A Level Maths: Moments

Sense of rotation

When undoing a nut it is turned in an anticlockwise direction. Moments can be taken in a clockwise and anticlockwise direction and this direction must be clearly stated. 

Example 1: 

Find the moment of the force at C at the points A, B and C and state the direction of the moment.

A: 3 \times 6=18 \mathrm{Nm}

B: 3 \times 2=6 \mathrm{Nm}

C: 3 \times 0=0

Sum of moments

If you are given a number of forces acting on a body then their moments can be added together provided their direction is taken into account. 

Example 2:

Three forces have moments of 15 Nm clockwise, 10 Nm anticlockwise and 13 Nm clockwise. What is the sum of the moments and their direction?

Solution:

Taking clockwise to be positive: sum of moments = 15 – 10 + 13 = 18 Nm clockwise. 

If P = 6N and Q = 5N find the combined turning effect about O. 

If P = 7N and Q = 3.5N find the combined turning effect about O.

Solution: taking clockwise to be positive: 

Solution: taking clockwise to be positive

a) (5 x 4) clockwise + (6 x 2) anticlockwise = 20 Nm – 12 Nm = 8 Nm clockwise

b) (3.5 × 4) clockwise + (7 x 2) anticlockwise = 14 Nm – 14 Nm = 0

Forces which are parallel but act in the same direction are called like forces. 

Forces which are parallel but act in opposite directions are called unlike forces. 

Example 4:

Two forces of 5N act at the ends of a rod as shown. Find the sum of the moments of the forces about A and B. 

In this example it should be noted that because the forces are opposite they have no “translational effect” on the rod i.e. the forces are not causing the rod to move in a straight line.

Couple

If two unlike forces of equal magnitude that are not acting in the same line are said to form a couple. This will have a turning effect but there will be no translatory effect. 

If the magnitude of each force forming a couple is P and the perpendicular distance between their lines of action is “a” then the moment of the couple is (P x a) Nm

Example 5:

Show that the system of forces shown forms a couple and find the moment of this couple. 

Solution:

Resolving the forces vertically we have 7 – 2 – 5 = 0. Because the forces balance out there is no translatory effect. We must now see if there is a turning effect. 

Moments about A: (7 x 0) clockwise + (2 x 1) clockwise + (5 x 4) clockwise = 22Nm clockwise. Because there is a turning effect we must have a couple. 

Using a resultant force

There are some instances where parallel forces do not reduce to a couple and in such cases these can be reduced to a single force. This single resultant force must be parallel to the already given forces and have the same rotational effects as the given forces. 

Example 6:

Two forces of 2 N and 6 N acting in the same direction are 21 m apart. What is the resultant force, its direction and the line of action?

Solution:

Suppose the initial scenario is as follows:

To find the resultant, the direction and line of action, suppose we have the following new scenario: 

Resolving vertically: R = 2 + 5 = 7 N

Moments about A (initial scenario): (2 x 0) + (5 x 21) = 105 Nm

Moments about A (new scenario): Rx

• 7x = 105 N 🡺 x = 15 m

The resultant is 7 N, the direction is the same as the two forces given and it acts 15 m from the line of action. 

Forces in equilibrium

Forces in equilibrium cannot be a couple. If this was the case there would be a turning effect. 

A resultant force that replaces any forces in equilibrium would be zero and so the moment about any point would also be zero. 

Example 7:

The diagram shows a uniform beam AB of length 2m and mass 4kg. Masses of 3kg and 1kg are attached are each end. What is the position of the support at C from A if the beam is to stay horizontal?

Solution:

The diagram above shows how you should draw a diagram to help with a question like this. And because the beam is uniform the length from each end to the midpoint will be 1 m. 

Moments about C: (3 g \mathbf{x} x)=4 \mathrm{~g}(1-x)+1 \mathrm{~g}(2-x)

Moments of a force acting at an angle

If we take a force P as shown in the diagram above then in the x direction we have P \cos \theta and in the y direction we have P \sin \theta

Remember when finding moments it is the perpendicular distance that is required as the following example demonstrates. 

Solution:

First it is important to resolve the force perpendicular to the rod. This is highlighted by the red arrow above and this is 40sin50°. 

Moments about O: 40sin50° x 1.5 = 46.0 N (2 sf) anticlockwise.

You can see that there is a lot to know in terms of doing moments in A Level mechanics maths but the main point to take away is that it is the force multiplied by the perpendicular distance. 

Remember to define your positive direction and to give yourself plenty of room to work as the methods needed can get a little complicated sometimes and you may also need to draw a very detailed diagram so that you can take moments effectively.