A Practical Approach

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Tirivabaya Kwirirai
Courage Madondo
Learner’s Book
 A Practical Approach
to

Physics
Form 3

Learner’s Book

Tirivabaya Kwirirai Courage Madondo

 CONTENTS
Chapter 1: Measurements and physical quantities...................................................... 1
Unit 1.1 Measuring physical quantities.......................................................................................... 1
Unit 1.2 Density...................................................................................................................................... 13
Unit 1.3 Scalars and vectors............................................................................................................... 15
Chapter 2: Kinematics..................................................................................................... 20
Unit 2.1 Speed, velocity and acceleration..................................................................................... 20
Unit 2.2 Graphs of motion, bottom of form................................................................................. 22
Unit 2.3 Motion under gravity........................................................................................................... 30
Chapter 3: Forces............................................................................................................ 38
Unit 3.1 Effects of force on shape and size of materials........................................................... 38
Unit 3.2 Effects of force on motion.................................................................................................. 46
Unit 3.3 Friction and circular motion.............................................................................................. 48
Unit 3.4 Turning effects of force....................................................................................................... 52
Unit 3.5 Centre of mass / centre of gravity................................................................................... 55
Examination 1: Paper 1 ........................................................................................................ 63
Examination 1: Paper 2 ........................................................................................................ 70
Examination 1: Paper 3 ........................................................................................................ 73
Chapter 4: Pressure......................................................................................................... 76
Chapter 5: Machines....................................................................................................... 89
Unit 5.1 Simple machines................................................................................................................... 89
Chapter 6: Mechanical structures.................................................................................. 97
Unit 6.1 Beams........................................................................................................................................ 97
Unit 6.2 Trusses....................................................................................................................................... 102
Unit 6.3 Joining materials................................................................................................................... 105
Unit 6.4 Large structures..................................................................................................................... 108
Examination 2: Paper 1 ........................................................................................................ 117
Examination 2: Paper 2 ........................................................................................................ 123
Examination 2: Paper 3........................................................................................................ 128
Chapter 7: Work energy and power............................................................................... 130
Unit 7.1 Work........................................................................................................................................... 130
Unit 7.2 Energy........................................................................................................................................ 132
Unit 7.3 Power......................................................................................................................................... 137
Chapter 8: Thermal physics............................................................................................ 142
Unit 8.1 The kinetic theory of matter.............................................................................................. 143
Unit 8.2 Thermal properties............................................................................................................... 152
Unit 8.3 Heat transfer............................................................................................................................ 163

Chapter 9: Internal combustion engines...................................................................... 173

Unit 9.1 Four stroke petrol engine................................................................................................... 173
Unit 9.2 Social and economic considerations of using fuels.................................................. 176

Chapter 10: Waves............................................................................................................. 179

Unit 10.1 Mechanical properties of waves...................................................................................... 179
Unit 10.2 Sound........................................................................................................................................ 185
Unit 10.3 Electromagnetic radiation................................................................................................. 187
Examination 3: Paper 1 ........................................................................................................ 192
Examination 3: Paper 2 ........................................................................................................ 202
Examination 3: Paper 3 ........................................................................................................ 205
 Chapter MEASUREMENTS AND

1
PHYSICAL QUANTITIES

Chapter objectives
By the end of this chapter, learners should be able to:
• measure physical quantities; read an instrument scale to the nearest fraction of a division.
• determine density of regular and irregular objects.
• express quantities in terms of S.I. units.
• derive other units from base units.

Introduction
A standard unit must be chosen before any form of measurement is done. The standard international
units (SI units) are the common units used internationally. It is a decimal system in which units are
divided or multiplied by ten to give smaller or larger units. In physics there are three basic quantities to
be measured, these are length, mass and time. The units for other quantities are derived from the base
quantities.

UNIT 1.1 MEASURING PHYSICAL QUANTITIES

Table 1.1 Base quantities and SI units

Base quantity Name of unit Symbol for unit
Length Metre M
Mass Kilogram Kg
Time Second S
Temperature Kelvin K
Electric current Ampere A
Luminous intensity Candela Cd
Amount of substance Mole Mol

1
 Table 1.2 Derived quantities and their units
Derived quantity Relationship between base Symbol for unit Special name
and derived quantities
Area Length × width m²
Volume Length × width × height m³
Density Mass ÷ volume Kg/m³
Speed Distance ÷ time m/s
Acceleration Change in velocity ÷ time m/s²
Force Mass × acceleration Kgm/s² Newton (N)
Charge Current × time As Coulomb (C)
Heat capacity Energy ÷ change in temperature J/K

Length
Length is measured in metres (m). It is a distance travelled during a specific time interval. There are many
instruments used to measure the lengths, such as ruler, metre rule, tape measure and others. The correct
way of reading the length using a ruler or metre rule as in figure 1.1. Your eye must be directly over the
mark on the scale to avoid parallax error because the thickness of the ruler causes that.

1
10 = one tenth of a centimeter
3 =three tenth of a centimeter
10 3
= five tenth of a centimeter
10
8 = eight tenth of a centimeter
10
1 = one centimeter
0 1

1mm
1cm
Fig. 1.1 Metre rule

When taking readings from the metre rule, make sure that the line of vision is perpendicular to the scale
so as to avoid parallax error.

Eye sees 2,2cm (incorrect)

Eye sees 1,9cm

(correct)
3
2

Object

Fig. 1.2 Measuring length

2
Area
Table 1.3 Geometric formulas

PERIMETER
NAME FIGURE AREA
CIRCUMFERENCE
M
Triangle A=bxh
h 2 P = MN+NP+PM

N b I
D G
Parallelogram h A=bxh P = DE+EF+FG+GD
E b F

Rhombus h A=bxh P = b+b+b+b

P = 4b
b

P = L+W+L+W
Rectangle W
A=LxW P = 2L+2W
P = 2(L + W)
L

l P = l+I+l+l
Square A = l2
P = 4l
l

M B R
A = 12 (MR + NP) h
Trapezoid h P = MN+NP+PR+RM
A = (B+b) x h
N b P 2

Circle r
d A = ∏r2 C = 2∏r = ∏d

The SI unit of area is the square metre (m²) which is the area of a square with sides 1m long.

For a rectangle, the area is determined by area = length × width.

For a triangle, it is given by ½ × base × height.

The area of trapezium is given by ½ (a+b) h.
The area of a circle given by: лr² where л= 22/7.

3
Use of calipers
A pair of Vernier calipers is used for measuring distances on solid objects where an ordinary rule cannot be
applied directly. It consists of a pair of hinged steel jaws which are closed until they touch the object in the
desired position. It can also be used to measure the internal diameter of tubes.
The Vernier enables us to obtain accurately the second decimal place in centimetre measurements
without having to estimate fractions of a division by an eye.
How to read a Vernier caliper
Fig 1.3 shows a vernier calipers and how it is used in taking readings. The vernier calipers consists of
inside and outside jaws. It also have the main and the vernier scale on which readings are taken.

Inside jaws

Main Scale

0 1 2 3 4 5 6 10
cm
0 5 10
Depth bar

Vernier scale

Outside jaws

Fig. 1.3 Reading a Vernier caliper

When you are using a vernier scale, there are five steps that are to be followed, these are:
Check for zero error
This is done by closing the jaws and check that the zero mark on the main scale coincides with the zero
mark on the vernier scale. If there is no zero error, the reading will be as below.

0
Main scale 1

Vernier scale
0

Grip the object

Open the jaws and put the object between the jaws and grip it.
Read the main scale
Read the main scale opposite the zero mark on the vernier scale. On fig 1.3 the main scale reading is
2.1cm.

4
Read the vernier scale
Read line on the vernier scale which exactly coincides with the mark on the main scale. The vernier scale
reading is 0.05cm.
Adding main and vernier reading
2.1 + 0.05 = 2.15cm this is the final reading on the vernier calipers.
However, it is not always that the vernier scale will not have, a zero error. It might have in that case, it
must be dealt with accordingly. There are two types of zero errors, the positive and negative. Table below
will summarise on identification and correction of these errors.

Zero error Observed reading Corrected reading

Positive zero error
0 1 2 3

0 5 10
0 5 10
The zero mark on the Vernier Scale Observed reading = 2.24cm
lies on the right of the zero mark on Corrected reading
the main scale. = observed minus zero error
= (2.24 – 0.02)cm
The second marking on the Vernier
= 2.22cm
Scale coincides with the marking on
the main scale.
Zero error = + 0.02cm

Negative zero error

0 1 3 4

0 5 10 0 5 10

Zero mark on the Vernier scale lies to Observed reading = 3.03cm Corrected reading
the left of the zero mark on the main
= observed – zero error
scale.
= [3.03 – (– 0.01)]cm
The first marking from the "10" mark
on the Vernier scale coincides with a = 3.04cm
marking on the main scale.
Zero error = – 0.01cm

5
Volume of regular objects
Regular objects are those objects with specific dimemensions. Their volume can be calculated using the
formulae.
Cuboid

l
Its volume can be calculated by length × width × height
V = lwh
Cylinder
L

Its volume can be calculated by cross sectional area × length V = Al = πr2l

∏d2
V = L
4

Volume (as a quantity of space an object takes up)

• 1ml = 1cm³
• 1 000 ml = 1 litre= 1 000cm³
• 1 000 000 ml = 1m³ = 1000 litres = 1 000 000 cm³
The SI unit of volume is the cubic metre (m3) but this is rather large, for most purposes the cubic
centimetre (cm3) is used. The volume of a liquid can be measured through the use of measuring cylinder.
The volume of solid regular objects can be found through the use of linear dimensions. The volume of
irregular solids can be measured by submerging the solid in water in a measuring cylinder. The volume
of displaced water is the volume of the irregular solid.

7
When taking readings from the measuring
wrong
cylinder, the bottom of the water meniscus is read
horizontally at the eye level to avoid parallax error
as in figure 1.5. Take note of the following when 40
using a measuring cylinder: correct
• place the measuring cylinder on a flat
surface all the times you use it and wait
for the liquid to settle before taking any 30
wrong
measurement.
• always avoid parallax error by taking the
reading just below the meniscus. 20
Fig. 1.4 Readings from the measuring cylinder
Mass

The mass of a body is the quantity of matter it contains. The SI units of mass is the kilogram. The beam
balance is used to measure mass.
It has the following effects:
(i) All objects are attracted to the earth. The greater the mass of an object, the stronger the earth's
gravitational pull on it.
(ii) All objects resists attempts to make them go faster, slower, or in a different direction. The greater the
mass, the greater is resistance to change in motion.

Fig. 1.5 Triple beam balance Fig. 1.6 Digital scale balance

• 1 kg = 1000 grams (g)

• 1 tonne (t) = 1000 kg
• 1 g = 1000 milligrams (mg)
• 1g = 1000 000 micrograms (µg)

8
Time
Time is measured in seconds. The devices used to measure time include stop watches, ticker-tape timers
and electronic light gate timers.
Stopwatch
Scientists prefer to use stopwatch because it is best especially if you want to measure short time intervals.
Stopwatches are of two type namely: digital stop watch and analogue stop watch but the digital stopwatch
is the best to use as compared to analogue stopwatch because its more precise as it can measure time in
intervals of 0.01 s whereas analogue stopwatch can only measure time in intervals of 0.1 s.

Fig. 1.7 Analogue stopwatch Fig. 1.8 Digital stopwatch

Ticker-tape Timer
It is an electrical device which makes use of the oscillations of a steel strip to mark short intervals of time.

ticker tape
magnet carbon paper disc

Power input

coil

Steel strip

Fig. 1.9 Ticker-tape Timer

How a ticker tape timer measures time

The vibrating bar vibrates 50 times a second and make 50 dots in a second on the paper under the carbon
disc. The space between two consecutive dots represent a time interval of 1 s = 0.02 s. On fig 1.9 there are
50
4 dots and there are 3 spaces on a piece of paper, the time taken for the tape to pass through the time is
(3 × 0.02) = 0.06 s.

9
 Activity 1.1 Experiment
Aim: To measure time intervals.
Materials: stopwatch, metal bob, string, stand, metre rule, protractor.
Method:
1. Displace the pendulum to an angle about 15° as shown on figure below.

Point of
release

m2 Equilibrium
position

2. Release the bob and after at least one oscillation start the stop watch when the bob is at one extreme. Count
from zero the number of oscillations.
3. For at least 15 oscillations, read and record the time.
4. Repeat procedure 1-3, maintaining the same angle and record your results on the table below.

Time for 15 Average time/s for 15 Time for one oscillation (period)
oscillations oscillations in seconds
T₁ T₂ T₃ T average T

Time for 15 oscillations

L/cm t1/s t2/s tav/s T/s

5. Vary the length of the pendulum and repeat the experiment. Check if you get the same results.
6. Use the bob with a bigger mass and compare your results.

Activity 1.2 Experiment

Aim: To measure mass.

Materials: spring balance, retort stand, boss and clamp, mass hanger, masses labelled m1 and m2 and digital
balance.

11
 UNIT 1.2 DENSITY

There are many materials, both man-made and natural, and it would be useful to be able to compare
different materials scientifically. The simplest comparison is that of a materials’ density. The mass per unit
volume of a substance is density.
Density = mass ÷ volume
Density of regular objects
To calculate the density of a regular solid, you need mass and volume.
The mass of the solid can be measured using a beam balance. The volume of the solid is calculated using
the formula:
• Volume of a rectangular block = length × width × height
• Volume of a cylinder = ∏ × (radius)2 × height
The density of a regular object such as cube can easily be found mathematically by first measuring the
mass of the object and then divide it by the calculated volume.
Density of irregular objects
For solids that are irregular-shaped such as stones, the displacement method is used determine its
density.

Activity 1.3 Experiment

Aim: To determine the density of irregular object.
Materials: stone, water, measuring cylinder, beam balance.
Method

70 70

60 60
measuring cylinder eye
50 50

40 40
eye
30
30 30
stone
20
20 20

10
10 10

1. Measure the mass of the stone using a beam balance.

2. Half fill a measuring cylinder with water and record the level of the water.
3. Put the stone in the measuring cylinder and note the new volume.
4. Find the difference between the two levels of water to get the volume of stone that means volume before
putting the stone and after putting the stone.
5. Use the formula density = mass ÷ volume in g/cm³ or kg/m³
6. ρ = M/V where ρ is the density.
Volume of rectangular block = length × width × height

13
 UNIT 1.3 SCALARS AND VECTORS

All quantities are classified as either scalars or vectors. Table 1.4 shows some examples of scalars and
vectors.
Table 1.4: Scalar and vector quantities

Scalars Vectors
Speed Velocity
Time Acceleration
Mass Weight
Density Force
Energy Moment of a force
Addition of scalars
Scalars are added using simple mathematics of adding. For example, 10˚C plus 15˚C gives the answer
25˚C.
Addition of vectors
Addition of vectors is not as that of scalars since the direction of the vector quantities also needs to
be considered. Vectors are usually represented by arrows and the length of the arrow represents the
magnitude of the vector. The addition of two or more vectors of the same kind produces a resultant
vector. The resultant vector is the combined effect of the original vectors. The simplest vector addition
involves parallel vectors or vectors along a straight line.
Example
Calculate the resultant of two forces of magnitude 5N acting in the same direction.
5N
5N
Solution
The resultant is obtained by adding the two forces
5 + 5 = 10N.

Exercise 1.3

1. Calculate the resultant force in each of the following cases;

(a) 7N 7N

(b) 3N 7N

Vectors at an angle
The resultant of vectors at an angle can be obtained graphically using parallelogram method and vector
triangle.

15
Parallelogram Method
The parallelogram law of vector addition states that if two vectors acting at a point are represented by
the sides of a parallelogram drawn from that point, their resultant is the diagonal which passes through
the point of the parallelogram.
Example
The illustration below shows two forces, F1=6N and F2=3N acting on a body with an angle of 40°
between them.What is the resultant force R?

3N

3N R
F2
40°
6N Ø 6N
F1

Assignment
Work in pairs to discuss on how the vector method can be used to find the resultant of two vectors
shown below. The lines are drawn to scale.

Summary
Scalar quantities have magnitude only.
Vector quantities have both magnitude and direction.
Resultant of scalars can be calculated by direct adding while that of vectors can be calculated either
using parallelogram method or vector method.

Exercise 1.4

1. (a) State the difference between velocity and speed.

(b) In what way is velocity and speed similar?
2. Determine the resultant force in each of the following situations.
(a) (b) 3N 7N
6N 2N

6N 3N
(c) (d)

4N 7N 4N

3. Two forces of magnitude 6N and 9N act on a body. What are the maximum and minimum resultant
forces that act on the body?

16
Interesting facts

1. Density and buoyancy are interconnected. Low density objects will float on a material that is higher
in density (such as water).

Summary of the chapter

• Metre rule is accurate to 1mm.

• Stop watches are accurate to 0.1s or 0.01s.
• Ticker-timers are accurate to 0.02s.
• Base units are: metre; second; kilogram; kelvin; ampere, cd; mol.
• All other quantities are derived quantities, for example, speed, area, volume and density.
• Scales are either analogue or digital and for analogue scales, the smaller the range, the more
accurate they are.
• Density is the mass of the object divided by the volume of the object; units are kg/m³.
• An object with lower density than that of a liquid float in that liquid.

Glossary of terms

Physical quantity – a characteristic or property of an object that can be measured or calculated from
other measurements.
S.I. units – international system of units that scientists in most countries have agreed to use.
Derived units – units that can be calculated using algebraic combinations of the fundamental units.
Density – mass per unit volume.
Measurement – a process of detecting an unknown physical quantity by using standard quantity.
Mass – is a measurement of how much matter is in an object.
Vernier calliper – is a device used to measure linear dimensions.
Volume – is a quantity of 3-dimensional space occupied by a liquid, solid or gas.
Period – is the interval of time between successive occurrences of the same state in an
oscillatory or cyclic phenomenon.
Accuracy – the quality or state of being correct or precise.

Revision Exercises
Multiple Choice Questions
1. The SI unit for density is _____________.
A. g/cm³ B. g/cm2
C. kg/m³ D. m/s²
2. The SI unit for temperature is ______.
A. millilitres B. degrees celcius
C. ohmmeter D. kelvin
3. Which of the following instruments is ideal for measuring thickness of a water pipe?
A. Forcemeter B. 30 cm ruler
C. Vernier calipers D. String
4. The time taken by a simple pendulum to cover 40 complete oscillations is 50 seconds. What is the
period of the pendulum?
A. 1.25 seconds B. O.5 minutes
C. 0.8 seconds D. 0.1 minutes

17
5. Which is a base measurement?
A. Density B. Volume
C. Pressure D. Time
6. Which step is not necessary in the determination of the volume of an irregular solid object using a
measuring cylinder?
A. Making sure the object is at the bottom of the cylinder.
B. Lowering in the object gently.
C. Using a thin string to lower the object.
D. Making sure the object is fully submerged.
7. A simple pendulum takes 13.8 s to move from X to Y and back again 10 times . What is the period of
the pendulum?
A. 0.69s B. 1.38s
C.13.9s D. 2.3s
8. Which of the following is an SI unit?
A. ml B. km/hr
C. g/cm³ D. kg/m³
9. A cuboid has sides 2 cm, 3 cm and 4 cm. The material of which it is made of has a density of 7g/cm³.
What is the mass of the cuboid?
A. 168g B. 290g
C. 240g D. 340g
10. The diagram below shows Vernier calipers.
10cm 11cm

0 5 10

What is the reading shown?

A. 10.2cm B. 11.1cm
C.10.9cm D. 11.4cm
Structured Questions
1. (a) Explain how you would use a stopwatch to measure the pulse rate in beats per minute of a
student. [6]
(b) There are two types of stopwatch, digital and analogue watch. Which of the of the two is best to
use? Explain why. [4]
2. A student used a digital stop watch three times to time 10 oscillations of a pendulum. The timings
were 13.52s, 13.64s and 13.58s.
(a) Calculate (i) the average time for 10 oscillations, (ii) the period of the oscillations. [5]
(b) Estimate the accuracy of the timings, giving a reason for your estimate. [5]
3. (a) A book has 100 pages and its thickness excluding covers is 4cm, calculate the thickness of each
sheet of paper. [5]
(b) Three liquids A, B and C have different densities. Given that X has the least density and Y the
greatest, and that the liquids are immiscible, show how they would all settle in a beaker. [3]
(c) Convert 10km/hr to metres per second. [2]

18
4. (a) Copy and complete the table shown below:

Quantity Instrument SI Unit

Metre rule Metres
Force newton
Temperature Kelvin
Voltmeter Volts
Time Stopwatch
[5]
(b) What instrument would you use to measure the following?
(i) mass of a pin. [2]
(ii) internal diameter of a cup. [2]
(iii) thickness of a sewing needle. [2]
Practical Questions
1. Describe an experiment you would do to;
(a) measure the density of a stone.
(b) measure the density of oil.
(c) measure your own mass. [10]
2. In an experiment in taking the voltmeter and ammeter readings, the following set up of the circuit
diagram was done:
Voltmeter
v
Ammeter
A
Connecting
wire
Cell
+ _

(a) Discuss and comment on the full-scale deflection of the metre and its accuracy. [5]
(b) Explain on what would be the difference (if any) in your results; if you were to use a digital
ammeter or voltmeter. [3]
(c) What precautions should be taken when taking those readings? [2]

19
 Chapter
KINEMATICS
2
Chapter objectives
By the end of this chapter, learners should be able to:
• define displacement, speed, velocity and acceleration.
• plot, draw and interpret graphs of motion.
• define free-fall.
•• determine acceleration of free fall.

Introduction
Kinematics is a branch of classical mechanics that describes the motion of points, bodies and systems
of bodies without considering the forces that caused the motion. It is often referred as the “geometry of
motion” and is occasionally seen as a branch of mathematics.

UNIT 2.1 SPEED, VELOCITY AND ACCELERATION

Displacement
Distance is the total length of travel irrespective of the direction of motion. It is a scalar quantity. The
distance travelled by a car is indicated by an odometer.

Fig. 2.1 Odometer

20
But your velocity would be zero since your position didn’t change between the beginning and the
end of the interval. There was no displacement seen at the end of the time period. You would have an
instantaneous velocity if it were taken at a point where you had moved from your original position. If
you go two steps forward and one step back, your speed is not affected, but your velocity would be.
Acceleration
Acceleration (symbol: a) is defined as the rate of change of velocity. It is thus a vector quantity with
dimension length/time². In SI units, acceleration is measured in metres/second² using an accelerometer. To
accelerate an object is to change its velocity, which is accomplished by altering either its speed or direction
(like in case of uniform circular motion) in relation to time. In this strict mathematical sense, acceleration
can have positive and negative values (deceleration). Any time that the sign (+ or -) of the acceleration is
the same as the sign of the velocity, the object will speed up. If the signs are opposite, the object will slow
down. Acceleration is a vector defined by properties of magnitude (size or measurability) and direction.
When either velocity or direction are changed, there is acceleration (or deceleration).

Exercise 2.1
1. Explain the following terms:
(a) displacement.
(b) speed.
(c) velocity.
(d) acceleration.
2. State the mathematical expression of the terms in question 1.

UNIT 2.2 GRAPHS OF MOTION, BOTTOM OF FORM

Distance-time graphs
Distance is the total length travelled by an object. The standard unit is the ‘metre’. A distance-time graph
shows how far an object has travelled in a given time. Distance is plotted on the Y-axis (left) and Time
is plotted on the X-axis (bottom). Below you can see that the object represented by the blue line has
travelled 10m in 2s whereas the object represented by the red line has only travelled 4m in this time and is
therefore travelling more slowly.

10 steady speed

9
8
7
distance (m/s)

stationery
6
5
4
3
2
1
0
1 2 3 4 5 6
time in (s)
Fig. 2.3 Distance-time graph

22
‘Straight lines’ on a distance-time graph tell us that the object is travelling at a constant speed. Note that
you can think of a stationary object (not moving) as travelling at a constant speed of 0 m/s. On a distance-
time graph, there are no line sloping downwards. A moving object is always ‘increasing’ its total length
moved with time.

Increasing speed Decreasing speed

(acceleration) (deceleration)
distance

distance
Time Time

Fig. 2.4 Acceleration and deceleration

‘Curved lines’ on a distance time graph indicate that the speed is changing. The object is either getting
faster = ‘accelerating’ or slowing down = ‘decelerating’. You can see that the distance moved through each
second is changing.
Calculating speed from a distance-time graph
40 +

35
+ +

30 20m
+
distance in (m)

25 The gradient (slope) of a distance-time

+
graph indicates the objects speed
20 +
5s

15 + Speed = change in distance = rise

+
change in time run
10
+ *Note that if the graph slopes downward
5
+ you will get a negative value indicating the
object is travelling back towards it`s origin
0
1 2 3 4 5 6 7 8 9 10
time in s
Fig. 2.5 Calculating speed from a distance-time graph

The average speed can be calculated for any part of a journey by taking the change in distance and dividing
by the change in time for that part of the journey. You can even do this for a curved line where the speed
is changing, just remember that your result is the average speed in this case. You may also notice that the
formula for calculating speed is sometime written with small triangles Δ (the Greek letter delta) in front
of d (distance) and t (time). The Δ is just short hand for “change in”. Therefore, Δt means “change in time”.

23
 d
V=
t
distance

time

Fig. 2.6 Calculating speed from a distance-time graph

Displacement
Displacement is the length between start and stop positions and includes a direction. Displacement is a
vector quantity. If an object goes back to where is started in certain time, then its displacement is zero. Its
distance would be the total length of the journey. A displacement-time graph is able to show if an object
is going backwards or forward. Usually, a line with a negative gradient would indicate motion going
backwards. This cannot be shown on a distance-time graph.
Describing motion of an object
In most mechanical problems, we are asked to determine the connection between speed, position and
time. Will two cars crash if they are heading towards each other as they apply brakes at a certain time?
To describe the position of a moving object, you have to specify its position relative to a particular point or
landmark that is understood by everyone. Along a straight line, you only need the position of the landmark
and how far the object is from the landmark left or right (or east or west). 5 metres from the door does not
mean anything without giving some indication of direction (inside or out for example).
Now left or right is not a good distinction as not everyone can agree with it. In Physics, we specify the origin
landmark at 0 and the points either side of it are either positive or negative numbers (in units of metres).

10
9
stationery
8 B
displacement in (m)

7
6
ity
loc

5
Ve

C
t
an

4 A
nst
Co

3
constant velocity
2
1
0
1 2 3 4 5 6 7 8 9 10
time in (s)

When describing the motion ofFig.

an 2.7
object try to be
Describing as detailed
motion as possible. For instance, ...
of an object.

24
When describing the motion of an object, try to be as detailed as possible. For instance:
• during ‘Part A’ of the journey the object travels + 8m in 4s. It is travelling at a constant
velocity of +2ms-1.
• during ‘Part B’ of the journey, the object travels 0m in 3s. It is stationary for 3 seconds.
• during ‘Part C’ of the journey, the object travels -8m in 3s. It is travelling at a ‘constant velocity’
of ‘-2.7ms-1’ back to its starting point, our reference point 0.
Why can we use ‘velocity’ instead of ‘speed’? Because by labelling our two directions + and -, we now
know which way our object is moving in 1-dimension, forwards or backwards.
Velocity-time graphs

steady speed
10
9
constant velocity
8 B
7
velocity in (m/s)

Con
ion
rat

st
A C

ant
ela
acc

dec
nt

ela
sta

rati
con

on
2
1
0
1 2 3 4 5 6 7 8 9 10
time in (s)

Fig. 2.8 Calculating speed from a velocity-time graph

On a velocity-time graph, a horizontal (flat) line indicates the object is travelling at a constant speed.
A straight diagonal line indicates the objects velocity is changing. In the graph on the left (figure 2.8),
the line sloping upwards shows the object is accelerating and the line sloping downwards in this case
towards v = 0, shows it is decelerating. The negative value of the gradient gives the negative value for the
acceleration (or deceleration). But a negative value does not always mean slowing down.
The general rule of thumb is if the object is going from a high speed to a low speed, it is decelerating.

25
Calculating acceleration from velocity-time graph

20

15

∆v
a = ∆v
velocity (m/s)

10
∆t

5
∆t

0
0 10 20 30 40 time / s

Fig. 2.9 Calculating acceleration from velocity-time graph

The average acceleration can be calculated for any part of a journey by taking the change in velocity and
dividing by the change in time for that part of the journey.
Calculating the distance travelled from a velocity-time graph

30
B
Velocity m/s

A
20
C
10

0 10 20 30 40 50 60 70
Time (in seconds)

Fig. 2.10 Calculating the distance travelled from a velocity-time graph

The total distance travelled by an object can be determined by calculating the area underneath the
velocity time graph. Start by dividing the graph into sections that consist of simple triangles and
rectangles as shown above in blue and red. Calculate the area of each shape as shown in the following
diagrams.

26
You can then simply add the areas together and the total area represents the total distance travelled. For
example, the graph above has a total area of:
(1/2 × 10 × 20) +(20 × 20) + (1/2 × 40 × 20) = 900
Therefore, the total distance travelled was 900m.

Area = ½ base × height

height

Base

Base

height Area = base × height

Fig. 2.11 Formulas for calculating areas

Acceleration – time graphs

They display the motion of a particle by showing the changes of velocity with respect to time. For example;

a/ms-2

a1

t (time)/s

Fig. 2.12 Acceleration – time graph

Above acceleration – time graph represents an object moving with uniform acceleration.

t2
t1

t (time)

Fig. 2.13 Acceleration – time graph

27
Above acceleration – time graph represents particle going at uniform acceleration in positive direction until, t = t1,
when it starts moving with uniform velocity. From t = t2, the object has negative acceleration and its velocity starts
to retard. Velocity – Time graphs are also called a − t graphs.

Example 1
The graph below shows a velocity time graph of a toy car.
v(m/s)

20

0 20 30
time/s

Calculate the distance covered during the 30s.

Solution: Distance = area under the graph = 1 (30s + 20s) 20m/s = 500m
2

Activity 2.1 Experiment

Aim: To find the average acceleration with a ticker-timer.
a ( acceleration )

20m/s

t ( time )
0 5s 10s 15s

Materials: trolley, elastic cords for accelerating trolley, rod for attaching elastic cord to trolley, ticker-timer with
power supply unit, ticker-tape, sellotape.
Method
1. Thread a length of ticker-tape through a ticker-timer and attach the end to a trolley.
2. Pull a trolley with a fixed force along a bench. Loop one end of the elastic cord around a rod attached to the
trolley. Keep the force constant by making sure that the cord is always stretched by the same amount as the
trolley moves. Practice doing this.
3. Choose and cut through a dot near to the start of the tape. Do this when the trolley is travelling quite slowly
but the dots are far enough apart to clearly distinguish one from another.
4. Count ten dot-to-dot spaces and cut the tape, through a dot, again. You have cut a ‹ten-tick-tape›.
5. Count 40 more dot-to-dot spaces along the tape. Then cut the next 10 dot-to-dot spaces to make another
ten-tick-tape.

28
00
UNIT 2.3 MOTION UNDER GRAVITY

When a body falls towards the earth, we can notice that its velocity increases as it approaches earth.
Hence the falling body is accelerating. Galileo performed several experiments and showed that falling
bodies with different mass from same height reach the earth at the same time and hence they fall with
constant acceleration. This acceleration is called acceleration due to gravity and is denoted by symbol g.
Experiments have shown that the value of g is the same at the same place and varies slightly from place to
place. A fairly correct value of g is 9.81m/s².
A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being
acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion
characteristics that are true of free-falling objects:
• Free-falling objects do not encounter air resistance.
•• All free-falling objects (on earth) accelerate downwards at a rate of 9.8m/s².
Terminal velocity
In general, the air resistance on any falling object increases as it gains speed. If it continues to fall, the
increasing air resistance causes it to reach a constant velocity, referred to as its terminal velocity. At this
velocity, the air resistance on the object opposes the force of gravity with an equal force.
Acceleration of free-fall
Free fall is defined as the falling of an object when the only force acting on it is gravitational force. This
may be achieved by allowing an object to fall in a vacuum. An object falling through air or through any
fluid is not under free fall since it will be experiencing a drag force. The gravitational acceleration for free
fall of a body near the earth is approximately constant. Its value is approximately 9.81 m/s² as indicated by
experiment before.
Determine acceleration of free fall - vertically downward motion
A particle moving vertically downwards from height h goes downward with an acceleration . If we take
the downward directions as positive and replace s by h and a by g in equations of motion, we get:
v2 = u2 + 2gh
1
h = ut + 2 gt²
v² = u² + 2gh

Vertically upward motion

A particle thrown vertically upward moves with a retardation . Its velocity slowly diminishes and becomes
zero when the particle is at its maximum height. The body is then at rest for an instant and immediately
begins to fall with acceleration . We take upward direction as positive and replace with then equations
of motions become;
v = u – gt
1
h = ut – 2 gt²
v2 = u2 – 2gh

30
Greatest height and time of ascent when particle is thrown upward vertically
Let H be the maximum height attained in time t. Since velocity is 0 at the maximum height, 0 = u–gt and
0 = u2–2gH
Time to reach the top: t = u
g
Maximum height reached: H = u2
2g
Time of flight (T)

When the particle returns to the earth, the vertical displacement is zero. If T is the time taken to come
back,
1
0 = uT –
2gT2
T = 2u this gives the total time of flight.
g
Time of flight is twice the time of ascent.
Time taken to reach a height of h

From
h = ut – 1 gt2
2
When, u2 > 2gh, t has two real values which shows that particle reaches the height twice once on its way
up and once on its way down after it has reached the greatest height.
t = u ± u – 2gh
2

g
u2
If u2 < 2gh, t becomes imaginary. It shows that the particle does not go higher than
2g
Velocity at a height (h)
From v2 = u2– 2gh we have v = ± √u2 – 2gh

The positive value represents upward velocity and negative value represents downward velocity at the
same height h. It shows that magnitude of the velocity is the same when the particle is at the same height
whether it is ascending and desending.
h = ut + 12 gt2

The effect of air resistance on a falling object

An object thrown down from a certain height above the earth, descends to the ground at a constant speed
because air resistance on it opposes the force of gravity on it with an equal force. That air resistance is
sometimes referred to as the drag force.
So, the air resistance on any falling object increases as it gains velocity. If it continues to fall, the increasing
air resistance causes it to reach a constant velocity, known as its terminal velocity. At this velocity, the air
resistance on the object opposes the force of gravity with an equal force.

31
Example 2
A ball is thrown vertically upwards with a velocity of 40m/s. Calculate;
(a) the maximum reached.
(b) the total time the ball is in the air.
(g =10m¯²)
Solution
Use then equation v² = u² + 2 as ( a = -g since the object is opposing the gravitational field)
At maximum height v = 0
0 = 40² +2(-10)h
40²
h = 20
=80m
Example 3
A particle is released from rest and takes 5s to reach the ground. What is its velocity on reaching the ground?
Solution
v = u + at
v = 0 + 10(5)
v = 50m/s

Activity 2.2 Experiment

Aim: Measurement of acceleration due to gravity (g) using the freefall method.
Materials: millisecond timer, metal ball, trapdoor and electromagnet.

switch
electro-magnet

trap door

0.0

Method
1. Set up the apparatus as shown. The millisecond timer starts when the ball is released and stops when the
ball hits the trapdoor.
2. Measure the distance s using a metre stick.
3. Flick the switch to release the ball and record the time t from the millisecond timer.
4. Repeat for different values of s.
5. Calculate the values for g using the equation s = (g/2) t2. Obtain an average value for g.
6. Draw a graph of s against t2 and use the slope to find a value for g (g will be twice the slope).

32
Results

s (m)
t(s)
t2 (s2)
g (m s-2)
gavg =

Conclusion
When we plotted the graph, the slope worked out to be 4.85, which resulted in a value for g of 9.9 m s-2, which
is pretty damn close to the theoretical value of 9.8 m s-2. Conclusion? We rock!
Precautions / sources of error
1. For each height s repeat three times and take the smallest time as the correct value for t.
2. Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release.
3. Remember to convert from milliseconds to seconds.
NB: Both points 1 and 2 above are associated with the problem that even though you switch off the power
for the electromagnet (and in so doing switch on the timer) it will not lose its magnetism immediately,
therefore the ball will not fall straight away. This means that the reading on the timer will always be
(slightly) longer than the time for which the ball was dropping.

Exercise 2.3

1. Objects under free fall near the earth’s surface experience a constant acceleration. What is the value
of this acceleration?
2. Why is it important to release the objects simultaneously?

Interesting facts

1. Two moving objects can have the same speed but different velocities.
2. An object moving steadily round in a circle has a constant speed. Its direction of motion continues
changing as it goes around so its velocity is not constant.

Summary of the chapter

• The steeper a distance-time graph is, the greater the speed it represents.
• The steeper the displacement-time graph is, the greater the velocity it represents.
• The slope on a distance-time graph represents speed.
• The slope on a displacement-time graph represents velocity.
• Speed is the rate of change of distance.
• Velocity is the rate of change of displacement.
• The area under velocity-time graph represents displacement.
• The SI units of speed and velocity is m/s.
• Acceleration is the rate of change of velocity.
• The unit of acceleration is m/s².

33
 • The slope of velocity time graph represents acceleration.
•• A free-falling object has a constant acceleration.

Glossary of terms

Speed – is the rate of change of distance.

Velocity – is the rate of change of displacement.
Distance – is the total length travelled by an object.
Displacement – is the length between start and stop positions and includes direction.
Acceleration – is the rate of change of velocity.
Free fall – is the falling of an object when the only force acting on it is gravitational force.
Gradient – is the steepness or sloppiness of graph.
Terminal velocity – is when the force of gravity is equal to air resistance.

Revision Exercises
Multiple Choice Questions
1. The area under a speed-time graph represents ___________.
A. acceleration B. speed
C. distance travelled D. time taken
2. The gradient of distance-time graph represents ___________.
A. time taken B. acceleration
C. speed D. distance travelled
3. The speed-time graph shows the movement of a car.
The graph shows that the car is ____________.
speed (v)

time (t)

A. accelerating B. stationary
C. moving at a constant speed D. moving up a hill
4. A motorcar is travelling at 80km/h. What is its speed in m/s?
A. 22.2 B. 1.33
C. 0.022 D. 1333
5. Which of the following statements is not true?
A. Free fall implies that gravity is the only force acting on an object.
B. The gradient of velocity time graph gives acceleration.
C. The gradient of a displacement-time graph give velocity.
D. The area under acceleration-time graph gives the distance travelled.

34
6. An object is falling in air until it reaches terminal velocity. What happens to its acceleration?
A. Remains constant. B. Decreases to zero.
C. Decreases to a lower value. D. Increases.
7. A boy takes 20 s to run 100 m and another 30 s to run the next 120m. What is his average speed?
A. 5.0m/s B. 4.0m/s
C. 10.0m/s D. 4.5m/s
8. A car drive wants to take a total of 2 hours to make a journey of 80km. She drives at a speed of
60km/h for the first half hour. She has a coffee break for half an hour after that. At what average speed
must she travel during the rest of the journey if she wants to complete a journey in 2hrs?
A. 20km/h B. 100km/h
C. 50km/h D. 70km/h
9. A ball falls freely under gravity near the surface of the earth. Which quantity remains constant?
A. Displacement B. Speed
C. Velocity D. Acceleration
10. A student conducts an experiment to determine the acceleration of a trolley. He measures the
velocity of the trolley. At one instant, the velocity of the trolley is 1.0m/s. Two seconds later, the
velocity is 5.0m/s. What is the acceleration of the trolley?
A. 2.0m/s² B. 6.0m/s²
C. 2.5m/s² D. 3.0m/s²
Structured Questions
1. The table shows the readings obtained by a student during an experiment in which a toy car runs.

Distance/m Time/s
0 0
0.2 0.50
0.4 0.95
0.6 1.35
0.8 1.70
1.0 2.00
1.2 2.25
1.4 2.50
1.6 2.75
1.8 2.90
2.0 2.90
(a) Draw a distance-time graph using the readings in the table. [4]
(b) State what the graph tells you about the motion of the toy car:
(i) During the first second.
(ii) From 1.0 to 1.6 seconds.
(iii) From 1.8 to 2.0 seconds. [6]

35
Practical Questions
1. In an experiment to determine the acceleration of free fall g a student attached a length of ticker
tape to a metal ball. When the ball was released, it fell to the ground pulling the tape through a ticker
timer that printed 50 dots per second on the tape. The student measured lengths of tape to find the
speed of the ball as it falls.
(a) Calculate the time taken for the timer to print 10 dots on the tape. [2]
(b) The student cut the tape into lengths with 10 dots each, starting at the end attached to the ball.
Would you expect the lengths to increase, decrease or stay the same as the ball fell? [2]
(c) The student’s results showed that after 0.6s the speed of the ball was 5.2m/s. Calculate the value
for the acceleration of free fall g using the student’s results. [5]
(d) The value obtained in part (c) is less than the accepted value. Suggest a reason for this difference,
assuming that the experiment was carried out with care. [1]
2. Assuming air resistance is negligible, describe an experiment to show that objects fall at the same
rate. [10]

37
EXAMINATION 1

Paper 1: Multiple Choice Questions

There are forty questions in this paper. Answer all questions. For each question, there are four possible
answers, A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the
separate answer sheet.
1. T he diagram below shows micrometre screw gauge used to measure the thickness of a wire. What is
the reading shown?

A. 0.288mm B. 2.88 mm
C. 0.88mm D. 2.538mm
2. A stopwatch shows the following reading 01 20 32. How is this value recorded in SI units?
A. 1:20:32 s B. 20.32 s
C. 120.32 s D. 80.32 s
3. What is the value of the density of metal block of mass 60kg and volume 3m3?
A. 180kgm–3 B. 18kgm–3
C. 20kgm–3 D. 63kgm–3
4. What is the volt expressed in base units?
A. kgm2s–3 A¯1 B. kgm2s–2 A¯1
C. kgm2s A¯1 D. kgm2s–1 A¯1
5. Which of the following is not a symbol for an SI unit?
A. m B. ˚С
C. kg D. N
6. What is the mass of a block of density 2.3 kgm–3 and its length is 0.5m width 0.3m and its height is
0.2m?
A. 0.69kg B. 0.069kg
C. 0.345kg D. 1.38kg
7. Which of the following correctly defines velocity?
A. Rate of change of distance B. Rate of change of distance with time
C. Rate of change of displacement with time D. Rate of change of displacement
8. The gradient on a velocity time graph represents the __________.
A. speed B. displacement
C. acceleration D. area

63
12. The diagram below shows a velocity time graph.

40

v/ms¯1 20

0 t/s
2 4 6 8 10

What is the distance covered in 10 seconds?

A. 80m B. 360
C. 280m D. 400m
13. A kangaroo jumps to a maximum height of 1.5m in 5seconds. What is its take off speed?
A. 7.5m/s B. 0.3m/s
C. 3.3m/s D. 5.4m/s
14. A stone and a feather are released at the same time. Which statement is correct if air resistance is
neglected?
A. The stone reaches the ground first B. The feather will never reach the ground
C. They reach the ground at the same time D. The feather reaches the ground first
15. A student carried out an experiment to determine the value of the acceleration of free fall g he
obtains the following sets of results; 9.83,9.90, 10.0 What is the accurate value of g?
A. 10.0ms–2 B. 9.95ms–2
C. 9.83ms–2 D. 9.93ms–2
16. Which graph shows the variation of speed with time for a body allowed to fall in a fluid until it
reaches terminal velocity?
A. B.

v/ms¯1 v/ms¯1

t/s
t/s

C. D.

v/ms¯1
v/ms¯1

t/s
t/s

17. Which statement is true for a body undergoing circular motion at constant speed?
A. There is no acceleration B. There is an acceleration towards the centre
C. There is no resultant force acting on it D. The velocity remains the same

65
24. Which graph shows the velocity – time graph of a rubber ball dropped from the hand?
A. B.

v/ms¯1 v/ms¯1

t/s
t/s

C. D.

v/ms¯1 v/ms¯1

t/s
t/s

25. A spring of spring constant 2N/cm is loaded with a mass 1000g. What is its extension (g=10m¯2)?
A. 0.5cm B. 5cm
C. 500cm D. 50cm
26. The diagram below shows the extension–force graph of a certain material that behaves in an unusual
manner.

s
Extension/mm

Force/N

Which statement is correct about the material?

A. The material becomes easier to extend after s
B. The material becomes harder to extend after s
C. The force and extension are always directly proportional
D. The force and extension are always inversely proportional
27. Which of the following equations correctly defines mass?
weight
A. mass = B. mass = weight gravity
gravity
C. mass = weight D. mass = force + gravity
28. A force of 10N acts on a mass of 2kg as shown below.
10N
Surface
The acceleration was expected to be 5m¯2, but it was found to be less. What is the possible
explanation?
A. The block was so heavy for such a small force.
B. The surface offered a frictional force which reduced the effective force.
C. The force should act at an angle for the block to move.
D. The force metre used was faulty.

67
PAPER 2: THEORY
Answer all questions in Section A and B.

Section A
1. Fig 1.1(a) shows a Vernier scale when the two jaws touch each other and fig 1.1(b) when there is an
object whose diameter is to be measured.

0 1
4 5cm

0 5 10 0 10
Fig 1.1(a) Fig 1.1(b)

(a) (i) Name the condition illustrated on 1.1(a)

(ii) What is the diameter of the object? [4]
(b) Describe how you can improve accuracy when measuring the diameter of a wire using a micro-
meter screw gauge. [2]
(c) A metal block is 50mm in length, 30mm wide and has a thickness of 3.0mm. Its mass is 0.015kg.
Calculate the
(i) volume of metal block.
(ii) density of the block. [4]
2. (a) Cooking oil is added to water, certain observations are made. Explain the following terms referring
to cooking oil and water.
(i) Flotation
(ii) Sinking [2]
(b) Distinguish between speed and velocity.
(c) A body starts with a velocity of 5m/s and accelerated uniformly in 8.0s to a velocity of 20m/s.
Calculate;
(i) – the average velocity of the object
(ii) – distance covered at the end of this period
(iii) – the acceleration [6]
(d) (i) Explain why an object undergoing circular motion will be accelerating at a constant speed.
(ii) State the direction of the acceleration. [4]
3. (a) Friction can be both a nuisance and useful.
State one situation in which friction is:
(i) nuisance.
(ii) useful.
(b) State two ways of reducing friction. [4]
(c) (i) Explain what is meant by the moment of a force and show how it is calculated.

70
Paper 3: Practical Test
Answer all questions.
For each of the questions. You are expected to record all your observations as soon as these observations
are made. All your answers should be written in the spaces provided on the question paper. An account
of the method of carrying out the experiments is not required.
1. In this experiment you will determine how the speed of a metal ball depends on the angle of inclination
θ.

rolling without slipping

(a) Measure and record the length, L, marked on the run way.
L=
(b) (i) I ncline the runway at an angle θ and release the ball from the top of the runway and
simultaneously start the stop watch. Measure the time taken for the glass ball to move a
distance L.
(ii) R epeat step (a) increasing the angle of inclination by 5 until you have six sets of readings.
Complete the table of result below.

θ t₁/s t₂/s tav/s v/

20
25
30
35
40
25
(c) Plot a graph of velocity against θ [4]
(d) It is given that v = k θ + 3p
(i) Calculate the gradient of the graph.
(ii) Deduce the value of k and its appropriate unit. [4]
(e) (i) Deduce the value of the y intercept and hence the value of p.
(ii) State source of error in this experiment. [4]

73
 Chapter
MECHANICAL

6 STRUCTURES

Chapter objectives
By the end of this chapter, you should be able to:
• define a beam.
• describe a beam by its cross-sectional area.
• compare the strength of beams.
• explain the effects of push and pull forces.
• explain how stress is distributed in a loaded beam.
• construct trusses.
• explain the use of triangles in truss.
• explain the advantages of trusses over beams.
• explain how a load can be distributed throughout a truss.
• identify struts and ties in truss.
• explain the design of a roof truss.
• describe methods of joining materials.
• compare the strength of joints.
• identify materials used in large structures.
• compare properties of construction materials.
• explain the design and materials used in different types of bridges.
• explain the use of arches in construction of large structures.
• explain composition and shape of dam walls.

Introduction
Physics is applicable in everyday life and should not be taken for granted. Buildings, bridges and dams are
a result of a branch of physics called physics of materials. Engineers who designed such structures need to
put into cognisance of which structure is strongest, lightest and cheapest so as to assist builders to come
up with the best structure with these three attributes. How best can a structure which is stronger, lighter
and cheaper can be chosen? This chapter has the solutions.

UNIT 6.1 BEAMS

A beam is a structural element capable of withstanding load by resisting against bending force. A beam
can also be defined as a solid material supported at one or two ends and in turn support a load.

97
The load must be at right angle to the length of the beam. A beam supported at both ends or at the middle
is called a simply supported beam. This is illustrated in figure 6.1.

Load
C
X
W

Fig. 6.1 An unloaded and loaded beam

A beam on which one of its ends is built into a rigid support is called a cantilever. This has been
illustrated in fig 6.2. The clamp is acting like a rigid point like a wall.

unloaded beam

T
C
Clamp
loaded beam

Fig. 6.2 Cantilever

Types of beams
Beams are described according to their cross-sectional shape in some cases the cross-sectional shapes
take the shape of letters of the alphabet so those are named according to these letters.

T-beam Z- beam H-beam

98
 Activity 6.3 Experiment
Aim: Comparing strength to mass ratio of beams and trusses.
Materials: A beam and truss of equal mass, weights, mass hanger, retort stand
Method:
1. Set up the apparatus as shown below.

beam

weights

clamp

2. Load weights to the beam until it breaks and not the total weight at breaking point.
total weight at breaking
3. Calculate the strength to mass ratio of the beam. ( mass of beam ).
4. Repeat steps 1 and 2 using a truss of same mass as that of the beam.
5. Compare the strength to mass ratio of the two.
Observations
Which of the two carries more load before breaking?
Conclusion
Which one has large strength to mass ratio?
Advantages of trusses over beams.
1. They are stronger.
2. They have a large strength to mass ratio.
3. They are economy.
Trusses like beams can support a load. The load can be distributed throughout a truss. When loaded, some
members of the truss experience compressive force and others tensile force. Members under tension are
referred to as ties while those under compression are referred to as struts.

Activity 6.4 Experiment

Aim: To determine the members of a truss under tension and those under compression
Materials: Maize stalk, optic pins, string, weights/loads, retort stands with bosses and clamps
Method:
1. Fig 6.6 (a) was constructed using maize stalks joined by optic pins.
2. Load the truss with some weights.

103
The sideways down ward forces needs to be balanced by an opposite upward force thus compression acts
in these members. The horizontal beam is acted upon by tensional forces.

load

Deflection
This member is
in compression
Bearing

Reaction This member is Reaction

in tension
Fig. 6.8 A roof truss with several members

The type of force acting on members of truss depends on the position of the load and this should be in
such a way as to prevent the structure from collapsing. The reaction forces shown are due to the walls of
the building.

UNIT 6.3 JOINING MATERIALS

Most structures are made up of two or more beams or members. These members are to be joined to
make structures. It is of great importance to choose the most appropriate method of joining materials. The
method of joining to be used depends on the type material, there are some methods which are suitable
for joining wood yet not suitable for joining metals or plastics.
There are two mechanisms for joining materials, these are pinning and surface contact.
Pinning
This is joining materials by gripping from the exterior using binding devices. There are several binding
devices that can be used for pinning, these include nailing, screwing, riveting and bolting.
Nailing
Nailing is mainly used on soft woods because they do not crack during the nailing. A hammer is used to
strike the nail into wood. Hammer with a longer handle is to be used in order to use a small effort.
Bolting
This consists of bolts and nuts. A hole is drilled in the
material to be joined and a bolt is passed through.
The head of the bolt must be larger than the hole
drilled to prevent the bolt from falling into the hole
that has been drilled. The bolt has some thread at
its end, the nut is fixed on these threads to pin the
materials together. Washers are at times placed
before the nut is threaded in.
Fig. 6.9 Bolt and nut

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Metal • durable • can corrode if not
• malleable coated
• ductile • Expensive
• Strong under tension
Stones • Cheap. • heavy
• Strong under compression.
• beautiful
concrete • strong in compression. • heavy
• does not decay • can crack under tension
• expensive
Reinforced • strong in both compression • heavy
concrete and tension • expensive
• durable
When choosing materials to use on large structures, there are some factors that are to be considered these
are cost and durability. Materials which are cheaper are preferred to those that are expensive. Durable
materials are those that can last long before they need to be replaced.
Bridges
Bridges are constructed to enable animals and
vehicles to cross rivers. In its simplest form, bridges
consists of a beam called bridge deck supported
at its ends, figure 6.12. Bridge deck is the path
where animals and vehicles moves on. It is made of
materials which can withstand both compression
and tensile stresses. They also need to be fire and
water resistant.

Early bridges were built from stones and then steel,

but nowadays reinforced concrete is very common.
When concrete and steel are used, the design is such
that steel bars are at points under tension whereas Fig. 6.12 Simple beam bridge
concrete is used where compression is experienced.
Beam and pier bridge
As well as having supports at the two ends, it has
one or more pillars or piers in the middle. The piers
stop the bridge deck from bending too much and
make the bridge to be much stronger. A load on the
bridge puts the pier under compression but being
made of stone, brick or concrete, they can with
stand this.

Fig. 6.13 Beam and pier bridge

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Summary of the chapter

• A beam can be supported at one or to ends. The one with one end rigidly fixed is called a
cantilever.
• When loaded a beam can experience compressional, tensional or shear forces.
• There are no forces acting on the part of the beam called the neutral axis.
• Trusses are made of ties and struts.
• Triangular shapes are used in trusses since they are stable.
• On trusses the load is distributed among members.
• Trusses have a large strength to mass ratio.
• Some materials like concrete are weak in tension but strong in compression.
• A bridge is a structure that spans a gap. The four main types of bridges are simple beam, beam
and pier, arch and suspension bridge.
• Dams are designed to hold back water. There are two types of dam namely earth and concrete
dam.
• Dam walls have broader bases and narrow apex and some are curved or arch shaped.

Glossary of terms

Abutment – ends of the bridge where it is supported.

Beam – supported bar that bears a load.
Compression – force that tends to squash a material.
Strength – ability of material to carry a load.
Strength to mass ratio – measure of strength of material compared to its mass.
Stressed – loaded.
Struts – member under compressional force.
Tension – force that tends to pull a material part.
Tie – member under tensional force.
Truss – structure made of many members joined.

Revision Exercises
Multiple Choice Questions
1. The diagrams show beams that are joined. Which is the strongest structure?

A. B.

C. D.

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2. Rectangular gates prevent shearing by having ____________.
A. parallel cross bars B. perpendicular cross bars
C. diagonal cross bars D. curved cross bars
3. The diagram shows loaded cantilever.
Which row correctly shows forces that are acting at P and Q?

unloaded beam

T Q
C
Clamp P
loaded beam

P Q
A. Tension Compression
B. Shear Compression
C. Compression Tension
D. Shear Tension
4. Why do engineers prefer hollow cylindrical beams to solid cylindrical beams?
A. They are weak. B. They are lighter.
C. They have a smaller strength to mass ratio. D. They have more material hence stronger.
5. The diagram shows a roof truss.

1
2

5 3
4

Which parts are to be removed so that only members under tension are to remain?
A. 3, 5 and 4 B. 1, 2,3 and 4 C. 1, 3, 4 and 5 D. 1, 2 and 5
6. Which methods are used to join metals by surface contact?
A. Soldering, brazing and screwing. B. Riveting, welding and brazing.
C. Brazing, soldering and glazing. D. Brazing soldering and brazing.
7. Which expression is to be considered when designing a dam wall?
A. Pressure = Force/Area. B. Area = length × width.
C. Pressure =density × gravity × height. D. Volume = length × width × height.

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 (iv) R
 epeat steps (ii) and (iii) for different values of x and breaking loads until you have six sets of
readings.
(b) Record the value of x and load (L).

Overlap/cm Load at breaking/N

5.0
10.0
15.0
20.0
25.0
30.0
(c) Use your values in table above to plot a graph of load (y-axis) against overlap (x-axis).
(d) (i) Calculate the gradient of your graph.
(ii) Identify one source of error in the experiment.
(e) (i) What conclusion can be drawn from this experiment?
(ii) State two factors to be maintained constant in this experiment. [20]
2. A thin solid cylindrical beam was loaded with six different masses and the extension was measured
until it breaks as shown in the table below.

Mass/g 0 100 200 300 400 500 600

Extension/mm 0 5 8 15 19 25 30
(a) (i) Comment on the relationship between mass and extension.
(ii) Explain why the extension is not uniform although the increase in mass is uniform.
Plot a graph of extension (y-axis) against mass (x-axis).
Using your graph deduce the mass which will cause an extension of 27mm.
(b) (i) On the same axis sketch a graph for a material with a small strength to mass ratio clearly mark
the breaking point.
(ii) Explain the nature of your graph in relation to the graph in (a). [20]

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EXAMINATION 2

Paper 1: Multiple Choice Questions

There are forty questions in this paper. Answer all questions. For each question there are four possible
answers, A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the
separate answer sheet.
1. W hat happens to the centre of gravity on the body at equilibrium position when it is rotated in a
clock wise circular manner?
A. It lowers B. It rises
C. It moves to the left D. It remains at the same height
2. Chipo moved a bucket towards her, but only lifts one end of the bucket while the other was
supported. When she released the bucket, it went back to the original position. What could be the
reason for this?
A. The centre of gravity has been raised.
B. The bucket has a moment about the centre of gravity which makes it retain to the original position.
C. The vertical line passing through the centre of gravity lies out of the base.
D. It has got the moment about the pivot allowing it to return to the original position.
3. Why do wooden rocking chairs often have diagonal members nailed on them?
A. To resist bending B. To make it stable
C. To resist tensional and compressional forces D. To make triangles
4. The roof of a round thatched hut often do not include ties. Why do the walls of such a hut not
collapse?
A. It resist tensional force B. It is strong
C. It resist compressional forces D. It resist torsional forces
5. Why do trusses have a greater resistance to load than beams?
A. Because they have a large structure
B. Requires a greater labour
C. Resist both compressional and tensional force
D. It is expensive
6. The diagram shows a roof truss.

3 1

load
2
Which members are under compression?
A. 2 and 1 B. 3 and 4
C. 4 and 1 D. 2 and 1

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