AQA AQA 2019-style practice: Paper 2, Section B. A uniform rod AB of length 4 m and weight 80 N rests horizontally on two supports, one at A and one at C, where C is 3 m from A. A load of 50 N hangs from B. Take the rod as uniform. (a) Find the reaction at the support C. (b) Find the reaction at the support A.

The rod is uniform, so its 80 N weight acts at the midpoint, 2 m from A. The load of 50 N acts at B, 4 m from A. Let the reactions be R_A at A and R_C at C. Take moments about A to eliminate R_A: R_C × 3 = 80 × 2 + 50 × 4 = 160 + 200 = 360, so R_C = 120 N. Resolve vertically: R_A + R_C = 80 + 50 = 130, so R_A = 130 - 120 = 10 N. Markers reward placing the weight at the midpoint, taking moments about a support to remove one reaction, and resolving vertically for the other.

EnglandMathsSyllabus dot point

How do you decide whether a rigid object will balance or turn, and where its supports must act?

The moment of a force about a point, the principle of moments, equilibrium of a rigid body under coplanar forces, reactions at supports, and modelling uniform and non-uniform rods.

A focused answer to the AQA A-Level Mathematics moments content, covering the moment of a force, the principle of moments, equilibrium of a rigid body, reactions at supports, and modelling uniform and non-uniform rods.

Generated by Claude Opus 4.810 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
Equilibrium of a rigid body
Uniform and non-uniform rods
A reliable method for equilibrium problems

What this dot point is asking

AQA wants you to calculate the moment of a force about a point, apply the principle of moments, analyse the equilibrium of a rigid body under coplanar forces, find reactions at supports, and model uniform and non-uniform rods. Moments extend the forces topic from particles to rigid bodies, where the position at which a force acts now matters.

The moment of a force

The word "perpendicular" is doing real work here: it is the perpendicular distance from the pivot to the line along which the force acts. If a force is applied at an angle, either use the perpendicular component of the force or the perpendicular distance to its line of action; both give the same moment.

Equilibrium of a rigid body

Because the moment condition holds about any point, you choose the point cleverly. Taking moments about a point where an unknown force acts removes that force from the equation (its moment is zero there), leaving one equation in fewer unknowns.

Reactions on a uniform rod with a hanging load

A uniform rod $AB$ of length $6$ m and weight $40$ N rests on supports at $A$ and at $D$ , where $D$ is $4$ m from $A$ . A weight of $20$ N hangs at $B$ . Find both support reactions.

Step 1: Place the forces

The rod is uniform, so its $40$ N weight acts at the midpoint, $3$ m from $A$ . The $20$ N load acts at $B$ , $6$ m from $A$ . Reactions $R_A$ at $A$ and $R_D$ at $D$ act upward.

Step 2: Take moments about $A$ to remove $R_A$

$R_D \times 4 = 40 \times 3 + 20 \times 6 = 120 + 120 = 240$ .

Step 3: Solve for $R_D$

$R_D = \dfrac{240}{4} = 60$ newtons.

Step 4: Resolve vertically for $R_A$

$R_A + R_D = 40 + 20 = 60$ , so $R_A = 60 - 60 = 0$ newtons. The rod is on the point of tipping about $D$ .

Uniform and non-uniform rods

A uniform rod has evenly distributed mass, so its weight acts at its geometric centre (the midpoint). A non-uniform rod has its centre of mass at a stated point, often found by taking moments; you place the whole weight there. If a rod is described as light, its weight is negligible and you ignore it. A common exam twist asks for the point at which a rod is about to tip, which is when one support reaction falls to zero.

A reliable method for equilibrium problems

Rigid-body equilibrium problems follow a fixed routine. First, draw a diagram marking every force with its position: the weight at the centre of mass, any loads at their points, and the unknown reactions at supports or hinges. Second, write the vertical equilibrium equation, total upward force equals total downward force. Third, take moments about a well-chosen point, ideally where an unknown reaction acts, so it contributes no moment and drops out, leaving one equation in one unknown. Fourth, solve for that reaction, then substitute back into the vertical equation for the remaining unknown.

Choosing the pivot cleverly is the single biggest time-saver. If two unknown reactions act at the two ends of a beam, taking moments about one end removes that reaction entirely, giving the other directly. The tipping condition is then easy to spot: the beam is on the point of tipping about a support exactly when the reaction at the other support reaches zero, so set that reaction to zero and solve for the critical load position.

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 20197 marksPaper 2, Section B. A uniform rod

AB

of length

4

m and weight

80

N rests horizontally on two supports, one at

A

and one at

C

, where

C

3

m from

A

. A load of

50

N hangs from

B

. Take the rod as uniform. (a) Find the reaction at the support

C

. (b) Find the reaction at the support

A

Show worked answer →

The rod is uniform, so its $80$ N weight acts at the midpoint, $2$ m from $A$ . The load of $50$ N acts at $B$ , $4$ m from $A$ . Let the reactions be $R_A$ at $A$ and $R_C$ at $C$ . Take moments about $A$ to eliminate $R_A$ : $R_C \times 3 = 80 \times 2 + 50 \times 4 = 160 + 200 = 360$ , so $R_C = 120$ N. Resolve vertically: $R_A + R_C = 80 + 50 = 130$ , so $R_A = 130 - 120 = 10$ N. Markers reward placing the weight at the midpoint, taking moments about a support to remove one reaction, and resolving vertically for the other.

AQA 20216 marksPaper 2, Section B. A non-uniform plank

PQ

of length

5

m and weight

W

rests on supports at

P

and

Q

. The centre of mass is

2

m from

P

. The reaction at

P

is found to be

90

N. (a) By taking moments about

Q

, form an equation in

W

. (b) Hence find

W

and the reaction at

Q

Show worked answer →

The weight $W$ acts $2$ m from $P$ , hence $3$ m from $Q$ . Take moments about $Q$ to eliminate the reaction at $Q$ : $R_P \times 5 = W \times 3$ , so $90 \times 5 = 3W$ , giving $450 = 3W$ and $W = 150$ N. Resolve vertically: $R_P + R_Q = W$ , so $R_Q = 150 - 90 = 60$ N. Markers reward correctly locating the centre of mass relative to both supports, taking moments about $Q$ , and resolving for the second reaction.

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Sources & how we know this

AQA A-level Mathematics (7357) specification — AQA (2017)