A uniform beam of weight 40\ N and length 2.0\ m is pivoted at its centre. A 10\ N weight hangs 0.8\ m from the pivot on one side. How far from the pivot must a 20\ N weight hang on the other side to balance it?

Two parallel forces of 15\ N act in opposite directions, separated by 0.30\ m, forming a couple. Find the torque. [2 marks]

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When is an object in equilibrium, and how do moments and forces balance?

Forces as vectors, the moment of a force and the principle of moments, couples, centre of gravity, and the conditions for the equilibrium of a body.

A CCEA A-Level Physics answer on forces as vectors, the moment of a force and the principle of moments, couples and torque, centre of gravity, and the two conditions a body must meet to be in equilibrium.

Generated by Claude Opus 4.811 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 answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to treat forces as vectors, define and calculate the moment of a force, apply the principle of moments, describe a couple and its torque, locate the centre of gravity, and state and use the two conditions for a body to be in equilibrium. A "take moments" beam calculation appears in almost every paper.

The answer

Moments and the principle of moments

The principle of moments states that when a body is in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment about that point. This lets you find an unknown force or distance on a balanced beam; choosing a pivot through an unknown force removes it from the equation.

Couples and centre of gravity

The centre of gravity is the single point at which the entire weight of a body appears to act. For a uniform body it lies at the geometric centre. An object is stable when its centre of gravity is low and its base is wide, because a small tilt keeps the line of action of the weight inside the base, so the weight provides a restoring moment.

Conditions for equilibrium

Worked example: a hinged sign

Tension in a supporting cable

A uniform horizontal beam of weight $80\ \text{N}$ and length $1.5\ \text{m}$ is hinged at a wall. A cable attached to its far end runs up to the wall at $30^\circ$ above the beam, holding it horizontal. Find the tension in the cable.

step Identify the forces and their moments

Take moments about the hinge to eliminate the hinge force. The weight acts at the centre, $0.75\ \text{m}$ out; the cable's vertical component $T\sin 30^\circ$ acts at the far end, $1.5\ \text{m}$ out.

step Apply the principle of moments

T\sin 30^\circ \times 1.5 = 80 \times 0.75.

step Solve for the tension

T \times 0.5 \times 1.5 = 60 \Rightarrow 0.75\,T = 60 \Rightarrow T = 80\ \text{N}.

The cable must pull with

80\ \text{N}

to hold the sign horizontal.

Examples in context

Example 1. A tower crane. A counterweight is placed on the short arm so its moment about the central mast balances the moment of the load on the long arm. As the load moves out along the jib, the operator adjusts the counterweight position so the net moment stays near zero, keeping the crane in equilibrium and preventing it toppling.

Example 2. A tightrope walker's pole. A long, heavy pole lowers the walker's combined centre of gravity and increases the moment of inertia, so any tilt produces a slow, correctable rotation. Holding the pole low keeps the centre of gravity over the wire, the same stability principle as a wide, low base.

Try this

Q1. A uniform beam of weight $40\ \text{N}$ and length $2.0\ \text{m}$ is pivoted at its centre. A $10\ \text{N}$ weight hangs $0.8\ \text{m}$ from the pivot on one side. How far from the pivot must a $20\ \text{N}$ weight hang on the other side to balance it? [2 marks]

Cue. $20 \times d = 10 \times 0.8$ , so $d = 0.4\ \text{m}$ .

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

Cue. Resultant force is zero, and resultant moment about any point is zero.

Q3. Two parallel forces of $15\ \text{N}$ act in opposite directions, separated by $0.30\ \text{m}$ , forming a couple. Find the torque. [2 marks]

Cue. Torque $= F \times d = 15 \times 0.30 = 4.5\ \text{N m}$ .

Exam-style practice questions

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

CCEA 20206 marksA uniform horizontal beam of weight 120 N and length 4.0 m rests on two supports, one at the left end and one 1.0 m from the right end. A load of 200 N hangs 0.50 m from the right end. By taking moments, calculate the upward force provided by each support.

Show worked answer →

The beam's weight acts at its centre, 2.0 m from the left end. The right support is at 3.0 m from the left end. Let the left support force be $R_L$ and the right support force be $R_R$ .

Take moments about the left support (so $R_L$ has no moment):

Clockwise: weight at 2.0 m gives $120 \times 2.0 = 240$ ; the load at $(4.0 - 0.50) = 3.5$ m gives $200 \times 3.5 = 700$ .

Anticlockwise: $R_R \times 3.0$ .

$R_R \times 3.0 = 240 + 700 = 940$ , so $R_R = 313$ N.

Vertical balance: $R_L + R_R = 120 + 200 = 320$ N, so $R_L = 320 - 313 = 7$ N.

Markers reward taking moments about a sensible point, both moment contributions, and using vertical equilibrium to find the second force.

CCEA 20184 marksDefine the moment of a force and explain what is meant by a couple. State the two conditions necessary for a rigid body to be in equilibrium, and explain why a wide, low object is more stable than a tall, narrow one.

Show worked answer →

The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force, measured in newton metres.

A couple is a pair of equal, opposite, parallel forces whose lines of action do not coincide; it produces a turning effect (torque) but no resultant force.

For equilibrium: (1) the resultant force in any direction must be zero, and (2) the resultant moment about any point must be zero.

A wide, low object has its centre of gravity low and a large base. When tilted, the line of action of the weight stays inside the base over a larger angle before it falls outside and topples, so it is more stable.

Markers reward the perpendicular-distance definition, the couple description, both equilibrium conditions, and the centre-of-gravity-over-base stability argument.

Related dot points

Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)