OCR OCR 2021-style practice: A non-uniform plank PQ of length 6 m and weight 200 N rests on supports at P and Q. The reaction at P is 120 N. Find the reaction at Q and the distance of the centre of mass from P. Take g = 9.8 m s^-2.

Resolve vertically: R_P + R_Q = 200 (M1), so R_Q = 200 - 120 = 80 N (A1). Let the centre of mass be a distance d from P. Take moments about P (M1): the weight gives 200d and R_Q gives 80 × 6 = 480 N m (A1). So 200d = 480, giving d = 2.4 m (M1, A1). Markers reward resolving for R_Q, taking moments about P with the weight at the unknown centre of mass, and solving for the distance.

EnglandMathsSyllabus dot point

How do you analyse the turning effect of forces and the conditions for a rigid body such as a beam to balance?

The moment of a force about a point, the principle of moments, the equilibrium of a rigid body, and problems involving uniform and non-uniform rods, beams and reactions at supports.

A focused answer to the OCR A-Level Mathematics A moments content, covering the moment of a force about a point, the principle of moments, the equilibrium of a rigid body, and problems involving uniform and non-uniform rods, beams, reactions at supports, and tilting.

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

OCR wants you to find the moment of a force about a point, apply the principle of moments for a body in equilibrium, combine moments with the resolution of forces to find unknown reactions, and solve problems with uniform rods (weight at the centre) and non-uniform rods (weight at an unknown centre of mass), including when a beam is on the point of tilting about a support.

The answer

The moment of a force

The moment of a force measures its turning effect about a point. It depends on the force and the perpendicular distance from the point to the line of the force.

A moment has a sense: clockwise or anticlockwise. Choose one as positive and be consistent.

The principle of moments

For a rigid body in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment, and (separately) the forces balance in every direction.

Uniform and non-uniform rods

A uniform rod has its weight acting at its midpoint. A non-uniform rod has its weight at an unknown centre of mass, which is often what the question asks you to locate using moments.

Examples in context

Finding two reactions

The standard technique is to take moments about one support, which eliminates its (unknown) reaction and gives the other directly, then resolve vertically for the first. Choosing the pivot wisely turns two unknowns into one.

On the point of tilting

A beam on two supports is about to tilt about one support when the reaction at the other support falls to zero. Setting that reaction to zero and taking moments gives the limiting load or position. This is a favourite extension, testing whether you understand what "about to tip" means physically.

Try this

Q1. A force of $20$ N acts perpendicular to a spanner $0.3$ m from the pivot. Find the moment. [1 mark]

Cue. $20 \times 0.3 = 6$ N m.

Q2. A uniform rod of weight $50$ N and length $2$ m is pivoted at its centre. A $30$ N weight hangs $0.4$ m from the pivot. How far on the other side must a $40$ N weight hang to balance? [3 marks]

Cue. The rod's weight acts at the pivot (no moment), so $40d = 30 \times 0.4 = 12$ , giving $d = 0.3$ m.

Exam-style practice questions

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

OCR 20197 marksA uniform rod

AB

of length

5

m and weight

120

N rests horizontally on supports at

A

and at

C

, where

C

4

m from

A

. A particle of weight

40

N is placed at

B

. Find the reactions at the two supports. Take

g = 9.8

m s

^{-2}

Show worked answer →

The rod's weight ( $120$ N) acts at its midpoint, $2.5$ m from $A$ ; the $40$ N particle acts at $B$ , $5$ m from $A$ (B1).

Take moments about $A$ to eliminate $R_A$ (M1): $R_C \times 4 = 120 \times 2.5 + 40 \times 5$ (A1).

So $4R_C = 300 + 200 = 500$ , giving $R_C = 125$ N (A1).

Resolve vertically: $R_A + R_C = 120 + 40 = 160$ (M1), so $R_A = 160 - 125 = 35$ N (A1).

Check by taking moments about $C$ : consistent (A1).

Markers reward placing the weights correctly, taking moments about a support, resolving vertically for the second reaction, and the two values.

OCR 20216 marksA non-uniform plank

PQ

of length

6

m and weight

200

N rests on supports at

P

and

Q

. The reaction at

P

120

N. Find the reaction at

Q

and the distance of the centre of mass from

P

. Take

g = 9.8

m s

^{-2}

Show worked answer →

Resolve vertically: $R_P + R_Q = 200$ (M1), so $R_Q = 200 - 120 = 80$ N (A1).

Let the centre of mass be a distance $d$ from $P$ . Take moments about $P$ (M1): the weight gives $200d$ and $R_Q$ gives $80 \times 6 = 480$ N m (A1).

So $200d = 480$ , giving $d = 2.4$ m (M1, A1).

Markers reward resolving for $R_Q$ , taking moments about $P$ with the weight at the unknown centre of mass, and solving for the distance.

Related dot points

Sources & how we know this

OCR A Level Mathematics A (H240) specification — OCR (2017)