Moments, Couples and Equilibrium

This lesson covers AQA Spec 3.4.1.1 (Moments) in full depth, including the principle of moments, centre of mass, toppling versus sliding, and coplanar force problems.

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1 Moment of a Force

Key Definition: The moment of a force about a point (or pivot) is the product of the force and the perpendicular distance from the line of action of the force to the point.

Moment = F × d

where d is the perpendicular distance. SI unit: N m.

If the force acts at an angle θ to the line joining the point of application to the pivot:

Moment = F × r × sin θ

where r is the distance from the pivot to the point of application.

A moment can be clockwise or anticlockwise about the pivot.

Worked Example 1 — Moment of a Force

A spanner is 0.30 m long. A force of 50 N is applied at the end, perpendicular to the spanner. Find the moment about the nut.

Solution:

Moment = F × d = 50 × 0.30 = 15 N m

Worked Example 2 — Force at an Angle

The same 50 N force is applied at 60° to the spanner handle. Find the moment.

Solution:

Moment = F × r × sin θ = 50 × 0.30 × sin 60° = 50 × 0.30 × 0.866 = 13.0 N m

Exam Tip: Always use the perpendicular distance to the line of action. If the force is at an angle, either resolve the force or use r sin θ.

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2 Couples and Torque

Key Definition: A couple consists of two equal and opposite parallel forces whose lines of action do not coincide.

A couple produces pure rotation — no resultant translational force.

Torque of a couple = F × d

where d is the perpendicular distance between the lines of action of the two forces.

Note: The torque of a couple is the same about any point. You do not need to specify a pivot.

Worked Example 3

A steering wheel has a diameter of 0.40 m. A driver applies equal and opposite forces of 12 N at opposite ends of a diameter. Find the torque.

Solution:

Torque = F × d = 12 × 0.40 = 4.8 N m

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3 The Principle of Moments

For a body in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.

Σ Clockwise moments = Σ Anticlockwise moments

This applies about ANY chosen pivot point. A wise choice of pivot simplifies the calculation by eliminating unknown forces that pass through the pivot.

Worked Example 4 — Simple Beam

A uniform beam of length 4.0 m and weight 200 N is supported at its ends, A and B. A 500 N load is placed 1.0 m from end A. Find the reaction forces at A and B.

Solution: The beam's weight (200 N) acts at its centre, 2.0 m from A.

Taking moments about A (to eliminate the reaction at A):

Clockwise moments = Anticlockwise moments

500 × 1.0 + 200 × 2.0 = R_B × 4.0

500 + 400 = 4.0 R_B

R_B = 900/4.0 = 225 N

Resolving vertically: R_A + R_B = 500 + 200 = 700 N

R_A = 700 − 225 = 475 N

Check: Take moments about B:

R_A × 4.0 = 500 × 3.0 + 200 × 2.0 = 1500 + 400 = 1900

R_A = 1900/4.0 = 475 N ✓

Worked Example 5 — Multiple Loads

A uniform plank of length 6.0 m and mass 30 kg is supported at points P (1.0 m from the left end) and Q (1.0 m from the right end). Loads of 200 N and 400 N are placed at the left end and right end respectively. Find the reactions at P and Q.

Solution: Weight of plank = 30 × 9.81 = 294.3 N, acting at the centre (3.0 m from left end).

Taking moments about P (to find R_Q):