Physics
Double Award
Double Award
Cambridge IGCSE
TOPIC 1A: FORCES and MOTION
1.5b 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.
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
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.
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?
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.
Questions.
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 |
| 0.6 |
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:
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.
The next questions are common examples of the principle of moments:
Questions:
3. Two children (with unusual body shapes!) are on a 'see-saw' in a playground.
The 'see-saw' is balanced. What is the weight, W, of the child on the right-hand side?
If the 'see-saw' is balanced, then:
the anti-clockwise moment = clockwise moment
The child on the left (L) is turning the
'see-saw' 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 |
| 1.2 |
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?
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 |
| 9 |
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.
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).
For regular shapes like spheres, cubes and cylinders, the centre of gravity is simply the centre of the shape, which should be easy to determine. It is harder to find for irregular shapes, as shown here:
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.
Figure 8. Stability and the centre of gravity.
Now test your understanding using these quick, 10 minute questions on moments from Grade Gorilla:
