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Cambridge IGCSE
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1.5c Turning Forces (Moments)

Car wheels need to be fastened really tightly - you would not want one falling off as the car is moving.
However occasionally the wheel needs changing, and loosening the nuts and bolts can be difficult:

Figure 1. Using a long lever to undo the wheel of a car

It does not matter how hard you push, sometimes if the lever is not long enough the nut will not move. The trick to getting a large 'turning force' is to use a large force AND a long lever.

In physics, the idea of a 'turning force' is given the name of a moment. A car wheel nut needs a very large moment to make it turn. The spanner used to undo the nut will need to produce a large moment by applying a large force, and this force needs to be a long way from the centre of rotation, called the pivot.

moments on a nut

Figure 2. Finding the moment on a wheel nut.

The formula for working out the moment is:

moment = force x perpendicular distance (from the pivot)

moment = F x d

Learn this formula!

Note: There is no standard letter for 'moment' in this course - just use the name.

The units used for moments are just the force in newtons x the distance in metres, to give Nm ("newton metres").

For example, in figure 2, if the force applied is 60N and the distance from the pivot is 40 cm, the moment is calculated as follows:
40 cm = 0.4 m,
moment = force x distance, so:
moment = 60 x 0.4
moment = 24 Nm.


Combining moments

In this section we look at what happens when there are more than one force acting on a 'system'. Let's have a closer look at this spanner example, Here there are two forces pushing on the spanner, but in opposite directions.

Which way will the spanner move?

Two forces on a spanner

Figure 3. Combining moments.

the trick here is that you do NOT just add the forces together. You need to work out the moment of each one separately, and then combine the moments.
Have a go at the questions here - see if you can do it without looking at the solution, to get a better understanding of how we combine moments.


1. In the diagram shown in figure 3, there are two forces acting, F1 and F2:

a) Force F2 is pushing upwards on the left-hand side, so is trying to turn the spanner in a clockwise direction.

b) The moment of F2 = force x perpendicular distance to the pivot
So moment = 40 x 0.4 = 16 Nm clockwise

c) The moment of F1 = force x perpendicular distance to the pivot
So moment = 30 x 0.6 = 18 Nm anti-clockwise

d) The anti-clockwise moment is larger (18 Nm compared to 16 Nm) so the spanner will rotate anti-clockwise.

e) The anti-clockwise moment is larger by 2 Nm, so therefore the total resultant moment is 2 Nm anti-clockwise.


2. Look at the spanner in figure 3.
If we increase or decrease the size of F1, calculate the force F1 needed to balance the spanner so that it does not rotate.

The moment from F2 is 16 Nm. Therefore to make the two moments balance, the moment from F1 also needs to be 16 Nm.
We know moment of F1 = force x perpendicular distance to the pivot:
Therefore 16 = F x 0.6,

F = 16

F = 26.66 N or 26.7 N to 3 sig figs. (27 N is also OK here)

How did you get on with these questions? If you understood these ideas, you are well on your way to understanding the next section:


The principle of moments

This is a very simple rule that says that - for a system that does not rotate - the moments in a clockwise direction must balance the moments in an anti-clockwise direction.

You may see the word 'equilibrium' used to describe a system that is not moving and where all of the moments are balanced.


The principle of moments:

"A system will not rotate ( is in equilibrium) if the clockwise moments equal the anti-clockwise moments".

This means that we can work out the missing force or distance in a system needed to make it balance and not rotate, just like in question 2 earlier.

A very common example of this is the see-saw or a similar system, with a force acting on each side of the pivot.

Required Practical: demonstrate that there is no resultant moment on an object in equilibrium

This is simple to do with a long ruler and a pivot placed in the centre:


showing the principle of moments - practical
Figure 4. Demonstrating the principle of moments


You should find that the clockwise moments and anti-clockwise moments are equal, or very close together. By repeating this for a range of different masses, you can show that thew two moments calculated should, in fact, be equal.

Example data:

mass (g) weight (N)
distance from
pivot (cm)
moment (N cm) mass (g) Weight (N) Distance from
pivot (cm)
moment (N cm)
80 0.78 20 15.6 50 0.49 32 15.7
Weight =mass (kg) x 9.8

Note: In rounding up our answers as well as through possible measurment error, the results are not an exact match, but 15.6 N cm is almost the same as 15.7 N cm.

The next questions are common examples of the principle of moments:



3. Two children (with unusual body shapes!) are on a 'seesaw' in a playground.
The seesaw is balanced. What is the weight, W, of the child on the right-hand side?

the moments of children on a seesaw

If the seesaw is balanced, then:
the anti-clockwise moment = clockwise moment
The child on the left (L) is turning the seesaw anti-clockwise, so we can write:
FL x d L = FR x d R
(Here, we are using the notation FL for the force on the left, and a similar notation for all the forces and distances)
The weight W of the child on the right is called FR in this formula, so:
300 x 1.5 = W x 1.2
450 = W x 1.2

W = 450

W = 375 N

4. Two boxes are balanced on a long beam as shown. What is the missing distance labelled d - the distance of the centre of the right-hand box from the pivot?

The moments on a balance

If the system is balanced, then the anti-clockwise moment = clockwise moment,

So again,
FL x d L = FR x d R ,
We will leave distances in cm for this question:
6 x 30 = 9 x d
180 = 9 x d , so

d = 180
d =20 cm (or d = 0.2 m)


Bridges and beams

Take a look at this diagram showing a bus of weight 40 kN in the middle of a bridge:

A bus on a bridge with supports

Figure 5. A heavy load on a light beam.

The bridge is being held up by the two supports, which are pushing upwards on the road with force A on the left and force B on the right. The weight of the bridge is so small we can ignore it for this question.
What do you think the values of force A and force B will be?

The answer is that the weight is split between the two supports, and as the bus is in the middle of the bridge, each support takes 20 kN. Did you guess this?


5. What about the situation where a heavier 60 kN bus is closer to the right?

A bus on a bridge with supports

Which of these answers do you think will correctly give the sizes of forces A and B?

The answer is (c) Force A = 20 kN, force B = 40 kN.
See below for an explanation of why this is so.

As a heavy weight like the bus moves closer to the right - the right-hand side support takes most of the load. The upwards forces must still balance the downwards forces, so force A + force B must still equal 60 kN.

This distribution of the weight is true for any bridge or light beam supporting a heavy load.

Calculating the Forces on a Bridge / Beam

To find the magnitude of the forces acting on a bridge or a beam is quite a difficult question for GCSE: It relies on the idea that as the bridge is stable, in equilibrium, then the principle of moments can be applied to either supporting point. In the example below shown in figure 6, we imagine that instead of the bridge resting on the support, it is acting as a pivot as shown:

A bridge with a pivot

Figure 6. Calculating the force from a bridge support

The weight of the bus is trying to rotate the beam clockwise, and force Fs from the support is preventing this, producing an anti-clockwise moment.
If the system is balanced, then the anti-clockwise moment = clockwise moment.
Fs x ds = Fbus x d bus
Substituting in the numbers from the diagram gives:
Fs x 15 = 60 000 x 10
Fs x 15 = 600 000,
So Fs = 600 000/15,
Fs = 40 000 N or 40 kN
(The same answer as question 5)



6. A heavier bus of weight 80 KN is now 12 m from the left-hand support on the same light bridge, where the weight of the bridge itself can be ignored.

A bridge with a pivot

a) Using the principle of moments, calculate the force from the right-hand support FR.
b) Using your answer from (a), calculate the force on the left-hand side support FL.

a) Again, we imagine that the left-hand side acts as a pviot as shown here:

A bridge with a pivot

The weight of the bus is trying to rotate the beam clockwise, and force FR from the support is preventing this, producing an anti-clockwise moment.
If the system is balanced, then:
anti-clockwise moment = clockwise moment
FR x dR = Fbus x d bus
Substituting in the numbers from the diagram gives:
FR x 15 = 80 000 x 12
Fs x 15 = 960 000,
So FR= 960 000/15,
FR = 64 000 N or 64 kN

b) This is much easier, and there is no need to repeat the maths above for the left-hand support: We know that the weight of the bus is 80 kN. If the right-hand support produces a force of 64 kN, then the remaining weight is supported by the left-hand side, producing 80 kN - 64 kN = 16 kN force.

FL = 16 000 N or 16 kN


Centre of gravity

Did you notice in question 6 above, that the arrow showing weight (the force of gravity) is always drawn in the centre of the object?

Gravity is pulling down on every part of our bodies - head, arms, legs - every atom in our bodies. However it is not realistic to draw all of these - it is much easier to draw a single arrow representing all of the forces acting at the centre. This point is where a single mass with the same weight would be placed, so that it has exactly the same effect. The point is called the centre of gravity.

centre of gravity

Figure 7. The centre of gravity.

The weight of a body acts through its centre of gravity

This is why we always draw a single arrow to represent the weight of any body.
The centre of gravity position can be thought of as the point which is the average location of all the atoms in the object - some will be above, some below, some to the left and some to the right. (Assuming all the atoms are similar).

Required Practical: determine the position of the centre of gravity of an irregularly shaped plane lamina

This is a standard practical and is simple to do with some easily obtainable apparatus:

The video here shows how to carry out this simple practical:

Youtube: vt.physics



For an object to be resting on a table and be completely stable, the centre of gravity has to be directly above the base of the object. If the centre of gravity is to the left or right, the object will topple over. This is why sumo wrestlers keep their feet apart. A wide base makes them more stable and less likely to be pushed over.
This idea is shown in figure 8 below: The three shapes are almost identical, but the base changes width. In diagram (C), the centre of gravity is not vertically above the base (which is narrow), so the shape will topple over.

centre of gravity

Figure 8. Stabilty and the centre of gravity.







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