OCR A Level Mathematics A: Moments and statics of rigid bodies (H240) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover moments and statics of rigid bodies in OCR A Level Mathematics A (H240), content references 3.04a to 3.04c of the Cambridge OCR specification, Version 3.1 (August 2026). All of this content is Stage 2, so it is A Level only. Take g = 9.8 m s⁻² throughout. If a method is unfamiliar, start with the study guide or the revision notes. See also the course hub, the checklist and the free diagnostics.

Questions

1. A force of 25 N acts at the end A of a horizontal rod OA, where OA = 0.4 m. The force makes an angle of 50° with the rod.

(a) Find the moment of the force about O. [2] (b) How far is the line of action of the force from O? [1]

2. A rectangular lamina PQRS has PQ = 1.2 m horizontal and QR = 0.7 m, with P bottom left, Q bottom right, R top right and S top left. These forces act in its plane: 8 N vertically downwards at R; 6 N from R towards S; 7 N vertically upwards at Q; 20 N along RP, from R towards P. Find the total moment about P, giving its sense. [4]

3. A light rod AB has length 1.5 m. It rests horizontally on a pivot at C, where AC = 0.5 m. Particles of mass 5 kg and 2 kg are attached at A and at D, where AD = 0.9 m. A particle of mass m kg hangs from B, and the rod stays horizontal.

(a) Find m. [3] (b) Find the magnitude of the force exerted by the pivot. [1]

4. A uniform beam AB has length 4.5 m and mass 30 kg. It is supported horizontally at C and D, with AC = 0.5 m and DB = 0.5 m. A person of mass 70 kg, modelled as a particle, stands 2 m from A. Find the reactions at C and D. [5]

5. A uniform beam AB has length 6 m and is supported horizontally at C and D, with AC = 1.0 m and DB = 1.2 m. When a particle of mass 12 kg is placed at B, the beam is about to tilt.

(a) Find the mass of the beam. [3] (b) The particle is removed. Find the greatest mass of a particle that can be placed at A without the beam tilting. [3]

6. A non-uniform rod AB has length 3 m and mass M kg. Its centre of mass G is x m from A. With a particle of mass 1.5 kg attached at A, the rod rests horizontally on a pivot 1.2 m from A. With a particle of mass 3 kg attached at A instead, it rests horizontally on a pivot 1.0 m from A. Find M and x, assuming that G is further from A than each pivot. [6]

7. A uniform rod AB has length 2.4 m and mass 9 kg. It is hinged at A to a vertical wall and rests in equilibrium with B above A, the rod making 35° with the horizontal. A light horizontal string joins B to the wall.

(a) Find the tension in the string. [3] (b) Find the force exerted on the rod by the hinge, giving its magnitude and its angle to the horizontal. [4]

8. A uniform ladder of length 4.8 m and mass 15 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes 65° with the horizontal.

(a) Show that the ladder cannot rest in equilibrium unless the coefficient of friction is at least 0.233. [4] (b) The coefficient of friction is 0.4. A person of mass 60 kg climbs the ladder. Find how far along the ladder from its foot the person can climb before it starts to slip. [4]

9. A uniform rectangular lamina of mass 3 kg and length 0.9 m rests horizontally on a table, long sides perpendicular to the edge, with 0.3 m overhanging. A particle of mass m kg is placed at the end of the overhang.

(a) Find the greatest value of m for which the lamina does not tilt. [3] (b) For this value of m, state the magnitude of the force exerted by the table on the lamina. [1]

10. A uniform plank AB has length 8 m and mass 30 kg. It hangs horizontally from vertical ropes at C and D, where AC = DB = 1 m. A person of mass 60 kg, modelled as a particle, stands x m from A (0 ≤ x ≤ 8). Each rope can bear a tension of at most 735 N.

(a) Show that the tensions in the ropes at C and D are 9.8(85 − 10x) N and 9.8(5 + 10x) N. [5] (b) Find the set of values of x for which neither rope breaks. [3]

Answers

1. (a) Moment = force × perpendicular distance = 25 × 0.4 × sin 50° [1] = 7.66 N m (3 s.f.) [1] (b) d = 0.4 sin 50° = 0.306 m (3 s.f.) [1] Examiner insight: Writing the product before evaluating it makes your method easy to credit. Part (a) needs the unit N m.

2. Take anticlockwise as positive and moments about P. 8 N down at R: perpendicular distance 1.2 m, clockwise, −9.6 [1] 6 N from R towards S: its line of action is 0.7 m above P, and it acts leftwards, so anticlockwise, +4.2 [1] 7 N up at Q: perpendicular distance 1.2 m, anticlockwise, +8.4 [1] 20 N along RP: the line of action passes through P, so the moment is zero. Total = −9.6 + 4.2 + 8.4 = 3.0 N m anticlockwise [1] Examiner insight: State the sense of the resultant as well as its size, and say why the 20 N force contributes nothing.

3. (a) Moments about C, with D 0.4 m and B 1.0 m on the other side. Clockwise: 2g × 0.4 + mg × 1.0 [1] Anticlockwise: 5g × 0.5 [1] 2.5 = 0.8 + m, so m = 1.7 [1] (b) The rod is light, so the pivot force is the total weight: (5 + 2 + 1.7) × 9.8 = 85.3 N (3 s.f.) [1] Examiner insight: A light rod has no weight, so list the forces first and check you have not added one.

4. Distances from C (0.5 m from A): centre of beam 1.75 m, person 1.5 m, D 3.5 m. [1] Moments about C: R_D × 3.5 = 30g × 1.75 + 70g × 1.5 [1] R_D = 157.5g ÷ 3.5 = 45g = 441 N [1] Vertical: R_C + R_D = 100g = 980 [1] R_C = 980 − 441 = 539 N [1] Examiner insight: Moments about one support avoid simultaneous equations, and a check about the other support catches a wrong distance.

5. (a) Tilting about D, so R_C = 0. The centre is 1.8 m from D (D is 4.8 m from A, the centre 3 m). [1] Moments about D: 12g × 1.2 = Mg × 1.8 [1] M = 8 kg [1] (b) The beam tilts about C when the particle is at A. The centre of the beam is 2.0 m from C and A is 1.0 m from C. [1] Moments about C: mg × 1.0 = 8g × 2.0 [1] m = 16 kg [1] Examiner insight: Say which support the beam tilts about, and that the other reaction is zero, before taking moments.

6. Moments about the first pivot: 1.5g × 1.2 = Mg(x − 1.2), so 1.8 = M(x − 1.2) [2] Moments about the second pivot: 3g × 1.0 = Mg(x − 1.0), so 3 = M(x − 1) [2] Subtracting gives 1.2 = 0.2M, so M = 6 [1] Then 3 = 6(x − 1), so x = 1.5 [1] Examiner insight: Label which pivot each of your two equations uses.

7. (a) Moments about A; B is 2.4 sin 35° above A. [1] T × 2.4 sin 35° = 9g × 1.2 cos 35° [1] T = 4.5g ÷ tan 35° = 63.0 N (3 s.f.) [1] (b) Horizontal: H = T = 62.98 N, directed away from the wall. [1] Vertical: V = 9g = 88.2 N, upwards. [1] Magnitude = √(62.98² + 88.2²) = 108 N (3 s.f.) [1] Direction = tan⁻¹(88.2 ÷ 62.98) = 54.5° above the horizontal, pointing away from the wall. [1] Examiner insight: Take moments about the hinge so its unknown force drops out, and describe the direction in words as well as the angle.

8. (a) Resolve vertically: R = 15g. [1] Moments about the foot: S × 4.8 sin 65° = 15g × 2.4 cos 65° [1] Resolve horizontally: F = S = 7.5g ÷ tan 65° [1] F ≤ μR gives μ ≥ 0.5 ÷ tan 65° = 0.2331…, so μ ≥ 0.233 (3 s.f.) [1] (b) R = 75g, so greatest friction = 0.4 × 75g = 30g, and S = 30g at slipping. [1] Person d m up the ladder; moments about the foot: 30g × 4.8 sin 65° = 15g × 2.4 cos 65° + 60g × d cos 65° [1] 144 tan 65° = 36 + 60d [1] d = 4.55 m (3 s.f.) [1] Examiner insight: For a “show that”, include every step from the equilibrium equations to the inequality, and state the unrounded value before the given one.

9. (a) Tilting about the table edge. The centre is 0.45 m from the free end, so 0.15 m inside the edge. [1] Moments about the edge: mg × 0.3 = 3g × 0.15 [1] m = 1.5 kg [1] (b) The table supports all the weight: (3 + 1.5) × 9.8 = 44.1 N [1] Examiner insight: Identify the table edge as the pivot before writing any distances.

10. (a) Moments about D. Distances from D: centre of plank 3 m, C 6 m, person 7 − x. [1] T_C × 6 = 30g × 3 + 60g(7 − x) [1] T_C = g(90 + 420 − 60x) ÷ 6 = 9.8(85 − 10x) N [1] Vertical: T_C + T_D = 90g [1] T_D = 90g − g(85 − 10x) = 9.8(5 + 10x) N [1] (b) T_C ≤ 735 = 75g gives 85 − 10x ≤ 75, so x ≥ 1. [1] T_D ≤ 75g gives 5 + 10x ≤ 75, so x ≤ 7. [1] Both tensions are positive here, so 1 ≤ x ≤ 7. [1] Examiner insight: A range needs both ends: find each rope’s condition separately, then combine them.

Where marks are usually lost

• Measuring along the rod when the force is angled.
• Leaving out the weight of the beam or rod, or putting it at an end instead of the midpoint.
• Taking moments about the wrong support in a tilting problem.
• Giving a hinge force as one component, or a magnitude without a direction.
• Using F = μR when the ladder is not about to slip.
• Rounding part-way through and losing accuracy in the final answer.
• Omitting the unit N m or the sense of a resultant moment.

Next steps

• Recap with the revision notes or the study guide.
• Go to the course hub and the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Cambridge OCR Level 3 Advanced GCE in Mathematics A (H240), Version 3.1 (August 2026), first assessment 2018, published by OCR. Content references 3.04a to 3.04c.