1.

5 Forces
1.5.1 Balanced and unbalanced forces
FOCUS POINTS
★ Understand that the size, shape and velocity of objects can be altered by forces.
★ Identify different types of force and use free-body diagrams to show the magnitude and direction of all the
forces that act on an object.
★ Become familiar with load–extension graphs for an elastic solid and describe an experiment to show how
a spring behaves when it is stretched.
★ Understand that when several forces act simultaneously on an object that a resultant can be determined.
★ Know that, unless acted upon by a resultant force, an object will remain at rest or will continue moving
with a constant speed in a straight line.
★ Define the spring constant and the limit of proportionality on a load–extension graph.
★ Apply the equation = P to calculate force and acceleration.
★ State Newton’s third law of motion.

A gravitational force causes a freely falling object to accelerate and keeps a satellite moving in a
circular path. Clearly a force can change the speed or direction of travel of an object. A force can also
change the shape or size of an object. If you stand on an empty paper carton it will change its shape
and if you pull on a spiral spring it will stretch. Several forces may act on an object at once and it is
useful to calculate a resultant force to predict their combined effect; both the size and direction of
the forces are needed for this. You have already learnt how to quantify some of these changes and in
this topic you will learn more ways to do so.

Types of force
A force is a push or a pull. There are different types These forces do not require contact between
of forces. You have already found that weight is a objects. A force can cause an object at rest to move,
gravitational force and will learn later that there are or if the body is already moving it can change its
magnetic forces between magnets (Topic 4.1) and speed or direction of motion.
electrostatic forces between charges (Topic 4.2).

▲ Figure 1.5.1 A weightlifter in action exerts first a pull and then a push.
30
 1.5.1 Balanced and unbalanced forces

A force can also change a body’s shape or size. For in relative motion. Other contact forces are the
example, a spring (or wire) will stretch when loaded force experienced by an object at rest on a surface
with a weight. (Figure 1.5.4 on p. 33) and the tension that occurs
Contact forces occur between objects that in a string or spring (Figure 1.5.2) being stretched
are touching each other. These include drag and (elastic force).
air resistance which are resistive forces caused Thrust is a sudden force (driving force) in a
by the motion of an object through a fluid such particular direction.
as a liquid or air. They act against the direction The forces on an object can be represented in a
of motion of the object to slow it down as does free-body diagram (Figure 1.5.4, p. 33).
friction, which occurs between two solid surfaces

Elastic deformation
Practical work

Stretching a spring them by small crosses and draw a smooth line

Safety through them.
● Eye protection must be worn (in case the
spring snaps). steel
spring
Arrange a steel spring as in Figure 1.5.2.
Read the scale opposite the bottom of the
hanger. Add 100 g masses one at a time (thereby
increasing the load by steps of 1 N) and take
readings from the scale after each one. Enter the 10

readings in a table for masses up to 500 g.

20
hanger
Note that at the head of columns (or rows) in data
30
tables it is usual to give the name of the quantity mm
scale
or its symbol followed by / and the unit. 90

Load/N Scale reading/mm Total extension/mm

▲ Figure 1.5.2

Sometimes it is easier to discover laws by 1 What is the shape of the graph you plotted?
displaying the results on a graph. Do this on 2 Do the results suggest any rule about how the
graph paper by plotting total extension along spring behaves when it is stretched?
the x-axis (horizontal axis) and load along the 3 What precautions could you take to improve
y-axis (vertical axis) in a load–extension graph. the accuracy of the results of this experiment?
Every pair of readings will give a point; mark 4 How could you test if the extension of the
spring is proportional to the stretching force?

Extension in springs Using the sign for proportionality, ∝, we can write

Springs were investigated by Robert Hooke just over W Q RQ ∝ W W QJ IR
350 years ago. He found that the extension was It is true only if the limit of proportionality of the
proportional to the stretching force provided the spring is not exceeded.
spring was not permanently stretched. This means
that doubling the force doubles the extension, Key definition
trebling the force trebles the extension, and so on. Limit of proportionality the point at which the load–
extension graph becomes non-linear

largest stress value beyond which stress is no longer proportional 31

to strain
1.5 FORCES

The graph of Figure 1.5.3 is for a spring stretched Load–extension graphs similar to Figure 1.5.3 are
beyond its limit of proportionality, E. OE is a straight obtained. You should label each axis of your graph
line passing through the origin O and is graphical with the name of the quantity or its symbol followed
proof that the extension is directly proportional to by / and the unit, as shown in Figure 1.5.3.
the stretching force over this range. If the force for The limit of proportionality can be defined as the
point A on the graph is applied to the spring, the point at which the load–extension graph becomes
proportionality limit is passed and on removing the non-linear because the extension is no longer
force some of the extension (OS) remains. proportional to the stretching force.

A Worked example
E
A spring is stretched 10 mm (0.01 m) by a weight of 2.0 N.
stretching force / N

Calculate
a the spring constant
b the weight of an object that causes an extension of
80 mm (0.08 m).
2.0 N
a k= F = = 200 N/m
x 0.01m

O
b = W W QJ IR
S total extension/mm = ×
= 1P× P
▲ Figure 1.5.3 = 1

Now put this into practice

Test yourself
1 Calculate the spring constant of a spring which is
1 In Figure 1.5.3, over which part of the graph does a stretched 2 mm by a force of 4 N.
spring balance work? 2 A 2 N force is applied to a spring which has a spring
constant of 250 N/m. Calculate the extension of the spring
in mm.
Spring constant
The spring constant, , is defined as force per unit
extension. It is the force which must be applied to a Test yourself
spring to cause an extension of 1 m. 2 State two effects which a force may have on an
If a force produces extension then object.
3 Make a sketch of a load–extension graph for a
k= F spring and indicate the region over which the
x extension is proportional to the stretching force.
or 4 Calculate the spring constant of a spring which is
stretched 4 cm by a mass of 200 g.
force
spring constant = 5 Define the limit of proportionality for a stretched
extension spring.
Rearranging the equation gives
=
Forces and resultants
Key definition Force has both magnitude (size) and direction. It is
Spring constant force per unit extension represented in free-body diagrams by a straight line
with an arrow to show its direction of action.
Proportionality also holds when a force is applied Usually more than one force acts on an object.
to an elastic solid such as a straight metal wire, As a simple example, an object resting on a table
provided it is not permanently stretched. is pulled downwards by its weight and pushed
upwards by a contact force due to the table
supporting it (Figure 1.5.4). Since the object is
32
 1.5.1 Balanced and unbalanced forces

at rest, there is no resultant force. We say the forces 3N

are balanced, i.e. = .
R 1N 2N

1N

2N 3N

▲ Figure 1.5.6 The resultant of forces acting in the same

straight line is found by addition or subtraction.

If the resultant of two or more forces is not zero, we

say the forces are unbalanced.

W
Going further
▲ Figure 1.5.4

In structures such as a giant oil platform Practical work

(Figure 1.5.5), two or more forces may act at the
same point. It is then often useful for the design
Parallelogram law
engineer to know the value of the single force,
i.e. the resultant force, which has exactly the Safety
same effect as these forces, If the forces act in ● Take care when using the mass in
the same straight line, the resultant is found by case it drops.
simple addition or subtraction as shown in
Arrange the apparatus as in Figure
Figure 1.5.6. If the forces act in different directions,
1.5.7a with a sheet of paper behind it
the vectors are added by taking account of their
on a vertical board. We have to find the
direction. This was described in Topic 1.1 for the
resultant of forces and .
resultant of two forces at right angles.
Read the values of and from the
spring balances. Mark on the paper the
directions of , and as shown by the
strings. Remove the paper and, using a
scale of 1 cm to represent 1 N, draw OA,
OB and OD to represent the three forces
, and which act at O, as in Figure
1.5.7b. ( = weight of the 1 kg mass =
9.8 N; therefore OD = 9.8 cm.)

force meter
(0–10 N)
string
P Q
O
W

1 kg

▲ Figure 1.5.7a
▲ Figure 1.5.5 The design of an offshore oil platform
requires an understanding of the combination of many
forces.

33
1.5 FORCES

Test yourself
and together are balanced by and 6 Jo, Daniel and Helen are pulling a metal ring.
so their resultant must be a force equal Jo pulls with a force of 100 N in one direction
and opposite to . and Daniel with a force of 140 N in the opposite
direction. If the ring does not move, what force does
Complete the parallelogram OACB. Helen exert if she pulls in the same direction as Jo?
Measure the diagonal OC; if it is equal 7 A boy drags a suitcase along the ground with a
in size (i.e. 9.8 cm) and opposite in force of 100 N. If the frictional force opposing the
direction to then it represents the motion of the suitcase is 50 N, what is the resultant
resultant of and . forward force on the suitcase?
8 A picture is supported by two vertical strings.
C If the weight of the picture is 50 N, what is the
force exerted by each string?
9 Using a scale of 1 cm to represent 10 N, find the
B
size and direction of the resultant of forces of 30 N
and 40 N acting at right angles to each other.
A
Q
P
O Newton’s first law
Use the scale 1 cm 1N
W
Friction and air resistance cause a car to come to
rest when the engine is switched off. If these forces
9.8 cm

were absent, we believe that an object, once set in

D
motion, would go on moving forever with a constant
speed in a straight line. That is, force is not needed
▲ Figure 1.5.7b Finding a resultant by the to keep a body moving with uniform velocity
parallelogram law
provided that no opposing forces act on it.
The parallelogram law for adding two This idea was proposed by Galileo and is summed
forces is: up in Newton’s first law of motion:
,I W R IR W QJ W SR QW QR M W W W W R RQW Q WR PR Q
S QW G Q QG G W RQ W W J W O Q W RQ W QW S G QO W G RQ
G RI S OO ORJ P G Q I RP W
SR QW W OW QW S QW G Q It seems that the question we should ask about a
QG G W RQ W G JRQ O RI W moving body is not what keeps it moving but what
S OO ORJ P G Q I RP W SR QW changes or stops its motion.
5 List the equipment you would need The smaller the external forces opposing a
for this experiment. moving body, the smaller is the force needed to
6 What quantity would you vary to test keep it moving with constant velocity. A hover
the law under different conditions? scooter, which is supported by a cushion of air
(Figure 1.5.8), can skim across the ground with little
frictional opposition, so that relatively little power
is needed to maintain motion.
A resultant force may change the velocity of an
object by changing its direction of motion or speed.

Key definitions
Newton’s first law of motion an object either remains at
rest or continues to move in a straight line at constant
speed unless acted on by a resultant force
Resultant force may change the velocity of an object by
changing its direction of motion or its speed

34
 1.5.1 Balanced and unbalanced forces

belts are needed. The reluctance of a stationary

object to move can be shown by placing a large
coin on a piece of card on your finger (Figure 1.5.9).
If the card is flicked sharply the coin stays where it
is while the card flies off.
coin

card

▲ Figure 1.5.8 Friction is much reduced for a hover scooter.

Mass and inertia

Newton’s first law is another way of saying that all
matter has a built-in opposition to being moved if
it is at rest or, if it is moving, to having its motion
▲ Figure 1.5.9 Flick the card sharply
changed. This property of matter is called inertia
(from the Latin word for laziness), see Topic 1.3. The larger the mass of a body, the greater is its inertia,
Its effect is evident on the occupants of a car i.e. the more difficult it is to move it when at rest and
that stops suddenly: they lurch forwards in an to stop it when in motion. Because of this we consider
attempt to continue moving, and this is why seat that the mass of a body measures its inertia.

Practical work

Effect of force and mass on acceleration

Safety tickertape timer
(or motion sensor)
trolley stretched elastic

● Take care when rolling the trolley down the

ramp. Ensure it is clear at the bottom of the
ramp and use a side barrier to prevent the
trolley from falling onto the floor.
The apparatus consists of a trolley to which a
force is applied by a stretched length of elastic
(Figure 1.5.10). The velocity of the trolley is
found from a tickertape timer or a motion sensor, ▲ Figure 1.5.10
datalogger and computer.
First compensate the runway for friction: raise (a) Force and acceleration (mass constant)
one end until the trolley runs down with constant Fix one end of a short length of elastic to the rod
velocity when given a push. The dots on the at the back of the trolley and stretch it until the
tickertape should be equally spaced, or a other end is level with the front of the trolley.
horizontal trace obtained on a speed–time graph. Practise pulling the trolley down the runway,
There is now no resultant force on the trolley and keeping the same stretch on the elastic. After a
any acceleration produced later will be due only few trials you should be able to produce a steady
to the force caused by the stretched elastic. accelerating force.

35
1.5 FORCES

Repeat using first two and then three identical first one trolley, then two (stacked one above
pieces of elastic, stretched side by side by the the other) and finally three. Check the friction
same amount, to give two and three units of force. compensation of the runway each time.
If you are using tickertape, make a tape chart Find the accelerations from the tape charts or
for each force and use it to find the acceleration computer plots and tabulate the results.
produced in cm/ten-tick2. Ignore the start of the
Mass ( )/(no. of trolleys) 1 2 3
tape (where the dots are too close) and the end
(where the force may not be steady). If you use a Acceleration ( )/cm/ten-tick2 or m/s 2
motion sensor and computer to plot a speed–time
graph, the acceleration can be obtained in m/s2 7 For part (a), does a steady force cause a
from the slope of the graph (Topic 1.2). steady acceleration?
8 Do your results in part (a) suggest any
Put the results in a table.
relationship between acceleration and
Force ( )/(no. of pieces of elastic) 1 2 3 force ?
9 Do your results for part (b) suggest any
Acceleration ( )/cm/ten-tick2 or m/s 2
relationship between and P?
10 Name the two independent variable quantities
(b) Mass and acceleration (force constant)
in experiments (a) and (b).
Do the experiment as in part (a) using two 11 How could you use the results to verify the
pieces of elastic (i.e. constant ) to accelerate equation = P ?

Newton’s second law This is Newton’s second law of motion. When using
it, two points should be noted. First, is the resultant
The previous experiment should show roughly that (or unbalanced) force causing the acceleration in the
the acceleration is same direction as . Second, must be in newtons,
(i) directly proportional to the applied force for a P in kilograms and in metres per second squared,
fixed mass, i.e. ∝ , and otherwise is not 1. The law shows that will be
(ii) inversely proportional to the mass P for a fixed largest when is large and P small.
force, i.e. ∝ 1/P.
Combining the results into one equation, we get Key definition
F Newton’s second law of motion
a∝ resultant force = mass × acceleration ( = P )
m
or
You should now appreciate that when the forces acting
∝P on a body do not balance there is a net (resultant)
Therefore force which causes a change of motion, i.e. the
body accelerates or decelerates. The force and the
= P acceleration are in the same direction. If the forces
where is the constant of proportionality. balance, there is no change in the motion of the body.
One newton is defined as the force which However, there may be a change of shape, in which
gives a mass of 1 kg an acceleration of 1 m/s2, case internal forces in the body (i.e. forces between
i.e. 1 N = 1 kg m/s2, so if P = 1 kg and = 1 m/s2, neighbouring atoms) balance the external forces.
then = 1 N.
Substituting in = P , we get = 1 and so we
can write
=P
or
resultant force = mass × acceleration

36
 1.5.1 Balanced and unbalanced forces

Worked example 13 A block of mass 500 g is pulled from rest on a

horizontal frictionless bench by a steady force
A block of mass 2 kg has a constant velocity when it is pushed and reaches a speed of 8 m/s in 2 s. Calculate
along a table by a force of 5 N. When the push is increased to a the acceleration
9 N what is b the value of .
a the resultant force
b the acceleration?
When the block moves with constant velocity the forces
acting on it are balanced. The force of friction opposing its
Newton’s third law
motion must therefore be 5 N. If a body A exerts a force on body B, then body B
a When the push is increased to 9 N the resultant exerts an equal but opposite force on body A.
(unbalanced) force on the block is (9 − 5) N = 4 N This is Newton’s third law of motion and states
(since the frictional force is still 5 N). that forces never occur singly but always in pairs
b The acceleration is obtained from = P where
as a result of the action between two bodies. For
= 4 N and P = 2 kg.
example, when you step forwards from rest your
a=
F 4 N 4 kg m/s 2
= = = 2 m/s 2
foot pushes backwards on the Earth, and the Earth
m 2 kg 2 kg exerts an equal and opposite force forward on you.
Two bodies and two forces are involved. The small
Now put this into practice force you exert on the large mass of the Earth gives
1 A box of mass 5 kg has a constant velocity when it is no noticeable acceleration to the Earth but the
pushed along a table by a force of 8 N. When the push is
equal force it exerts on your very much smaller mass
increased to 10 N calculate
a the resultant force causes you to accelerate.
b the acceleration.
2 A force produces a constant acceleration in a straight Key definition
line of 0.5 m/s2 on a block of mass 7 kg.
Newton’s third law of motion when object A exerts a
Calculate the value of .
force on object B, then object B exerts an equal and
opposite force on object A

Test yourself Note that the pair of equal and opposite forces
10 Which one of the diagrams in Figure 1.5.11 shows do not act on the same body; if they did, there could
the arrangement of forces that gives the block of
never be any resultant forces and acceleration would
mass M the greatest acceleration?
be impossible. For a book resting on a table, the
A B
book exerts a downward force on the table and the
10 N 12 N 2N 4N
M M table exerts an equal and opposite upward force on
the book; this pair of forces act on different objects
C D
and are represented by the red arrows in Figure
20 N 40 N 20 N 30 N
M M 1.5.13. The weight of the book (blue arrow) does
not form a pair with the upward force on the book
▲ Figure 1.5.11 (although they are equal numerically) as these two
forces act on the same body.
11 In Figure 1.5.12 if P is a force of 20 N and the object
moves with constant velocity, what is the value of An appreciation of the third law and the effect
the opposing force F? of friction is desirable when stepping from a rowing
boat (Figure 1.5.14). You push backwards on the boat
P F
object and, although the boat pushes you forwards with
an equal force, it is itself now moving backwards
▲ Figure 1.5.12 (because friction with the water is slight). This
12 a What resultant force produces an acceleration reduces your forwards motion by the same amount –
of 5 m/s2 in a car of mass 1000 kg? so you may fall in!
b What acceleration is produced in a mass of 2 kg
by a resultant force of 30 N?

37
1.5 FORCES

push of table
contact on book
force pair pull of Earth
push of book
on book
on table
gravitational
force pair

pull of book
on Earth

▲ Figure 1.5.13 Forces between book and table

NE
WT
O N III

▲ Figure 1.5.14 The boat moves backwards when you step forwards!

1.5.2 Friction
FOCUS POINTS
★ Understand that friction between surfaces acts to slow an object and produces heating.
★ Describe the motion of objects falling with and without air resistance or drag.
★ Explain how an object reaches terminal velocity.
★ Describe how the overall stopping distance of a car is affected by its speed and by factors that affect the
friction forces or factors that affect the driver’s reaction time.

Friction between a moving object and its surroundings is important as it acts to reduce the speed of
the object. A steady braking force applied to a car produces a uniform deceleration. The car slows
down and stops. The distance the car moves while it is braking depends on the size of the resultant
force, which depends on the force applied by the brakes and the size of the friction force.
A falling object accelerates due to the pull of gravity, however drag force caused by air resistance
reduces the acceleration. If the size of the drag force increases so that it is balanced by the object’s
weight, then the resultant force is zero and the object will move at constant speed.

Friction is the force that opposes one surface tend to when we walk on ice. On the other hand,
moving, or trying to move, over another. It can be engineers try to reduce friction to a minimum in the
a help or a hindrance. We could not walk if there moving parts of machinery by using lubricating oils
was no friction between the soles of our shoes and and ball-bearings.
the ground. Our feet would slip backwards, as they

38
 1.5.2 Friction

block spring balance Driving and car safety

In order to bring a moving car to rest, the brakes
must be applied over a certain distance known as
the braking distance. The thinking distance is
the distance travelled while the driver is reacting
F P before applying the brakes. For a driver travelling
at a constant speed, the thinking distance is
proportional to the reaction time W (distance = YW);
▲ Figure 1.5.15 Friction opposes motion between surfaces
in contact. the longer the reaction time, the further the car
travels before the brakes are applied. The thinking
When a gradually increasing force is applied through distance has to be added to the braking distance
a spring balance to a block on a table (Figure 1.5.15), to obtain the overall stopping distance, in other
the block does not move at first. This is because an words
equally increasing but opposing frictional force acts
where the block and table touch. At any instant and WRSS QJ G W Q = W QN QJ G W Q N QJ G W Q
are equal and opposite.
Key definitions
If is increased further, the block eventually
Braking distance distance over which brakes applied
moves; as it does so has its maximum value,
before vehicle brought to rest
called starting or static friction. When the block Thinking distance distance travelled during reaction time
is moving at a steady speed, the balance reading is of driver (before brakes applied)
slightly less than that for starting friction. Sliding Stopping distance total distance travelled in the time it
or dynamic friction is therefore less than starting takes to stop a vehicle; equals thinking distance + braking
or static friction. distance
Placing a mass on the block increases the force
pressing the surfaces together and increases friction. Typical values are given in Table 1.5.1 for different
When work is done against friction, the speeds. The greater the speed, the greater the
temperatures of the bodies in contact rise (as stopping distance for a given braking force. The
you can test by rubbing your hands together); data shows that the total stopping distance for a
kinetic energy is transferred to thermal energy by car travelling at 90 km/h is twice as long as for a
mechanical working (see Topic 1.7). car travelling at 60 km/h if the braking force is the
Solid friction can be described as the force same. (To stop the car in a given distance, a greater
between two surfaces that may impede motion and braking force is needed for higher speeds.)
produce heating. ▼ Table 1.5.1
Friction (drag) acts on an object such as a vehicle
or falling leaf, moving through gas (air resistance), Speed/km/h 30 60 90 120
which opposes the motion of the object. Similarly, Thinking distance/metres 6 12 18 24
friction (drag) acts on an object moving through Braking distance/metres 6 24 54 96
a liquid. Drag increases as the speed of the object
Total stopping distance/metres 12 36 72 120
increases, and acts to reduce acceleration and slow
the object down.
Thinking distance depends on the driver’s reaction
time – this will vary with factors such as the
Test yourself driver’s degree of tiredness, use of alcohol or drugs,
14 a Explain the conditions under which friction eyesight and the visibility of the hazard. Braking
occurs. distance varies with both the road conditions and
b Name two effects resulting from solid friction. the state of the car; it is longer when the road
15 A car is moving at a constant speed along a is wet or icy, so when friction between the tyres
straight road. Describe how the forces acting and the road is low, than when conditions are dry.
on the car influence the speed of the car.
Efficient brakes and deep tyre tread help to reduce
How is a constant speed achieved?

39
1.5 FORCES

the braking distance. The braking distance becomes object, like a raindrop, or an object with a large
longer when the load carried by the vehicle is surface area, such as a parachute, has a low terminal
increased. velocity and only accelerates over a comparatively
short distance before air resistance equals its
Air resistance: terminal weight. A skydiver (Figure 1.5.16) has a terminal
velocity of more than 50 m/s (180 km/h) before the
velocity parachute is opened.
Objects falling in liquids behave similarly to
In the absence of air resistance, a falling object has a
those falling in air.
constant acceleration as shown in the distance–time
graph of Figure 1.2.9 on p. 19. However, we cannot
usually ignore the effect of air resistance. As the
object accelerates, the air resistance opposing its
motion increases as its speed rises. This reduces its
acceleration; the acceleration is no longer constant.
Eventually, air resistance acting upwards equals
the weight of the object acting downwards. The
resultant force on the object is then zero since the
gravitational force balances the frictional force. The
object falls at a constant velocity, called its terminal
velocity, whose value depends on the size, shape and
weight of the object. Figure 1.2.3c on p. 14 shows the
speed–time graph for an object that is decreasing its
acceleration and approaching a terminal velocity.
A small dense object, such as a steel ball-bearing,
has a high terminal velocity and falls a considerable
distance with a constant acceleration of 9.8 m/ s2
before air resistance equals its weight. A light ▲ Figure 1.5.16 Synchronised skydivers

1.5.3 Circular motion

FOCUS POINT
★ Describe motion in a circular path and understand the effect on force if speed, radius or mass change.

There are many examples of bodies moving in circular paths: rides at a funfair, clothes being spun
dry in a washing machine, the planets going round the Sun, and the Moon circling the Earth. When
a car turns a corner, it may follow an arc of a circle. ‘Throwing the hammer’ is a sport practised at
Highland Games in Scotland (Figure 1.5.17), in which the hammer is whirled round and round before
it is released.

To keep an object moving in a circular path requires for a whirling hammer, it is the force exerted on the
a force to act towards the centre of the circle. In handle of the hammer by the athlete. The size of the
the case of a satellite orbiting the Earth, that force force depends on a number of factors.
is provided by the Earth’s gravitational attraction;

40
 1.5.3 Circular motion

its speed does not, then its velocity has changed.

A change of velocity is an acceleration, and so
during its whirling motion the ball is accelerating.
It follows from Newton’s first law of motion
that if we consider a body moving in a circle to
be accelerating, then there must be a force acting
on it to cause the acceleration. In the case of the
whirling ball it is reasonable to say the force is
provided by the string pulling inwards on the ball.
Like the acceleration, the force acts towards the
centre of the circle and keeps the body at a fixed
distance from the centre.
A larger force is needed if
● the speed Y of the ball is increased, with mass
▲ Figure 1.5.17 ‘Throwing the hammer’
and radius constant
the radius of the circle is decreased, with mass
Centripetal force ●
and speed constant
In Figure 1.5.18 a ball attached to a string is being ● the mass P of the ball is increased, with speed
whirled round in a horizontal circle. Its direction and radius constant.
of motion is constantly changing. At A, it is along
the tangent at A; shortly afterwards, at B, it is This force, which acts towards the centre and keeps
along the tangent at B; and so on. It can be seen a body moving in a circular path, is called the
that motion in a circular path is due to a force centripetal force (centre-seeking force).
perpendicular to the motion. Should the force be greater than the string can
bear, the string breaks and the ball flies off with
steady speed in a straight line along the tangent,
string i.e. in the direction of travel when the string broke
A (as Newton’s first law of motion predicts). It is not
force in string pulls thrown outwards.
ball into a Whenever an object moves in a circle (or circular
B circular path
arc) there must be a centripetal force acting on
▲ Figure 1.5.18
it. In throwing the hammer it is the pull of the
athlete’s arms acting on the hammer towards the
Velocity has both size and direction; speed has only centre of the whirling path. When a car rounds a
size. Velocity is speed in a stated direction and if bend, a frictional force is exerted inwards by the
the direction of a moving body changes, even if road on the car’s tyres.

Going further
Satellites Communication satellites
For a satellite of mass P orbiting the Earth at radius Communication satellites circle the Earth in orbits
with orbital speed Y, the centripetal force, , is the above the equator. Geostationary satellites have an
Earth’s gravitational force on the mass. orbit high above the equator (36 000 km); they travel
with the same speed as the Earth rotates, so appear
To put an artificial satellite in orbit at a certain height
to be stationary at a particular point above the Earth’s
above the Earth it must enter the orbit at the correct
surface – their orbital period is 24 hours. They are used
speed. If it does not, the force of gravity, which
for transmitting television, intercontinental telephone
decreases as height above the Earth increases,
and data signals. Geostationary satellites need to be
will not be equal to the centripetal force needed
well separated so that they do not interfere with each
for the orbit.
other; there is room for about 400.

41
1.5 FORCES

Mobile phone networks use many satellites in much

lower equatorial orbits; they are slowed by the Earth’s
atmosphere and their orbit has to be regularly adjusted
by firing a rocket engine. Eventually they run out of fuel
and burn up in the atmosphere as they fall to Earth.

Monitoring satellites
Monitoring satellites circle the Earth rapidly in low
polar orbits, i.e. passing over both poles; at a height of
850 km the orbital period is only 100 minutes. The Earth
rotates below them so they scan the whole surface at
short range in a 24-hour period and can be used to map
or monitor regions of the Earth’s surface which may
be inaccessible by other means. They are widely used
in weather forecasting as they continuously transmit
infrared pictures of cloud patterns down to Earth
(Figure 1.5.19), which are picked up in turn by receiving
stations around the world. ▲ Figure 1.5.19 Satellite image of cloud over Europe

Test yourself
16 An apple is whirled round in a horizontal circle on 17 Is the gravitational force on a satellite greater or
the end of a string which is tied to the stalk. It is less when it is in a high orbit than when it is in
whirled faster and faster and at a certain speed a low orbit?
the apple is torn from the stalk. Explain why this
happens.

1.5.4 Turning effect of forces

FOCUS POINTS
★ Describe and give everyday examples of the turning effect of a force (its moment) and use the appropriate
equation to calculate the moment of a force.
★ Apply the principle of moments to different situations.
★ Recall the conditions for an object being in equilibrium.
★ Be familiar with an experiment showing that an object in equilibrium has no resultant moment.

A seesaw in a children’s playground can be balanced if the two children have similar weights or if
the lighter child sits further from the pivot than the heavier child. Each child exerts a turning effect
on the seesaw, either clockwise or anticlockwise, which depends not only on their weight but also on
their distance from the pivot. Forces act in different ways depending on their orientation. In this topic
you will discover that the turning effect of a force (its moment) depends on both its magnitude and
the perpendicular distance from the pivot point. This means that a small force at a large distance can
balance a much larger force applied closer to the pivot. When the combination of all the forces acting
on a body is such that there is no net force or turning effect, the body is in equilibrium (the seesaw is
level) and will not move unless additional forces are applied.

42
 1.5.4 Turning effect of forces

Moment of a force a
3m
The handle on a door is at the outside edge so
hinge (fulcrum)
that it opens and closes easily. A much larger force O
would be needed if the handle were near the hinge.
gate
Similarly, it is easier to loosen a nut with a long
spanner than with a short one. F 5N
The turning effect of a force is called the moment
of the force. It depends on both the size of the force b
and how far it is applied from the pivot. It is measured
O
by multiplying the force by the perpendicular distance 1.5 m 1.5 m
of the line of action of the force from the pivot.
The unit is the newton metre (N m). F 5N
PRP QW RI IR = IR × S S QG O G W Q ▲ Figure 1.5.20
I RP W S RW
Balancing a beam
Key definition
To balance a beam about a pivot, like the ruler
Moment of a force moment = force × perpendicular
distance from pivot
in Figure 1.5.21, the weights must be moved
so that the clockwise turning effect equals the
anticlockwise turning effect and the net moment on
In Figure 1.5.20a, a force acts on a gate at its
the beam becomes zero. If the beam tends to swing
edge, and in Figure 1.5.20b it acts at the centre.
clockwise, P can be moved further from the pivot
In Figure 1.5.20a:
to increase its turning effect; alternatively, P can
PRP QW RI R W2= 1× P= 1P be moved nearer to the pivot to reduce its turning
In Figure 1.5.20b: effect. What adjustment would you make to the
position of P to balance the beam if it is tending
PRP QW RI R W2= 1× P= 1P to swing anticlockwise?
The turning effect of is greater in the first case. This
agrees with the fact that a gate opens most easily when
pushed or pulled at the edge furthest from the hinge.

Practical work

Principle of moments Balance a half-metre ruler at its centre, adding

Plasticine to one side or the other until it is
d1 d2 horizontal.
Hang unequal loads P and P from either side
of the pivot and alter their distances and
from the centre until the ruler is again balanced
pivot (nail through
m1 hole in ruler) m2 (Figure 1.5.21). Forces F and F are exerted by
gravity on P and P and so on the ruler; the
▲ Figure 1.5.21 force on 100 g is 0.98 N. Record the results in a
table and repeat for other loads and distances.
m /g F /N d /cm F × d /N cm m /g F /N d /cm F × d /N cm

43
1.5 FORCES

is trying to turn the ruler anticlockwise and 12 Name the variables you will need to measure
× is its moment. is trying to cause in this experiment.
clockwise turning and its moment is × . 13 Calculate the moments of a force of 5 N acting
When the ruler is balanced or, as we say, at a perpendicular distance from the pivot of
in equilibrium, the results should show that a 10 cm
the anticlockwise moment × equals the b 15 cm
clockwise moment × . c 30 cm.

Principle of moments Key definition

The principle of moments (also called the law of Principle of moments when a body is in equilibrium, the
the lever) is stated as follows: sum of the clockwise moments about any point equals the
sum of the anticlockwise moments about the same point
Q RG Q T O P W P RI W
OR N PRP QW R W Q SR QW T O W
P RI W QW OR N PRP QW R WW P Test yourself
SR QW 7 QR OW QW PRP QW RQ Q R M W Q
18 A seesaw has a weight of 40 N placed 1 m from the
T O P pivot and a weight of 20 N is placed on the opposite
Another way of saying this is that the sum of the side of the pivot at a distance of 2 m from the pivot.
Is the seesaw balanced?
moments is zero when the body is in equilibrium. 19 A half-metre ruler is pivoted at its mid-point and
balances when a weight of 20 N is placed at the
10 cm mark and a weight is placed at the 45 cm
Worked example mark on the ruler. Calculate the weight .

The seesaw in Figure 1.5.22 balances when Shani of weight

320 N is at A, Tom of weight 540 N is at B and Harry of
weight is at C. Find . Levers
3m 3m A lever is any device which can turn about a pivot.
A B O C In a working lever a force called the effort is used
pivot
to overcome a resisting force called the load.
1m
The pivotal point is called the fulcrum.
If we use a crowbar to move a heavy boulder
(Figure 1.5.23), our hands apply the effort at one
320 N 540 N W
end of the bar and the load is the force exerted by
▲ Figure 1.5.22 the boulder on the other end. If distances from the
Taking moments about the pivot, O: fulcrum O are as shown and the load is 1000 N (i.e.
QW OR N PRP QW = 1× P 1× P
the part of the weight of the boulder supported by
the crowbar), the effort can be calculated from the
= 1P 1P
law of moments. As the boulder just begins to move,
= 1P we can say, taking moments about O, that
OR N PRP QW = × P
OR N PRP QW = QW OR N PRP QW
By the law of moments,
IIR W × P= 1× P
OR N PRP QW = QW OR N PRP QW
∴ × P= 1P 10000 N cm
IIR W = = 50 N
∴ 1500 Nm 200cm
= = 500 N
3m
Examples of other levers are shown in Figure 1.5.24.

44
 1.5.4 Turning effect of forces

How does the effort compare with the load for fulcrum
scissors and a spanner in Figures 15.24c and d?
B
OA 10 cm effort
OB 200 cm

effort load

▲ Figure 1.5.24d Spanner

A
O
fulcrum Conditions for equilibrium
Sometimes a number of parallel forces act on an
object so that it is in equilibrium. We can then say:
load
(i) The sum of the forces in one direction equals the
▲ Figure 1.5.23 Crowbar sum of the forces in the opposite direction.
effort (ii) The law of moments must apply.
When there is no resultant force and no resultant
moment, an object is in equilibrium.

Key definition
Equilibrium when there is no resultant force and no
resultant moment

fulcrum As an example, consider a heavy plank resting on two

load trestles, as in Figure 1.5.25. In Topic 1.5.5 we will see
that the whole weight of the plank (400 N) may be
▲ Figure 1.5.24a Wheelbarrow taken to act vertically downwards at its centre, 2. If
and are the upward forces exerted by the trestles on
the plank contact forces, then we have from (i) above:
= 1
effort
1m 3m 2m 2m
P Q
biceps O
muscle
trestle trestle C plank
400 N
▲ Figure 1.5.25
fulcrum load Moments can be taken about any point but if we take
them about C, the moment due to force is zero.
▲ Figure 1.5.24b Forearm
OR N PRP QW = × P
fulcrum
QW OR N PRP QW = 1× P
= 1P
effort
Since the plank is in equilibrium we have from (ii)
above:
× P= 1P
800 Nm
load ∴ = = 160 N
5m
From equation (1)
▲ Figure 1.5.24c Scissors = 1
45
1.5 FORCES

Test yourself
20 The metre ruler in Figure 1.5.26 is pivoted at 21 A seesaw has a weight of 60 N placed 0.5 m from the
its centre. If it balances, which of the following pivot and a weight of 20 N is placed on the opposite
equations gives the mass of ? side of the pivot at a distance of 1.5 m from the pivot.
A + 50 = 40 + 100 Is the seesaw balanced? Justify your answer.
B × 40 = 100 × 50 22 The beam shown in Figure 1.5.27 is balanced with
C × 50 = 100 × 40 weights of 160 N, 120 N and in the positions shown.
D /50 = 40/100 Calculate the value of .

50 cm 40 cm 3m 3m
A B O C

1m pivot

M 100 g

160 N 120 N W
▲ Figure 1.5.26
▲ Figure 1.5.27

1.5.5 Centre of gravity

FOCUS POINTS
★ Define centre of gravity and the effect its position has on the stability of an object.
★ Be familiar with an experiment determining the position of the centre of gravity of an irregularly shaped
plane lamina.

Why are tall vehicles more likely to topple over on a slope than less tall ones? The answer lies in
the position of the centre of gravity. In the presence of gravity an object behaves as if its entire
mass is concentrated at a single point, the centre of gravity. The object’s weight appears to act at
this point. For a symmetrical object, such as a ball, the centre of gravity will be at its centre. In this
topic, you will learn that when an object is suspended so that it can swing freely, it comes to rest with
its centre of gravity vertically below the point of suspension. This enables the centre of gravity of
unsymmetrical objects to be located. You will discover that it is the position of the centre of gravity
that controls stability against toppling. If the centre of gravity remains within the footprint of the
base of the object, it remains stable.

Centre of gravity any other point it topples because the moment of its
weight about the point of support is not zero, as
An object behaves as if its whole mass were in Figure 1.5.28b. The centre of gravity is sometimes
concentrated at one point, called its centre of also termed the centre of mass.
gravity even though the Earth attracts every part of
it. The object’s weight can be considered to act at
Key definition
this point. The centre of gravity of a uniform ruler
Centre of gravity the point through which all of an
is at its centre and when supported there it can be object’s weight can be considered to act
balanced, as in Figure 1.5.28a. If it is supported at

46
 1.5.5 Centre of gravity

a b A tightrope walker has to keep their centre of

0 50 100 0 50 100
gravity exactly above the rope. Some carry a long
pole to help them to balance (Figure 1.5.30).
ruler support The combined weight of the walker and pole is then
W spread out more and if the walker begins to topple
▲ Figure 1.5.28 to one side, they move the pole to the other side.
Your centre of gravity is near the centre of your body
and the vertical line from it to the floor must be
within the area enclosed by your feet or you will fall
over. You can test this by standing with one arm and
the side of one foot pressed against a wall (Figure
1.5.29). Now try to raise the other leg sideways.

▲ Figure 1.5.30 A tightrope walker using a long pole

raise this leg
The centre of gravity of a regularly shaped
object that has the same density throughout is
at its centre. In other cases, it can be found by
▲ Figure 1.5.29 Can you do this without falling over? experiment.

Practical work

Centre of gravity of an irregularly shaped the lamina from a third hole. Also try balancing it
lamina at its centre of gravity on the tip of your forefinger.
Suppose we have to find the centre of gravity A
of an irregularly shaped lamina (a thin sheet) of hole nail clamped
in stand
cardboard. lamina C
centre of gravity
Make a hole A in the lamina and hang it so that
it can swing freely on a nail clamped in a stand. D B
It will come to rest with its centre of gravity plumb line
vertically below A. To locate the vertical line
through A, tie a plumb line (a thread and a
weight) to the nail (Figure 1.5.31), and mark its ▲ Figure 1.5.31
position AB on the lamina. The centre of gravity 14 a How could you make a plumb line?
lies somewhere on AB. b Explain the purpose and use of a plumb
Hang the lamina from another position, C, and line.
mark the plumb line position CD. The centre of 15 When an object is suspended and allowed to
gravity lies on CD and must be at the point of swing freely, where does its centre of gravity
intersection of AB and CD. Check this by hanging lie when it comes to rest?

47
1.5 FORCES

Stability The stability of a coach is being tested in

Figure 1.5.33b. When the top deck only is fully
The position of the centre of gravity of an object laden with passengers (represented by sand bags
affects whether or not it will be stable when tilted. in the test), it must not topple if tilted through an
This is important in the design of such things as angle of 28º.
tall vehicles (which tend to overturn when rounding Racing cars have a low centre of gravity and a
a corner), racing cars, reading lamps and even wide wheelbase for maximum stability.
drinking glasses.
An object topples (falls over) when the vertical
line through its centre of gravity falls outside its
base, as in Figure 1.5.32a. Otherwise it remains
stable, as in Figure 1.5.32b, where the object will
not topple.
a b

centre of
mass
base

▲ Figure 1.5.32 ▲ Figure 1.5.33a A fire truck under test to find its centre of
gravity
Toppling can be investigated by placing an empty
can on a plank (with a rough surface to prevent
slipping) which is slowly tilted. The angle of tilt
is noted when the can falls over. This is repeated
with a mass of 1 kg in the can. How does this affect
the position of the centre of gravity? The same
procedure is followed with a second can of the same
height as the first but of greater width. It will be
found that the second can with the mass in it can
be tilted through the greater angle.
The stability of a body is therefore increased by
(i) lowering its centre of gravity, and
(ii) increasing the area of its base.
In Figure 1.5.33a the centre of gravity of a fire
truck is being found. It is necessary to do this when
testing a new design since fire trucks are often
driven over sloping surfaces and any tendency to
overturn must be discovered. ▲ Figure 1.5.33b A coach being tilted to test its stability

48
 1.5.5 Centre of gravity

Going further
Balancing tricks and toys
Some tricks that you can try or toys you can make
Three terms are used in connection with stability. are shown in Figure 1.5.35. In each case the centre
of gravity is vertically below the point of support
Stable equilibrium and equilibrium is stable.
An object is in stable equilibrium if when slightly
displaced and then released it returns to its previous needle
position. The ball at the bottom of the dish in Figure
1.5.34a is an example. Its centre of gravity rises when cork
it is displaced. It rolls back because its weight has a
fork
moment about the point of contact that acts to reduce
the displacement.

a
can
dish ball
point of contact

centre weight
card cork
of mass

b
centre of mass
card

spar thick
wire

bull-dog clip
point of matchsticks
contact
iron nut
bar

weight ▲ Figure 1.5.35 Balancing tricks

A self-righting toy (Figure 1.5.36) has a heavy base

c and, when tilted, the weight acting through the
centre of gravity has a moment about the point of
contact. This restores it to the upright position.
▲ Figure 1.5.34 States of equilibrium

Unstable equilibrium
An object is in unstable equilibrium if it moves
further away from its previous position when slightly
displaced and released. The ball in Figure 1.5.34b
behaves in this way. Its centre of gravity falls when
it is displaced slightly because there is a moment
which increases the displacement. Similarly, in Figure
1.5.28a the balanced ruler is in unstable equilibrium.

Neutral equilibrium
An object is in neutral equilibrium if it stays in its new
position when displaced (Figure 1.5.34c). Its centre
of gravity does not rise or fall because there is no
moment to increase or decrease the displacement.
▲ Figure 1.5.36 A self-righting toy

49
1.5 FORCES

Test yourself
23 Where does the centre of gravity lie for a
a uniform ruler
b sphere of uniform density? A B C
24 a When does an object topple?
b How can the stability of an object be increased?
25 Figure 1.5.37 shows a Bunsen burner in three
different positions. State the type of equilibrium
when it is in position
i A ▲ Figure 1.5.37
ii B
iii C.

Revision checklist
After studying Topic 1.5 you should know and ✓ combine forces acting along the same straight line
understand: to find their resultant
✓ the significance of the term limit of proportionality ✓ recall the equation resultant force = mass ×
✓ Newton’s first law of motion acceleration =P and use it to solve problems
✓ friction as the force between two surfaces that ✓ define the thinking distance, braking distance
impedes motion and results in heating and that and stopping distance for a vehicle when a driver
friction also acts on an object moving through applies the brakes and explain the factors that
the air or a liquid affect thinking and braking distance
✓ that Newton's third law of motion describes pairs ✓ explain how an object falling in a fluid reaches
of forces of the same type acting on different terminal velocity
objects ✓ describe qualitatively motion in a circular path
✓ the conditions for equilibrium due to a perpendicular force and recall that the
✓ that an object’s weight acts through the centre of force required to maintain circular motion changes
gravity. when the speed, radius of orbit or mass changes
✓ define the moment of a force about a pivot and
After studying Topic 1.5 you should be able to:
give everyday examples; recall the principle of
✓ recall that a force can cause a change in the
moments and use it to solve problems, including
motion, size or shape of a body
the balancing of a beam
✓ identify and show different types of force on
✓ recall that an object behaves as if its whole mass
free-body diagrams
acts through its centre of gravity
✓ describe an experiment to study the relation
✓ describe an experiment to find the centre of gravity
between force and extension for springs; plot and
of an object and connect the stability of an object
draw conclusions from load–extension graphs
to the position of its centre of gravity.
✓ define the spring constant and use the equation
spring constant = force/extension = to solve
problems

50
 Exam-style questions

Exam-style questions
1 a Describe how you would investigate the 3 Two forces of 5 N and 12 N act at a point.
variation of the extension of a spring when a The two forces first act in opposite directions.
different loads are applied. Mention two i Make a sketch showing the direction
precautions you would take to obtain of the forces. [2]
accurate results. [6] ii Calculate the resultant of the forces
b The table below shows the results obtained and mark its direction on your sketch. [2]
in an experiment to study the stretching b The two forces then act at 90° to each other.
of a spring. Copy the table and fill in the Calculate the magnitude and direction of the
missing values. What can you say about the resultant force by calculation. [6]
relationship between the extension of the [Total: 10]
spring and the stretching force? [4] 4 Starting from rest on a level road, a girl can
reach a speed of 5 m/s in 10 s on her bicycle.
Stretching Scale Extension/ a Calculate the acceleration. [2]
Mass/g force/N reading/mm mm
b Calculate the average speed during the 10 s. [2]
0 20.0 0 c Calculate the distance she travels in 10 s. [2]
100 20.2 d Eventually, even though she still pedals as
200 20.4
fast as she can, she stops accelerating and
her speed reaches a maximum value. Explain
300 20.6 in terms of the forces acting why this
400 20.8 happens. [2]
500 21.0
[Total: 8]
5 Explain the following using = P .
[Total: 10] a A racing car has a powerful engine and is
2 The spring in Figure 1.5.38 stretches from 10 cm made of strong but lightweight material. [3]
to 22 cm when a force of 4 N is applied. b A car with a small engine can still
a Calculate the spring constant of the accelerate rapidly. [3]
spring. [3] [Total: 6]
b If the extension is proportional to the 6 A rocket has a mass of 500 kg.
stretching force when a force of a Calculate the weight of the rocket
6 N is applied, calculate on Earth where = 9.8 N/kg. [1]
i the new extension length of the b At lift-off the rocket engine exerts an
spring [2] upward force of 25 000 N.
ii the final length in cm of the spring. [1] i Calculate the resultant force on the
rocket. [2]
ii Calculate the initial acceleration of
the rocket. [3]
10 cm
22 cm [Total: 6]

4N

▲ Figure 1.5.38

[Total: 6]

51
1.5 FORCES

7 A car rounding a bend travels in an arc of a 0 cm 10 40 50 100 cm

circle.
a What provides the force to keep the car
travelling in a circle? [2]
b Is a larger or a smaller force required if 3N
i the car travels faster [1]
▲ Figure 1.5.40
ii the bend is less curved [1]
iii the car has more passengers? [1] 10 a Describe how you could find the centre of
c Explain why racing cars are fitted with tyres gravity of an irregular lamina. [5]
called ‘slicks’, which have no tread pattern, b A heavy box with a square base and a height
for dry tracks and with ‘tread’ tyres for wet twice the length of a side is to be transported
tracks. [2] by a lorry. Explain how the stability of
[Total: 7] i the lorry [2]
8 Figure 1.5.39 shows three positions of the pedal ii the box [2]
on a bicycle which has a crank 0.20 m long. will be affected if the box lies on its side in
The cyclist exerts the same vertically downward the van rather than its base.
push of 25 N with his foot. Calculate the turning [Total: 9]
effect in 11 Figure 1.5.41 shows the forces acting on a
a A [2] raindrop which is falling to the ground.
b B [2]
B
c C. [2]
[Total: 6]
raindrop
force

chain
A
crank
▲ Figure 1.5.41

a i is the force which causes the raindrop

to fall. Give the name of this force. [1]
ii is the total force opposing the motion
A B of the drop. State one possible cause of
this force. [1]
b What happens to the drop when
force = force ? [2]
[Total: 4]
C 12 A car is travelling at a constant speed of 20 m/s.
The reaction time of the driver is 0.7 s and there
▲ Figure 1.5.39 is a road block 50 m ahead.
9 The weight of the uniform bar in Figure 1.5.40 a Calculate the driver’s thinking distance. [2]
is 10 N and it is 100 cm long. b The car comes to a halt at the road block.
a Calculate the clockwise moment about the Calculate the braking distance. [1]
pivot. [3] c State two factors that can affect
b Calculate the anticlockwise moment about i thinking distance [2]
the pivot. [3] ii braking distance. [2]
c Does the beam balance, tip to the right or [Total: 7]
tip to the left? [2]
[Total: 8]

52
 Alternative to Practical

Alternative to Practical
13 A physics class is asked to investigate the 15 In an experiment to investigate the law of
extension of a stretched spring. moments, a half-metre ruler is balanced at its
You will be supplied with a spring, a clamp stand, centre as shown in Figure 1.5.42.
a half-metre ruler, a set square and a hanger with
d1 d2
100 g weights and sticky tape.
a Describe how you would carry out the
experiment. [5]
b Suggest any precautions you would take to
fulcrum (nail through
achieve good results. [3] m1 hole in ruler) m2
[Total: 8]
▲ Figure 1.5.42
14 a The table below shows the extension of a
spring for increasing stretching forces. Masses of 50 g, 100 g and 150 g are placed in turn
at the positions given in the table below.
Stretching force/N 0 1 2 3 4 5 a Complete the table, filling in values for
Extension/mm 0 2 4 6 8.5 12 i the units at the head of each column [1]
ii force ( ) [2]
i Plot a graph with extension/mm along iii distance from pivot ( ) [2]
the -axis and stretching force/N on iv moment about pivot ( × ). [2]
the -axis. [4] b State which combinations of two different
ii Draw the best line through the points; masses could be used to balance the beam. [3]
mark the region over which
proportionality holds. [2] Mass/g Force/ Ruler reading/cm / × /
iii Indicate the limit of proportionality. [1] 50 5 A
b Calculate the gradient of the graph. [2] 50 10 B
c Determine the spring constant . [1]
50 15 C
[Total: 10]
50 20 D
100 30 E
100 35 F
100 40 G
150 20 H
150 35 I

[Total: 10]

53