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A
u
Direction of projection
Vx =U
Height of
cliff =H
Vy = w
v
Vx =U
Ground
level
Horizontal range R
B
UNIT 5
Fig. 4.5: Horizontal projection from the top of a cliff
You know that the vertical component of the motion has zero initial
velocity, and acceleration g. This means that in time T, the vertical
distance travelled, which is the same as the height of the cliff H is given
by:
H = ½ gT2
Therefore, the time of flight is
T ==
2H and R = u 2H
g
g
(4.3)
You should note that the last equation relates the horizontal range to the
height of the cliff and the initial horizontal velocity u.
3.4
Resultant Velocity of a Projectile
You know that the velocity of the projectile during flight is made
up of a
horizontal component, u, and a vertical component ω. This
is shown in Fig.
4.5.
The vertical component ω at a time t after the start of the motion
is given by ω = gt
The resultant velocity v = (u2 +ω2 )
√ (u2 + g 2t2 )
(4.4)
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The angle φ , between the direction of motion of the body and the
horizontal is given by
Tan φ = ω = gt
u u
i.e. φ = tan-1
3.5
gt
u
(4.5)
Projection at an angle to the Horizontal
Here we consider the more general case of a body projected at an
angle to the horizontal from the ground. The body moves along a
parabolic path, reaches a maximum height and returns to the
ground as shown in Fig. 4.6
U Sin θ
U
H
θ
O
U Cos θ
Horizontal range
Fig. 4.6: Projecting at an angle to the horizontal
If you project the projectile with an initial velocity u at an angle θ to the
horizontal, then the initial horizontal component of the velocity,
Vx = U Cos θ
Initial vertical component of the velocity Vy = U Sin θ
You will recall that the horizontal component of the velocity remains
constant because no force acts in the x - direction. The vertical
component of velocity changes with time because it is subject to gravity.
At any instance, t, the vertical component Vy is given by
Vy = U Sin θ - gt
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when the ball is moving upwards, but,
Vy =U Sinθ + gt
When the ball is moving downwards.
The height attained by the projectile at time t is given by
y = U Sin θ. t -½ gt2 (from s = ut -½ gt2)
=Ut Sinθ - ½ gt2.
(4.6)
Worked examples
•
A body projected vertically upwards with a velocity of 30ms-1.
How high does it travel before it comes to rest momentarily at the
highest point of its motion? What is the duration of the entire
flight? (Take g = 9.8 ms-2).
Solution
Initial velocity of projection, u = 30 ms-1 Acceleration of free fall due to
gravity, g = 9.8 ms -2
At time t after projection upwards, height reached is given by
V2 = u2 - 2gh
Where v = velocity at that instant, and v = 0 at the maximum height
⇒ h =H = u2 = 302m2s-2 = 45.92 m
2g 2 x 9.8
Also, v = u-gt
= 0 at the maximum height
∴t =t1 = u2 = 30mls
= 3.1s
-2
g 9.8 ms
•
Duration of the entire flight T = 2t1
=
3.1s x 2
6.2s
•
A boy throws a stone horizontally with a velocity of 40ms ' from
the top of a building 70 m high. How far from the foot of the
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building and with what speed does the stone strike the ground?
(Take g = 9.8 ms-2)
Solution
Horizontal velocity of projection, u = 40 ms-1 Height of building, H = 70
Duration of flight,
T = √
2H
g
Horizontal range of projectile, R = uT = u
.'. R = 40 ms-1
2H
√ g
2 x 70m = 40 x 3.78
√ 9.8 ms-2
= 151 m
At impact, vertical component of velocity = gT
Horizontal component is constant = u
=> speed of stone at impact
v = √ (u2 + g2T2)
=
u
√(u2 + 2g H)
= √ (402 + 2 x 9.8 x 70
= √ 2972
=
54.52 ms-1
SELF ASSESSMENT EXERCISE 1
1.
2.
What is a projectile motion?
A body is projected from the ground with a velocity of 35 ms - 1 at
an angle of 60° to the horizontal. Taking the acceleration of free
fall due to gravity as 9.8ms2, calculate:
(i)
(ii)
(iii)
76
the time of flight
the horizontal range
the velocity with which the body strikes the ground
PHY 001
3.
A young girl fired a stone from a catapult at an angle of 45° to
the horizontal with a velocity of 90 ms-1. Calculate:
(i)
(ii)
(iii)
3.6
ACCESS PHYSICS
the time it will take to reach the highest point;
the maximum height reached; and
its horizontal distance from the point of projection at this
instant.
Derivations of the Equations of Projectile Motion under
Gravity
3.6.1 Time of Flight (T)
You will recall from our earlier work that the height attained by the
projectile at time t is given by
y = ut Sin θ-½ gt 2
If time of flight is T, then,
y = Ut Sin θ-½ gT2
•
(4.7)
Time of flight (T) is the time required for a projectile to return to
the same level from which it was projected.
This implies as you know that after time T, y = 0.
Hence, from equation (4.7), you obtain
0 = UT Sinθ-½ gT2
T = 0 or T = 2u Sin θ
g
T =
2u Sin θ
g
(4.8)
You may use an alternative method of obtaining the formula for time of
flight by considering the vertical velocity at the maximum height.
Using V = U Sin θ- gt gives you
0 = U Sin θ-gt1,
Because at maximum height v = 0, tl is the time to reach the maximum
height.
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Hence,
t 1 = u Sin θ
g
(4.9)
You know that the time taken to rise from level of projection to
maximum height is equal to the time taken to return from the height to
the level of projection. Therefore, the total time of flight T is equal to 2t1
or
T = 2u Sin θ
g
3.6.2
•
The Maximum Height (H)
The maximum height (H) is the highest vertical distance attained
as measured from the horizontal projection plane.
For you to find H, you will use the equation of motion under gravity V2
= u2 -2gh
recalling that the vertical component of velocity of projectile is Vy = U
Sin θ
Hence,
V2 = (U Sin θ)2 - 2 gh
At maximum height, h = H, v = 0
Hence
0 = u2 Sin2 θ - 2 gH
2
2
H = u Sin θ
2g
.....(4.10)
3.6.3 The Range
•
Range is defined as the horizontal distance from the point of
projection to the point where the projectile hits the projection
plane again.
You will recall that the time taken to travel this horizontal
distance is T, the time of flight. Since the horizontal
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component of the velocity of
the projectile remains U Cos θ
throughout the flight, we have that:
R = U Cos θ x T
= U Cos θ x 2u Sin θ
g
2
Hence; R = 2u Sin θ Cos θ
from trigonometry, you know that Hence:
2 Sin θ Cos θ = Sin 2θ
2
R = u Sin 2 θ
g
(4.11)
The range is maximum when Sin 2θ of the above equation is maximum.
The Sine of an angle is maximum when the angle is 90°.
Thus, the
maximum value of Sin 2θ equals 1 when 2θ equals 90° or θ = 45°. This
means that R =
Rmax at θ = 45° and
Rmax = u2
g
(4.12)
That is, the maximum horizontal range is attained when the projection
angle is 45°.
3.7
Applications of the Equations of Motion under Gravity
in Solving Numerical problems
3.7.1 With an initial velocity of 36 ms-l, a gold player hits a ball at an
angle of 40° from the horizontal thus making the ball to move in
a vertical plane. Find the
(a)
(b)
(c)
(d)
(e)
(f)
time taken by the ball to its highest point,
maximum height of the ball;
horizontal range of the ball;
total time spent by the ball in air;
velocity of the ball as it strikes the ground (Take value of
g as 9.8 ms-2).
Show that the motion of the ball is symmetrical
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Solution
(a)
The time taken by the ball to its highest point
U Sin θ: u = 36m/s, θ = 400, g = 9.8 m/s2
g
t1 =
⇒ t1 =
(b)
36 Sin 400 = 2.36sec
9.8
Maximum height of the ball
H=
u = 36 m/s, θ = 40o g = 9.8 m/s2
Sin 2 40o = 0.413
362 Sin2 40o = 27.32 m
2 x 9.8
=
(c)
u2 Sin2 θ
2g
Horizontal range of the ball
R=
u2 Sin 2θ : u = 36 m/s, θ = 40o, g = 9.8 m/s
Sin (2 x 40)o = Sin 80o = 0.95
= 362 Sin (2 x 40) = 130.24 m
9.8
(d)
Total time spent by the ball in air
u = 36 m/s, θ = 40o, g = 9.8 m/s2
2U
T =Sin θ,
g
Sin 40o = 0.643
=
80
2 x 36 Sin 40o = 4.72 sec.
9.8
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(e)
Velocity of the ball as it strikes the ground
V =√ vx + vy =
√ u2 + w 2
= √ (Cos θ)2 + 2gH
= √ (36 Cos 40o)2 + 2 x 9.8 x 27.32 : u = 36m/s; θ = 40o
g = 9.8, H = 27.3 m
Cos 40o = 0.766
= √ 760.52 + 535.472
= √ 1296 = 36 m/s
OR V = √ (36 Cos 40o)2 + (36 Sin 40o)2 = 36 m/s
(f)
To show that the motion of the ball is symmetricalies
direction of velocity
tan θ =
=
vy
vx
√ 535.472
=
√ 760.52
23.14
27.58
tan θ = 0.839
∴θ = tan-1 0.839 = 39.997o ≅ 40o
This shows that the motion of the ball is symmetrical as illustrated in
Fig. 4.7.
U Sin θ
Fig. 4.7: Symmetrical motion of a projectile
3.8
Applications of Projectiles
From your knowledge and study of projectiles you can now deduce that
the principle of projectiles is used in warfare for aiming accurately at
targets, obtaining maximum height and obtaining maximum ranges. For
instance, you know that for a bomber from an aeroplane to hit a target,
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the bomb must be released when the target appears at a certain angle of
depression ~ given by:
tanφ =
1
u
gh
2
You also know that at certain angle of projection (0 = 45°) a projectile
would attain a maximum range.
In sports, the knowledge of projectile motions could be useful since
maximum ranges can be obtained in sports in javelin, short put, football,
golf, baseball etc. Also, attaining maximum height, maximum time of
flight are sometimes required.
SELF ASSESSMENT EXERCISE 2
1
2
3
Describe two situations where the knowledge of projectile motion
is found useful.
What conditions are necessary for an object thrown into the air to
describe a parabola?
A bullet leaves a gun with a velocity 450ms-1 at 38° to the
horizontal. Find the:
(a)
(b)
(c)
(d)
(e)
4.0
time taken by the bullet to its highest point;
the greatest vertical height reached by the bullet;
the horizontal range of the bullet
total time spend by the bullet in air
velocity of the bullet as it strikes the target.
CONCLUSION
In this unit you have learned how to identify a projectile motion as
motion under gravity. You have also learned how to derive the equations
of projectile - range, maximum height and time of flight. Further more,
you have learned how to apply the equations of motion under gravity in
solving numerical problems and the importance of the study of
projectiles in our everyday life.
5.0
SUMMARY
•
A projectile is an object which given an initial velocity (usually
obliquely or horizontally) is then allowed to move under the
action of gravity and air resistance. Usually, in order to simplify
the analysis air resistance is neglected.
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•
•
•
ACCESS PHYSICS
The flight of a projectile involves two motions - a constant
horizontal motion and a vertical motion under gravity.
The path or trajectory of a projectile is a parabola.
Time of flight is the time required for a projectile to return to the
same level from which it was projected.
2USinθ
g
The maximum height attained is the highest vertical distance
attained by the projectile as measures from the horizontal
projection plane.
T=
•
H=
•
U2Sin2θ
2g
The range is defined as the horizontal distance from the point of
projection to the point where the projectile to return to the same
level again.
R=
U2Sin 2θ
g
•
The range is maximum when θ equals 45°.
6.0
TUTOR-MARKED ASSIGNMENT
1.
Define the range of a projectile motion and write down an
expression for it. At what angle of projection is the range of a
projectile maximum?
2.
Given that the height attained by a projectile at a time t is Y = ut
Sin θ -½ gt2. Show that the time of flight is
T = 2USinθ
3.
g
A body is projected from the ground with a velocity of 38 ms-I at
an angle of 60° to the horizontal. Taking the acceleration of free
fall due to gravity as 9.8 ms2, calculate:
(i)
(ii)
(iii)
the time of flight,
the horizontal range, and
the velocity with which the body strikes the ground.
83
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REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). New School Physics for Senior Secondary
Schools Onitsha: Africana - FEP Publishers Limited.
Awe, O. and Okunola, O. O. (1992). Comprehensive Certificate Physics.
lbadan: University Press PLC
Okpala, P. N. (1990). Physics Certificate Year Series for Senior
Secondary School. lbadan: NPS Educational.
84
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UNIT 5
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SIMPLE HARMONIC MOTION (SHM)
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Definition of Simple Harmonic Motion
3.1.1 Example of Simple Harmonic Motion – Circular
Motion
3.2
Simple Harmonic Motion from Circular Motion
3.3
Amplitude, Period and Frequency of SHM
3.3.1 Amplitude (A)
3.3.2 Period (T)
3.3.3 Frequency (f)
3.3.4 Relation between f and T
3.4
Speed and Acceleration of SHM
3.4.1 Speed of SHM
3.4.2 Acceleration of SHM (Using Motion of Q)
3.4.3 Acceleration of SHM (Using Motion of P)
3.4.4 Relation between T, ω and f
3.4.5 Angular Acceleration (aθ)
3.5
Energy in Simple Harmonic Motion
3.5.1 Energy in an Oscillating Simple Pendulum
3.5.2 Energy in an Oscillating Loaded Spiral Spring
3.5.3 Period of an Oscillating Loaded Spiral Spring
3.5.4 Period of an Oscillating of a Simple Pendulum
3.6
Forced Vibration and Resonance
3.6.1 Free, Force and Damped Vibration
3.6.2 Resonance in a Vibrating System
Conclusion
Summary
Tutor- Marked Assignment
References/ Further Readings
1.0
INTRODUCTION
In units 3 and 4 of this first Module we studied the motion of a body
moving with uniform acceleration under the action of a constant force. It
was easy and useful to derive expressions for its velocity, acceleration or
displacement from a given point at any given time. In this unit, we will
discuss the motion of a body under the action of a resultant force which
is not constant but changes during the motion. Like the force, the
acceleration also changes at each instant. We shall particularly consider
the situation in which the body always returns to an equilibrium position
when displaced because a "restoring force" acts on it. The restoring
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force causes the body to oscillate ‘to and fro’ past the equilibrium
position.
2.0
OBJECTIVES
At the end of this unit, you should be able to:
•
•
•
explain and define simple harmonic motion
show the relationship between:
*
linear and angular speed,
*
linear and angular acceleration.
show the relation between period and frequency of the oscillating
body
calculate the energy in the system
explain forced vibration and resonance.
3.0
MAIN CONTENT
3.1
Definition of Simple Harmonic Motion
•
•
Simple Harmonic Motion (SHM) is the periodic motion of a body or
particle along a straight line such that the acceleration of the body is
directed towards a fixed point (or centre of motion) and is also
proportional to its displacement from that point.
3.1.1
Examples of Simple Harmonic Motion
Examples of Simple harmonic motion include the:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
86
Motion of a simple pendulum
Heart beat
Motion of a loaded test tube in a liquid
Motion of a mass suspended from a spiral spring
Motion of the balance wheel of a watch
Motion of the strings in a musical instrument e.g. guitar strings
Motion of the pistons in a motor vehicle engine
Motion of the prongs of a sounding tuning fork
Motion of a child's swing
Motion of the diving board in a swimming pool.
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Simple Harmonic Motion from Circular Motion
rω
θ
A
D
Fig. 5.1: SHM from circular motion
Consider a particle, P moving round a reference circle with centre 0 and
radius r with a uniform speed v. If the angular velocity of the radius
vector OP is ω, then a = rω. Let us draw PQ perpendicular to the
diameter AB at Q so that Q is the projection of P on AB. As P moves
round the circle, Q moves ‘to - and – fro’ along AB, being monetarily at
rest at A and B and moving with maximum speed as it passes through 0,
the centre of the path AB. Thus the motion of Q along AB is simple
harmonic while the particle P moves with a constant speed 6 round the
circle ADBCA.
From Fig. 5.1, x = r Cos θ, where x is the horizontal distance of Q from
the centre 0 at that point in time. As the particle moves from B through
C, A, D and back to B (i.e.round the circle), the angle 0 swept through
by the radius vector from B to P varies from 0° at B to 90° at C, to 180°
at A, to 270° at D and to 360° or 0° back to B. Hence Cos θ varies from
1 to 0 to -1 to 0 and back to 1 as x varies from r to 0 to -r and to r again.
Thus the value of x, the horizontal displacement from the central
position 0, varies from the value r to -r and we can say that -r ≤ χ ≤ r.
x
A
٥
r
0
٥
B
r
Fig. 5.2: Simple harmonic motion in a straight line
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This is simple harmonic motion from the geometrical construction of the
motion of a particle in simple circular motion. The circular path of the
particle is called "reference circle" and, as we have seen, is quite useful
for studying simple harmonic motion.
3.3
Amplitude, Period and Frequency of SHM
3.3.1 Amplitude (A)
The maximum value of x, the displacement from the equilibrium
(central) position 0, is known as the amplitude (A) of the SHM. In Fig.
5.2, A = r.
Definition: The amplitude (A) of a simple harmonic motion is the
maximum displacement of the body performing simple harmonic motion
from its equilibrium or central position 0.
3.3.2 Period (T)
The total time taken by the point Q to move from B to A and back to B
is called the "period" of the SHM. It is also the same time taken to move
from 0 to B then to A and back to 0 (Fig. 5.1). This is the time taken by
the body in SHM to make one complete oscillation from one point back
to that same point but in the same direction.
Definition: The period (T) of a simple harmonic motion is the total time
taken by an oscillating body to make one complete oscillation or cycle
or revolution about a central reference point.
3.3. 3 Frequency (F)
If a body makes n oscillations in a given time t seconds then the number
of oscillations per second is called the frequency of the SHM and is
given by
f=
n
t
Definition: The frequency (f) of a simple harmonic motion is the
number of complete oscillations per second made by the vibrating body.
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3.3.4 Relationship between f and T
n complete oscillations or cycles are made in t seconds
∴
Number of complete oscillation or cycle in 1 second,
f=
n
t
Time for complete oscillation or cycle, T = t
n
= 1 = 1
n
f
t
1
∴T =
f
3.4
Speed and Acceleration of SHM
3.4.1 Speed of SHM
From Fig. 5.1, we note that as P moves round the circle once, it sweeps
through an angle of θ = 360° or 2π radians in the period T, the period of
motion. The rate of change of the angle θ with time t is called the
angular velocity ω. That is, angular velocity (ω) is defined as follows:
ω =
angle turned through by the body
time taken
∴ω =
θ
t
∴θ = ωt
θ is measured in radians (rad), where 2π rad = 360°.
θ
ω is measured in radians per second (rad/s), since ω
t =
As the angle θ is changing (Fig. 5.1) with time, the arc length S = BP is
also
changing with time. S= θ ∴S = rθ
2πr
2π
By definition θ (in radians) = s and hence S = rθ
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where r = amplitude (A) = radius of the circle.
The angular velocity (ω) is given by
ω=
=
θ
t
=
S
/r
t
s.1
t r
But s/t = v, the linear velocity of the particle
Hence, ω = v. 1/r
∴v = rω = Aω
Thus the linear speed is equal to the product of the angular speed and the
radius or amplitude of motion.
Examples:
1.
A small metal object is tied to the end of a string and whirled
round a circular path of radius 30 cm. The object makes 20
oscillations in 5 seconds. Calculate the angular and linear speeds
of the object.
Solution
20 complete oscillations = 360° x 20
= 2π x 20 rad.
= 40π rad
Angular speed, ω = θ = 40π rad = 8π rad/sec
t 5s
Linear speed, v = r ω = 30 x 8π cm
s
2.
90
A loaded test tube floating upright in water oscillates vertically. If
it makes 25 complete oscillations in 5 seconds, calculate the
period and frequency of the motion.
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Solution
time taken
number of vibrations
Period, T =
=
5s
25
frequency f =
= 0.2s
1 = 1 s-1 or Hz
T 0.2
= 5 HZ
3.4.2 Acceleration of SHM (Using Motion of Q)
From Fig. 5.1, the linear velocity v at the point Q whose distance from
the central point 0 is x, is given by
v = ωPQ
ω√ r2 – x2
The maximum velocity Vm corresponds to the velocity at 0 where x = 0.
Hence Vm=rω
Thus the maximum velocity of SHM, occurs at the equilibrium or
central position (x = 0) while the minimum velocity of SHM occurs at
the extreme position of motion (x = r) and is numerically equal to zero.
An expression relating linear acceleration with the angular velocity can
be obtained using differential calculus as follows:
From F ig. 5.1, x = r Cos θ
Where x is the displacement of the particle from the centre of motion
(same as the distance of Q from 0).
But θ = ωt, hence
X = rCos ωt
Where ω is the angular speed and r is the amplitude of vibration.
Differentiating, we have
dx
dt
= rω Sin ωt
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But dx = rate of change of displacement with time
dt
= linear velocity v
∴
dx = v = - rω Sin ωt
dt
We differential v in order to get the acceleration, as follows:
a=
dv= - rω2 Cos wt since x = r Cosωt.
dt
= - ω2 x
Hence, acceleration, a = - ω2 x
Thus the linear acceleration of simple harmonic motion is the product of
the square of the angular speed and the displacement, x of the particle
from the centre of motion.
The negative sign in the acceleration equation indicates that the
acceleration ‘a’ is always directed towards the centre of motion while
the displacement is measured from the centre outwards, i.e. the two
vectors always act in opposite directions at any point in time. This is an
essential feature of SHM, namely that (i) the (linear) acceleration of the
vibrating object and hence the force tending to restore the object to its
equilibrium position are always in a direction opposite to the
displacement of the body, and (ii) the acceleration and force are always
proportional to the displacement of the body from the equilibrium
position.
3.4.3 Acceleration of SHM (Using Motion of P)
The equation (a = - rω2 x) for the linear acceleration of SHM can also be
obtained by considering the circular motion of (the particle) P around
the reference circle (Fig..5. l ) as follows:
The centripetal acceleration `a' of the particle is along PO (the radius of
the circle) and its value is given by:
Centripetal acceleration, a =
-v2
r
Where r = radius of circle = amplitude of SHM
But v = rω, hence
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Acceleration, a = - v2 = - r2ω2
r
r
2
= rω
∴ a = - rω2
Note that the component of this centripetal acceleration along OB (ax) is
given by ax, = aCosθ = -ω2rCosθ
∴ax = -ω2x, since x = r Cos θ
centripetal acceleration = v2 = - w2r, and
r
Thus: (i)
Linear acceleration = - w2 x.
(ii)
3.4.4 Relation between T, w and f
It has been shown (section 3.4.1) that by definition, ω = θ/t
When t = T, then θ = 360° or 2π radians. Hence
ω=
Or
T=
2π
T
2π
ω
Similarly, the frequency f is given by
f= 1 = 1 .
T
2π/ω
T
Hence f =
2π⁄ ω
ω
2π
Or ω = 2πf
Example:
A body in simple harmonic motion oscillates with a frequency of 40Hz
and amplitude of 2 cm. Calculate.
(i)
(ii)
the period.
the acceleration at the middle and at the end of the path of
motion.
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(iii)
(iv)
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the velocities at the middle and at the end of the path of motion.
The velocity and acceleration at a distance of 1.0 cm from the
centre of motion.
Solution
(i)
T=
1
f
1
= 0.025sec
-1
40s
=
=
(ii)
1
40
s
Acceleration, a = -ω2 χ
(a)
(b)
At middle of path, x = 0
∴a = -ω2 x 0 = 0
At end of path, χ = amplitude = 2 cm
∴ a = -ω2 x 2
but ω = 2πf = 2π x 40s-1 = 80π rad
s
2 2
2
∴a = (80π rad ) x 2cm = 128 π x 102 cm
s2
s2
= 128 x 103 cm-2
= 1.28 x 103ms-2
(iii)
V = ω√r2 – x2
(a)
At χ = 0, v = ω√r2 – 0 = ωr = 80π x 2 ms-1
100
∴V = 1.6πms-1
(iv)
(b)
At χ = r, v = ω√r2 – r2 = ω x 0 = 0
(a)
V = ω√r2 – x2
= ω√22 – 12 (x = 1cm)
= 80π√3cm s-1
= 0.8π √3ms-1
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(b)
a = -ω2x
= (80π)2 x 1.0 cms-2
= (80π)2 x 1 ms-2
100
= 6400π2 x 1 ms-2
100
-2
= 640 ms
3.4.5 Angular Acceleration (aθ)
Definition: The angular acceleration (aθ) of a body in simple harmonic
motion is the time rate of change of its angular velocity (ω).
Unit of aθ: Angular acceleration (aθ) is expressed in radians per second
(rad s-2)
Suppose the angular velocity of a body in SHM changes uniformly from
ω1 to ω2 in t seconds, then angular acceleration (aθ) is given by
aθ = change in angular velocity
Time taken for the change
= ω2 - ω1
t
But ω = v or v = rw, hence
r
V2 – V1
aθ = r
r
t
= 1 (v2 – v1)
r
t
=
a
r
where a is the linear acceleration of the body
∴ a = raθ
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Thus linear acceleration (a) is equal to the product of angular
acceleration (aθ) and the radius (r) (or the displacement of the particle
from its equilibrium or central position, x).
3.5
Energy in Simple Harmonic Motion
When a body is in SHM, there is always a restoring force which tends to
return the body to its central or equilibrium position. Since "force" and
"displacement" are involved, it follows that "work" and "energy" are
also involved in SHM. In simple harmonic motion, the total energy of
the system is made up of only two types of energy, namely Potential and
Kinetic energies. At any instant of the motion, the energy of the system
may be only Potential energy (PE) or only Kinetic energy (KE) or a
combination of both PE and KE. Thus in SHM, there is a constant
transformation of energy from Potential energy to kinetic energy and
from Kinetic energy to potential energy as the body oscillates about its
equilibrium position. The total energy of a body in SHM is always
constant and at any point in time is given by
Total energy of SHM = Potential energy (PE) + Kinetic energy (KE)
(a)
(b)
Fig. 5.3: Kinetic and potential energy variation
Fig. 5.3 (a) shows how the PE and KE vary with the displacement x
from the centre of motion r is the maximum displacement or amplitude.
Fig.1.5.3. (b) shows how the P.E. and KE vary with time t, when t is
measured from the centre of motion 0. If T is the period, then T/4 is the
time when the object is at the end of its oscillation, T⁄2 is the time when
it returns to the centre, and so on. The graphs have the shape of sine
curve.
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We shall illustrate energy in SHM using two SHM system as examples:
(i)
(i i)
motion of a simple pendulum, and
motion of a loaded spiral spring.
3.5.1 Energy in an Oscillating Simple Pendulum
R
Q
P.E. = mgh
(P.E. is max.)
P
h
P.E. = mgh
(P.E. is max)
V = 0, K.E. = 0
0
h = 0, P.E. = 0
K.E = ½ mv2
K.E. is max., V is min
Fig. 5.4: Energy transformations of a simple pendulum
Fig.5.4 shows the energy transformations in a system comprised of an
oscillating simple pendulum. The energy at the extreme positions R and
Q is all PE and is equal to mgh where m is the mass of the bob, h its
height above the lowest level and g is the acceleration due to gravity. At
0, h = 0 and PE = 0, but velocity of the moving bob is maximum (Vm).
Hence the KE is
2
maximum and is equal to ½m √m
Thus,
mgh =½mVm2
or am = √2gh
At any instant of the oscillation of the simple pendulum,
Total energy = Potential energy + Kinetic energy
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i.e. Total energy = mgh + ½ Mv2 = mgh =½mVm2
where hm = maximum height reach of by the pendulum bob and
Vm = maximum velocity attained by the bob.
3.5.2 Energy in an Oscillating Loaded Spiral Spring
equilibrium
position
e
y
Fig. 5..5: Energy in a vibrating loaded spring
If a mass suspended from the end of a spiral spring is stretched
vertically downwards and released, it oscillates up and down in SHM.
The force that tends to restore the spring to its centre of motion as it
oscillates is the elastic (restoring) force and is given by
F = -ky
where k is the force constant, y is the displacement of the spring from
the equilibrium position or centre of motion. The total work (W) done in
stretching the spring through the distance y is given by
W = average force x displacement
= ½ ky x y = ½ ky2
Thus the maximum total energy stored in the spring is given by
W = ½ kr2
where r is the amplitude of motion or maximum displacement from
equilibrium position. This maximum energy is conserved throughout the
oscillation of the SHM system.
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At any point in time during the oscillation, the total energy is
W = ½ kr2 = ½ mv2 + ½ ky2
(KE)
(PE)
where v is the velocity of the suspended mass at a distance y from the
equilibrium position.
3.5.3 Period of Oscillation of a Loaded Spring
We have seen that the total energy (W) in a loaded spring in SHM is
given by
W = ½ kr2 = ½ mv2 + ½ ky2
∴ v2 = k (r2 – y2)
m
Thus the linear velocity (v) of the suspended mass is given by
V=
k (r2 – y2)
m
Comparing the two equations above, we have that
w=
k
m
w2 = k
m
Hence the period T is given by
T = 2π = 2π
w
∴ T = 2π
m
k
m
k
The period (T) therefore depends only on the mass (m) and the force or
elastic constant (k) of the spring. It does not depend on the amplitude of
vibration or on the value of g at the place of vibration. The value of k
can be obtained from Hooke’s law using F = mg = ke
where e is the extension produced in the spring by the mass m. Hence
K = mg
e
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In obtaining the formula for the period T of the oscillating loaded
spring, the mass (ms) of the spring has been assumed negligible. If the
mass of the spring is taken into account, then the true value of the period
T, is given by
T = 2π m + ms
k
3.5.4 Period of Oscillation of a Simple Pendulum
P
θ
T
l
L
x
0
Q
Mg sin θ
θ
Fig. 5.6: Theory of simple pendulum
To derive the formula for the period (T)of an oscillating simple
pendulum, all that is needed is to obtain the value of the angular velocity
(w) of the system in terms of the parameters of the system, since
T = 2π
ω
Consider a simple pendulum oscillating in such a way that at a time t,
the bob is at Q where OQ = x and angle OPQ – ө ( Fig.5 .6) With the
bob at Q the restoring force which pulls it towards O, the centre of
motion is equal to mgsinθ. PQ is perpendicular to the tangent at Q,
hence the tension in the
spring has no component along QO.
Therefore.
Restoring force = ma = - mgsinθ
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where a is the acceleration along the arc OQ and the negative sign
indicate that the force towards and the displacement x of the bob from O
are in opposite directions.
If ө is very small, then sinθ = θ in radians. Again from Fig.5.1; we have
that
Using the sector OPQ
x
= θ; =⇒θ =
2πl
2π
x
l
Hence the above equation for the restoring force becomes:
Restoring force = ma = -mgSinө = - mg x
C
∴
a
∴
= - gx
l
= - ω2x
ω2 =
g l
Or
ω =
√g l
The period of oscillation (T) of the simple pendulum is given by
T = 2π
ω
Hence
T = 2π
= 2π √ L/g
√g/l
∴
∴
2π √L/g
Thus since g is constant at a particular place the period of a simple
pendulum depends only on the length l of the pendulum. Since g varies
from place to place, the period (T) of a simple pendulum will also vary
from place to place.
Example
A mass of 25 g is suspended from the end of a spiral spring whose force
constant is 0.5 Nm-1.
If the system is set into simple harmonic motion with amplitude 20cm,
calculate the
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(i)
(ii)
(iii)
(iv)
(v)
(vi)
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Period of the motion,
Frequency of the motion,
Angular speed,
Total energy of the system,
Maximum velocity of the motion, and
Maximum acceleration of the system
Solution
(i)
T
=
=
2π 2π
ω
2π
=
(ii)
F
=
√ m/k
√ 0.025
0.5
1.41 sec
1
=
T
0.71 Hz
=
1 .
1.41
(iii) w = 2πf = 2π x 0.71 = 4.46 rad S-1
(iv) Total energy, E = ½ kr2
= 1/2x 0.5x (0.2)2
J
= 0.01J
(v) Let maximum velocity of motion be Vm, then
½ mvm2 = ½ kr2 = 0.01
:vm2
= 2x0.01
m2
0.025
s2
:(vi)
102
Vm
=
Vm
=
m2
s2
0.89 ms-1
√0.08
Maximum acceleration a is given by
a
=
- ω2r
=
(4.46)2 x 0.2 m
s2
= 3.98ms-2
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Force Vibration and Resonance
3.6.1 Free, Forced and Damped Vibration
When a system is performing simple harmonic motion, it gradually loses
its energy owing to friction within its parts as well as air resistance. The
amplitude of such a motion reduces gradually but steadily until it
eventually decreases to zero. Such a motion is known as “damped”
motion. If there is no damping, the motion is said to be free and the
amplitude remains constant with time. If the free oscillation of a body in
SHM is to be maintained for a long time, an “external period force”
would be required to force the body to continue its oscillatory motion
inspite of the damping. The oscillation of such a body is then said to be
a “force vibration”.
Time
Free oscillation
Time
Damped oscillation
Fig.5.7 above illustrates the sketch displacement – time graphs of a body
performing (a) free simple harmonic oscillations, and (b) damped simple
harmonic oscillation.
Definition
Forced vibrations are oscillations resulting from the action of an external
periodic force on an oscillating body.
3.6.2 Resonance in a Vibrating System
Every vibrating object possesses a frequency (f) of vibration. The
frequency (fo) with which the object would oscillate if it is left
undisturbed after being set into vibration is called the “natural
frequency” of vibration. If the object is subjected to an external periodic
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force, the frequency of such an external force is called the “forcing
frequency (fe)”.
If the frequency of a vibrating system which acts on another system
coincides with the natural frequency of the second system, then the
second system is set into vibration with a relatively large amplitude.
This phenomenon is known as “resonance”
Definition
Resonance in a system performing SHM is said to occur when the
forcing frequency (f) of an external periodic force coincides with the
natural frequency (fo) of a body with which it is in contact, causing the
body to vibrate with a large amplitude.
At resonance, maximum energy is transferred from the periodic external
force to the natural vibrations of a system.
Examples of Resonance in vibration systems
(i)
(ii)
(iii)
(iv)
A vibrating tuning fork, pressed against a table top, is heard to
give a louder sound.
Divers make use of resonance to obtain a maximum lift –off from
diving boards. They do this by tuning the frequency of their
bounce on the diving board with the natural frequency of the
diving board.
Glass windows may vibrate as a result of the high- pitched notes
emitted form a nearby radio set.
Car bodies rattle at certain car speeds as a result of the resonant
vibrations, induced in parts of the car when the natural frequency
of vibration coincides with the frequency of vibration of the car
engine.
SELF ASSESSMENT EXERCISE 1
1.
What do you understand by simple harmonic motion? Give three
examples of simple harmonic motion.
2. Explain the meaning of (i) amplitude, (ii) complete oscillation of an
object in simple harmonic motion.
3. An object in simple harmonic motion has an amplitude of 0.05m and
a frequency of 30 Hz. Calculate
(i)
(ii)
104
the period of oscillation ,
the accelerations at the middle and at the end of an
oscillation,
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(iii)
The velocities at the middle and at the end of an
oscillation.
4. Describe the energy transformations of the bob of a simple pendulum
Oscillating about its equilibrium position .
Calculate the period of oscillation of an object which vibrates at 10
cycles per second.
4.0
CONCLUSION
In this unit you learned about simple harmonic motion – Definition,
example and terms associated with simple harmonic motion. You also
learned speed and acceleration of simple harmonic motion and energy in
simple armonic motion. Finally, you learned forced vibration and
resonance in simple Harmonic Motion. (SHM).
5.0
SUMMARY
1.
Simple harmonic motion is characterized by regularity and
repetition of vibrational (to – and –fro) motion.
In SHM the acceleration is always directed towards a fixed point
(the centre of motion) and is directly proportion to the distance of
the particle from the fixed point.
The following relations hold in SHM
2.
3.
(i)
(ii)
(iii)
(iv)
(v)
Acceleration, a = - w2x , a max = - w2r
Period , T
= 2π
w
Frequency, f
= 1/T = w
2π
Linear velocity, v = + ω √ r2 - x2
v = + ωr
Angular velocity, w = θ/t
where r = amplitude of motion, and v is the velocity when
the displacement from the centre of motion is x.
(vi)
4
a
b.
Linear acceleration a = aө r, where aө
acceleration.
is the angular
For a simple pendulum of length l, the period of oscillation
is given by T
= 2π
= 2π
√ l/g
w
For a loaded spring in SHM
T
=
2π √ m/k
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where K is the spring constants
5 Resonance occurs in a vibrating system when the forcing frequency
of an external periodic force coincides with the natural frequency of
a vibrating body.
6.0
TUTOR-MARKED ASSIGNMENT
1.
By clearly defining each term, distinguish between, (i) angular
velocity and linear velocity, (ii) angular acceleration and linear
acceleration of a body in simple harmonic motion.
Define simple harmonic motion. A body in simple harmonic
motion has an amplitude of 10 cm and a frequency of 100Hz.
Calculate:
2.
(a) The period of oscillation,
(b) The acceleration at the maximum displacement
(c) The velocity at the centre of motion.
3.
Write the equation for the total energy of an oscillating loaded
spring in terms of the potential energy (PE) and the Kinetic
energy (KE) at any stage of the motion, explaining all the
Symbols used.
Hence or otherwise, derives an expression for the period, T of the
System.
7.0
REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). Physics for Senior Secondary Schools.
Onisha: Africana- FEP Publishers Ltd.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics,
SSCE Edition, Ibadan University Press Plc.
Ndupu, B.L.N. and Okele, P.N. (2000). Round - Up physics for West
African Senior School Certificate Examination: A Complete
Guide. Lagos: Longman Nigeria Plc.
Nelkon, M. (1986). Principles of Physics for Senior Secondary School.
(9th Edition).England: Longman Group Ltd.`
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MODULE 2
Unit 1
Unit 2
Unit 3
Unit 4
Force, Centre of Gravity and Equilibrium
Friction in Solids and Liquids
Linear Momentum
Simple Machines
UNIT 1
FORCE, CENTRE
EQUILIBRIUM
OF
GRAVITY
AND
CONTENTS
1.0
2.0
3.0
Introduction
Objectives
Main Content
3.1
Force
3.1.1 Concept of Force
3.1.2 Definition and Nature of Force
3.2
Types of Force
3.2.1 Contact Forces
3.2.2 Non- Contact Forces or Force Fields
3.3
Concepts and Types of Equilibrium
3.3.1 Concepts of Equilibrium
3.3.2 Types of Equilibrium
3.3.3 Resultant and Equilibrium Forces
3.3.4 Equilibrium of Three Forces Acting at a Point
3.4
Moment of a Force
3.4.1 Principles of Moments
3.4.2 Clockwise and Anticlockwise Moments
3.5
Conditions of Equilibrium
3.5.1 Conditions of Equilibrium under the Action of
Parallel Coplanar Forces
3.5.2 Conditions of Equilibrium under the Action of
Non- Parallel Coplanar Forces
3.5.3 Equilibrium under the Action of Three Non-Parallel
Forces (Special case)
3.6
Centre of Gravity (CG)
3.6.1 Position of CG for some Regular Uniform Bodies
3.7
Types of Stability of Equilibrium
3.7.1 Stable Equilibrium
3.7.2 Unstable Equilibrium
3.7.3 Neutral Equilibrium
3.7.4 Effect of Height of CG on Stability of Equilibrium
3.7.5 Application of the Effect of Height of CG on
Stability of Object
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3.8
4.0
5.0
6.0
7.0
Equilibrium of Bodies of Liquids
3.8.1 Concept of Upthrust
3.8.2 Archimedes Principles
3.8.3 Verification of Archimedes Principles
3.8.4 Principle of Floatation
3.9
Density and Relative Density
3.9.1 Measurement of Density and Relative Density
3.9.2 The Hydrometer
3.9.3 The Practical Hydrometer
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In units 3 and 4 of Module 1, the concepts of motion, types of motion,
and the parameters used in describing motion were considered. No
mention was made of what causes motion. This unit discusses the cause
of motion, its nature and types. The unit also treats the concepts of
centre of gravity, equilibrium of bodies and moment or turning effect of
a force about a point.
2.0
OBJECTIVES
At the end of the unit you should be able to:
• define force and explain its nature
• identify, explain and distinguish between various types of force.
• explain the meaning of centre of gravity and identify its position for
some regular uniform and irregular bodies.
• explain the concept of equilibrium and distinguish between (a) static
and dynamic equilibrium (b) resultant and equilibrant forces
• explain the conditions that must be satisfied if an object is to be kept
in equilibrium under the action of:
-
parallel
non-parallel forces.
• explain what is meant by moment of force about a point
• take moments of given forces about any point and show their
directions
• solve simple problems on objects kept in equilibrium by a number of
given forces
• explain the effect of cg on the stability of a body
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• identify the forces acting on a body partially or completely immersed
in a liquid and state the conditions for the body to float in the liquid.
3.0
MAIN CONTENT
3.1
Force
3.1.1 Concept of Force
Force is the agent which causes objects to move. It usually manifests in
the from of a “push” or a “pull”.
An object, e.g a book which is placed on a table is (macroscopically) at
rest unless it is caused to move visibly by
i.
ii.
iii.
Pushing it
Pulling it or
Lifting it.
When a body is moving in a straight line with uniform velocity, it will
continue in that manner unless a push or a pull is applied to it to change
its speed or direction, or both speed and direction.
Force may also change (or tend to change) the shape of a body. It is
difficult to describe force because it is not a visible entity. However, its
effects can be seen and therefore be described.
3.1.2 Definition and Nature of Force
Definition of Force
Force may be defined as a push or a pull which changes or tends to
change the shape or state of rest or uniform motion of a body in a
straight line.
Nature of Force
The nature of force may be summarized as follows:
i.
ii.
iii.
iv.
Force is a push or a pull
It is the thing or entity that causes objects to move.
It is a vector quantity:-it always has both magnitude and
direction.
Force is not a visible entity and can be described through its
effects which are easy to see.
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v.
When a body A exerts a force (action) on a body B, B also exerts
an equal force (reaction) in the opposite direction.
3.2
Types of Force
There are two main types of forces:
i.
ii.
Contact forces
Non-contact forces or force fields.
3.2.1 Contact Forces
This is a force which is exerted when two bodies or surfaces are in
contact with each other, directly or indirectly. Examples include a push,
a pull, reaction forces, friction, tension forces, thrust forces.
3.2.2
Non-contact Forces or Force Fields
This is a force which is exerted between objects which are not in contact
with each other. A force field makes itself felt without the presence of
any material connection between the bodies involved. Force fields are of
the action at a distance type. Suppose an object X sets up force field, it
is observed that within a fixed region or neighbourhood of X, the
influence of X on Y can be felt or detected and we talk about Y being in
the field of X. In some cases, it can also be considered that Y sets up a
force field and that a force is accordingly exerted on X in the field of Y.
Examples of force fields are gravitational, magnetic, electrostatic or
electric force fields and nuclear force fields. A brief description of each
of these force fields is presented below.
a.
Gravitational Force Field
Every object in the universe attracts other objects with a gravitational
force. This gravitational force is responsible for the orbital motion of the
moon round the earth or the planets round the sun. it was discovered by
Sir Isaac Newton (1642-1727).
b.
Electrostatic or Electric Force Field
Electrostatic force is the force between two electrical charges. A force of
attraction is experienced if the two charges are unlike (positive and
negative) while a force of repulsion is experienced if the charges are
alike (positive and negative or negative and negative). The law
governing the force between electrical charges was discovered by a
French scientist, Charles Coulomb (1736-1806).
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Magnetic Force Field
A permanent magnet or an electromagnet (a wire or coil carrying
current) sets up a force field in a region or neighbour hood surrounding
it. Another magnet or electromagnet at any point within this region or
neighbourhood experiences a force of attraction or repulsion due to the
magnet/electromagnet. Such a force is called a magnetic force field. A
moving charge also sets up a magnetic force field in addition to the
electric field which surrounds it. A second charge in these two combined
fields will experience a force due to the magnetic field only if it is in
motion.
d.
Nuclear Force Field
There exists, in the nucleus of an atoms, the strongest force in nature
which holds together the protons and neutrons that make up the nucleus.
If this force were not there, then the electrostatic forces of repulsion
between the protons which carry positive electrical charges would cause
the nucleus to disintegrate. A very important feature of nuclear forces is
that they are short ranged- two nuclear force between two nucleons is
negligible if the separation between the nucleons is greater that about
10-15m.
Another important feature or property of nuclear forces is that they are
independent of charge, i.e the nuclear force between two protons is the
same as that between two neutrons and the same as that between a
proton and a neutron.
3.3
Concepts and Types of Equilibrium
3.3.1 Concepts of Equilibrium
A body is said to be in equilibrium if its acceleration is zero, i.e there is
no net force acting on it in any direction. Therefore, the body may be at
rest or moving with constant velocity.
Definition
A body is said to be in equilibrium if:
i.
The body as a whole remains at rest
ii.
The body is moving in a straight line with constant velocity and
iii.
The body is rotating at a constant angular velocity.
3.3.2 Types of Equilibrium
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There are two types of equilibrium: static equilibrium and dynamic
equilibrium.
i.
Static Equilibrium
If a body has zero acceleration by virtue of being at rest, the body is said
to be in static equilibrium.
An example is a book resting on a table.
ii.
Dynamic Equilibrium
If a body is moving with a constant velocity in a straight line or is
rotating with a constant angular velocity about a fixed point or axis
through its centre of mass, the body is said to be in dynamic or kinetic
equilibrium.
An example is a car whose engine is running at constant speed for a
considerable length of time.
3.3.3 Resultant and Equilibrium Forces
Resultant (R)
R
P
Q
O
S
Equilibriant
Fig.1.1: Resultant and equilibrium forces
Consider the force board arrangement shown in Fig 1.1 above. Suppose
the force P, Q and S are in an equilibrium state. The resultant R of P and
Q acting at O can be obtained by the principle of parallelogram of
forces. The resultant R would be found to be equal in magnitude but
exactly opposite in direction to the third force S which keeps the point O
in equilibrium. This third force is known as the equilibrant force and
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counterbalances the other two forces P and Q while R is called the
resultant (force) of P and Q.
Definitions
i.
ii.
A resultant force is that single force which, acting alone, will
have the same effects in magnitude and direction as two or more
forces acting together.
The equilibrium of two or more forces is that single force which
will balance all the other forces taken together. It is equal in
magnitude but opposite in direction to the resultant force.
Note: If three forces F1, F2, F3 acting at a point are in equilibrium (Fig.
1.2), the resultant of two of these forces is equal but opposite in
direction to the third force.
F1
F2
F3
Fig. 1.2: Three forces in equilibrium
Any of these forces is said to be the equilibrant of the other two. Again,
the lines of action of F1, F2 and F3 must all pass through common point
(O in this case).
Worked example
Two forces 5N each are inclined to each other at 120 o. Find the single
force that will
a.
b.
Replace the given force system
Balance the given force system
R
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5N
120
∝
5N
E
The resultant force (R) is given by
R2
=
52 + 52 + 2 x 5 x 5 Cos 120o
:. R
=
=
=
=
52 + 25 + 50 Cos (90o + 30o)
25 + 25 – 50 Sin 30
25
5N
Using the Sin rule, we have
5
Sin α
=
R
Sin60
=
5
Sin 60
:.
5
=
Sin α
5
Sin 60
:.
:.
Sin α =
α
=
Sin 60
60o
The resultant (R) is a 5N force at 60o with each of the given forces.
3.3.4 Equilibrium of Three Forces Acting at a Point
(Parallelogram and Triangle of Forces)
i.
Parallelogram of Forces
The principle of parallelogram of forces states that if two inclined forces
are represented in magnitude and direction by the adjacent sides of a
parallelogram, their resultant is represented in magnitude and direction
by the diagonal of the parallelogram passing through the point of
intersection of the two sides.
ii.
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Triangle of Forces
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The principle of triangle of forces states that if three forces are in
equilibrium, they can be represented in magnitude and direction by the
three sides of a triangle taken in order.
B
C
Q
R
C
(a)
E
O
Q
(b)
E
P
A
O
P
A
Fig.1.3: Parallelogram and triangle of forces
Fig.1.3 (a) Shows three forces P, Q and E which keep the point O in
equilibrium. OACB is the parallelogram of forces in which OA
represents P in magnitude and direction and OB represents Q in
magnitude and direction. The diagonal of the parallelogram OC
represents the resultant R of P and Q in magnitude and direction. R is
counterbalanced by the equilibrant force E.
Triangle OAC in Fig 1.3 (b) is the triangle of forces for P, Q and E
(Note that the arrows are taken in a cyclical order.
The resultant R of P and Q is given by
R2
=
P2 + Q2 + 2PQ Cos Q
where Q is the angle at which P and Q are inclined.
Worked examples
1.
Two forces of 3N and 4N acting at point O are inclined at an
angle of 60o with each other. Find the resultant forces.
Solution
By scale drawing
Choose a suitable scale and draw OA and OB to represent 3 units and 4
units at an angle of 60o with each other.
B
C
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3 units
60o
O
4 units
A
Fig.1.4
Complete the parallelogram by drawing AC parallel to OB and BC
parallel to OA. Join and measure OC which represents the resultant in
magnitude and direction. Measure also angle AOC (α), the angle
between the resultant force and the 4N force. You will find that R is
approx 6.1 N and α = 25o 18.
By analytical method
B
C
3N
R
o
60
∝
O
120o
A
4N
Using the cosine rule, we have that
R2
=
32 + 42 + 2 x 3 x 4 Cos 60
R
=
9 + 16 + 12 =
:.
R
=
37
6.08N
Using the sine rule, we have that
3
Sin α
=
3 x Sin 120o
6.08
= Sin-1(0.4273) = 250 181
: .Sin α
:.
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α
6.08
Sin 120o
=
=
0.4273
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A mass of 2.5 kg is supported by two chords which make angles
of 30o and 50o with the vertical. Calculate the tensions in the two
chords.
30
o
50
T2
o
30
o
25N
o
50
(a)
0
100o
T1
25N
Fig. 1.5
Solution
Analytical method
The problem is illustrated in Fig.1.5 (a)
Mass of 2.5 kg = 25N force
Sketch the triangle of forces
Fig. 1.5 (b) (not to scale), then use the sine rule as follows:
25
Sin 100
Sin 100o
:.
T1
T2
=
=
T1
Sin 30o
Sin (180o-80o)
=
=
=
25Sin 30
Sin 80
25Sin 50
Sin 80
T2
Sin 50
= Sin80o
=12.69N
=
19.45N
(You should attempt the solution by graphical method and compare the
values obtained by the two methods).
Triangle OAC in fig. 2.5 (b) is the triangle of forces for the forces P, Q
and E (Note that the arrows are in a cyclical order around the triangle).
In fig 2.3 (a) O A C B O is the parallelogram of forces. OC is resultant
force.
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Moment of a Force
The turning g effect of a force is called its moment. Examples of the
turning effects of forces in everyday life include:
i.
ii.
iii.
iv.
v.
vi.
Opening a door about its hinges by exerting a force on the door
handle or door edge
Turning a tap on or off
Force exerted on the pedals when riding (i.e pedaling) a bicycle
Opening of the metal cap of a soft drink bottle
Driving a screw in or out of wood with a screw driver
Tightening or loosing a nut with a spanner etc.
Definition
The moment of a force about a point or axis is the turning effect of the
force about that point or axis. It is equal to the product of the force and
the perpendicular distance of its line of action from the point or axis.
Mathematically,
Moment = Force x perpendicular distance of pivot to the line of action
of the force.
Unit of moment
Since force is in newtons (N) and distance is in metres (M), the S1 unit
of moment is Newton-metre (Nm).
θ
C
Q
d
P
d
C
P
(a)
(b)
Fig. 1.6: Moment of a force
In Fig. 1.6 above, the moment of the force P about the point C is given
by
Moment
Moment
118
=
=
P x d (Fig 1.3 (a)
P x d sin Q (Fig.1.3 (b)
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Worked examples
Weight of 20 N and 30 N are hung on a light beam as slow in Fig.1.7.
Find the
moments of the forces about 0.
y
X
0
20N
30N
Fig. 1.7
Solution
Moment of 20N force = 20 ×X = 20x Nm
Moment of 30N force = 30× y = 30y Nm
3.4.1
Principles of Moments
The principle of moments states that if a body is in equilibrium under
the action of a number of coplanar forces, then the sun of the clockwise
moment about any point on the body is equal to the sum of the
anticlockwise moment about same point.
3.4.2 Clockwise and Anticlockwise Moments
•
C
P
R
In Fig 1.8: The forces PA do to turn the body in anticlockwise direction.
In Fig. 1.8 the force P tends to turn the body in anticlockwise direction.
Its moments are called anti clockwise moment. On the other hand the
forces ∝ and R tend to turn the body in clockwise direction. Their
moments about C are thus clockwise moments. If clockwise moments
are taken to be positive, and then anticlockwise moments must be taken
as negative and vice versa.
Its moment is called anticlockwise moment. On the other hand the force
α and R tends to turn the body in a clockwise direction. Their moments
about C are thus clockwise moments.
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3.4.3 Couples
Definition
A couple is a pair of equal but oppositely directed and parallel forces
which are not acting in a straight line.
Characteristics of a couple
(i)
(ii)
(iii)
(iv)
(v)
(vi)
The couple has a zero resultant force but has a turning effect on
the body on which it acts.
A couple can only cause a body on which it acts to rotate but not
to move linearly.
The moment of a couple is the product of one of the forces and
the perpendicular distance between the lines of action of the two
forces.
The moment of a couple is also called a “torque”.
The distance between the two equal parallel forces of the couple
is called the “arm of the couple”.
A couple can be councelled or neutralized by applying an equally
similar couple of opposite moment to the body experiencing the
couple.
Examples (Application) of couples
Uses, application or examples of (the effect of) couples in everyday life
include:
(i)
Turning a tap on or off. It is easier to turn a tap on or off by
applying two equal and opposite parallel forces (couple) rather
than by applying a single force.
(ii)
It is easier to turn a steering wheel of a vehicle by applying a
couple with our two hands instead of applying a single force with
one arms.
(iii)
Loosing a bolt. It is easier to loosen or tighten a bolt by applying
a couple at both ends of the bolt using two spanners.
10N
20N
d
dd
120
10N
=2m
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1.5m
(a)
20N
(b)
Fig.1.9: Moments of a couple
In Fig. 1.9: Above, the moments of the couples are:
(i)
In Fig. 1.9 (a) moments = 10Nx2m = 20Nm
In Fig. 1.9 (b) moments = 20Nx 1.5m= 30Nm
3.5
Conditions of Equilibrium
3.5.1 Conditions of Equilibrium under the Action of Parallel
Coplanar Forces
Coplanar forces are forces that lie in the same plane. Parallel forces are
forces whose lines of a action are all parallel to each other.
Consider a body which is under the action of several forces which are
coplanar and parallel. For the body to be in equilibrium (i.e. it does not
move or rotate), then the following two conditions must be satisfied:
(i)
Forces: The algebraic sum of the forces in any given direction
must be zero. Thus means the sum of forces acting in one diction
must be equal to the sum of forces working in the opposite
direction.
(ii)
Moments: the algebraic sum of the moments of all forces about
any point on the body must be zero, or the total clockwise
moments of the forces about any point on the body must be equal
to the total anticlockwise moments of the forces about the same
point.
Worked examples
1.
the
Fig. 1.10 shows a beam balanced at P under the action of
forces shown. Calculate the value of the weight W.
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25cm
15cm
25cm
P
W
1N
4N
Fig. 1.10: A beam balanced at P
Solution
For equilibrium,
Anticlockwise moments about P = clockwise moment about P
2.
∴
∴
(W x 25)
(W x 25)
∴
W
= (4 x 15) + (1 x 40)
= (60 + 40)
=
1
25
=
100
25
= 4N
In Fig.1.11 - a beam 1m long is balanced at its midpoint.
Calculate the magnitude of the force P required for the beam to
be in equilibrium
40cm
25cm
20cm
P
0.6N
0.4N
Fig.1.11: A 1m- long beam balanced at its mid point
Solution
For equilibrium,
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Clockwise moments about O = anticipation moments about O
∴
P x 0.35 + 0.4 x 0.2 =
0.6 x 0.4
∴
(0.6 x 0.4)-(0.4 x 0.2)
P
=
0.35
P
= 0.5N
Measurement of weight of a meter rule using principle of moments
A
G
B
B
O
W
P
Fig. 1.12: Determining weight of a metre rule
Find the rough position of the centre of gravity (G) of the ruler by
balancing it on the knife edge O. Suspend the ruler at a whole number
O(e.g. 70cm) near one end. Hang a suitable weight P on the opposite
side of O and adjust the position until the ruler balances horizontally. By
taking moments about O, we have that,
W x Go = P x Bo
∴
W= P x Bo
Go
The experiment can be repeated using different values of P and
obtaining
different balance points. An average value of W is then
calculated.
3.5.2 Conditions of Equilibrium under the Action of NonParallel Coplanar Forces
A set of non-parallel coplanar forces can act on a body to keep it in
equilibrium. The non-parallel forces can easily be resolved into
horizontal and vertical components, giving rise to two parallel forces at
right angles to each other. The problem can then be treated as in section
3.1.5(a) with the method of that section applied separately to the
horizontal and vertical compo nets of the forces. The two condition of
equilibrium for a body under such non-parallel forces would then be:
(i)
Forces
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The vector sum of all the forces acting on the body must be zero. This
means that the algebraic sum of the forces or components of the forces
acting on the body on any direction must be zero, i.e. ΣFx = 0 or any
algebraic sum of the horizontal must be zero and ΣFy =0 or the algebraic
sum of the vertical components must be zero.
(ii)
Moments
The algebraic sum of the moments of all the forces about any point on
the plane or axis perpendicular to the plane of the forces must be zero.
Worked examples
A force of 100N in equilibrium by two ropes as shown in Fig.1.13. If
one rope
pulls the body in a horizontal direction and the other in a
direction of 400 with the vertical, calculate the tension in each rope.
T2
40o
T1
100N
Solution
Method 1: Resolving into vertical an d horizontal components
Let T1, T2 be the tensions in the rope. For equilibrium sum of horizontal
components are equal
∴
∴
∴
T2 Cos (90-40)0 –T1 == 0
T2 Sin 400 – T1 = 0
T1 = T2 Sin 400
(1)
Also sum of vertical components are equal for equilibrium,
∴
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T2 Cos 400 – 100 = 0
(2)
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∴
T2 = 100 = 130.54N
Cos 40
From (i), T1 = T2 Sin 400 = 130.54x Sin 400
= 83.91N
Method 2: Using triangle of forces
Sketch the triangle of forces for the system (not to scale)
From the Sin rule,
100
Sin 50
=
∴
T1
T1 =
T2 =
=
T2
Sin 40
Sin 90
100 Sin 40 = 83.91N
Sin 50
100sin90 = 130.54N
Sin 50
(You should also solve the problem by graphical method and compare
the values of tension obtained).
3.5.3 Equilibrium under the Action of Three Non-Parallel
Forces (Special Case)
For such a body to be in equilibrium, the following conditions must be
satisfied:
(i)
(ii)
(iii)
The three forces must lie in a plane (i.e. they must be coplanar
forces.
The line of action of the forces must intersect in a common point.
The vectors representing the three forces can be arranged to form
a closed triangle with side respectively parallel to the direction
and proportional in length to the magnitude of the forces.
3.6 Centre of Gravity
Definitions
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(i)
(ii)
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The centre of gravity (C.G) of a body is defined as the point
through which the line of action of the weight ofthe body always
passes irrespective of the point at which the entire weight of the
body appears to be concentrated.
The centre of mass of a body is defined as the point at which the
total mass of the body appears to be concentrated.
Note: For small objects the centre of gravity (C.G) and the centre of
mass coincide.
3.6.1 Position of C.G for some Regular Uniform Bodies
(i)
The C.G of a uniform rod is at the midpoint (G) of the rod. (Fig
1.14)
(ii)
The C.G of a uniform circular plate is at its centre
(Fig.1.14 (b))
(iii)
The C.G. of a uniform circular ring is at its centre
(Fig. 1.14( c) )
(iv)
The C.G. of a uniform square or rectangle sheet is at
the point of intersection of its diagonal (Fig. 1.14(d)).
(v)
The C.G. of a uniform triangle plate is at the
intersection of the Medians (Fig. 1.14 (e))
O
0
G
b. Uniform
circular plate
d.
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G
c. Uniform
Uniform rectangular sheet
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Uniform rod
uniform triangular plate
Describe an experiment to locate the C.G. of a lamina of any shape
Fig. 1.14 C.G. of objects
3.6.2 An Experiment to Locate the Centre of Gravity of a
Lamina to locate the C.G. of a Lamina
B
0
C
lamina
0
A
Plumb line
Fig 1.15: Locating C.G. of a lamina
You will make use of a plumb line (i.e. a thread with a small heavy
weight tied to one end). Make holes near the edge of the lamina at about
three different locations A, B, C, as shown in fig. 1.15 Hang the lamina
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from a smooth horizontal support at A so that it can swing freely.
Suspend the plumb line from the same point A on the lamina Repeat the
procedure by suspending the lamina at B and C. if you have done the
experiment carefully you will observe that the three lines intersect at a
point which is the C.G. of the lamina. This is because the lamina will
always come to rest with its C.G. vertically below the point of
suspension when it is free to move.
3.7
Types of Stability of Equilibrium
There are three types of stability of equilibrium which a body at rest
may experience:
(a)
(b)
(c)
Stable equilibrium
Unstable equilibrium
Neutral equilibrium.
3.7.1 Stable Equilibrium
Definition
A body is said to be in stable equilibrium if it tends to return to its
original position when slightly displaced.
Examples of bodies in stable equilibrium
(i)
A rectangular block resting in a horizontal surface.
(ii)
A cone resting on its base
(iii) A racing car with wide base and low C.G.
(a)
RG
W
Fig. 1.16:
R
G
(b)
W
Stability of rectangular block
Consider a rectangular block resting on a horizontal surface (Fig 1.16).
In Fig.1.16 (a) the weight of the block W and the reaction R of the
surface of the table are equal but opposite in direction and the same line
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of action. In Fig. (b) the block has been given a slight tilt. The line of
action of W still falls within the base of the block. W and are still equal
and opposite but act along different lines as shown. The two forces
therefore constitute a couple which acts in a clockwise sense restore the
block to its original position. Hence the block is in stable equilibrium.
Requirements for Stable Equilibrium
Generally,
(i)
(ii)
a low C. G; and
a wide base
3.7.2 Unstable Equilibrium
Definition
A body is said to be in unstable equilibrium if it tends to move further
away from its original position or topples over when slightly displaced.
Examples of bodies in unstable equilibrium
(i)
(ii)
(iii)
a cone resting on its apex
an egg resting on its pointed end.
A tight-rope walker (i.e. a person walking on a tight rope).
(a)
G
W
R
G
R
(b)
W
A
A
Fig. 1 .17: Unstable equilibrium
Consider a cone resting on its apex A (Fig.1.17) A. In Fig. 1.17 (a) W
and R are equal and opposite and have the same line of action. As the
conies slightly displaced (Fig 1.17(b)), the line of action of W quickly
falls out of the apex. W and R now constitute a couple with clockwise
moment whose effect causes the cone to topple over or move further
away from its original position. Hence the cone is in unstable
equilibrium.
3.7.3 Neutral Equilibrium
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Definition
A body is said to be in neutral equilibrium if it tends to come to rest in
its new position when slightly displaced.
Examples:
(i)
(ii)
A cone, or cylinder or an egg resting on its side.
A ball or sphere on a smooth horizontal table (or surface).
D.
R
A.
C
B.
W.
a.
ball
B
A
B
c.
W
Fig. 1.18
b
Consider the cylinder or cone resting on its side or the ball resting on the
ground (Fig.1.18). When the ball is slightly displaced, the weight of the
ball has no moment about the point of contact with the ground. The
position of the C.G. and the height of the C.G. above the ground remain
unchanged. The bodies are therefore in neutral equilibrium.
3.7.4 Effect of Height of C.G. on Stability of Equilibrium
As already pointed out, the C.G. of a body above the ground level and
the width of the base of the object affect the stability of the object.
D
C
C
D
G2
B
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A
(a)
B
A
(b)
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•
•
Fig.1.19: Height of C.G. and stability of an object
y
Consider a body in the form of a rectangle ABCD resting on a table
(Fig.1.19 (a)). Suppose the body is slightly lifted as in Fig.1.19 (b) if the
body has a low C.G. at G1, the body will not topple over because the line
of action of the weight of the body passes within the base AB. On other
hand, if the body has a high C.G. at G2, a different situation will result if
the body is slightly titled about A. the line of action of the weight of the
body (W) falls outside its base of support. Therefore a wide base and
low C.G. enhance the stability of an object.
3.7.5 Application of the Effect of Height of C.G. on Stability of
Object
(i)
(ii)
(iii)
(iv)
(v)
3.8
Racing cars are built low and with a wide wheel base to prevent
them from overturning while negotiating bends at highspeed
Cars engine are always located at the bottom of the car and not on
the roof.
A tight rope walker often carries a weighted pole or umbrella
Lorries that carry heavy loads usually have the loads stacked on
the floor but not near the roof or on top of the roof.
Certain toys which are difficult to topple are designed in such a
way that they have low C.G. and large base area.
Equilibrium of Bodies in Liquids
3.8.1 Concept of Upthrust
Common experience shows that when a box is totally or partially
immersed in a liquid such as water, the body appears to be lighter. Such
an object appears heavier when completely out of water. This
suggested there s an upward force exerted by a liquid on an object
completely or partially immersed in it. As a result of this upward force
there is thus an apparent loss in weight of the immersed object. This
upward force is called the ‘upthrust’ of the liquid on the object.
.
Beaker
u
Liquid (water)
W
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Fig. 1.20: Upthrust of liquids on an object
Fig 1.20 shows an object completely immersed in water. The weight of
object in air is W, while weight in water is W2. The upthrust of the water
U is given by U= W1-W2. Hence the apparent weight in water W2 is
given by W2 = W, -U. Hence W2 is less than W1. The value of the
upthrust depends on the volume of the object immersed
and
the
density of the liquid in which it is immersed. The magnitude of the
upthrust is given by Archimedes principle.
3.8.2 Archimedes Principle (statement)
Archimedes principle states that when a body is totally or partially
immersed in a fluid (liquid or gas), it experiences an upthrust which is
equal to the weight of fluid displaced.
Formula for upthrust
An object completely immersed in a liquid displaces a volume of liquid
equal to its own volume. The upthrust is the weight of this
displaced
volume of liquid and is given by = volume of object x density of liquid
x g where g is acceleration due to gravity. If upthrust is U, volume of
object V and density of liquid ρ, then
U = Vρg
where, V = volume of object submerged, ρ = density of the fluid and g
= acceleration of free fall.
If the object is partially immersed in the liquid (e.g. only 2 of the object
is immersed in the liquid), then
3
U = 2V x ρ x g
3
∴
U = 2Vρg
3
3.8.3 Verification of Archimedes Principle
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Fig. 1.21: Verification of Archimedes principle
Fig. 1.21 shows the apparatus for an experiment to verify Archimedes
principles. A solid object (e.g. a piece of stone) is freely suspended from
a spring balance and its weight in air (W1) read off. The object is then
gently lowered into the water in the Eureka can filled to the level of the
spout as shown.
The weight of the object in water (W2) is read from the spring balance.
Water displaced by the
object is collected in the beaker which has
been previously weighed. The beaker with water is reweighed to obtain
the weight of water displaced. The readings are recorded as follows:
Weight of object in air = W1
Weight of object in water = W2
Weight of empty beaker =W3
Weight of beaker + water displaced = W4
Apparent loss of weight of object = W1-W2
Weight of water displaced = W4-W3
It will be found by comparing the values that
W1-W2
=
W4=W3
Thus the apparent loss in weight of object (upthrust) is equal to weight of
water displaced showing that an Archimedes principle is true. The
principle will be similarly confirmed if the experiment is repeated with
the object only partially immersed in the water.
Worked example
A spring balance reads 12.0N when a metal block is suspended from it
and 10.0N when the block is completely immersed in water. Calculate
the upthrust on the block, the mass of water displaced, and the volume
of the block.
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Solution
(i)
upthrust on block = loss in weight in water
= 12.0N- 10.0N =2.0N
(ii)
weight of water displaced = upthrust = 2.0N
∴
mass of water displaced = weight
g
=
10
2.0
kg
=0.2
kg
= 200 g.
(iii)
volume of block = volume of water displaced
=
=
mass of water displaced
density of water
0.2 kg
1000 kg
=
2x10-4 m3
m3
= 200 cm3
3.8.4 Principle of Floatation
The Principle of floatation states that an object will float in a fluid
(liquid or gas) when the upthrust exerted by the fluid in which it floats is
equal to the weight of the object.
From the principles of floatation it can be said that:
(1)
A floating body displaces its own weight of the fluid in which it
floats.
(2)
Any floating body is in equilibrium under two forces, namely its
own weight acting downwards and the upthrust of the liquid
acting upwards.
A body will float in a liquid if the density of the object is less
than the density of the liquid.
A body denser than a liquid can still float in the liquid, if the
body is shaped in such a way that its volume can displace its own
weight of the liquid. For this reason metal ship floats in water
because the large volume of the ship displaces a large volume of
liquid whose weight counter balances the huge weight of the ship.
(3)
(4)
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Worked example
A cube of wood of side 10 cm floats in water with 4.5 of its depth below
the
surface, and with its sides vertical. What is the density of the
wood?
Solution
Volume of wood submerged = (10 x10 x4.5) cm3
= 450 cm3
∴
volume of water displaced = 450 cm3
∴
mass of water displaced = 450 cm3
(Density of water =1 gcm-3)
Since a floating object displaces its own weight of fluid, then
∴
Mass of wood
Volume of wood
density of wood
=
=
=
450 g
1000 cm3
450 g
1000 cm3
=0.45g cm-3
3.9
Density and Relative Density
Equal volumes of different substances have different masses or weights.
This is due to differences in the property of the substances known as
density.
Definition
The density of a substance is defined as the mass per unit volume of the
substance.
S.I unit of density is kilogram per metre cubed (kg/m3 or kg m-3) and its
symbol is ρ ( rho).
Definition
The relative density (R.d) of a substance is defined as:
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R.d. =
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mass (or weight) of a substances
Mass (or weight) of equal volume of water
Alternative definition of relative density is:
R.d. =
density of substances
Density of water
Relative density has no unit since from the two definitions above, mass
or weights are compared.
3.9.1 Measurement of Density and Relative density
1(a)
The density of a regular solid (e.g. wooden block) can easily be
obtained by measuring its mass using a chemical balance. The
volume is obtained by measuring its dimensions.
Density
(a)
(b)
=
Mass
Volume
The density of an irregular solid (e.g. a piece of stone) can be
obtained by first measuring the mass using a chemical balance.
The volume is obtained by immersing the solid completely in a
measuring cylinder containing water. The difference in the levels
of water before and after the immersion of the solid gives the
volume of the solid.
Relative density of a solid which floats in water (e.g. cork) is best
measured by means of Archimedes principles as follows:
Fig. 1.22: Relative density of a solid which floats
The cork cannot sink in water of its own accord, therefore a “sinker’ is
used to make it sink completely in water when immersed. The sinker
(e.g. a glass stopper or brass weight) is first suspended from the hook of
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a spring balance and its weight when completely immersed in water is
noted from the spring balance (Fig 1.22(a)). The cork is then tied to the
thread carrying the sinker in such a way that the cork does not touch the
water. The weight of sinker in water
and cork in the air is obtained
(Fig. 1.22(b)). The cork is then tied near the sinker and the weight a
sinker and cork, both completely immersed in water is noted. The
relative density of cork is calculated as follows:
Weight of sinker in water =W1
Weigh of sinker in water and cork in air =W2
Weight of cork in air =W2-W1
Upthrust on cork = Apparent loss in weight of cork
? = W3-(W2-W1)
Relative density of cork
=
Weight of cork in air
Weight of equal volume of water
Weight of cork in air
Upthrust on cork
=
3
4
W2-W1
W2-W3
(a)
Relative density of solid in form of particles or powder
(e.g. sand) is best measured using a relative density bottle
and a chemical balance. This should be read up from
physics texts by students.
(b)
Relative density of solid particles or powder which
dissolves in water (e.g. table salt) is similarly measured
using a relative density bottle, a chemical balance and a
liquid in which the solid (powder) does not
dissolve
but whose density is know. This should also be read up
from physics texts by students.
(a)
Relative density of a liquid using relative density bottle.
Liquid
.
Fig. 1.23: Relative density of liquid
The relative density bottle is cleaned, dried and weighed empty. The
bottle is filled with the liquid whose relative density is required, cleaned
and reweighed. The bottle is often emptied of the liquid, cleaned, dried
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and reweighed when filled with water. The relative density of the liquid
is calculated as follows:
Mass of empty bottle
=
Mass of bottle + liquid
=
Mass of bottle + water
=
Mass of liquid
=
Mass of equal volume of water =
M1
M2
M3
M2-M1
M3-M1
Relative density of liquid =
mass of liquid
Mass of equal volume of water
=
M2-M1
M3-M1
Relative density of a liquid can also be measured easily using
Archimedes principle. In this, case a solid (e.g. glass stopper) is used.
The solid is suspended from the hook of a spring balance and weighed
in air. It is then weighted when completely immersed in a liquid whose
relative density is required. The solid is then cleaned, dried and
reweighed when completely immersed in water. The relative density if
the liquid is calculated as follow:
Mass of objects in air
=
Apparent mass of objects in liquid =
Apparent mass of object in water =
M1
M2
M3
Upthrust in liquid = weight of liquid displaced by the objects
=
M1 –M3
Upthrust in the water = weight of water displaced by the object
= M1-M3
Relative density of liquid =
Worked examples
138
upthrust in liquid
Upthrust in water
M1-M2
=
M1-M3
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A body weighs 0.52N in air. Totally immersed in water it weighs
only 0.32N while its weight when immersed in another liquid is
0.36N. The density of water is 1000
kg/m3. What is the
density of the other liquid?
Solution
Upthrust in water = loss of weigh in water
= 0.52N -0.32N= 0.20N
Mass of water displaced = upthrust
g
= 0.20N =0.02 kg
10N kg.
Volume of water displaced = Mass
Density
= 0.20N kg
=
0.20 kg
1000 kg.
∴
Volume of water displaced = Mass
Density
= 0.02 kg =
0.02x10-3 m3
1000 kg
Upthrust in liquid = loss of weight in liquid
= 0.52N-0.36N = 0.16N
∴
Mass of liquid displaced = 0.16N 0.016 kg
10N
kg
Volume of liquid displaced = volume of water displaced
= 0.02x10-3 m3
∴
2.
Density of liquid
= mass of liquid displaced
Volume of liquid displaced
= 0.016 kg
0.02x10-3m3
= 0.8x103 kg/m3
= 800 kg/m3
An empty relative density bottle has a mass of 15.0 g. When
completely filled with water its mass is 39.0 g. What will be its
mass if completely filled with acid of relative density 1.20?
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Solution
Mass of water which fills bottle = 39g-15 g
= 24 g
Volume of water which fills bottle = mass
(Density of water = 1gm /cm3) density
= 24g = 24 cm3
1g
cm3
Relative density of liquid =
mass of liquid
Mass of equal volume of water
∴
∴
mass of liquid which fills bottle =1.20x24 g
=28.8 g
∴
mass of bottle completely
filled with liquid (acid) =28.8+15.0 g
= 48.3 g
3.9.2 The Hydrometer
140
1.20 = mass of liquid
24 g
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Fig. 1.24: Simple hydrometer
The hydrometer is an instrument used to measure the relative density of
liquids.
The Principle of the Hydrometer
Consider a test tube weighted by means of lead shots so that it can float
upright in a beaker of water as shown in Fig. 1.24. If the tube is floated
in different liquids it will be noted that the length of the tube submerged
in liquid is different for different liquids. This is due to the differences in
the relative densities of the liquids.
Suppose the weight of the text tube (with the lead shots) is W. each time
the tube floats in a liquid, the weight of liquid displaced is W Newton’s
∴
Volume of liquid displaced
=
mass
Density
=
W
ρg
where ρ is the density of the liquid and g is the acceleration due to
gravity. If the cross-sectional area of the tube is A and the length, of the
tube submerged is l, then:
Volume of liquid = Axl
∴ Al = W
ρg
∴L =
W
Aρg
∴L∝L
ρ
Since W, A and g are constants.
The above relationship implies that the length of the test-tube
submerged in the liquid is inversely proportional to the density of the
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liquid, i.e. the higher the density of liquid, the shorter the length of the
tube immersed and vice versa.
The working of the hydrometer is used on this principle, that is, the less
the
density of a liquid, the more the hydrometer sinks in it.
3.9.3 The Practical Hydrometer
Fig. 1.25: A battery hydrometer
Fig. 1.25 shows a practical hydrometer which is commonly used in daily
life to measure the relative densities of liquids. The real hydrometer is
housed (as shown) in an endorser with a glass chamber into which the
liquid (whose relative density is required) can be drawn so that the
hydrometer can float in it and the value of the relative density can be
read directly.
Uses of the practical hydrometer
(i)
(ii)
(iii)
Checking the relative density of batteries
Testing the purity of liquids whose densities are know
Testing the quality of milk
SELF ASSESSMENT EXERCISE
1.
142
Explain what you understand by the term “force”. Define the
terms “resultant” and “equilibrant” with respect to the forces
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acting at a point and explain the differences between the two
terms. A force of 10N acts in the direction E 600 N while another
force of 15N in the easterly direction if both forcesa act at a
point, point, find (i) the resultant, and (ii) the equilibrant of the
two forces.
2.
State the law of triangle of forces and briefly describe an
experiment to illustrate it. An object of mass 700 g hangs from
the end of a long sting attached to a support. A horizontal sting
attached to the object pulls it in such a way that the first string
makes an angle 300 with the vertical. Find the tensions in the two
strings.
3
Explain what is meant by the “moment of a force”. State the
principle of moments and give two examples of the practical
application of the principle in everyday life.
A uniform plank AB has length 5 m and mass 80 kg. It is
supported horizontal on two points C and D where C is 1m from
the end A and D is 1 m from the end B of the plank. Find the
magnitude and direction of the minimum force which would be
applied at; A (i) to lift the plank just off the pivot at C, (ii) to lift
the plank just clear of the pivot at D
4
State the Principle of Archimedes and explain what is meant by
“upthrust” exerted by a fluid on an object immersed in it.
How is a submarine made (i) to sink to a point below the surface ofwater
(ii) to rise to the surface of water from beneath it?
A floating crane has mass 20600 kg and floats in sea water of density
1.03 g/cm3. If the base of the crane is a rectangular block 4m square and
2.5m deep, what is the maximum mass which can be lifted by the krane
when the top of the base is just awash with water.
4.0
CONCLUSION
In this unit you learned about cause of motion – force. You also learned
resultant force and forces in equilibrium. Furthermore, you learned
about laws governing bodies in equilibrium. In addition you learned
Archimedes principle, the principle of floating and relative density.
Finally, you learned the applications of the laws and principles in
solving numerical problem.
5.0
SUMMARY
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1.
Motion is caused by a “force” which is essentially a “push” or
“pull”, acting on the moving object. There are four major groups
or fields of force, namely; gravitational, electric or electrostatic,
magnetic and nuclear force.
2.
The “resultant force” is that simple force which, acting alone will
have the same effect in magnitude and direction as two or more
forces acting together.
3.
A body is said to be in equilibrium if
(a)
(b)
The body as a whole either remains at rest or moves in a
straight line with constant speed, or
The body is rotating with a constant angular velocity or is
not rotating at all.
4.
The “equilibrant force” is that single force which will balance all
the other forces taken together. The equilibrant force is equal in
magnitude but opposite in direction to the resultant force.
5.
The "law of triangle of forces” states that when three forces are in
equilibrium, they can be represented in magnitude and direction
by the three sides of a triangle taken in order.
6.
The “moments of a force” about a point or axis is the turning
effect of the force about that point or axis. It is equal to the
product of the force and the perpendicular distance of the line of
action of the points from the point.
7.
For a body in equilibrium, the sum of the clockwise moments at a
point equals the sum of the anticlockwise moments about the
same points. This is also called the “principle of moments”.
8.
Two equal but opposite parallel forces which are not acting in a
line is called a “couple”
9.
The “centre of gravity” of a body is the point through which the
line of action of the weight of the body always acts, or it is the
point at which the entire weight of the body appears to be
concentrated.
10.
A body is said to be:
(i)
144
In “stable equilibrium” if it tends to return to its original
position when slightly displaced.
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(ii)
(iii)
In “unstable equilibrium” if when slightly displaced, it
tends to move further away from its original position.
In “neutral equilibrium” if when slightly displayed, its
tends to come to rest in its new position
11.
For adequate stability, a body requires (i) a wide base of support,
and a low centre of gravity.
12.
“Archimedes principle” states that when a body is totally or
partially immersed in a fluid, it experiences an upthrust which is
equal to the weight of fluid displaced
13.
“The principle of flotation” states that an object floats in a fluid
when the upthrust exerted upon it by the fluid is equal to the
weight of the object.
14
“ Relative density (R.d)” of a substance is defined by:
R.d
=
mass of substance
Mass of equal volume of water
=
density of substance
density of water
=
upthrust in liquid
upthrust in water
Relative density has no units. It can be measured with a “hydrometer”.
6.0
TUTOR- MARKED ASSIGNMENT
1.
What do you understand by the “resultant” of two or more forces
acting at a point? Explain briefly how such a resultant may be
obtained. Three forces of magnitude 5√2N, 15√2N and 20√2N
act at a point in the west, north and east direction respectively.
Another force of magnitude 10N acts at the same point in a
direction S 450W, that is 45o to the 5√2 N force. Draw a sketch of
the arrangement and determine the result of the system in
magnitude and direction.
2.
Explain briefly the meaning of (i) moment of force,(ii) a couple.
A uniform plank 600 cm long rests on two supports A and B
which are 100 cm apart. Weights of 100N and 80N are hung from
the ends of the beam near A and B. Calculate the reaction forces
at the supports A and B.
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State Archimedes principle and describe how you will use the
principle to determine the relative density of a given liquid. An
Object weighs 0.30N in air, and 0.25N when fully immersed in
water. If the object weighs 0.27N in a liquid, calculate (i) its
relative density (ii) the relative density of the liquid
7.0
REFERENCES/FURTHER READINGS
Anyokoha, M.W. (2000). Physics for Senior Secondary Schools.
Onistha: Africana – FEP Publishers ltd.
Awe, O. and Okunola O.O. (1992). Comprehensive Certificate Physics,
SSCE Edition. Ibadan: University Press Plc.
Ndupu, B.L.N. and Okeke, P.N. (2000).Round-Up Physics, for West
Africa Senior School Certificate Examination: A Complete
Guide. Lagos: Longman Nigeria Plc.
UNIT 2
CONTENTS
146
FRICTION IN SOLIDS AND LIQUIDS
PHY 001
1.0
2.0
3.0
ACCESS PHYSICS
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1. Concept of Friction
3.2. Advantages and Disadvantages of Friction
3.2.1 Advantages ofFriction
3.2.2 Disadvantages of Friction
3.3
Methods of and Reasons for Reducing Friction.
3.4
Types of Friction
3.5
Laws of Friction and Coefficient of Static Friction
3.5.1 Laws of Friction
3.5.2 Coefficient of Friction
3.6
Determination of the Coefficient of Static Friction Using
3.6.1 Horizontal Plane
3.6.2 Inclined Plane
3.7
Viscosity
3.7.1 Concept of Viscosity
3.7.2 Effects of Viscosity
3.8
Laws of Viscosity and Coefficient of Viscosity
3.8.1 Laws of Viscosity.
3.8.2 Coefficient of Viscosity
3.9
Motion of a Ball in a Viscous Liquid
3.10 Similarity and Difference between Viscosity and Friction
3.10.1 Similarities
3.10.2 Differences
Conclusion
Summary
Tutor -Marked Assignment
References/Further Readings
1.0.
INTRODUCTION
In this unit you will learn about friction in solids and liquids. Whenever
one moves or tends to move over another surface, frictional forces are
called into play. Friction is unavoidable because no surface is perfectly
smooth. When one object rubs against another, the roughness of their
surfaces prevents them from sliding freely over another. Frictional
forces in liquids (viscosity) explain why some liquids flow easily and
others do not.
2.0.
OBJECTIVES
By the end of this unit, you should be able to:
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•
identify a force resisting the motion between two surfaces in
contact and moving relative to each other.
•
define friction and viscosity
•
state the advantages and disadvantages of friction.
•
reduce friction in a given situation
•
classify liquids according to their viscous properties.
•
give at least two applications of viscosity.
3.0
MAIN CONTENT
3.1.
Concept of Friction
You are aware that when a solid is in contact with another, the
roughness of their surfaces prevents them from sliding freely over one
another. Such action is brought about by the effect of friction.
Frictional forces act tangential to the surface of separation between two
bodies in contact. It does not appear unless there is relative motion or a
force tending to produce motion, then it always appears to oppose the
motion or the force tending to produce motion.
You have to overcome some frictional force before an object can move
over another. The frictional force continues to act as long as one body
moves over another. When you withdraw the force producing the
motion, friction slows down the moving body until it eventually comes
to rest.
 Friction is defined as a force which acts at the surface of separation
between two objects in contacts and tends to oppose the motion of
one over the other.
3.2.
Advantages and Disadvantages of Friction
3.2.1. Advantages of Friction
We desire friction for the following reasons:
i.
148
Without friction, it would be impossible for us to walk or to stop
walking after getting started. Friction protects us from slipping
while walking.
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ii.
Friction enables the automobile tyres to make a firm grip with the
road thus aiding their movement. The threading pattern of vehicle
tyres is designed to give the tyre a good grip on the road.
iii.
Friction enables the locomotive engine to pull the train on the
nail.
iv.
Friction is utilized in fan belts used over wheels or pulleys in
machinery. The belting that drives the wheels in machines
depends on friction to make the wheels turn without slipping.
v.
Friction enables a screw or a nail to remain in place after being
screwed into position. Such nails thus hold pieces of wood
together because of the friction between the nails and the wood.
vi.
Friction is used in the grindstone to sharpen knives and chisels.
The grindstone’s rough surface wears away the metal surface to
sharpen them.
vii.
Friction enables the brakes to stop the car and train
3.2.2. Disadvantages of Friction
i.
Friction leads to wear and tear on the moving parts of machines.
ii.
Friction reduces the efficiency of machines because it causes loss
of energy. Much useless work is therefore done by machine in
overcoming friction which opposes their motion.
iii.
It causes heating of engines. Energy that must be used to
overcome friction is converted into heat. This results in the
bearing of machines getting hot when they are running.
Sometimes, the energy is converted to sound.
3.3.
Methods and Reasons for Reducing Friction
Here we consider three major ways of reducing friction. You may have
been using these methods in reducing friction in the past and you may
also know other methods. These are:
i.
The use of lubricants like oil, grease, air and graphite. You can
use lubricants to keep two metal surfaces from coming into direct
contact. You can also do this by maintaining a layer of air or
other gases between them.
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ii.
The use of ball or roller bearing. Where a wheel is turning on an
axle, the use of ball or roller bearings reduces friction. These are
like wheels within wheels and are precisely shaped and have
specially hardened surfaces. In roller bearings, the surfaces do not
slide but roll over the others and rolling friction is much less than
sliding friction.
iii.
The streamlining of body shapes of moving objects.
You can reduce friction by shaping a body to the streamlines of
the fluid through which it is moving. This reduces the frictional
force or retarding force. You will notice that modern cars, ships
and aero- planes are made in such shapes as to lesson the friction
between them and the fluid or the medium through which they
move.
We try to reduce friction in machines for two reasons
i.
ii.
it causes wear and tear
it uses up energy and reduces efficiency.
3.4.
Types of Friction
There are two types of frictions which we are going to define here.
These are:
i.
ii.
Static or Limiting Friction
This is the maximum force that must be overcome before a body
can just start to move over another.
Kinetic or Dynamic Friction
This is the force that must be overcome so that a body can move with
uniform speed over another body.
•
Usually dynamic friction is slightly less than the static friction.
3.5 Laws of Friction and Coefficient of Static Friction
3.5.1
Laws of Friction
At this point you will learn the laws of friction. They are:
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i.
Friction opposes the relative motion between two surfaces in
contact. It acts in a direction opposite to that of the motion.
ii.
The force of friction increases to the same extent as the force
which tends to start the motion.
iii.
Frictional force depends on the two surfaces in contact- the
nature of the substance and the conditions of the surfaces such as
a rough, smooth, polished, wet, dry and others.
iv.
It is independent of the area of the surfaces in contact.
v.
It varies directly with the normal force pressing the surfaces
together. That is, it is proportional to the normal reaction R.
3.5.2 Coefficient of Friction
Results of earlier experiments show that the frictional force F is roughly
proportional to the normal reaction R between the two surfaces in
contact.
That is:
F
Or
&
F
R
=
MR
---------
(1)
The constant M is known as the coefficient of friction.
M
=
F
R
----------- (2)
Equation (2) defines
•
The coefficient of friction (M ) as the ratio
M =
Frictioned force F
Normal reaction between the two surfaces in
contact R
You will note that:
i.
For static friction, the frictional force F required to determine M
in the above equation is the maximum force required to start the
body moving and M is called the coefficient of static or limiting
friction.
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ii.
For kinetic friction, the frictional force used is the force required
to keep the body in uniform motion and M in the above equation
is known as coefficient of dynamic or kinetic friction.
iii.
The coefficient of static friction is greater than the coefficient of
dynamic friction.
iv.
From experience you know that more force is required to start a
body sliding along a surface than is required to keep it in motion
once it is started.
3.6 Determination of the Coefficient of Static Fiction Using:
1.
Horizontal Plane.
To determine the coefficient of static friction using a horizontal plane
you will need the apparatus set up as shown in Fig. 2.1.
You will now follow these procedures.
•
•
•
•
•
•
•
•
•
M
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Weigh the block of wood B with a spring balance and record its
weight (WB).
Place the block of wood on the horizontal board and connect it to
a cord adjusted in such a way that it pulls parallel to the surface
of the board.
Pass the cord over a pulley and attach it to a scale pan to which
known weights can be added.
Suspend known weights in the scale pan until the block is just
about to move.
Tap the block gently to prevent sticking each time a weight is
added into the pan.
Place a known weight on the block B and increase the weight on
the scale pan until the block is just about to move again after
gentle tapping.
Repeat the experiment five more times by varying M and noting
the corresponding values of W in the scale pan.
The total weight pulling the block also includes the weight of the
scale pan (Wo)
Tabulate your readings as shown in Table 2.1.
R = WB + M
F = W + Wo
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Table 2.1
•
Plot a graph of force (W + Wo) against normal reaction R = (WB
+ M) and determine slope of the graph which is the coefficient of
static friction.
The graph of an a earlier experiment is as shown in Fig. 2.2 below.
M
B
Pulley
Inextensible string
Scale pan and weight
W
F
* *
* *
* *
*
R
Fig 2.1: Measurement of coefficient of friction. Fig. 2.2: Graph of F
against
ii.
Inclined Plane
You can also determine the coefficient of static friction M by using
inclined plane.
Here, you will obtain the value of M for the two surfaces by placing the
block A on B as shown in Fig. 2.3 and gradually increasing the angle 0
of inclination of B to the horizontal unit A is just about to move down
the plane.
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A
h
θ
Fig. 2.3: Block A on an inclined plane with angle Q
•
Measure the values of x and h using your metre rule. You will
then obtain the value of M (the coefficient of friction) by
evaluating the expression relating M to the values of x and h.
That is:
tan θ = h/x = M
(3)
Precautions
You will take the following precautions while determining the
coefficient of friction
•
•
•
•
•
Take repeated readings.
The block of wood should be started at the same spot each time.
The inclined plane should be gradually increased.
The weight in the scale pan should be gradually increased.
It is more accurate to measure θ by measuring h and x.
Worked example
1.
A brass block weighs 25N. The coefficient of limiting friction
between the block and the horizontal surface of a table is 0.61
a.
b.
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What is the horizontal force required to just move the
block?
If another 25N brass block is stacked on this block, what
is the minimum horizontal force that must be applied to
move the blocks?
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Solution
a. Coefficient of limiting friction, M = 0.61 Normal reaction, R =
weight of block. = 25N
Using M
=
F
=
F
R
M R = 0.61 x 25N = 15.25N
The minimum horizontal force required to move the block is 15.25N.
b.
A 25 N brass block is stacked on the given block coefficient of
limiting friction remaining the same.
Normal reaction, R = 25N + 25N = 50N
Using F =
F=
2.
M R
0.61 x 50N
=
30.5N
A 250N weight can just rest on a plane inclined at 35o to the
horizontal when a 49N force, parallel to the plane, is applied on
it.
Calculate:
a.
b.
the value of the frictional force present.
the coefficient of limiting friction.
Solution
R
P
F
N
5
250
s3
Co
35
35
0
25
25
os
0C
Fig. 2.4: Forces on weight resting on an inclined plane
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Let the frictional force be F, and the applied force of 49N be P (Fig.
2.4).
Restoring the 250N weight into two perpendicular directions, we have
Component weight parallel to the plane
=
250N sin 35o =
143⋅79N
Component weight perpendicular to the plane
= 250 N cos 35 o = 204.79 N
∴Normal reaction R =204.79N
Since the block is at rest,
∴F
P+F= 143.39N
= 143.39-P= (143.39-49) N =94.39N
The friction force is 94.39N
(b)
using M =
M
F
R,
= 94.39N
=0.46
204.79N
The coefficient of limiting friction is o.46
SELF ASSESSMENT EXERCISE 1
1
a.
b.
c.
2.
State
i.
ii.
iii.
iv.
3.
156
What is friction?
Name and define the two types of friction.
Distinguish between coefficients of static and dynamic
friction.
the laws of friction between the surfaces of two bodies in
contact
two advantages of friction
two disadvantages of friction
methods of reducing friction
A body of mass 5 kg rest in an inclined place whose angle of
inclination to the horizontal can be varied. If the coefficient of
friction between the body and the surface of the inclined place is
0.4, at what angle does the body start to slide down the plane?
PHY 001
4.
A block of wood weighing 5N is placed on a horizontal table. It
is then pulled by means of a spring balance attached to one of its
ends. The block just begins to move when the spring balance
records a horizontal force of 2.5N.
i
ii
3.7
ACCESS PHYSICS
what is the coefficient of static friction?
what is the frictional force when a horizontal force of 2.0N
is recorded by the spring balance.
Viscosity
3.7.1 Concept of Viscosity
Frictional forces also exist between different layers of liquid surfaces.
We refer to such frictional force as viscosity. The viscosity of a liquid is
low if the frictional force is low and vice versa. For instance, if you
throw an object into water and a similar object into palm oil, the object
moving through water will experience less resistance than when it
moves through palm oil because the viscosity of water is Lower than
that of palm oil. This makes the object to move faster in water.
The attraction between molecules in neighboring layers of a liquid is
responsible for the viscosity of the liquid.
Viscosity can be defined as the property of a fluid which tends to
prevent motion of one part over another.
It is the internal friction between layers of a liquid or gas in motion.
The viscosity of a liquid rapidly decreases as its temperature rises. You
will have noticed that tar which is used for making roads is first heated
so that it will flow readily.
You can compare the viscosity of liquids roughly by dropping a small
steel ball –bearing into a tall wide cylindrical vessel filled with the
various liquids whose viscosity you want to compare. You will then
measure the time it will take the ball bearing to fall through the liquids.
You will notice that with water, the ball bearing fall rapidly. It falls
fairly slowly in oil, and very slowly in glycerine (Fig. 2.5).
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Viscous force
+upthrust
Weight
A- Water
B- Engine oil
C-Glycerine
Fig. 2.5: Motion in a viscous liquid
3.7.2 Effects of Viscosity
Here we look at some important applications of viscosity. You may have
identified some of this yourself in our earlier discussions although you
may not have categorized them under this heading.
The viscosity of oil is very important in selecting suitable lubricants for
the various sections of the machines. We use more viscous oils for
heavier machines than in light ones.
Oils, grease and air are used as lubricants because of their viscosity.
We use engine oils in lubricating engines to keep metal surfaces from
rubbing against each other.
Water is not used as a lubricant because it has low viscosity.
The viscosity of oils and greases decrease with temperature. Therefore,
their lubricating effects are lowered at high temperature.
The viscosity of the air makes the bob of a swinging pendulum come to
rest more quickly with a cord attachment to the string than without a
cord.
The wider the area of the cord the greater the viscous force opposing the
pendulum’s motion and the pendulum bob comes to rest quicker.
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Laws of Viscosity and Coefficient of Viscosity
3.8.1 Laws of Viscosity
•
•
•
•
It opposes motion (like solid friction).
It depends on the area of the surface in contact (unlike solid
friction).
It depends on the relative motion between the layers (unlike solid
friction).
It depends on the nature of the liquid.
It is independent of normal reaction (unlike solid friction).
3.8.2 Coefficient of Viscosity
The coefficient of viscosity of a fluid is a measure of the viscosity of the
fluid.
To define this coefficient, let us consider a thin layer of liquid between
two flat parallel plates (Fig. 2.6). If the lower plate is held in a fixed
position,
F
A
d
Fig. 2.6: Viscosity of a fluid
Each layer moves with a small velocity relative to an adjacent layer.
a force F, is required to move the upper plate at a constant speed. This
force is observed to be proportional to the area A, and the speed V, of
the moving plate, and inversely proportional to the thickness d, of the
liquid between the plates, that is,
Or
F
∝
F
=
AV
d
η AV
d
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Where η, the constant of proportional, is the coefficient of viscosity of
the liquid,
We have that
η= fd = F/A
AV
V/d
The ratio; i, F/A is force per unit area (pressure)
Unit: Pa
Ii, V/d is velocity gradient
Units: S-1
∴ S. I .unit of coefficient of friction is Pascal second
(PaS).
Table 2.2: Typical values of viscosity
Temperature
0
C
0
20
40
60
80
100
Coefficient of Viscosity
Castor oil Pas Water 10-2 Pas Viscosity Air
10-6 Pas
53.00
1.792
171
9.86
1.005
181
2.31
0.656
190
0.80
0.469
200
0.30
0.357
209
0.17
0.284
218
Table 2.2 shows that the viscosities of substances vary with temperature,
but while those of liquid decrease with temperature, those of gases rise
as the temperature rises.
3.9
Motion of a Ball in a Viscous Liquid
When you throw a steel ball to fall through a viscous liquid, the ball is
subject to three forces:
•
•
•
160
Its weight (w) acting downwards;
The upthrust (u) of the liquid on the stone acting upwards, and
The viscous force acting opposite to the motion of the ball i.e.
upwards. This is shown in Fig. 2.7 below.
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Fig. 2.7: A ball falling through a viscous liquid
The ball accelerates at first but finally attains a uniform velocity known
as terminal velocity.
At first the equation of motion is given by mg – v – u = ma
Where v = viscous force, u = upthrust, m = mass of the ball, g = accn
due to gravity and a = acceleration of the ball in motion.
When the terminal velocity is reached, a = 0,
The equation becomes;
Mg - V - U = 0
⇒ mg = V + U.
Or
V = W-U
(Mg = w)
This constant velocity (v) is termed the terminal velocity.
Fig. 2.8 shows the variation of velocity with time as an object dropped
inside a viscous liquid attains termined velocity.
←Terminal velocity
←Acceleration
Time
Fig. 2.8: Viscous fluid terminal velocity in viscous fluid
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3.10 Similarities and Differences between Viscosity and
Friction
You must have observed that the following similarities and differences
exist between viscosity and friction.
Similarities
•
•
Both forces oppose relative motion between surfaces.
Both depend on the nature of the materials in contact.
Differences
•
•
•
•
Friction does not depend on area of surfaces in contact; viscosity
depends on areas of surfaces in contact.
Friction is dependent on normal reaction, viscosity is not.
Friction occurs in solids, viscosity takes place in liquids and
gases (fluids).
Friction does not depend on the relative velocities between two
layers; viscosity depends on the relative velocity between layers.
SELF ASSESSMENT EXERCISE 2
1
Explain the terms
i.
Viscosity
ii.
Terminal velocity
2
Draw a diagram to show the forces acting on steel ball falling
through a viscous liquid.
3.
Sketch the graph of velocity against time of the steel ball and
explain its nature.
4.
What are the differences and similarities between friction and
viscosity?
4.0
CONCLUSION
In this unit you learned about friction in solids and liquids. You also
learned about its advantages and disadvantages; why it is necessary to
reduce friction as well as the laws governing it. You also learned how to
determine the differences between friction and viscosity, and the
applications of viscosity.
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5.0
SUMMARY
•
Friction is the force which acts to oppose motion at the surface
separating two objects in contact.
•
Static or limiting friction is the maximum force that must be
overcome before a body can just start to move over another.
•
Kinetic or dynamic friction is the force that must be overcome so
that a body can move with uniform speed over another.
•
Coefficient of static friction is the ratio of the frictional force (F)
to the normal reaction R.
•
•
•
•
m = F
R.
Frictional force is independent of the areas of surfaces in contact.
Friction has both desirable and undesirable effects.
Friction is reduced by lubrication, ball bearing and by
streamlining
Viscosity is the internal friction between layers of a liquid or gas
in motion.
Viscosity decreases with temperature.
•
Viscous oils and air under pressure are used as lubricants.
6.0
TUTOR-MARKED ASSIGNMENT
1
Define :
i,
Friction.
ii,
Viscosity.
2
List the three major ways of reducing friction.
3
State :
i,
four differences,
ii,
two similarities
between friction and viscosity.
4
Draw a diagram to show the forces acting on steel – ball falling
on a viscous liquid.
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5
A metal box weighs 50N. The coefficient of limiting friction
between the metal box and the horizontal surface of a table is 0.3.
i.
What is the horizontal force required to just move the
metal box?
ii.
If another 50N metal box is stacked on this metal box,
what is the minimum horizontal force that must be applied
to move the metal boxes?
7.0
REFERENCES/FURTHER READINGS
Anyakoya, M.W. (2000). New School Physics for Senior Secondary
School.Onisha Africana –FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physic.
Ibadan: University Press.
Ndupu, B.L.N. and Okeke, P.N. (2002). Round- Up Physics. Ikeja:
Longman Nigeria
Okpala, P.N. (1990). Physics. Certificate Year Series for Senior
Secondary Schools. Ibadan: N.P.S Educational.
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LINEAR MOMENTUM
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Momentum and Impulse
3.1.1 Momentum
3.1.2 Impulse
3.2
Newton’s Laws of Motion
3.2.1 Newton’s First Law of Motion
3.2.2 Newton’s Second Law of Motion
3.2.3 Newton’s Third Law of Motion
3.3
Conservation of Linear Momentum
3.3.1 Statement of Principle of Conservation of Linear
Momentum
3.3.2 Illustration of the Principle
3.3.3 Elastic and Inelastic Collision
3.4
More Applications (examples) of Newton’s Laws and the
Satellites and weightlessness
3.4.1 Conservation of Linear Momentum Law
3.4.2 How Walking is Possible
3.4.3 Jet and Rocket Propulsion
3.5
The Lawn Sprayer
3.5.1 Inertial Mass and Weight
3.5.2 Inertial Mass
3.5.3 Weight
3.5.4 Weight of a Body inside the Lift
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit we will discuss the influence of unbalanced forces on bodies,
the motion of objects, and the laws governing such objects. We will also
study the collision between bodies that are virtually free from forces
other than those which they exert on each other at impact and the laws
governing such collisions.
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OBJECTIVES
At the end of this unit, you should be able to:
•
•
State and explain in your own words, each of Newton’s laws of
motion
Show that Newton’s second law of motion gives
o a measure of force as the product of mass and acceleration;
o an operational definition of force as rate of change of
momentum.
Solve simple problems based on Newton’s laws of motion and
the principle of momentum.
State and explain the meaning of the law of conversation of linear
momentum.
Solve simple problems based on conservation of linear
momentum.
Explain
o Why walking is possible
o Why a gun recoils when fired
o How a rocket is propelled
o How a jet plane is propelled
Explain inertial mass and the relationship between mass and
weight.
Explain the variation in the weight of a body from place to place.
3.0
MAIN CONTENT
3.1
Momentum and Impulse
•
•
•
•
•
•
3.1.1 Momentum
Definition
The momentum of a body is defined as the product of its mass and its
velocity.
If a body of mass m moves with a velocity v, then its momentum p is
given by:
P = mv
Unit of momentum is kilogram metre per second (kgms-1).
Momentum is a vector, its magnitude is the numerical value of the
product m x v and its direction is the direction of v.
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Bodies moving in a straight line possess “linear momentum” while
rotating bodies have “angular momentum”.
3.1.2
Impulse
Definition
Impulse is defined as the product of the average force on the particle and
the time during which it acts.
If a force F acts for a short time t, the impulse I is given by
I=Fxt
Impulse is a vector with magnitude equal to the product F x t, and
direction same as that of the force F.
The unit of impulse is Newton – second (Ns).
Worked examples
1
A body of mass 4.0 kg moves with a velocity of 15 ms -1.
Calculate the momentum of the body?
Solution
P = mv
:. p = 4kg x 15m
S
= 60 kg ms-1
2
An average force of 30N hits a stationary ball for a time of 0.025.
What is the impulse experienced by the ball?
Solution
I = Ft
:. I = 30 N x 0.02s
= 0.6 Ns
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Newton’s Laws of Motion
3.2.1 Newton’s First Law of Motion
Statement
Newton’s law of motion states that every object continues in its state of
rest or of uniform motion in a straight line unless acted upon by an
external force. Importance of Newton’s first law
(i)
The law identifies inertia as an inherent property of a body at rest
or moving with a constant velocity.
Inertia is thus a property of matter and may be defined as the
reluctance of a body to change its state of rest or of uniform
motion in a straight line unless an external force acts on it.
The mass of a body is a measure of its inertia. The greater the
mass of a body, the greater its inertia and vice versa.
(ii)
Newton’s first law explains why passengers in a fast moving
vehicle tend to move forward when the vehicle stops suddenly, or
jers backwards when the vehicle suddenly speeds off, since there
is little or no force to restrain them. For this reason motorists are
advised to use seat belts (or safety belts) while traveling in their
cars.
(iii)
Newton’s first law also explains what force does, although it does
not suggest how it should be measured.
3.2.2 Newton’s Second Law of Motion
Statement
Newton’s second law states that the rate of change of momentum is
proportional to the force causing motion and takes place in the direction
of that force.
Importance of Newton’s Second Law
(i)
It enables us to define an absolute unit of force which remains
constant under all conditions.
(ii)
It gives a measure of force as product of mass and acceleration.
(iii)
It gives an operational definition of force as the rate of change of
momentum.
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According to Newton’s Second Law,
Fα
change in momentum
time taken for the change
i.e.
F α mv – mu
t
where m, u, v, t are mass, initial velocity, final velocity and time
respectively of motion of the body acted upon by a force, F.
∴
∴
F α mv – mu or f m(v – u)
t
t
F α ma, where v – u = acceleration, a.
t
F = k ma, where k = constant.
In the S.I. unit system, F is in Newton, m in kilogram, and the
acceleration, a in meter per sec squared (ms-2). The Newton is chosen
such that K has a value of 1.
∴
F = ma
Definition
The Newton is defined as the unit of force which gives a mass of 1kg an
acceleration of 1ms-2.
Worked examples
1.
A mass of 5.0 kg is acted upon by an unbalanced force of 25N.
What acceleration does the force give the mass?
Solution
F = ma
25N = 5 kg x a
∴ a = 25N/5 kg
= 25kgms-2/5 kg
= 5ms-2
2.
A stone of mass 900 g is pushed along a floor by a horizontal
force of 20N. A frictional force of 8N opposed the motion of the
stone. Calculate the acceleration given to the stone.
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Solution
Unbalanced force = (20 – 8) N = 12N
F = ma
12N = 900 kg x a
1000
∴ a = 12 x 1000 ms-2
900
= 13.33 ms-2.
3.
A steel ball of mass 0.1 kg fell from a height of 8 m onto a hard
surface and bounced back to a height of 2 m. Calculate the
change of momentum. If the ball was in contact with the hard
surface for 0.15s, find the force exerted on the surface (g = 10
ms-2)
Solution
Velocity v of the ball on hitting the floor is given by
i.e.
v2
=
u2 + 2gh
v
=
√0+2 x 10x8 m
∴
v
=
=
√160 m2
s2
12.65 ms-1
let the velocity with which the ball left the surface after impact be U1,
At maximum height the (final) velocity is zero. Hence
170
0
=
u12-2gh
∴
0
=
u12-2 x10 m x 2 m
s2
∴
u12
=
40m2
s2
i. u1
=
√40m2
=
s
2
6.32 ms-1
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Change in momentum
=
mv + mu1, i.e (mv-(-mu,))
= 0.1(12.65+6.32)
= 1.897 kgms-1 (or Ns)
(Note that if the momentum before hitting the surface is positive, then
momentum after hitting the surface is negative. Hence change in
momentum is mv-(-mu,) = mv + mu1 )
Force
= Change in momentum
time
= 1.897 Ns
0.15s
=12.646N
=12.6N
3.2.3 Newton’s Third Law
Statement
Newton’s third law of motion states that to any action there is an aqual
and opposite reaction.
Example of Newton’s Third Law in operation
1.
If you hit your head against the wall, you exert some force on the
wall and the wall in turn exerts an equal and opposite force on
your head. If the force you exerted on
the wall is sizeable,
then pain you would feel on your head is a confirmation of
Newton’s third law. This implies that forces are always paired.
No single force exists in isolation.
R
Block of wood
Surface of table
W
Fig.3.1: Wooden block resting on a table
ii.
If you place an object (e.g. a wooden block) on a table, it will
remain at rest at the spot you left it (Fig.3.1). The block exerts on
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the table a force equal to its weight. The table exerts on the block
a reaction force, R, equal and opposite of W. since the two forces
balance, the block remains at rest on the table.
iii.
The recoil of a gun, If you fire a bullet from a gun, you will
observe that the gun jerks backwards or recoils (Fig.3.2). The
force propelling the bullet forward (action) must be equal to the
recoil force (reaction) acting on the gun (i.e. opposite). The
jerking of the gun (which you will fell in your hand) is actually
due to the conservation of momentum.
Fig.
3.3
2.3.2 Recoil of a gun
Conservation of Linear Momentum
3.3.1 Statement of Principle of Conservation of Linear
Momentum
The principle of conservation of linear momentum may be stated in
different ways as follows:
i.
In any system of colliding objects, the total momentum is always
conserved provided that there is no net external force acting on
the system.
OR
ii.
The total momentum of an isolated or closed system of colliding
bodies remains constant
OR
iii.
If two or more bodies collide in a closed system, the total
momentum after the collision is equal to the total momentum
before the collision.
Closed System:
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A closed or isolated system is a system on
which no external forces act. The only forces
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acting in a closed system of colliding bodies
are those due to the collisions.
The principle of conservation of linear momentum follows from
Newton’s seconds and third laws of motion.
3.3.2 Illustration of the Principle
U1
A
U2
F1
F2 = -F1
A
B
Before collision
B
At collision
V1
V2
BB
A
After collision
Fig.3.3: Conservation of linear momentum
Consider the bodies A and B (Fig..3.3). of masses m1, m2 respectively
moving towards each other with velocities u,u2 respectively. Let their
accelerations be a1, a2 respectively when the bodies collide, the force, f1
on A is equal and opposite to the force, f2 on B. from Newton’s third
law
F1 = - F2 ………………………………..
(1)
From Newton’s second law, F=ma
∴
M1a1, = m2a2
……………………………………….
(2)
We recall that acceleration, a = v-u, hence
t
∴
m1, v1-u2 = - m2 v2 – u2 ....................... (3)
t
t
where t is the contact time and u is the initial velocity before and v is the
final velocity after impact. Eliminating t from equation (3) we have
m1 (v1 –u2) = m2 (v2-u2)
ie.
m1,u1,+m2u2 = m1,v1,+m2v2 …………………………………………….
(4)
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Since we have defined momentum as mass x velocity equation (4)
means that the total momentum of the system before the impact is
equal to the total
momentum after the impact.
Worked examples
1.
A bullet of mass 10 g is fired from a gun of mass 5 kg at a
velocity of 250 m s-1. Find the velocity of recoil of the gun.
Solution
Mass of gun, m1 = 5 kg
Mass of bullets, m2 = 10g = 0.01 kg
Let final velocity of gun =v1 (in ms1)
Final velocity of bullet = v2 =250 ms-1
Total momentum of gun and bullet before firing is zero since both were
at rest.
Total final momentum = m1v1 + m2v2
Since momentum is conserved,
m1v1 =m2v2 = 0
ii. v1 = - m2v2
m1
- 0.01 x 250 ms-1
=
5
= - 0.5 ms-1
The negative sign indicates that the velocity of recoil of gun is in the
opposite direction to the velocity of the bullet.
A body of mass 5 kg moving with a velocity of 30 ms-1 collides with
another body moving in the opposite direction with a velocity of 20 ms-1.
if both of them now move in the direction of the first body (of mass 5
kg), at a common velocity of 10 ms-1. Calculate the mass of the second
body.
Solution
A sketch of the diagram will help us
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u2 =20ms-1
u1=30ms-1
A
B
m1= 5 kg
m2 (kg)
Before collision
V2 = 10 ms-1
V2 = 10 ms-1
.
B
A
After collision
Fig.3.4
Let mass of B = m2 kg
Total momentum before collision =
=
(5 x 30) + (-m2 x 20)
150 – 20 m2
Total momentum after collision
=
(5 x 10) = m2 x 10
=
50 + 10 m2
Since linear momentum is conserved, we have
150 -20m2 = 50 +10m2
30m2
= 100
i.e (150-50 = 10m2 + 20ms)
m2 = 3.33 kg
⇒ 30m2 = 100
m2 = 100
30
3.3.3 Elastic and Inelastic Collision
There are two major types of collision:
(ii)
(iii)
elastic collision and,
inelastic collision
(a) A collision is said to be “perfectly” elastic if both of the momentum
and the kinetic energy (k.e) of the system are conserved i.e.
constant ,
Hence,
(i)
m1 u1 + m2 u2 =
m1 v1 + m2 v2; and
(ii)
1 m1, u2 + 1 m2 u22 = 1 m, v2 + 1 m2 v22
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2
b
2
2
2
In an inelastic collision only the linear momentum is conserved.
The kinetic energy after collision is less than the kinetic energy
before collision
Hence,
(i)
(ii)
m1u1 + m2u2 =
m1v1 + m2v2
Ek1 > Ek2
Where Ek1 is the total K.E before impact and Ek2 is the total K.E after
impact.
OR
1 m1 U12 + 1 m2 U22 > 1 m1 v12 + 1 m2 v22
2
2
2
2
i.e ∑ KE before > ∑ KE after
where
∑ EK before is the total Kinetic Energy before impact
and
∑ EK after is the total Kinetic Energy after impact
In inelastic collision, the colliding bodies usually stick together and
move as a unit after collision, so as that v1 = v2 = v.
Hence
Total kinetic energy before= 1 m1u12 +1m2 u2 2 and
2
2
Total kinetic energy after = 1 m1 v2 + 1 m2 v2
2
2
or EK2 =1 (m1m2)v
2
2
1 m1 u1 +1 m2 u22 > 1 (m1 +m2) v
2
2
2
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SELF ASSESSMENT EXERCISE 1
Study explosion and compare it with elastic and inelastic collision
3.4
More Applications (Examples) of Newton’s Law and the
Conservation of Linear Momentum Law
3.4.1 How Walking is Possible
A person walks by pushing with his/her foot against the ground and the
ground exerts an equal and opposite force on the person (Fig.3.5)
Fig...3.5:
How
walking is possible
It is this reaction force of the ground on the person walking which
moves him or her forward.
Thus a person walking is actually pushed forward by the reaction force
of the ground on him and not by his on her own push.
3.4.2
action
forces
Jet and Rocket Propulsion
Fig. 3.6: Jet
propulsion:
and
reaction
The
principle
of
conservation of linear momentum and Newton’s third laws are also in
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the propulsion of jet aircraft and the rockets used for launching
satellites. Gases are burnt in the combustion chambers of the engine.
When jets of hot gases are expelled downwards through the tail nozzle
at high speeds, from rocket or aircraft, an equal opposite momentum is
given to the rocket or aircraft causing it to move.
3.4.3
The Lawn Sprayer
Fig.3.7: the lawn sprayer
The lawn sprayer is a device used to water a lawn or garden (Fig.3.7). It
combines Newton’s third law, the law of conservation of momentum
and the principles of a couple in its operation. Water issues from the
forward and backward nozzles of the sprayer at the same speed. The
reaction forces due to the thrust of the water from the two jets constitute
a couple whose moment causes the sprayer to rotate and so water the
lawn or garden.
Worked example
A rocket burns fuel at the rate of 200 g s-1 and is ejecting all the gas in
one direction at the rate of 400 ms-1. Find the maximum weight the
rocket can have if it is going to move vertically upwards.
Solution
Mass of gas per second = 200 kg s-1
1000
Velocity of expulsion = 400 ms-1
Momentum change per second = 200 x400 kg ms-2
1000
From Newton’s second law, we have
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F=w
200 x 400
1000
∴ F = 80N
F=
3.5
Inertial Mass and Weight
3.5.1 Inertial Mass
Mass was defined by Newton as the “quantity of matter in a body”. We
have seen from Newton’s first law that “inertia”, the reluctance of a
body to change its state of rest or uniform motion in a straight line, is an
inherent property of matter. The mass of a body can also be defined as
“a quantitative measure of the inertia of a body”. The more the mass a
body has, the greater the force required to change its state of or uniform
motion in a straight line and the greater the force required to give an
acceleration M ∝ F . Thus
a
this inherent property of matter is also called “inertial mass”.
Definition
The inertial mass of a body is a property of matter which represents the
resistance of the body to any kind of force whatever.
The mass of a body gives a quantitative measure of its inertial mass.
3.5.2 Weight
Definition
The weight of a body is the force acting on the body due to the earth’s
gravitational pull.
The weight (w) of a body is given by
W = mg
where m is the mass of the body and g is the acceleration due to gravity.
3.5.3 Weight of a Body Inside a Lift
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Fig.3.8: Weight of an object in a lift
Suppose you are standing in a lift or elevator.
Two forces act on you:
(i)
(ii)
your true weight (w = mg) acting vertically downwards, and
the reaction R of the floor of the elevator on you acting upwards.
We will use two forces to consider the apparent variation in your weight
(which you will feel) as the lift:
(a)
(b)
(c)
(d)
(a)
is stationary
accelerates upwards
accelerates downwards, or
falls freely under gravity, if the cable is cut.
When the elevator is stationary or moves with constant
velocity (Fig..3.8a), the acceleration, a = o and we
have that
W = mg = R……………………………..
(1)
If you are standing on a spring scale, it will record your true weight.
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When the lift accelerates upwards with an acceleration a
(Fig.3.8.b), you will also be pulled up with an acceleration
a while still standing on the floor of the lift.
The unbalanced force (F) acting on you is given by
∴
F = ma =R1, = mG
R1, = ma + mg = m (a+g)
Therefore your apparent weight W1, when the elevator moves upwards
with acceleration a (and which is balanced by R1,) is given by
∴
(c)
F = ma = R, -mg
R, = ma + mg = m (a+g)
Therefore your apparent weight W, when the elevator moves
upwards with acceleration a (and which is balanced by R) is
given by
W1= R1 = m (a+g) ……………………….
(2)
It is clear from (1) and (2) that. W1 > W.
You will therefore feel yourself pressing downwards (R1,) on the
floor with a larger force than when the lift is at rest. Your
apparent weight (W1,) as recorded by the scale will be greater
than your true weight (W).
(c)
When the elevator accelerates downwards with acceleration, a
(Fig.3.8c), the unbalanced force on you is
F= ma = mg = R2
iii. W2 = R2 = m (g-a) …………………… (3)
The scale on which you stand will read this value W 2. From (2)
and (3) it is clear that W>W2, hence your apparent weight, W2
will be less than your true weight and you will feel ligher.
(d)
In this case, the lift cable is cut and it falls freely with an
acceleration, a = g (Fig.3.8d). You will also fall freely with a = g.
when this happens, we see from (3), that
W3 = m (g-a) =o
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Thus you will appear to have no weight (i.e. weightless). You will exert
no force on the floor and the floor will exert no force on you.
The scale reads zero.
3.5.4 Satellites and Weightlessness
Without
gravity
Earth
With gravity
Fig.3.9: Free-fall of satellites:
Weightlessness
Persons and objects inside a satellite orbiting close to the earth are said
to experience weightlessness. This weightlessness is similar to the one
you would feel in a freely falling elevator as already discussed.
When a satellite moves round in an orbit close to the earth, the force of
gravity forces it to “fall” out of its natural straight line path (Fig.3.9).
While orbiting round the earth, the earth’s gravitational pull is just
enough to provide the centripetal force mv2 required for the orbital
movement, where m
r
is mass of the satellite, v its speed and r the radius of the orbit.
Thus mv2 = mg
r
The satellite therefore falls all the time. It does not strike the ground,
however, because its initial horizontal velocity causes it to miss the
surface of the earth. As a result of this free fall situation, every person
and object within the satellite is also falling freely under the gravity.
Hence such persons and objects within the satellite appear to be
weightless (.i.e experience seeming weightlessness)
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Worked example
An object of mass 70 kg rests on a spring weighing machine inside a lift.
The lift ascends with an acceleration of 2.5 ms-2. Calculate the reading
of the weighing machine. What will the reading of the machine be if the
velocity of the lift is constant? (g = 10 ms-2)
R
a = 2.5ms-2
W
Fig. 3.10
Solution
(i)
The unbalanced force F is given by
∴
F = ma = R- mg
R = m (g+a)
= 70x (10+2.5)
(ii)
=875N
When the velocity is constant, acceleration is zero
R = m (g+o)
= mg
=70x10
=700N
SELF ASSESSMENT EXERCISE 2
1.
Distinguish between momentum and impulse and state the
relationship between them. An object of mass 5 kg moving with
a velocity of 30 ms-1 is suddenly hit by another body B moving
in the same direction. If the velocity of A is changed to 50 ms-1
in the same direction, calculate the impulse received by it.
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2.
State Newton’s laws of motion. Show how the second law
enables us to define an absolute unit of force which remains
constant under all conditions. A body of mass 0.2 kg is dropped
from a height of 10 m onto a hard floor so that it bounces back
to a height of 2 m. Calculate the change in
momentum. If the
body is in contact with the floor for 0.1 sec. find the force
exerted on the body.
3.
Explain what is meant by the term “inertia” of a body.
A mass of 5 kg on a horizontal table surface with coefficient of
friction 0.2 is connected to a second mass of 10 kg over a
frictionless pulley. If g= 10 ms-2, calculate the acceleration of the
system when released.
4.
Why does a body in a lift appear to be weightless when the lift is
A body of mass 10 kg is to be given an acceleration of 20 ms-2.
Find the force required when the acceleration is (i) vertically
upwards,(ii) horizontal and the body moves over a surface whose
coefficient of friction is 0.5. (Take g=10 ms-2)
5.
State the principle of conservation of linear momentum. A ball of
mass 50 kg moving with a velocity of 10 ms-1 collides with
another ball of mass 60 g moving with 5 ms -1 in the opposite
direction. If the two balls stick together after collision, find the
magnitude and direction of their common velocity.
4.0
CONCLUSION
In this unit you learned linear momentum when we discussed the
influence of unbalanced forces on bodies, the notion of objects and the
laws governing such objects. You also learned the collision between
bodies and the laws governing various types of collisions.
5.0
SUMMARY
1.
Momentum (p) of a body is the product of its mass and its
velocity
(p = mv)
2.
Impulse (I) is the product of the force and the time during which
the force acts (I =n). Impulse acts only for a short time.
3.
Newton’s laws of motion are as follows:
First law:
184
Every object continues in its state of rest or of uniform
motion in a straight line unless acted upon by an external
force. This law leads to the idea of “inertia” which is the
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tendency of a body to preserve its state of rest or uniform
motion.
Second law:
Third law:
The rate of change of momentum is proportional to
the impressed force and takes place in the direction
of the force. This law leads to the law F = ma.
Action and reaction are equal and opposite.
4.
Principle of conservation of linear momentum states that when
two or more bodies act on one another their total momentum
remains constant that is conserved provided there are no external
forces acting on the bodies. Such a system of bodies is called an
isolated or closed system.
5.
In elastic collisions, both the linear momentum and kinetic
energy are conserved. In inelastic collision, linear momentum is
conserved but kinetic energy decreases.
6.
Mass is the quantity of matter in a body. It is also a measure of
the inertia of a body. Mass is constant all over the earth.
7.
“Inertial mass” is a measure of the tendency of a body to resist a
change in momentum. According to Newton’s second law of
motion, F ∝ a, the constant of proportionality is the inertial mass.
8.
Weight of an object is the force with which the earth pulls the
body to the earth’s centre. Weight varies from place to place
because the acceleration due to the earth’s gravity (g) varies from
place to place and weight of an object w=mg.
6.0
TUTOR- MARKED ASSIGNMENT
1
(a)
Explain what you understand by the expression “linear
momentum”
(b)
State the principle of conservation of linear momentum
and mention the condition under which the principle holds.
(c)
A body P of mass 100 g moving with a velocity of 10 ms-1
collides with another body Q of mass 200 g moving with a
velocity of 2 ms in the opposite direction. If P and Q stick
together after the collision, calculate their common
velocity v in the direction of P. Comment on your answer.
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2
State and explain the principle on which the jet aeroplane
functions. Mention two other useful applications of the same
principle. A jet expels gas at the rate of 0.5 kg s-1. If the force
produced by the jet is 300N, find the velocity with which the gas
is expelled.
3
State Newton’s laws of motion
A man whose mass is 80 kg stands on a spring weighting
machine inside an elevator. What is the reading of the weighting
machine when.
(ix)
(x)
(xi)
7.0
The elevator ascends with an acceleration of 2.0 ms-2?
The elevator is moving with a uniform velocity?
The elevator is coming to rest with a retardation of 4.0
ms-2?
REFERENCES/FURTHER READINGS
Anyokoha, M.W. (2000). Physics for Senior Secondary Schools.
Onisha: Africana – FEP Publishers ltd.
Awe, O. and Okunnola O.O. (1992). Comprehensive Certificate Physics,
SSCE Edition. Ibadan: University Press Plc.
Ndupu, B.L.N. and Okeke, P.N. (2000).Round-Up Physics, for West
Africa Senior School Certificate Examination: A Complete
Guide. Lagos: Longman Nigeria Plc.
Nelcon, M. (1986). Principles of Physics for Senior Sec. Schools (9 th
Edition).England: Longman Group Ltd.
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UNIT 4
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SIMPLE MACHINES
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
1.0
Introduction
Objectives
Main Content
3.1
Concept, Definition and Examples of Machines
3. 2.1 Definitions of (i) Force Ratio (Mechanical
Advantage), (ii) Velocity Ratio, (iii) Efficiency
3.2.2 Relationship between Force Ratio, Velocity Ratio
and Efficiency
3.3
Numerical Problems in Force Ratio, Velocity Ratio and
Efficiency of Simple Machines
3.4
Force Ratio, Velocity Ratio and Efficiency of Simple
Machine
3.4.1 Lever
3.4.2 Inclined Plane
3.4.3 Pulleys
3.4.4 Wheel and Axle
3.4.5 Screw
3.4.6 Wedge
3.5
Simple Machine as Components of Complex Machine –
Bicycle
3.6
Effects of Friction on Simple Machine and Methods of
Reduction
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
INTRODUCTION
In this unit, you will learn about machines, their types and modes of
action including their efficiencies and the effect of friction on them. You
will also see a machine as a system which enables work to be done more
easily or more quickly. You will learn about the working principles of a
machine before studying some simple machines.
2.0
OBJECTIVES
By the end of the unit you will be able to:
•
•
define a machine and list at least five simple machines.
define (a) force ratio (b) velocity ratio (c) efficiency and write
down the mathematical relationship between them.
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•
•
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•
relate (a) force ratio (b) velocity ratio and efficiency of a machine
calculate force ratio, velocity ratio and efficiency for a simple
machine.
draw (a) an inclined plane (b) wheel and axle, (c) a lever (d) a
pulley (e) a screw (f) a wedge and where possible do this to
achieve a specified velocity ratio.
identify the simple machines that make up a given complicated
machine such as a bicycle.
apply simple machines to do work.
3.0
MAIN CONTENT
3.1
Concept, Definition and Examples of Simple Machine
•
•
A machine enables us to overcome a large resistance or load by applying
a small effect. A machine enables us to do work more easily and
conveniently than could be done without it. Its purpose is to exert force
on an object which is usually applied to the machine. Sometimes the
force which the machine exerts is in a different direction from that of
the applied force.
3.1.2 Definition
A machine is a device or tool which allows a force (or effort) applied at
one point to overcome a resisting force (or load) at another point.
3.1.3 Examples of Simple Machines
You already know some of the simple machines which you come across
in your everyday life. These machines are the lever, pulleys, pliers
wheelbarrows, nut-crackers, the wedge, wheel and axle, inclined plane,
screw jack and many others.
3.2.1 Definitions of (1) Force Ratio (ii) Velocity Ratio (iii)
Efficiency of a Machine
•
Force ratio or mechanical advantage
We define efforts as the force applied to a machine and load as the force
or resistance overcome by the machine. The ability of a machine to
overcome a large load through a small effect is seen as Force Ratio or
Mechanical Advantage (MA). This is given by:
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Force ratio or mechanical advantage
MA = LOAD
EFFORT
The load you will take as the output force and the effort as the input
force. You can therefore define mechanical advantage or force ratio by:
Mechanical Advantage or Force Ratio =Output force
Input force
i.e. MA or FR =Output force
Input force
You will note that the mechanical advantage of a machine is influenced
by friction in the parts. You will use part of the effort applied to
overcome friction and part to overcome resistance or lift a load.
•
Velocity Ratio (VR)
We define Velocity Ratio (VR) as the ratio of the distance moved by the
effort and the load in the time interval
VR =
distance moved by effort (e)
distance moved by load (l)
The Velocity Ratio depends on the geometry of the machine. It is
independent of friction. For an ideal machine, that is, a machine that has
no friction, also called a perfect machine, worked done by machine =
work done on machine. Therefore, Load distance moved by load = effort
× distance moved by effort.
or
load
effort
=
distance moved by effort (e)
distance moved by load (l)
= Velocity Ratio
Hence for an ideal or perfect machine mechanical advantage = velocity
ratio.
•
Efficiently (Ef)
We define the Efficiency (Ef) of a machine as:
Ef = Useful work done by the machine
Work put into the machine.
× 100%
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Usually, the efficiency of a machine is expressed as a percentage. That is,
EF = useful work done by the machine
Work put into the machine
x 100%
Note that work (W) =force x distance
You know that a perfect or ideal machine has a 100% efficiency. This
means that all the work done by the effort is wholly used to overcome
the load. You also know that in practical machines; the efficiency is
usually less than 100% because of friction in the moving parts of the
machine. Part of the effort applied is used to overcome frictional forces
which are always present. Hence, the useful work done by the machine
is less than the workdone by the effort on the machine.
3.2.2 Relationship between Mechanical Advantage (Force
Ratio), Velocity Ratio and Efficiency
For any machine, we know that:
Efficiency (Ef) = work done on load × 100%
Work done by effort
=
=
⇒
Load (L) x
Effort (E)
L
E
×
L
E
distance moved by load (L) x 100%
distance moved by effort (e)
l
e
× 100% =
×
l
L ÷
E
e . ×
l
100%
× 100% ....(1)
e
where l and e are the distance moved by the load L and the effort E
respectively in the same time interval.
For mechanical advantage (MA) and Velocity Ratio (VR), we have the
following rotationships:
Mechanical Advantage, M. A. = L .......(2)
E
Velocity Ratio, V. R. = e .......... (3)
l
You will substitute equations (2) and (3) into equation (1) to obtain
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Efficiency (
Ef) = L
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x
l
x 100%
E
e
=
=
÷ e
L
E
x 100%
l
MA x 100%
VR
This implies that
Efficiency (Ef) = MA x 100%
VR
Mechanical Advantage, Velocity Ratio and Efficiency are ratios of
similar quantities and so have no units.
3.3
Numerical Problems and Force Ratio, Velocity Ratio and
Efficiency of Simple Machines
Worked examples
1
A simple machine has a VR of 5 and an efficiency of 90%. It is
used to raise a load of 2.kg through a height of 10 m in 5s.
Calculate:
(a)
(b)
(c)
The force ratio (FR) of the machine
The work done (W) on the load, and
The power (P) developed. (Take g = 10 s-2 )
Solution
(a)
∴
Efficiency (Ef) = FR × 100%
VR
FR (MA) = Ef × VR
100
=
90 × 5
100
=
450 = 4.5
100
The Force Ratio of the machine is 4.5
(b)
Load = 2.5 kg × 10 ms-2 = 25N
Work done on load = load × distance moved by load
= 250N × 10 m = 250 J.
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The work done on the load is 250 J
(c)
Power developed = Work done
Time taken
= 250 = 50W
5s
The power developed is 50 W
2.
A machine whose efficiency is 80% is used to lift a load of
1000N. Calculate the effort put into the machine if it has a
velocity ratio of 6.
Solution
Effiency (Ef) = MA (FR) 100% = 80%
VR
MA = 80 .
6
100
⇒
MA or FR = 6 × 80
100
= 480
100
= 4.8
But
FR= load =
Effort
∴ 4.8 =
1000
Effort
1000
Effort
Effort = 1000 =208.33 N
4.8
the effort put into the machine is 208.33
3
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The efficiency of a machine is 70%. Find in joules the work done
by a student using this machine to raise a load of 75 kg through a
vertical distance of 3 m (g = 10 ms-2)
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Solution
Efficient % = Work done by the machine
Work done on the machine
⇒
∴
70 =
100
75 10 x 3
work done by student
work done by student
x 100%
x 100
100
=
75 x 10 x 3 x100
70
=
3214.29 Joules
The work done by the student by using the machine to
raise the load is 3214.29J.
SELF ASSESSMENT EXERCISE 1
1
(a)
(b)
2
Define the terms: Force Ratio, Velocity Ratio and Efficiency as
applied to a machine. State the relationship between these terms.
3.
The efficiency of a machine is 80%. Determine the work done by
a person using this machine to raise a load of 250kg through a
vertical distance of 5.0 m (take g = 10 ms-2).
4.
A simple machine has a VR of 5, and an efficiency of 95%. It is
used to raise a load of 2 kg through a height of 15 m in 10 s.
Calculate (a) the FR of the machine (b) the work done on load,
and (c) the power developed.
3.4.
Force Ratio, Velocity Ratio and Efficiency of Simple
Machines
3.4.1
Define a machine? Give 5 examples.
What do you understand by the statement that the force
ratio of a machine is 7 and its velocity ratio 12?
Lever
The lever is one of the simplest machines we know. You can overcome
a large resistance by the application of a small force with it.
The lever consists of a stiff bar or rigid rod supported at the fulcrum or
pivot about which it can rotate. If you apply an effort (E) at one point
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of the lever, it will lift a load (L) or overcome a resistance at some
other point. The diagram is shown below in Fig. 4.1
a
b
E
Fulcrum
or
Pivot
L
Fig. 4.1: The lever
The lever operates on the principle of moments. According to this
principle, we can state from Fig. 4.1 that:
E x a = Lx b ..... (4)
where: a is the distance of the effort from the fulcrum; b is the
distance of the load from the same fulcrum.
From equation (4),
L =
E
a
b
That is
Force Ratio = Velocity Ratio
where
a/b is ratio of the two arms of the lever
That is: efficiency (Ef) =
FR = 1
VR
•
Types of Levers
We have three classes of levers: first order, second order and third order.
This classification is dependent on the relative position of the effort,
load and fulcrum.
•
First Order Levers
Here, the fulcrum (F) or pivot is between the load and the effort.
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F
E
F
Fig. 4.2: lst order lever
Examples
The crawbar, a pair of scissors or pincers, claw hammer and pliers.
The velocity ratio is usually greater than 1 but could be less than or
equal to 1.
•
Second order levers
In the second order lever, the load (L) is between the effort (E) and the
fulcrum or pivot.
E
F
L
Fig. 4.3: 2nd order lever
Examples
Wheelbarrow and nutcrackers
The mechanical advantage and velocity ratio are always greater than 1.
•
Third order levers:
Here, effort (E) is between load (L) and the fulcrum (F).
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E
Fulcrum
F
Load (L)
3rd order lever
Fig. 4.4:
Examples: forceps, tongs, forearm.
The force ratio and velocity ratio are less than 1.
3.4.2 The Inclined Plane
You have seen lorry drivers use this type of machine to raise heavy
loads such as drums of oil, up a sloping plank to the high floors of L
lorries. The sloping plank is an example of an inclined plane. As shown
in Fig. 4.5, it is a rigid plane which is kept inclined to the horizontal at a
certain angle, θ
(a)
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B
E
(a)
h
L
θ
90o
A
O
(b)
Fig. 4.5: The inclined plane (a) loading in lorry through an inclined
plane (b) schematic of the inclined plane with load
As the applied force or effort E moves along OB, at an angle o to the
horizontal, the load L is lifted up through a vertical height AB. Thus, the
velocity ratio is given by:
VR =
distance moved by effort
distance moved by load
= OB =
AB
VR
1
sin θ
=
1
Sin θ
If friction forces are absent, you will have FR= VR =
1. .
sinθ
But friction along the plane affects the efficiency and the FR will be less
than VR.
3.4.3
The Pulleys
The pulley system is one of the most widely used types of machine. You
often see pulleys being used at construction sites to raise or lower heavy
loads. As you can see, a simple pulley is a fixed wheel hung on a
suitable support with a rope passing round its groove as shown in Fig
4.6.
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Rope
Fixed wheel
So
F
Rope
L
F
E
So
L
Fig. 4.6 :A simple pulley
E
E
Fig. 4.7: Pulley system
(a) 4 pulleys
(b) 5 pulleys
A load L is attached at one end of the rope and an effort E is
applied at the other end. If you neglect the weight of the rope and
the frictional forces, the effort (E) applied to the rope will be
equal to the load lifted because the tension on the rope will be
same throughout.
Hence, L = E and FR = 1
Also,
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The distance moved by the effort So = the distance moved by load So.
Hence,
VR =1
You will realize that in practice, the effort has to lift the load (L) and
also overcome the friction which hinders the pulleys from turning freely.
Thus, there may be a mechanical disadvantage. However, you will still
have the advantage in the use of such a pulley in that one can stand on
the ground and use one’s body weight in addition to his muscle to pull
downwards on the rope and raise the load upwards. You know that this
is easier than lifting the load directly upwards. You can also make use of
such a single pulley to change the direction in which a force is to be
exerted.
You can always determine the VR of a pulley system from the number of
pulleys. The VR of a pulley system is to equal to the number of pulleys.
For instance, in the block and tackle system of pulleys shown in figure
4.7, the VR of (a) having 4 pulleys is 4 while that of (b) having 5 pulleys
is 5.
You will also note that
•
The MA (FR) increases as the number of pulleys increases
•
If efficiency is 100%, MA is equal to the VR, however, owing to
friction and weight of pulleys, the efficiency is always less than
100%.
•
Efficiency gets less as the number of pulley increases.
•
With a particular pulley system, the greater the load to be lifted,
the greater the efficiency. This is because the weight of the
pulleys attached to one load is less in proportion to the load.
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3.4.4
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Wheel and Axle
a
b
E
L = Weight
Fig. 4.8: Wheel and axle
Do you know that the wheel and axle is equivalent to a lever? The effort
arm is the radius of the wheel a, and the load arm, the radius of the axle
b in this
case,
E x 2πa
=
L x 2πb
From which
Mechanical advantage
=
L
=
a
E
b
Here, you note that as the effort arm completes one revolution, the load
is raised by the circumference of the axle.
Therefore,
Velocity ratio =
2 πa
2 πb
=
a
b
Here again, in the ideal case, the efficiency is 100%. However, owing to
friction, its value is less than 100% in a practical case.
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The Screw Mechanisms
Fig. 4.9: The Screw mechanisms
You are familiar with the screw jack used to lift a car so as to change
/one of its wheels. They are examples of screw mechanism (Fig. 4.9).
We call the vertical distance P between successive threads, the pitch.
When you move the load through a distance P, the effort moves through
a distance equal
to the circumference of the handle of th screw
driver in Fig. 4.9(a) or through the circumstances of the circle traced out
by the lever arm.
Fig. 4.9(b)
Therefore,
Velocity ratio =
2πr
P
With a high value of r compared with p, it is possible to obtain a high
velocity ratio and therefore a big mechanical advantage.
In the jack, friction reduces the mechanical advantage considerable such
that the efficiency is usually below 50%.
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3.4.6 The Wedge
The wedge is a combination of two inclined planes. You may have
noticed some people use it to split wood or rock into piece or to widen
an opening. Examples of wedge – type of machines are axes, chisels,
knives, and other cutting tools usually used by builders to separate
bodies which are held together by large forces. This is shown in Fig.
4.10.
E
χo = t
χo
Axe
B
C
χ1 = L
R
R
Handle
θ
A
(a)
(b)
(c)
(d)
Fig. 4.10: The wedge
We noticed that when the users drive in the wedge by the effort force, a
much larger force is exerted to each side by the wedge surface. Thus, as
the wedge is driven in a distance x1, the surfaces being separated are
moved apart through a distance χ as shown in Fig. 4.10.
If we take the length of the wedge to be L and its thickness to be t, then
the mechanical advantage (neglecting frictional forces) is given by
M.A =
load
Effort
M. A. =
=
χ1 = L
χ2 t
Slant height of wedge
Thickness of wedge
You will note that a thin wedge has a higher mechanical advantage than
a short thick one; or the smaller than angle O between the slant heights,
the greater the M.A.
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3.5
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Simple Machines as Components of Complex Machine –
The Bicycle
3.5.1 The gear wheels
belt
A
shafts
B
(b)
(a)
B
Fig. 4.11: Principle of gears
Principle of gears
Gear wheels are commonly used in cars, bicycles and cranes. In these
machines you will apply the principle of the wheel and the axle to the
two wheels (often toothed wheels) of different radii connected by a belt
and rotating on separate shafts.
Here both the load and the effort are applied to the shafts and velocity
ratio is given by.
number of teeth on driven wheel (A)
Number of teeth on driving wheel (B
VR =
3.5.2 The Bicycle
C
2b
2a
Hub wheel
Pedal
Back wheel
Front wheel
Sprocket wheel
Fig. 4.12: The bicycle
You are aware that the bicycle uses two cog wheels, the sprocket wheel
which is rapidly attached to the back wheel of the cycle. The cogs on the
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two wheels are evenly separated. When the pedals and sprocket wheel
complete a revolution, the hub wheel rotates through 2πa revolutions.
2πb
and the back wheel does the same.
You will notice that the back wheel has moved through a distance of
2πC x a
b when the pedals complete a revolution.
We regard the friction between the tires and the ground as constituting
the load while the effort is applied on the pedals.
If we take L to be length of the pedal crank, and the distance moved by
effort in one revolution is 2πl, then, velocity ratio equals
2π ι =
bι
2πac
ac
b
The velocity ratio and the mechanical advantage (or force ratio) is
considerably less than 1. But a rapid movement of the load compared
with the effort is achieved.
3.6
Effect of Friction on Simple Machines and Methods of
Reduction
In this unit, you have noted how frictional forces add to the load which
has to be overcome by effort. The presence of friction between the
moving parts of machines as mentioned in Module 2 unit 3.3 makes it
impossible for any machine to have an efficiency of 100%. This is
because part of the effort applied is used to overcome frictional forces
and the remainder is utilized to do useful work. The work done in
overcoming friction is ‘useless’ or wasted energy’ and it appears as heat
in the moving parts of machines.
In order to improve the efficiency of machines you will try to reduce the
friction present between the moving parts of the machine by :
1
2
204
Lubricating which prevents metal surfaces from rubbing against
each other and;
Using ball bearings or roller bearings between the two surfaces in
contact.
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3.7
Worked Examples
1.
A lever of velocity ratio 25 overcomes a resistance of 2.0x 103
N with a force of 100N.
Calculate:
(i)
(ii)
the mechanical advantage
the efficiency of the lever
Solution
(i)
Mechanical advantage
resistance (load)
Effort
=
2.0 x 103 N
100N
=
(ii)
Efficiency (Ef)
=
mechanical advantage
Velocity ratio
=
2.
= 20
20 =
25
=
0.8
80%
A force parallel to an inclined plane is used to raise a load of
500N through a height of 3.5m while the load moves a distance
of 35m along the plane.
(i)
(ii)
(iii)
How much work is done on the load?
If the efficiency is 80%, determine the effort.
How much work is done by the effort?
Solution
(i)
Work done on the load = increases in P.E. of load
(mgh = wh)
= 500N x 3.5 = 1750J
(ii)
Efficiency
=
work output
Work input
Work input
=
Effort (E) x distance moved by effort
=
E x 35 m
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Efficiency (Ef)
=
∴ Effort (E) =
=
(iii)
80
100
=
1750J .
E x 35m
1750 J x 100 =
80 x 35
175000
2800
62.5 N
Work done by effort =
=
Work output
Efficiency
1750 =
2187.5N
0.8
=
2.19 x 103 N
SELF ASSESSMENT EXERCISE 2
1
2.
4.0
(a)
A screw jack whose pitch is 2 mm is used to raise a motor
car of mass 900 kg through a height of 20.0 cm. The
length of the tommy bar of the jack is 40 cm. If the jack is
60% efficient, Calculate the:
(i)
The velocity ratio
(ii)
Mechanical advantage of the jack
(iii) Effort required
(iv)
Work done by the effort (g = 10 ms-2)
(b)
Explain why the efficiency of a screw jack is usually less
than 100%.
By using a pulley system with a velocity ratio of 5, it is found
that a load of 500N can be raised at a uniform speed by an effort
of 125 N.
a
What is the mechanical advantage of the machine for this
load?
b
If the efficiency of the pulley system is 80%, what effort
will be required to lift a load of 10 N.
CONCLUSION
In this unit you learned about simple machine and its working
principles. You learned the definition and meaning of terms such as
effort, load, force ratio, or mechanical advantage, velocity ratio,
efficiency used in connection with machines.
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The simple machines you studied in the unit include the lever, the
pulley, the screw, the inclined, plane, wheel and axle, the wedge.
You also learned about a simple machine as a component part of a
complicated machine – the bicycle.
Finally, you solved problems arising from the unit and looked at
practical effect of friction and its reduction in machines.
5.0
SUMMARY
•
A machine is any device which allows a large weight or
resistance to be overcome by a small effort.
•
Mechanical Advantage or Force Ratio is defined by M.A
Load
Effort
•
Velocity Ratio V.R
=
distance moved by effort
distance moved by load
Efficiency
=
=
•
•
•
Work output
Work input
M.A x 100%
VR
=
in the same time
x 100%
M.A depends on friction but V.R is independent of friction.
In this lever machine, V.R is the ratio of the lengths of two arms
of the lever. The M.A increases with an increase in this ratio.
In an inclined plane with θ as the angle of inclination to the
horizontal,
V.R = . 1 .
Sin θ Hence V.R increases when θ decreases.
•
In the Block and Tackle pulley system, V.R is equal to the
number of pulleys.
•
In the screw – jack, V.R. =
2 πa
p
where a is the length of the tommy bar and p is the pitch.
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•
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In the wheel and axle
V.R. =
radius of wheel = R
Radius of axle r
•
In the absence of friction MA= VR.
•
Friction reduces the efficiency of machines = friction is reduced
by lubrication and by the use of ball or roller – bearings.
6.0
1.
TUTOR-MARKED ASSIGNMENT
Define the terms: force ratio, velocity ratio and efficiency as
applied to a machine. State the relationship between them.
2. By using a pulley system with a velocity ratio of 5, it is found that a
load of 500N can be raised at a uniform speed by an effort of 125N.
(a) What is the mechanical advantage of the machine for this
load?
(b)
If the efficiency of the pulley system is 80% what effort
will be required to lift a load of 10 newtons?
3.
7.0
A wheel and axle is used to raise a mass of 50 kg by the
application of an effort of 250N. If the radius of the wheel and
axle is 300 mm and 100 mm respectively, determine the
efficiency of the machine. (Take g = 10 ms-2).
REFERENCES/ FURTHER READINGS
Anyakoha, M.W.(2000). New School Physics for Senior Secondary
Schools. Onisha: Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics.
Ibadan: University Press.
MODULE 3
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Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Mechanical Energy
Elastic Properties of Solids
Elastic Properties of Solids
Measurement of Temperature
Transfer of Heat and Heat Capacities
Latent Heat and Evaporation
Expansion of Gases
UNIT 1
MECHANICAL ENERGY
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Forms of Energy and World Energy Sources
3.1.1 Concept of Energy
3.1.2 Energy Sources
3.1.3 Form of Energy
3.2
Concept and Measurement of Work
3.2.1 Concept of Work
3.2.2 Measurement of Work
3.2.3 Work Done in a Force Field
3.3
Concept and Measurement of Energy
3.3.1 Concept of Mechanical Energy
3.3.2 Potential Energy (P-E)
3.3.3 Kinetic Energy (K.E)
3.3.4 Measurement of Energy
3.4
Concept and Measurement of Power
3.5
Transformation and Conservation of Mechanical Energy
3.5.1 The Law of Conservation of Energy
3.5.2 The Principle of Conservation of Mechanical
Energy
3.5.3 The Conservation of Mechanical Energy of a
Falling Body
3.5.4 The Conservation of Mechanical Energy involved
in a Swinging Pendulum
3.6
Applications of Mechanical Energy
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
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This unit introduces you to the concepts of work, energy, power and
their measurements. You will learn the forms and sources of energy.
Furthermore, you will learn in depth about mechanical energy, its types
and transformation. Finally, you will learn to treat mechanical energy,
work and power quantitatively.
2.0
OBJECTIVES
By the time you finish this unit, you will be able to:
•
•
•
explain work, energy and power and give examples of each
calculate
 the work done, given a force and displacement it produces in
its direction
 the gravitational potential energy at a height h above a given
reference plane
calculate the power in watts given an applied force and the time it
takes to produce a displacement
identify the type of energy possessed by a body under given
conditions
distinguish between kinetic and potential energy
identity energy transformation from one form into another
state the law of conservation of energy
calculate the kinetic energy of a body
calculate the potential energy of a body
show that the total energy is conserved for a given set of data on
the energy of a particle in a conservative field
verify the conservation of energy principle.
3.0
MAIN CONTENT
3.1
Forms of Energy and World Energy Sources
•
•
•
•
•
•
•
•
3.1.1 Concept of Energy
Energy is the mainspring of all life and of all the activities of mankind.
It may be defined as an ability or capacity to perform work. You use
energy in your home in the following ways:
•
•
210
Chemical energy stored in the food you eat helps you in walking,
playing, doing household work and other duties.
Chemical energy in wood or load or kerosene is converted by
combustion to heat energy for cooking your food, and heating
purposes.
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•
•
•
Light energy that falls on your eyes enables you to see things;
Sound energy that enters your ears makes you hear;
Electrical energy operates most appliances such as fan, TV sets,
radios, electric pressing irons, refrigerators and other appliances
in your home.
3.1.2
Energy Sources
You know that the sun is the main source of energy in the universe.
However, we can identify other sources which we have classified under:
•
•
•
Natural sources which include food, sun, wood, coals, natural
gas, oils and fats, tides/waves, wind, waters.
Manufactured sources include energy from batteries, generators,
electricity from the mains.
Petroleum products include that from kerosene, gasoline and
refinery gases (methane, ethane, propane and butane).
3.1.3 Forms of Energy
You are aware that energy can exist in different forms. Among these
are:
•
•
•
•
•
Chemical energy (stored in food and other fuels)
Kinetic energy (associated with motion).
Gravitational potential energy (associated with height above the
ground).
Elastic potential energy (stored in springs, rubbers, stretched
bows, catapults).
Nuclear energy (stored in nucleus of atoms).
Heat energy (flows from one point to another as a result of
temperature difference between the point.
Sound energy (produced by vibrating objects – violins, guitars,
piano, drums).
Light energy (produced by light bulbs)
Mechanical energy (associated with machines at work)
Electrical energy (taken in by TV set and other electrical
appliances).
Magnetic energy (associated with magnetism)
Solar energy (associated with the sun).
3.2
Concept and Measurement of Work
•
•
•
•
•
•
•
3.2.1 Concept of Work
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In every day life, different people apply the term work to any form of
activity that requires the exertion of physical or mental effort. However,
in Physics (or in mechanics), we use the term in a very specific sense.
We say that work is done whenever a force is applied to cause a body to
move.
When you push a car a certain distance, you are said to do work. When
no movement takes place after you have applied a force to a body such
as a car, no work is done.
Work is said to be done whenever a force moves its point of application
a distance in the direction of the force.
3.2.2 Measurement of Work
You will measure work by obtaining the product of force moving a body
and the distance moved by the body in the direction of the force. That
is, if a force F moves a body a distance S in the direction of dthe force,
the work done is given by:
i.e.
Work =
W
=
Force x displacement
FxS
-------(1)
The S.I. unit of work is Joule (J)
The Joule is equivalent to Newton – metre (Nm).
In general, when a constant force F at an angle θ = with the direction of
motion causes a body to move a distance, S, we define the work done by
W = F Cos θ x S
-------------- (2)
where F cos θ is the component of F in the direction of motion.
F
θ
F Cos θ
S
Fig .1.1:
Work is a scalar quantity
W = F Cos θ x S
You can also obtain equation (2) by resolving S in the direction of force
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S Cos θ
⇒
W = F Cos 0 x S or F x S Cos θ
(3)
When the force and the motion are in the same direction, θ = 0;
W = F x S Cos 0
=FxS
(4)
But, if the force is at right angles to the displacement, it has no
component in the direction of the displacement. Hence,
F Cos 90 = 0; work done = 0.
3.2.3
Work Done in a Force Field
(a)
(b)
Lifting a body
Falling bodies
(a)
Lifting a body
You know that to lift a body through a height h, a pulling force must be
applied to overcome the weight of the body. When an object is lifted
upwards vertically, work is done against the force of gravity or against
the weight of the body. You will obtain the magnitude of the work done
from
Work = force x distance
= mg x h
= mgh
(5)
where m = mass of the body,
g = acceleration due to gravity
h = height
(b)
Falling bodies
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When a body falls freely under gravity, we say that the earth’s
gravitational force does work on the body. For instance, if a body of
mass m, falls through a vertical distance h, the work done by gravity on
the body is obtained from.
W = mgh.
Also, if a body of mass m rolls down a hill of height h and length l, the
work done is mgh. In this case, you will still use the vertical height (h)
in the calculation and not the length of the slope ℓ
Worked examples
1. A body of weight 200 N falls freely through a height of 10 m, what
is the work done?
Solution
Work done = mgh
Weight (force) = 200N, height = 10 m
Work done = 200N x 10 m = 2000 J
(2)
∴
A boy lifts a load of weight 200N through a vertical height of 1 m
before putting it on his load. When he finds that the load is too
heavy to carry him quickly drops the load in its original place.
What is the work done by the boy ?
Solution
Work done = zero.
Reason: No displacement.
(3)
A load is pulled 7 m along a horizontal floor by a constant force
of 30N which acts at 400 to the horizontal.
Calculate the work done by the force.
Solution
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Work done = F Cos θ x S
F = 300N; θ = 40o ; S = 7 m
. . Work done = 30 Cos 40o x 7
= 30 x 0 .766 x 7
= 160.87J
3.3
Concept and Measurement of Energy
3.3.1 Concept of Mechanical Energy
We have already said that anything capable of doing work has energy.
This means that a person pushing a car along the road is doing work on
the car. So far as the car moves some distance. A mango fruit falling
from the top of the tree does work and therefore possess energy.
Can you mention other situations in which the application of force cause
a body to move. All of them possess energy.
There are many forms of energy but here, you will concentrate on the
study of mechanical energy.
Mechanical energy is classified into two types namely Potential Energy
and Kinetic Energy.
3.3.2 Potential Energy (P.E)
Potential energy is simply ‘stored energy’ or energy possessed by a
body by virtue of its position or state.
The stored energy is used to do work when the body is free to move
Examples of potential energy are:
•
•
•
•
•
a magnet at rest in a magnetic field (magnetic potential energy).
An electric charge at rest in an electric field (electric potential
energy).
a coiled spring when stretched or compressed (elastic potential
energy.
petrol, wood, and other fuel sources burn when they burn,
chemical potential energy is released,
a body at rest in a gravitational field e.g. a mango fruit on a
mango tree, gravitational potential energy.
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3.3.3 Kinetic Energy (K.E)
Kinetic energy is the energy possessed by a body by virtue of its motion.
Examples of kinetic energy are;
-
a student running a race.
an object falling freely under gravity.
electrical charges in motion.
any object in motion.
3.3.4 Measurement of Energy
(i)
(ii)
Potential energy (P.E)
Kinetic energy (K.E)
(i)
Measurement of Potential Energy (P.E)
Gravitational potential energy as you know depends on the weight and
height of an object above the surface of the earth. To lift an object at a
constant speed under gravity, you must exert a force, which is equal to
its weight. The work you do when you lift a body of mass m at a
uniform speed through a height h is mgh. The work done in lifting the
body to a height h, is stored and can be recovered in the form of kinetic
energy by allowing the body to fall through a distance h. The potential
energy of a body is not an absolute quantity. You will measure its
magnitude relative to some reference position.
To measure the gravitational potential energy P.E of a body, you will
measure the mass of the body and its height above a reference level h.
This is given by
P.E = mgh.
The potential energy of an object in other force fields which act on the
object is calculated in much the same way. First, you will select a
reference point and then obtain the potential energy at some other point
as the work done in moving from the reference point to the position of
the body in the force field.
ii
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Measurement of Kinetic Energy (K.E)
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The kinetic energy of a body in motion depends on both its mass m and
speed v. To obtain the kinetic energy of a body in motion you will use
the formula
K.E = 1/2 mv2 ........................…(6)
For you to obtain a value for the kinetic energy of a body you will know
its mass and speed. For instance, suppose an object of mass m moves
with a horizontal velocity u and is gradually brought to rest in a distance
S by a constant horizontal force F acting on it. The kinetic energy of the
body has been used in doing work against the opposing force F.
Therefore, kinetic energy originally possessed by the body
= work done against F
=FxS
But F = ma
where a = retardation of the object.
Hence F
F x S = Mas
You will recall from the relation
V2 = U2 + 2as
That since v = 0, a is negative (i.e. retardation)
Therefore,
V2 = U2 + 2as becomes 0 + u2 – 2as
⇒
U2 + as
2
Also,
KE + F x S = mas
You can now substitute ½ u2 for as in the above. When you do that,
K.E becomes
K.E = F x S + mas = ½ mu2
ie
K.E = ½ mu2
kinetic energy = ½ x mass x
..................................…(7)
(velocity)2
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Worked Examples
1.
A bullet of mass 0.5 kg was fired with a speed of 300 m/s.
Calculate its kinetic energy.
Solution
K.E = ½ mv2
(m = 0.5 kg, v = 300ms-1)
= ½ x 0.5 x (300)2
= 22500J
=====
A body of mass 50 kg is raised to a height of 15 m. What
is the potential energy of the body (g = 9.8 m/s-2).
2.
Solution
P.E = mgh
= 50 x 9.8 x 15 (m = 50 kg, g = 9.8 ms-1 h = 15m)
= 7350J
====
3.4
Concept and Measurement of Power
Let us consider two boys Chika and Ali who are carrying an equal load
mb each from the ground floor to the first floor of a building. The
amount of work done by Chika and Ali in lifting is exactly the same.
However, if Chika accomplishes the task before Ali, we say that Chika
works at a faster rate than Ali. That is, Chika has more power than Ali.
Also, you do the same amount of work in lifting your weight when you
climb up a staircase 10 metres high either in 15 seconds or in 25
seconds. But, you do not work at the same rate. You work faster when
you climb it in 15 second ss than in 25 second. In Physics, the work
done in unit time is called power.
-
218
Power is defined as the rate of doing work or using energy.
Power = Work done or energy expended
Time taken
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Usually, work is measured in joules and time in seconds, so, power is
expressed in joules per second. One joule per second = 1 watt.
Therefore, work is measured in watts (W).
A power of one watt is too small for practical use. So, we usually
measure power in a larger unit called a kilowatt (KW).
Worked Example
1.
A boy of mass 25 kg walks up a flight of stairs 8 metres high in
32 seconds, calculate the power of the boy (g = 9.8 ms-2)
Solution
Power = Work done
Time taken
Work done = mgh = 25 kg x 9.8 ms -2 x 8 m
Time taken =
∴ Power
t = 32 s.
= 25 x 9.8 x 8 = 1960
32
32
=
61.25
W
======
3.5
Transformation and Conservation of Mechanical Energy
3.5.1 The Law of Conservation of Energy
The law of conservation of energy states that in an isolated or closed
system, the total amount of energy is always constant, although energy
may be changed from one form to another.
When we talk about an isolated or closed system, we mean a group of
objects that neither receives energy from nor gives energy to objects
outside the system.
For mechanical energy being considered here, the law simply shows that
the sum of the P.E. and K.E is always constant for a given body, but the
energy may change from P.E to K.E or from K.E to P.E.
3.5.2 The Principle of Conservation of Mechanical Energy in a
Conservative Field
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You will note that the above principle is true in a conservative field. An
important aspect of a conservative force is that the work done by such a
force is recoverable. For example, if you do a work against a
gravitational force in raising a body to a height h above the ground, that
work is recovered when the body falls a distance h back to its original
position.
The earth’s gravitational field is an example of field containing
conservative forces. In such a conservative field, total mechanical
energy is conserved.
Friction is an example of a non-conservative force. We noticed that
when friction acts between the moving parts of a machine, some
mechanical energy is lost but it reappears in the form of heat and the
total energy which is now mechanical plus thermal energy remains
constant.
The general principle of the conservation of energy states that the total
energy in a given system is always constant or energy can neither be
created nor destroyed although it can be transformed from one form to
another.
3.5.3 The Conservation of Mechanical Energy of a Falling
Body
A
B
h
l
C
Fig.1. 2: Conservation of mechanical energy
Here, let us consider a body of mass m at rest at a point A, of height h
above the ground level. As the body falls, its PE changes into KE.
What it loses in PE is equal to what it gains in KE
At A the total energy is all PE and is given by mgh. There is no KE
because the body is at rest.
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Total energy at A = mgh
If the ball falls to another point B at height l above the ground,
P.E at B = mgl
and
K.E at B = ½ mvB2
V B 2 = u2 + 2 gS
= 0 + 2 g (h-l)
∴ K.E = ½ mv B 2 = ½ m x 2 g (h-l) = mg (h-l)
Total energy, P.E + K.E =mgl + mg (h-l)
= mgh
At C, the lowest level, h = 0
P.E = mg x 0 = 0
K.E = ½ mv C 2
Usually v C 2 =u2 + 2 bs, we have
V2c = 0 + 2 gh
K.E = ½ m x 2 gh = mgh
Total energy = P.E + K.E = mgh
As you have seen, the total energy of the body at each point A, B, or C
is a constant (mgh) although the energy is all potential at A, partly
potential and partly kinetic at B and all kinetic at C. The total
mechanical energy is therefore constant or conserved in a conservative
field.
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3.5.4
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The Conservation of Mechanical Energy Involved in a
Swinging Pendulum
A
K.E = O
P.E. = mgh
B
h
D
C
P.E. = O
K.E = ½ mv2
(K.E. is max. at C )
K.E = O
(P.E. = mgh
P.E is max.)
Fig. 1.3: Energy transformation in the motion of a simple pendulum
Now we will consider energy transformation in the motion of a
swinging pendulum. A simple pendulum consists of a bob attached to a
string the other end of which is suspended from a firm, a support. When
the pendulum swings from end to end, the energy of the system changes
from potential to kinetic and vice versa. At each stage of the swing, the
total energy remains constant.
The pendulum swings from the highest point A through the centre of the
swing C to the other highest point B,
At C, the bob is at the lowest position. Therefore, the P.E of the system
is zero. Also at C, the speed of the pendulum is maximum and so the
K.E is maximum at this point. That is,
At C,
P.E = 0
K.E = ½ mv2
(maximum).
As the bob moves from C to B, the K.E at C is gradually transformed to
P.E with the P.E becoming maximum at B which is at a height h above
C.
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At the B, the total energy is potential energy and is equal to mgh. This
is also the P.E at A which is at the same height above C as B. That is,
At B and A,
P.E = mgh (max)
K.E = 0
You can find the maximum velocity of the swinging bob that occurs as
the bob passes through C. If you take this Velocity to the v max, since
total energy is always conserved, you will have:
½ mv2 max = mgh
(K.E at C) = (P.E at A or B)
Therefore,
V2 max = 2 gh
V max = √ 2 gh ..............................................… (8)
You have seen that at A and B, the energy is all P.E. At C, energy is all
K.E. At any intermediate point such as D, the energy is partly potential
and partly kinetic. At each point however, the total energy (P.E + K.E)
is constant and is equal to mgh.
Here also, you have seen that the total energy of the body remains
constant. Although there is transformation of energy from P.E to K.E,
but, there is no loss or gain in its total energy. This confirms the law of
conservation of energy.
However, you will observe from experience that a pendulum bob left
swinging on its own will eventually come to rest after some time. This
is because its energy is gradually used up in moving the air around it.
That is, in doing work against air resistance.
Worked example
A stone of mass 500 g is thrown vertically upwards with a velocity of 30
ms-1
Find (a) the potential energy possessed by the stone at its greatest
height; (b) the kinetic energy of the stone just before it hits the ground.
(g = 9.8 ms-2)
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Solution
(a)
Using v2 = u2 + 2 gS
where v = 0 ms-1 (stone momentarily at rest).
g = - g (because it acts in opposite direction to that of u)
When you substitute for v and g in the equation above, you
obtain
0 = u2= + 2( - g) S
0 = u2 – 2g S
2g S = u2
S = u2
2g
U = 30 ms-1, g = 9.8 ms-2
and
S = (30 ms-1)2
2 x 9.8 ms2
=====
= 45.92 m
This is the greatest height reached by the stone. Therefore, the potential
energy of the stone at a height, h, of 45.92 m is given by
Potential energy = mgh
= 0.5 kg x 9.8 ms-2 x 45.92 m
= 225 J
====
The potential energy possessed by the stone at its greatest height is 225J
The kinetic energy of the stone just before it hits the ground equals the
potential energy possessed by the stone at its greatest height i.e. 225 J.
This is also the potential energy of the stone in falling through a height
45.92 m to the ground.
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3.6
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Pendulum
3.6.1 Applications of Mechanical Energy
Mechanical energy is applied in the working principles of machines
where forces move over distances. You learned about this extensively
in Module 2 Unit 4.
SELF ASSESSMENT EXERCISE 1
1
2
Define the terms work, energy and power.
a.
b.
State the principle of the conservation of mechanical
energy.
How is the law illustrated by
(i)
(ii)
3.
A cable car is pulled up a slope by a constant force of 5000 N at a
uniform speed of 6 m/s. It takes 4 minutes to complete the
journey.
(a)
(b)
(c)
4.0
the swing of a simple pendulum?
the energy of a falling orange?
How much work is done in getting the car to the top of the
slope?
How much work could be done if the speed were 12 m/s
(the force remaining the same)?
How does the power developed in (a) compare with (b)?
CONCLUSION
In this unit you learned about work, energy and power.
learned in depth mechanical energy and its conservation.
5.0
You also
SUMMARY
You have learned in this unit that
•
•
•
Work is Force (F) x Displacement(S) in the direction of force. If
F and S are inclined at an angle θ, then work done (W) = Fs Cos
θ. It is measured in joules.
Energy is capacity to do work. It is measured in joules.
Mechanical energy can be classified as kinetic energy and
potential energy
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•
•
•
•
•
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Potential Energy is the energy due to position. P.E = mgh;
Kinetic energy is the energy due to motion. K.E = ½ mv2
Power is the rate at which work is done. It is also the rate at
which energy is expended. It is measured in watt.
The total energy in the universe is constant.
The principle of the conservation of mechanical energy states that
in a closed or isolated system, the total mechanical energy is
always constant, although energy may change from P.E to K.E or
vice versa.
6.0
TUTOR-MARKED ASSIGNMENT
1.
What kind of energy is possessed by:
(a)
(b)
(c)
(d)
(e)
2
(a)
the wheel of a moving motor car?
a man standing on the roof of a car?
cooking gas?
a ball rolling down a slope?
Music
In every day life we say that you have great power if you
have great strength or much authority such as a senate
president. Is the scientific meaning of power the same as
its everyday meaning? Give reasons for your answer.
(b) Friction is often used to slow down objects. Is friction used
for this purpose doing work? Give a reason for your answer.
3.
A boy weighing 200 N climbs a staircase that is 20 m long and
reaches a height of 8 m. How much work has he done to raise his
height. What happens to his kinetic energy at the end of the
climb? (Take g = 9.8 ms-2).
4.
5
If you weigh 75 kg and in 5s you can run up a flight of stairs
consisting of 36 steps, each 15 cm high, what is your power?
(Take g = 9.8 ms-2)
(a)
State the Principle of Conservation of Mechanical Energy
(b)
On what factors do the measurement of
(i)
kinetic energy
(ii)
Potential energy of a body depend.
7.0
REFERENCES/FURTHER READINGS
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Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools. Onitsha: Africana FEP..
Nelkon, M. (1986). Principles of Physics for Senior secondary Schools.
England: Longman.
Ravi, K.; George, K.O., Hui, T.C. (1987). New School Physics
Certificate Science Series. Singapore: FEB International Private.
Ewelukwa, G.O., Odunze, G.A., Anozie, A.C. (1995). Revision
Integrated Science for JSC Examinations Onitsha: Africana FEP.
CESAC (1971). Nigerian Secondary School Science Project Physics
Textbook as Teachers Guide.
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UNIT 2
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ELASTIC PROPERTIES OF SOLIDS
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Concept of Elasticity
3.1.1 Hook’s Law
3.1.2 Statement of Hook’s Law
3.2
Verification and Demonstration of Hook’s Law
3.3
Graph of Extension against Load for a Permanently
Strained Elastic Material
3.4
Young’s Modulus of Elasticity
3.5
Work Done in Spring and Elastic String
3.6
Elastic Potential Energy
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In Module 1 unit 2, we studied the particle nature of matter and stated
that particles that make solids are closely packed together and held in
relatively fixed positions by strong intermolecular forces. As a result of
these forces, it is not easy to dislodge the molecule of a solid because
any external force applied on the solid is stiffly resisted by the
intermolecular forces. In this unit, we shall study the behaviour of those
solids which change their shape or size when a sufficiently strong
external force is applied to them and return to their original shape or size
when the applied force is removed. Specifically we shall study Hooke’s
law which governs the relationship between the stretching force and the
extension produced by it in an elastic material. We shall also discuss
Young’s modulus of elasticity and study the work done in stretching or
compressing spring and elastic strings.
2.0
OBJECTIVES
By the end of this unit you will be able to:
•
•
•
228
explain the elastic properties of substances
state Hooke’s law and explain the meaning of the terms such as
stress, strain, elastic limit, young’s modulus
describe and carry out an experiment to verify Hooke’s law
PHY 001
•
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draw a graph of load versus extension for an elastic material
 before the elastic limit is reached, and
 beyond the elastic limit
•
solve simple problems on Hooke’s law and on work done in
stretching or compressing a spring and elastic string.
3.0
MAIN CONTENT
3.1
Concept of Elasticity
Solids tend to change their shape or size when sufficiently strong
external forces are applied to them and to return to their original shape
or size after the forces causing the change are removed. Solids which
retain their shape or size after the force causing the change has been
removed are said to be ‘elastic’, and this property of solids is called
elasticity.
Definitions
(1)
Elasticity is the ability of a substance to regain its original shape
and size after being distorted by an external force.
(ii)
An elastic material is one that regains its original shape and size
after the distorting external force has been removed.
3.1.1 Hooke’s Law
3.1.2 Statement of Hooke’s Law
Hooke’s law states that, provided the elastic limit of an elastic material
is not exceeded, the extension, e, of the material is directly proportional
to the applied force, F.
Mathematically,
i.e
Fαe
F = K e …………………………
(1)
where k is the constant of proportionality called elastic constant or force
constant or stiffness of the material.
From (1),
K = F …………………………………..
e
(2)
If F is in newtons and e in metres, then K is in newtons per metre (Nm-1)
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Definition:
Elastic constant or stiffness of an elastic material is the force required to
produce unit extension of the material.
The working of the spring balance is based on Hooke’s law. In this case
F = mg, the weight of the body which is proportional to e, the extension
of the spring.
3.2
Verification or Demonstration of Hooke’s Law
Extension
L
O
Figure 2.1a
Load
Fig. 2.1b: Graph of extension against load.
Hook’s law
Fig.2.1: Verification or demonstration of Hook’s law
You can verify or demonstrate Hooke’s law using experimental set up as
shown in Fig.2.1 as follows:
(i)
Obtain a spiral spring and attach one end of it to a firm support.
To the other end attach a scale pan which has been previously
weighed.
(ii)
Attach a pointer, P, to the spring with plasticize and arrange
vertical scale over which the pointer movement can be read. Set
up the whole apparatus as shown in the Fig. 2.1.a
(iii)
Note the initial pointer reading on the scale and call it ℓ.
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(iv)
Add weights to the scale pan in equal steps of about 0.05 kg and
note and record the pointer reading for each increasing load. Note
that the total load stretching the spring in each instance is the
weight in the scale pan plus the weight of the scale pan.
(v)
Remove the load from the scale pan in equal steps also and note
the record the corresponding pointer readings.
(vi)
Record your readings as shown in the table below:
Load
(vii)
Scale reading
Load
increasing
Load
decreasing
Average
reading
Extension
Plot a graph of extension against load. If the total maximum load
has not extended the spring permanently, the graph obtained will
be a straight line graph passing through the origin (Fig.2.1b).
Since the graph is a straight line, it means that the extension of
the spring is directly proportional to the load. Hence, Hook’s law
is verified.
SELF ASSESSMENT EXERCISE 1
1
Explain what is meant by (a) elasticity, and (b) elastic material?
2
State Hooke’s law and show how the law leads to the idea of
stiffness of the elastic material. State the unit (S.I.) in which
stiffness is measured, showing clearly how this unit may be
derived.
3
Describe an experiment to verify Hooke’s law using a graphical
method.
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4
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Suppose you performed Hooke’s law experiment and obtained a
set of readings of loads and their corresponding extensions.
Sketch the graph of;
(a)
(b)
Extension against load,
Load against extension.
State two important similarities and one difference between the two
graphs.
3.3
Graph of Extension against Load for a Permanently
Strained Elastic Material
Extension
Load
(a)
B
P•
P
•
•Y
•E
E
•
•Y
O
O
Load
Extension
P = Proportionality Limit
E = Elastic Limit
Y = Yield point
OE = Elastic Region
EB = Plastic Region
Fig. 2.2:
Graphs of (a) extension against load (b) load against
extension up to breaking point
If, in the Hooke’s law experiment, you continue to increase the load
until the spring wire snaps or breaks, the graph of extension against load
will be as shown in figure 2.2 above. The graph is a straight line up to
OP, followed by the curved part PEYB which rises slowly initially and
then sharply.
Along OP, the straight portion of the graph, Hooke’s law is obeyed and
the extension is proportional to the load.
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P is called the ‘proportionality limit’. However, up to the point E
(slightly above P) which is called the ‘elastic limit’. The spring will
return to its original length if the applied force is removed.
Definition
The elastic limit of an elastic material is the limit of force beyond which
the stretched material does not return to its original length when the
stretching force is removed.
Beyond the elastic limit, E, the material loses its elasticity. The region
OE from the origin to the elastic limit E is called the ‘elastic region’.
The difference between the ‘proportionality limit’ P and the ‘elastic
limit’ E is that, strictly speaking, Hooke’s law holds up to P, but
elasticity holds up to E.
Beyond E, the wire stretches rapidly with increasing load and then
enters the ‘elastic region’ which begins at E. In this ‘plastic region’ the
wire does not return to its original length upon removal of the stretching
force. At Y, the ‘yield point’ the material is said to have ‘yielded or lost
all elasticity permanently and has become ‘plastic’.
Definition
Yield point is the point beyond the elastic limit in which the elastic
material has yielded all its elasticity permanently and has become
plastic.
The difference between the “elastic region” (OE) and the “plastic
region” (EB) is that in a latter, a sudden and rapid increase in the
extension for any slight increase in load. B is the ‘breaking point’ where
the material may finally ‘break’ or ‘snap’. It corresponds to the
maximum possible extension that can be produced in the material.
Worked examples
1.
A force of 10N is required to extend the length of a spiral spring
by 4 cm. Find the extension of the spring when a mass of 360 g is
suspended from its end (g = 10ms-2).
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Solution
By Hook’s law, F = ke
∴ k= F
e
=
10N
0.04 m
when a body is suspended, the weight w should produce an extension, e
∴ W = mg = k x e
∴ e = mg
k
2
=
360kg x 10N
1000
kg
=
1000 N
4 m
=
=
1.44 x 10-2 m
1.44 cm.
An elastic material is stretched 45 mm by a force of 3 N.
Calculate the additional force which will stretch the material
50mm, if the elastic limit is not exceeded.
Solution
By Hook’s law,
F = k.e
∴ k=F = 3N
e 45 x 10-3m
Let the force stretching the material 50 mm be F
∴ F = k.e
=
k x 50 x 10-3m
=
∴
234
3N
45 x 10-3m
Additional force
x 50 x 10-3 m
1
=
3.33N
=
=
(3.33 – 3.00) N
0.33 N
PHY 001
3.4
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Young’s Modulus of Elasticity
Suppose a wire of length ℓ (m) and cross-sectional area A (m2) is
extended through e (m) by a force F (N).
(a) The ratio of the force to the area, F/A is called the stress or ‘tensile’
of the elastic material.
Stress =
F
…………………………(1)
A
(b)
The ratio of the extension, e to the original length, ℓ of the wire
i.e e is called the tensile strain of the wire.
ℓ
∴ Strain = e …………………………………… (2)
ℓ
From (1)
F = stress x A ……………………………….(3)
From (2)
e = Strain x ℓ…………………………………(4)
By Hooke’s law, F = ke
∴ Stress x A = k x Strain x ℓ
∴ Stress = kℓ x strain
A
∴ Stress = k1 strain where k1 =
∴
Stress
Strain
constant = kℓ
A
1
=
k …………………………………… (5)
OR
Stress α
Strain
Hence Hooke’s law can also be stated as follows:
The tensile stress of the material is directly proportional to its tensile
strain provided the elastic limit is not exceeded.
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The constant of proportionality, k1 (see equation (5) above) is called
Young’s modulus of elasticity and is represented by the symbol, ૪.
∴ Young’s modulus (૪) = Stress
Strain
∴
૪
=
F A
e
ℓ
The unit of ૪ is Nm (Newton per square metre) the same unit as
-2
stress, since strain has no unit.
Dimension of
૪
=
Dimension of stress
Dimension of strain
236
=
Dimension of stress
=
Dimension of force
Dimension of Area
=
MLT-2
L2
=
ML-1T-2
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3.5
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Work Done in Springs and Elastic String
Natural
ι length
ι+e
Compression - e
p
e P1
load
m
F = mg (N)
Fig. 2.3: Work done in extending or compressing a spring
Consider a spring of unstretched length, ℓ (m). Suppose a force F
(N) causes an extension (or compression) e (m). The force in the
wire has increased from an initial value of zero at P to a value of
F at ‘P’. The work (W) done in stretching the spring through the
extension is given by:
W
=
Average force
=
Initial force + final force x extension
2
∴W =
But
x extension
O+Fxe
2
∴W =
1/2 Fe ………………………………… (1)
F
ke
=
∴W =
(Hooke’s law)
1/2 k.e2 ……………………………… (2)
This work done is stored as potential energy in the material which can
do work when the stretching force is removed and the material regains
its original length.
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This type of stored energy is applied in the working of many spring
clocks and toys. The energy is stored in the clock spring (or toy spring)
which is compressed by winding.
3.6
Elastic Potential Energy
Definition
The elastic potential energy of a stretched or compressed material is the
ability of the material to do work.
Elastic potential energy arises because work is done in stretching or
compressing the material as we have seen in the previous section
W
=
1/2 Fe
=
1/2 ke2
Where F is the maximum stretching (or compressing) force, e is the
extension (or compression) and k is the force constant or stiffness of the
material.
Example (or application) of elastic potential energy:
When you stretch the rubber of a catapult and project a stone, the elastic
potential energy stored in the rubber is converted into the kinetic energy
of the flying stone according to the law of conservation of energy.
Worked examples
1.
An elastic material is stretch 1.2 cm by a mass of 96g suspended
from one end of it. What is the elastic constant of the material?
Solution
Weight of object
Extension
F
238
=
stretching force
=
960 x 10 N
1000
=
9.6 N
=
1.2 10-2 m
=
=
ke
k x 1.2 x 10-2 N
(W = mg)
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∴
2.
k
=
9.6
Nm-1
1.2 x10-2
=
800 Nm-1
A spiral spring is compressed by an external force in such a way
that 0.08J of energy is stored in it. If the elastic constant of the
spring is 400Nm-1, calculate the compression of the spring.
Solution
ke
Energy stored in the spring, W
=
Fe
=
W
=
1 k e2
2
1 x 400N x e2
2
1/2
2
∴
0.08J =
∴
e2
=
0.08 x 2 m2
400
∴
e
=
0.02m
SELF ASSESSMENT EXERCISE 2
1.
Explain the terms (i)elasticity, (ii) elastic limit of a material.
A mass of 300 g is suspended from the lower end of an elastic
material. If the material extends by 6.0 cm, find the force needed
to produce an extension of 1.0 cm.
2.
(a
(b
(c
3.
(a
(b
Define elasticity and state two reasons why this property is
useful in daily life.
Describe an experiment to illustrate the behaviour of an
elastic wire which is steadily loaded and forced to extend
its length beyond its elastic limit.
Sketch a graph of the relation between the extension of the
wire and the load attached to it gradually up to and beyond
the elastic limit.
States Hook’s law and mention the condition under which
it is obeyed.
Describe an experiment to test its validity in the laboratory
for a spiral spring.
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PHY 001
(c
(d
4.
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Explain briefly how the working of a spring balance
depends on Hooke’s law.
A helical spring is extended through 3.0 cm by a force of
40 N. Find (a)
the force which will stretch it through
5.0 cm,
(b)
the length through which a force of 100N will
extend it.
Defined the terms strain, stress and Young’s modulus.
One end of the steel wire of length 10.0 m and uniform crosssectional area 1.0 x 10-6m fixed to a rigid support. A force of
100N applied to the other end of the wire produces in it an
extension of 1.0 x 10-2 m. If Hooke’s law is obeyed, calculate (i)
the stress, (ii) the strain, (iii) the young’s Modulus for the metal.
4.0
CONCLUSION
In this unit you learned about elastic properties of solids. Specifically,
you learned about Hooke’s law, Young’ modulus of elasticity and their
definitions. You also learned about work done in spring/ string and
elastic potential energy
5.0
SUMMARY
1.
‘Elasticity’ is the property of a material which enables it to regain
its original shape and size after the force causing the deformation
has been removed.
2.
‘Hook’s law’ states that the extension of an elastic material is
directly proportional to the load or applied force causing the
extension, provided the elastic limit of the material is not
exceeded.
3.
‘Elastic limit’ is the maximum stretching force beyond which a
stretched elastic material would not return to its original length
when the force is removed.
4.
‘Yield point’ is the point at which the elastic material loses its
elasticity permanently and becomes plastic.
5.
Work done in stretching or compressing a material is the same as
the energy stored in the material or its elastic potential energy and
is given by;
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W
=
=
Also W
=
1 Fe
2
1 (maximum force) x extension
2
1 ke2, since F = ke, from Hook’s law
2
‘Young’s modulus’ Y
=
=
F A Nm-1
e ℓ
6.0
TUTOR-MARKED ASSIGNMENT
1.
(a
(b
2.
(a)
(b)
Stress
Strain
State Hook’s law and explain what is meant by elastic
limit.
A load of 2N produces an extension of 3.0 cm in a spiral
spring. Find the additional force which will stretch the
material by 3.5 cm.
State the formula for the energy stored in an elastic
material in terms of; the force producing an extension in
the material.
An electric material is compressed by 2.0 cm. If the elastic
constant of the material is 400 Nm-1, determine the elastic
potential energy stored in the material.
3.
A stone of mass 5 g is projected with a rubber catapult. If the
catapult is stretched through a distance of 7 cm by an average
force of 70 N, calculate the instantaneous velocity of the stone.
7.0
REFERENCES/FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools Onisha: Africana FEP Publishers Limited.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics
SSCE Edition, Ibadan: University Press.
Ndupu, B. L. N and Okeke, P. N. (2000). Round-up Physics for West
African Senior School Certificate Examination: A Complete
guide. Lagos. Longman Nigeria Plc.
Ndupu, B. L. N and Okeke, P. N. and Ladipo, O. A. (2000). Senior
Secondary Physics Book 1. Lagos: Longman Nigeria Plc.
241
PHY 001
UNIT 3
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ELASTIC PROPERTIES OF SOLIDS
CONTENTS
1.0
2.0
3.0
Introduction
Objectives
Main Content
3.1
Concept of Heat and Temperature
3.2. Effects of Heat
3.3
Kinetic Molecular Theory Statement of Hooke’s Law
3.3.1 Kinetic Theory Explanation of Temperature.
3.3.2 Kinetic Theory Explanation of Expansion.
3.3.3 Kinetic Molecular Theory Explanation of Change
of State
3.4
Expansivity
3.4.1 Linear Expansivity (α)
3.4.2 Determination of Linear Expansivity
3.4.3 Area Expansivity (β)
3.4.4 Volume or Cubic Expansivity (૪)
4.0
5.0
6.0
7.0
3.4.5 The Relationship between Linear Expansivity, Area
Expansivity and Cubic Expansivity
3.5
Some Applications of Expansion
3.6
Expansion of Liquids
3.6.1 Real and Apparent Expansivity
3.6.2 The Anomalous Expansion of Water
3.6.3 The Importance of Anomalous Expansion of Water
in Nature
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
This unit will introduce you to the concepts of heat, temperature and
molecular theory. You will learn the effects of heat and change of state
and expansion, their explanation using the kinetic molecular theory.
Also, you will learn expansivity and explanation of temperature changes
using molecular theory. Finally, you will learn the applications of
expansivity in your daily life.
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2.0
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OBJECTIVES
By the time you finish this unit, you will be able to:
• explain temperature, expansion, change of state and vaporization
using the kinetic molecular theory
• solve simple problems involving linear, area and volume
expansivity.
3.0
MAIN CONTENT
3.1
Concept of Heat and Temperature
When you touch a kettle of boiling water, you will feel a burning
sensation. You will say that the water is hot. Also, if you hold ice block,
your hand feels chilly and you conclude that the water is cold. We
describe an object which has been given heat energy as hot. When we
place a kettle of cold water on a heating stove, after a few minutes, the
water feels hotter because it has been given heat energy. Similarly, we
describe as cold a body from which heat energy has been removed. For
instance, keeping a bottle of water in a refrigerator, you must have
experienced that heat always flows from a hot object to a cold object.
Heat energy is the energy that is transferred from a hot object to a cooler
object as a result of their difference in temperature.
Temperature is the degree of hotness or coldness of an object.
Temperature is a property of an object which decides which way heat
will flow when it is placed in contact with another object. Usually heat
flows from a body of higher temperature to one at lower temperature.
The instrument for measuring temperature is the thermometer.
3.2
Effects of Heat
Here we look at some effects of heat on an object. You may have
identified some of this yourself in your everyday life. These effects are
summarized as follows:
Addition of heat to the body will cause:
(a)
(b)
(c)
Change in temperature of the body except during a change of
state.
Change of state of the body solid to liquid, liquid to gaseous
state.
Expansion of the body.
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PHY 001
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(f)
(g)
Change in the physical property of a body such as the electrical
resistance, magnetic properties conductivity, elasticity, density
and colour of the body.
Themionic emission- the emission of electrons from the surface
of the metal.
Change in chemical properties of the body.
Changes in pressure and volume of gases.
3.3
Kinetic Molecular Theory
(e)
We consider kinetic molecular theory and its assumptions. In addition,
we use it to explain major concepts in this unit. The molecular theory of
matter assumes the matter is made up of atoms that aggregate in
molecules.
The molecule is a group of atoms of the same or different elements
joined together in a simple proportion. These molecules are under the
influence of two types of forces:
(i)
(ii)
Attractive forces which prevent the molecule from moving apart
and
repulsive forces which prevent the molecule from moving closer.
There is normally a balance between these forces in a substance.
In solid substances, the attractive forces between molecules are so
strong that the molecules do not move about freely. The molecules only
vibrate about their mean positions maintaining a fixed shape and
volume.
In liquids, the molecules are loosely held together by weak attractive
forces. The molecules are free to move about within the liquid and are
always in a state of random motion. Molecules of solids and liquids are
held together by intermolecular forces.
In gases, the force of attraction between the molecules is very weak. The
molecules are therefore in constant motion having overcome the
intermolecular force. They move very freely at a very high speed. They
are always in the state of random motion and take up the shape and
volume of their container.
The kinetic molecular theory assumes that:
(i)
244
Every substance is made up of tiny particles called molecules
PHY 001
(ii)
(iii)
(iv)
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The molecules are in a constant state of random motion, colliding
elastically with one another and changing their direction as a
result.
There is always an attractive force between the molecules.
The volume of the gas molecule is negligible compared with the
volumes of the gas container.
3.3.1 Kinetic Theory Explanation of Temperature
We say that the molecules of a substance are in constant motion,
therefore, they possess kinetic energy.
The temperature of the body is a measure of the average kinetic energy
of its molecules.
When you add heat to a substance, the motion of the molecules
increases resulting in an increase in the average kinetic energy of the
molecules of the substance, thus, an increase in its temperature. On the
other hand, when you remove heat from a body, the motion of its
molecule decreases resulting in reduction in their average kinetic energy
and a decrease in temperature.
It follows therefore, that the temperature of a body is related to the
average kinetic energy of its molecules.
3.3.2 Kinetic Theory Explanation of Expansion
When we heat a substance, it increases in size, we say it expands. When
the substance cools, it contracts i.e reduces in size.
According to the kinetic molecular theory, when an object is heated the
molecules acquire more kinetic energy which enables them to overcome
their intermolecular forces resulting in increase in vibration of the
molecules and their displacement about their mean positions. The
average distance between the molecules of the substance becomes larger
leading to an increase in the size of the substance. The increase depends
on the strength of the intermolecular forces. If these forces are strong,
the expansion will be small and vice versa.
The intermolecular forces are stronger in solids than in liquids and
weakest in gases. Therefore, when heat is applied, gases expand more
than liquids and liquids expand more than solids.
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3.3.3 Kinetic Molecular Theory Explanation of Change of
State
You are aware that substances exist in any of the three states of matter;
solid, liquid and gas. With the help of heat energy, it is possible to
transform a substance from one state to another. Change of state refers
to the process of transforming a substance from one state of matter to
another.
When a solid is heated its temperature increases until it reaches a certain
maximum temperature at which the molecules acquire maximum kinetic
energy. At this maximum temperature, further heating does not increase
the kinetic energy of the molecules but the heat energy (latent heat) is
used to break down the intermolecular forces binding the molecules of
the solids in a regular pattern. The molecules then melt and move about
freely as they are in the liquid state. The maximum temperature is the
melting point of the solid. Also, the change from liquid to gas takes
place in a similar fashion. The liquid temperature increases as it is
heated and reaches its maximum temperature at which the molecule
acquires maximum kinetic energy.
Further heating (latent heat) supplies the needed energy to overcome the
forces of attraction between the liquid molecules and changes the liquid
to vapour without temperature change. The molecules are then
practically independent of each other and exist as a gas.
The energy needed in this case is much greater than from solid to liquid
and the latent heat is much greater. Removal of heat energy from a
substance results in a reverse process.
The heat required to break down the intermolecular forces of attraction
in a solid is the latent heat of fusion.
The latent heat of vaporization is the heat needed to overcome the forces
of attraction between the molecules of the liquid.
3.4
Expansivity
Solids expand when heated and contract when cooled. You can observe
this in simple laboratory experiments such as:
•
246
Ball and ring experiment:
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When cold, a metal ball can just pass through a metal ring. On being
heated strongly for some minutes, it expands and can no longer pass
through the ring. When it cools, it contracts and passes through the ring
again.
(a) Ball passes through the ring (before heating) (b) Ball does not pass
through the ring
(after heating)
Fig. 3.1
•
Bar and guage experiment
•
•
Bar
Bar
Guage
•
•
guage
Before heating
Fig. 3.2:
After heating
Bar and guage
The bar in Fig. 3.2 just fits into the guage at ordinary room temperature.
After heating, the heated bar expands in length when hot and can no
longer fit into the guage. When cooled to the room temperature, the bar
contracts and fits once more into the guage.
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•
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Bimetallic strip;
Different substances expand by different amounts for the same
temperature range. You can observe this in the heating of bimetallic
strip shown in Fig. 3.3.
Iron
Wooden handle
brass
(a)
Before heating
Fig.3.3:
Expansion of a
solid (bimetallic strip)
Before heating the bimetallic strip of brass and iron riveted together has
equal length. After heating, the strip bends as shown above because
brass expands more than iron for a given temperature rise.
3.4.1 Linear Expansivity
You can determine the extent which a unit metal substance changes in
length when it temperature changes by one degree. This quantity is
known as linear expansivity of the metal substance. It is represented by
the symbol α (pronounced alpha).
•
The linear expansivity (α) of a substance is defined as the increase in
length per unit length per degree rise in temperature.
In symbols, this is equivalent to:
α
(i)
=
ℓ2 - ℓ 1
ℓ1 (θ2-θ1)
α
ℓ1
ℓ2
=
=
=
linear expansivity
length of metal at temperature θ1
length of metal at temperature θ2
=
e
ℓ1 θ
……………..
Where:
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θ
e
α
=
=
=
temperature rise given by θ2 - θ1
expansion or increase in length ℓ2- ℓ1
Increase in length
original length x temperature rise
Unit per oC or per k (k-1)
•
The statement that the linear expansivity of brass is 0.000018 k-1 or
0.000018/oC means that a unit length of brass expands by 0.00018
units when it is heated through 1k (or 1oC) rise in temperature.
3.4.2 Determination of Linear Expansivity
Fig. 3.4:
Measurement of linear expansivity
The apparatus shown in Fig. 3.4 is used in the determination of linear
expansivity of a solid.
Carefully measure the original length ℓ1 of a metal rod (about 50 cm)
with a metre rule. Enclose the rod in a fixed jacket between a fixed stop
A and a micrometer guage B, so that no movement of the rod can take
place at A while the other end at B is free to move. Run cold water
through the jacket. Adjust the micrometer screw until it just touches the
B end of the rod. Record the guage’s first reading (e1) and the initial
temperature ( θ1) of he rod from the thermometer placed inside the
jacket.
Unscrew the micrometer screw guage from the end B of the rod to leave
enough space for the rod to expand. Pass hot steam through the jacket
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for several minutes to allow the rod expand as a result of this heating.
Screw the guage to make contact with the rod again and take a new
reading of the guage (e2). Repeat the process until the reading of the
screw guage becomes constant. Take the final steady temperature (θ2) of
the rod from the thermometer inserted into the steam jacket.
Calculate the linear expansivity from the equation;
Linear expansivity (α)
=
α
expansion (or change in length)
Original length x temperature change
=
e2 – e
ℓ1(θ2 - θ1)
=
ℓ1 θ
e
3.4.3 Area Expansivity (β)
When we heat a solid, it expands in all directions that is the length,
breath and height. This results in an increase in area as well as in
volume of the solid. This increase in area when the body is heated is
known as area or superficial expansion. It is represented by the symbol β
(pronounced beta).
Definition
The area or superficial expansivity, β, of a solid is the increase in area
per unit area per degree Kelvin increase in temperature or the fractional
increase in area per Kelvin rise in temperature.
Area expansivity β =
β
Where:
a
A2
A1
θ
change in area
.
Original area x temperature rise
=
A2 –A1
=
………………(2)
A1 x (θ2 -θ1)
a
A1 θ
=
=
=
=
increase in area A2 –A1
area at temperature θ2
area at temperature θ1
θ2 -θ1
3.4.4 Volume or Cubic Expansivity (૪)
An increase in volume when a body is heated is known as cubic or
volume expansion. We denote cubic expansivity with the symbol ૪
pronounced gamma.
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Definition
The volume or cubic expansivity ૪, is the increase in volume of a
substance per unit volume per Kelvin rise in temperature or the
fractional increase in volume per Kelvin rise in temperature.
Cubic expansivity (૪)
=
change in volume
Original volume x temperature rise
૪
=
V2 – V1
=
V1 (θ2 - θ1)
v
………… (3)
V1θ
Where
v
V2
V1
θ
=
=
=
=
increase in volume V2 – V1
volume at temperature θ2
volume at temperature θ1
θ2 - θ1
3.4.5 The Relationship between Linear Expansivity (α), Area
Expansivity (β) and Cubic Expansivity (૪)
If α is the coefficient of linear expansivity of a metal, its area
expansivity (β) and volume expansivity (૪) are related (α) by:
β
૪
=
=
2α
3α ………………………………. (4)
We can also write the following:
ℓ2
A2
V2
=
=
=
ℓ1 (1 + αθ)
A1 (1 + βθ) =
V1 (1 + ૪) =
(A1 + 2αθ)
(V1 + 3αθ)
where the symbols retain their previous meaning.
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Worked Examples on Expansivity
1
In an experiment to determine the linear expansivity of a solid
rod, the following readings were obtained:
Initial temperature of rod (θ1)
Original length of rod (ℓ1)
Final temperature of rod (θ2)
Final length of rod (ℓ2)
=
=
=
=
18oC
50.0 cm
98oC
50.8 cm
Calculate the (a) linear, (b) area and (c) volume expansivity of the rod.
Rise in temperature = 98oC – 18oC = 80oC
Expansion of rod = Increase in length = 50.8 cm – 50.0 cm = 0.8cm
Solution
(a)
linear expansivity
=
=
expansion
original length x temperature change
0.8
50 x 80
=
0.0002 per oC
(b)
area expansivity (β) =
2α
=
=
0.0002 x 2
0.0004 per oC
(c)
cubic expansivity (૪)
=
3α
=
3 x 0.0002
=
0.0006 per oC
2
The coefficient of linear expansivity of a metal is 1.8 x 10-5 cm
per degree Celsius. A cube of the metal has each length as 2.5 m.
Find the area of each surface and the volume of the cube if it is
heated through a temperature of 60oC
Solution
Coefficient of linear expansivity α = 1.8 x 10-5
Coefficient of area expansivity β = 2α = 2 x 1.8 x 10-5 = 3.6 x 10-5
Coefficient of volume expansivity ૪ = 3α = 3 x 1.8 x 10-5 = 5.4 x 10-5
Area of each surface A2
=
=
252
A1 (1 + 2 αθ)
6.25 (1 + 3.6 x 10-5 x 60)m
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=
6.26m2
The volume of the cube V2
3.5
=
V1 (1 + ૪)
=
=
15.63 (1 + 5.4 x 10-5 x 60)m
15.69 m3
Some Applications of Expansion
Here are some of the applications:
•
Gaps are left between rails in railway lines to allow for free
expansion and contraction of the rails. Without the gaps, the rail
joints will swell up on hot days, the railway line will buckle and
trains would be derailed.
• The end of the steel structure of long bridges rests on rollers which
allow for the expansion and contraction of the bridge without
weakening the structure.
• Telegraph wires are allowed to sag when fixed in the warm, raining
season, so that they do not snap when they contract in the cold
harmattan.
• A bimetallic strip is used in the thermostat, a device for maintaining
steady temperature. Thermostats are used in electric laundry irons,
immersion heaters for hot water tanks, refrigerators and
airconditioners.
• Balance wheel of watches are also made of bimetallic strips without
which increase of temperature will increase the diameter of the
balance wheel thereby weaking the elasticity of the hair spring and
causing the watch to lose time.
• A thick glass tumbler cracks when hot water is poured into it because
the inside of the tumbler expands more rapidly than the outside and
causes a strain in the glass. A type of glass known as pyrex is used
for making laboratory beakers and flasks to avoid the above effects.
Pyrex has low thermal expansivity.
• Other applications of expansion of solids is in:
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-
riveting
the fixing of metals tyres on the metal wheels in
locomotives;
in the expansion loops in steam pipes.
-
3.6
Expansion of Liquids
Liquids like solids expand on heating and contract on cooling. You can
observe liquid expansion with the following activity using the apparatus
shown in Fig. 3.5.
Capillary tube
Rubber bung
Flask
Coloured
water
Gauge
Heat
Fig. 3.5: Liquid expansion on heating
Fill the flask with coloured water to the brim. Fit the rubber bung with a
good length of the capillary tube projecting outside the flask as shown in
Fig. 3.5. Heat the flask gently with the bunsen burner and observe the
coloured liquid in the narrow tube. Remove the liquid and observe the
movement of the liquid level in the tube.
You will observe that the liquid level initially falls slightly and then
rises slowly and steadily as more heat is applied. This is because on the
application of the heat, the glass first expands faster than the liquid,
hence, the fall of the liquid level. As more heat is applied, the expansion
of the liquid increases more than that of the glass and the liquid rises in
the narrow tube. On removal of heat the liquid contracts and the level of
liquid in the tube is seen to fall.
By using different liquids, following the same process above but using
the apparatus shown in Fig. 3.6, you will also observe that liquids
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expands by different amounts when heated through the same
temperature ranges.
Fig. 3.6
: Different liquids expand differently
You will observe that in the three flasks, the level of liquids in the tube
first falls and then rises steadily. The level of alcohol is the highest
followed by benzene and least is water.
3.6.1 Real and Apparent Expansivity
In the last activity, we observe that when a liquid in a container is
heated, both the liquid and the container expand. The expansion of the
liquid as observed is the apparent expansion. This is always less than the
true or real expansion of the liquid. It is therefore necessarily to
distinguish between the real and the apparent cubic expansion of the
liquid.
Real expansion = Apparent expansion + expansion of the container
૪ real
=
૪
apparent
+
૪ container ......................... (5)
•
The real (or absolute) cubic expansivity (૪ r) of a liquid is the
increase in volume per unit volume per degree rise in temperature.
•
The apparent cubic expansivity (૪ a) of a liquid is the increase in
volume per degree rise in temperature when the liquid is heated in an
expansible vessel.
The apparent expansivity of a liquid is obtained by weighing a density
bottle empty and then fills with the liquid. The bottle with the contents is
then heated in a warm water bath. As the temperature increases, the
glass and the liquid expand and the liquid overflows. The bottle is then
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taken out of the water bath, cleaned and reweighed to determine the
mass of liquid expelled. The apparent expansivity is calculated from:
૪ apparent =
mass of liquid expelled
……….(6)
mass of liquid remaining x rise in temperature
3.6.2 The Anomalous Expansion of Water
We know from the foregoing that most liquids expand when heated and
contract when cooled. Water is an exception to this rule as it behaves in
an anomalous way when cooled. It contracts on cooling from a higher
temperature until the temperature falls to 4oC. At 4oC, it starts to expand
if cooled further below this temperature to 0oC. That is water expands
when cooled from 4oC to 0oC. As a result of this, a given mass of water
has its least volume and its highest density at 4 oC. This anomalous
behaviour of water is illustrated in an experiment known as Hope’s
experiment.
3.6.3 The Importance of Anomalous Expansion of Water in
Nature
The anomalous expansion of water is important to us because it makes
ponds, lakes and rivers to freeze from the top surface rather than from
the bottom. As a result, Marine lives can survive during winter since ice
forms at the surface of the water while the bottom of the lake remains at
4oC a temperature warm enough for aquatic animals such as fishes.
SELF ASSESSMENT EXERCISE 1
1
Which gives you greater burn, water at 100oC or steam at 100oC?
Why?
2
Two identical cans are placed in contact, one containing hot
water and the other cold water. Why is it that the temperature of
the cold water rises and that of the hot water falls and not the
other way round?
3
(a)
What is meant by the statement, the linear expansivity of a
solid is 1.0 x 10-5 K-1.
3
(b)
i
ii
256
Describe an experiment to determine the linear
expansivity of a steel rod.
Steel bars each of length 3 cm at 29oC are to be
used for constructing the rail line. If the linear
expansivity of the steel is 1.0 x 10-5K-1, Calculate
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the safety gap that must be left between successive
bars if the highest temperature expected is 41oC.
(c)
State three advantages and two disadvantages of thermal
expansion of solids.
4
Distinguish between real and apparent cubic expansivity of a
liquid.
5
A glass bottle full of mercury has mass 500 g. On being heated
through 35oC, 2.43 g of mercury are expelled. Calculate the mass
of mercury remaining in the bottle. Cubic expansivity of mercury
is 1.8 x 10 -4 K-1, linear expansivity of glass is 8.0 x 10-6 K-1.
4.0
CONCLUSION
In this unit, you have learned a number of important concepts which
relate to the effects of heat and their explanation using kinetic molecular
theory. You also learned linear, area and volume expansivity of a
substance and the applications of expansion. In addition, you learned
about liquid expansion and the importance of anomalous expansion of
water.
5.0
SUMMARY
• All substances are made of molecules. When a substance is heated, it
molecules begin to move faster.
• Heat is a form of energy called thermal energy. It is the energy
communicated from one body at higher temperature to another at
lower temperature.
• Temperature is the degree of hotness or coldness of a body.
• According to kinetic molecular theory, the temperature of a body
depends on the average kinetic energy of its molecules
•
The linear expansivity α of a substance is the fractional increase in
its length per degree change in temperature.
•
Area or superficial expansivity, β is the fractional increase in area
per unit change in temperature.
•
The volume or cubic expansivity ૪ is the fractional increase in
volume per degree change in temperature.
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•
β =
•
Real volume expansivity
=
apparent volume expansivity +
volume expansivity of container; ૪ r
=૪a + ૪
•
Water unlike other liquids, contract when it is heated from 0oC to
4oC. Water has a maximum density at 4oC.
•
Advantages and disadvantages of expansion.
•
The advantages of expansion are utilized in fitting of wheels on
metal rims, fire alarm, thermostat, bimetallic thermometer,
compensated pendulum.
2α;
=
3α
• The disadvantages of expansion are the sagging of over head cable,
the buckling of railway lines, in the gaining or losing of time as a
result of expansion of balance wheels of watches and clocks.
6.0
TUTOR -MARKED ASSIGNMENT
1.
State three advantages and two disadvantages of thermal
expansion of solids.
2.
Which gives you a greater burn, water at 100oC or steam at
100oC? Why?
3.
The linear expansivity of iron is 0.000012 K-1. What does this
statement mean?
4.
The linear expansivity of a cube is 0.000018 K-1. If the length of
each side of the cube is 10 cm, find the area of one face of the
cube and the volume of the cube when its temperature is raised
by 30 k.
5.
A container is completely filled with 95.1 kg of a liquid at 36 oC.
When the temperature of the liquid was raised to 78oC, it is found
that 1.2 g of the liquid overflows. Determine the apparent cubic
(volume) expansivity of the liquid.
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REFERENCES/FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools Onisha: Africana FEP Publishers Limited.
Ndupu, B. L. N. and Okeke, P. N. (2000). Round-up Physics. Ikeja
Lagos: Longman Nigeria Plc.
Okpala, P. N. (1990) .Physics. Certificate Year Series for Senior
Secondary Schools Ibadan. NPS Educational.
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MEASUREMENT OF TEMPERATURE
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Heat and Temperature
3.2
Principle Used in Temperature Measurement
3.3
Temperature Scales
3.3.1 Fixed Temperature Points of a Thermometer
3.3.2 Types of Temperature Scales
3.4
Types of Thermometers
3.4.1 Liquid – in – Glass Thermometer
3.4.2 Table Comparison of Mercury and Alcohol as
Thermometric Liquids
3.4.3 Gas Thermometers
3.4.4 Platinum Resistance Thermometer
3.4.5 Thermoelectric Thermometer
3.4.6 Optical Pyrometers
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In the previous unit (Module 3, unit 3) we discussed the concepts of heat
and temperature and stated the difference between them. We noted that
heat is a form of energy (thermal energy) while temperature is a measure
of the degree of hotness or coldness of a body. In this unit we shall
restate the difference between heat and temperature and discuss the
various methods of measuring temperature. The relationship between
pressure and temperature of a gas, the different types of scales used in
temperature measurement and types of thermometers will be treated.
2.0
OBJECTIVES
At the end of this unit, you should be able to:
•
•
Construct a device for measuring the temperature of the body
Use the variation of
o Pressure of a gas with temperature
o Expansion of solid, liquid or gas with temperature
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o Electrical resistance of a material, to measure the temperature
of the body.
•
•
distinguish between heat and temperature and between
temperature points and temperature intervals
select those liquids which are suitable for use in liquid –in-glass
thermometers from a given list of liquids and their properties.
describe the absolute scale of temperature and explain the
meaning of the absolute zero of temperature
convert a given temperature in the Celsius scale to a temperature
in the Kelvin scale and vice versa
describe the kinetic molecular model of temperature.
3.0
MAIN CONTENT
3.1
Heat and Temperature
•
•
•
Heat is a form of energy which can be transformed into other forms of
energy. It is also called “thermal energy” and always flows from a body
at a higher temperature to a body at a lower temperature, if the two
bodies are in contact. It is a measure of the internal energy of a body.
The quantity of heat contained in a body depends on the mass of the
body and its temperature. The unit of heat is the joule (J).
Temperature is the degree of hotness or coldness of a body. It is related
to the energy of movement (kinetic energy). It is also a measure of the
average kinetic energy of the molecules of a body. The unit of
temperature is degree Celsuis (0C) or Kelvin (K).
3.2
Principle Used in Temperature Measurement
In measuring the temperature of a body we use the principle that certain
physical properties of some substances vary with temperature. Any
physical property of a substance which varies in a known way and can
easily be measured is employed in an instrument called “thermometer”
to measure temperature. The substance whose physical property that is
used is called a “thermometric substance”. Table 4.1 below presents a
summary of the types of thermometers and their thermometric
substances.
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Table 4.1: Thermometers and Thermometric Substances
Types of
Thermometers
1. Liquid-in-glass
thermometer
Thermometric Physical Property
substance
Mercury
or Change in volume of
Alcohol
liquid with temperature
2. Gas thermometer
Gas
Change of gas pressure
at constant volume with
temperature
3. Resistance thermometer
Resistance wire
Change in the electrical
resistance of wire with
temperature
4. Thermocouple
Two dissimilar
metals (e.g.
copper and
constantan)
Change in electric
potential difference (or
current) between two
metal junctions at
different temperature.
5. Bimetallic thermometer
Two dissimilar The differential
metals (e.g. iron expansion of the two
and copper)
metals of the bimetallic
strip.
3.3
Temperature Scales
3.3.1 Fixed Temperature Points of a Thermometer
A thermometer has two reference temperatures or “fixed points”,
namely:
(a)
(b)
The upper fixed temperature point
The lower fixed temperature point.
Definitions
(a)
(b)
262
The upper fixed point of a thermometer is the temperature of
steam from pure water boiling at standard atmospheric pressure
of 760 mm of mercury.
The lower fixed point of a thermometer is the temperature of pure
melting ice at the standard atmospheric pressure of 760mm of
mercury.
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The fundamental interval (or temperature interval) is the
difference in temperature between the upper and lower fixed
points of a thermometer.
The fundamental interval of a thermometer is usually calibrated to give
readings of temperature values between the upper and lower fixed
points. The calibration of this interval depends on the scale of
temperature chosen for the thermometer.
3.3.2 Types of Temperature Scales
Three types of scales that are in common use are as follows
(i)
(ii)
(iii)
Fahrenheit scale
Celsius scale
Kelvin (absolute or thermodynamic) scale.
Fig. 4.1 shows the three scales with their upper and lower fixed points
and their fundamental intervals.
Fig. 4.1
Fig. 4.1: Fahrenheit, Celsius and Kelvin scales of temperature
(a)
Fahrenheit Scale (oF)
(i)
(ii)
Lower and upper fixed points are 32oF and 212oF
respectively.
Fundamental interval is divided into 180 equal parts
(i.e. 212 – 32 = 180) each of which defines one degree
Fahrenheit (1oF) in this scale.
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Celsius Scale (oC)
(i)
(ii)
(c)
Lower and upper fixed points are 00C and 1000C
respectively.
Fundamental interval is divided into 100 equal parts or
degree, each equal to 10C.
Kelvin or Absolute Scale (K)
(i)
(ii)
Lower and Upper fixed points are 273K and 373K
respectively.
Fundamental interval is divided into 100 equal parts or
units each of which is called one Kelvin (K).
You should note well that in this scale temperatures are not measured in
degrees but in units called Kelvin (K). Hence the unit symbol K is
written without the degree sign. The zero on the Kelvin scale is equal to
-273oC (i.e. OK = -273oC). It is called the “absolute zero”
Definition
The absolute zero temperature is the temperature at which the volume of
a fixed mass of gas at constant pressure, which is steadily cooled below
0oC, would theoretically become zero.
Conversion Formulas
(i)
Conversion from Fahrenheit to Celsius:
Scale suppose X of corresponds to a temperature of yoC.
212oF
x
320F
100oC
y
00C
Fig. 4.2
Fahrenheit
Scale
264
Celsius
Scale
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By proportions:
X – 32 =
Y–0
212 – 32
100 – 0
x – 32 =
y
180
100
Y
=
100 ( x -32)
180
Y
=
5 (X – 32)
9
o
where X is in F and Y is in 0C.
SELF ASSESSMENT EXERCISE 1
(i)
Give two examples of conversion from Celsius to Fahrenheit and
Fahrenheit to Celsius.
Conversion from Celsius to Kelvin Absolute Scale
We have seen that
OK = -273oC or 00C = 273 K
Hence, if t oC corresponds to T on the absolute scale, then
T
=
t + 273
You should note that although the fixed points on the Celsius and Kelvin
scales are different, a temperature change of 1oC (on the Celsius scale) is
equivalent to a temperature change of IK (on the Kelvin scale).
Determination of Temperate on a Celsius Scale
You can determine an unknown temperature 0 in the Celsuis scale as
follows:
1.
2.
3.
4.
Identify a physical property of a substance which varies in a
known way with temperature and which can easily be measured.
Measure the value of the physical property at the lower fixed
point, 00C.
Measure the value of the physical property at the upper fixed
point, 1000C
Finally measure the value of the physical property at the
unknown temperature 00C
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X100
1000C
X0
θ0C
X0
00C
Fig. 4.3
Let value of physical property at 00C
=
X0
Let value of physical property at 1000C =
X100
Let value of physical property at θ0C
X0
=
If change in physical property is directly proportional to change in
temperature, then using proportions, we have
X θ - xo
x100 – xo
0
100
0
=
=
=
θ – 00C
100 - 0oC
xo - xo
x100 - xo
xo - xo
x 100
x100 - xo
This method can be used to measure an unknown temperature where a
calibrated thermometer is not available.
Worked example
The upper and lower fixed points of a certain thermometer are 30cm
apart. On a certain day, the length of mercury thread in the thermometer
is 9 cm above the ice point. What is the temperature recorded by the
thermometer on (a) Celsuis scale (b) Kelvin scale?
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Solution
1000C
X100
373K
30cm
0
0C
Xt
T
00C
X0
273K
Fig. 4.4
(a)
Using proportions, we have
θ–0
100 - 0
0
100
0
(b)
9
= 30
=
=
9
30
9 x 100
30
300C
=
Again using proportions, we have
T - 273
373 - 273
=
9
30
T - 273
100
=
9
30
T - 273 = 9 x 100
30
T
=
= 30
30 + 273 = 303K
Alternatively for (b), once 0 has been obtained in (a), we can
apply the conversion formula as follows:
T
T
T
=
=
=
t + 273
θ + 273
30 + 273
=
303K
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(c)
Constant Volume Gas Thermometer Scale
(i)
Upper fixed point is the triple point of water.
Definition
The triple point of water is the temperature at which ice liquid water and
water vapour coexist in equilibrium.
(ii)
Lower fixed point is the absolute zero or Ok (-273oC).
The scale is usually calibrated at these two fixed points and other
temperatures are obtained by measuring the pressure of the gas at the
unknown temperature and making use of the definition of temperature in
the scale given below.
Determination of Temperature
Thermometer Scale.
on
the
Constant
volume
Gas
Suppose the unknown temperature = T
Let temperature of the triple point of water = Ttr = 273.16K
Let pressure of gas at the unknown temperature = PT
Let pressure of gas at triple point of water =
Ptr
Using
T1
=
P1
,
we have
T2
P2
T
Ttr
=
Pr
Ptr
T
=
273.16
PT
Ptr
T
PT x 273.16
Ptr
For elementary work, where ice point is used instead of the triple point,
we have
T
=
PT x 273.16
Pice
(d)
=
Standard Thermometer Scale
Early experiments measured temperature with mercury thermometers
and often found that different instruments gave different readings for the
same temperature. Other types of thermometers such as the platinum
resistance or the gas thermometers which were used later had the same
problem. It became clear that one type of thermometer based on one
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scale of temperature would have to be taken as standard. Gas
thermometers, either the constant volume or constant pressure types,
showed the closest agreement over a wide range of temperatures. They
were also very sensitive, accurate and highly reproduceable. It was
observed that as the pressure was lowered (approaching zero), all gas
thermometers indicate the same reading.
Lord kelvin suggested that the standard scale of temperature should be
based on an imaginary ideal gas with properties that were those of real
gases at very low i.e it obeyed Boyle’s law. He proposed that the
product of the pressure and volume of this ideal gas should be used as
the thermometric property of the gas. Thus if P1, P2 are the pressures and
V1, V2 the volumes of the gas at temperatures T1 T2, then
T1
T2
=
P1V1 ……………….(1)
P2V2
If the volume is kept constant, then we have
T1
T2
=
P1
P2
……………..(2)
Which gives a definition of temperature on the constant – volume gas
thermometer scale. If however, pressure is kept constant, we have
T1
=
V1
…………………..(3)
T2
V2
which gives a definition of the temperature on the constant-pressure gas
thermometer scale.
For practical reasons (among which were the extreme sensitivity of the
steam point to pressure changes) another upper fixed point, the triple
point of water, was chosen to replace the steam point as the upper fixed
point on the proposed standard temperature scale. The absolute zero
(OK) was chosen as the lower fixed point.
As it has already been pointed out either the constant-volume or
constant-pressure gas thermometer could be reliably used for measuring
temperatures in this scale since at low pressures all gases (including the
imaginary ideal gas) give the same reading. However, the constant –
volume gas thermometer has been chosen and hence the standard
thermometer scale “is the same as the constant-volume gas thermometer
scale” It is also called “ideal gas scale “or thermodynamic temperature
scale.
We have already defined unknown temperature on this scale as follows:
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T
=
PT
Ptr
x 273.16
where T is the unknown temperature, PT is the pressure of the gas at the
unknown temperature, T, Ptr is the pressure of the gas at the triple point
of water and 273.16 is the triple point of water.
(d)
International Practical Temperature Scale (IPTS)
Measuring temperatures on the standard thermometer scale (or ideal gas
scale) using the constant –volume gas thermometer is a tedious exercise.
The international temperature scale (ITS) was therefore proposed and
adopted as a more practical scale for general use. The scale consists of
eleven primary fixed points ranging from the triple point of hydrogen
(13.81K) to the freezing point of gold (1337.58K), which have been
accurately determined on the ideal gas scale by a constant volume gas
thermometer and also some other secondary fixed points. Particular
types of thermometer are specified for measuring temperatures over a
certain range using agreed formulas. At the fixed points, there is
agreement between the IPTS and the Kelvin scale, the differences at the
intervening temperatures being usually negligible.
3.4
Types of Thermometer
3.4.1 Liquid – in – Glass Thermometer
This is the most common type of thermometer. Temperature is measured
by measuring the volume of a fixed mass of liquid inside the
thermometer owing to change in temperature. Mercury and alcohol are
the common liquids used in this type of thermometer but mercury
thermometers are more commonly in use. Mercury is preferred because.
(i)
(ii)
(iii)
(iv)
It is easily seen
It expands almost uniformly with increase in temperature,
It does not stick to or wet glass
Its temperature range, -39oC to 357oC, is convenient for normal
use.
Figure .4.5 shows a mercury-in-glass thermometer with fixed points
indicate in the Celsius and Kelvin scale of temperature.
The thermometer is made sensitive by:
(a)
270
using thin glass for the bulb to enable the liquid in the bulb to
assume the temperature of its surrounding very quickly,
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using a narrow capillary tube of uniform bore to make it possible
for small temperature changes to cause large changes in volume
of mercury (or length of mercury thread).
using a liquid of high expansivity such as mercury (note the
properties of mercury listed above earlier).
Fig. 4.5: Mercury –in – glass thermometer
Water as a thermometric liquid
Water is unsuitable as a thermometric liquid for the following reasons:
(i)
(ii)
(iii)
(iv)
It has a small range of expansion – it freezes at 0oC and boils at
1000C
It does not expand uniformly and between 00C and 40C it
contracts when heated and expands when cooled (anomalous
expansion).
It wets glass (unlike mercury it leaves traces of water as it moves
on the glass.
It is colourless, making the mercury in glass difficult to read.
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3.4.2 Comparison of Mercury and Alcohol as Thermometric
Liquids
The following (Table 4.2) shows the comparison of mercury and alcohol
as thermometric liquids.
Table 4.2: Comparison of Mercury and Alcohol as Thermometric
Liquids
Mercury
Alcohol
1.
Has greater conductivity
and expands rapidly and so
indicates a temperature
change quickly
Has much less conductivity and
expands slowly and so does not
respond quickly to temperature
change
2.
Has silvery surface which Needs to be coloured since it is
makes it opaque and hence courless
can be seen easily.
3.
Does not wet glass because
of its convex meniscus and
thus enables the meniscus
to give accurate readings of
temperature
Wets glass because of its concave
meniscus and so tends to cling to
the walls of stem of thermometer
resulting to inaccurate readings
especially when the temperature is
falling
4.
Does not vaporize easily
Vaporises easily
temperatures
5.
Can be used to measure Boils at 78oC and so cannot be used
higher temperatures since it to measured temperatures above
boils at 357oC
78oC
6.
Freezes at -39oC and so not
suitable for measuring
temperature lower than this
value
7.
Has
much
lower Has a much higher expansivity.
expansivity for the same Expands, six times, as much as
272
even
at
low
Freezes at -115oC and so is suitable
for
measuring
much
lower
temperatures
than
mercury
thermometers.
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rise in temperature.
mercury for
temperature.
some
rise
in
Points 1-5 above represent the advantages of mercury over alcohol as a
thermometric liquid while points 6 and 7 represent the advantages of
alcohol over mercury (or the disadvantage of mercury over alcohol).
Other types of liquid – in – glass thermometer. Apart from the type of
mercury –in-glass thermometer commonly used in schools and
hospitals, we also have
(a)
Clinical Thermometer
This type of thermometer is used in hospital for measuring the
temperature of the human body. It has range of 35oC to 43oC because the
normal temperature of a healthy person is about 37oC but may rise to
about 41oC during high fever. This is illustrated in Fig. 4.6 below:
Fig. 4.6: Clinical thermometer
(b)
Maximum and Minimum Thermometer
This type of thermometer is used by weathermen in the meteorological
department to record maximum and minimum temperature for a day in
meteorological stations. . The information is used to predict weather.
Six’s combined maximum and minimum thermometer named after the
inventor, James Six, is most commonly used. Fig. 4.7 illustrates the
major features of the thermometer. Two steel indices A1 and A2,
floating in the alcohol in both aims of the tube, are used to read the
minimum and maximum temperatures respectively.
When temperature rises, the alcohol expands pushing the mercury round
the thermometer. As a result the steel index A2 is pushed upwards. Its
lower end indicates the maximum temperature, which is read from the
attached scale. Decrease in temperature cause the alcohol to contract so
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that mercury moves up the minimum arm, pushing the steel index A1
upwards.
Alcohol vapour
Alcohol
Steel index
Mercury
Fig. 4.7: Six’s maximum and minimum thermometer
The lower end of this index (A1) indicates the minimum temperature
reached. Tiny springs attached to A1 and A2 enable them to remain in
positions to which they have been pushed. A magnet is used to reset the
indices by bringing them in touch with the mercury column more.
3.4.3 Gas Thermometers
Liquid-in-glass thermometers are not very suitable for very accurate
measurements of temperature. This is mainly because the glass
containing the mercury or alcohol also expands with the liquid in an
irregular manner. Again such thermometers have relatively small
temperature range determined by the boiling and freezing points of the
liquids used.
For accurate temperature measurements, gas thermometers are used.
They are usually of two main types: constant-pressure gas thermometers
in which the pressure or volume respectively of a fixed mass of gas
increases linearly as temperature increases. Constant volume gas
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thermometers are more commonly used than their constant – pressure
counter – parts.
Constant Volume Gas Thermometer
Fig. 4.8 shows the main features of a constant volume gas thermometer.
It consists of a flask containing dry air or nitrogen connected to a
manometer tube by means of a capillary tube. The mark A is the
constant volume mark. In order to measure the pressure of the gas in the
flask , the height of the mercury reservoir B is adjusted until the mercury
level in the capillary tube reaches the level A, the
Fig. 4.8: Constant volume gas thermometer
h
= difference in mercury levels in the manometer at A and B
(measurement in cm),
H
=
height
of
mercury
barometer
(atmospheric pressure) then the
pressure (p) of the gas in the flask is
given by the total pressure on the gas
P
=
the total pressure on the gas (H + h)
cm mercury where H = atmospheric
height.
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ho
=
h100 =
then
height of mercury at ice point
height of mercury at the steam point,
hθ
=
height of mercury at the temperature
θ to be measured,
θo C
=
Pθ
P100
(H +
(H +
θ
=
hθ - ho
h100 - ho
- Po x 100 0C
- Po
ho ) - ( H - ho)
h100) - (H - ho)
3.4.4 Platinum Resistance Thermometer
Fig. 4.9 shows the main features of a platinum resistance thermometer
which makes use of the fact that the electrical resistance of a metallic
conductor changes proportionally with its temperature. If the
temperature increases, the resistance also increases.
Fig. 4.9: Platinum resistance thermometer
The thermometer consists of a long thin platinum wire wound (nonconductively) round a small spool made of mica or silica. The ends of
the wire are connected to a resistance measuring device such as the
metre bridge. If a given platinum wire has a resistance R o at 0oC, R100 at
1000C and Rθ at an unknown temperatureθ, then the unknown
temperature is given by
θ
276
=
Rθ - Ro
R100 - Ro
x 100oC
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In most cases, the temperature scale may be calibrated to read the
temperature directly.
Platinum resistance thermometers are useful for measuring very low and
very high temperatures, usually of range – 1800 to 1,1500C.
3.4.5 Thermoelectric Thermometer
Fig 4.10: Thermoelectric thermometers
A thermoelectric thermometer is essentially a thermocouple and works
on the principle of the thermometer. It consists of two different metals,
e.g. copper and constantan joined at the ends. One end, the hot junction,
is heated (or cooled) while the other end, the cold junction, is kept
constant at the temperature of melting i.e. so that a current (usually in
milliamperes) flows along the metals. The thermometer is connected in
series to a potentiometer which measures the e.m.f across the ends of the
metals in millivolts. The value of the e.m.f depends on the difference
between the temperature of the hot and cold junction.
Suppose Eo
E100
=
Eo
Then θ
=
emf at 00C (ice point)
emf at 1000C (steam point)
=
emf at the unknown temperatureθ,
Eθ - Eo x 1000C
E100 - Eo
Thermoelectric thermometers are particularly useful for measuring very
high temperatures in industry.
=
Advantages of the thermometer are as follows:
(i)
(ii)
(iii)
It is very sensitive and can measure rapidly changing or
momentary temperatures
It can measure high temperature up to 5000C.
It is very small in size and so can measure the temperature at a
point.
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3.4.6 Optical Pyrometers
When it is required to make accurate measurements of very high
temperature, optical pyrometers are used. This is done by observing the
radiation from the hot body and the process is called pyrometer and the
instrument used a radiation pyrometer or optical pyrometer.
Optical pyrometers make use of the principle that very hot substances
change colour at certain known temperatures. For example the
temperature of steel, when it is below red heat, can be judged by its
colour. Temperatures below red heat can also be estimated by the use of
paints which change colours at known temperatures.
Radiation pyrometers can be used to measure temperatures (e.g. of
furnaces) only above red-heat which is about 600oC. They fall into two
major classes:
(i)
Optical radiation pyrometers, which respond only to visible light
and
(ii)
Total radiation pyrometers , which respond to the total radiation
from the hot body, heat and light
Fig. 4.11:
Optical radiation pyrometer
The principle of the commonest type of optical radiation pyrometer also
known as the “disappearing filament” optical pyrometer is illustrated in
Fig. 4.11. Light from the hot source and the tungsten filament lamp is
made to pass through a red filter of a known wavelength range before it
reaches the eye. The eyepiece of the instrument (not shown in the figure
but between the eye and the filter) is focused upon the filament of the
tungsten light .The hot source, S whose temperature is required is then
focused by the objective lens, L so that its images also lies in the plane
of F. The light from both the hot source and the filament passes through
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the red filter (of red glass) before reaching the eye. Both of them (source
and filament) therefore appear red.
If the body is brighter than the filament, the filament appears dark on a
bright background. If the filament is brighter than the source, it appears
bright on a dark background. By adjusting the current through it, the
temperature of the filament is adjusted until it merges as nearly as
possible into its background. It is then as bright as the body or source
and is said to have “disappeared’. The rheostat R which adjusts the
current as well as the ammeter A which measures the current are both
mounted on the body of the pyrometer.
The ammeter is usually calibrated to read the temperature directly in
degrees Celsius. Radiation pyrometers of this type can measure
temperatures up to 10000C (i.e range between 6000C and 10000C but
more elaborated pyrometers can be adjusted to measure much higher
temperatures. Generally the range of a radiation pyrometer can be
extended by introducing a green filter between objective lense L1 and
the filament lamp. This reduces the brightness of the red filter. When
this is done, a second scale on the ammeter is provided for reading the
temperature accurately when such a green filter is inserted
Summary of
Thermometers
Ranges,
Advantages
and
Disadvantage
of
Table 4.3 presents a summary the ranges, advantages and disadvantages
of the common types of thermometers discussed.
Table 4.3: A summary of types of thermometers
Types of thermometers Advantages
and temperature range
Mercury-in-glass
(i) Easily and quickly
thermometer -390C to
read – temperature
0
357 C
read directly.
(ii) Very portable.
(iii) suitable for normal
use.
Constant volume gas (i)
Sensitive.
thermometer
(ii) Accurate
-2700C to 1 5000C
(iii) Large range
(iv) Easily
constructed.
(v)
Used for
calibrating other
types of
thermometers.
Disadvantages
(i) Limited range.
(ii) Less accurate
(iii) Easily broken.
(i)
Difficult to use and
time consuming.
(ii) Volume of flask is
large, therefore
unable to measure
temperature at
isolated confined
space.
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(vi)
Used as standard
thermometer for
temperatures
below -1830C.
Platinum
resistance (i)
Accurate.
(i) Not suitable for
thermometer
(ii) Wide range.
measuring
rapid
-1800C to 1 1500C
(iii) Most suitable for
changes in temperature
measuring
because it takes time to
constant
attain
thermal
temperatures
equilibrium with the
accurately.
surroundings.
(iv)Used as standard
thermometer for
temperature
between -1850C
and 6300C.
Thermoelectric
(i)
Sensitive.
(i)Less
accurate
if
0
thermometer ----250 C (ii)
Wide range
millivoltmeter is used
0
to 1 150 C
(iii) Most suitable for
to measure emf. A
rapid changes of
potentiometer used for
temperature.
measuring emf would
(iv) Able to measure
be more accurate.
temperature of
small
and
isolated places.
(v)Used as standard
thermometer for
temperatures
between 6300C
and 0630C.
Radiation
pyrometer (i)
Accurate
(i) Not suitable for
0
above 600 C
(ii)
Sensitive
temperatures below
(iii)
Wide range
6000C
(iv) Suitable for very
high
temperatures
(e.g
furnaces)
without risk or
danger
to
persons
involved.
4.0
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CONCLUSION
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In this unit you learned about heat and temperature. You also learned
about the principle used in temperature measurement, temperature scales
and types of thermometers.
5.0
SUMMARY
1.
Temperature is the degree of hotness or coldness of a body. It is
measured by an instrument called thermometer and it is a scalar
quantity.
2.
Heat is a form of energy (thermal energy) which flows from one
point to another as a result of temperature difference between the
points.
3.
Thermometers employ any physical property of a substance
which varies in a known way with temperature and is easily
measurable.
4.
Five major types of thermometers are in use, namely: liquid-inglass thermometers, gas thermometer, resistance thermometers,
thermoelectric thermometers and radiation pyrometers. Each type
of thermometer has associated with it, a particular level of
accuracy sensitivity, range, advantages and disadvantages.
5.
The following temperature scale exists:
(i)
Fahrenheit scale (no longer in common practical use)
(ii)
Celsuis Scale (also called Centigrade scale) on this scale,
absolute zero is -2730C or more accurately -273.15; ice
point is 0.000C; steam point is 100.000C and triple point of
water is 0.010C.
(iii)
Kelvin scale (also called “absolute scale”, “standard scale
“ or thermodynamic scale”). The so called “ ideal gas
scale” or the “Constant volume gas thermometer scale”
also belong to this class; On this scale absolute zero is
0.00K, ice point is 273.15K, steam point is 373.15K and
triple point of water is 273.16K.
(iv)
International practical temperature scale (IPTS) made up
of eleven primary fixed points ranging from the triple
point of hydrogen (13.81K) to the freezing point of gold
(1337.58K) which have been accurately determined on the
ideal gas temperature scale.
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6.
Kelvin scale of temperature is related to the Celsius scale by
T (K) = 273 + θ 0C
A more accurate relationship is
T (K) = 273.15 + θ 0C
Change of I K = change of 10C.
The absolute zero of temperature = 0 (K) or -2730C (or
more accurately -273.150C)
7.
Temperature θ on the Celsius scale is defined by
θ
Xθ - X0 x 100 0C
X100 - X0
where X is the physical property varying with temperature,
Xθ = value of X at θ 0C
X0 = value of X at 00C, and
X100 = value of X at 1000C.
=
SELF ASSESSMENT EXERCISE 2
1
2
(3)
(a)
Distinguish between heat and temperature
(b)
Name three physical properties used in the construction of
thermometers and state how each property varies with
temperature.
(a)
Give reasons why water is not suitable as a thermometric
liquid.
(b)
State four advantages and two disadvantages of mercury
as a thermometric liquid.
Explain each of the following with respect to a thermometer:
(i)
(ii)
fixed point
Fundamental interval.
A faulty thermometer reads 0.50C at the ice point and 99.20C at
the steam point. What is the correct temperature when the
thermometer reads 200C and at what temperature will its reading
be exactly correct?
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(a)
Name five different types of thermometer and describe the
working of any two of them.
(b)
List five different scales of temperature and briefly
describe each one of them.
(5)
A fixed mass of gas at constant pressure has a volume of 200.0
cm3 at the ice point, 273.2 cm3 at the steam point and 525.1 cm3 at
the melting point of a solid .What is the melting point of the solid
on the constant pressure gas scale of temperature?
6.0
TUTOR-MARKED ASSIGNMENT
1.
Define an unknown temperature θ on the Celsius temperature
scale using a platinum resistance thermometer. State the principle
on which such a thermometer functions. The resistance of a
platinum resistance thermometer wire 10.40 ohms at 00C and
10.71 ohms at 1000C. Calculate the temperature when the
resistance is 9.61 ohms.
2.
Explain the meaning of “fixed points” and “fundamental internal”
of a thermometer. The ice and steam points of a thermometer is a
marked 40 and 120 respectively. What is
3.
4.
(i)
The temperature in oC when the reading on this is 60,
(ii)
The thermometer reading when the temperature is 600C?
The fixed point of a standard thermometer is the triple point of
water which is given the value 273.16K
(i)
What is meant by the fixed point of a thermometer?
(ii)
Why is the triple point temperature given the value
273.16K
Suggest a suitable thermometer to measure the following
temperatures:
(a)
A pool of molten lead (melting point of lead is about
600K)
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PHY 001
(b)
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A liquid nitrogen bath at its normal boiling point (about
80K). For each case, outline the working principle of the
thermometer.
REFERENCES/FURTHER READINGS
Anyakoha, M. W. (2000). Physics for Senior Sec. School. Onitsha:
Africana – FEP Publishers Ltd.
Awe, O and Okunola O.O (1992). Comprehensive Certificate Physics,
SSCE Edition. Ibadan: University Press Plc.
Duncan, T. (1977). Advanced Physics: Fields, Waves and Atoms.
London: John Murray.
Nelkon, M. (1986). Principles of Physics for Senior Secondary Schools
(9th Edition). England. Longman Group Ltd.
Nelkon, M and Parker (1982). Advanced Level Physics (5th Edition).
London: Heinemann Educational Books Ltd.
Ndupu, B. L. N. Okeke, P.N. and Ladipo, O.A. (2000). Senior
Secondary Physics Book 2 Lagos: Longman Nigeria Plc.
Yong, P.L; Anyakoha M.W. and Okeke, P.N. (2002). University
Physics. Also for Polytechnics and Colleges, Volume 1. Onitsha:
Africana – FEP Publishers Ltd.
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UNIT 5
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TRANSFER
CAPACITIES
OF
HEAT
AND
HEAT
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
286
Introduction
Objectives
Main Content
3.1
Conduction of Heat
3.1.1 Good and Bad Conductors of Heat
3.1.2 Experiment to Compare the Thermal Conductivities
of Materials
3.1.3 Explanation of Conduction in Terms of Kinetic
Molecular Theory
3.1.4 Thermal Conductivity of Liquids and Air
3.1.5 Practical Applications of Good and Bad Conductors
3.2
Convection of Heat through Liquids and Gases
3.2.1 Convection Currents in Liquids
3.2.2 Explanation of Convection in Terms of Kinetic
Molecular Theory (KMT)
3.2.3 Convection Current Artificial/Natural Current
3.2.4 Practical Applications of Convection Current
3.3
Radiation of Heat
3.3.1 Emission and Absorption of Radiation by Different
Surfaces
3.3.2 Applications of Radiation
3.4
Heat Capacity and Specific Heat Capacity
3.4.1 Heat Capacity – Definition, Symbol and Unit
3.4.2 Specific Heat Capacity –Definition, Symbol and
Unit
3.4.3 Quantity of Heat
3.5
Measurement of Specific Heat Capacity of a Substance
3.5.1 Method of Mixtures
3.5.2 Electrical Method
3.6
Newton’s Law of Cooling
3.6.1 Demonstration of Newton’s Law of Cooling
3.6.2 Correction for Heat Losses in Calorimetry
Conclusion
Tutor-Marked Assignment
Summary
References/Further Readings
PHY 001
1.0
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INTRODUCTION
This unit will introduce you to the different modes (conduction,
convection and radiation) of transferring heat energy from one place to
another. You will learn to explain the modes of heat transfer using
kinetic molecular theory. In addition, you will learn the important
concepts of heat energy – the heat capacities, their measurements and
determination. Finally, you will learn Newton’s law of cooling and
cooling correction.
2.0
OBJECTIVES
By the time you finish this unit, you will be able to:
•
•
•
explain conduction, convection and radiation in terms of kinetic
molecular theory
know the difference between good and bad conductors of heat
using metals water and air as examples
explain the relationship between the heat supplied to a substance
and
o its temperature change at constant mass
o Its mass at constant temperature changes
•
•
•
explain the terms specific heat capacity and thermal capacity
explain why there is unequal rise in temperature for different
substances of the same mass supplied with the same quality of
heat
state the assumptions underlying the experiments involving the
measurements of heat capacities
calculate unknown qualities using the relation h = mc ∆ t when
no change of state is involved
state Newton’s law of cooling
explain the Newton’s law of cooling; and the cooling correction.
3.0
MAIN CONTENT
3.1
Conduction of Heat
•
•
•
It is an everyday experience that when you dip a silver spoon into a hot
tea, the handle of the spoon very quickly feels hotter than it was before.
The heat from the hot tea has been transferred along the metal handle to
the other end of the spoon by the process of conduction.
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Conduction of heat is a process by which heat energy is transferred
through a material, the average position of the particles of the material
remaining the same.
Heat energy is always transferred when different parts of a solid body
are at different temperatures. The direction of heat transfer is always
from the hotter to the cooler parts of the solid.
3.1.1 Good and Bad Conductors of Heat
Most metals such as copper, aluminum, silver, iron, allow heat energy to
pass through them very easily. We refer to such materials as good
conductors of heat. They are used where heat is required to travel
quickly (eg the cooking utensil). On the other hand, we refer to those
materials which do not allow heat to pass easily through them as poor
conductors of heat or insulators. Examples of such materials are most
non-metals such as water, air, wood, plastic, cloth, cotton wool, cork.
The handle of a cooking utensil is made of wood or plastic which are
insulators so that heat cannot be conducted quickly through them.
The ability of a metal to conduct heat is known as its thermal
conductivity.
3.1.2 Experiment to compare the thermal conductivities of
materials
Co
ppe
r
Bra
ss
Wate
Tank r
Woo
d
Iro
n
Lea
d
Fig. 5.1: Comparison of thermal conductivities
An experiment to compare the conductivity of different substances is
illustrated Fig. 5.1.
A metal container has rods of different materials but of the same length
and diameter fitted into holes in one of the sides by means of corks as
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shown. The rods are first dipped into molten paraffin wax; when they
are withdrawn, a coating of wax solidifies on them. Boiling water is then
poured into the container so that the ends of the rods are heated to the
same temperature. The wax on the rods begins to melt from the hot ends
and after some time, it is found that the wax has melted to different
distances along the rods, with the copper rod having the longest length,
followed by aluminum, iron, lead while the wooden rod hardly melts at
all. This indicated that the different substances have different thermal
conductivities. Thus, in this experiment, copper has the greatest while
wood has the least thermal conductivity.
3.1.3 Explanation of Conduction in Terms of Kinetic
Molecular Theory
Layers of molecules
Boiling
Water
ABCD
Direction of heat transfer
Fig. 5.2: Molecular explanation of conduction
Heat is transferred through solids mainly by conduction because of the
nature of solid molecules which are very close together. When we heat a
portion of solid directly as shown in Fig. 5.2, the solid molecules that
are directly heated (layer A) absorb the heat and vibrated faster than
before the heat was applied. They do not move from place to place
because of the strong forces of attraction between them. As they vibrate,
they pass on some of their heat energy to their neighbours (layer B)
which is turn vibrate faster and pass on some of the heat to the next
layer of molecules (c), and so on.. The heat continues to move in this
way until it reaches the end of the solid. All the molecules of the solid
eventually vibrate more rapidly about their fixed positions. We say that
thermal energy is transferred along the solid although the average
position of the molecules remains unchanged. When heat is transferred
in this way, it is said to be conducted along the body.
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3.1.4 Thermal Conductivity of Liquids and Air
Boiling water
Heat
Ice wraped
in wire gaize
Fig. 5.3 Water, a bad conductor of heat
Water and other liquids are poor conductors of heat. The only exception
is mercury which is a metal in liquid form.
You can verify that water is a poor conductors of heat by the following
simple experiment.
You will wrap a piece of ice in wire gauze to make it sink to the bottom
of a test tube of water. Heat the top of the water as shown in Fig. 5.3 (to
prevent convection).
You will observe that the ice does not melt even when the water at the
top of the tube boils. This shows that water is a bad conductor of heat.
To show that air is a poor conductor of heat, hold your hand very close
but not touching the side of a kettle of boiling water. You can feel the
radiation but your hand is not burned although separated from the kettle
by a thin layer of air.
3.1.5 Practical applications of Good and Bad Conductors
Here, we will consider the following practical applications of good and
bad conductors
•
Cooking pots and frying pans –made of metals to ensure quick
transfer of heat from the fire to the food being cooked. The
handles of cooking utensils are made of insulators so that the
utensils when hot can be held comfortably by the handles.
•
Home cooling in the tropics – asbestos ceiling and thatched-roof
houses prevent heating up of houses because their materials are
insulators and do not conduct the heat from the sun into the room.
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•
The use of cloth to keep warm – clothes keep us warm by
trapping air (poor conductors) around the body. Clothing made
from woollen and fur materials are worn in the cold climates to
retain heat and keep the body warm.
•
Use of rugs on floors – Heat transferred from one’s foot to the
floor is conducted away quickly by the floor and the foot feels
cold but the rug does not conduct away heat rapidly. It heats up to
the temperature of the foot.
•
Cotton wool and air used as insulators to reduce to the minimum
the loss of heat in calorimeters. This is called lagging.
3.2
Convection of Heat through Liquids and Gases
You have seen that liquids and gases are generally poor conductors of
heat. However, heat is effectively transferred in them by convection.
•
Convection is the process by which heat is transferred in a liquid
or gas by the actual movement of the heated fluid from the hotter
to the colder parts.
3.2.1 Convection Current in Liquids
You can observe convection in liquids by performing this simple
activity.
You will fill about two-thirds of a large round bottomed flask with
water. Drop a tiny crystal of potassium permanganate, using a long glass
tube, at the bottom of the flask. Heat the bottom of the flask just below
the crystal gently and observe the movement of the coloured liquid.
You will observe that the colour streaks of the potassium permanganate
rises upwards from the bottom of the flask where the heat was applied.
The streaks curve downwards at the top of the water as shown in figure
5.4 below.
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Drop crystal
of potassium
permanganate
Water
Convection
current
Potassium permanganate
crystal
Heat
Fig. 5.4: Convection current in liquid
The upward movement of the coloured streaks is due to convection and
the circulation of the liquid is called convection current. The movement
of the convection current is explained as follows:
The heated liquid at the bottom of the vessel expands and rises up
because warm water has a smaller density than cold water. The
ascending water carries some heat with it and the colder, denser water
falls to the bottom where it gets heated itself and also rises. This sets up
convection current which circulates continually.
3.2.2 Explanation of Convection
Molecular Theory (KMT)
in
Terms
of
Kinetic
We explain convection in terms of KMT as follows:
When the water is heated below, its molecules at the point of heating
gain more kinetic energy, expand by vibrating farther away from their
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fixed position, become less dense and therefore rise. The denser
molecules from the top sink to the place of those that have risen above.
These molecules in turn get heated and rise. With the heating, upward
warm currents and downward cooler currents are established until the
water boils.
3.2.3 Convection Current Artificial/Natural Current
Gases expand much more readily than liquids. As a result, convection
currents are easily set up in gases when their temperature rises. Here,
you will study the diagrams (Fig. 5.5) illustrating convection current in
nature: Land and sea breezes, showing convection current in gases.
Fig. 5.5: Land and sea breezes
3.2.4 Practical Applications of Convection Current
Convection current is important to us in the following ways:
•
Land and sea breezes:
In daytime, the sun warms the land to a higher temperature than the sea,
because land has a lower specific heat capacity than sea-water and the
sea is always in motion. The warm air above the land rises, and cooler
air from the sea moves in and takes its place.
The convection current is completed by the air in the upper atmosphere
moving down in the opposite direction, and we have sea breeze.
At night, the temperature of the land drops much faster than that of the
sea. As a result the sea is warmer than the land. Convection current is set
up in the opposite direction to the daytime. We call the breeze moving
from the land to the sea land breeze. The process is illustrated in Fig. 5.5
above.
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•
Convection current is used in house ventilation, in motor car
cooling systems in domestic hot water systems, and air
conditioning of rooms.
•
Glider pilots use convection current as air lift for their gliders.
•
Winds are caused by convection currents in the atmosphere,.
Near the equator hot air rises, and cooler and denser air from the
Polar Regions flows in and takes its place.
3.3
Radiation of Heat
You experience radiant heat (heat energy given out by radiation) in your
daily life heat from the sun and by warming yourself by the fire side. It
requires no material medium.
Radiation is the process by which heat is transferred from a hotter to a
cooler place without heating of the intervening medium.
We refer to radiation as electromagnetic wave of the type called intra
red. It travels with the velocity of light (3 x 108m51). It can be detected
by a radiometer and a thermopile.
3.3.1 Emission and Absorption of Radiation by Different
Surfaces
We know that different types of surfaces at the same temperature unit or
radiate heat at different rates depending on the nature of the surface. For
a given temperature, a dull black surface radiates most and a highly
polished surface radiates least. You can verify this by performing this
simply experiment.
Dull black
surface
Cork falls
as war
melts
Wax melts
first
Highly polished surface
Cork holds firm
Wax does not melt
Bunsen burnes
Fig. 5.6: Comparison of heat absorption by two surfaces
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You will use two sheets of thin plates, one painted black and the other
highly polished as shown in Fig. 5.6. Attach a cork to the reverse side of
each plate using a candle wax. Place a hot object like Bunsen burner or
burning candle half way between the two surfaces fixed vertically facing
each other and a few centimetres apart.
After a few minutes, you will notice that the wax behind the dull black
surface begins to slide off the plate. The wax behind the polished
surface does not melt and the cork remains fixed.
From this activity you can conclude that the dull black surfaces are
better absorbers of radiant heat than polished surface. Also, good
emitters of heat are good absorbers.
Heat energy falling on a body is partly absorbed and partly, reflected.
Black surfaces are good absorbers and poor reflectors; poor absorbers
are good reflectors of heat.
3.3.2 Applications of Radiation
•
Heat from the sun gets to us by the process of radiation.
•
House roofs in the tropics as well as factory roofs are brightly
painted in shiny aluminium colour to reduce absorption of heat
during the day. It also reduces the heat loss by radiation at night.
•
It is not advisable to wear a dark coloured jacket in hot sunshine
because it will absorb heat and make the wearer feel hot and
uncomfortable. Also, a brightly painted car is preferred to a
black-painted car in Nigeria and other hot tropical countries. The
black-painted car will absorb and retain heat from the sun and
inside the car will be very hot.
•
Petrol storage tanks are sprayed with silver paints to reflect heat
rays falling on it.
•
A teapot has a silvery surface and an electric iron has a silvered
surface base in order to make each capable of retaining its heat
for a long period.
*
A fire-fighting suit is bright and shiny so that it does not absorb
much heat to burn the fire man.
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SELF ASSESSMENT EXERCISE 1
1.
Vacuum flask is a good application of our knowledge of the three
modes of heat transfer. Look carefully at the vacuum flask in Fig.
5.7 and answer the question below.
Fig. 5.7 :Vacuum flask
(i)
(ii)
(iii)
(iv)
Why is it that a vacuum surrounds the liquid?
Why are the glass surfaces silvered?
Why is it that a cork or plastic stopper is used? (two reasons)
How will the heat escape if the vacuum seal is damaged.
2.
An electrician is fitting an electrical heater 3.4 into a water tank.
Explain whether he should put the heater at the top or at the
bottom in order to heat the water in the tank.
3.
Two motor cars of the same type are left in the sun. One is
painted black and the other white. Which will be hotter inside?
Why?
4.
Explain why a tile floor feels colder to the feet than does a rug in
the same room.
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3.4
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Heat Capacity and Specific Heat Capacity
3.4.1 Heat capacity – definition, symbol and unit
As you know, the supply of heat to a substance increases the average
kinetic energy of the molecules of the substance hence, an increase in
temperature of the substance. The average kinetic energy gained by the
molecules per degree Celsius rise in temperature is the heat (thermal)
capacity of the substance.
Definition
The heat capacity of a substance is defined as the heat energy required to
raise the temperature of the substance by 10C (or 1K).
Symbol
C; S .I. unit: joule per Kelvin (J/K).
3.4.2 Specific Heat Capacity – Definition, Symbol and Unit
From simple experiment, we know that equal masses of different
substances require different quantities of thermal (heat) energy to raise
them through the same temperature range. To take care of these
differences and to make it possible to calculate the amount of heat
energy required to raise the temperature of a substance through a
definite temperature range, the idea of specific heat capacity was
introduced.
Definition
Specific heat capacity of a substance is the amount of heat required to
raise the temperature of unit mass of the substance by 10C (or IK).
Symbol
C; S.I unit joule per kilogram per Kelvin (Jkg-1K-1)
You will note that
Heat capacity of a substance
C
=
Mass of the substance x specific heat capacity of the
substance.
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When the temperature of a substance changes, heat energy given out (or
received).
Q
=
mass x specific heat capacity x temperature change.
In symbols:
and
C
Q
=
=
mc …………………….(1)
mc (θ2 - θ1) …………….(2)
where
C
=
Q
=
M
=
C
=
θ2
=
θ1
=
(θ2 - θ1) =
heat capacity of the substance
heat energy
mass of substance
specific heat capacity of the substance
higher temperature
lower temperature
temperature change)
We call Q the quantity of heat – it is sometimes denoted by H.
3.5
Measurement of Specific Heat Capacity of a Substance
We call the instruments designed for experiments to measure quantities
of heat and specific heat capacities calorimeters. The simple calorimeter
used in school laboratories is usually made of copper which is a good
conductor of heat. This enables it to reach the same temperature as its
contents quickly.
The calorimeter is placed inside a larger container and the two are
separated by layers of lagging material- a poor conductor of heat. To
ensure an even temperature, a stirrer is provided while the lid prevents
evaporation of the liquid in the calorimeter. Here, we are going to
consider two methods of determining the specific heat capacities of
solids and liquids.
3.5.1 Method of Mixtures
(i)
298
Specific heat capacity of solids by the method of mixtures.
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Thermometer
String
Boiling
water
Solid of
unknown heat
capacity
Heat Source
Thermometer
Stirrer
String
Outer container
Lagging
material
Calorimeters
Solid
Calorimeter
Fig. 5.8: Determination of specific heat capacity by the method of
mixtures
In this experiment, the solid (eg a metal block is first weighed and its
mass recorded . it heated in boiling water as shown in Fig. 5.8 for some
minutes to allow it assume the temperature of the boiling water. The
solid is then quickly transferred to a copper calorimeter which had been
previously weighed empty and when about half-full of water. The initial
temperature of the water in the calorimeter is taken before the hot solid
is transferred into it.
The mixture of solid and water in the calorimeter is gently stirred to
attain even temperature. The highest temperature attained is recorded.
The specific heat capacity is calculated as follows:
Mass of calorimeter
Mass of calorimeter + water
Mass of solid
Initial temperature of water and calorimeter
Temperature of solid in boiling water
Temperature of mixture
Mass of water
Specific heat capacity of water
Specific heat capacity of copper
Let specific heat capacity of metal
=
=
m1(kg)
=
m2(kg)
=
m3(kg)
=
θ01C
=
1000C
=
θ02C
=
(m2 - m1) kg
=
4200Jkg-1K-1
=
420 J kg-1 K -1
CJ kg -1 K-1
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Heat lost by hot metal
Heat gained by water
Heat gained by calorimeter
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=
m3c (100 - θ2) J
= (m2 – m1) x 4200 x (θ2 - θ1) J
= m1 x 420 x (θ2 - θ1) J
Assuming no heat is lost to the surrounding and no heat is equally
gained, we have that:
Heat lost by the solid = heat gained by the calorimeter and water.
M3C [(100 - θ2) = (m2 - m1 ) x 4200 + (m1 x 420)] (θ2 - θ1)....(3)
You can calculate the value of c for the solid from equ (3) if the values
of the other variables are known.
Precautions
You will take the following precautions when performing the above
experiment in the laboratory.
•
•
Lag the calorimeter to prevent heat exchange with the
surroundings.
Transfer the hot metal quickly to the calorimeter without
splashing
Stir the mixture gently
(ii)
Specific heat capacity of liquids by method of mixtures.
•
We carry out this experiment as above. But here, we use a liquid of
unknown specific heat capacity instead of water in the calorimeter. We
also use a solid of known specific heat capacity. The specific heat
capacity is again calculated as follows:
Mass of calorimeter
=
m1 (kg)
Mass of calorimeter + liquid
=
m2 (kg)
Mass of solid
=
m3 (kg)
Initial temperature of liquid + calorimeter
=
θ10C
Temperature of mixture
=
θ02C
Temperature of hot solid
=
1000C
Specific heat capacity of copper calorimeter
=
420Jkg -1 K -1
Known specific heat capacity of metal
=
C1 J kg -1 K -1
Let specific heat capacity of liquid
=
C2 J kg -1 K -1
Heat lost by hot metal
= M3 C1 (100 - θ2)
Heat gained by calorimeter = M1 x 420 x (θ2 - θ1)
Heat gained by liquid = (M2 - M1) x C2 x (θ2 - θ1)
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Since,
heat gained
=
heat lost
We have
M1 x 420 x (θ2 - θ1) + (m2 - m1) x C2 x (θ2 - θ1)
=
M3 C1 ( 100 - θ2) ………(4)
The specific heat capacity of the liquid C2 would then be calculated from
equ. 4 when all the other variables are known.
3.5.2 Electrical Method
Specific heat capacity of solid by electrical method
Fig. 5.9: Measurement of specific heat capacity of solid by electrical
method.
Two holes are bored in the block whose specific heat capacity is to be
measured as shown in Fig. 5.9. The block is weighed .A thermometer
and an electrical heater are respectively inserted in each of the holes. A
little oil is dropped in each hole to help establish a good thermal contact
with the block. The metal block is surrounded with a lagged jacket to
reduce heat losses.
The initial temperature of the block is noted and the electrical heater is
switched on. The current is allowed to flow for some time until the
temperature rises by about 150C. The exact time of flow of a known
current is measured with a stop-watch and the initial and final
temperatures of the metal block are measured by the thermometer.
The specifc heat capacity is calculated as follows:
Let:
Mass of block
=
Value of steady current
=
Value of p.d. across heater =
m (kg)
1 (A)
V(v)
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Temeprature rise of block =
θ0 C
Time of flow of currnt
=
t (secs)
Specific heat capacity of block = C(Jkg -1K-1)
Heat supplied by current = H = 1Vt joules.
Heat required to raise the temperauter of block
By
θ0 C
=
mc θ
Assuming no heat loss to the surroundings, we have.
1Vt = Mc θ
1vt J kg -1 K -1 …………….5
mθ
C
=
(iii)
Specific heat capacity of a liquid by the electrical method
Fig. 5.10: Determination of specific heat capacity of liquid by electrical
method
In this experiment, the apparatus is set up as shown in Fig. 5.10. First,
we weigh a plastic container and then reweighed when about two –thirds
full of the liquid (e.g. water). We then connect the ammeter, voltmeter
and heating coil as shown in the circuit. A plastic stirrer and a
thermometer are then fitted into the container through holes in the
wooden lid. The initial temperature of the liquid is read and recorder.
The current is now switched on end the rheostat adjusts to give a
suitable steady current for about 3 minutes.
This is followed by stirring the liquid gently as it is being heated. The
current is then switched off and the final temperature read.
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We calculate the specific heat capacity of the liquid as follows:
Mass of liquid
=
mkg
0
Initial temperature of liquid=
θ1 C
Final temperature of liquid
=
θ20C
Current flowing in circuit
=
IA
Voltage across heating coil
=
V volts
Time of flow of current
=
t seconds
Let the specific heat capacity of liquid = CJkg-1K-1
Rise in temperature
Heat gained by liquid = mc (θ2 - θ1) joules Electrical energy supplied
to liquid = 1vt joules
Heat gained by liquid = electrical energy supplied
This implies that :
MC (θ2 - θ1) = 1Vt
C
Precautions
=
1 vt ………………...7
M(θ2 - θ1)
•
Cooling the water with ice to about 100Cbelow room temperature
to reduce heat lost to the surrounding.
•
Heating water until its final temperature is about 100C above
room temperature for above reason.
3.6
Newton’s Law of Cooling
•
Newton’s law of cooling stats that for a small difference of
temperature between body and its surrounding , the rate at which
a body loses heat is proportional to the temperature difference
between the body and its surroundings.
The law is approximately true in still air only for a temperature
excess of about 200oC or 300oC. It is true for all excess
temperature in conditions of forced convection of the air, i.e. in a
draught. At low excess temperature less than 10oC, radiation is
the major contributing factor to the rate of cooling of an object.
3.6.1 To demonstrate Newton’s law of cooling, we plot a temperature
(θ) – time (t) cooling curve for hot water in a calorimeter placed
in a draught (Fig. 5.11(a))
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If QR is the room temperature, then the excess temperature of the water
is (θ - θR). At various temperature such as θ in Fig. 5.11 (b), we may
draw targets such as APC to the curve. The slope of the target in degrees
per second, gives us the rate of fall of temperature when the water is at
the temperature θ.
If we then plot these rates against the excess temperature θ - θR, as in
Fig. 5.11 ( c ), we will obtain a straight line passing through the origin.
The result of the experiment shows that the heat lost per second by
water and calorimeter is proportional to the rate of fall of the
temperature. This is in agreement with Newton’s law of cooling.
3.6.2 Correction for Heat Losses in Calorimetry
When a hot body is transferred to a liquid in a calorimeter, the
temperature of the liquid and the calorimeter rises rapidly above that of
the surrounding. From Newton’s law of cooling, you can deduce that an
amount of heat is lost to the surrounding because allowance is not made
for this when reading the final temperature of the mixture. As a result,
the final temperature recorded is less than the actual final temperature
and correction has to be made for this.
Newton’s law of cooling enables us to estimate the heat lost in an
experiment in the method of mixtures.
In doing this experiment, we take the temperature of the mixture at half
minutes intervals and plot it against time as shown in Fig. 5.12.
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temp
Qc
p
c
QR
t1
Time
Fig. 5.12: Cooling correction
The broken line shows how we would expect the temperature to rise if
no heat were lost, we have therefore to estimate the difference, P,
between the plateau of this imaginary curve and the crest of the
experimental curve, C. P is known as the cooling correction. That is, the
temperature to be added to the observed final temperature to compensate
for loss of heat (the cooling correction).
As a precaution, you may cool the liquid initially to an amount of
temperature (about 100C) below room temperature so that the
temperature rise above room temperature at steady state should be equal
to the fall in temperature of the liquid below room temperature before
transferring the solid into it.
Worked examples
A lump of copper of unknown mass is heated to 1000C. If the final
temperature is 550C when dropped in 250 g of water at 40oC, calculate
the mass of copper, (specific heat capacity of water is 4200 Jkg-1K-1,
copper is 380J kg-1 K-1)
Solution
Let the mass of copper in kilogram = Mc
Heat in joule lost by copper = Heat in joule gained by water
Mc x C x (θ2 - θ1 ) = m x c x ( θ2 - θ1)
Mc x 380 x (100 – 55) = 0.25 x 4200 x (55 - 40)
⇒
mc
=
0.25 x 4200 x 15
380 x 45
The mass of copper is
=
0.92 kg
0.92 kg
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(2)
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An electric heat rated 20v 60w, fitted into a metal block supplied
heat to the block of mass 2kg, and specific heat capacity of 460 J
kg-1 K -1. Calculate the temperature rise in the block if the current
flows for 15 minutes
Solution
Power of heater = 1v = 60w
Heat produced by heater in 15 minutes = power x time (in seconds)
=
60 x 15 x 60J = 54000J.
Let temperature rise in metal block = θK
Heat gained by metal block = 2 x 460 x θJ.
Heat produced by heater = Heat gained by metal block
=
θ
54000 = 2 x 460 x θ
=
54000 = 58.7k or 58.7oC
2 x 460
The temperature rise in the block is 58.7 K or 58.70C.
SELF ASSESSMENT EXERCISE 2
1. Distinguish between heat capacity and specific heat capacity of a
substance. Explain how the specific heat capacity of a substance
could be determined.
2.
The specific heat capacity of zinc is 384Jkg -10C-1. What does the
statement mean? A 60 watt heating coil is used for 3 minutes to
heat a metal block of 700 g and specific heat capacity of 1.0J /
g0C. What is the temperature rise?
3.
The hot water of a bath delivers water of 950C at a rate of 5 kg
min-1. The cold water tap of the bath delivers water at 350C at the
rate of 15 kg min-1. If both baths are left on for 5 mins, calculate
the final temperature of the bath water, ignoring heat losses to the
surrounding.
4.0
CONCLUSION
In this unit you learned about the various modes of heat transfer
(conduction, convection and radiation), their molecular explanations and
applications. You also learned about heat capacity and specific heat
capacity. The determination of specific heat capacity for solids and
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liquids using method of mixtures and the electrical method. Finally, you
learned about Newton’s law of cooling and cooling correction.
5.0
SUMMARY
•
Conduction is a method of transferring heat with the aid of
molecules which vibrate about their mean position.
•
Metals are good conductors while non metals are usually poor
conductors.
•
Convection is a method of transferring heat which involves actual
movement of molecules from one position to another.
•
Convection currents could be observed in liquids and gases.
•
Radiation is a method of transferring heat which does not require
any type of material medium.
•
Heat transferred by radiation travels with the speed of light.
•
Good absorbers of heat (eg dull black surfaces ) are also good
radiations.
•
Poor absorbers of heat (eg. shinny bright surfaces) are also poor
radiations.
•
Heat (thermal) capacity of a substance is the amount of heat
required to raise the temperature of the substance by 10C.
•
Specific heat capacity of a substance is the amount of heat
required to raise the temperature of 1kg of the substance by 10C.
•
In activities involving measuring of heat quantities, it is assumed
that heat lost by a hotter substance is equal to heat gained by a
colder substance.
•
Newton’s law of cooling states that for a small difference in
temperature between a body and its surrounding, the rate at which
a body loses heat is proportional to the temperature difference
between the body and its surroundings.
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6.0
TUTOR -MARKED ASSIGNMENT
1.
Name the three modes of heat transfer. State the major difference
between them.
2.
What is the final temperature of a mixture if 2 kg of water at 800C
is added to 2.5 kg of cold water at 100C. (Neglect heat absorbed
by the container specific heat capacity of water is 4200J kg-1 k-1).
3.
4 kg of a certain metal is heated to 86oC and then dropped in 1.5
kg of water at 26oC. If the final temperature of water is 40oC,
calculate the specific heat capacity of the metal (specific heat
capacity of water = 4200 J kg -1 K-1).
7.0
REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools. Onitsha Africana FEP.
Anyakoha, M.W. (1991). Physics at a glance. Onitsha. Africana FEP.
Awe, O and Okunola, O.O. (1992). Comprehensive Certificate Physics.
Ibadan University Press.
Okpala P.N. (1990). Physics Certificate Year Series for Senior
Secondary School Ibadan NPS Educational.
UNIT 6
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LATENT HEAT AND EVAPORATION
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CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Latent Heat
3.1.1 Concept of Latent Heat
3.1.2 Molecular Explanation of Latent Heat
3.1.3 Types and Definitions of Latent Heat
3.2
Numerical Problems on Latent Heat
3.3
Measurement of Specific Latent Heat
3.3.1 Experimental Determination of Specific Latent
Heat of Fusion of a Solid (Ice)
3.3.2 Experimental Determination of the Specific Latent
Heat of Vaporization (Steam)
3.4
Application of Latent Heat
3.5
Melting and Boiling Point
3.5.1 Melting Point
3.5.2 Boiling Point
3.5.3 Effects of Impurities and Pressure on the Melting
and Boiling Points of a Substance
3.6
Evaporation and Boiling
3.6.1 Concept of Evaporation
3.6.2 Factors that affect the rate of Evaporation
3.6.3 Concept of Boiling
3.6.4 Evaporation and Boiling
3.6.5 Differences between Evaporation and Boiling
3.7
Vapour and Vapour Pressure
3.7.1 Concept of Vapour
3.7.2 Concept of Vapour Pressure
3.7.3 Saturated and Unsaturated Vapour
3.8
Concept of Sublimation
3.9
The Working Principle of some Common Devices
3.10 Humidity
3.11 Humidity and Weather
3.12 Formation of Dew, Mist, Fog and Rain
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
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In unit 5 of this last Module you learnt about transfer of heat and heat
capacities. In this unit, you will learn about latent heat – its concept and
definitions. You will also solve numerical problems on latent heat. In
addition, you will learn to determine and measure the specific latent heat
of fusion of ice and of vaporization as well as applications of latent heat.
Furthermore, you will be introduced to Evaporation. You will learn the
concepts of evaporation, vapours, boiling, sublimation and the effect of
pressure and impurities on melting and boiling points. Finally, you will
learn about relative humidity, dew point, humidity and weather.
2.0
OBJECTIVES
By the end of this unit you will be able to:
•
•
•
determine the melting point of a solid
determine the melting point of a given liquid
list the effects of impurities and pressure in:
o the melting point of solid (ice)
o the boiling point of a liquid
•
explain the terms:
o latent heat of fusion
o latent heat of vaporization
•
•
•
•
solve simple problems involving latent heat
distinguish between evaporation and boiling
explain sublimation and dew point
explain the working principles of such common devices as
o refrigerator
o airconditioner
o pressure cooker
•
explain the effects of humidity on personal comfort.
3.0
MAIN CONTENT
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PHY 001
3.1
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Latent Heat
3.1.1 Concept of latent heat
You must have observed that when sufficient heat is given to a solid it
melts into liquid. If heat continues to be supplied, the temperature
remains constant until all the solid has changed to liquid. The heat
supplied to melt the solid is called latent heat because the temperature
gave no indication of the heat required to change from a solid to a
liquid.
Also, if the liquid is heated further, its temperature rises until it reaches
a constant temperature. Then, it begins to change into vapour or gas
Again this heat is called latent heat because of change of state from
liquid to gas which gave no indication of heat required to change the
substance from liquid to gas.
That is during change of state,
Solid
(heat)
Liquid
(heat)
Gas
(Constant temperature)
(Constant Temperature)
Temperature remains constant although heat is being supplied.
The heat supplied or removed, which causes a change of state without a
change in temperature is called latent heat. It is called latent (meaning
hidden) heat because it is not made apparent by a change in temperature.
3.1.2 Molecular Explanation of Latent Heat
You already know that the molecules of a solid are kept in average fixed
position by strong forces of attraction. Latent heat represents the energy
used in breaking down the forces which hold the molecules of a solid in
a regular pattern. In liquid, it is the energy required to overcome the
forces of attraction between molecules of a liquid. The molecules are
then practically independent of each other and exist as gas.
3.1.3 Types and Definitions of Latent Heat
There are two types of latent heat namely:
i.
Latent heat of fusion (melting) (from solid to liquid)
ii.
Latent heat of vaporization (gas or vapour) (from liquid to gas)
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Definitions
i.
Latent heat of fusion of a solid is the heat energy that is used to
change a solid at its melting point to liquid at the same
temperature (the energy is stored in the hot liquid)
ii.
The specific latent heat of fusion of a solid is the energy required
to melt 1kg of the solid at its melting point.
It is equal to the energy given out by 1 kg of the liquid in order to
solidify at the freezing point.
If Q heat energy is supplied to melt mass M of solid at its melting point,
then the specific latent heat L is given by
L = Q
M
Q = ML
where: L = specific latent heat
Q = heat energy
M = mass of solid at melting point
Unit Jg-1 or Jkg -1
iii.
Latent heat of vaporization of a liquid is the heat energy that is
used to change a liquid in its boiling point to liquid at the same
temperature (the energy is stored in the vapour)
iv.
The specific latent heat of vaporization of a substance is the heat
required to change unit mass of it from liquid at its boiling point
to vapour without a change in temperature.
3.2
Numerical Problems on Latent Heat
i.
A bucket has a mass of 1 kg and specific heat capacity of 378 kg-1
k-1. It contains 4 kg of water at 40oC. When 0.91 kg of ice at 0oC
is added and completely melted the temperature of water is 18.2
o
C. What is the specific latent heat of fusion of ice?
(Assume no heat exchange with the surroundings)
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Solution
Since no heat is lost to the surrounding:
Heat lost by water + heat lost by bucket
= Heat used for melting ice + heat gained by iced water from 00C to
18.20C . Eqn…….(1)
Recall that heat lost:
where
Q =
Q =
C =
M=
θ2 =
θ1 =
MC (θ2 - θ1)
heat energy
heat capacity of the substance
mass of substance
higher temperature
lower temperature
Hence:
Heat lost by water = 4 x 4200 x (40 -18.2) J
Heat lost by bucket = 1 x 378 x (40 -18.2) J
Heat gained by iced water = 0.91 x 4200 x (18.2 – 0)0C
Heat used to melt 0.91 kg of ice = 0.91L
where L = the specific latent heat.
To obtain the specific latent heat of the fusion of ice we will
substitute the above values in ‘eqn 1’ and solve for L.
That is
Heat lost by water + heat lost by bucket
= Heat used for melting ice + heat gained by iced water from 00C to
18.20C
⇒ MC (θ2 - θ1) for water + MC (θ2 - θ1) for bucket
= ML for ice + MC (θ2 - θ1) for iced water
⇒ [4 x 4200 x (40 – 18.2)J)]+ [(1x 378 x (40 – 18.2)J)]
= 0.91L + [0.91 x 4200 x (18.2 - )J]
0.91L = [4 x 4200 x (40 – 18.2) J)] + [(1 x 378 x 40 – 18.2)J]
- [0.91 x 4200 x (18.2 – 0)J]
= 366240 + 8240.4 – 69560.4
= 374480.4 – 69560.4
∴L
= 304920
0.91
= 335076.92 Jkg-1
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ii.
ACCESS PHYSICS
A copper calorimeter of mass 170.0 g contains 100.0 g of water
at 19.50C. Naphthalene (melting point 80.00C) is melted in a test
tube, cooled at 800C, and then poured into the calorimeter. If the
highest temperature reached by the calorimeter after stirring is
30.00C and the final mass of the calorimeter and its content is
185g. Calculate the latent heat of fusion of naphthalene (specific
heat capacity of copper 0.4 kg-1, of naphthalene 1.3 kJ kg-1, water
4.2 kJ kg-1).
Solution
Mass of naphthalene = 185.0 – 170.0 g = 15 g
Heat lost by naphthalene
= heat gained by water and calorimeter
⇒ 15L +[15 x 1.3 x (80.0 – 30.0)J]
= [100 x 4.2 x (30.0 – 19.5)]+ [170x 0.4 x (30.0 – 19.5)]
⇒ 15L = (100 x 4.2 x (30.0 – 19.5)) + (170 x 0.4 x (30.0 – 19.5))
- (15. x 1.3 x (80.0 – 30.0)
L
3.3
= 4410 + 714 – 975
= 5124 – 975 = 4149
= 4149
15
= 276.6 Jkg-1
≅ 276.6 kJkg-1
Measurement of Specific Latent Heat
3.3.1 Experimental Determination of Specific Latent Heat of
Fusion of a Solid (Ice)
i.
By method of mixture
T h e r m o m e te r
Ic e
b lo c k s
W a te r +
ic e
L a g g in g
In s u la to r
Fig. 6.1: Latent heat of fusion by method of mixture
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To determine the specific latent heat of fusion of solid (ice) by method
of mixture, we use the apparatus as set up in the diagram above. The
calorimeter is weighed empty and reweighed when it is about half-filed
with water whose temperature is about 50C above room temperature.
Some small ice blocks are dried with blotting paper and then added to
the water in the calorimeter. The mixture is gently stirred and more dry
ice added until the temperature of the mixture is about 5 0C below room
temperature. The lowest temperature of the mixture is noted after all the
ice particles has melted. The calorimeter is reweighed to find the mass
of ice added. The result is then calculated as follows:
Assuming
Mass of calorimeter
Mass of calorimeter + water
Mass of calorimeter + water + ice
=
=
=
M1
M2
M3
=
=
M2 - M1
M3 - M2
=
=
=
θ10C
θ20C
4200(Jkg-1)
Then:
Mass of water
Mass of ice
If:
Original temperature of water
Final temperature of mixture
Specific heat capacity of water
Specific heat capacity of calorimeter +
Material
Specific latent heat of fusion of ice
Heat required to melt the ice
=
C(Jkg-1)
=
l(Jkg-1)
= (M3 - M2) l(joules)
Heat required to raise the temperature of melted ice from 0C to θ20C
=(M3 - M2) x 4200 x θ2 (joules)
Heat lost by the calorimeter and water in cooling from θ1 to θ2
= (M1, C + (M2 - M1) x 4200)] (θ1 - θ2)J
Since:
Heat gained
=
Heat lost
(M3 - M2) l + ((M3 - M2) x 4200) (θ1 - θ2)
From the above equation, the latent heat of fusion l can be calculated.
Precautions
You will take the following precautions when performing the above
experiment in the laboratory:
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Use only dry pieces
Add small quantity of ice at a time
Stir the mixture gently to obtain even temperature.
ii By electrical method
Electrical
immersion heaters
(12 v supply)
Pieces of ice
Filter funnels
Retort stands
Beakers
(a) Experimental
(b) Control
Fig. 6.2: Latent heat of fusion by electrical method
To determine the specific latent heat of fusion of solid (ice) by electrical
method, we use the apparatus as set up in the diagrams above (Fig. 6.2)
A 50W immersion heater is placed in a glass filter tunnel and some
pieces of ice are packed round the heater. The heater is connected to a
12v supply with the circuit as shown in Fig. 6.2a. Since ice will melt at
room temperature the same set-up is arranged as in figure 6.2b with the
heater not connected to the 12v supply. This forms a control for the
experiment. Two similar beakers of the same mass are used to collect
drip water by placing one below each beaker.
Quickly empty the beakers when the drip rates from the funnels appear
to be the same. Then replace them and switch on the current. Ensure
that the heater is in good thermal contact with the ice throughout the
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experiment. Allow current to flow for about 200 seconds and note the
value of the current and the potential difference. When the drip rate
appear to be constant again, remove the beakers and weigh them with
their contents. Use your result to calculate the specific latent heat of
fusion in the following manner:
Assuming these results are obtained:
Mass of the experimental beaker + water
Mass of the control beaker + water
Mass of the ice melted by electrical heating
Current
Potential difference
Time of flow of current
Energy supplied to ice
=
=
=
=
=
=
=
M1
M2
M2 - M1
I
V (12v)
t
VIt
This implies that:
Energy supplied to melt 1 kg of ice
=
Energy supplied to ice
Mass of ice melted by electrical heating
=
V
x
I
x
M2 - M1
t
=
Vit
Jkg-1
M2 - M1
You should note that:
1.
If Q heat energy is supplied to melt mass m of solid at its melting
point then the specific latent heat L is given by
L = Q
M
Q = ML
2.
The immersion heater should never be switched on unless it is in
a liquid or a block of metal as it will overheat in air.
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3.3.2 Experimental Determination of Specific Latent Heat of
Vapourisation (Steam)
i.
By method of mixture
Set up the apparatus as shown in Fig. 6.3. Weigh the calorimeter empty
and reweigh it when it is about half filled with water.
Outlet
Fig. 6.3 Latent heat of vaporization by method of mixture
Then place it in an insulating jacket. Note the temperature of the water.
Pass dry steam into the water until the water temperature rises by about
250C above room temperature. Note the final temperature and find the
mass of steam by reweighing the calorimeter and its content. Tabulate
your readings as shown below:
Mass of calorimeter
Mass of calorimeter + water
Mass of calorimeter + water + steam
Initial temperature of water + calorimeter
Final temperature of mixture
Temperature of steam
318
=
=
=
=
=
=
M1
M2
M3
θ10C
θ20C
1000C
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Let:
Specific latent heat of vapourisation of steam = L(Jkg-1kg-1)
Mass of water
Mass of steam
Specific heat capacity of water
Specific heat capacity of calorimeter material
Heat lost by steam in condensing
=
=
=
=
=
M2 - M1 (kg)
M3 - M1 (kg)
4200(Jkg-1k-1)
c(Jkg-1k-1)
(M3 - M2)joules
Heat lost by condensed steam in cooling from
1000C to θ20C = (M3 - M2) x 4200 x (100-θ2)J
Heat gained by water and calorimeter
= (M2 - M1) x 4200 x (θ2 - θ1) + M1 x C (θ2 - θ1)
Since
Heat gained
=
Heat lost
(M2 - M1) x 4200 x (θ2 - θ1) + M1C (θ2 - θ1)
= (M3 - M2)L + (M3 - M2) x 4200 x (100 - θ2)
Hence
L, the specific latent heat of vaporization can be calculated from the
above equation.
Precautions:
While performing the above experiment you should take the following
precautions:
Use only dry steam
Take care to prevent the calorimeter from receiving or losing heat from
or to its surroundings.
Cool the water to about 100C below room temperature before passing
the steam. Continue to pass the steam until the temperature of mixture
is about 100C above room temperature.
Stir the mixture continuously throughout the experiment.
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To power unit
i.
Resort
stand
Heater
Beaker + water
Lead
shot
Balance
Fig. 6.4:
Determination of specific latent heat of vaporization by
electrical method
To determine the specific latent heat of vaporization by electrical
method, you will set up the apparatus as shown in the diagram above.
Place a polystyrene (polyphenylethene) cup (beaker) on one side of a
balance and partly fill it with water. Suspend an immersion heater in the
water so that it is clear of the sides and bottom.
Switch on the current. When the water is boiling, pour leadshots into
the other pan until balance is almost obtained. As the water boils off,
the two sides will balance. At this instance start a stop-watch. Do not
allow any condensed water vapour to drop back into the cup. Now,
place a 20 g mass on the pan beside the cup. When 20 g of water boils
off, the two sides will again balance. Note the time when this happens.
State your results as shown below:
Power rating of heating coil
Time noted
Energy supplied (electrical)
Mass of water boiled off
320
=
=
=
=
p
t
pxt
0.020 kg
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Calculate the specific heat of vaporization from:
L =
pxt
0.02
3.4
Applications of Latent Heat
From the foregoing, you must have realized that whenever a liquid is
converted into vapour, heat is absorbed from the liquid or any object in
contact with it. The heat is the latent heat used in causing the
transformation from liquid to vapour. The phenomenon is utilized in the
working principles of some common household devices such as: (i)
Pressure cooker (ii) The refrigerator (iii) The air conditioner.
Later in this unit, we would discuss the working principle of these
devices.
SELF ASSESSMENT EXERCISE
1.
Explain the terms:
(i)
(ii)
(iii)
(iv)
Latent heat of fusion
Specific latent heat of fusion
Latent heat of vaporization
Specific latent heat of vaporisation
2.
A bucket has a mass of 100 g and specific heat capacity 420
Jkg-1k-1. It contains 400 g of water at 200C. How many grammes
of steam at 1000C are needed to raise the temperature of the water
and bucket to 950C? Specific latent heat of steam L = 2.26 x 10 6
Jkg-1.
3.
(a) Define latent heat (b) Describe the measurement of the
specific latent heat of (i) fusion of ice (ii) vaporization of water
under school laboratory conditions.
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3.5
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Melting and Boiling Point
3.5.1 Melting point
You know that the melting point of a substance is the temperature at
which the substance changes from solid to liquid state (without further
change in temperature)
To find the melting point of a solid, you should follow this procedure
listed below:
i.
Put a solid substance (e.g naphthalene) in a tube and melt it by
heating the test tube through a beaker of boiling water. The
molten naphthalene should half-fill the test tube.
Retort stand with
clamps
Thermometer
Beaker
Water
Naphthalene
Source of heat (bunsen
burner)
Fig. 6.5: Determination of melting point of Naphthalene
ii.
Insert thermometer so that its bulb is midway in the naphthalene.
iii.
Remove the test tube from the flame and allow the naphthalene to
solidify with the thermometer inside it.
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iv.
Warm the test tube in a beaker of water and record the
thermometer readings at interval of 30 seconds.
v.
Stop taking readings 4 minutes after the naphthalene has
completely melted.
vi.
Plot a graph of temperature against time.
vii.
Remove the test tube of molten naphthalene from the water bath
and allow it to cool.
viii.
Record the readings of the thermometer at the intervals of 30
seconds.
ix.
Stop taking of readings 10 minutes after the naphthalene has
completely solidified.
x.
Tabulate your readings and use it to plot a graph of temperature
against time.
a.
Table of Values
Temperature (0C)
(b)
Time (sec)
Graph
Temperature (0C)
Melting
Point
(a)
-----------
Time
Temperature (0C)
Freezing
Point
(b)
----------
Time
Fig. 6.6 : Heating and cooling curves of naphthalene
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Observation and Results
You would observe that as the solid is heated, the temperature rises. At
a certain stage, the temperature remains constant until all the solid has
changed to liquid. This is shown by the flat portion (plateau) of the
graph (Fig. 6.6a) after which the temperature starts to rise again.
This temperature at which the naphthalene melts is constant and is
called the melting point of the solid naphthalene.
When the naphthalene is cooling, energy is abstracted from it. At a
stage, the liquid naphthalene starts to solidify. This is shown in the flat
(plateau) portion of the cooling curve (Fig. 6.6b) after which the
temperature starts to fall.
This temperature at which a liquid changes to a solid is called the
freezing point of the liquid. You should note that the melting point and
the freezing point of a substance are the same.
3.5.2 Boiling point
The boiling point of a substance as you know is the temperature at
which the substance begins to boil and changes from liquid to gaseous
state (without further change in temperature).
To find the boiling point of a substance (Water).
For you to determine the boiling point of a liquid, you should follow
those steps below:
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Retort stand
with clamps
Thermomete
r
Conical flask
Water
o o o
o o
o
Bunsen burner
(Source of heat)
Fig. 6.7: Determination of the boiling point of liquid (water)
i.
Put the liquid substance (eg water) in a conical flask containing a
thermometer.
ii.
Use a bunsen burner to heat the water
iii.
Take the water temperature at 1 minute intervals until the water
has boiled for seven minutes.
iv.
Plot a graph of temperature against time to obtain a graph of the
type shown overleaf using your tabulated table of values.
a.
Table of Values
Temperature (0C)
Time (sec)
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(b)
Graph
Temperature (0C)
Boiling
Point
-----------
Time
Fig. 6.8: The boiling point of a liquid (water)
Observation and result
You must have observed that as you heat the water in the conical flask,
its temperature rises and the liquid evaporates. As the heating is
continued, the temperature is found to rise until the water begins to boil.
During boiling bubbles of air formed inside the liquid rise to the surface.
The temperature of the liquid remains steady during boiling. This is
indicated in the flat portion of the graph (plateau) Fig. 6.8 this constant
temperature is called the boiling point of the liquid.
3.5.3 Effects of Impurities and Pressure on the Melting and
Boiling Points of a Substance
The melting and boiling points of a substance are affected by pressure
and impurities. You can verify these by investigating these scientific
observations.
i.
Effect of pressure on boiling point
Increase in pressure increases the boiling point of a liquid. The pressure
cooker is a practical application of the effect of pressure on boiling
point. Increased pressure of the trapped air (gas) above the liquid in a
pressure cooker raises the boiling point of the liquid. Thus a high
cooking temperature is reached very fast. Hence, food is cooked faster
in the pressure cooker.
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ii.
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Effect of pressure on the melting point of ice (or freezing
point of water)
Increased pressure lowers the melting point of ice (or the freezing point
of water). This can be shown by passing a piece of wire with heavy
weights attached to both ends through a block of ice as shown in Fig. 6.9
below:
Wire cut through
ice
Freezing
above wire
Ice block
Wire
Weight
Fig. 6.9: Effect of pressure on melting point
The piece of wire cuts through the block of ice but the block remains
solid behind the wire. The ice directly below the wire melts because the
increased pressure lowers the freezing point. The wire passes through
the ice. Above the wire, decreased pressure raises the freezing point of
the melted ice and the water freezes again. The above process is known
as regelation.
iii.
Effect of impurities on the boiling point of a substance
Impurities raise the boiling point of a pure solvent. Thus, the boiling
point of a solution of salt in water is higher than that of pure water.
iv.
Effect of impurities on the melting point of a substance
The melting point of a pure solid is lowered by the presence of
impurities. For instance, a freezing mixture of ice and salt has a lower
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temperature than 00C, the freezing point of pure water (or melting point
of ice)
3.6
Evaporation and Boiling
3.6.1 Concept of evaporation
You are familiar with the concept of evaporation. You know that
evaporation is the change of liquid to vapour without the application of
heat to the liquid. It is a slow and quiet process which takes place at all
temperatures. This phenomenon can often be seen in our daily lives.
For instance;
i.
when a piece of wet cloth is spread out in the open during
harmattan, it dries up quickly;
ii.
liquids left in shallow open containers soon disappear;
iii.
when some liquid perfume is poured into a cup and left
uncovered, the cup is found to be empty after few hours.
3.6.2 Factors that affect the rate of evaporation
Can you guess the factors that affect the rate of evaporation? Compare
your answers with these factors that affect the rate of evaporation. They
are:
i.
Temperature – Evaporation is more rapid if the surrounding
temperature is higher. This is because the process of evaporation
involves absorbing the latent heat of vaporization from the
surroundings. The warmer the surroundings, the more the amount
of heat supplied by it to provide the latent heat of vaporization.
ii.
Pressure – The greater the pressure, the slower the evaporation
and vice versa. If an open container of water is placed inside a
bell-jar from which air has been pumped out (reduced pressure),
evaporation takes place at faster rate.
iii.
Area of liquid surface exposed – The greater the surface area of
liquid exposed, the more rapid will be the evaporation.
iv.
Dryness of the air – The drier the air (i.e the less water vapour it
contains) the quicker the evaporation. Water droplets on a watchglass evaporate faster if left in the open than when covered with a
bell jar. As the water evaporates, the air in the bell jar gets filled
with water vapour and this slows down the evaporation. In the
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case of the watch glass in the open, the water vapour quickly
escapes into the open air and more evaporation takes place. This
is the reason why wet clothes dry faster in dry weather than in
wet weather.
v.
Motion of the Air – Wet clothes placed in the draft from a fan
will dry up sooner than if placed in still air. The air stream or
wind removes the vapour and brings fresh and drier air into
contact with the liquid. Hence, evaporation is more rapid.
vi.
Nature of the Liquid – Different liquids evaporate at different
rates. The lower the boiling point of a liquid, the greater will be
the rate of evaporation. For example, mercury with a boiling
point of 3570C hardly evaporates while either (boiling point
350C) evaporates very rapidly. Liquids like ether which
evaporates rapidly are called volatile liquids.
3.6.3 Concept of boiling
Boiling occurs when a substance in the liquid state is thrown into violent
agitation as a result of continuous supply of heat to the liquid molecules.
3.6.4 Evaporation and boiling
Whether a liquid changes rapidly into its gaseous state by boiling or by
slow evaporation, it requires the same amount of latent heat to do so. In
the case of boiling, this heat is supplied by the heat source. In the case
of evaporation where there is no particular heat source, the liquid
obtains its latent heat of vaporization from the surroundings.
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i
Evaporation
This is the change from liquid to
vapour state at temperature below
normal boiling point
ACCESS PHYSICS
Boiling
This is the change from
liquid to vapour state at the
boiling point of the liquid.
ii
Evaporation takes place at all Boiling takes place at a fixed
temperatures
temperature for a particular
liquid.
iii.
It occurs when molecules escape It takes place throughout the
from the surface of a liquid
volume of the liquid
iv.
It is influenced by pressure,
It is influenced by pressure
and
temperature, dryness of the the amount of impurities
atmosphere, wind and amount of present in the liquid
exposed area of the liquid
v.
It takes place slowly and proceeds It occurs at a constant rate for
more rapidly as the temperature a particular temperature
increases.
vi.
It is a silent phenomenon
It is a noisy phenomenon
vii.
It causes cooling
It does not cause cooling
3.7
Vapour and Vapour Pressure
3.7.1 Concept of Vapour
As you know a liquid changes into vapour if sufficient heat energy is
supplied to it. The change from liquid to vapour takes place throughout
the entire liquid at a particular temperature.
3.7.2 Concept of Vapour pressure
If a liquid is allowed to evaporate in a closed container, the vapour
formed will exert a certain pressure. This is known as vapour pressure.
3.7.3 Saturated and unsaturated vapour
i.
330
Saturated vapour – A saturated vapour is one that is in contact
with its own liquid in a closed space. The pressure exerted by
such a vapour is the saturated vapour pressure (S.V.P).
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The saturated vapour pressure varies with temperature as shown in Fig.
6.10.
S.V.P
Fig. 6.10: Saturated vapour pressure (S.V.P) vs temperature
The saturated vapour pressure of water is approximately equal to the
external pressure when water boils. It does not obey the gas laws.
100oC
ii.
Unsaturated vapour – An unsaturated vapour is one that is not
in contact with its own liquid in a closed space. The gas laws are
approximately obeyed in such a vapour.
According to the kinetic theory, the number of molecules leaving the
liquid during evaporation is greater than those returning to the liquid in
an unsaturated vapour. On the other hand, there is equal number of
molecules leaving and returning to the liquid in a saturated vapour.
At higher temperatures, the energy of the molecules increases and more
molecules are then able to leave the liquid and the saturated vapour
pressure will thus increase.
3.8
Concept of Sublimation
This is a process whereby some solid substances when heated, change
directly into gaseous state without the intermediate liquid state. For
example, solid iodine and solid carbon dioxide changes from solid state
to gaseous state without changing to the intermediate liquid state.
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3.9
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The Working Principle of Some Common Devices
You have seen that whenever liquid is converted into vapour, heat is
absorbed from the liquid or any object in contact with it. This heat is the
latent heat used in causing the transformation from liquid to vapour.
The absorption of the latent heat from the liquid brings about a fall in its
temperature. The faster the evaporation, the greater is the fall in
temperature. This phenomenon is utilized in the working principle of
these common household devices.
i.
The Refrigerator
The cooling effect of evaporation is used in a refrigerator. The volatile
liquid used is usually liquid ammonia or Freon. Ammonia evaporates
inside coiled copper tubes surrounding the freezing compartment,
assisted by a pump which reduces the pressure. As it evaporates in these
coils, the ammonia absorbs heat from the surrounding air, thus cooling
the inside of the refrigerator and its contents. The vapour produced is
pumped away and expressed in a condenser where it condenses to liquid
ammonia. The heat released during this condensation is quickly
dissipated by an arrangement of cooling fans at the back of the
refrigerator.
The liquid is recycled through the evaporator coil (Fig. 6.11). The
process is repeated. This sets up a continuous circulation of liquid and
vapour.
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Freezing
compartment
Vapour
Pump
Liquid
Cooling fins
Fig. 6.11: Household electric refrigerator
ii.
Control of humidity (Air conditioning)
The humidity of an enclosure could be controlled to suit our personal
comfort. The basic principle of the working of an air conditioner is
illustrated in Fig. 6.12 below.
Evaporator
Cool air moves
into the room
Warm air in the
room
Fig. 6.12 : Domestic air conditioner
Warm air from
outside the room
Condenser
Hot air
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The air conditioner contains a powerful fin which draws air from outside
to inside the enclosure (e.g a room, a car). It first cleans the air by
removing dust particles from it. It further humidifies the air to a
specified value (to make it cold) before circulating it within the
enclosure. The humidity value of the circulating air is usually controlled
by setting the air conditioner to certain convenient levels. Air
conditioners are generally found in cars, public buildings, offices, ships
and so on.
iii.
Pressure Cooker
The basic principle of the working of the pressure cooker is based on the
fact that increased pressure raises the boiling point of water. Substances
inside a pressure cooker are heated at about a pressure of 2 atmosphere.
Such a high pressure can raise the boiling point of water to about 1200C.
Food items will cook more quickly when subjected to such a high
temperature condition inside a pressure cooker (Fig. 6.13)
Variable
cooling valve
Safety valve
+
Handle
Water boils at
120oC
Source of
heat
Fig. 6.13: A pressure cooker
Pressure cookers are found very useful by people and explorers in
mountainous areas of the world where atmospheric pressure is generally
lower than normal, where water can boil at 910C. This makes cooking
to be expensive in terms of time and fuel. Pressure cookers are usually
used to remedy the situation.
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3.10 Humidity
The atmosphere contains varying amounts of water vapour. Humidity is
the amount of water vapour in the atmosphere.
i
Relative humidity
If the air is saturated or nearly saturated with water vapour, it is
described as very humid. If the air is highly unsaturatged it is said to be
dry. Humid weather makes one feel uncomfortable as perspiration does
not evaporate easily from the skin into the already unsaturated air. Also,
very dry air evaporates the moisture from the skin too rapidly. This is
harmful to the skin. A comparison between the amount of moisture
actually present in the air with the amount of moisture it can hold when
saturated is termed relative humidity. This is expressed as a ratio which
indicates the extent to which air is saturated with water vapour.
Mass of water vapour in a given volume of air
Relative Humidity =
Mass of water vapour required to saturate the
Same volume of air at the same temperature
When the atmosphere is fully saturated with water vapour, the relative
humidity reaches 1. also, since the pressure exerted by the water vapour
is proportional o its mass per unit volume, the relative humidity can also
be expressed thus:
Pressure of water vapour present in the air
Relative Humidity =
Saturated vapour pressure at the same air pressure
ii
Dew point
This is defined as the temperature at which water vapour present in the
air is just sufficient to saturate it. It is dependent upon prevalent
atmospheric conditions like temperature, wind and the amount of water
vapour in the atmosphere. For instance, when ice blocks are kept inside
glass tumbler, moisture would form on the outside surface of the
tumbler after sometime. This is because the cold glass surface cools the
air around it to temperature at which the water vapour contained in the
air is sufficient to saturate the air.
Saturated vapour pressure at dew point
Relative Humidity = Saturated vapour pressure at original
Temperature of air
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3.11 Humidity and Weather
The need for effective weather forecasting is becoming very important
in all countries of the world. However, to achieve a good level of
accuracy in such weather forecasting, it requires accurate determination
of the humidity level of the air. This job in most cases is left for the
meteorologist (specialists in aspects of atmospheric physics). Their
reports concerning humidity and weather are found very useful in
scheduling or flights in airports, navigation, radio transmission,
factories, food storage, farming etc.
Also, the construction of certain household equipment such as
refrigerators, air conditioners and the like require knowledge of how to
keep the relative humidity of an enclosure at a specified level. The
formation of clouds, rain, fog, hailstone and snow depends among other
things on the relative humidity of the atmosphere.
3.12 Formation of Dew, Mist, Fog and Rain
i
Formation of dew
It is a common experience that in the evening, the ground and the air in
contact with it begins to cool as they lose heat through radiation. A
clear sky and still air may cause temperature air to fall even lower so
that the air becomes nearly saturated with water vapour. Furthermore,
the surfaces of some objects particularly grass and other plants, give off
water vapour, This adds to the amount of vapour in the air. When
compared with the surrounding air, these surfaces are good conductors
and lose heat sooner than the air. As the temperature of the ground and
the grass falls below the dew point, the moisture in the surrounding air
will condense and be deposited in their surfaces as dew.
If the dew point is below 00C, the vapour changes into the solid form of
ice crystals called frost.
ii
Formation of mist and fog
If the air near the surfaces of the land cools until it reaches dew point,
the water vapour begins to condense around tiny dust particles in the
atmosphere. These particles have surfaces on which the condensed
vapour collects. Such a collected of water particles is known as mist.
A thick mist is called a fog. The water particles of the mist or fog are
small enough to float in air. When they become larger, they fall as rain.
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Clouds and rains
A thick mist formed high up in the atmosphere is referred to as a cloud.
During the day, the land mass is warm and the warm air containing
water vapour rises above the land. As the air moved upwards, it
expands (due to reduced pressure at high altitudes) and becomes cooler.
This cooling process, if continued long enough, causes the air to become
saturated and its water vapour eventually condenses to form tiny
droplets of water. When this happens, a cloud begins to form. The tiny
droplets of water combine to form larger drops which fall as rain.
You should note that not all clouds are formed as water droplets. Some
clouds like the fleecy cirrus clouds which form at great heights are
composed of ice crystals because the water vapour in the air condenses
at a height where the temperature is below freezing point.
Raindrops that are carried by strong convection currents to greater
altitudes becomes frozen as they are cooled below 00C by the cold air at
such heights. These frozen raindrops fall as hail.
SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.
5.
6.
Explain why yam cannot be easily cooked in an open pot on top
of a high mountain.
When two ice blocks are pressed together, they stick together.
Explain this observation.
Distinguish between evaporation and boiling.
List and discuss the factors that influence the rate of evaporation.
What do you understand by vapour, vapour pressure and
saturated vapour pressure?
Define the terms:
i.
ii.
iii.
7.
Relative humidity
Dew point
Sublimation
Write short notes on the formation of the following:
i.
ii.
iii.
iv.
v.
Dew
Clouds
Fog
Mist
Rain
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CONCLUSION
In this unit you learned about latent heat – its concept and definitions.
You also learned how to solve numerical problems involving latent heat.
In addition, you learned how to determine and measure the specific
latent heat of fusion of ice and of vaporization as well as applications of
latent heat. You also learned about evaporation vapours, boiling
sublimation and the effects of pressure and impurities on melting and
boiling points. Finally, you learned about relative humidity, dew point
and humidity and weather.
5.0
SUMMARY
The heat supplied or removed from a body which causes a change of
state without a change in temperature is called latent heat.
Latent heat of fusion of solid is the heat energy that is used to change a
solid at its melting point to liquid at the same temperature.
The specific latent heat of fusion of a solid is the energy required to melt
1 kg of the solid at its melting point.
Latent heat of vaporization of a liquid is the heat energy that is used to
change a liquid in its boiling point to liquid at the same temperature.
The specific latent heat of vaporization of a substance is the heat
required to change unit mass of its liquid at its boiling point to vapour
without a change in temperature.
Increased pressure lowers the freezing point and raises the boiling point
of water.
Dissolved substances in a liquid lower the freezing point and raise the
boiling point of the liquid.
The factors that affect the rate of evaporation are:
i.
ii.
iii.
iv
v
vi
Temperature
Pressure
Area of liquid surface exposed
Dryness of the air
Motion of the liquid
Nature of the liquid
The vapour contained in an enclosed space above its liquid is said to be
saturated when the number of molecules escaping from the liquid per
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unit time is equal to the number returning to the liquid per unit time.
The pressure it exerts is called the saturated vapour pressure of the
liquid.
The boiling point of a liquid is the temperature at which its saturated
vapour pressure is equal to the atmospheric pressure.
Mass of water vapour in a given volume of air
Relative Humidity =
Mass of water vapour required to saturate the
Same volume of air at the same temperature
6.0
TUTOR-MARKED ASSIGNMENT
1.
2.
3.
Define latent heat?
Explain why a pressure cooker boils food more quickly than on
ordinary pot.
What are the differences between evaporation and boiling?
7.0
REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools Onisha: Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics
Ibadan: University Press.
Okpala, P.N. (1990). Physics Certificate Year Series for Senior
Secondary Schools. Ibadan: NPS Educational.
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EXPANSION OF GASES
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Expansion of Gases
3.1.1 Kinetic Molecular Theory of Gas and its
Assumptions
3.2
Boyle’s Law
3.2.1 Statement of Boyle’s Law
3.2.2 Verification of Boyle’s Law
3.2.3 Use of Kinetic Theory to Explain Boyle’s Law
3.3
Charles’ Law
3.3.1 Statement of Charles’ law
3.3.2 Verification of Charles’ Law
3.3.3 Cubic Expansivity of a Gas
3.3.4 Use of Kinetic Molecular Theory to Explain
Charles’s Law
3.4
Pressure Law of Gay – Lussac’s Law
3.4.1 Statement of Pressure Law
3.4.2 Verification of Pressure Law
3.4.3 Use of Kinetic Molecular Theory to Explain
Pressure Law
3.5
Ideal or the General Gas Law
3.5.1 The General Gas Law or Equation
3.5.2 The Ideal Gas Equation
3.6
Conversion to Standard Temperature and Pressure (S.T.P)
3.7
Intermolecular Energy and Force
3.8
Pressure Calculations
3.9
Introduction of Temperature
3.10 Deductions from the Ideal Gas Equation
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit you will learn how the three properties of a gas-the
temperature, the pressure and the volume that are necessary to describe
the state of the gas are related for a given mass of gas in the gas laws.
Also, you will learn how to state, verify and derive the general gas
equation for the ideal gas based on the laws. Finally, you will learn
about other laws governing the real gases.
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OBJECTIVES
By the time you finish this unit, you will be able to:
•
Explain, using the ideas of the kinetic theory of gases
o The variation of volume with temperature of a gas when the
pressure is kept constant
o The variation of pressure with volume of a gas when the
temperature is kept constant
•
•
•
•
•
Explain Charles’ and Boyle’s laws of gases
Deduce the general gas law from a given mass of gas which
obeys Charles’ law
Identify the limitations of the gas laws
Solve simple problems involving the gas laws
Explain
o Molar gas constant
o Intermolecular energy and forces
•
State
o
o
o
o
o
Assumptions of the elementary kinetic theory of gases
Graham’s law of diffusion
Avogradro’s law
Dalton’s law of partial pressures
Maxwell’s distribution law.
3.0
MAIN CONTENT
3.1
Expansion of Gases
We classify substances in the gaseous state as gases or vapours. You
know that a gas has no volume of its own but takes the volume of its
containers. The volume of a gas is changed by varying the pressure of
the gas or by changing the temperatures of the gas.
Thus, the pressure, volume and temperature of a gas may change when
any one of the quantities change. Hence, the thermal expansion of a gas
is complicated by the fact that the volume can be altered by a change of
temperature and of pressure. For us to study the behaviour of a gas with
respect to the three variables: volume, temperature and pressure, we
investigate the relation between two of the variables while the third one
is kept constant.
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3.1.1 Kinetic Molecular Theory of Gas and its Assumptions
We use the kinetic molecular theory of gases to explain the effect of
temperature on the volume and pressure of a gas. The theory makes the
following assumptions.
•
A gas is considered to be made up of a large number of molecules
acting like elastic spheres.
•
The molecules are in a state of constant motion knocking against
each other and the walls of the containing vessel.
•
The particles exert attractive forces on one another.
•
The particles possess kinetic energy due to their motion.
•
The volume of the molecules of a gas is negligible compared with
the volume of the container.
3.2
Boyle’s Law
Here, we will consider the relationship between the pressure and the
volume of a fixed mass of a gas given by Boyle’s law in 1660.
3.2.1 Statement of Boyle’s law
Boyle’s law states that the pressure of a fixed mass of gas is inversely
proportional to its volume, provided that the temperature is kept
constant.
In symbols, Boyle’s law is written as
P
=
1
V
or
PV
=
constant
Where
P
=
pressure
V =
Volume
of a fixed mass of gas.
3.2.2 Verification of Boyle’s law
We can verify Boyle’s law in the laboratory using simple apparatus
shown in Fig. 7.1.
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Air
hcm
Closed glass
glass tube B
Tube C
Open
Fig. 7.1: Verification of Boyle’s law
To do this, we entrap a fixed mass of dry air with mercury is a closed
glass tube of uniform cross-section attached by rubber tubing to a
mercury reservoir. The volume V of the dry air is proportional to the
length L of the air column because the tube is of uniform cross-sectional
area. The pressure and volume of the dry air are varies by the open limb
vertically up or down. The length which we read from the metre scale is
taken as the volume of the gas while the atmospheric pressure (Po) is
read from a fortin barometer.
When the mercury surface in B is lower than that in C, the pressure (P)
of gas is given by:
P = (Po + h) cm Hg
But, when the mercury surface in B is higher than that in C, the gas
pressure (P) is given by.
P = (Po + h) cm Hg.
By lowering or raising the tube B we would obtain a set of readings of L
and h. A table of these readings is then made as shown below in Table
7.1.
l cm
hcm
(Po ± h)cm
1
l
cm -1
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The values obtained in an earlier experiment when used in plotting
graphs give the following result as shown in Fig. 7.2
•
Graph of P against V (ie Po + h against L) gives a curve as shown
in Fig. 7.2a.
•
Graph of P against 1 (ie Po + h against 1 )
V
l
is a straight line graph passing through the origin as shown in Fig. 7.2b.
The straight line graph indicates that the pressure is inversely
proportional to the volume of air or any gas at constant temperature.
P
(a)
V
p
x
x
x
x
x
(b)
1
/V
Fig. 7.2: Graph of Boyle’s law
You will take the following precautions when performing this
experiment.
•
Ensure that the entrapped air is dry by adding a small quantity of
concentrated sulphuric acid into the tube before adding mercury.
•
Ensure that there is no air bubbles in the mercury.
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•
Ensure constant temperature by allowing the system to attained
the temperature of its surrounding. You will do this by waiting
for a few minutes after each variation of h and L.
•
Take the readings when the mercury levels are steady and avoid
parallax error of the metre rule while reading it.
Worked example
The pressure of a fixed mass of gas is 96 cm Hg when the volume of
the gas is 5.0 m3. Find the pressure of the gas when its volume becomes
7.0 m3. Assume the temperature is constant.
Solution
From Boyle’s law
PV = constant at constant temperature
=
P1 V1 = P2 V2
96 x 5 = P2 x 7
P2 = 96 x 5
7
=
68.57 cm Hg
3.2.3 Use of Kinetic Theory to Explain Boyle’s Law
Now, let us consider a given mass of gas at constant temperature. At this
constant temperature, the average velocity of the gas molecules is
constant and the number of collisions they make per unit area of the wall
of the containing vessel is also constant. Hence, the gas exerts a certain
constant pressure, P, on the walls of the vessel.
If we now double the original volume of the gas, the gas molecules will
be spread out and it will take longer time to bombard the walls and
hence, fewer impact per second. As a result, the pressure of the gas will
be halved. Thus, at constant temperature, volume of a gas increases as
pressure decreases.
On the other hand, if we halve the original volume of the gas, the gas
molecules will be more closely pack and then it will take less time to
bombard the walls of the containing vessel. The gas pressure increases
(double) as more impact per second are made. Hence, at constant
temperature, decrease in volume of a gas leads to increase in pressure.
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This means that the pressure of a given mass of gas varies inversely with
its volume at constant temperature in accordance with Boyle’s law.
This is illustrated diagrammatically in Fig. 7.3
P2 P23
P1
1
P4
3
4
Fig. 7.3: A representation of Boyle’s law
3.3
Charles’ Law
Here we will consider the connection between the change in volume
with temperature change of a fixed mass of gas kept at constant pressure
published by Charles in 1787 and independently by Gay-Lussac in 1802.
3.3.1 Statement of Charles’ Law
Charles’ law states that the volume of a fixed mass of gas increases by
1/273 of its volume at 00C per degree Celsius rise in temperature
provided its pressure remains constant.
We can also state Charles law using Kelvin temperature scale. You are
aware that a temperature toC on the Celsius scale is related to an
absolute or Kelvin temperature T by the equation
T (K) =
273 + toC
Using the Kelvin scale therefore, Charles’ law can be stated as
The volume of a fixed mass of gas is directly proportional to its absolute
temperature if its pressure remains constant.
In symbols, we express Charles’ law as
V
V
Or V
T
⇒
V1
T1
where
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∝
=
=
T
; if pressure is kept constant
KT
K
=
V2
T2
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V = volume; T = Kelvin temperature; K = a mathematical constant;
V1 = volume of gas at temperature T 1; V2 = volume of gas at
temperature T2 of gas at constant pressure.
3.3.2 Verification of Charles’ Law
We can verify Charles’ law in the laboratory using the apparatus shown
in Fig. 7.4 below.
S
Thermometer
t
Metre rule
Capillary tube
A
Vair
B
Heat
Fig. 7.4: Investigation of volume - temperature change (constant
pressure)
To do this verification, we trap a column AB of air by a length of
sulphuric acid which dries the air or by a length of mercury in a
capillary tube T of uniform diameter. The capillary tube is sealed at the
lower end and is attached to a halt-metre rule S by rubber bands.
The whole arragement is now placed inside a deep vessel filled with
enough water to cover most of the tube T as shown. The water is now
warmed with a burner, stirred gently with a thermometer suspended in
the vessel of water. After a suitable temperature rise the temperature t is
recorded as well as the length AB of the air colum from the reading as
the rule S. The length AB of the air column is proportional to the
volume V of the air. We can obtain more readings of t and V by heating
the water further to higher tempratures. The readings are as shown in
Table 7.2.
Table 7.2 : Readings of temperature and the corresponding volume
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t/oC
V
When values obtained in an earlier experiment were used to plot a graph
of volume V against temperature toC, using zero (0oC) on the
temperature axis, the resulting graph was as shown in Fig. 7.5 below.
Volume (V)
*
*
** *
* *
vo vt
W*
Z
273
0
80
Temp. 10C.
Fig. 7.5 :Volume – temperature change graph (constant pressure)
You can see from Fig. 7.5 above that we obtained a straight line graph.
Because we have performed this experiment at constant pressure equal
to the atmospheric pressure plus that due to the height of the liquid
column, we conclude that the volume of a fixed mass of gas increases
linearly as the temperature rises when the pressure is kept constant
If you are to perform this experiment, you have to take these
precaustions.
•
Take temperature when the thermometer reading is steady
•
Stir the water continuously to ensure even temperature.
•
Ensure that the bore of the capillary tube is clean and dry.
•
Use only dry air to gas.
•
Carry out the experiment slowly to allow heat time to pass
through the thick wals of the capillary tube and bring the gas to
the same temperature as the water in the vessel.
3.3.3 Cubic Expansivity of a Gas
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You can also use Charles’ law experiment to obtain the value of the
cubic expansivity of a gas at constant pressure. This is defined as:
Increase in volume per unit volume at 00C per degree Celsius rise in
temperature.
That is,
if
૪
=
cubic expansivity of a gas
Vo = volume at temperature 00C
Vt
= volume at temperature t oC
Then
૪
=
vt - vo
Vo x t
We have from careful experiments obtained the fraction ૪ in the
equation above to be same for all gases and is equal to
1
/273.
You know the cubic expansivity of a gas unlike that of solids and liquids
is the same irrespective of the nature of the gas. Also, the cubic
expansivity of a gas is calculated from the fractional increase of its
volume at 00C?
Worked example
1.
A fixed mass of gas is heated at contnat presure from 25 0C to
80oC. If the volume at 250C is 150 cm3, what volume will it
occupy at 800C.
Solution
From Charles’ law
V1
=
V2
T1
T2
V1 = 150 cm3 ; V2 = ?
T1
= (273 + 25 ) K = 298 K
T2 =
( 273 + 80 ) K = 353K
Substituting in the above equation gives us
150
=
V2
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298
353
150 x 353 cm3
298
=
177. 68 cm3
The volume to be occupies by the gas at constant pressure and at 800C is
177.68 cm3.
=
2.
V2
=
The volumes of a fixed mass of gas at constnat pressure are 150
cm3 at 00C and 178 cm3 at 800C. Calculate the cubic expansivity
of the gas .
Solution
You will recall that the cubic expansivity (૪) of a gas is given by the
equation
૪ =
Increase in volume from 00C
Volume at 00C x temperature rise
That is,
R
=
V1 - Vo
Vo x t
=
=
=
178 - 150
150 x 80
28
12000
2.33 X 10-3 /0C or 2.33 X 10-3/k
3.3.4 Use of Kinetic Molecular Theory to Explain Charles’
Law
To do this, let us consider a fixed mass of a gas confirned in a vessel at
constant pressure. Now, if we heat this gas, the molecules acquire more
K. E., move faster and collide more often with the walls of the vessel.
As a result, increases the pressure they exert. For us to maintain a
constnat pressure, we then increasae the volume of the containing vessel
to allow the same mass of gas to travel a longer distance before striking
the walls of the container. Thus an increase in temperautre leads to an
increase in the gas volume. This is in agreement with Charles’s law.
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Pressure Law or Gay- Lussac’s Law
Here, we will consider the connection between the pressure change with
temperature of a fixed mass of gas maintained at constant volume. This
connection was given by Gay-Lussac in 1802 and these are two
alternatives of stating the relation (law).
3.4.1 Statement of Pressure Law
•
We can now state the law in the two alternative ways thus:
The law states that the presure of a fixed mass of gas at constant volume
increase by ½73 of its pressure at 00C for every degress Celsius (or
Kelvin) rise in temperature.
We write the above in symbols as
Pt
=
Po 1 + t
273
where Po = pressure at temperature 00C
Pt = Pressure at temperature t0C.
The alternative way of stating the pressure law is:
•
The pressure of a fixed mass of gas at constant volume is
proportional to the absolute temperature of the gas.
In symbols, we state the law as P α
Or
That is, P1
T1
P
T
T
= constant
=
P2
T2
3.4.2 Verification of Pressure Law
Fig. 7.6 below shows one of the forms of apparatus we use to verify
Pressure law. It consists of a large bulb F filled with dry air; a mercury
manometer with one side BC open to the atmosphere. This is connected
to F by a very narrow tube D.
We place the bulbs F inside a large vessel and cover it completely
with water as shown in Fig. 7.6.
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Fig. 7.6: Pressure – temperature change apparatus (constant volume)
First, we record the initial temperature t of the gas. We then raise or
lower the open side BC of the manometer until the top of the mercury on
the other side reaches the level L. We then read and record the height h
of the mercury column above L and the atmospheric pressure A.
We warm the air in the bulb F by heating the water round the bulb and
stirring it gently. At a suitable temperature rise of between 15 to 200C,
we take away the burner and raise BC until the mercury level on the
other side again reaches L. We then read and record the new difference
in levels, h, between B and L; and the temperature t.
We repeat these experimental procedures with higher temperature for
the air. In each case, BC is raised until the mercury level on the other
side is at L. Thus, maintaining the volume of the air constant. We read
and record the corresponding values of h and t. These readings are as
shown in Table 7.3 below.
Table 7.3: Readings of temperature and the corresponding values of h
and p
t/0C
h / mm Hg
P ( = A + h or A – h)
Using the values from the table, we plot a graph of the pressure P
against temperature t at constant value. If B is above L, p = A + h but if
B is below L, P = A – L.
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In an earlier experiment, we obtained a straight line AB which showed
that at constant volume, the pressure of a gas varies uniformly or
linearly with temperature. A typical graph of an earlier experiment is
shown in Fig. 7.7 below.
Pressure (P)
*
B*
*
* * *R*
* *
Q* Po
Po
Pt*Pt
P
Z
0
40
temp. /0C
Fig. 7.7: Graph of pressure againt temperature (volume constant)
From the result of the above experiment, we can deduce that for all
permanent gases and dry air, the pressure increases by about 1/273 of its
pressure at 00C for each degree Celsius rise in temperature. Hence, the
pressure expansivity of gases is the same as the volume expansivity.
Therefore, we define the pressure expansivity of the gas as
૪ = Increase in pressure from 00C
Pressure at 00C x temperature rise
Symbolically,
૪
=
pt - po
Po x t
or Pt = Po 1 + t
273
Worked example
The pressure of a car type at 100C is 0.20N /mm2. What would be the
pressure if the temperature rises to 250C.
Solution
Recall that from Pressure law,
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P α T at constant volume
Therefore,
=
P1 = T1
P2
T2
0.20
P2
=
273 + 10 = 283
273 + 25
298
P2 = 0.20 x 298
283
=
0.21 N/ mm2
3.4.3 Use of Kinetic Molecular Theory to Explain Pressure
Law
Do you remember that heat is a form of energy and that when a gas is
heated the molecules gain kinetic energy and move about faster. As a
result, the momentum changes produced by the molecules at the walls
per impact is greater. This also leads to increase in the number of
impacts of each molecule per second because of the increase in speed.
Thus, the pressure increases with rise in temperature at content volume.
This is statement of Gay-Lussac’s law or Pressure law.
SELF ASSESSMENT EXERCISE 1
1
(a)
State and describe experiments to verify:
(i)
(ii)
(iii)
(b)
Boyle’s law
Charles’s law
Pressure law.
Use kinetic molecule theory to explain the above laws.
2.
Describe as fully as you can the kinetic picture of a gas and show
how its accounts for the change in pressure when the temperature
of a gas is increased at constant volume.
3.
The pressure of a fixed mass of gas at constant volume increase
from 60 cm of mercury at 100C to 95 cm of mercury at 800C.
What would be the pressure of the air at 00C?
4.
A fixed mass of air occupying 500 cm3 at 150C was heated to
850C. What was its new volume?
3.5
Ideal or the General Gas Law
3.5.1 The General Gas Law or Equation
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From the gas laws we know that the volume of a gas depends on both its
temperauture and are pressure. We can sum up the relationship between
these three variable – volume, temperature and pressure to obtain the
general gas law. We have that.
From Boyle’s law,
PV =
constnat if T is constant from,
From Charles’ law,
V
= constant if P is constant
T
From the Gay – Lussac’s or pressure law,
P
= constant at contact volume
T
We can combine any two of these three equations to obtain the general
gas equation.
That is,
PV
T
=
K
where K is a constant for a fixed mass of gas.
⇒
p1 V1
=
P 2 V2
T1
T2
Where
P1 V1 are the gas pressure and volume at temperature T1 and P2 V2are
the gas pressure and volume at temperature T2.
We can use the general gas equation to find any of the variable in the
equation when the other two are known. For instance, if we know the
initial volume of a gas, we can determine the final volume when both its
pressure and temprature changes.
That is,
V2
=
P1 V1 T2
P2 T1
You have to note that
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• The law does not hold good when a gas is near its liquefying
temperature.
•
The relationship ‘PV/T = constant’ is a law for an ideal gas. It
assumes that the attraction between the gas molecules and the
volume they occupy are both neglible. But, this is not the case for a
real gas.
3.5.2 The Ideal Gas Equation
The ideal gas equation relate generally presure, volume and absolute
temperature of a fixed mass of gas as we have already discussed in the
general gas equation of section 3.5.1 above .
That is,
PV
T
=
constant
However, for an ideal gas, at standard temperature and pressure, for 1
mole of gas we have
PV
=
RT
where R is a constant known as the molar gas constant and its value is
8.31J mol -1 K -1
The above equation is refrred to as the Ideal gas equaton. It holds true
for all gases.
For n moles, however, we have
Or
PV
PV
T
=
=
nRT
nR
The above equation is the general form and is called the equation of
states of an ideal gas.
3.6
Conversion to Standard Temperature and Pressure
(S.T.P)
You have seen that the volumes of gases changes remarkely with
changes in temperature and pressure. For this reason, it is not easy to
compare result of experiments between countrie or towns across the
world having temperate and tropical climates. This observation made
scientist to choose 00C or 273K and 760 mm Hg or 1.01 x 105 Nm-2 to be
the standard temperature and pressure (s. t. p. ) at which gas volumes are
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given. This makes it easy and possible for us to compare volumes of
different masses of a gas.
3.7
Intermolecular Energy and Forces
We know from common experience that forces exist between molecules.
These intermolecular forces arise from two main causes:the potential
energy (interactions with surrounding molecules and is electrical in
nature) and thermal energy (the kinetic energy which depends on the
temperature of the substance concerned) of the molecules. The
intermoleular forces between molecules decrease rapidly as the distance
between molecules increases.
Thus for small distances between molecules (solid or liquid) the
intermolecular forces are repulsive and increases rapidly as the
separation is reduced.
The forces between molecules vary in the form shown in Fig. 7.8
below . /I/*t shows the variation of the potential energy V between two
molecules at a distance r apart.
V+
F = - dV
dt
A
Repulsive
r < ro
O
Equilibrium
R = roRepulsive
M
r < ro
ro
B
Attractive
r > ro
D
C
X
G
Y
Fig. 7.8 : Molecular potential energy and force
At the point B, the intermolecular energy is zero. For distances shorter
than OB, the force is negative. Along the part AB, the P.E is positive.
The force between the molecules is always given by F = -dv/dr = potentital gradient. At distances greater than OB, the force is attractive.
It increases with distance, reaches a maximum and then decrease
rapidly. Along CD, the force is attractive and it decreases with distance
r according to an inverse – power of r. Along ABC, the force is
repulsive. This variation of F with r is illustrated in figure 7.8. At C, the
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minimum P.E point of the curve, the molecules are at their normal
distance apart in the absence of thermal energy. At this equilibrium
distance, the attractive and repulsive forces balance each other. If the
molecules are closer ( r < ro), they repel each other. If they are further
apart (r > ro), they attract each other.
For gases molecules, the molecules are far apart that the intermolecular
attractions are small. As the molecules are brought closer by
compressing the gas, the force of attraction increases, passes through the
maximum and then decreases to zero at B where it exists in liquid phase.
Small cohesive force exist in gases. The intermolecular attraction
between molecules of liquids is larger than that of gaes because they are
closer together. This makes liquid to have definite volume and higher
cohesion than gases.
The intermolecular forces between solid molecules are very strong and
are much more than that of liquids and gases. This makes them to have
definite shape their atoms and molecules have definite pattern or lattice.
Thus, solids are less compressible than liquids.
3.8
Pressure Calculations
Here, suppose a cubical box of side 1 contains N molecules of gas each
of mass M. A typical molecule of this gas will have a velocity C at any
instant with components of U, V and w respectively in the direction of
the three perpendicular axes 0x, 0y, 0z as shown in Fig. 7.9.
Hence C2 = U2 + V2 + W2.
Z
w
L
v
L
O
c
u
+
X
L
Fig. 7.9 :Calculation of gas pressure
Suppose a molecule of the gas in this box moves and collide with the
box on the face x due to the component of the velocity u.
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Momentum before impact = mu
Momentum after impact = -mu
Therefore, momentum change on impact
=
mu - ( - mu )
=
-2 mu
Time taken for the molecule to move across the cube to the opposite
face and back to x is 2l/u
Therefore
Momentum change per second
momentum change
time
2mu = mu2
21/u
l
=
=
force on x
=
mu2
l
You know that
Pressure on x = force
area
This implies that pressure on x
=
force
=
mu2
area
=
mu2 …….(i)
l x l2
l3
For N molecules in the cube, in the x direciton we have
P
=
mu21 + mu22 + mu23 + ……………+ mu2N
l3
l3
l3
l3
P
=
m ( u1 + u22 + u32 + ……….+ u2N) ……(ii)
l3
where
U12, U22, u32 …….U2N are components of velocity of various magnitudes
in the OX direction.
If U2 represents the average or mean value of all the squares of the
velocity components in nthe OX direction, that is,
U2
=
u12 + u22 + u32 + ………+ u2
N
Which implies that
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N u2 = u12 + u22 + u23 +………u2N
N
Then equation (ii) becomes
P = N mu2 ………………….(iii)
l3
Now, if we consider a larger mumber of molecules of varying speed in
randon motion, the mean square of the component speed in any of the
three axes is the same.
That is
2
U
2
V
=
2
w
=
For each molecule, however
C2 = u2 + v2 + w2
Thus, the mean square C-2 is given by
2
= U2 + V2
+
C
Therefore
2
2
=
U
1/3 C
From equation (iii)
P = 1 NmC
2
w
2
3 l3
The number of molecules per unit volume
n = N/l3
Hence,
P = 1/3 nmC-2 ………………..(i)
where n is in molecules per metre3, m in kilogram and c in metre per
second. Unit of pressure: newton per metre2 (Nm-2)
3.9
Introduction of Temperature
We can move a step further by considering the volume of gas containing
N molecules. The number of molecules per unit volume is
n
=
N
V
or
n
=
N
l3
Thus, the pressure of a gas from equ (1) becomes
P
=
1/3 nmC
2
= 1 N mc
2
3V
Therefore
PV
=
1/3 NmC
2 …………………….(2)
Equation (2) is similar to the equation combining Boyle’s law and
Charles’ law
PV =
RT
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We can by making kinetic theory consistent with the observed behaviour
of a gas, write
1/3 N mc-2
=
RT ……………….(3)
Here, we assume that the mean square speed of the molecules
proportional to the absolute temperature of the gas.
C
2
is
The kinetic energy of a molecule moving with a speed c is ½ MC2.
Thus, the kinetic energy of translation of the random motion of the
molecule of a gas is
½ mc2.
Relating ½ mc2 to the temperation enables us to put eqn (3) in the form
RT = 1/3 N mc2 = 2N
3
(1/2 mc-2)
where
½ mc2 =
3
R T…………(4)
2
N
Thus, the average kinetic energy of translation of a molecule is
proportional to the absolute temperature of the gas.
We note that the ratio R/N is a universal constant . R is the gas
constant .while N is the number of molecule.
R
N
=
K
where k is the gas constant per molecule and also universal constant
called Boltzmann’s constant.
Substituting K in eqn. (4) gives us
½ Mc2 =
3KT ……………………..(5)
2
Boltzmann’s constant relates energy to temperature and is usually
given in joules per degree.
K
=
½ mc2
3 T
2
The value of k is 1.38 x 10-23 Jk -1.
3.10 Deductions from the Ideal Gas Equation.
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Determination of the value of root mean square (r ms ) speed
( √C2 )
You will recall from equation (1) that
P
=
1/3 nm
C
2
…………….(i)
where n = number of molecules per unit volume ( N/V).
Nm = total mass M of the gas per volume
Substituting the above in equ (i)
gives us p = 1/3
m
C
2
…………………(6)
V
You know that density ρ is mass per unit volume. Therefore, when we
substitute ρ for m/v in eqn (6) we have
P = 1/3 e C-2 ………………….(7)
= 1 C2
3
Or p
ρ
which gives us
C-2
=
3p
ρ
…………..(8)
The square root of C 2 is called the root – mean – square speed.
That is
C-2
But
p
=
3p
ρ
=
RT
V
Hence root-mean –square (rms) speed
C
2
=
3RT …………………(9)
M
Since ρ = m/v and v = 1 unit volume
To find the value of the root-mean-square.
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C
2
, we subatitute
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known values of ρ and p for hydrogen at STP.
At STP ρ for hydrogen = 0.09 kgm-3
P = gρ H
g
= acceleration due to gravity 9.81 ms-2
ρ =
density of mercury = 13600 kgm3
H
=
barometer height = 760 mm = 0.76 m
Substituting the above in eqn (8) gives us
Where
C
2
=
3p
ρ
=
3xgρ H
ρ of hydrogen
=
3 x 9.81 x 13600 x 0.76
0.09
=
304188. 48
0.09
=
3.37 x 106 m2S-2
Therefore, the root mean square, C 2 is
C2
3.37 x 106
= 1838. 44
=
1.84 x 103 ms -1
Table 7.2 values of C 2 at 0oC for the common gases
ii
=
Gas
Hydrogen
Helium
Nitrogen
Oxygen
Carbon dioxide
Graham’s Law of Diffusion
C2
1.839 x 103mS-1
1.310 x 103 mS-1
0.493 x 103 mS-1
0.461 x 103 mS-1
0.392 x 103 mS-1
Graham, an English Chemist, in 1933 discovered that a less dense
gas diffuse through a porous material much faster than a denser
gas. He carried out an experiment to investigate this relationship
and the result of his investigtion is commonly referred to as
Graham’s law of diffusion.
Graham’s law of diffusion states that at constant temperature
and pressure the rate of diffusion of a gas through porous
material is inversely proportional to the square root of its
density.
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In symbols,
Rate of diffusion
iii
α
1
ρ
Avogadro’s law
Avogadro (1776 – 1856), an Italian professor put forward a
hypothesis that illuminated Chemistry as Newton illuminated
mechanics. This hypothesis he puts forward was accepted in the
middle of 19th century anad is now known as the Avogadro’s law.
Avogadro’s law states that equal volumes of all gases at the same
temperature and pressure contained equal number of molecules.
We recognize 1cm3 of gas at STP as containing 2.6 x 1019
molecules.
Avogadro,s law enables comparison of gas masses by chemical
methods.
iv
Daltons’ Law of Partial Pressure
According to Dalton in 1801, the total pressure of a mixture of
gases is the sum of the pressures which each individual gas
would exert if it were confined alone in the volume occupied by
the mixture. This statement first credited to Dalton is generally
referred to as Dalton’s law of Partial Pressure.
The law states that the total pressure of a mixture of gases is
equal to the sum of the pressures of the individual gases
assuming each gas occupies the volume of the mixture at the
same temperature.
We express the law mathematically as
Ptotal
=
where
PA + PB + PC ………….Pz
total = total pressure of the mixture and PA, PB, Pc ……..pz are the
partial pressures exerted separately by the individual gases A, B, C,
…………Z that make up the mixture.
v
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Maxwell’s Distribution Law
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You are aware that the speed of individual molecules vary over a wide
range of magnitude. Therefore, there is a characteristic distribution of
moleculsr speeds for a given gas which depends on the temperature.
Even gases which have the same initial molecular speed v later change
their speeds because of collisions. Many of these molecules do not have
speeds near zero. (ie << Vrms) or much more than Vrms (>> rms).
Because these extreme speeds require an unlikely sequence of
preferential collisions. C Maxwell solved this problem of distribution of
speeds in a large number of molecules of a gas in his law – Maxwell’s
Distribution law
Maxwell’s molecular speed ditribution law states that for a sample of
gas containing N molcules, that
NV = 4π N (m/ 2 π KT ) 3/2 V2 e –mv2kT
Where
N (v)dv is the number of molcules in the gas sample having speeds
between v and v + dv. That is
N
∫∞o N (v) dv
=
Unit molecules / ( cm/s)
T is the absolute temperature, k is boltmann’s constant and m is the
mass of a molecule.
Nv molecules per unit speed
intervals 103 molecules /(m/s)
For a given gas, the speed distribution depends only on temperature.
6.0
N = 106 oxygen molecules
5.0
4.0
Speed intels
3.0
T = 2000C (= 73K)
T = 00C (=273k)
2.0
1.0
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0
v speed, m/s
200
400
vp
600
800
1000 1200
Vrms
Fig. 7.10: The Maxwellian distribution of speed for oxygen molecules
Figure 7.10 above shows Maxwell distribution of speeds for molecules
of oxygen at two different temperatures. The area under the curve
repesents the number of molecules at speed between V1 and V2.
SELF ASSESSMENT EXERCISE 2
1.
State the equation of state of an ideal gas.
2.
Derive an expression for the root-mean-sqaure (rms) speed (C2)
State your derivation from the equation expressing pressure of a
gas as
P = 1/3 nmC 2
All symbole retain their usual meaning.
3.
State
(a)
(b)
(c)
(d)
4.0
Graham’s law of diffusion
Avogadro’s law
Dalton’s law of partial pressure
Maxwell’s distribution law
CONCLUSION
In this unit you learned about expansion of gases. You also learned
about the kinetic molecular theory of gas and its assumptions.
Furthermore, you learned about Boyle’s law, Charles’ law, Pressure law,
their verifications and explanation using kinetic molecular theory. In
adition you learned the general gas law, the ideal gas equation,
intermolecular energy and forces, pressure calculations and introduction
of temperature. Finally, you learned deductions
from the ideal gas equation – the value of root-mean-square (C 2) speed,
Graham’s law of diffusion, Avogadro’s law, Dalton’s law of partial
pressure and Maxwell’s distribution law.
5.0
SUMMARY
• Boyle’s law states that the pressure of a fixed mass of gas is
invensely proportional to its volume, provided that the temperature is
kept constant.
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•
•
•
•
P = 1 or PV = constant
V
Where
P
=
pressure
V
=
volume
•
of a fixed mass of gas.
•
Charles’ law states that the volume of a fixed mass of gas
increases by 1/273 of its volume at 00C per degree Celsius rise in
temperature, provided its presure remains constant.
•
OR. The volume of a fixed mass of gas is directly proportional to
its absolute temperature if its pressure remains constant.
•
V
=
k
T
Where V = volume, T = kelvin temperature
•
Pressure or Gay – Lussac’s law states that the pressure of a fixed
mass of gas at constnat volume increases by 1/273 of its pressure
at 00C for every degree Celsius (or kelvin) rise in temperature.
•
Pt = Po ( 1 + t/273).
•
Or The pressure of a fixed mass of gas at constant volume is
proportional to the absolute temperautre of the gas.
•
P
T
•
The General Gas Equation / law
•
PV
T
•
•
The ideal Gas Equation
PV = nRT
•
Root - mean - sqaure (rms) speed
•
•
•
•
=
constant
=
constant
C
Or
2
C
2
=
3RT/M
=
3P/ρ
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Graham’s law of diffusion states that at constant temperature and
pressure the rate of diffusion of a gas through porous material is
proportinal to the square root of its density
•
Rate of diffusion
α
1/ ρ
•
Avogadro’s law states that equal volume of all gases at the same
temperature and pressure contained equal number of molecules.
•
Dalton’s law of partial pressure states that the total pressure of a
mixture of gases is equal to the sum of the presures of the
individual gases assuming each gas occupies the volume of the
mixture at the same temperature.
•
Ptotal = PA + PB + PC +……….Pz
•
•
Maxwell’s molecular speed distribution law states that for a
sample of gas containing N molecules, that
Nv = 4π N (m/2 π Kt )3/2 V2 e-mv2kT
6.0
TUTOR -MARKED ASSIGNMENT
1.
State Boyle’s law and express the law mathematically. What are
the precautions to be taken when verifying this law in the
laboratory?
2
(a)
(b)
Mention five assumptions of the kinetic molecular theory
of gases
Use kinetic molecular theory to explain Charles’ law.
3.
(a)
(b)
Write down the ideal gas equation
A fixed mass of gas occupies 10-x m3 at S.t.P What
volume does it occupy at 270C if its pressure is 62 on of
mercury?
4.
Given the equation
p = 1/3 nmC-2
where
n = number of molecules per unit
nm = total mass M of the gas
per unit volume M of the gas
per unit volume
Show that the root mean square speed is given by the expression
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C2
7.0
3RT
Mo
REFERENCES/FURTHER READINGS
Anyaoha, M. W. (2000). New School Physics for Senior Secondary
Schools, Onitsha. Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics.
Ibadan University Press Plc.
Nelkon, M. (1986). Principle of Physics for Senior Secondary Schools.
MODULE 4
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Reflection of Light
Reflection at Curved (Spherical) Mirrors
Refraction of Light
Refraction at Curved Surfaces (Lenses)
Applications of Light Waves
Dispersion of Light and Colours
UNIT 1
REFLECTION OF LIGHT
CONTENTS
1.0
Introduction
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2.0
3.0
Objectives
Main Content
3.1
Concept of Light
3.1.1 Nature of Light
3.1.2 Sources of Light
3.1.3 Properties of Light
3.1.4 Transmission of Light
3. 1.5 Rays and Beams of Light
3.2
Rectilinear Propagation of Light
3.2.1 Shadows
3.2.2 Eclipses
3.2.3 The Pinhole Camera
3.3
Numerical Examples
3.4
Reflection of Light at Plane Surfaces
3.5
Laws of Reflection
3.6
Verification of Laws of Reflection
3.6.1 The Principle of Reversibility of Light
3.6.2 Formation of Images at Plane Mirror Surfaces
3.7
Locating Images in Plane Mirror
3.8
Method I
3.8.1 Method II
3.8.2 Image of Large Object – Lateral Inversion
3.8.3 Characteristics of Images Formed by Plane Mirror
3.8.4 Deviation of Reflected Ray by Rotated Mirror
3.9
Images Formed by Inclined Mirrors
3.10 Images Formed by Parallel Mirrors
3.11 Application of Reflection of Light at Plane Mirror
Surfaces
4.0
5.0
6.0
7.0
Conclusion
Summary
Tutor- Marked Assignment
References/ Further Readings
1.0
INTRODUCTION
In this Module, starting from this unit, you will be introduced to light
waves. Specifically, you will learn about the concept, nature, sources
and properties of light. Also, you will learn about transmission and
reflection of light at plane mirror surfaces, laws of reflection and
formation of images by plane mirrors. Furthermore, you will study the
deviation of light at plane mirror surfaces and of reflected ray by rotated
mirror. Finally, you will learn about the applications of reflected light at
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plane mirror surfaces especially in simple periscope, kaleidoscope and
mirror galvanometer.
2.0
OBJECTIVES
At the end of this unit you will be able to:
•
•
•
•
•
list some sources of light you come across in your local environment
determine the angle of reflection for a given angle of incidence
draw ray diagrams to show the formation of images by plane mirrors
explain some practical applications of plane mirrors
solve simple quantitative problem involving plane mirrors.
3.0
MAIN CONTENT
3.1
Concept of Light
Light (optics) is a form of energy which causes the sensation of vision.
This energy association with light is known as luminous energy and it is
the energy which causes the sensation of vision when it falls on our
eyes.
3.1.1
Nature of light
As you know, there are two views about the nature of light namely;
(i)
Light as particle (Corpuscular theory);
(ii)
Light as matters (Wave theory);
According to the corpuscular theory light consists of a stream of minute
weightless particles given off by the source. On the basis of this theory,
interference and diffraction of light cannot be explained while the
photoelectric effect could be explained satisfactorily by assuming that
light exists in discrete particles known as photons.
On the other hand, the wave theory considered light as a form of
disturbance spreading out from its source. On the basis of this theory,
the phenomena of interference and diffraction of light could be
satisfactorily explained while that of the photoelectric effect cannot be
explained by it.
On the basis of these experimental observations, we accept that light
sometimes behaves like waves and at others times likes particles. Thus
the wave properties of light are used to explain phenomenon such as
interference and diffraction while the particle nature of light is used to
explain the photoelectric effect. Hence, we speak of the wave – particle
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duality of light. These two aspects of light (wave and particle) cannot be
observed in a single experiment.
In recent times, however, the wave nature of light has been accepted as a
better option. In this regard, light is perceived to consist of waves – like
packets of energy called photons with a traveling speed of 3 x 108 ms-1
3.1.2
Sources of Light
Visible light is a form of energy which our eyes can detect. Visible light
is divided into two sources namely the luminous sources and the nonluminous sources.
(i)
Luminous sources: These are light sources that generate and emit
light by themselves (give out light of their own). Examples are
the sun, stars, electric lamp, burning flames, TV screen, fire –
flies, some deep – sea fishes and other artificial light sources. The
light they produce enter our eyes.
(ii)
Non - luminous sources: These are light sources that do not give
out light on their own. They depend on the natural or artificial
light sources to illuminate them. They reflect light from luminous
objects for them to be seen. Examples include moon, wood,
pages of book, road sign. Luminous sources illuminate them and
make them appear luminous.
3.1.3
Properties of Light
When light strikes an object, it may be, depending on the nature of the
material:
(i)
(ii)
Reflected, absorbed or transmitted;
Partly absorbed, party reflected or partly transmitted.
3.1.4 Transmission of Light
Opaque materials – They partly absorb and partly reflect light energy
but do not transmit light at the surface. Examples are wood, brick wall
or a sheet of cardboard.
Transparent Objects – They transmits most of the light energy falling on
it such that the object can be seen. It reflects and absorbs very little.
Examples are glass and clear water.
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Translucent Object – They allow light to energy to pass through them in
such a way that the object cannot be seen. Examples include glass sheet
used for certain windows and waxed sheet.
3.1.5
Rays and Beams of Light
The direction or path along which light energy travels is called a light
ray. They are indicated in diagrams by thin lines with arrow heads
which indicate the direction of travel of the light. A collection of rays is
called a beam. There are three types of beams namely:
(i)
(ii)
(iii)
Parallel beams – A parallel beam is one in which the rays of light
are parallel to one another. Examples of parallel rays are rays
from search light.
Convergent beam – A convergent beam is one in which all the
rays of light converge or meet at a point. Convergent beam can
be produced using hand lens.
Divergent Beam – A divergent beam is one in which all the rays
of light come from a point and spread out or diverge from that
source. Lights from lamps are examples of divergent beam.
(i) parallel beam
(ii) convergent beam
Fig. 1.1:
3.2
(iii) divergent beam
Rectilinear propagation of light
Rectilinear Propagation of Light
You know that rectilinear propagation of light simply means that light
travels mainly in a straight line. This is illustrated in Fig. 1.1.
Eye
(a)
Candle light can be seen by the eye
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Eye
(b)
Candle light cannot be seen by the eye
Fig. 1.2:
Rectilinear propagation of light
Two natural effects result from the rectilinear propagation of light.
These are the formulation of shadows and eclipses. Also, the principle
of operation of the pin – hook camera depends on the fact that light
travels in straight lines.
3.2.1
Shadows
A shadow is an area in which light rays from a source cannot reach. It
occurs because light travels in a straight line. Shadows are produced by
the obstruction of light by an opaque object. For instance, when light
rays arrives at an opaque obstacle (one that absorbs Light), the rays just
graze the edges of the obstacle and produce the outline of a shadow.
The kind of shadow obtained depends on the size of the luminous object
(source of light) sending out the rays.
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Shows how shadows are formed
Umbra
Small source
-ray –box lamp
Fig. 1.3:
Shadows (a) small source (b) Large source
In Fig. 1.3 a, we used a small light source (point source) of light to
produce a sharp shadow of an opaque solid ball B on a screen placed
behind B. The shadow is uniformly dark. When the lamp is replaced by
a large source such as a pearl lamp P the shadow produced is no longer
uniformly dark. This is illustrated in Fig. 1.3b. Light does not reach the
region of full darkness (umbra), the inner most shadow region. The edge
of the umbra corresponds to the rays NM and OP which graced B at the
edge. As we move away from the edge the darkness diminishes in
intensity. Here also, there is an outer light or grey area known as the
penumbra or partial shadow. In this region, the light is partially blocked
by the opaque object. At point in the screen, the ray PQX indicates that
X receives light from small part NP of the pearl lamp while point Y
receives light from another small part OQ. This is indicated by the ray
QRY. Thus the regions X and Z are in partial shadow (penumbra) as
shown by the grazing rays OMS and NRZ. Other points above S and R
on the screen are illuminated by the whole of the lamp and as such no
shadow is obtained there.
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Eclipses
Do you know that an eclipse is formed as a result of the shadow cast by
one heavenly body on another? That is, the eclipse of the sun and of the
moon are direct result of the formation of shadows caused by the
revolution of the Earth round the Sun. As we have earlier discussed, the
sun is a luminous object while the Moon and the Earth are both nonluminous objects. At certain times during these movements, the three
bodies will lie in a straight line. This results in the formation of
shadows.
(i)
Eclipse of the Sun
An eclipse of the sun occurs when the Moon, an opaque body, comes
between the Sun and the Earth. This is illustrated in Fig. 1.4.
b
a
d
e
c
(a)
(b)
Fig. 1.4: Eclipse of the sun
At this instance, the shadows of the Moon will be cast on the Earth’s
surface. People living in the part of the Earth directly in the Moon’s
shadows are cut off from the Sun’s view. They are in the Umbra region
of the Moon’s shadow. This experience, for those at the Umbra (Total
darkness) region of the Moon’s shadow is called an Eclipse of the Sun
or Solar Eclipse and it lasts for just a few minutes.
Those living at the penumbra (partial darkness) regions of the Moon’s
shadow experience partial Eclipse because at these regions, the sun is
only partially covered. The appearance of the sun from various parts of
the Earth is illustrated in Fig. 1.4b.
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A person at a, as shown in Fig. 1.4a, sees a total Eclipse, another at b
and d sees a partial Eclipse while another at c or e sees no eclipse.
Annular Eclipse
On some occasions, the Earth and the Moon may be in positions such
that the rays at the moon’s edge intersect before reaching the Earth (Fig.
1.5), thus forming an annular eclipse.
a
a
Moon
a
Anmilar eclipse
Fig. 1.5.: Annular eclipse
That is, a ring of light is formed around the shadow of the moon. An
observer in a position a of the Earth would experience the annular
eclipse as a ring of light round the moon.
(ii)
Eclipse of the Moon
An Eclipse of the moon occurs when the Earth comes directly between
the Sun and the moon. This is illustrated in Fig. 1.6.
Fig. 1.6:
Eclipse of the moon
The Earth, being an opaque body, casts a shadow on the moon. That is,
the Moon is in the Earth’s shadow. This leads to the formation of the
Eclipse of the Moon or Lunar Eclipse just like the Solar Eclipse, Lunar
eclipse can be total or partial. Also, since the Moon’s orbit round the
Earth does not lie in the same plane as that of the Earth’s orbit round the
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Sun, partial Eclipses of the Moon do not occur every time the Moon and
the Sun are on the opposite sides of the Earth.
3.2.3
The Pinhole Camera
Are you aware that the pinhole camera is the earliest camera? It makes
use of the fact that light travels in straight lines. This camera consists of
a light proof box, one end of which has a small hole made with a pin or
needle point.
The opposite end has a screen made of tracing paper or ground glass.
Light from an object in front of the pinhole passes through it and forms
an image on the screen. If the screen is replaced with a photographic
paper or film, a picture of an object can be taken.
You can make a simple pinhole camera by removing the cover from a
closed box such as the cardboard boxes used for packing sugar cubes or
bournvita tin. Remove one face of it and replace it with either semitransparent or translucent paper to serve as screen thereby sealing it
again from outside light. Make a pinhole at the centre of the face of the
box (tin) opposite the screen (semi-transparent or translucent paper).
Paint the inside of the box (tin) black so as to absorb stray light due to
reflection in order not to fall on the screen. Fig. 1.7 represents a simple
pinhole camera and how it forms images.
Fig. 1.7: Pinhole camera and images formed
In Fig. 1.7, OB represents the object in front of the camera. Light enters
through the pinhole P and an inverted image IM is formed on the screen.
If the hole is very small, the image will be very sharp, clear, but dim due
to limited amount of light passing through it. On the other hand, if the
whole is large, the image will be bright but blurred.
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If the screen is replaced by a photographic film or plate, the pinhole
camera can then be used to take photograph of still objects but it
requires a long time of exposure to enable the image to be developed on
the film or plate. In recent times, however, converging lens of
appropriate local length is used to bring all rays from many pinholes or
wide hole on an object to a single image by bringing all rays from a
point on an object to a single point image on the screen even when the
rays are incident on different parts of the lens. This gives rise to a sharp
image of the object on the screen.
Linear magnification produced by the pinhole camera
Definition
Magnification is the ratio of the size (or height) of the image to the size
(or height) of the object.
You can compare the size of an object and its image by using the ratio:
Size of image
Size of object
=Linear magnification
Or
Height of the image = Linear magnification
Height of the object
Or
Distance of the image from pinhole
Distance of the object from pinhole
= Linear magnification
Or
Length of camera
Distance of object from pinhole
= Linear magnification
Symbolically,
M
= hi = v
ho
u
Where
M
hi
ho
v
=
=
=
=
u
=
linear magnification
height of the image
height of the object
perpendicular distance of the image to the
pinhole
perpendicular distance of the object to the pinhole
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3.3
Numerical Examples
(i)
The height of an object placed at a distance of 30 cm from the
hole of a pinhole camera is 20 cm. If the length of the camera is
15 cm, calculate the magnification produced by the camera and
the height of the image.
Solution
Magnification
=
length of camera
distance of object from pinhole
=
height of image (hi)
height of object (ho)
Length of camera
Distance of object from pinhole
Height of image
Height of object
⇒
Magnification
= 15
30
M
=
hi
ho
=
1
2
⇒
=
hi
20
= 20
= 20
2
=15 cm
= 30 cm
= hi = hi
= ho = 20 cm
= 1 = 0.5
2
Also
hi
20 cm
i.e
(ii)
2hi
hi
= 10 cm
The distance between the pinhole and the screen is 8 cm,
calculate the minimum distance from the pinhole of a body 2.0 m
tall. If a full length picture of him is required on a photographic
plate of length 25.0 cm.
Solution
Magnification (m) = height of the image (hi)
Height of the object (ho)
=
380
distance of the Image from pinhole (v)
distance of the object from pinhole (u)
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Image distance
Object distance
Image height
Object height
v = 8cm
u = ?
hi = 25 cm
ho = 2.0 m = 200 cm
Magnification
=
⇒
hi
25 cm =
200 cm
v
ho
hi
ho
= v
u
u
8 cm
u
25u
u
= 200x 8
= 200 x 8
25
= 64 cm
SELF ASSESSMENT EXERCISE 1
(1)
(2)
List at least five light sources you come across in your immediate
environment.
With the aid of diagrams explain how total and partial eclipse of:
(i)
(ii)
(3)
Construct a simple pinhole camera using improvised box or
closed can. Use the camera to focus on a named object and
answer the following questions.
(i)
(ii)
(iii)
(4)
the Sun
the Moon
are formed
Is the image of the object erect?
How sharp is the image?
It is observed that the pinhole camera is not generally used
in taking photographs of objects. Why?
An object stands 15 m away from a pinhole camera which is 10
cm long if the height of the object is 25 m, what is the height of
its image?
3.4Reflection of Light at Plane Surfaces
Earlier in this unit, you have learned that we see non-luminous objects
because of the light they reflect. As you know, the light originates from
some luminous object such as the Sun, an electric lamp or other
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luminous sources. When light rays fall on the surfaced of an object, it is
absorbed, transmitted or reflected. Sometimes, however, a combination
of the above processes may occur. When light strikes on a surface, the
way it is reflected depends on the nature of the surface. There are two
types of reflections namely Regular and Diffused reflection resulting
from two types of surfaces namely smooth or polished surface and
rough or irregular surfaces. The two types of reflection are illustrated in
Fig. 1.8 below.
Incident ray of
light
Incident
Plane mirror Rays of light
(smooth
surface)
Reflected rays of light
(a)
Regular reflection
Paper (rough
surface)
Diffusely
reflected rays
of light
(b) Diffused reflection
Fig. 1.8: Regular and diffused reflection
Regular Reflection
When parallel beam of light strikes smooth or polished surfaces such as
plane mirror, the rays are reflected parallel to one another in one definite
direction from the point where they strike the surface as a result of the
nature of the surface. This is termed Regular Reflection (Fig.1.8a).
Diffused Reflection
When parallel beam of light strikes rough surfaces such as paper,
person’s face or clothing materials, the rays are scattered because of the
lack of smoothness or grain of the surface. Parallel rays of light incident
on such Surfaces are reflected in different directions. This reflection is
termed scattered, irregular or diffused reflection (Fig. 1.8b).
3.5Laws of reflection
There are two laws of reflection at plane surfaces. These are:
(i)
(ii)
382
The incident ray, the reflected ray and the normal at the point
of incidence all lie in the same plane.
The angle of incidence is equal to the angle of reflection.
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The laws are illustrated in Fig. 1.9
Incident ray
Normal
Reflected ray
Angle of
Incidence
i
r
Plane mirror
Point of
incident
Angle of
reflection
Fig.1.9: Laws of reflection
Plane mirror surfaces consist of plane glasses silvered at the back-Light
rays are reflected at the silvery at the back by preventing light rays
that strikes the glass from passing through the glass.
Plane mirrors are used in investing the laws of reflection in school
laboratories.
(xiii) Verification of the first law of reflection
The apparatus is set up as shown in Fig. 1.10.
Fig. 1.10: Verification of the first law of reflection
Direct the incident ray from a light source (ray box or candle light) to
traverse along the surface or the plane of the board. You will observe
that it is only when the incident ray is visible on the board that the
reflected ray will also be distinctly visible on the board. This shows that
the incident and reflected rays are in the same plane. Since the mirror
is placed perpendicular to the board, the normal to the mirror also lies
in the same plane as the incident and reflected rays.
(xiv) Verification of the Second Law of Reflection
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Using Optical pins
Fix a sheet of paper firmly on a drawing board. Mount a plane
mirror vertically along a line AB on the paper and support it with
plasticine or wooden block (Fig. 1.11a).
Eye
Fig. 1.11a:
Verification of the second law of reflection using optical
pins
Draw a normal ON. That is a line ON perpendicular to the mirror at N to
biset AB at right angles. Draw lines (one at a time) on the paper making
angles such as 200, 300, 400, 600 and 700 to O N.
Fix two pins P and Q in front of the mirror along the first line making
angle 200 to ON (see Fig. 1.11a). Use two other pins RS to fix at points
in front of the mirror where the eye sees them in line with the images P1
and Q1 of P and Q in the mirror. The line PQ represents the reflected
ray.
Remove the pins and produce lines PQ and RS to meet at a point N on
AB measure the angle of incidence ί (the angle between the incident ray
and the normal) and the angle of reflection r (the angle between the
reflected ray and the normal).
Repeat the experiment for angles 300, 400, 600, and 700, measure and
record the angles of incidence ί and the corresponding angles off
reflection r in each case. Tabulate your result.
You will observe that each set of angles, angle of incidence ί and the
corresponding angle of reflection r, are always equal.
-
384
Using a Ray Box
set up the apparent as before (Fig.1.11a)
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Fig. 1.11b: Verification of the second law of reflection using a ray box
Place a ray box in such a way that a single ray follows the drawn line for
angle 200, line AN. Mark the reflected ray BN with two dots. Use the
edge of the incident ray and the corresponding edge of the reflected ray
if the ray is thick to obtain accurate result. Draw other lines for angles
300, 400, 600 and 700 in turn, using the ray box, direct the ray (incident
ray) along the drawn lines and mark each reflected ray direction as you
did for angle 200.
Measure and record the angles of incidence l and the corresponding
angles of reflection γ. tabulate your readings.
You will also observe here that in all cases, the angle of incidence is
equal to the angle of reflection r.
Note that the angles of incidence and reflection are always measured
from the normal ON (perpendicular) to the mirror.
3.5.1
The Principles of Reversibility of Light
You must have observed in the experimental verification that a light ray
directed to wards a mirror say AN will be reflected along BN in
accordance with the laws of reflection of light. This observation is
illustrated in Fig. 1.12. If light ray is incident along BN, it will emerge
along AN after reflection.
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Fig. 1.12: Principle of reversibility of light
Thus, the path of the light ray is seen to be reversible. This observation
is a general principle in optics and is known as the principle of
reversibility of light.
Definition
The principle states that the path of light ray is reversible It therefore
follows that if a person standing at point A can see the image of an
person standing at point B, the person at point B, can likewise see
the image of the person at A.
3.6
Formation of Images at Plane Mirror Surfaces
If you look at the front of a plane mirror, you will see the image of
yourself. Your image appears to be behind the mirror but it is not
actually there, it cannot be caught in a screen.
Now consider a situation where a candle light is placed in front of a
plane mirror to serve as an object (Fig. 1.13).
Fig. 1.13: Image of an object O observed in a plane mirror
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The upright image of the candle light appears to an observer as if it is
located at I behind the mirror from which the light rays enter the eye.
This observation can be explained thus: Two incident rays OA and OC
from the tip O, of the candle light strike the plane mirror at A and C and
are reflected as AB and CD respectively in accordance to the laws of
reflection. After reflection, the rays appear to the observer’s eye as if the
rays are coming from the tip of the image I of the candle light in the
mirror. You can see that I is the point of intersection of the two reflected
rays produced backwards. Also a ray ON normal to the mirror is
reflected back along NO. This ray meets the other two rays at I when
produced backward and the distance ON is equal to NI.
Actually, no rays come from I because AB and CD are diverging from
each other according to the laws of reflection. The image of the candle
light in the mirror is said to be a virtual image and cannot be received on
a screen.
•
•
•
•
An image is a picture of an object formed by a reflecting surface.
A real image can be received on a screen.
It is produced by actual intersection of rays of light.
A virtual image cannot be received on a screen it is produced by
apparent intersection of rays of light.
3.7
Locating Images in Plane Mirrors
3.7.1
Method I
The image of an object O such as a pin in a plane mirror M can be
located as described below:
(i)
Mount a plane mirror on a flat board and erect an optical pin O in
front of the mirror. View the image of the object pin O in the
mirror (Fig. 1.14)
Erect two other pins A and B in line with the image I of the pin
observed in the mirror. Repeat with two more pins in a different
position such as C and D. After taking the pins away, produce the
lines AB and CD until they intersect. This point of intersection
behind the mirror is the position of the image I. Measure the
distances ON and IN. You would see that ON is equal to IN
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I
image
N
D
A
CO
Object B
Fig. 1.14: Locating images in plane mirrors
3.7.2
Method II
It is also possible to locate the image position by the no – parallax 1
method. You can do this by using two optical pins P1, and P2 mounted
on a cork as shown in Fig. 1.15.
P2
Search pin
image
P1
Object
Fig. 1.15: No parallax method
3.7.3
Image of Large Object –Lateral Inversion
The image of large object is drawn by taking one point after another on
the object and locating its corresponding image using either method I or
Method II above.
Lateral Inversion
This is the effect of plane mirror on large objects placed in front of it.
The image appears like a reversal of the object (Fig. 1.16).
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A
Mirror
B
AB
Fig. 1.16: Image of large object showing lateral inversion
Lateral inversion is due to the fact that the object is perpendicularly
opposite its image behind the mirror.
3.7.4
Characteristics of Images Formed by Plane Mirror
From the forgoing you must have noted the following characteristics of
images formed by plane mirror.
(i)
(ii)
(iii)
(iv)
(v)
It is the same size as the object;
Each part is the same distance behind the mirror as the object is
in front of it;
It is laterally inverted (ie turned around);
It is erect (ie upright)
It is virtual
3.8
Deviation of Reflected Ray by Rotated Mirror
If a mirror is turned through an angle, the reflect ray of any light
incident on it is turned through twice that angle. That is, if the direction
of an incident ray on a mirror is kept constant and the mirror is rotated
through an angle, the reflected ray will be rotated through an angle, the
reflected ray will be rotated through twice that angle. This is illustrated
in Fig. 1.17. The mirror initially in the position MM is rotated through
an angle θ to the position M1M1
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The incident ray AP is normal to MM and is reflected back along its
own path. When the mirror is rotated to the position M1M1, the normal
to the mirror becomes PN at that point of incidence P. This makes an
angle of θ with PA. Thus the angle of incidence on M1M1 is θ as well as
the angle of reflection. Hence, the direction PA to the direction PB
through an angle of rotation equal to 2θ.
3.9
Images Formed by Inclined Mirrors
The number of images formed by inclined mirrors depends on the angle
of inclination of the mirrors. For instance, when two mirrors are
inclined at a angle θ, the number of images, n, formed by the two
mirrors is given by the formula.
n
Where
and
3600
n
=
=
3600 -1
θ
n = number of images formed;
θ =
angle between the two mirrors;
is always approximated or rounded up to the nearest
higher whole number when a fraction is obtained.
For example, if two mirrors are inclined at an angle of 900, then, the
number of images formed by the mirrors is
n
= 3600 -1
= 3600 -1
i.e (4-1) images = 3 images
This is illustrated in Fig. 1.18.
Fig. 1.18: Images formed in two mirrors inclined at right angles
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As illustrated in Fig. 1.18 above, two mirrors are inclined at right angles
to each other and a candle stick is placed between than. As you can
see, the number of images formed is three and these are:
(a) The image (I1) of the candle formed by mirror AC;
(b) The image (I2) of the candle formed by mirror BC;
(c) The coincident Image (I3), which is a combination of the images of I1
and I2 formed by both mirrors.
3.10
Images Formed by Parallel Mirrors
Have you ever noticed that when two mirrors A and B are placed
parallel to each other and an object O is place between then that multiple
images are formed? This is illustrated in Fig. 1.19.
B
A
X2
X1
O
Y1
Y2
Fig. 1.19: Multiple images by two parallel mirrors
As illustrated, when an object O is placed between two parallel mirrors
A and B, multiple images of the object will be formed. In the Fig. 1.19,
X1 is the image of the object O formed by A and Y1 is the image of the
object O formed by B. X2 is the image of image Y1 formed by mirror A
and similarly, Y2 is the image of image X1 formed by mirror B. That is,
as the images act as objects for further formation of images in the
mirrors, an infinite number of images are formed by the parallel mirrors.
However, as the number of images increases, the images become fainter
because a little light is diffused and absorbed at each reflection.
.
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3.11
Applications of Reflection of Light at Plane Mirror
Surfaces
Some of the applications of the reflection of light at plane mirror
surfaces are:
(i)
(ii)
(iii)
(iv)
Driving mirrors – Plane mirrors are used as driving mirrors in
vehicles
Looking glass - The common looking glass is a plane mirror.
The Sextant – This is an instrument used by navigators for
finding the angle of elevation of the Sun and other celestial
bodies it depends on
the constant deviation of rays striking
two mirrors inclined at a fixed
angle.
Kaleidoscope – This is a toy which makes use of two plane
mirrors inclined at an angle of 6o0 to each other. It produces
different image patterns each time it is shaken. Designers use it in
looking for new color pattern.
Periscope
The instrument consists of two mirrors fixed facing each other at an
angle of 450 to the line joining them. A ray from an object enters the eye
of an observes by successive reflections from the mirrors A and B. This
enables the observers to see an image of the object.
A
Object
Plane
mirror
B
Eye
Fig. 1.20: A simple periscope
Simple periscope is used in submarine to enable men below the water
surface see what is happening above the water surface. It is also used to
see over peoples head or spy over a tall wall or around corners. That is,
it enables the use to see over obstacles.
(vi)
Uses of Plane Mirrors in Meters
A plane mirror placed below the pointer in the same plane as the scale of
commercial meters helps to overcomes errors in reading due to parallax
between the point and the scale
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A plane mirror is also used to measure the small angle of rotation of the
coil in a mirror galvanometers used for the measurement of very small
electric current.
SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.
With the aid of the ray diagrams, distinguish between regular and
diffuse reflections.
State the laws of reflection. How would you verify the second
law in the laboratory?
With the aid of ray diagram, illustrate the formation of image by
a plane mirror. What are the characteristics of images formed by
such mirrors?
The angle between a plane mirror and an incident ray is
400. What is the;
(i)
angle of incidence
(ii)
angle of reflection
(ii)
total angle through which the ray is deviated?
5.
Explain why
(i)
You cannot see the image of yourself in this papers page.
(ii)
Inside a barber’s shop, Ada, observed the image of her
new hairstyle formed by a plane mirror kept 2 m away
from her eyes.
4.0
CONCLUSION
In this unit you studied reflection of light at plane surfaces. You learned
about sources of light, shadows, eclipses and the pinhole camera. You
also learned how to draw ray diagrams to show the formation of images
by plane mirrors as well as the laws of reflection. In addition, you
learned about the deviation of light at plane mirror surfaces and of
reflected ray by rotated mirror and how to solve quantitative problems
involving plane mirrors. Finally, you learned some practical applications
of plane mirrors.
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5.0
SUMMARY
•
Light is perceived to consist of wave like packets of energy
called photons with a traveling speed of 3x108 ms-1
Visible light is a form of energy which our eyes can detect. It is
divided into two sources namely the luminous source and the
non–luminous sources.
Luminous sources generate and emit light by themselves. Non
luminous sources do not give out light on their own but depend
on natural or artificial light sources to illuminate them.
When light strikes an object, it may be, depending on the nature
of the material:
•
•
•
(i)
(ii)
•
•
•
A ray is the path taken by light energy. A beam is a collection of
rays.
Shadows and eclipse occur because light travels mainly in a
straight line.
The laws of reflection are:
(i)
(ii)
•
•
•
•
•
•
394
reflected, absorbed or transmitted;
partly absorbed, partly reflected or partly transmitted.
The incident ray, the reflected and the normal at the point
of incidence all lie in the same plane.
The angle of incidence is equal to the angle of reflection.
An image is a picture of an object formed by a reflecting surface
A real image can be received on a screen and is produced by
actual intersection of rays of light.
A virtual image cannot be received on a screen and is produced
by apparent intersection of rays of light.
The image formed in a plane mirror is virtual, laterally inverted,
the same size as the object and as far behind the mirror as the
object is in front.
The number of images formed by inclined mirrors depends on the
angle of inclination of the mirrors
When two mirrors are inclined to each other at an angle θ, the
number of images in formed by the two mirrors is given by the
formula.
n
=
360o -1
θ
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6.0
TUTOR-MARKED ASSIGNMENT
1.
The heights of an object placed at a distance of 30 cm from the
hole of a pinhole camera is 20 cm. If the length of the Camera is
15 cm, calculate the magnification produced by the Camera and
the height of image
2.
State the lens of reflection of light and illustrate the laws with the
aid of a diagram.
3.
An object is placed 5 cm in front of a plane mirror. With the aid
of ray diagram, locate and state the characteristics of the image
formed.
7.0
REFERENCES/ FURTHER READINGS
Anyakola, N.W (2000). New School Physics for Senior Secondary
Schools, Onitsha: Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics
Ibadan: University Press.
Nelkon, M. (1999). Principles of Physics for Senior Secondary Schools.
Ibadan: Longman Group Limited.
Okpala, P.N. (1990). Physics, Certificate Year Series for Senior
Secondary Schools. Ibadan: NPS Education.
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REFLECTION
MIRRORS
AT
CURVED
(SPHERICAL)
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Types of Curved Mirror
3.1.1 Terms Used in Curved Mirrors
3.1.2 Principal Rays for Locating Images in a Curved
Mirror
3.2
Images Formed by Curved Mirrors
3.2.1 Images Formed by Concave Mirror
3.2.2 Images Formed by Convex Mirror
3.2.3 Distinction between Real and Virtual Images
3.3
Uses of Curved Mirrors
3.4
Measurement of Focal Length of a Concave Mirror
3.4.1 Quick but Approximate Method
3.4.2 The Mirror Equation Method
3.5
The Mirror Formula and Sign Conventions
3.5.1 The Sign Convention
3.5.2 Mirror Formulae
3.6
Defects of Spherical Mirrors
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit you will focus attention on curved mirrors and learn about
formation of images by curved mirrors. Also, you will, learn how to
draw ray diagrams to show how images are formed by concave mirrors,
use ray diagrams to locate image position at different object positions.
Furthermore, you will learn practical applications of curved mirrors
especially in car headlamps and driving mirrors. Finally you will learn
about the sign convention, the mirror formula and its application.
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OBJECTIVES
By the time you finish this unit, you will be able to:
•
•
•
•
Draw ray diagrams to show the formation of images by curved
mirrors;
Explain some practical applications of curved mirrors;
Locate image positions by using curved mirror;
Investigate the relationship between object distance (u), image
distance (v) and focal length (f) of the curved mirror.
3.0
MAIN CONTENT
3.1
Types of Curved Mirror
We have two types of curved mirrors namely the concave and the
convex mirrors. They are made silvering a glass surface which is part of
a sphere.
If the outside surface of the spherical part is silvered and the inside
surface of the reflecting part, the resulting mirror is called a concave
mirror. On the other hand, if the inside surface is silvered and the
outside surface is the reflecting part, the resulting mirror is called
convex mirror. This is illustrated in Fig. 2.1
Reflecting
surface
B
Principal
axis
F
c
Reflecting
Surface
B
p
P
v
Principal
Axis
f
c
v
A
Silvered
surface
A
Silvered
Surface
Fig. 2.1: Curved mirrors
(a) Concave or converging mirror
(b) Convex or diverging mirror
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Terms Used in Curved Mirrors
We use the following terms illustrated in Fig. 2.1 in describing the
essential parts of a spherical mirror.
•
•
•
•
•
•
•
•
The aperture: the width (AB) of the mirror
The pole (p): the centre of the reflection surface of the curved
mirror.
The Centre of Curvature (c) = the centre of the sphere of which
the mirror forms a part
The Radius of curvature (r) = the distance CP. It is the radius of
the sphere of which the mirror forms a part.
The Principle Axis: the line PC from the pole to the centre of
curvature.
The Principal Focus (f): The principal focus (f) of a curved mirror
is that point on the principal axis to which incident rays parallel
and close to the principal axis converge or from which they
appear to diverge after reflection.
The Principal Focus of a concave mirror is said to be a real focus
because the reflected rays pass through it while that of a convex
mirror is a virtual focus because the reflected rays do not pass
through it. This is illustrated in Fig. 2.2.
The Focal Length (f): the distance (f-p) from the principal focus
to the pole. It is half the radius of curvature. That is f = r/2.
Incident parallel rays
c
f
Reflected
converging rays
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p
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(a)
Concave mirror
Incident parallel
rays
p
f
c
Reflected
diverging rays
Fig. 2.2: Principal focus of concave and convex mirrors
Principal rays for locating images in a curved mirror
When ever you want to construct ray diagrams, you will use the
following principal rays to locate the image position of your object.
(i)
For a concave mirror
Rule1
•
Incident rays parallel to the principal axis are reflected through
the principal focus.
Rule1
C
P
F
Rule21 Incident rays passing through centre of curvature meet the mirror
at right angle and so are reflected back along their own paths.
Rule2
p
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Rule3: Incident rays passing through the principal focus are reflected
parallel to the principal axis.
c
p
f
Any two of the rays are sufficient for you to locate an angle.
(ii)
For a convex mirror
Rule1: Incident rays parallel to the principal axis are reflected so that
they appear to come from the principal focus.
F
P
P
Rule2: Incident rays traveling towards the centre of curvature meet the
mirror at right angles and so are reflected back along their own paths.
F
3.2
C
Image Formed by Curved Mirrors
The size, nature and position of an image formed by a curved mirror
depend on the position of the object from the pole of the mirror. You
can study there by the use of an illuminated object, a concave mirror
and a screen.
To do this, you will place the concave mirror on a stand and allow light
from the illuminated object to fall centrally on it by the adjustment of
the height of the mirror. The mirror is slightly turned to one side so as to
form an image where possible on the screen. You can do this for various
object positions.
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If you use a convex mirror, you will obtain an erect and virtual image
smaller than the object position.
Also, you can always obtain the images formed by curved mirrors by
scaled ray diagrams. To do this, you will make use of the ‘principal
rays’ discussed in 3.1.3. The point I to which these reflected rays
intersect represents the required image of O, the top of the object lies on
the axis, since the ray XCP strikes the mirror normally at p and is
reflected straight back. This is illustrated in Fig. 2.3 for concave mirror
and 3.4 for convex mirror.
3.2.1
(a)
Images Formed by Concave Mirror
Object beyond c
B
O
I
P
X
C
F
E
Image is: between F and C
 Real
 Inverted
 Diminished
(b)
Object at C
O
B
F
C
X
I
Image is:



P
D
at C
Real
Inverted
Same size as object
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Object between C and F
O
B
C
F
x
P
I
D
Image is:
•
•
•
•
Beyond c
Real
Inverted
Magnified
(d)
Object at F
E
O
C
F
Image is:
p
X
•
•
•
•
at infinity
Real
Inverted
Magnified
(e)
Object between f and p
F
I
E
Image is:
•
•
•
•
B
O
behind the mirror
virtual
erect
magnified
C
f
X
B
p
Fig. 2.3: Images formed by concave mirror
You will observe that in the ray diagrams, OB represents rays parallel
and close to the principal axis (rilr1) which after reflection passes
through f.
Ray OC passes through the centre of curvature C (rule2) and is reflected
back on its on path.
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Images Formed by Convex Mirror
-*-*Fig. 2.4: Image formed by concave mirror
From Fig. 2.4 you see that convex mirror produced only virtual erect
and diminished images. The image is formed at the intersection of the
reflection rays produced backwards (in broken lines) and is formed
behind the mirror (virtual).This is the case for all positions of the object
in front of the mirror.
3.2.3 Distinction between Real and Virtual Images
You will now note the following distinction between real and virtual
images.
3.3
Real Image
Virtual Image
(a) can be received on a
Screen.
(b) Produced by actual
inter
-section of rays of
light
Cannot be received on a
Screen
Produced by apparent
intersection of rays of
light.
Uses (Applications ) of Curved Mirrors
Here, we are going to discuss some practical applications of images
formed by curved mirrors at various objects by curved mirrors at various
object positions.
•
Object between F and P of a concave mirror
The concave mirror, at this object position is used as dentist
mirror, make – up mirror and showing mirror because the image
formed in this position of object is enlarged, erect and virtual.
•
Object at f of a concave mirror
At the object position, the concave mirror is used for torch. Also,
parabolic (not spherical) concave mirrors are used in car
headlamps and searchlight. They produce a wide parallel beam of
light of constant intensity when a small light source is placed at
its focus. Spherical concave mirrors are not used because they do
not provide parallel beam of constant intensity (Fig. 2.5).
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Parabolic mirror
Parallel beam
of light
Fig. 2.5: Parabolic mirror
•
Object between f and c
At the object position, concave mirrors are used in some kind of
flood light
•
Object at c
Concave mirror is used as projector lamp in this object position.
•
Concave mirrors are also used as reflectors in reflecting telescope
and microscopes.
•
Convex mirrors are used as driving mirrors because they always
give an erect image of an object behind the driver. In addition,
they provide a wider field of view than a plane mirror of the same
aperture. This makes it possible for the driver to see objects with
in a large angle.
Wide Field of view
Eye
Convex mirror
Fig. 2.6: Field of view with a convex driving mirror
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SELF ASSESSMENT EXERCISE 1
(1)
With the aid of ray diagrams, illustrate the characteristics of
image formed by a concave mirror when an object is placed
(i)
Between F and P
(ii)
at F,
(i)
between F and C
(ii)
at C;
(iii) beyond C, and
(iv) at infinity
(2)
In what situations would the image formed by a convex mirror be
useful?
Distinguish between real and virtual images.
What is meant by the radius of curvature and the principal focus
of a concave mirror?
(3)
(4)
3.4
Measurement of Focal Length of a Concave Mirror
3.4.1
Quick but Approximate Method
A simple and quick method of finding an approximate value of the focal
length of a concave mirror is to focus the image of a distant object on a
screen. To do this, you will hold a concave mirror with its reflecting
surface facing a distant object such as a widow pane at an angle (Fig.
2.7). You will also hold a screen, may be a piece of cardboard sheet.
Move the screen in front of the mirror until the sharpest image of the
mirror window forms on it. Measure the mirror distance between the
screen and the mirror. This value is the approximate value of the focal
length of the mirror.
Fig. 2.7: Focal length by quick and approximate method
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Since the mirror is at a considerable distance away from the windows,
the rays from it may be considered as parallel rays when they strike the
mirror. They will therefore come to a focus after reflection at the
principal focus.
3.4.2
The Mirror Equation Method
In this method you will make use of the relationship between the object
distance u, the image distance v and the focal length of a spherical
mirror the relationship is expressed in the equation.
I +
U
I
V
= I
F
- - - - - - -- -
(1)
In using this equation, you will first obtain the approximate focal length
of the concave mirror using the method in paragraph 3.4.1 above. You
will need an illuminated object (the cross wire of a ray box or a candle
stick), a screen, and a concave mirror. Set up the apparatus as shown in
Fig. 2.8
Fig. 2.8: Focal length by the mirror equation method
Place the illuminated object at distance of about 3/2 f in front of the
mirror and adjust the screen position until a sharp image is obtained.
Measure and record the object distance (distance from the object to the
concave mirror), u and the image distance (distance from the screen to
the concave mirror), v. Repeat the produce for at least six object
distance, u, and obtain the corresponding values of image distance, v.
Tabulate your readings as in the Table 2.1 below.
Table 2.1: Values of u and the corresponding values of v
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u
1
2
v
uv U+V uv
u+v
U
V
1/u 1/v 1/u +1/
v
1
2
3
3
4
4
5
6
5
6
Table B.
Table A.
You can now obtain the value of F from your readings in Table 2.1 in
the following ways.
•
Add the values of uv / u+v in Table A and find the average. The
result is the experimental value of f.
•
Add the values of
1/u +
1/v in Table B and find the
average. The result is 1/f from which you will the determined
the value of F. F is the reciprocal od the value btained.
•
Plot a graph of uv against u + v You will obtain a straight
line graph. The slope of this graph is equivalent to f, see Fig. 2.9
(a)
•
Plot a graph of 1/v against 1/u.You will obtain a graph with
intercepts on both axes. Each of the intercepts gives you a value
of 1/f. From this value, find the reciprocal to obtain the value. of
F (Fig. 2.9b).
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uv(cm)
Slope
= f
(u+ v)cm
Fig. 2.9(a): Graph of uv against (u + v)
1 cm-1
v
1/f
1
1/ucm -1
F
Fig. 2.9(b): Graph of 1/v against 1/u.
3.5
The Mirror Formula and Sign Conventions
You can solve any problem on spherical mirror of small aperture by
drawing to scale ray diagrams as in paragraph 3.2 above.
You can also use an equally valid method, the algebraic formulae which
connect the object distance u, image distance v, focal length f, and
magnification m.
For a concave mirror,
1/v + 1/u
= 1/f
- - - - - - (1)
from experimental determination of focal length it was found that
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1/v + 1/u = 1/f
Also the magnification
m = v/u ----------- (2)
For us to extend the usefulness of the above equations to virtual images
as well as to convex mirrors, it is necessary to adopt what is called a
sign convention. The sign convention tells us when to add a position (+)
or a negative (-) sign to each of the distances. Two commonly used type
sign conventions are presented below for you to study and use.
3.5.1 The Sign Convention
Real is positive
New Cartesian
• Real objects and real images •
are considered to be a
positive distance from the
mirror.
•
• Virtual images are at a
negative distance from the •
mirror.
Distances measured to the right
of the mirror are positive.
Distances measured to the left of
the mirror are negative.
Focal length of convex mirror is
negative
• Focal length of a concave •
mirror is positive
Focal length of a convex mirror
is positive.
• Focal length of a convex •
mirror is negative
Distances are measures from the
Pole of the mirror. The mirror is
positioned with its face to the
left.
3.5.2 Mirror Formulae
The mirror equations are:
(i)
1/u + 1/v = 1/f (Real is positive sign convention)
or
1/v - 1/u = 1/f (New Cartesian sign convention)
Where
u
v
f
(ii)
= object distance from the mirror
= object distance from the mirror
= focal length of the mirror.
m = height of image
= v/u
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height of object
where
M
(iii)
f
=
Linear magnification
= r/2
Where
r
=
Radius of curvature of mirror
f
=
Focal length of mirror.
Application of the Mirror Formulae in Solving Numerical Problems
Work Examples
(1)
An object 3 cm tall is placed 12 cm from a concave mirror of
focal length 8 cm. find the image position and its height.
Solution
u = 12, f = 8.
Using the mirror formula
1/u + 1/v = 1/f (Real is positive convention).
Substituting in the above equation gives
1/12 + 1/v = 1/8
1/v = 1/8 - 1/12
1/v = 1/24
v
= 24
The image position is 24 cm.
Magnification
M= v/u = 24/ 12 = 2
Height of the image = (3x2) cm
(2)
An Object is magnified five times by a concave mirror of radius
of curvature 30 cm. What is the object position if the image is (a)
real and (b) virtual?
Solution
(a)
410
= 6 cm.
When the image is real
M = v/u = 5/1
If u = x
V = 5x (image is real).
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Using the mirror formula:
1/u + 1/v = 1/f
r = 30cm
⇒ f = ૪/2 = 30/2 = 15cm
Substituting in the formula gives
1/x + 1/5x = 1/15
5+1 = 1 .
5x
15
6 / 5x = 1 /15
5x
= 6 x 15 = 90
x
= 90 / 5 = 18 cm.
The object position when the image is real is 18 cm.
(b)
When the image is virtual
m
f
= v / u = 5 /1
= r /2 = 30 / 2 = 15cm
If
u =x
V = -5x (image is virtual).
Substituting the values in the mirror formula we have
1/x
- 1 / 5x = 1/ 15
5+1 = 1 .
5x
15
4/5x
= 1/15
5x
= 4 x 15 = 60
x
= 60/ 5 = 12cm
The object position when the image is virtual is 12 cm.
3.6
Defects of Spherical Mirrors
You will observe spherical aberration when the aperture of a spherical
mirror is large. You will notice that a beam of light parallel to the axis
does not converge to the principal focus. Such a mirror brings rays close
to the principal axis to a focus further away from the pole of the mirror
than the focus. In addition, it is found that the reflected rays are tangents
to a curve known as a caustic curve. The inability of such spherical
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mirror of wide aperture to bring all parallel rays to the same focus is
known as spherical aberration. In order to avoid this phenomenon,
spherical mirrors of small apertures are usually used. Also, parabolic
mirrors, where all parallel rays are brought to a single focus are used.
To avoid blurred images due to spherical aberration, parabolic mirrors
are used to produce parallel beams in search lights, torch lights and car
headlamps.
SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.
4.0
Describe an experiment to measure the focal length of a concave
mirror.
Explain the phenomenon spherical aberration.
How is it corrected in practice?
A small object is placed 24 cm from a concave mirror of focal
length 16 cm. What is the position of the image?
An object is placed 50 cm in front of a concave mirror and a real
image is produced 20 cm from the mirror.
(a)
What is the magnification produced?
(b)
What magnification will be produced if the object is
placed 30 cm from the mirror?
CONCLUSION
In this unit you studied reflection of light at curved surfaces. You
learned about types of curved mirrors, terms used in curved mirrors, and
principal rays for locating images in a curved mirror. Also, you learned
how to locate images formed by curved mirrors by scale drawings and
calculations. In addition you learned how to find the focal length of
concave mirrors experimentally. Finally, you learned the applications of
curved mirrors and the defects of spherical mirrors.
5.0
SUMMARY
•
A concave mirror converges incident parallel rays while a convex
mirror diverges incident parallel rays.
•
A concave mirror has a real focus while a convex mirror has a
virtual focus.
•
Concave mirrors produce different images when the object is
placed in different positions.
•
Convex mirrors always produce virtual diminished and erect
images.
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•
The mirror equations are:
(i)
(ii)
(iii)
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1 / u + 1 / v = 1/f
m = v/u
f = r/2
Some practical applications of;
(a) Concave mirrors are:
(i)
As shaving mirror, dentist mirror, make up mirror,
projector lamp and some kind of flood lights.
(ii)
Parabolic mirrors are used in searchlights, torches
and car headlamps
(b) Convex mirrors are used as driving mirrors.
6.0
TUTOR-MARKED ASSIGNMENT
1.
With the aid of ray diagram illustrate the characteristics of the
image formed by a concave mirror when an object is placed
between c and f.
2.
What is the defect of a spherical mirror? How is it corrected?
3.
Which of the types of curved mirrors is used as a driving mirror
and why?
4.
An object is placed 20 cm in front of a convex mirror and an
image is produced 4 cm behind the mirror. Calculate the focal
length of the mirror
7.0
REFERENCES/FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools Onitshas: Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics.
Ibadan: University Press.
Okpala, P.N.(1990). Physics. Certificate Year Series for Senior
Secondary Schools. Ibadan: NPS Educational.
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UNIT 3
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REFRACTION OF LIGHT
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Refraction of Light
3.1.1 Concept of Refraction
3.1.2 Refraction of Light through Rectangular Glass
Block
3.1.3 Refraction of Light through Triangular Prism
3.2
Laws of Refraction
3.2.1 Statements of Laws of Refraction
3.2.2 Verification of Snell’s Law
3.3.1 Real and Apparent Depths
3.3.2 Relation of Refractive Index to Real and Apparent
Depths
3.4
Measurement of Refractive Index
3.4.1 Measurement of Refractive Index of Solid (glass)
3.4.2 Measurement of Refractive Index of Liquid (water)
3.4.3 Determination of the Refractive Index Glass by the
Method of Minimum Deviation of Light through a
Glass Prism
3.5.1 Critical Angle and Total Internal Refraction
3.5.2 Relationship between Critical Angle c and Total
Internal Reflection
3.5.3 Applications of Total Internal Reflection
3.6
Numerical Examples
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit, you will learn the refraction of light through glass blocks
and prisms. You will also learn the laws of refraction – the statements of
the law and their verifications. Furthermore you will learn about real and
apparent depths in solids and liquids and the relationship between them.
In addition, you will learn the measurement of refractive index of glass
and liquid using Snell’s law and ‘real and apparent depths’. Finally, you
will learn about critical angle and total internal reflection as the
applications of total internal reflection.
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OBJECTIVES
By the time you finish this unit, you will be able to:
•
•
•
•
•
explain how the direction of light changes as it travels from one
medium into the other
measure angles of incidence and refraction and hence deduce a
value for the refractive index of a given material
explain the meaning of critical angle and total internal reflection
stating the conditions under which they occur
establish the relationship between critical angle and refractive
index and apply it to the solution of simple problems
trace light rays through a triangular prism and obtain graphically
the value of the angle of minimum deviation.
3.0
MAIN CONTENT
3.1
Refraction of Light
3.1.1 Concept of Refraction
If you consider the path taken by a ray of light as it travels from one
medium to another say from air to water, you will observe that the
direction of the ray changes instantly at the boundary separating the
two media (air and water). This instant change in the direction of light
ray at the boundary of two transparent media is referred to as the
refraction of light. This phenomenon called refraction is evident in the
following every day observations:
i.
ii.
iii.
The bottom of clear pond appearing to be nearer the surface than
it really is;
A rod or spoon appears bent or broken when it is partially
immersed in water;
When a thick glass block is placed on a printed page, the letters
under the block appear raised relative to those not under the
block.
A rod appears bent in water
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A rod
Water
(a) a rod appears
bent in water
( b ) P h y s ic s is a b r a n c h o f s c ie n c e
Glass block
Fig. 3.1: Demonstration of refraction effects
Definition
Refraction is the bending of a light ray as it crosses the boundary
between two media of different densities thereby causing a change in its
direction.
Refraction is due to the difference in the speed of light in different
media.
If we consider a ray of light traveling from a medium A to another
medium B as shown in Fig. 3.2
P
Incident ray
Angle of incidence
0
Medium A
Medium B
Angle of refraction
r
Refracted
R
Fig. 3.2: Ray of light traveling from one medium to another
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The incident ray PO is the path along which the light travels in the first
medium.
The angle of incidence (i) is the angle which the incident ray makes with
the normal to the surface at O. The incident ray is refracted as it enters
the second medium and it travels along the path OR. This ray OR is
called the refracted ray. The angle between the refracted ray and the
normal to the surface at O is called the angle of refraction(r).
Speed of light is more in medium of less optical density. Rays of light
traveling from low optical dense medium to light optical dense medium
bend towards the normal. On the other hand the rays of light traveling
from high
optical dense medium bend away from the medium.
3.1.2 Refraction of Light through Rectangular Glass Block
Let us study the refraction of light at air glass boundary using a
rectangular glass block. To do this, you will place a thick rectangular
glass block (or slab) horizontally on a cardboard sheet and draw the
outline of the block with a pencil. Direct a light ray from a ray box into
the glass block surface AB along a line PO inclined at an acute angle to
the normal ON. The light ray is refracted in the glass CD. Mark two
points along the path of this ray. Join the point by a line SR and produce
it to meet the glass – air boundary at the point R, Remove the glass
block. Using a pencil, draw a line to join the points R and O. This line
RO represents the path of the light ray in the glass block.
N
Incident ray
A
A ir
Glass
D
O
B
Refracted ray
R
C
R
Emerge ray
E
1
N
S
Fig.3.3: Refraction through a rectangular glass block
You will notice that the line RS is parallel to the line PO (Fig. 3.3). The
line PO is known as the incident ray, OR is the refracted ray while RS is
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the emergent ray. The angle (i) between the incident ray and the
normal at O is the angle of incidence. The angle (r) between the
refracted ray OR and the normal at O is the angle of refraction. The
angle (e) between the emergent ray RS and the normal at R is the angle
of emergence.
You may have also noticed that when the light travels from air to glass,
it is refracted towards the normal; when it travels from glass to air, it is
refracted away from the normal. This is a general phenomenon earlier
mentioned and is due to the different in densities of the two media in
which the incident and refracted rays travel. That is:
•
•
When a ray of light travels from a less dense medium to a denser
medium, the ray will be refracted towards the normal.
A light ray passing from a denser medium to a less dense medium
will be refracted away from the normal.
Note that if the incident ray is along the normal NO- at an angle of 900 to
the side AB, ( i = O ). It will pass straight through the glass and emerges
on the surface CD without a change of direction.
Note also that there is no deviation or change in the direction of the
emergent ray when compared with the incident ray.
P
incident ray
Emergent ray
produced to
incident ray
produced
A
i
N
o
r
Air B
Glass
d
RC
lateral
displacement
Emergent ray
S
Fig. 3.4: Lateral displacement
The above figure shows that the incident ray produced is parallel to the
emergent ray produced; thus resulting in the lateral displacement d.
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3.1.3 Refraction of Light Through Triangular Prism
Using a pencil, draw the outline ABC of a 600 prism placed on a white
sheet of cardboard paper. Remove the prism and use a protractor to
draw a normal ON to AB at a point O between A and B. Now, consider
the passage of light rays through this triangular glass prism (Fig. 3.5).
A
Refracting
prism
angle
of
Angle of deviator (d)
N
Incident ray
O
d
i
e
r
emergent ray
B
C
Fig. 3.5: Refraction through triangular prism
Consider one such ray PQ incident on the prism side AB. The incident,
the refracted and the emergent rays are as shown in Fig 3.5. Here, the
emergent ray is not parallel to the incident ray but it is at an angle to it.
The prism deviates the incident ray through an angle d and is known as
the angle of deviation. It is the angle between the incident ray and the
emergent ray of light passing through the prism. This angle of deviation
depends on:
(i.)
(ii)
(iii)
The angle of incidence
the refracting angle (A) of the prism and
the refractive index of the material of the prism.
It has been proved from experimental results that the angle of deviation
(d) varies with the angle of incidence (i) as shown in Fig. 3.6
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(a)
A
dm = D
e
r
(b)
d
D
Fig. 3.6:
i=e
i
At minimum deviation ¡ = e
Figure 3.6 shows that there is a minimum angle of deviation (D) which
occurs when the incident and emergent rays are equally inclined to their
respective surfaces. Also, the ray passes symmetrically through the
prism and the angle of incidence equal angle of emergence (i = e).
At minimum deviation, the refractive index (n) of the glass, the
refracting angle (A) of the prism and the angle of minimum deviation
(D) are related by the equation.
n
3.2
=
Sin ½ ( A + D )
sin ½
A
Laws of Refraction
The phenomenon of the refraction of light was first observed in 300 BC
by Euclid, a Greek mathematician. An Egyptian Ptolomy was the first to
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record the values of the angles of incidence and refraction but he could
not find the relationship between these angles. However in 1621 a
mathematician, willebrod Snell formulated the exact relationship
between the angles of incidence and refraction. Hence, the law is known
as Snell’s law of refraction.
3.2.1 Statements of Laws of Refraction
There are two laws governing the refraction of light. These are:
i
ii
The incident ray, the refracted ray and the normal at the point of
incidence all lie on the same plane.
The ratio of the sine of the angle of incidence to the sine of the
angle of refraction is a constant for a given pair of media.
The second law is the Snell’s law and it may be expressed as:
Sin i
Sin r
= constant
For a given pair of media
where
ί
r
n
=
=
=
angle of incidence
angle of refraction
the refractive index of the second medium with
respect to the fist medium.
The refractive index n is a number which gives a measure of the
refraction or bending of light as it travels from one medium to another.
When light travels from air to glass, the refractive index of glass is given
by
= Sine of angle of incidence in air
a ηg
Sine of angle of refraction in glass
When light travels from glass to air then the refractive index gηa is given
by
η
g a
= Sine of angle of incidence in glass
Sine of angle of refraction in air
From the principle of the reversibility of light we have that
a η g = 1/gηa
Also, because refraction is due to the change in the speed of light as it
travels from one medium to another, the refractive index is given by
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η = speed of light in air (vacuum)
speed of light in glass.
a g
3.2.2 Verification of Snell’s Law
Snell’s law can be verified by performing an experiment using a
rectangular glass
block. To do this, place the glass block on a sheet
of paper and trace the outline with a pencil. Draw a normal ON and the
draw lines making angles of 300, 400, 500, 60o and 700 with then normal
ON. Direct a single ray of light from a ray box along one of the lines
making angles of say 300 with the normal.
This is now the incident ray PO. Mark two point TS where the light
emerges from the glass block with a pencil. This ray TS is the emergent
ray – join the two points and produced the line to cut the glass block at
R (fFg. 3.7). Remove the glass block and join the points.
OR. Draw a normal YRZ at point R where the emergent ray line or
represents the refracted ray in the glass.
P
N
ί
Air
O
A
B
r
Glass block
D
e
e e
e
C
Air
Z
S
Fig.3.7: Verification of laws of refraction
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Measure the angle of incidence ί and the angle of refraction r. Repeat the
experiment for other angles of incidence and obtain the corresponding
angles of refraction. Measure and tabulate the angle of incidence (ί) and
their corresponding angles of refraction (r) as shown in Table 3.1 below.
Table 3.1: Experimental values of the incident angle (i) and their
corresponding angles of refraction (r) for a glass block
Angle of incidence Angle of refraction Sin ί Sin r
(i)
(r)
0
30
400
500
600
700
Evaluate and record sin ί and sin r. Use the values to plot sin ί on the
vertical axis and sin r on the horizontal axis. Obtain the slope of the
graph as shown in Fig. 3.8 below.
A
C
Slope = AB
CB
= Change in sin ί
change in sin r
B
=
Sin r
Fig. 3.8.: Graph of sin ί against sin r
Ŋ
When sin ί is plotted against sin r, a straight line graph passing through
the origin is obtained as show in Fig. 3.8 which proves that the ratio sin ί
sin r
is a constant. That is sin ί / sin r = constant. Hence,
Snell’s verified. The slope of the graph gives the refractive index for the
glass block used.
(ii.)
We can also verify Snell’s law with optical pins in the experiment
instead of a ray box. Here optical pins are used to represent the
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incident ray while sighting pins are used to locate the emergent
ray. Two pins MN are fixed on the line normal MO making angle
300 with the normal NO to represent the incident ray. A sighting
pin X is moved along CD until it appears to be in same straight
line as MN when viewed through the glass block. A second pin
Y is fixed so that X and Y appear coincident with the image of M
and N in the glass block.
This is illustrated in Fig. 3.9
m
N
N
A
D
O
B
R
C
X
y
Fig. 3.9: Verification of Snell’s law
The glass block is then removed and the incident ray is drawn through
M and N while the emergent ray is drawn by joining the point O at
which the line MN meets the glass surface to the point R where the
emergent ray leaves the glass surface. The angles of incidence and
refraction are measured and the produce and repeated for various angles
of incidence.
This is followed by plotting the graph of sin ί against sin r. Again, the
graph is a straight line passing through the origin indicating sin ί / sin r
is a constant and the slope gives the value of the refractive index. Thus,
Snell’s law is also verified and found to be true.
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3.3.1 Real and Apparent Depths
There appears to be an upward displacement of the bottom of a vessel
containing water when viewed upward. An object inside water appears
to be closer to the surface than it actually is. That is, the depth appears to
be shallower than the actual depth. This apparent depth is caused by
refraction (Fig. 3.10)
Eye
Air
A p p a re n t
dept
Water
Real depth
Fig. 3.10: Real and apparent depths
The rays from an object O under water do not travel straight to the eye
but are refracted away from the normal as they leave the water surface.
When these rays are produced backwards, they meet at I where the rays
appear to be coming from –the virtual image of O. Thus, the observer
sees the object as though it was at I.
The apparent depth is the distance of I from the surface of the water
while the distance of the object O from the surface is called the real
depth.
In the same way, if a thick glass block is placed on a printed page, the
letters under the block will appear raised relative to those not under the
block. This is already illustrated in Fig. 3.1b.
3.3.2 Relation of Refractive Index to Real and Apparent
Depths
We can show how the real and apparent depths are related to the
refractive index of the medium. You know that the refractive index for
rays of light traveling from air to water is given by the relation
n =
Sin ί (in air)
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Sin r (in water)
Suppose an incident ray from an object O in water bends away from the
normal as it travels from water to air in accordance to refraction effects.
This is illustrated in Fig. 3.11.
B
P
Q
Air
I
Water
N
O
Fig. 3.11: Relation between refractive index and real and apparent
depths (Points Q and A are close to each other)
From the principle of reversibility of light, light traveling in the opposite
direction from air to water would follow the same path.
Recall that:
Refractive index, η
= Sin i
Sin r
This implies that
η
But
and
Sin ∠ PAB
Sin ∠ OAN
∠ PAB
= ∠ QIA (corresponding angles)
∠ OAN = ∠ QOA (alternate angles)
Therefore
η
=
=
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Sin ∠ QIA
Sin ∠ QOA
QA / AI
QA/ AO
=
QA x AO
AI QA
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AO
AI
Since the eye sees both rays OQ and OA, they must be close to each
other.
Hence
AL = QI and AO = QO
=
⇒
η
=
η
=
QO
QI
That is,
Real depth
Apparent depth.
The above is a general formula and can be used to calculate the apparent
depth if the refractive index and the real depth of a medium are known.
3.4
Measurement of Refractive Index
3.4.1 Measurement of Refractive Index of Solid (glass)
Using
(i)
(ii)
(i)
Snell’s law
Real and apparent depths
By Snell’s law
This follows same procedure as the verification of Snell’s law already
discussed and illustrated in section 3.2.2.
(ii)
By real and apparent depths
The refractive index of glass can be determined by the real and
apparent depth method. This is illustrated in Fig. 3.12.
Sliding Cork
Retort Stand
Search pin
Object Pin P
Image Pin P’
P’
Glass block
d
D
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Fig. 3.12: Finding the refraction index of glass block by real
and apparent depth methods
This is done by fixing an object pin P with plasticine vertically touching
one side of a rectangular glass block mounted on a cardboard paper
placed on a laboratory table. A search pin on a sliding cork is then
mounted above the rectangular glass block using retort stand and damp
as shown in Fig. 3.12. The object pin P is then viewed through the block
from the opposite side of the pin as the search pin is moved back and
forth until there is no parallax between the object pin P and its image P 1
as seen through the glass block. The real thickness (depth) D and the
apparent thickness (depth) d of the glass block are measured. The
refractive index n of the glass is the then calculated using the formula:
Refraction index, η =
real thickness(depth)
apparent thickness (depth)
=
D
d
3.4.2 Measurement of Refractive Index of Liquid (water)
Using
(i)
(ii)
Snell’s law
Real and apparent depths
(i)
By Snell’s Law
A straight – sided transparent Perspex box of water is placed on a sheet
of cardboard paper and it outline ABCD is drawn using a sharp pencil.
A ray box is then adjusted to give a narrow beam of light. The narrow
beam of light is directed to strike the box of water surface AB at point O
an acute angle such that the incident ray PO makes an angle
of
0
incidence i say 60 to the normal ON.
The light beam comes out into the air at R in the same direction as the
incident beam at O but is displaced sideways. The point Q and R are
marked on the emergent ray. The water box is then removed and the
points PORQ are joined as shown in Fig. 3.13.
.
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Incident ray
P
I
¡
O
Angle of incidence
A
B
Angle of refraction
r
R
Refracted ray
Perspex box of water
D
E
C
Emergent angle
Emergent ray
Fig. 3.13: Finding the refractive index of water using Snell’s law
The angle of incidence i and the corresponding angle of refraction r are
measured and recorded. Following r are same procedure, four different
values of the angles of incidence say 500, 400, 300 and 200 and their
corresponding angles of refraction are obtained. The result is than
tabulated in a composite table such as Table 3.2 below.
Table 3.2: Experimental values of the incident angle (i) and their
corresponding angles of refraction (r) for a box of water
Incident Refracted Sin i
Sin r
0
0
Angle i Angle r
600
500
400
300
200
Average
η
n = sin i
sin r
The refraction index is obtained by either plotting a graph of sin i
against sin r as in section 3.2.2 or finding the average of η .
(ii)
By real and apparent depths.
We can also find the refractive index of a liquid (water) by the real and
apparent depth methods illustrated in Fig. 3.13
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Measuring Cylinder
Retort Stand
Water
Search Pin
Cork
Apparent
depth
Real Depth
Large pin P’
Object pin P
Fig. 3.14:
Finding the refractive index of ware by real and apparent
depth methods
In this method an optical pin is placed at the bottom of a beaker to serve
as the object P. Water is then poured into the beaker until it is about
three – quarters full. A search pin is attached horizontally to a sliding
cork and it is moved up and down until it coincides with the apparent
position of the pin as seen from above the liquid. At this point there is
no parallax between the object pin p and the image pin P1. The real and
apparent depths of the pin from the surface of the water are than
measured and the refractive index of the liquid (water) is calculated
using the formula:
Refractive index, η = real depth
apparent depth
3.4.3 Determination of the Refractive Index of Glass by the
Method of Minimum Deviation of Light through a Glass
Prism
In this experiment, the path of a single ray of light through a triangular
prism is traced using either the optical pins no – parallax method or a
ray box. The angle of incidence i of say 300 of the ray of light at the face
AB and the corresponding angle of emergent e at the face AC are
obtained, measured and recorded by drawing normal at G and L
respectively. This is done by joining the emergent ray HK and
producing it to meet the incident ray FG produced at a point M. the
angle of deviation d is also measured and recorded. This illustrated in
Fig. 3.14.
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F
¡
m
G
r
d
e
L
k
H
B
C
Fig. 3.15: Determination of refractive index of glass using triangular
prism
The above procedure is repeated for several other values of angles of
incidence ¡ say 400.450, 500, 550 and 600, aid the corresponding values of
emergent angles e and angles of deviation d measure and recorded in
each case. The result is tabulated in the headings shown in Table 3.3
below:
Table 3.3: Experimental values for determining refractive index using
triangular prism
Angle
of
0
incidence ί ( )
Angle
of Angle
of
0
emergent e ( )
deviation d (0) (d – e) (0)
30
40
45
50
55
60
A graph of (d – e) on the vertical axis against ί on the horizontal axis is
then plotted from which the intercepts on both axes are determined. The
mean of these intercepts is determined, using the magnitude of the
intercepts (ie disregarding the negative sign). This gives the value of A,
the refracting angle of the triangular prism. That is, the mean of the
values of the magnitude of the intercepts on both axes is equal to the
refracting angle A the triangular prism.
y
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(d- e)(0)
Intercept on x -
axis
X
Intercept on y axis
Fig. 3.16: Graph of (d-e) against i
A graph of d on the vertical axis against I on the horizontal axis is also
plotted. From this graph (Fig.3.17), the minimum value of d (ie D) is
determined and the value of I at this minimum deviation D is read from
the graph.
d
Dm = D= 370
¡ =e
i
Fig. 3.17: Graph of d against i showing minimum deviation D
These values of A and D are then substituted in the equation
η = Sin ½ (A + D)
Sin ½ A
From which the value of the refraction index n of the glass is then
calculated.
The result of experiments show that as the angle of incidence increases,
the angle of deviation, d, decreases, until at a certain minimum value,
370, the angle of deviation begins to increase for any further increase in
the angle of incidence. Hence the graph is a curve.
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In this experiment the refractive index n of the prism material can also
be worked out from the values of angles of incidence I and the
corresponding angles of refraction r from the relation.
Refractive Index, η = Sin ¡
Sin r
Either by plotting the graph of sin ¡ against sin r as in section 3.2.2 or by
finding the mean value of the ratio of Sin ¡ for each set of reading as in
Sin r
section 3.4.2
3.5.1 Critical Angle and Total Internal Reflection
When ever light is incident on the interface between two media, there is
usually a weak reflected ray in addition to the refracted ray (Fig. 3.18)
Incident
ray
r
Air
ted
frac
Re
ra y
Weak reflected
ray
l l
Air
Plane bounding
Water or glass
Water or glass
r
l l
Refracted ray
Incident
ray
Weak reflected
ray
Fig. 3.18: Refraction and partial reflection
A striking change occurs when the situation is reversed optically so that
light passes from glass to air.
When light passes through an optical dense medium to meet the surface
of less dense one, as the angle of incidence I is increased, the angle of
refraction r also increases until it is 900. In that case, the refracted ray is
along the interface (surface) of the two media.
•
Critical angle c is the angle of incidence in the denser medium for
which the angle of refraction in the less dense medium is
900. The
value of the critical angle C depends on the nature
of the two media or their relative refractive indices. (It is about
420 for crown glass the most common type of optical glass).
•
Total internal reflection occurs when the incident angle I is more
then the critical angle C. At this point all incidents light is
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reflected back to give strong reflected ray. There is no refraction.
Hence total internal reflection (Fig. 3.19).
Weak reflected ray
Weak reflected ray
Incident
ray
l
l
Glass
Air
Glass
Air
(a) l < c
l
Strong
reflected
ray
Strong reflected
ray
Strong
reflected
ray
Glass
Air
No refraction
total reflection
(b) l= c
(c) l> c
Fig. 3.19: Illustration of total internal reflection
Total internal reflection cannot occur when light is incident in air on a
glass block. This is because the angle of refraction in the glass is always
less than the angle of incidence. Therefore, even when the angle of
incidence in air is nearly 900, a refracted ray is obtained.
Reflection always occurs at air- glass boundary but this reflection is
only partial because most of the light passes into the glass (water).It is
therefore important to note that: For total internal reflection to occur
•
•
3.5.2
Light must travel from an optically denser medium to an
optically less dense medium;
The angle of incidence in the denser medium must be greater than
the critical angle.
Relationship between Critical
Internal Reflection
Angle
C and Total
You will recall that from Snell’s law we have for a ray passing from:
•
air to glass, a Ŋ g =
Sin ί
Sin r
Therefore, applying the principle of reversibility of light, for a ray
passing from:
•
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Glass (denser medium) to air (less dense medium),
g Ŋ a = Sin r
Sin ί
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That is, the refractive index n of the denser medium is given by
η = Sin r
Sin I
When ί = C, the critical angle
r = 900
Hence η = Sin 900 =
Sin C
1
.
Sin C
Since η = 1/ sin C
ie
η sin C = 1
Sin c = 1/n
In general, refractive index
η = 1/ sin (critical angle)
⇒
Sin C = 1/η
Generally, the critical angle in the denser medium is given by
Sin C = ηa/ ηb
Where nb is the absolute refractive index for the optical denser medium.
na -
is the refractive index for the less dense medium.
Since C cannot be greater than unity ηb must always be greater than na.
3.5.3
Applications of Total Internal Reflection
Total internal reflections are the basis for explaining the working
principle of some of the phenomena we experience in nature and some
optical instruments. For instance the working of prism periscope, prism
binoculars, bicycle reflectors, reflectors on road , light guides optical
fiber are explained on the basis of total internal reflection. Also, the
phenomena of image, fish and cat’s eyes view are some natural
observations that are explained on the basis of total internal reflection.
Now let us consider some of them:
(i.)
Optical fiber
The principle of total internal reflection is used in optical fibers. The
optical fiber (Fig.3.20) is made up of a fine strand of glass coated with
another type of glass of lower refractive index. It is constructed in such a
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way that light rays falling on the interphase between the two types of
glass always undergo total internal reflection as shown in Fig. 3.20.
Low refractive
High refractive
index
Fig. 3.20: Optical fiber action
When light enters one end of the fibre at an angle less than say 60 0, it is
refracted towards the normal by the inner glass. The refracted ray
impinges on the boundary between the two types of glass and undergoes
total internal reflection. The light emerges at the other end after
undergoing many repeated total internal reflections at the walls of the
optical fibre.
Uses
•
•
•
(ii)
They can be used as light guides to enable the image of an object
at one end to be seen at the other.
They can be fitted with an objective and an eyepiece and inserted
into the stomach or throat to study the internal structure.
They are used for electrical and engineering purposes such as in
telecommunications and TV transmission, the electrical impulses
are converted into light signals which are then transmitted via
these fibres.
Mirage
Mirage is one of the natural phenomenons that can be explained on the
basis of total internal reflection. This occurs at the boundary of a hot and
a cold gas. On a hot day for instance the air in contact with the ground
gets very much hotter than the layer above the ground. The hotter air is
optically less dense than the colder air above the ground. The refractive
index of air depends on its intensity. Hence, the lower hotter layers have
a lower refractive index than the layers above then. Rays of light passing
through these different layers of air are refracted due to the gradual
change in density. Therefore a ray from part of the sky passing from
colder to warmer air layer (from denser to a less dense medium) bends
gradually from the incident direction until it enters a layer of air near the
ground at B where total internal reflection occurs. This ray is then
reflected upwards into the dense air as illustrated in Fig. 3.2.
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Hot air layer varging n
Sky
C
Observer
B
Pool
Total internal reflection
Fig. 3.21: Image
After undergoing refraction in a gradual upward direction, the ray enters
the eye of an observer at C as if it comes from A. It gives the illusion of
a reflecting pool of water in the road some distance away. I am sure you
must have experienced this illusion on a hot day while on a car moving
on a tarred road.
People may sometimes see illusions of inverted images of trees in a pool
of water below the actual tree for the reason explained above especially
around hot deserts because of the intense heat there.
3.6
Numerical Examples
(i.)
The apparent depth of a liquid in a big jar is 6 m when its real
depth is 8 m. What is the refractive index of the liquid?
Solution
Refractive index of liquid = Real depth
Apparent depth
ie
n
=
real depth
apparent depth
=
=
8/6
4/3
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The angle of incidence of a ray of light striking an equilateral
triangular prism is 450. If the refractive index of the prism is 1.5,
and the refracting angle is 600, calculate
(a)
the angle of refraction at the first face,
(b)
the angle of emergence
Solution
(a) let r be the angle of refraction at the first face, AB
using n
= sin ί
sin r
= sin 450
sin r
1.5
Sin r = sin 450
1.5
= 0.7071
1.5
= 0.4714
= Sin-1 0.4714
∴r
Angle of refraction r = 28.130
(b)
600
M
¡
G
x
e
r
N
B
Let the emergent angle be e as shown above.
x = 900 - r (MN normal to AB)
= 900 - 28-130
= 61.870
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90o + y = 600 + x
(exterior angles of triangle AFG)
y = 600 + 61. 870 - 900
= 31.870
Note that the incident ray at the second face AC is FG and the refracted
ray is GH. Thus applying the principle of reversibility of light, a light
ray incidence on AC along HG will be refracted along GF.
Hence
Refractive of index prism
ie
1.5
Sin e
Angle of emergence
=
Sin e
Sin y
=
Sin e
Sin y
=
1.5 x Sin 31.870
=
e =
1.5 x 0.5280
sin-1 0.7920
e
52.37o
=
SELF ASSESSMENT EXERCISE 1
1.
2.
3.
4.
5.
4.0
What do you understand by the term refraction of light?
With the aid of a diagram explain the refraction of light through
rectangular glass block.
State the laws of refraction and explain how you can measure the
refractive index of water using Snell’s law.
What do you understand by the concepts? Critical angles and
total internal refraction? What is the relationship between the two
concepts?
Mention five applications of total internal reflection and explain
any two of them.
CONCLUSION
In this unit you studied the refraction of light. You learned the
refraction of light at both the rectangular and triangular glasses. Also
you learned about the laws of refraction and how to verify them. In
addition, you learned about real and apparent depths and how to use
them in finding the refractive index of both water and glass block.
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Finally, you learned about the critical angle and total internal reflection
as well as the applications of total internal reflection.
5.0
•
•
•
•
•
•
SUMMARY
When light passes from one medium to an optically denser
medium it is refracted towards the normal. When it passes to an
optically less dense medium it is refracted away from the normal.
Laws of refraction
(i)
The incident ray, the refracted ray and the normal at the
point of incidence all lie in the say place.
(ii)
The ratio of the sine of angle of incidence to the sine of the
angle of refraction is a constant for a given pair of media.
(Snell’s Law).
The ratio
Sin i
is called the infractive index η.
Sin r
n = Sin ¡ Sin r
The ratio real depth / apparent depth is equal to the refractive
index n at near normal incidence.
Critical angle C is the angle of incidence in the denser medium
for which the angle of refraction in the less dense medium is 900.
Total internal reflection occurs when the incidents angle ¡ is more
than the critical angle C and light travels, from denser to less
dense medium.
6.0
TUTOR -MARKED ASSIGNMENT
1
State the laws of refraction and illustrate the second law with the
aid of diagram.
With the aid of diagram explain the meaning of real and apparent
depth.
Defined the following terms
(i)
Critical angle
(ii)
Total internal reflection.
2.
3.
7.0
REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools Onitsha: Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics
Ibadan: University Press.
Nelkon, M. (1999). Principles of Physics for Senior Secondary Schools
Ibadan: Longman Group Limited.
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REFRACTION
(LENSES)
AT
CURVED
SURFACES
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
Refraction through Lenses
3.1.1 Types of Lenses
3.1.2 Terms used in Lenses
3.1.3 Rules for Constructing Images Formed by
3.2
Formation of Images by Lenses
3.2.1 Formation of Images by Convex (converging) Lens
3.2.2 Formation of Images by Concave (diverging) Lens
3.3
Practical Applications (use) of Convex (converging) Lens
3.4
Practical Applications (use) of Concave (diverging) Lens
3.5
Lens Formulae and Sign Conventions
3.5.1 Sign Conventions
3.5.2 Linear Magnification
3.5.3 Lens Formulae
3.6
Determination of the Focal Length of a Converging Lens
by the Measurement of Object and Image Distances
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit, you will learn about lenses, how to obtain images through
lenses, the relationship between object distance, image distance and the
focal length of a lens. Also, you will learn the derivation and use of the
lens formula and how to solve problems on lenses both graphically and
analytically.
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OBJECTIVES
By the time you finish this unit, you will be able to:
•
obtain images due to light rays through converging and diverging
lenses using:
(a) ray tracks; and
(b) ray tracing methods.
•
•
•
derived and use the lens formula to solve numerical problems on
lenses
use the ray – box to show relationship between object distance
and the focal length of a lens
solve problems on lenses both graphically and analytically.
3.0
MAIN CONTENT
3.1
Refraction through Lenses
Refraction through lenses involves the change in direction of light rays.
A lens is any transparent material with two faces of which at least one is
curved. It causes a beam of light passing through it to either converge or
diverge.
3.1.1 Types of Lenses
We have two types of lenses namely convex lens and concave lens.
•
•
A convex lens is a converging lens. It makes a parallel beam of
light falling on it to converge. It is thicker at its centre than at its
edges.
A concave lens is a diverging lens. It makes parallel beams of
light falling on it to appear to diverge from a point. It is thinner at
its centre
than at the edges.
We have various forms of convex and concave lenses but the most
commonly used ones in school laboratory experiments are the biconvex
and the biconcave lenses. Plano- convex and Plano – concave lenses are
used in optical instruments while concavo – convex and convexo –
concave are used as contact lenses to correct defective vision. Most
lenses are made from glass, a few are made with quartz or plastic. The
eye has a crystalline biconvex lens. These are shown in Fig. 4.1.
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B.
D.
C.
E.
Convex lenses
Concave Lenses
(Converging Lenses)
(Diverging Lenses)
(a)
(b)
(c)
biconvex
Plano – convex
Concavo - convex
Fig. 4.1:
(d)
(e)
(f)
F.
biconcave
Plano - concave
convexo – concave
Types of lenses
3.1.2 Terms Used in Lenses
We have some important parameters associated with a lens. They are
defined as follows:
•
•
•
•
•
•
•
•
The optical centre (c) (Fig. 4.2) of a lens is the point through
which rays of light pass without being deviated by the lens.
The principal axis is the line passing through the optical centre of
the lens and joining the centre of curvature of its surfaces.
The principal focus (f) of a converging or convex lens is the point
to which all rays parallel and close to the principal axis converge
after refraction through the lens.
The principal focus of a diverging or concave lens is the point
from which all rays parallel and close to the principal axis
appaear to diverge after refraction through the lens.
A lens has two principal foci because light may pass through a
lens from either direction.
Focal length (f) is the distance between the optical centre and the
principal focus of the lens.
The power (p) of a lens is equal to the reciprocal of the focal
length and is measured in diapers when f is in meters. The shorter
the focal length f the greater will be the power of the lens.
The principle of reversibility of light states that, the path of a
light ray is reversible. Because of the principle of reversibility of
light, rays
originating from f and passing through the convex
lens will emerge
parallel to the principal axis.
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The principal focus of a converging lens is real, that of a
diverging lens is virtual.
Lenses whose thicknesses are small in comparism with their focal
lengths are referred to as thin lenses. For such lenses, the two foci
are symmetrically placed with respect to the optical center of the
lens.
•
The terms associated with lenses are illustrated in Fig. 4.2 below.
A
F
C
F
B
A
F
C
F
B
Reversibility of light
A
B
F
C
F
C= Optical center
F= principal focus
F1F1 = principal foci
A
B
F
C
F
CF = focal length
AB= principal axis
Fig. 4.2: Terms associated with lenses
3.1.3 Rules for Constructing Images Formed by:
(i)
(ii)
Convex (Converging) lens
Concave (Diverging) lens
(i)
Rules for convex (converging) lens.
You will obey the following rules when constructing images formed by
convex (converging) lens.
Rule 1:
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Incident rays parallel to the principal axis are refracted
through the principal focus.
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F′
Rule 2:
C
F
Incident rays passing through the optical center are
undeviated
F′
Rule 3:
C
F
Incident rays passing through the principal focus
emerge parallel to the principal axis after being
refracted through the lens.
F
C
f
You can use any two of the three rays to locate an image.
(ii)
Rule for concave (diverging) lens.
The following rules will guide you when constructing images formed by
concave (diverging) lens.
Rule 1:
Incident rays parallel to the principal axis are refracted
away from the principal focus.
F
C
F
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Rule 2:
F
3.2
C
Incident rays passing through the optical centre are
undeviated.
F
Formation of Images by Lenses
You should note that light rays incident on a converging lens converge to
form a real image on the other side of the lens after passing through the
lens. For a diverging lens, the rays from an object diverge to form a
virtual image of the object. When you know the distance of an object
from the lens and the focal length of the lens, you can obtain the position
and nature of the image by the construction of ray’s diagrams. To do this,
you will employ the rules for constructing images formed by lenses
discussed in 3.1.3 above.
3.2.1 Formation of Images by a Convex (Converging) Lens
You will locate the position and determine the nature of the image
formed by a convex lens by following these construction steps.
(i)
(ii)
Choose a suitable scale for the construction
Draw a straight vertical line MW (Fig. 4.3.) to represent the
position of the lens and mark the position of the focus F
according to the scale. Represent the focal length with CF.
(iii) Represent the object by drawing an arrow line OA perpendicular
to OC,OC should correspond to the distance of the object from
the lens
(iv) and OA, the height,
(v)
Draw a ray AP parallel to the principal axis from the top of the
OA,
According to rule 1, ray AP is refracted through F1
(vi) Draw another ray AC through the centre of the lens C. This ray
passes straight through without changing direction (rule 2).
(vii) The point of intersection B of the two emerging rays PB and CB
represents the top of the image.
(viii) Now, complete the image by drawing B1 normal to the principal
axis.
(ix) Measure the image distance from C and the image height B1,
convert to the actual values using your scale.
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Ray diagrams for various object positions are presented in Fig. 4.3.
(a) Object between F and Lens image is:
•
•
•
•
(b)
Erect
Magnified
Virtual
Beyond the object on same side as object
Object at F: image is:
•
•
•
•
At infinity
Real
Inverted
Magnified
Im
ag
(c)
e
at
in
f
in
i
ty
Object between F and 2F image is:
•
Inverted
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•
•
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Magnified
Real
Beyond 2f
Object at 2F Image is
•
•
•
•
Inverted
Real
Same size as object
At 2F the other side of the lens.
I
F
(e)
Object beyond 2f Image is
•
•
•
•
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2F
Inverted
Diminished
Real
Between F and 2F on the other side of lens.
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I
F
(f)
2F
Object at infinity image is
•
•
•
•
Inverted
Diminished
Real
At F on the other side of lens.
Obje
ct at
infinit
y
2F
Fig. 4.3: Image formed by convex (converging) lens
3.2.2 Formation of Images by a Concave (Diverging) Lens
You will observe by drawing ray diagrams for a concave lens that the
image is virtual, erect and diminished. It is always on the same side of
the lens as the object and between the object and the lens. These image
characteristics are for all object positions. This is illustrated in Fig.
4.4.A concave lens is usually used in spectacles for the correction of
short sightedness because of its qualities.
(a)
Object beyond 2f image is:
•
•
Erect
Diminished
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•
•
Virtual
Between F and C
A
P
B
0
F
F I
2F
2F
(b) Object between F and C image is
•
•
•
•
Erect
Diminished
Virtual 2F
Between F and C
A
P
B
2F
F 0
I
F
2F
Fig. 4.4: Images formed by concave (diverging) lens
These ray diagrams were constructed using the rules listed in section
3.1.3 (ii).
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Practical Applications (uses) of the Convex (Converging)
Lens
We use the convex (converging) lens in many optical instruments. Some
of them are listed below.
•
•
•
•
3.4
When the object is between F and lens, the lens is used as eye
lens of many instruments, magnifying glass, spectacles for long
sight.
When object is at F, it is used at theatre and stadium spotlights.
When object is between F and 2F, it is used as objective lens of
microscopes and projectors.
When object is at 2F, it’s used in copying camera.
Practical Application (use) of the Concave (Diverging)
Lens
A concave lens is usually used in spectacles for short sight.
SELF ASSESSMENT EXERCISE 1
1.
Define
(i)
The optical centre
(ii)
The principal focus
(iii) The focal length of a converging lens.
2.
What do you understand by?
(i)
The principle of reversibility of light
(ii)
The power of a lens.
3.
List the similarities and differences between the action of a
concave mirror and a convex lens.
4.
What are the applications of convex and concave lenses?
5.
With the aid of rays diagrams, illustrate the characteristics of
images formed by a convex lens when the object is placed at:
(i)
F
(ii)
2F
(iii) Between F and 2F
6.
With the aid of ray diagrams illustrate the characteristics of
images formed by a concave lens when an object is placed at two
different positions from the lens.
3.5
Lens Formulae and Sign Conventions
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3.5.1 Sign Conventions
You have already learned the sign conventions in Module 4 Unit 2 for
the case of curved mirrors. A similar situation applies for lenses. We
have that for;
•
New Cartesian convention
i
The optical centre is taken as the origin for measuring all
distances.
ii
Distances measured against the incident light are negative.
iii
Distances measured in the same direction as the incident
light are positive.
If the object is always placed to the left of the lens, this convention
implies that all distances to the left of the lens are negative, while all
those to the right are positive.
•
Real – is – positive convention
(i)
(ii)
(iii)
The optical centre is taken as the origin for measuring all
distances.
Distances of real objects and images are positive.
Distances of virtual objects and images are negative.
When you use the real – is – positive convention, you will take the focal
length of converging lens to be positive because the lens has a real
principal focus. On the other hand, you will take the focal length of a
diverging lens to be negative because the lens has a virtual principal
focus.
3.5.2 Linear Magnification
You will recall that the linear magnification (m) produced by mirrors is
the ratio.
M
=
height of image
height of object
This is also true for a lens.
You will easily see this relationship from the following ray diagrams in
Fig. 4.5.
F
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C
F
F
B
U
V
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Converging lens
P
B
v
I
O
C
F
F
(b) Diverging lens
Fig.4.5:
Ray diagrams for lens formula
You can see from Fig. 4.5 (a) and (b) that triangles C1B and COA are
similar.
It therefore follows that:
M
=
height of image
height of object
=
1B = 1C
OA OC
That is,
M
=
distance of image from lens
distance of object from lens
In symbols,
M
=
V
U
Where,
M
V
U
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=
=
=
linear magnification
image distance
object distance
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3.5.3 Lens Formulae
We can derive the formulae relating u, v, m and f by either theoretically
from the geometry of ray diagrams or practically from the result of
experiments.
Here, our derivation is theoretical from the geometry of ray diagrams.
Also, the real – is positive sign convention is adopted.
Recall that for thin lenses, distances are measured from the optical
centre of a lens. Distances measured to real foci, images and objects are
positive, distances measured to virtual foci, images and objects are
negative.
From figure 4.5, you can see that:
triangles CIB and COA are similar
also,
triangles FIB and FCP are similar
Therefore,
for the converging lens,
The magnification
M
V
= IB
=
OA
U
because
∆F IB and ∆ FCP are similar.
Also,
IB
FI
M
=
=
CP
FC
because
∆S
FIB and FCP are similar
From figure 4.5 (a)
and
FI
FC
=v–f
= f
This implies that
M
=
v–f
F
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Therefore,
v
v–f
=
u
f
From the above, we have that
v
v- f
=
u
f
M =
vf
=
uv – uf
Dividing through by v gives us
vf
v
f
=
=
uv
v
- uf
v
u
_ uf .
v
Dividing through by u gives us
f
_
u _ uf
u
u
vu
f
_
1
-f .
u
v
Dividing through by f gives us
F
=1 -f .
uf
f
vf
1/u = 1/f -1/v
Rearranging, we have,
1/u + 1/v
= 1/f
The above relation is the lens formula.
For the diverging lens, you can also, by a similar procedure, using the
two pairs of similar triangles obtain two expressions for the
magnification m.
That is,
m
= -v/ u = - f + v
-f
m
=
or
v/u = - f + v
f
The above will lead you to
1/v + 1/u = 1/f
as in that of converging lens.
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Determination of the Focal Length of a Converging Lens
by the Measurement of Object and Image Distances
Fig. 4.6 (a):Image of distant object: Quick and approximate methods
Fig. 4.6(b): Focal length of a converging lens by u, v measurements
You can use the lens formula to obtain the focal length of a converging
lens, by measuring the object and image distances. You will first
obtain the
approximate value of f by the Quick but approximate
method (Fig. 4.6(a)). You will do this by focusing the image of a distant
object on a screen using a lens. Place the lens in front of the screen and
adjust it until a sharp image of a distant object (Fig. 4.6a) is obtained in
the screen. Measure the distance between the lens and the screen which
is the approximate value of the focal length. Rays coming from the
distant object are parallel rays and therefore converge to the principal
focus of the lens after refraction.
Now, having obtained the approximate focal length, f, of the lens, set up
the apparatus as shown in Fig. 4.6(b).
Place an illuminated object at a known distance, u, greater than f in front
of the lens. Adjust the position of the screen on the other side of the lens
until you obtain a sharp image of the object on the screen. Measure the
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distances u and v from the ray box to the lens (u) and the screen t o the
lens (v). Repeat the procedure for five other values of u and obtain the
corresponding values of v. Tabulate your readings as in Table 4.1.
Table 4.1: Values of u and the corresponding values of v
u
v
uv u+v uv/
+v
v
1/u 1/v
u
1
2
ֽ
ֽ
ֽ
6
Table A
u
1/u +1/v
1
2
ֽ
ֽ
ֽ
6
Table B
You can obtain the value of f from your readings in Table 4.1 in the
following ways:
•
•
Add the values of uv / (u + v) in table A and find the average.
The result
is the experimental value of f.
Add the values of 1/u + 1/v in table A and find the average. The
result is 1/f from which you will then determine the values of f. f
is the reciprocal of the value obtained.
Plot a graph of w against (u + v). You will obtain a straight line graph.
The slope of this graph is equivalent to f. Obtain it (Fig. 4.7(a)).
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Plot a graph of 1/v against 1/u.You will obtain a graph with intercepts
on both axes. Each of the intercepts gives you a value of 1/f. from these
values; find the reciprocal to obtain the value of f (Fig. 4.7(b)).
Fig. 4.7(a) Graph of uv against (u+v)
uv(cm)
Slope = f
(u+v)cm
Fig. 4.7(a): Graph of uv against (u + V)
1/vcm-1
1/f
1/f
1/ucm-1
Fig. 4.7(b): Graph of 1/v against 1/u.
You will always take the following precautions when you deal with light
rays and optical experiments:
•
•
•
•
•
Use sharp pencil in drawing ray diagrams to obtain neat traces.
Avoid parallax when measuring with the metre rule.
Ensure proper alignment of object, lens or mirror and the screen.
Ensure that a sharp image is properly focused on the screen.
Ensure that optical pins (if used) are vertical and well spaced.
Worked examples
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Find the nature and position of the image of an object placed 15
cm in front of a converging lens of focal length 30 cm which
produces a magnified image 16 cm high. What is the size of
the object.
Solution
Using the real – is – positive convention
1
1
= 1 .
v
u
f
1
v
1
15
1
v
1
30
-1
15
1-2
30
-1
30
1
v
⇒
1
30
v
=
-30 cm
The image is virtual and is 30 cm from the lens.
Also,
m
=
v .
u
height of image
height of object
⇒
v/u
height of image
height of object
Substituting gives us
30
=
16
15
height of object
∴
height of object
15x16
=
30
=
The object is 8cm high
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30
=8cm
-1
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An object is placed between a converging lens of focal length 10
cm and a plane mirror, being 12 cm distant from each. Find the
positions of two real images formed by the lens on the opposite
side from the object.
Solution
The two real images formed by the lens are:
(i)
(ii)
The one formed by the object directly,
The image of the object formed by the mirror which now
becomes a second object for the lens.
(i)
Image formed by the lens from the object directly.
Using the real – is – positive sign convention
1
v
1
v
1
v
1
v
⇒
v
+
+
=
1
u
=
1
12
1
10
= 1
60
=
_
1
f
1
10
1 =
12
6–5
60
=
1
60
= 60 cm
The image distance is 60 cm from the lens on the opposite side of the
object.
(ii)
The image of the object formed by the mirror which now
becomes a second object for the lens.
Using the real – is – positive sign convention
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1
v
1
v
1
v
360
⇒
1
u
+
1
36
+
=
=
=
1
f
1
10
1 _
10
1.
v
=
26 .
360
26v
=
360
v
=
360
26
1
36
=
=
36 – 10
360
=
26
13.85cm
The second image distance is 13.85 cm from the lens on the opposite
side of the object.
SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.0
Why is it advisable not to look directly at the sun? Give scientific
reason for your answer.
A convex lens forms a real image of an object magnified two
times.
The distance between the object and image is 60cm. Find the
focal length of the lens.
Describe how you would determine the focal length of a
converging lens by means of;
(i)
Quick but approximate method
(ii)
The measurement of object and image distances.
CONCLUSION
In this unit you learned about refraction of light through lenses. You
also learned how to locate images formed by convex (converging) and
concave (diverging) lens and their applications. Further more; you
learned how to determine the focal length of a convex lens using the
quick but approximate method and by the measurement of object
distance (u) and image distance (v). Finally, you learned the derivation
and use of the lens formula.
5.0
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SUMMARY
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•
Convex lenses are converging lenses. They converge parallel rays
of light to a real principal focus.
•
Concave lenses are diverging lenses. They diverge parallel rays
of light away from a virtual principal focus.
The principal focus of a lens is the point on the principal axis to
which rays parallel and close to the principal axis converge after
refraction through a converging lens or appear to diverge from
after refraction through a diverging lens.
•
•
Focal length of a lens is the distance from the lens to the principal
focus.
•
The power of a lens is the reciprocal of the focal length expressed
in metres.
•
Convex lenses produce different images when the object is placed
at different positions.
•
Concave lenses always produce virtual diminished and erect
image.
•
Lens formulae
(i)
(ii)
6.0
Real – is – positive
1
1
+
v
u
=
m
=
=
m
=
v
u
New Cartesian
1 _ 1 = 1
v
u
f
v
u
=
1
f
height of image
height of object
height of image
height of object
TUTOR -MARKED ASSIGNMENT
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1.
Define
(i)
The principal focus; and
(ii)
The focal length of a converging lens.
2.
(a)
3.
An object is placed on the principal axis of a lens and at a
distance of half the focal length from the lens. Show by means of
ray diagrams where the image is formed when the lens is;
(i)
a diverging lens,
(ii)
a converging lens
Find the nature and position of the image of an object placed 10
cm from a converging lens of focal length 15 cm.
4.
7.0
State the characteristics of images formed by concave lens
(b)
What is the application of such a lens and why?
REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools.Onitsha: Africana FEP.
Awe, O’ and Okunola, O.O.(1992). Comprehensive Certificate Physics.
Ibadan: University Press.
Okpala, P.N. (1990). Physics. Certificate Year Series for Senior
Secondary Schools. Ibadan: NPS Educational.
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UNIT 5
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APPLICATIONS OF LIGHT WAVES
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
The Simple Camera and Projector
3.1.1 The Camera- Features and Operations
3.1.2 The Projector- Features and Operations
3.2
The Human Eye
3.2.1 The Features and Operations of Human Eye
3.2.2 The Action of the Eye
3.2.3 The Defects of Vision and their Corrections
3.2.4 Comparison of the Human Eye and the Camera
3.3
Simple and Compound Microscopes
3.3.1 The Simple Microscope or Magnifying Glass
3.3.2 The Compound Microscope
3.4
The Telescope
3.4.1 The Astronomical Telescope
3.4.2 Reflecting Telescope
3.4.3 The Terrestrial Telescope
3.4.4 The Galilean Telescope
3.4.5 The Prism Binoculars
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
You have learnt about the reflection and refraction of light in units 1 to 4
of this Module. In this unit you will learn to apply the knowledge you
acquired from the earlier units to explain the working of cameras,
human eyes, projectors, microscopes and telescope. You will also learn
to describe these optical instruments with respect to the nature and
functions of their various parts.
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OBJECTIVES
By the time you finish this unit you will be able to:
•
•
•
construct a model of a box camera
explain the formation of images by the camera and the projector
by tracing rays of light through them
explain the roles played by some parts of the eyes in the
formation of image on the retina
compare and contrast the eye and the camera
state the defects of the eye and their causes
identify the type of lenses for correcting the various defects of
the eyes
trace the paths of light rays through simple and compound
microscopes and telescopes.
•
•
•
•
3.0
MAIN CONTENT
3.1
The Simple Camera and Projector
3.1.1 The Camera – Features and Operations
Fig. 5.1: A lens camera
The features and operations of the lens camera
-
Consist of a light – tight box which contains a convex lens in front
and a film at the opposite end
The lens converges the incident rays from objects.
the film serves as a screen on which the image is received
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characteristics of image formed
 Real
 Inverted
 Diminished
-
Focusing screw alters the distance of lens from film in order to
always focus the image on the film
Diaphram controls the amount of light entering the camera. This is
made possible by the variable aperture or hole.
Shutter controls the amount of light entering the camera. It has
variable speed and therefore can open for different lengths of time
(exposure time).
The extent to which the diaphragm and the shutter combine to determine
the amount of light entering the camera depends on:
 Sensitivity of the film;
 Brightness of the object; and
 Kind of effect that is wanted.
3.1.2 The Project –Features and Operations
The projector is an instrument for projecting on a screen an enlarged
image of transparent object such as a slide. That is a projector is used for
showing enlarged image of transparent slides and films on a white
screen.
Fig. 5.2: The slide projector
-
A powerful source of light is essential as the light on the screen has
to spread over many times greater than that on the slide (film) in
order to achieve the high magnification required. The camera
projector magnification could be as high as 100 times. The powerful
but small light source illuminates the object which is usually nonluminous A converging mirror placed behind the light source helps
to direct the light onto the film or slide. This source of light is at the
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focus of the first lens aid the second lens converges the parallel beam
upon the projection lens. The convergent beam illuminates the slide
(film).
-
The projection lens is adjusted so that the slide (film) is just beyond
its principal focus.
-
The slide (film) must be placed upside down so that the image on the
screen appears the right way up.
-
A device to absorb the intense heat from the source of light is usually
interposed between the condenser and the slide ( film) or a fan is
incorporated to send a cooling draught on the slide (film) to prevent
it from over heating
-
Characteristics of image formed
 Real
 Enlarged
The principle of the slide project is used in cine- projector, photographic
enlargers and several other devises.
3.2
The Human Eye
3.2.1 The features and operations of the human eye
L e n s
Lr
is
Aqu
eous
hum
our
R
P u p il
a
i n
e t
Ye
w
ll o
Sp
nd
B li
ot
Sp
ot
C o rn e a
Ci
lia
ry
Mu
sc
le
Optic nerve
V itr e o u s
Fig. 5.3: Optical features of the human eye
-
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The cornea is transparent and allows light rays to pass through to
meet the eye lens.
The eye lens focuses the image of the object on the retina.
PHY 001
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The sensitive retina sends message to the brain by way of optic
nerve.
The pupil is a hole in the middle of the iris. It appears black
because no light is reflected from it.
-
The iris is the coloured portion of the eye. It adjusts the size of
the pupil to vary the amount of light that enter the eye through the
pupil
-
The cilliary muscle alters the focal length of the eye lens inorder
to focus images on the retina (an action referred to as power of
accommodation of the eye). The contraction and relaxation of the
cilliary muscles vary the thickness of the lens (ie the focal
length).
-
The aqueous humour is a transparent liquid and fills the space
between the cornea and the lens
-
The vitreous humour is a jelly - like substance that fills the space
between the lens and the retina.
3.2.2 The action of the eye
The normal eye in the relaxed state is focused on infinity known as the
far point .
-
Near point and far point: the nearest point at which an object can
be clearly seen is called the near point. The farthest point at
which an object can be clearly seen is called the far point.
-
To see near objects clearly, the cilliary muscles increase the
curvature of the eye lens and so enable it to focus the divergent
light. Accommodation is effective upon the near point.
-
For young children, the near point may be as close as 7.5 cm.
-
The least distance of distinct vision for the normal eye is 25 cm.
(a)
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(b)
Fig. 5.4: Near point and far point
-
Accommodation: This is the ability of the eye to focus both near
and far objects on the retina through the action of the ciilliary
muscle by which the focal length of the lens is altered.
-
Binocular vision: This is the combination of the two overlapping views of an object seen by both eyes in such a way as to
create the right sense of distance and depth. By binocular vision
we get a proper perspective of an object.
-
Persistence of vision: A clear image formed on the retina gives a
sensation of vision which remains for a short but definite time.
The retention of the vision of the image is known as the
persistence of vision. if a light thicker at a very short interval of
about ten times per second, it appears to the eye as a continuous
source of light owing to the persistence of vision. This
phenomenon is made use of in the production and showing of
motion pictures. A rapid succession of separate pictures cast on
the screen appears to have continuity and can depict motion. If
the speed of the successive pictures is reduced, the continuity is
broken and the moment in the sequence becomes jerky or a
thickening effect observed.
3.2.3 The Defects of Vision and their Corrections
The principal defects of vision which can be corrected with lenses are
short sightedness ( myopia) , long sightedness ( hypermetropia), loss of
accommodation (presbyopia) and astigmatism.. These eye defects are
corrected by the use of suitable spectacles or by contact lenses which are
placed to fit closely on the cornea. We are going to consider these eye
defects and their corrections under these headings: meaning, cause and
corrections.
i.
Short Sightedness (Myopia)
Meaning: This is the ability of a person to see near objects clearly but
not distant objects. This eye defect is called short- sightedness.
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Cause: it is caused by a person’s eye ball being too long and so focus
distant object in front of the retina. The far point for the eye is much
nearer than the normal eye.
Correction: A short – sighted person should wear a concave (diverging)
spectacle lens. This spectacle lens (concave) is used to form a virtual
image of a distant object at the person’s far point as illustrated in Fig.
5.5.
Fig. 5.5: Short – sightedness and its correction
Parallel rays from a distant object appear to diverge from a far point
(the principal focus of the spectacle lens) and make them appear to
come from a near distance. The short – sighted person has no problem
of seeing near object closer to the eye than 25 cm because with the
spectacle, the near point is slightly further away from the eye.
ii.
Long – Sight (Hypermetropia)
Meaning: This is the ability of a person to see distant object clearly but
not near object. Common example is those who require glasses to read
but not to see distant object.
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Cause: it is caused by the person’s eye ball being two short or eye lens
being too thin or the optical system of the eye being too weak so that
rays from the normal near point ( 25 cm) are brought to focus beyond,
instead of on, the retina. That is, rays from near objects focus behind the
retina.
Correction: A convex (converging) lens is used for the correction of
long- sight. The focal length should be such that it would produce a
virtual image of an object placed 25 cm away at the person’s near point
(Fig. 5.6).
iii.
Loss of Accommodation (Presbyopia)
Meaning: This is the loss of elasticity of the eye-lens. That is, the eye –
lens tends to become inelastic and unable to accommodate. The far point
usually approaches the eye while the near point recedes.
Cause: It is caused by increasing age which makes the eye-lens to
become inelastic and unable to accommodate.
Correction: It requires two pairs for its correction. A concave pair for
viewing distant object and a convex pair for viewing near objects. The
two pairs may be combined in one frame with the top and bottom parts
having different powers (bifocal spectacles).
iv.
Astigmatism
Meaning: this is when the focusing power of the eye is different in
different planes. The horizontal lines may appear in focus while the
vertical lines are not seen distinctly: This is caused by the distortion of
the curvatures of the cornea or of the eye-lens.
Correction: It is corrected by shaping the spectacle lens so as to have
different curvatures in different directions.
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3.2.4 Comparison of the Human Eye and the Camera
i.
Human Eye
Camera
The Eye
Iris
Pupil
Lens
Retina
Eyelid
Cilliary Muscle
The Camera
Diaphram
Aperture ( in the diaphram)
Lens
Screen
Shutter
Focusing ring
The similarities
 The human eye is impregnated with black pigments within
 The camera has light tight box painted black inside
 The pupil of the eye serves the same function as the aperture of the
camera
 Both the human eye and the camera have converging lens to focus
rays from an external object.
 Both have light sensitive materials on which the rays are focused the
retina in the eye and the film in the camera.
 The iris in the human eye serves the same function as the diaphram
in the Camera, to regulate the amount of light entering the eye
(camera).
iii.
The differences
 The human eye has the power of accommodation. That is, can vary
its lens focal length. While the focal length of the camera is fixed.
 The distance between the lens and the retina is fixed in the human
eye. In the camera, the distance between the lens and the film can be
varied
 The eye can suffer from the defect of vision unlike the camera.
 The eye is a biological organ while the camera is mechanical device,
leading to other differences.
SELF ASSESSMENT EXERCISE 1
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In this exercise you would need the following materials:
Two cardboard boxes of about 25 by 15 cm such that one box can fit
into the other, a convex lens of about 10 cm , tissues paper or any
transparent paper, an illuminated object of height equal to half the
height of the box and plasticize.
Use the above materials and follow the illustration of Fig. 5.7. Construct
a box camera thus:
(a)
(b)
Fig. 5.7: Making a camera
Make a hole of diameter of about 9 cm on one face of the cardboard
box. Remove the end of the box opposite the hole. Remove both ends of
the smaller box and cover one and with tissue paper or transparent
paper. Arrange the given lens to cover the hole (or opening) and hold it
in position with plasticize. Fit the smaller box into the larger box with its
covered end towards the lens. Place an illuminated object (burning
candle stick) in front of the lens. Adjust the position of the small box
until its harp image appears on the tissue or transparent paper.
For you to obtain best result, you should carry out this exercise in a dark
room or a fairly dark corner of a room
Describe the characteristic of the image you see on the film of the box
camera which you have built.
2.
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With the aid of ray diagrams, illustrate
PHY 001
i.
ii.
iii.
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The ability of a short- sighted person to see near object;
his inability to see distant objects ; and
the correction of the eye defect using diverging spectacle
lens.
3.
With the aid of ray diagrams, illustrate
i.
the ability of a long- sighted persons to see distant objects;
ii.
his inability to see near object, and
iii.
the correction of the eye defect using converging spectacle
lens
4.
State the differences and the similarities between the human eye
and the lens camera
5.
With the aid of a labelled diagram illustrate the formation of
image by the projector by tracing rays of light through it.
3.3
Simple and Compound Microscopes
3.31 The Simple Microscope or Magnifying Glass
For viewing near object, the simple microscope or magnifying glass is a
convex lens which is used to produce magnified images of small objects.
The objects are usually placed nearer to the lens than the principal focus
so that an erect virtual and magnified images seen clearly at 25 cm from
the normal eye. Magnification of the object produced by the lens is
given by the formula
M= 1- D
f
=1+D
F ( image is virtual )
Where
M = angular magnification
D = least distance of distinct vision (for a normal
eye, D= 25
cm)
F = focal length of the lens.
Thus, the image subtends a larger visual angle than that produced when
the object itself is placed 25 cm form the unaided eye and therefore
looks bigger. This is illustrated in Fig. 5.8.
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Fig. 5.8: Simple microscope or magnifying glass
The simple microscope is used for reading small print and for studying
biological specimen in the laboratory
3.3.2 The Compound Microscope
For scientist to produce higher magnification than that produced from a
simple microscope, a combination of two convex lenses is used in an
arrangement called the compound microscope.
The compound microscope consists of two convex lenses fitted at the
opposite ends of a tube. The tube is mounted on a stand so that it can be
lowered or raised for focusing. The object is placed under and close to
the lower lens which has a very short focal length and is called the
objective lens. The final magnified image is viewed through the upper
lens which has a longer focal length. This lens is called the eyepiece.
The eyepiece has a comparatively larger focal length than the objective
lens. The compound microscope generally has a horizontal platform
under the objective lens on which the object to be viewed is placed.
Usually there is a converging micro which serves to concentrate light on
the object.
The object to be magnified is placed in front of the objective such that
the object distance you is greater than the focal length of the objectives.
This forms a magnified, inverted and real image of the object at I, as
illustrated in Fig. 5.9
Fig. 5.9: A compound microscope
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If the eye piece is adjusted correctly, I1, will fall between the eyepiece
and its principal focus fe as shown. Then a magnified and virtual image
of I1, will be produced at I 2. Hence, the eye piece behaves just like a
magnifying glass. The final image of an object produced by a compound
microscope is highly magnified and inverted. The observer can adjust
the eyepiece so as to obtain distance of about 25 cm (least distance of
distinct vision for normal eye).
The magnifying power M of the compound microscope is the product of
the magnification due to the two lenses. That is, magnification M = M1
X M2 where M, is the magnification produced by the objective lens and
M2 is the magnification produced by the eyeprice. Thus, a microscope
with a magnifying power of 300 will show an object 300 times as large
as it would appear to the naked eye from a distance of 25 cm.
3.4
The Telescope
The telescope is an instrument for viewing distant objects such as the
stars and planets. There are many types of telescopes in use but,
basically, a telescope is either designed to be used for observing objects
on the earth’s surface - the terrestrial telescope or for observing celestial
objects- the astronomical telescope.
All telescopes consists of objective convex lens or concave mirror of
long focal length because of the distance of the objects to be viewed and
an eyepiece lens of short focal length for producing magnified image.
3.4.1 The Astronomical Telescope
An astronomical telescope is used to view very distant objects like the
planets and the stars. It makes use of two convex lenses O and E as its
objectives and eyepiece respectively as earlier stated. The objective has
a long focal length and the eyepiece a short focal length. Owing to the
great distance of the object from the telescope, the light rays from any
point on the object are very nearly parallel on reaching the objective.
Astronomical telescope can either be adjusted such that its image is
formed at infinity (normal adjustment) or at a fairly near point. For
instance
i.
When the final image is formed at infinity as illustrated in Fig.
5.10 the telescope is said to be in normal adjustment.
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Fig. 5.10: An astronomical telescope in normal adjustment
In such an adjustment, the two lenses are arranged such that the distance
between them is equal to the sum of their individual focal lengths ( fo+
fe). In this case, the objective lens converges parallel rays from all
points in the object and forms a real, inverted and diminished image I 1
on its focal plane. The focus of the eyepiece coincides with that of the
objective. Therefore, the eyepiece forms a virtual image of this image
(which now serves as object) at the focal plane at infinity and this is the
image that will be seen by the eye. Hence the total length of the
telescope is the sum of the focal lengths of the objectives and the
eyepiece.
In other words, if the position of the eyepiece is adjusted such that I 1
(the real image) is made to fall on the principal focus of the eyepiece,
the final image will be formed at infinity
ii.
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On the other hand, an astronomical telescope can also be adjusted
such that its image is formed at a fairly near point to the eyepiece.
In this case, the position of the eyepiece is adjusted such that I,
(the normal image) falls between the eyepiece and its principal
focus fe, a magnified and virtual image of I, is then formed at I 2
as illustrated in Fig. 5.11.
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Fig. 5.11: Final image at near point astronomical telescope
Here, the eyepiece of the astronomical telescope performs exactly the
same function as the eyepiece of the compound microscope. You can
see from Figs. 5.10 and 5.11 that the final image of the object produced
by an astronomical telescope is inverted.
The magnification produced by the astronomical telescope is given by
the ratio of the angle subtended at the eye by the image of a distant
object when a telescope is used to the angle subtended at the unaided
eye by a distant object. That us, the angular magnification of the
telescope. This angular magnification is given by
M = a/a
where
a = h/fe
and
a = h/fo
where
fe and fo are the eyepiece and objective focal lengths respectively (Figs.
5.10 and 5.11) and h is the height of the real image I.
This implies that magnification
M= fo / fe.
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3.4.2 Reflecting Astronomical Telescope
This is another form of astronomical telescope. It makes use of a large
concave parabolic mirror with a polished coating of aluminum on the
reflecting surface. The concave parabolic mirror serves as the objective
of the telescope. It has a large diameter that enables it to gather a lot of
light and also a long focal length which enhances increase in
magnification of the image in order to reveal minor details in the object.
The use of these two properties – large diameter and long focal length of
concave mirrors are not easily obtained in convex lenses.
This telescope is very effective in the sense that it is able to collect as
much light rays as possible from distant stars and planets thereby
produces a fairly sharp image. Therefore, the most powerful telescopes
are the reflecting ones.
Parallel rays from a distant object are reflected at the objective concave
mirrors. These reflected rays are intercepted by a small plane mirror
before they form a real image I1 (Fig. 5. 12). This image is magnified in
the usual way by the eyepiece coverging lens
Fig .5.12:
Reflecting astronomical telescope
Telescopes that use concave mirror as objectives are referred to as
reflector or refectory telescopes while those that use lens as objectives
are called a refractor or refracting telescope.
3.4.3 The Terrestrial Telescope
This is a modified astronomical telescope for viewing distant objects on
the earth surface. The difference between the terrestrial telescope and
the astronomical telescope is that a convex lens is put in between the
objective and the eyepiece so that a final image in erect position, but
with the size unchanged is obtained.
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For instance, if a convex lens C of a suitable short focal length fc is
placed at a distance d= fo+ 2fc from the objective as illustrated in Fig.
5.13 a, magnified and erect image will be produced by the eyepiece.
Fig. 5.13: Terrestrial telescope
In the terrestrial telescope, the lens C acts as an erecting lens and reinverts the image I1, formed by the objective. The re- inverted image I2
falls at the distance 2fc from C and is of the same size as I 1. The
eyepiece is adjusted such that a magnified image of 12 is obtained. The
final image I3 observed is an erect image of the object.
The terrestrial telescope is longer than the astronomical telescope by
four times the focal length of the erecting lens. Also, distant objects
appear near and in their natural orientation (that is, upright) when
viewed through the terrestrial telescope.
3.4.4 The Galilean Telescope
The Galilean telescope uses convex lens of long focal length as the
objective and a concave lens of short focal length as the eyepiece.
Parallel rays from distant object refracted by the objective lens are
supposed to form a real image I1, at its principle focus, fo. But, before
this image could be formed, the concave eyepiece intercepts the
converging rays and makes them to diverge as illustrated in Fig. 5.14.
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Fig. 5.14: The Galilean telescope
Thus, the eyepiece forms an erect, virtual and magnified image at I2 . In
this situation, the distance between the two lenses, mounted on common
axes is equal to the difference between their focal lengths (ie, d = fo- fe).
Thus, the distance between the objective and the eyepiece in a Galilean
telescope is approximately the difference in their focal length (fo-fe) this
design is often mounted in pairs for low power binocular used as “opera
glasses” and night glasses. They give a very small field of view except
with very low magnifications, but as they only use two lenses to give an
erect image, the light loses within them are relatively small.
3.4.5 The Prism Binoculars
The prism binoculars consist of a pair of astronomical telescopes each of
which is fitted with two right – angled isosceles prisms. The
arrangement of the two prisms serves a similar function to that of the
erecting lens of a terrestrial telescope. That is, they revert the image
formed by the objective and present an erect image of the object to the
viewer.
That is, in the prismatic binoculars, each side consists of an
astronomical telescope effectively folded in three. One half of this
arrangement of astronomical telescope is illustrated in Fig. 5.15.
Fig. 5.15: Action of one half of prism binoculars
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The re- inversion of the image is done in two stages, by two lateral
inversions using 45o reflecting prisms which in addition to laterally
inverting as a mirror would do, also displace the rays sidewards so that
they return along the next limb.
The two prisms are set so that the right - angle edges are horizontal and
vertical in the figure- not parallel to one another
The first prisms P turns the real inverted image formed by the
objective in a horizontal direction (turns the image round and corrects
for lateral inversion ) while the second prism Q turns this image the
right way up). This erect image of the object is presented to the eyepiece
which subsequently magnifies it. Without the prisms the image would
be upside down and laterally inverted.
The major advantage of prism binoculars is that they can give wide field
of view with a reasonable high magnification.
SELF ASSESSMENT EXERCISE 2
1.
With the aid of labeled diagrams trace the path of light rays
through
a.
b.
c.
d.
e.
f.
g.
2.
State for each of the optical instruments listed in ‘1’ above
a.
b.
c.
4.0
Simple microscope
Compound microscope
The astronomical telescope
The reflecting astronomical telescope
Terrestrial telescope
The Galilean telescope
The prism binoculars
the advantage (s)
The disadvantage (s) and
The use (s)
if any
CONCLUSION
In this unit you studied the applications of lenses. You applied the
knowledge you acquired from earlier units in the study of the human eye
and other optical instruments. You learnt the formation of images by the
human eye, the camera, microscope and telescope. You also learned the
similarities and the differences between the human eye and the camera,
the defects of human eyes and the types of lenses for correcting the
various defects. Finally, you learned how to describe the projectors,
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microscopes and telescopes with respect to the nature and function of
their various parts.
5.0
SUMMARY
• A camera consists of a converging lens which forms real images of
objects on a film.
• In the file projector, a projector (convex) lens is used to project on a
screen, a magnified image of a small picture on a film.
• The human eye is able to see objects at different distances because it
has the ability for varying the focal length of its lens
(Accommodation).
• The defects of the eye- long sight and short – sight can be corrected
by using suitable lenses such as convex, and concave lenses
respectively
• A single convex lens can act as a simple microscope (magnifier), if
the object is placed within its focal length.
• The compound microscope consists of an objective lens of short
focal length and an eyepiece lens of longer focal length.
• The astronomical telescope consists of a converging objective of
long focal length and a converging eyepiece of short focal length.
• The terrestrial’ telescope, a modified version of the astronomical
telescope has a converging erecting lens between the objectives and
the eyepiece.
• The Galilean telescope has a convex lens as its objectives and a
concave lens as its eyepiece; it produces a final erect image.
• Prism binoculars consists of a pair of astronomical telescope each
of which is fitted with two right –angled isosceles prisms; it is
optically equivalent to a terrestrial telescope which is about three
times longer in length.
6.0
TUTOR-MARKED ASSIGNMENT
1.
Explain with a ray diagram how a converging lens of suitable
focal length may be used as a simple magnifier
2
a.
b.
3.
In using a slide projector, the image of the slide must be
i.
the right way up
ii.
magnified
Explain briefly how this is done.
484
State two advantages of binocular vision.
How does the eye adjust itself to deal with
i.
Light of different intensity?
ii.
Objects at varying distances?
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4.
Sketch diagrams indicating the types of lenses used and the
positions and nature of image formed by an astronomical
telescope in
i.
normal adjustment
ii.
at near point
5.
With the aid of ray diagrams illustrate
a.
The
ability of a long- sighted person to see distant
objects;
b.
His inability to see near objects;
c.
The correction of the eye defect using converging
spectacles lens
7.0
REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (2000). New School Physics for Senior Secondary
Schools. Onitsha, Africana FEP.
Nelkon, M. (1999). Principles of Physics for Senior Secondary Schools.
Ibadan Longman Group Limited
Rabi, k, George , K. O. and Hui, T. C. (1991). New School Physics .
Certificate Science Series. Singapore . FEP International Private
Limited
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UNIT 6
DISPERSION OF LIGHT AND COLOURS
CONTENTS
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Introduction
Objectives
Main Content
3.1
The Concept of Dispersion of Light
3.2
White Light Spectrum
3.3
Production of Pure Spectrum
3.4
Primary, Secondary and Complementary Colours
3.4.1 Primary Colours
3.4.2 Secondary Colours
3.4.3 Complementary Colours
3.5
Mixing of Coloured Pigments (paints)
3.6
Colours of Objects
3.7
Light Filters
3.8
Formation of Rainbow
3.9
The Electromagnetic Spectrum
Conclusion
Summary
Tutor-Marked Assignment
References/Further Readings
1.0
INTRODUCTION
In this unit you will learn about the dispersion of light and colours.
Specifically, you will learn about the white light spectrum, its
combination and production. You will also learn about the primary,
secondary and complementary colours. In addition you will learn about
mixing of colour pigments (paints), colours of object in white lights and
light filters.
2.0
OBJECTIVES
By the time you finish this unit, you will be able to:
• obtain the spectrum of white light
• show dispersion of white light by a triangular prism
• explain the formation of rainbow and why objects appear the way
they do.
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3.0
MAIN CONTENT
3.1
The Concept of Dispersion of Light
Dispersion as you know is the breaking up of or separation of white
light into its component colours when it is passed through a glass prism.
It is due to the fact that different colours of white light travel at different
speeds through the glass. As a result of which each colour is refracted in
a slightly different direction or angle through the glass. That is each
colour has its own wavelength, velocity and refractive index of
refraction. Thus, white light is split up into seven colours: Red, Orange,
Yellow, Green, Blue, Indigo and Violet. We remember these colours by
picking the first letter of each colour to form ‘ROYGBIV.‘
3.2
White Light Spectrum
When a ray of light is inclined obliquely on a glass prism, it emerges
deviated and separated into its component colours- Red, Orange,
Yellow, Green, Blue, Indigo and Violet (ROYGBIV), due to refraction.
These colours resulting from the separation (i.e. dispersion) of white
light make up the spectrum which is generally referred to as the white
light spectrum.
Sir Isaac Newton was the first to observe in an experiment that when
white light is passed through a prism, an elongated coloured patch of
light is obtained on a screen placed behind the prism. This coloured
pattern now referred to as the spectrum of white light consists of the
Red, Orange, yellow, Green, Blue, Indigo, Violet (ROYGBIV) earlier
mentioned. An observation of the spectrum shows that the red light is
deviated least while the violet light is deviated most. Because red is
deviated least from its original direction in air, its change of speed on
entering glass is least and as such its speed in glass is greatest .Violet is
deviated most on entering glass and it has the least speed has in glass.
The white light spectrum is illustrated in Fig. 7.1.
Fig. 7.1: Dispersion of white light resulting into the white light spectrum
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If another identical prism is placed to intercept the refracted rays as
shown in Fig. 7.2 a, the same arrangement of colours will energy on the
screen. This time, however, the colours will be more widely separated.
On the other hand, if the second prism is inverted as shown in Fig. 7.2b,
the colours will disappear and only a patch of white light will be visible
on the screen
Fig. 7.2 a: Dispersion of white light spectrum
Fig. 7.2b: Recombination of white light spectrum
The disappearance of the colours was due to their recombination to
reproduce white light. Hence white light spectrum is the band of colours
formed when white light passes through a prism.
SELF ASSESSMENT EXERCISE 1
Divide a circular disc into seven parts and paint each part with one of
the
ROYGBIV colours (Fig. 7.3).
Fig. 7.3: Newton’s colour disc
Spin the colour disc to make it rotate at high speed. You will observe
that the colours are mixed. These colours you can see seem to be white.
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Now slow down the rotation of the disc. As you do this, you will notice
the individual colours again.
Monochromatic light
You should note that if light of one wavelength (α) and colour
(monochromatic) is passed through a prism refraction; occurs without
dispersion. That is, if red light passes through a prism it will come out as
red light (Fig. 7.4).
prism
Red light
Red light
Fig. 7.4: No dispersion with monochromatic light
3.3
Production of Pure Spectrum
Have you noticed that the spectrum produced by a prism (Fig. 7.1) is an
impure spectrum because the different colours overlap. If you have not,
perform that simple experiment illustrated in Fig. 7.1 to observe the
spectrum.
A pure spectrum is that in which the colours are clearly separated or
distinct from each other. To produce such a pure spectrum, prism is used
in combination with converging lenses of as shown in Fig. 7.5 below.
Fig. 7.5: Production of pure white light spectrum
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The apparatus consists of these components – source of white light, slit,
two convex lenses, one triangular prism and a screen.
The narrow slit illuminated by a bright source of white light produces a
series of narrow coloured images which minimizes the chances of over
lapping colours.
This sources of white light is placed at the focal point of the first lens
(i) This enables the converging lens ( i) to produce a white parallel
beam of light as the light incident on it are refracted. As you already
know the prism disperses the beam of light into its component colours
while the converging lens (ii) collects the different colours and
converges them differently on the screen.
The screen is placed at the focal plane of the converging lens (ii) so that
the coloured bands of the spectrum will appear well defined while the
slit is placed at the principal focus, f, of the converging lens (i) to ensure
that parallel rays are produced by the lens. Production of clear spectrum
is also ensured by using a very fine slit and a bright light source.
This pure spectrum consists of Red, Yellow, Green, Blue, Indigo and
Violet. Red is least deviated while Violet is the most deviated.
3.4
Primary, Secondary and complementary Colours
3.4.1 Primary Colours
Primary colours are red, green and blue, they are called primary colours
because it is not possible to produce them by mixing other colours.
When three primary coloured lights are mixed they produce white light.
3.4.2 Secondary Colours
Secondary colours are those colours that can be obtained by mixing any
two primary colours. Examples are yellow, cyan and magenta. They are
produced by mixing two primary colurs such as
• Green + blue=
• Red + blue =
• Green + red =
cyan (secondary colour)
magenta (secondary colour)
yellow ( secondary colour)
This is illustrated in Fig. 7.6 below.
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Fig. 7.6: Additive colour - mixing of lights
3. 4.3 Complementary Colours
If two colours, one primary and the other secondary are such that when
mixed give white colour, then they are said to be complementary
colours.
Examples are:
• Yellow + blue
• Cyan + red
• Magenta + green
=
=
=
white
white
white
This is also illustrated in Fig. 7.5 above.
3.5
Mixing of Coloured Pigments (paints)
In contrast to the mixing of colored light, when blue and yellow paints
are mixed, the result is green not white. The colours of pigments (paints)
are the result of the kinds of light they reflect. For instance, yellow paint
reflects red, yellow and green light and absorbs the rest. On the other
hand, blue paint reflects blue and green light and absorbs the rest. But,
when yellow and blue paints are mixed the only colour reflected by both
components of the mixture is green, which is thus the colour of the
mixture.
This mixture of paints is known as colour mixing by subtraction (Fig.
7.7) it is different from mixing of coloured lights which are pure colours
called mixing by addition
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Fig. 7.7: Subtractive colour -mixing of paints
The primary colours for paints unlike lights are red, yellow and blue.
They are the basis for all other colours of paint and cannot be produced
by mixing. When a primary colour is mixed with a secondary colour on
the opposite side of the triangle, the result will be black as shown in Fig.
7.7.
3.6
Colours of Objects
The colours of objects in white and coloured lights depend on the
colours in the light incident or falling on them and also on the
absorption and reflection of these lights by the objects.
A white object appears white in daylight because it reflects equally all
the colours of the spectrum. Also the colours of opaque objects seen are
the colours of light that are reflected. The following are some examples:
• Rose flower appears red because it reflects red light and absorbs the
other colours
• White objects reflect all the colours of the spectrum and appear
white.
• Black objects absorb all the colours of spectrum and appear black.
That is, objects appear to be of a certain colour because they reflect
light of those colours and absorb the rest. Primary colours will reflect
or transmit only their respective colours. Other colours will reflect or
transmit other colours along with their own colours
Light filters are transparent- coloured sheets which allow light of the
colour of the filter to pass through while absorbing all other colours in
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the incident light they are made of gelatin and coloured with various
dyes.
3.7
Light Filters
Filter papers are used to absorb light of various colours from the
spectrum of white light. Generally, a filter paper absorbs all other
colours except that of the filter. For instance;
• When a red filter is placed on the path of white light, only red light
passes through ;
• When a green filter is placed on the path of a red light, no light
passes through. Hence, there is a black appearance.
An unusual situation occurs when a yellow filter is placed on the path of
white light. This time all other colours are absorbed except yellow,
green and red.
Subtractive colour- mixing also results from the super- position of two
colour filters. A combination of two filters results in the transmission of
only those colours which are common to both of them.
3.8
Formation of Rainbow
Have you seen a rainbow before? It is a natural phenomenon sometimes
seen after a rain. It is formed in the same way as a spectrum is formed
by a glass prism. Here, rain drops or drops of water suspended in the air
after rainfall act as prisms which disperse the white beam of light into
seven colours of the rainbow. The viewer can see the rainbow only if he
is in a position to receive the refracted rays of the sunlight. This is
illustrated in Fig. 7.8.
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Sometimes, two rainbows are seen namely the primary and the
secondary rainbows
i.
The Primary Rainbow
The primary rainbow is brighter than the secondary rainbow. It is caused
by light rays which undergo one internal reflection inside the rain drop.
When a horizontal light ray enters a water droplet at point A, dispersion
takes place (Fig. 7.9 a) At R1, the coloured rays undergo total internal
reflection and they exist from the water droplet at B where further
refraction occurs. The violet ray is refracted the most and the red ray the
least. The red ray which emerges from each droplet makes a larger angle
of about 42o with the horizontal than the violet ray that makes about
angle 40 o also with the horizontal.
a.
One internal
reflection inside
droplet
b, Two internal reflections
inside droplet
Fig. 7.9: Formation of rainbow- dispersion of sunlight by rain drops
Other droplets seen by the observer at the same angle with the horizontal
produce the same colour. Hence, an arc of the colour, red is seen by the
observer. Other coloured rays emerge at different angles to the
horizontal. Thus, an arc of coloured bands- the rainbow – is seen in the
sky. This band has red on its outer edge and violet on the inner edge
(Fig. 7.10).
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Fig. 7.10: Primary and secondary rainbows
ii.
The Secondary Rainbow
Secondary rainbows may occasionally be seen in addition to the primary
ones. This is caused by light rays which undergo two internal reflections
inside the rain drops. This causes the coloured bands of such rainbow to
be in the reverse order. That is, it has violet on its outer edge and red on
its inner edge ( Fig. 7.10) . In the secondary rainbow , the red rays are at
angle of 50o and the violet rays at 540 to the horizon
3.9
The Electromagnetic Spectrum
You have seen that the spectrum of white light obtained from sun- light
consists of seven colours. Beyond each end of this coloured spectrum
(visible light) are the invisible rays (spectrum) of radiant energy called
infra-red and ultraviolet rays. These rays are also emitted by the sun.
These rays cannot be seen but they cause certain materials to
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fluorescence and also affect photographic plates. They are known as
ultra-violet rays.
Radio waves, infra-red rays, the visible rays (from red to violet) ultraviolet rays, x-rays and gamma rays are all electromagnetic waves.
These waves make up the electromagnetic spectrum. They all travel
through a vacuum with exactly the same speed and differ only in having
different wavelengths.
SELF ASSESSMENT EXERCISE 2
1.
2.
3.
4.
5.
6.
7.
8.
4.0
What do you understand by the dispersion of light?
Draw a clear diagram to show the dispersion of light through a
prism
On what factors do the colours of objects depend?
Distinguish between primary and secondary colours. List the
colours under their various groups.
What do you understand by white light spectrum? With the aid of
a diagram describe how you can obtain a pure spectrum of white
light.
Name the primary colour for paints
Explain how a rainbow is formed. With the aid of diagrams
distinguish between primary and secondary rainbows.
List the radiations that make up the electromagnetic spectrum
CONCLUSION
In this unit you learned about the dispersion of light and colours, you
learned about the white light spectrum, its combination and production.
You also learned about primary, secondary and complementary colours
as well as light filters, colour pigments and the appearance of different
colours of objects in white light. Finally, you learned about the
formation of rainbow and radiation that made up the electromagnetic
spectrum.
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SUMMARY
• Dispersion is the separation of light into its constituent colours.
• White light is composed of seven colours red, orange, yellow, green,
blue, indigo and violet. These colours make up the white light
spectrum
• A pure spectrum of white light can be produced by using a bright
source of white light, a narrow slit, two converging lenses, a prism
and a screen.
• The three primary colours are red, green and blue. Secondary colours
are obtained by additive combination (mixing) of any two of the
secondary colours. The secondary colours are yellow, cyan and
magenta
• Paint (pigments) are mixed by a process of subtractive colourmixing. The primary colours for paint are red, yellow and blue.
• Light filters are transparent coloured sheets which allow light of the
colour of the filter to pass through while absorbing all other colours
in the incident light.
• The rainbow is caused by dispersion of sunlight by water droplets
left suspended in the air by the rain
• The electromagnetic spectrum is made up of radio waves, infrared
rays, the visible rays (from red to violet), ultra- violet rays, x-ray and
gamma rays.
6.0
TUTOR -MARKED ASSIGNMENT
1.
a.
b.
2.
With the aid of labelled diagram only illustrate the production of
pure white light spectrum
3.
What is a rainbow? What causes it? Name the two types.
On what factors do the colours of an object depend?
Why does a white object appear white in daylight?
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REFERENCES/ FURTHER READINGS
Anyakoha, M.W. (200). New School Physics for Senior Secondary
Schools Onitsha, Africana FEP.
Awe, O. and Okunola, O.O. (1992). Comprehensive Certificate Physics
Ibadan: University press
Nelkon, M. (1999). Principle of Physics for Senior Secondary Schools
Ibadan: Longman Group Limited
Ravi, K., Geroge, K.O and Hui, T. C. (1991). New School Physics.
Certificate Science Series. Singapore. FEP International Private
Limited.
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