AQA A-Level Mathematics: S: Moments (7357) – 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 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 places Section S in Paper 2. A calculator is required in every 7357 paper. Take g = 9.8 m s⁻² and give answers to 3 significant figures where they are not exact, unless the question says otherwise.

Revise first with the Moments study guide and the Moments revision notes. Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/. Not sure where you stand? Try the free 10-minute diagnostics.

Questions

1. A force of 40 N acts at a perpendicular distance of 35 cm from a point O. Find the moment of the force about O. [2]

2. A light rod AB has length 2 m and is pivoted at A. A force of 25 N acts at B, at 60° to the rod. Find the moment of this force about A. [3]

3. A light rod PQ of length 3 m is horizontal, with P on the left, and is pivoted at its midpoint M. A downward force of 12 N acts at P, a downward force of 20 N acts at R where PR = 2.4 m, and an upward force of 6 N acts at Q. Find the resultant moment about M and state which way the rod starts to turn. [3]

4. A uniform rod AB has length 2 m and mass 3 kg. It rests horizontally on a single pivot at C, where AC = 0.8 m. A particle of mass m kg is attached at A and the rod is in equilibrium.

(a) Find m. [3] (b) Find the reaction at the pivot. [1]

5. A uniform plank AB has length 4 m and mass 25 kg. It rests horizontally on a support at A and a support at C, where AC = 3.5 m. A box of mass 15 kg is placed on the plank 1 m from A. Find the reactions at A and at C. [5]

6. A non-uniform rod AB of length 3 m and mass 6 kg rests horizontally on supports at A and at B. The reaction at B is twice the reaction at A.

(a) Find the reactions at A and at B. [2] (b) Find the distance of the centre of mass of the rod from A. [3]

7. A uniform rod AB of length 2 m and weight W N rests horizontally on supports at C and D, where AC = 0.4 m and AD = 1.5 m. When a particle of weight 30 N is attached at B, the rod is on the point of tilting.

(a) Find W. [3] (b) Find the reaction at D. [1]

8. A uniform beam AB of length 8 m and mass 50 kg rests horizontally on supports at C and D, where AC = 2 m and AD = 6 m. A man of mass 80 kg walks along the beam from D towards B.

(a) Show that the beam tilts before the man reaches B. [4] (b) A load of mass M kg is fixed at A so that the man can stand at B without the beam tilting. Find the least value of M. [3] (c) State one modelling assumption you have made about the man. [1]

9. A uniform rod AB, of length 1.6 m and mass 5 kg, is hinged at A to a vertical wall. It is held horizontal by a light string attached at B and to a point on the wall vertically above A. The string makes 30° with the rod. A particle of mass 2 kg hangs from the rod at D, where AD = 1.2 m.

(a) Show that the tension in the string is 8g N. [3] (b) Find the magnitude and direction of the force exerted by the hinge on the rod. [5]

10. A uniform ladder AB, of length 6 m and mass 20 kg, rests with A on rough horizontal ground and B against a smooth vertical wall. The ladder makes an angle θ with the ground, where tan θ = 12/5. A man of mass 75 kg stands on the ladder at C, where AC = 4.5 m. The ladder is about to slip.

(a) Find the normal reaction at A. [1] (b) Find the reaction at B. [4] (c) Find the coefficient of friction between the ladder and the ground. [2] (d) The wall is in fact slightly rough. State, with a reason, how this would affect the friction needed at the ground. [2]

11. A diving board is modelled as a uniform rigid rod clamped at one end, and a diver is modelled as a particle.

(a) Explain what the word “uniform” allows you to assume in this model. [1] (b) Explain what “rigid” allows you to assume. [1] (c) Give one way in which the model of the diver as a particle may not match the real situation, and its effect on a calculated moment. [2]

Answers

1. 35 cm = 0.35 m, moment = 40 × 0.35 [1] = 14 N m [1] [2] Examiner insight: a bare 1400 (using centimetres) loses both marks here; the method mark depends on a distance in metres.

2. Perpendicular distance = 2 sin 60° (or component 25 sin 60° perpendicular to rod) [1] moment = 25 × 2 sin 60° [1] = 43.3 N m (3 s.f.) [1] [3] Examiner insight: 25 × 2 = 50 N m earns nothing, because it ignores the angle; the first mark is for the perpendicular distance or the perpendicular component.

3. Anticlockwise: 12 × 1.5 + 6 × 1.5 = 27 N m; clockwise: 20 × 0.9 = 18 N m [1] resultant = 27 − 18 [1] = 9 N m anticlockwise, so the P end goes down [1] [3] Examiner insight: the upward force at Q turns the rod the same way as the downward force at P; treating every force as clockwise on the right loses the method mark.

4. (a) Rod’s weight acts 1 m from A, so 0.2 m from C [1] moments about C: mg × 0.8 = 3g × 0.2 [1] m = 0.75 [1] (b) R = (3 + 0.75)g = 36.75 N (36.8 N to 3 s.f.) [1] [4] Examiner insight: in (a) g cancels, so leaving it in or substituting 9.8 both work; in this scheme (b) is marked follow-through from your m.

5. Plank’s weight 25g at 2 m from A [1] moments about A: R_C × 3.5 = 25g × 2 + 15g × 1 [1] R_C = 65g ÷ 3.5 = 182 N [1] resolve vertically: R_A + R_C = 40g [1] R_A = 392 − 182 = 210 N [1] [5] Examiner insight: each moment term must pair a force with its own distance; one wrong distance usually costs the method mark for that equation and both accuracy marks.

6. (a) R_A + 2R_A = 6g [1] R_A = 19.6 N, R_B = 39.2 N [1] (b) moments about A: 6g × x = 39.2 × 3 [1] 58.8x = 117.6 [1] x = 2 m from A [1] [5] Examiner insight: in (b) the weight goes at an unknown x; writing 6g × 1.5 assumes the rod is uniform and scores nothing.

7. (a) about to tilt about D, so the reaction at C is 0 [1] moments about D: W × (1.5 − 1) = 30 × 0.5 [1] W = 30 [1] (b) R_D = W + 30 = 60 N [1] [4] Examiner insight: in this scheme the first mark is for stating, not just using, that the reaction at C is zero; say it in words before the equation.

8. (a) about to tilt about D, so R_C = 0 [1] man at y m beyond D; moments about D: 50g × 2 = 80g × y [1] y = 1.25 m [1] so tilting starts when he is 7.25 m from A, which is less than 8 m, so the beam tilts before he reaches B [1] (b) with the man at B and R_C = 0, moments about D: 50g × 2 + Mg × 6 = 80g × 2 [1] 100 + 6M = 160 [1] least M = 10 [1] (c) The man is a particle, so his weight acts at a single point [1] [8] Examiner insight: in a “show that” the final mark needs the comparison written out (7.25 < 8 and a conclusion), not just y = 1.25.

9. (a) moments about A: T sin 30° × 1.6 = 5g × 0.8 + 2g × 1.2 [1] 0.8T = 6.4g [1] T = 8g N (78.4 N), as required [1] (b) horizontal: H = T cos 30° = 8g cos 30° [1] = 67.9 N, away from the wall [1] vertical: V + T sin 30° = 7g, V = 68.6 − 39.2 = 29.4 N upwards [1] magnitude = √(67.9…² + 29.4²) = 74.0 N [1] direction: tan⁻¹(29.4 ÷ 67.9…) = 23.4° above the horizontal [1] [8] Examiner insight: a “magnitude and direction” answer needs both; a correct 74.0 N with no angle leaves the last mark unearned.

10. (a) R = (20 + 75)g = 931 N [1] (b) sin θ = 12/13, cos θ = 5/13 [1] moments about A: S × 6 sin θ = 20g × 3 cos θ + 75g × 4.5 cos θ [1] S × 72/13 = 397.5g × 5/13 [1] S = 271 N (270.5…) [1] (c) F = S and F = μR [1] μ = 270.5… ÷ 931 = 0.291 (3 s.f.) [1] (d) A rough wall gives an upward friction force at B [1] which has a moment about A, so S, and hence the friction needed at A (F = S), would be smaller; the model overestimates the friction needed [1] [9] Examiner insight: using the 3 s.f. value 271 in (c) gives 0.291 as well, but rounding S to 270 gives 0.290; carry the unrounded value.

11. (a) The weight of the board acts at its midpoint [1] (b) The board does not bend, so distances along it stay fixed [1] (c) The diver’s weight is spread over the feet, not at one point [1] so the true moment about the clamp may be slightly different from the calculated one [1] [4] Examiner insight: in this scheme each mark needs a link to the diving board or diver; a textbook definition with no reference to the context is not credited.

Where marks are usually lost

• Using the distance along a rod, not the perpendicular distance, for a force at an angle.
• Forgetting the weight of a rod that is not described as light.
• Placing a non-uniform rod’s weight at the midpoint.
• Not stating that the far reaction is zero when a beam is about to tilt.
• Swapping sin θ and cos θ in ladder moments.
• Giving a hinge force without its direction.
• Rounding a tension or reaction before using it again.
• Leaving out units (N, N m) on final answers.

Next steps

• Recap the methods: Moments revision notes.
• Full explanations: Moments study guide.
• Units of moment: Quantities and units practice questions.
• Related practice: Forces and Newton’s laws and Kinematics.
• Modelling language: Overarching themes practice.
• Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

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) and Section 3.1.3 OT3: Mathematical modelling.