OxfordAQA International AS and A-level Mathematics: Moments, rigid bodies in equilibrium and centres of mass (9660) – 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, rigid bodies in equilibrium and centres of mass from section M2.3 of the OxfordAQA International AS and A-level Mathematics specification, Version 5.2. The specification places this content in International A-level Unit M2, so it is not examined at International AS. Take g = 9.8 m s⁻², as the specification does. Unless a question says otherwise, round answers to 3 significant figures.

Read the study guide and the revision notes first if you need them. Course pages: OxfordAQA A-level Mathematics hub and the printable checklist.

Questions

1. A uniform rod AB has length 0.8 m and is hinged at A. A force of 30 N acts at B, at 50° to the rod. Find the moment of this force about A. [3]

2. A uniform rectangular lamina has vertices at (2, 1), (10, 1), (10, 7) and (2, 7).

(a) Find the coordinates of its centre of mass. [2] (b) State the centre of mass of a uniform circular disc whose centre is at (3, -2). [1]

3. A uniform horizontal plank AB has length 4.5 m and mass 24 kg. The plank lies on two supports, C and D, with AC = 0.6 m and DB = 0.5 m. A painter of mass 75 kg stands on the plank 1.5 m from A.

(a) Find the reactions at C and D. [4] (b) The painter leaves and a man of mass 120 kg stands between D and B. Find the greatest distance from D at which he can stand without the plank tilting. [2]

4. A uniform horizontal beam AB has length 2.5 m and mass 8 kg. End A is hinged to a wall. A vertical cable is attached at C, where AC = 2 m, and a lamp of mass 5 kg hangs from B.

(a) Find the tension in the cable. [3] (b) Find the magnitude and direction of the force the hinge exerts on the beam. [2]

5. A uniform ladder of length 8.5 m and mass 20 kg rests with its top against a smooth vertical wall. Its foot is on rough horizontal ground, 4 m from the wall, and the top is 7.5 m above the ground. The ladder is in equilibrium.

(a) Find the normal reaction at the ground and the friction force at the foot. [4] (b) Find the least possible coefficient of friction between the ladder and the ground. [2]

6. The ladder in question 5 is used by a boy of mass 45 kg. The ground is rough, with coefficient of friction 0.35 at the foot. The ladder stays at rest. Find the greatest distance up the ladder, measured along it from the foot, that the boy can stand. [6]

7. Particles of mass 2 kg, 5 kg and 3 kg are at (1, 6), (4, 2) and (-3, -2).

(a) Find the coordinates of the centre of mass. [2] (b) A fourth particle of mass 5 kg is added so that the centre of mass of the four particles is at (2, 3). Find its coordinates. [3]

8. A uniform rod AB is 2.4 m long with mass 5 kg. Particles of mass 1 kg and 4 kg are attached at A and B.

(a) How far from A is the centre of mass of the loaded rod? [3] (b) The rod rests horizontally on a single pivot placed at this point. Find the force on the pivot. [1]

9. A uniform T-shaped lamina has a top bar and a stem. In cm, the bar is −6 ≤ x ≤ 6, 8 ≤ y ≤ 11 and the stem is −2 ≤ x ≤ 2, 0 ≤ y ≤ 8. Find the distance of its centre of mass from the base of the stem (the line y = 0). [5]

10. A uniform rectangular lamina occupies 0 ≤ x ≤ 30, 0 ≤ y ≤ 12 (cm). A circular hole of radius 3 cm is cut out with its centre at (22, 6). Find the coordinates of the centre of mass of the remaining lamina. [5]

11. A uniform rectangular lamina ABCD has AB = 16 cm, BC = 10 cm and mass 6 kg. A particle of mass 2 kg is fixed at C. Take A as the origin, AB along the x-axis and AD along the y-axis.

(a) Find the centre of mass of the lamina and particle together. [4] (b) The body is freely suspended from A and hangs in equilibrium. Find the angle AB makes with the vertical. [3] (c) The body is freely suspended from B instead. Find the angle BA makes with the vertical. [3]

Answers

1. Perpendicular distance from A to the line of the force = 0.8 sin 50° [1]. Moment = 30 × 0.8 sin 50° [1]. Moment = 18.4 N m [1]. [3] Examiner insight: Write the product with the sine before you evaluate it. A slip in the arithmetic then costs less than a missing sine factor, which shows the wrong method.

2. (a) x-coordinate = (2 + 10) ÷ 2 = 6 [1]. y-coordinate = (1 + 7) ÷ 2 = 4, so the centre of mass is (6, 4) [1]. (b) (3, -2) [1]. [3] Examiner insight: A one-line answer gains credit only when it is the right point. Quote the symmetry argument in a word or two if the command word is “find” rather than “state”.

3. (a) Vertical: R_C + R_D = 99g [1]. Moments about C: R_D × 3.4 = 24g × 1.65 + 75g × 0.9 [1]. R_D = 31.5g = 309 N [1]. R_C = 99g - 31.5g = 67.5g = 662 N [1]. (b) Tilting about D means R_C = 0. Moments about D, with the man y metres beyond D: 120g × y = 24g × 1.75 [1]. y = 0.35 m [1]. [6] Examiner insight: Say which support the plank turns about and what happens to the other reaction. A clear limiting condition makes each later line easy to credit.

4. (a) Moments about A: T × 2 = 8g × 1.25 + 5g × 2.5 [1]. T × 2 = 22.5g [1]. T = 11.25g = 110 N [1]. (b) Vertical: Y + T = 13g, so Y = 127.4 - 110.25 [1]. The force is 17.15 N upwards [1]; there is no horizontal component since every other force is vertical. [5] Examiner insight: The hinge force is unknown in size and direction, so take moments about the hinge first. Check the direction by asking whether the cable alone could support the whole weight.

5. (a) Vertical: R = 20g = 196 N [1]. Moments about the foot: S × 7.5 = 20g × 2 [1], so S = 40g ÷ 7.5 = 52.3 N [1]. Horizontal: F = S = 52.3 N [1]. (b) Equilibrium needs F ≤ μR, so μ ≥ 52.27 ÷ 196 [1]. The least coefficient is 0.267 [1]. [6] Examiner insight: The weight acts 2 m horizontally from the foot, not 4.25 m along the ladder. Writing the horizontal distance out as its own step is the safest way to avoid that slip.

6. Vertical: R = (20 + 45)g = 65g [1]. Friction available is at most 0.35 × 65g = 22.75g [1]. The boy is d m up, so his horizontal distance from the foot is 8d ÷ 17. Moments about the foot: 7.5S = 20g × 2 + 45g × (8d ÷ 17) [1], so 7.5S = 40g + (360g d) ÷ 17 [1]. Since S = F ≤ 22.75g: 170.6 ≥ 40 + 360d ÷ 17 [1]. d ≤ 6.17 m [1]. [6] Examiner insight: The cosine of the ladder angle is 8/17 because the foot is 4 m from the wall on an 8.5 m ladder. Using a ratio from the given lengths avoids rounded angles and keeps the final answer clean.

7. (a) Σm = 10. Σ m x = 2 + 20 - 9 = 13, so X = 1.3 [1]. Σ m y = 12 + 10 - 6 = 16, so Y = 1.6 [1]. (b) Total mass 15, so 15 × 2 = 13 + 5p [1]. p = 3.4 [1]. 15 × 3 = 16 + 5q gives q = 5.8 [1]. [5] Examiner insight: Set the equations up from X Σm = Σ m x before solving. That layout shows the method even if one sum is wrong.

8. (a) Total mass 10 kg. 10d = 5 × 1.2 + 1 × 0 + 4 × 2.4 [1]. 10d = 15.6 [1], so d = 1.56 m [1]. (b) The pivot supports the total weight: 10g = 98 N [1]. [4] Examiner insight: The uniform rod’s mass goes at its midpoint, 1.2 m from A. Putting it at an end is the usual way to lose the answer to part (a).

9. By symmetry the centre of mass is on x = 0 [1]. Bar area = 12 × 3 = 36, centre at y = 9.5. Stem area = 4 × 8 = 32, centre at y = 4 [1]. Total area 68. 68 Y = 36 × 9.5 + 32 × 4 [1]. 68 Y = 342 + 128 = 470 [1]. Y = 470 ÷ 68 = 6.91 cm [1]. [5] Examiner insight: A table of area and centre for each part keeps the working checkable. Use the same base line for every centre.

10. Rectangle area 360, centre (15, 6). Hole area 9π, centre (22, 6) [1]. (360 - 9π) X = 360 × 15 - 9π × 22 [1]. Remaining area = 331.7, so X = (5400 - 198π) ÷ (360 - 9π) [1]. X = 14.4 [1]. The lamina is symmetrical about y = 6, so Y = 6 [1]. [5] Examiner insight: Treat the hole as a negative area. Leave π in the working until the last line so you do not round part way through.

11. (a) Total mass 8 kg. Lamina centre (8, 5), particle at (16, 10) [1]. 8X = 6 × 8 + 2 × 16 = 80 [1], so X = 10. 8Y = 6 × 5 + 2 × 10 = 50 [1], so Y = 6.25. The centre of mass is (10, 6.25) [1]. (b) G is vertically below A, so AG is vertical [1]. tan α = 6.25 ÷ 10 [1]. α = 32.0° [1]. (c) BG is vertical. G is 6 cm from B horizontally (16 - 10) and 6.25 cm vertically [1]. tan β = 6.25 ÷ 6 [1]. β = 46.2° [1]. [10] Examiner insight: Draw the suspension point, G and the vertical in a sketch. Mark which angle you are finding and check that it is between the edge named in the question and the vertical.

Where marks are usually lost

• Using a length along a sloping rod as the perpendicular distance, with no sine or cosine.
• Forgetting a weight when taking moments, especially the weight of the beam or ladder itself.
• Placing a uniform rod’s weight at an end rather than at its midpoint.
• Taking moments about a point but not saying which point.
• Mixing up the friction force with the coefficient of friction.
• Using different origins for different parts of a composite body.
• Forgetting that a removed piece has negative area.
• Rounding the centre of mass early and giving a suspension angle that is out in the third figure.
• Measuring a hanging angle between the wrong edge and the vertical.

Next steps

• Revision notes
• Study guide
• OxfordAQA A-level Mathematics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

OxfordAQA, International AS and A-level Mathematics (9660), Version 5.2, section M2.3 (Statics and forces), part of International A-level Unit M2 (Mechanics).