Moments

This lesson covers moments (turning effects of forces) as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to be able to calculate moments about a point and apply the principle of moments to solve equilibrium problems involving rigid bodies.

What is a Moment?

The moment of a force about a point is the turning effect of that force about that point.

Moment = Force x Perpendicular distance from the point to the line of action of the force.

Moment of F about point P = F x d

where d is the perpendicular distance from P to the line of action of F.

Units: Newton-metres (N m).

Sign convention: Clockwise moments are usually taken as positive (or negative — just be consistent). In exam solutions, state your convention clearly.

The Principle of Moments

For a body in equilibrium (not rotating):

Sum of clockwise moments about any point = Sum of anticlockwise moments about the same point

Equivalently: the net moment about any point is zero.

Exam Tip: You can take moments about ANY point. Choose a point where an unknown force acts — this eliminates that force from the equation and makes the algebra simpler.

Worked Example 1: Uniform Rod

A uniform rod AB of length 4 m and mass 10 kg rests horizontally on two supports at points C and D, where AC = 1 m and AD = 3 m. Find the reaction forces at C and D.

The weight of the rod acts at the centre of mass (midpoint), 2 m from A.

Taking moments about C (to eliminate Rc):

Clockwise: Weight = 10g at distance (2 - 1) = 1 m from C.
Anticlockwise: Rd at distance (3 - 1) = 2 m from C.

10g(1) = Rd(2) Rd = 10g/2 = 5g = 49 N

Resolving vertically (equilibrium): Rc + Rd = 10g Rc = 10g - 5g = 5g = 49 N

By symmetry (the centre of mass is equidistant from C and D), the reactions are equal. This makes sense.

Worked Example 2: Non-Uniform Rod

A non-uniform rod AB of length 6 m and mass 8 kg is supported at A and B. Its centre of mass is 2 m from A. Find the reaction at each support.

Taking moments about A (to eliminate Ra): Rb(6) = 8g(2) Rb = 16g/6 = 8g/3 ≈ 26.1 N

Resolving vertically: Ra + Rb = 8g Ra = 8g - 8g/3 = 16g/3 ≈ 52.3 N

Tilting and Limiting Equilibrium

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

Example: A uniform plank AB of length 5 m and mass 20 kg rests on supports at C and D, where AC = 1 m and DB = 1 m. A person of mass m kg stands at B. The plank is about to tilt about D. Find m.

When the plank is about to tilt about D, the reaction at C becomes zero (Rc = 0).

Taking moments about D:

Clockwise: Person at B, distance 1 m from D. Moment = mg(1).
Anticlockwise: Weight of plank at midpoint (2.5 m from A = 0.5 m to the left of D since D is at 4 m from A). Wait — let me recalculate.

D is 1 m from B, so D is at 4 m from A. The centre of mass is at 2.5 m from A, which is 1.5 m to the left of D.

Anticlockwise: 20g(1.5) (plank weight, 1.5 m to the left of D).
Clockwise: mg(1) (person at B, 1 m to the right of D).

For tilting about D: mg(1) = 20g(1.5) m = 30 kg

Moments of Inclined Forces

If a force is not perpendicular to the rod, you must find the perpendicular distance from the pivot to the line of action of the force, or resolve the force into components.

Method 1: Moment = F x d sin θ, where θ is the angle between the force and the line from the pivot to the point of application.

Method 2: Resolve the force into components parallel and perpendicular to the rod. Only the perpendicular component produces a moment.

Suspended Bodies

If a body is suspended from a point and hangs in equilibrium, the centre of mass is directly below the point of suspension.

Summary

Moment = Force x Perpendicular distance from the pivot.
For equilibrium: sum of clockwise moments = sum of anticlockwise moments (about any point).
Choose the pivot point wisely to eliminate unknown forces.
For tilting problems, set the reaction at the support about to leave the surface to zero.
The weight of a uniform body acts at its geometric centre (midpoint for a rod).

A-Level Deep Dive: Moments

Spec mapping

Edexcel 9MA0-03 specification, section 11 — Moments covers moments in simple static contexts (refer to the official specification document for exact wording). The sub-strand sits inside Paper 3 (Statistics and Mechanics) and is examined alongside section 8 (Forces and Newton's laws) and section 10 (Equilibrium). Although the topic is conceptually narrow, it draws synoptically on section 7 (vectors in two dimensions) when forces act at an angle, and on section 1 of Paper 3 mechanics — modelling assumptions (rod as light, uniform, or non-uniform). The Edexcel formula booklet provides no moment-specific formulae — students must know that moment = force $\times$ perpendicular distance from pivot.

Worked example with full mark scheme

Question (8 marks): A uniform rod $AB$ of length $4\,\text{m}$ and mass $12\,\text{kg}$ rests horizontally on two supports at $P$ and $Q$ , where $AP = 0.5\,\text{m}$ and $AQ = 3\,\text{m}$ . A particle of mass $5\,\text{kg}$ is placed at $B$ .

(a) Find the magnitudes of the reaction forces $R_P$ and $R_Q$ at the two supports. Take $g = 9.8\,\text{m s}^{-2}$ . (8)

Solution with mark scheme:

Step 1 — draw a diagram and identify forces.

Forces acting on the rod: weight $12g$ at the midpoint ( $2\,\text{m}$ from $A$ , since the rod is uniform); particle weight $5g$ at $B$ ( $4\,\text{m}$ from $A$ ); reactions $R_P$ at $0.5\,\text{m}$ and $R_Q$ at $3\,\text{m}$ .

B1 — correct identification of all four forces with positions, including the centre-of-mass at $2\,\text{m}$ from $A$ (this requires the modelling assumption "uniform rod").

Step 2 — take moments about $P$ .

Choosing $P$ as pivot eliminates $R_P$ from the equation. Anti-clockwise moments about $P$ from $R_Q$ ; clockwise moments from the two weights (both lie to the right of $P$ ).

$R_Q \times (3 - 0.5) = 12g \times (2 - 0.5) + 5g \times (4 - 0.5)$

M1 — taking moments about a sensible pivot, with the correct perpendicular distances measured from that pivot.

A1 — correct equation: $2.5 R_Q = 12g \times 1.5 + 5g \times 3.5 = 18g + 17.5g = 35.5g$ .

Step 3 — solve for $R_Q$ .

$R_Q = \dfrac{35.5g}{2.5} = 14.2g = 14.2 \times 9.8 = 139.16\,\text{N}$

A1 — $R_Q \approx 139\,\text{N}$ (3 s.f.).

Step 4 — apply vertical force balance.

Total weight downward = $(12 + 5)g = 17g$ . So $R_P + R_Q = 17g$ .

M1 — resolving vertically and equating to total weight.

Step 5 — solve for $R_P$ .

$R_P = 17g - 14.2g = 2.8g = 2.8 \times 9.8 = 27.44\,\text{N}$

A1 — $R_P \approx 27.4\,\text{N}$ (3 s.f.).

B1 — final answers given to a sensible degree of accuracy (3 s.f.) with units. Many candidates lose this presentation mark by leaving answers as $2.8g$ without evaluating, or by writing absurd precision such as $139.1600000\,\text{N}$ .

Total: 8 marks (B1 M1 A1 A1 M1 A1 B1, with one A1 absorbed into B1 above).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A non-uniform rod $CD$ of length $5\,\text{m}$ and mass $8\,\text{kg}$ rests in equilibrium on a single support at $M$ , where $CM = 2\,\text{m}$ . A particle of mass $m\,\text{kg}$ is placed at $D$ and the rod remains in equilibrium when the centre of mass of the rod is at distance $d$ from $C$ .

(a) Show that $8(2 - d) = 3m$ . (4)

(b) Given that $d = 1.5\,\text{m}$ , find the value of $m$ . (2)

Mark scheme decomposition by AO:

(a)

B1 (AO3.3) — recognising that for equilibrium on a single support, the support must lie directly under the combined centre of mass, i.e. taking moments about $M$ gives zero net moment.
M1 (AO1.1b) — taking moments about $M$ : clockwise contribution from $mg$ at $D$ (distance $5 - 2 = 3\,\text{m}$ ); anti-clockwise from $8g$ at distance $2 - d$ from $M$ (assuming $d < 2$ ).
M1 (AO2.1) — equating moments: $8g(2 - d) = mg \times 3$ .
A1 (AO2.5) — cancelling $g$ to obtain $8(2 - d) = 3m$ .

(b)

M1 (AO1.1a) — substituting $d = 1.5$ : $8(0.5) = 3m \implies 4 = 3m$ .
A1 (AO1.1b) — solving: $m = \tfrac{4}{3} \approx 1.33\,\text{kg}$ .

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. The AO3 mark is the modelling step — recognising that single-support equilibrium forces the support to coincide with the combined centre of mass.

Synoptic links

Connects to:

Section 8 — Forces and Newton's laws: every moments problem first requires force identification. Weights act at centres of mass; tensions along strings; normal reactions perpendicular to surfaces. Misidentifying the line of action of a force (especially tension in a non-vertical string) propagates straight through to a wrong moment equation.
Section 10 — Statics and equilibrium: moments are one of the two equilibrium conditions. The full pair — $\sum \mathbf{F} = \mathbf{0}$ and $\sum M = 0$ about any point — must hold simultaneously. A common A* discriminator is recognising when force balance alone is insufficient (e.g. a rod on two supports has two unknowns; one equation cannot solve it).
Section 7 — Vectors in two dimensions: when forces act at an angle, the perpendicular component must be extracted before computing the moment. Equivalently, moment $= \mathbf{r} \times \mathbf{F}$ in vector form, with magnitude $|\mathbf{r}||\mathbf{F}|\sin\theta$ .
Modelling assumptions (cross-cutting): "light rod" means weight is negligible (no weight term in the moment equation); "uniform rod" places the centre of mass at the geometric midpoint; "non-uniform rod" requires the centre of mass to be given or deduced. Stating the assumption used is itself an AO3 mark on harder questions.
Engineering structures (applied modelling): ladders against walls, hinged beams, see-saws, and crane jibs are all rod-and-pivot problems. The same equilibrium conditions extend to trusses and frames in undergraduate statics.

Mark-scheme literacy

Moments questions on 9MA0-03 split AO marks more evenly than pure-procedural topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Computing moments correctly, applying force balance, solving the resulting linear system
AO2 (reasoning / interpretation)	25–35%	Choosing a sensible pivot, justifying sign convention, interpreting "rod remains in equilibrium" as a constraint
AO3 (problem-solving / modelling)	15–25%	Stating modelling assumptions, recognising when extra information is needed, translating "on the point of tipping" into a reaction-force-equals-zero condition

Examiner-rewarded phrasing: "taking moments about $P$ (anti-clockwise positive)"; "since the rod is uniform, the weight acts at the midpoint"; "on the point of tipping about $Q$ , the reaction at $P$ is zero". Phrases that lose marks: omitting the pivot when writing a moment equation; using slant rather than perpendicular distance for an angled force; assuming a uniform rod when the question specifies "non-uniform".

A specific Edexcel pattern: questions phrased "find the greatest mass that can be placed at $B$ without the rod tipping" require recognising that the limiting case is when the reaction at the other support becomes zero. This is the AO3 step that converts an inequality into an equality.

Grade-band model answers

3-mark question

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

Grade C response (~150 words):

The moment is force times distance, so $30 \times 2 = 60\,\text{N m}$ clockwise about $A$ .

Moments

Moments

What is a Moment?

The Principle of Moments

Worked Example 1: Uniform Rod

Worked Example 2: Non-Uniform Rod

Tilting and Limiting Equilibrium

Moments of Inclined Forces

Suspended Bodies

Summary

A-Level Deep Dive: Moments

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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