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AQA A-Level Mathematics: S: Moments (7357) – Revision Notes

Condensed notes on moments for AQA A-level Maths (7357) Section S: rigid-body equilibrium, tilting, non-uniform rods, hinges and ladders, with a self-test.

Subject
Mathematics
Level
A LEVELS
Topic
S: Moments
Updated

Aligned to AQA A Level Mathematics (7357), For first teaching 2017. Official specification .

Syllabus page (what it covers and how it is assessed): AQA A Level Mathematics.

Syllabus points this page covers

7357

  • 20 S: Moments (whole topic)

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These notes condense Section S: Moments (S1) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification lists Section S under Paper 2, and a calculator is required in every 7357 paper. For full explanations and six worked examples, use the Moments study guide.

Practise afterwards with the Moments practice questions. Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/. Check your gaps with the free 10-minute diagnostics.

S1 in one line

The specification says: Understand and use moments in simple static contexts. In practice that means turning effects of forces on rods, beams, planks, hinged rods and ladders that are at rest.

Definitions

Term Meaning
Moment of F about O F × perpendicular distance from O to the line of action of F
Unit newton metre, N m
Sense clockwise or anticlockwise; always state it
Resultant moment sum of moments in one sense minus the sum in the other
Line of action the line along which a force acts, extended both ways
Rigid body an object that keeps its shape; distances along it are fixed
Centre of mass the point where the weight of the body acts

Key results

Situation Result
Force at right angles, distance d moment = F d
Force at angle θ to the rod, distance d along rod moment = F d sin θ
Force whose line of action passes through O moment = 0
Rigid body in equilibrium resultant force = 0 and resultant moment about any point = 0
Beam about to tilt about support D reaction at the other support = 0
Weight of a uniform rod acts at the midpoint
Smooth contact reaction perpendicular to the surface, no friction
Rough contact friction F ≤ μR; F = μR only in limiting equilibrium

Appendix B of the specification lists formulae you must recall. For mechanics it includes weight = mass × g and F ≤ μR, both used in almost every moments question.

Modelling words

  • Light: no weight.
  • Uniform: weight at the midpoint.
  • Non-uniform: weight at an unknown point; find it with moments.
  • Particle: a person or load treated as a point.
  • Rigid: does not bend.
  • Smooth / rough: no friction / friction acts.

These come from OT3.5 (understand and use modelling assumptions). The Overarching themes guide covers the modelling cycle.

Related mechanics notes: Forces and Newton’s laws (resolving, friction) and Kinematics.

Method in steps: rigid body in equilibrium

1. Draw a clear diagram. Mark every force: weights, reactions,
   tensions, friction, hinge components.
2. Put each weight at the right point (midpoint if uniform).
3. Take moments about the point where the most unknowns act.
4. Resolve vertically and horizontally for any remaining unknowns.
5. Check with moments about a second point if time allows.

Method in steps: on the point of tilting

1. Identify the support it tilts about (call it D).
2. Set the reaction at the other support to zero.
3. Take moments about D.
4. Solve for the unknown mass, distance or force.

Method in steps: ladders

1. Smooth wall: one horizontal reaction S.
2. Rough ground: normal reaction R (up) and friction F (towards the wall).
3. Resolve vertically: R = total weight.
4. Resolve horizontally: F = S.
5. Moments about the foot: S × (height of top) = weights × (horizontal distances).
6. If about to slip, F = μR; otherwise F ≤ μR.

Small worked reminders

Plank with an overhang load. A uniform plank AB, 5 m long and of mass 12 kg, rests on supports at A and at C, where AC = 4 m. A 6 kg load sits at B. With g = 9.8 m s⁻²:

Moments about A:  R_C × 4 = 12g × 2.5 + 6g × 5 = 60g
R_C = 15g = 147 N
R_A = 18g − 15g = 3g = 29.4 N

Angled force. A 40 N force acts at 50° to a rod, 0.9 m from a hinge:

moment = 40 × 0.9 × sin 50° = 27.6 N m (3 s.f.)

Hinge with a vertical string. A uniform rod AB, 2 m long and of mass 6 kg, is hinged at A and held horizontal by a vertical string at B:

Moments about A:  T × 2 = 6g × 1
T = 3g = 29.4 N
Hinge: V + T = 6g, so V = 3g = 29.4 N upwards, and H = 0

With a vertical string there is no horizontal force at all. An angled string gives the hinge a horizontal component too.

Choosing where to take moments

Unknowns in the problem Take moments about
Two support reactions one support (the other reaction is then found by resolving)
A hinge force and a tension the hinge
Ladder: ground reaction, friction, wall reaction the foot of the ladder
Beam about to tilt the support it tilts about
Position of an unknown centre of mass the end you measure from, or a support

A second moments equation, about a different point, can replace the resolving step. It is also the quickest check on your answers.

Must-know distinctions

  • Particle vs rigid body: a particle needs only resultant force zero. A rigid body also needs resultant moment zero.
  • Distance along the rod vs perpendicular distance: they are equal only when the force is at right angles to the rod.
  • Horizontal vs vertical distance on an inclined rod: a vertical force (weight) uses the horizontal distance, d cos θ, where θ is the rod’s angle to the horizontal. A horizontal force (smooth wall) uses the vertical distance, d sin θ.
  • Uniform vs non-uniform: never assume the midpoint for a non-uniform rod.
  • F = μR vs F ≤ μR: equality only when slipping is about to happen.
  • Tilting vs slipping: tilting means a reaction becomes zero; slipping means friction reaches μR.

Quick self-test

  1. What is the SI unit of a moment?
  2. Find the moment of an 18 N force at a perpendicular distance of 25 cm from a point.
  3. A 50 N force acts 1.2 m along a rod from a pivot, at 30° to the rod. Find its moment about the pivot.
  4. State the two conditions for a rigid body to be in equilibrium.
  5. Where does the weight of a uniform rod of length 1.5 m act?
  6. A child of mass 28 kg sits 1.5 m from the pivot of a seesaw. How far from the pivot, on the other side, must a 35 kg child sit to balance it? Treat the seesaw as uniform and pivoted at its centre.
  7. A light rod AB of length 3 m rests on supports at A and B. A 60 N load hangs 1 m from A. Find both reactions.
  8. A plank is about to tilt about support D. What is the reaction at the other support?
  9. A non-uniform rod AB of length 2 m and weight 40 N has its centre of mass 0.8 m from A. It rests on supports at A and B. Find both reactions.
  10. A ladder rests against a smooth vertical wall. In which direction does the wall push on the ladder?
  11. Why do you usually take moments about a hinge?

Answers

  1. The newton metre, N m.
  2. 18 × 0.25 = 4.5 N m.
  3. 50 × 1.2 × sin 30° = 30 N m.
  4. Resultant force is zero, and resultant moment about any point is zero.
  5. At its midpoint, 0.75 m from either end.
  6. 28g × 1.5 = 35g × x, so x = 1.2 m.
  7. Moments about A: R_B × 3 = 60 × 1, so R_B = 20 N and R_A = 40 N.
  8. Zero: the plank is just losing contact there.
  9. Moments about A: R_B × 2 = 40 × 0.8, so R_B = 16 N and R_A = 24 N.
  10. Horizontally, perpendicular to the wall, away from it.
  11. The hinge force has unknown components; their moments about the hinge are zero, so they drop out.

Where marks are usually lost

  • Multiplying by the distance along the rod when the force is at an angle; you need d sin θ.
  • Swapping sin and cos for the distances on a ladder or an inclined rod.
  • Forgetting the weight of the rod itself when it is not described as light.
  • Placing a non-uniform rod’s weight at the midpoint.
  • Not setting the far reaction to zero in a tilting question, or setting the wrong one to zero.
  • Writing F = μR when the ladder is not stated to be about to slip.
  • Leaving the hinge force as a single vertical force when a string pulls at an angle.
  • Dropping the direction in a “magnitude and direction” answer.
  • Using centimetres in a moment and giving an answer 100 times too large.
  • Rounding a tension early, then using it to find a reaction.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.20 S: Moments (S1), with Section 3.19 R: Forces and Newton’s laws (R4, R6), Section 3.1.3 OT3: Mathematical modelling, and Appendix B: mathematical formulae and identities.

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