Moments

This lesson covers moments (turning effects of forces) at A-Level. A moment quantifies the tendency of a force to cause rotation about a point or axis. Understanding moments is essential for solving problems involving rigid bodies in equilibrium.

Definition of a Moment

The moment of a force about a point is:

$\text{Moment} = F \times d$

where:

$F$ is the magnitude of the force (in newtons)
$d$ is the perpendicular distance from the line of action of the force to the pivot point (in metres)

The unit of a moment is newton-metres (Nm).

Direction: Moments are either clockwise or anticlockwise about the pivot.

Exam Tip: The distance $d$ must be the perpendicular distance from the pivot to the line of action of the force. If the force is not perpendicular to the line from the pivot to the point of application, you must resolve the force or find the correct perpendicular distance.

The Principle of Moments

For a body in rotational equilibrium (not rotating or rotating at constant angular velocity):

$\text{Sum of clockwise moments} = \text{Sum of anticlockwise moments}$

Equivalently, the net moment about any point is zero.

Equilibrium Conditions

For a rigid body to be in complete equilibrium:

The resultant force is zero (translational equilibrium): $\sum F = 0$
The resultant moment about any point is zero (rotational equilibrium): $\sum M = 0$

Both conditions must be satisfied simultaneously.

Uniform Beams and Rods

A uniform beam has its weight concentrated at its centre of mass, which is at the midpoint of the beam.

Example: A uniform beam of length 6 m and weight 200 N rests horizontally on two supports at points A (left end) and B (right end). A 300 N load is placed 2 m from A. Find the reaction forces at A and B.

Taking moments about A (eliminates $R_A$ ):

$R_B \times 6 = 200 \times 3 + 300 \times 2$ $6R_B = 600 + 600 = 1200$ $R_B = 200 \text{ N}$

Resolving vertically: $R_A + R_B = 200 + 300 = 500$ $R_A = 500 - 200 = 300 \text{ N}$

Tilting Problems

A beam is on the point of tilting about a support when the reaction at the other support becomes zero.

Example: A uniform plank of length 8 m and mass 40 kg rests on two supports at 1 m from each end. A child of mass $m$ stands at one end. The plank is about to tilt. Find $m$ .

When tilting about the closer support, the reaction at the far support = 0. Take moments about the tilting support to find $m$ .

Non-Uniform Beams

For a non-uniform beam, the centre of mass is not at the midpoint. Its position is usually given in the question, and you treat the weight as acting at that point.

Moments with Angled Forces

If a force acts at angle $\theta$ to the beam, resolve the force into components:

Component perpendicular to the beam: $F\sin\theta$ — this creates a moment.
Component along the beam: $F\cos\theta$ — this does not create a moment about the pivot (zero perpendicular distance).

$\text{Moment} = F\sin\theta \times d$

where $d$ is the distance along the beam from the pivot to the point of application.

Summary

The moment of a force is $F \times d$ (force times perpendicular distance).
Principle of Moments: clockwise moments = anticlockwise moments in equilibrium.
For equilibrium: resultant force = 0 AND resultant moment = 0.
Take moments about a point where an unknown force acts to eliminate it from the equation.
Uniform beams have weight at the midpoint; non-uniform beams have weight at the specified centre of mass.
For tilting problems, the reaction at the support about to lose contact is zero.

Exam Tip: When choosing which point to take moments about, pick a point where an unknown force acts. This eliminates that unknown from the moment equation, simplifying your calculation.

A-Level Deep Dive: Moments

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section V (Year 2 content) covers moments in simple static contexts (refer to the official specification document for exact wording). The specification statement is brief but the examined content is wide: candidates must compute moments about a chosen pivot, apply the rotational equilibrium condition $\sum M = 0$ alongside the translational condition $\sum F = 0$ , model rods as either light (massless) or uniform (centre of mass at the geometric centre), handle reactions at supports and hinges, and resolve forces that are not perpendicular to a beam. Moments problems appear most often in Paper 3 mixed-mechanics questions and synoptically with vectors (Section IX), forces and equilibrium (Section IV) and kinematics where a body is initially at rest. The AQA formula booklet does not provide the moment definition — students must remember moment = force $\times$ perpendicular distance from the pivot.

Worked example with full mark scheme

Question (8 marks):

A uniform rod $AB$ of length $4\,\text{m}$ and weight $60\,\text{N}$ rests horizontally on two supports. The first support is at point $P$ , located $0.5\,\text{m}$ from end $A$ . The second support is at point $Q$ , located $1\,\text{m}$ from end $B$ . A particle of weight $40\,\text{N}$ is attached to the rod at end $B$ .

(a) Find the magnitudes of the reactions $R_P$ and $R_Q$ at the two supports. (6)

(b) The particle at $B$ is replaced by one of weight $W\,\text{N}$ . Find the largest value of $W$ for which the rod remains in equilibrium. (2)

Solution with mark scheme:

(a) Step 1 — set up a clear diagram and identify forces.

Place $A$ at the origin, with the rod along the positive $x$ -axis so that $B$ is at $x = 4$ . Then $P$ is at $x = 0.5$ , $Q$ is at $x = 4 - 1 = 3$ , and the centre of mass of the uniform rod is at $x = 2$ (the midpoint). The forces acting are: the weight $60\,\text{N}$ downward at $x = 2$ ; the particle weight $40\,\text{N}$ downward at $x = 4$ ; the reaction $R_P$ upward at $x = 0.5$ ; the reaction $R_Q$ upward at $x = 3$ .

B1 — correct positions for centre of mass ( $x = 2$ from $A$ ) and supports ( $x = 0.5$ , $x = 3$ ). Many candidates lose this mark by placing the centre of mass at the midpoint between the supports rather than at the geometric centre of the rod.

Step 2 — apply the rotational equilibrium condition.

Take moments about $P$ (at $x = 0.5$ ). Using the convention "anticlockwise positive":

$\sum M_P = R_Q \cdot (3 - 0.5) - 60 \cdot (2 - 0.5) - 40 \cdot (4 - 0.5) = 0$

$2.5 R_Q - 60 \cdot 1.5 - 40 \cdot 3.5 = 0$

$2.5 R_Q = 90 + 140 = 230$

$R_Q = 92\,\text{N}$

M1 — taking moments about a single point with the correct perpendicular distances. Choosing $P$ as the pivot is strategic: the unknown $R_P$ has zero moment arm about itself, so it disappears from the equation, leaving a single unknown $R_Q$ .

A1 — correct equation in $R_Q$ with the right signs and distances.

A1 — correct value $R_Q = 92\,\text{N}$ .

Step 3 — apply the translational equilibrium condition.

Vertically: $\sum F_y = R_P + R_Q - 60 - 40 = 0$ , so $R_P = 100 - R_Q = 100 - 92 = 8\,\text{N}$ .

M1 — using $\sum F = 0$ to find the second reaction.

A1 — correct value $R_P = 8\,\text{N}$ .

Total: 6 marks (B1 M1 A1 A1 M1 A1).

(b) The rod tips when the reaction at $P$ falls to zero (the rod is on the verge of rotating about $Q$ ). Setting $R_P = 0$ and taking moments about $Q$ (at $x = 3$ ):

$60 \cdot (3 - 2) - W \cdot (4 - 3) = 0$

$60 - W = 0 \implies W = 60$

M1 — correct identification that tipping corresponds to $R_P = 0$ and moments about $Q$ .

A1 — $W = 60\,\text{N}$ .

Total: 2 marks.

Question total: 8 marks (M4 A4).

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A non-uniform rod $CD$ of length $6\,\text{m}$ and weight $80\,\text{N}$ rests in equilibrium on two supports placed at $C$ and at the point $E$ which is $1.5\,\text{m}$ from $D$ . The reactions at $C$ and $E$ are $30\,\text{N}$ and $50\,\text{N}$ respectively. Find the distance of the centre of mass of the rod from $C$ .

Mark scheme decomposition by AO:

B1 (AO3.1b) — modelling the rod as a rigid body with a single point of application for its weight (the centre of mass). For a non-uniform rod the centre of mass is not at the geometric centre; the question is asking the student to find where it actually is.
M1 (AO1.1a) — applying $\sum F = 0$ vertically: $R_C + R_E - W = 30 + 50 - 80 = 0$ , confirming the system is consistent with equilibrium.
M1 (AO1.1b) — taking moments about $C$ : let $\bar{x}$ be the distance from $C$ to the centre of mass. Then $80 \bar{x} = 50 \cdot (6 - 1.5) = 50 \cdot 4.5 = 225$ .
A1 (AO1.1b) — solving: $\bar{x} = 225/80 = 2.8125\,\text{m}$ .
A1 (AO2.5) — stating the answer in a sensible form: $\bar{x} = 2.8125\,\text{m}$ or equivalently $2\tfrac{13}{16}\,\text{m}$ from $C$ .
A1 (AO3.2a) — interpreting the result: the centre of mass lies between the geometric centre (at $3\,\text{m}$ ) and the support $C$ , consistent with a rod that is heavier toward the $C$ end.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. The AO3 marks reward modelling judgement (recognising what "non-uniform" means) and physical interpretation, not new calculation.

Synoptic links

Connects to:

Section IV — Forces and Newton's laws (equilibrium): moments problems are an extension of $\sum F = 0$ to rigid bodies. For a particle, translational equilibrium is sufficient. For a rigid body that can rotate, both $\sum F = 0$ and $\sum M = 0$ are required. Students who jump straight to moments without first checking translational equilibrium often miss reactions entirely.
Section IX — Vectors: the perpendicular-distance interpretation of a moment is the magnitude of the cross product $\vec{r} \times \vec{F}$ . Although A-Level Mechanics does not require the cross product explicitly, the geometric intuition — that only the component of $\vec{F}$ perpendicular to $\vec{r}$ contributes — is the same.
Modelling assumptions: the contrast between "light rod" (no weight, no contribution to moments from the rod itself) and "uniform rod" (weight acts at the geometric centre) is a recurring modelling decision. Examiners expect candidates to read the question carefully and use the correct assumption — using "uniform" when the question says "light" introduces a phantom weight and corrupts every moment equation.
Statics of structures: ladders against rough walls, hinged beams supporting loads, and step-ladders with tie-ropes all reduce to the same three equations ( $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ) applied with carefully chosen pivots.
Engineering applications: the principles of moment equilibrium underlie the analysis of cantilevers, simply supported beams, trusses and cranes — the foundations of statics taught in first-year civil and mechanical engineering courses.

Mark-scheme literacy

Moments questions on AQA 7357 Paper 3 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Setting up moment equations correctly, applying $\sum F = 0$ and $\sum M = 0$ , solving simultaneous equations for reactions
AO2 (reasoning / interpretation)	15–25%	Justifying choice of pivot, explaining sign conventions, interpreting reactions physically (e.g. "negative $R$ means the support is pulling down, so the rod must be hinged not resting")
AO3 (problem-solving / modelling)	20–30%	Identifying when the rod is on the verge of tipping ( $R = 0$ at one support), choosing modelling assumptions, recognising when friction must be invoked (ladder problems)

Examiner-rewarded phrasing: "taking moments about $P$ , anticlockwise positive"; "since the rod is uniform, its weight acts at the midpoint"; "at the point of tipping, the reaction at $Q$ is zero". Phrases that lose marks: omitting "perpendicular distance" when forces are inclined; writing "moment = force $\times$ distance" without specifying the geometry; failing to state the chosen pivot; mixing units (using $\text{cm}$ in one moment and $\text{m}$ in the next within the same equation).

A specific AQA pattern to watch: when a question asks to "show that the rod is in equilibrium" with given reactions, candidates must verify both $\sum F = 0$ and $\sum M = 0$ — a single condition is not enough. Examiners reward explicit verification of both conditions.

Grade-band model answers

3-mark question

Question: A light rod $AB$ of length $2\,\text{m}$ is freely pivoted at $A$ . A force of $30\,\text{N}$ is applied perpendicular to the rod at $B$ . Find the moment of this force about $A$ .

Grade C response (~140 words):

The moment is force $\times$ perpendicular distance from the pivot. The force is $30\,\text{N}$ and the perpendicular distance from $A$ to the line of action is the length of the rod, $2\,\text{m}$ (since the force is applied perpendicular to the rod at $B$ ).

So moment $= 30 \times 2 = 60\,\text{Nm}$ .

Examiner commentary: Full marks (3/3). The candidate states the definition, identifies the perpendicular distance correctly, and gives the numerical answer with units. The phrase "since the force is applied perpendicular to the rod" shows they have understood why the rod's full length is the moment arm. A weaker candidate might assume the answer is "force $\times$ length" without thinking about why, and be caught out by a slight rotation of the question on a later paper.

Grade A response (~180 words):*

Moments

Moments

Definition of a Moment

The Principle of Moments

Equilibrium Conditions

Uniform Beams and Rods

Tilting Problems

Non-Uniform Beams

Moments with Angled Forces

Summary

A-Level Deep Dive: Moments

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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