AQA AQA 2018-style practice: A uniform beam of weight 60 N and length 4.0 m is pivoted at its centre. A 30 N weight hangs 1.5 m from the pivot. Calculate how far from the pivot, on the other side, a 45 N weight must hang to balance the beam.

The beam is uniform and pivoted at its centre, so its own weight acts at the pivot and produces no moment. Apply the principle of moments, equating clockwise and anticlockwise moments about the pivot: 30 × 1.5 = 45 × d. 45 = 45d, so d = 1.0 m. Markers reward a clear statement of the principle of moments, recognising the beam's weight gives no moment at the central pivot, and correct substitution of perpendicular distances.

EnglandPhysicsSyllabus dot point

What makes an object turn, and when does it stay balanced?

The moment of a force about a point, the principle of moments, couples and torque, the centre of mass, and the conditions for the equilibrium of a rigid body under coplanar forces.

A focused answer to AQA A-Level Physics 3.4.1.2, covering the moment of a force, the principle of moments, couples and torque, the position of the centre of mass, and the conditions a rigid body must satisfy to remain in equilibrium.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The moment of a force
The principle of moments
Couples and torque
Centre of mass and equilibrium
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What this dot point is asking

AQA specification point 3.4.1.2 wants you to calculate the moment of a force, apply the principle of moments to balanced systems, understand couples and torque, locate the centre of mass, and state and use the conditions for the equilibrium of a rigid body.

The moment of a force

If the force is not perpendicular to the lever, use the perpendicular distance from the pivot to the line of action, or equivalently the component of the force at right angles to the lever, giving $\text{moment} = Fd\sin\theta$ . This is why pushing a door near the hinge, or at a shallow angle, is far less effective than pushing the edge at right angles.

The principle of moments

This lets you solve for unknown forces or distances on balanced beams and see-saws. A powerful exam tactic is to take moments about the point where an unknown force acts: that force has zero moment about its own point of action, so it drops out of the equation, leaving fewer unknowns.

Couples and torque

The torque of a couple is $\text{torque} = F \times s$ , where $s$ is the perpendicular distance between the lines of action of the two forces. Turning a steering wheel or a tap with both hands applies a couple.

Centre of mass and equilibrium

The centre of mass is the single point at which the whole weight of a body can be taken to act. For a uniform symmetrical object it lies at the geometric centre. A rigid body is in equilibrium only when both the resultant force is zero and the resultant moment about any point is zero. An object topples when its centre of mass passes beyond its base of support, which is why a wide base and low centre of mass make an object stable.

Finding an unknown force on a loaded beam

A uniform plank of weight $120 \text{ N}$ and length $3.0 \text{ m}$ rests on two supports, one at each end. A child of weight $300 \text{ N}$ stands $1.0 \text{ m}$ from the left support. Find the upward force from the right support.

step 1: Take moments about the left support

This eliminates the left support force, which has zero moment about its own point.

step 2: Identify the clockwise moments

The plank's weight acts at its centre, $1.5 \text{ m}$ from the left: moment $= 120 \times 1.5 = 180 \text{ N m}$ . The child: $300 \times 1.0 = 300 \text{ N m}$ .

step 3: Set the anticlockwise moment equal

The right support force $R$ acts $3.0 \text{ m}$ from the left: $R \times 3.0 = 180 + 300$ .

step 4: Solve

$3.0R = 480$ , so $R = 160 \text{ N}$ .

Try this

Q1. Define the moment of a force about a point. [2 marks]

Cue. Force multiplied by the perpendicular distance from the pivot to the line of action.

Q2. State the two conditions required for a rigid body to be in equilibrium. [2 marks]

Cue. Zero resultant force and zero resultant moment about any point.

Q3. State what is meant by the centre of mass of an object. [1 mark]

Cue. The point at which the whole weight of the object can be taken to act.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20183 marksA uniform beam of weight

60 \text{ N}

and length

4.0 \text{ m}

is pivoted at its centre. A

30 \text{ N}

weight hangs

1.5 \text{ m}

from the pivot. Calculate how far from the pivot, on the other side, a

45 \text{ N}

weight must hang to balance the beam.

Show worked answer →

The beam is uniform and pivoted at its centre, so its own weight acts at the pivot and produces no moment.

Apply the principle of moments, equating clockwise and anticlockwise moments about the pivot: $30 \times 1.5 = 45 \times d$ .

$45 = 45d$ , so $d = 1.0 \text{ m}$ .

Markers reward a clear statement of the principle of moments, recognising the beam's weight gives no moment at the central pivot, and correct substitution of perpendicular distances.

AQA 20214 marksA uniform horizontal shelf of weight

24 \text{ N}

and length

0.80 \text{ m}

is supported by a hinge at one end and a vertical string at the other. Calculate the tension in the string.

Show worked answer →

Take moments about the hinge to eliminate the unknown hinge force. The shelf is uniform, so its weight $24 \text{ N}$ acts at its centre, $0.40 \text{ m}$ from the hinge.

The string acts vertically at the far end, $0.80 \text{ m}$ from the hinge. For equilibrium, the clockwise moment of the weight equals the anticlockwise moment of the tension: $24 \times 0.40 = T \times 0.80$ .

$9.6 = 0.80T$ , so $T = 12 \text{ N}$ .

Markers reward taking moments about the hinge, the weight acting at the centre at $0.40 \text{ m}$ , and the correct tension.

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Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)