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Moments practice problems Edexcel IAL Mechanics 1, Chapter 6

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 structured exam question

Moments of forces, uniform and non-uniform rods and beams in equilibrium on supports or strings, and rods on the point of tilting. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Exam question practice
  5. Common mistakes
  6. Keep going

Key equations for moments

moment = F × dd is the perpendicular distance from the pivot (N m)
total clockwise = total anticlockwiseabout any point, in equilibrium
upward forces = downward forcesresolve vertically as well
uniform rodweight acts at the midpoint

Take g = 9.8 m s−2, as the WME01 paper asks, and give answers to 2 or 3 significant figures. Answers within 2.5% are marked correct, so rounding along the way is fine. Revise the methods in our A level Maths mechanics guide.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
A force of 15 N acts at right angles to a rod, 0.4 m from a pivot. Find the moment of the force about the pivot.
N m

Hint

Moment = force × perpendicular distance.

Worked solution
  1. Moment = 15 × 0.4 = 6 N m.
Q2Warm-up
A rod is described as uniform. Where does its weight act?

Hint

Uniform means the mass is spread evenly.

Worked solution
  1. The centre of mass of a uniform rod is at its midpoint, so its weight acts there: B.
Q3Warm-up
A child of mass 30 kg sits 1.5 m from the pivot of a see-saw. How far from the pivot, on the other side, must a child of mass 45 kg sit to balance it?
m

Hint

Take moments about the pivot. g cancels.

Worked solution
  1. 30g × 1.5 = 45g × x.
  2. x = 45/45 = 1 m.

Exam-style questions

Questions at the level of the Edexcel IAL Mechanics 1 exam. The ones marked Challenge combine more than one idea.

Q4Exam-style
A uniform beam AB, of length 6 m and mass 20 kg, rests horizontally on supports at A and B. A man of mass 30 kg stands on the beam 2 m from A. Find the reaction at A.
N

Hint

Take moments about B so that the reaction at B does not appear.

Worked solution
  1. Moments about B: RA × 6 = 20g × 3 + 30g × 4.
  2. RA = 180g/6 = 30g.
  3. RA = 294 N.
Q5Exam-style
For the same beam, find the reaction at B.
N

Hint

Resolve vertically.

Worked solution
  1. RA + RB = 50g, so RB = 50g − 30g = 20g.
  2. RB = 196 N.
Q6Exam-style
A plank rests on two supports, C and D. A person walks along it past D until the plank is on the point of tilting about D. What is the reaction at C at that moment?

Hint

On the point of tilting, the plank is about to lift off one support.

Worked solution
  1. About to tilt about D means the plank is about to leave C.
  2. So the reaction at C is zero: A.
Q7Exam-style
A uniform plank AB, of length 5 m and mass 40 kg, rests on supports at C and D, where AC = 1 m and DB = 1.5 m. A boy of mass 60 kg walks from D towards B. How far beyond D can he go before the plank tilts?
m

Hint

The plank's weight acts 2.5 m from A, which is 1 m from D. Set the reaction at C to zero.

Worked solution
  1. D is 3.5 m from A, and the centre of the plank is 2.5 m from A, so 1 m on the other side of D.
  2. On the point of tilting, RC = 0. Moments about D: 40g × 1 = 60g × x.
  3. x = 40/60 = 0.667 m (3 s.f.), still on the plank since DB = 1.5 m.
Q8ChallengeNon-uniform rod
A non-uniform rod AB, of length 4 m and mass 12 kg, rests horizontally on supports at A and B. The reaction at A is twice the reaction at B. Find the distance of the rod's centre of mass from A.
m

Hint

Resolve vertically first to find each reaction, then take moments about A.

Worked solution
  1. RA + RB = 12g with RA = 2RB, so RB = 4g.
  2. Moments about A: 12g × x = 4g × 4.
  3. x = 16/12 = 1.33 m (3 s.f.).

Exam question practice

One original structured question in the style of the WME01 paper. Answer it in full, with working, before you open the mark scheme: M marks are for a correct method, A marks for accurate answers that follow from it, and B marks for a correct statement on its own.

Structured exam question style · 6 points · about 10 minutes

A uniform rod AB has length 4 m and weight 80 N. It hangs horizontally from two vertical light strings attached at C and D, where AC = 0.5 m and DB = 1 m. A particle of weight 30 N is attached to the rod at B.

  1. (a) Find the tension in the string at D.
  2. (b) Find the tension in the string at C.
  3. (c) Explain how you have used the fact that the rod is uniform.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Moments about C: TD × 2.5 = 80 × 1.5 + 30 × 3.5 = 225, so TD = 90 N.

Part (b)

Resolving vertically: TC + 90 = 110, so TC = 20 N.

Part (c)

Its weight acts at the midpoint, 2 m from A.

Your mark: 0 of 6

Common mistakes in moments

Keep going

Revise the ideas behind these questions:

Then try real exam questions: Edexcel IAL Mechanics 1 past papers, with official scoring guidelines. · All Edexcel IAL Mechanics 1 units