Pearson Edexcel International Advanced Level Mathematics: Centres of mass and rigid body equilibrium (YMA01) – 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 sections 2 and 5 of unit M2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics specification, Issue 3, April 2019: centres of mass, laminas in equilibrium, moments and statics of rigid bodies. The specification lists M2 as an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. Take g = 9.8 m s⁻² and give answers to 3 significant figures unless exact.

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Questions

1. A force of 15 N acts at the end A of a light rod OA of length 1.6 m, at 50° to OA.

(a) Find the moment of this force about O. [2] (b) A second force of 8 N acts at the midpoint of OA, perpendicular to the rod, and turns it the opposite way. Find the resultant moment about O and its sense. [2]

2. A light rod AB of length 2.4 m carries particles of mass 3 kg at A, 2 kg at a point 1.0 m from A, and 5 kg at B. Find the distance from A of the point where a single support would balance the rod. [3]

3. Particles of mass 2 kg, 3 kg and 5 kg are at the points (1, 3), (5, −2) and (−3, 4) respectively.

(a) Find the coordinates of the centre of mass. [3] (b) A fourth particle of mass 10 kg is added so that the centre of mass of all four is at (1, 3). Find its coordinates. [3]

4. A uniform circular lamina has centre O and radius 6 cm. A circular hole of radius 2 cm is cut out, with its centre 3 cm from O.

(a) Find the distance of the centre of mass of the remaining lamina from O, and say on which side of O it lies. [4] (b) Point B is on the rim, with OB perpendicular to the line through O and the centre of the hole. The lamina hangs freely from B. Find the angle OB makes with the vertical. [2]

5. A uniform lamina is a rectangle 6 cm wide and 4 cm tall with an isosceles triangle of height 3 cm on its top 6 cm edge.

(a) Find the distance of the centre of mass from the base of the rectangle. [4] (b) The lamina is suspended from one bottom corner A, with AB the base. Find the angle AB makes with the vertical. [3]

6. A uniform semicircular lamina has radius 0.9 m and diameter AB. It hangs freely from A. Find the angle AB makes with the vertical. [4]

7. A uniform rectangular lamina ABCD has AB = 1.2 m, BC = 0.8 m and mass 6 kg. It can turn freely in a vertical plane about a smooth horizontal axis through A. It is held with AB horizontal and C vertically below B by a horizontal force of magnitude P N applied at C, directed away from the line AD.

(a) Find P. [3] (b) Find the magnitude and direction of the force exerted on the lamina by the axis. [4]

8. A uniform rectangular lamina ABCD has AB = 0.7 m and AD = 0.4 m. It stands with AB on a rough plane that is slowly tilted, with coefficient of friction 0.45.

(a) Find the angle of the plane at which the lamina would be about to topple, if it did not slip. [3] (b) Show that the lamina slips before it topples. [2]

9. A uniform beam AB has length 5 m and mass 30 kg. It rests on supports at C and D, where AC = 1 m and AD = 3 m. A 5 kg load is placed at B.

(a) Find the reactions at C and D. [4] (b) The load at B is replaced by a heavier one. Find the mass at which the beam is about to tilt, and about which support. [2]

10. A uniform ladder AB has length 4 m and mass 12 kg. Its foot A is on rough level ground, where μ = 0.45, and its top B is against a smooth vertical wall. The ladder is at 55° to the ground. A man of mass 52 kg stands on it 1.5 m from A.

(a) Find the reaction at the wall. [3] (b) Show that the ladder does not slip. [2] (c) Find the greatest distance from A that the man can climb before the ladder slips. [3] (d) State one modelling assumption you made about the man. [1]

11. A uniform ladder AB, of weight W, rests with A on rough horizontal ground and B against a rough vertical wall. The coefficient of friction is 0.5 at the ground and 0.25 at the wall. The ladder is about to slip at both ends.

(a) Show that the normal reaction at the ground is 8W/9. [4] (b) Find the angle the ladder makes with the horizontal. [5]

Answers

1. (a) moment = 15 × 1.6 × sin 50° [1] = 18.4 N m [1] (b) moment of the 8 N force = 8 × 0.8 = 6.4 N m [1] resultant = 18.385 − 6.4 = 12.0 N m, in the sense of the 15 N force [1] [4] Examiner insight: show the perpendicular distance as 1.6 sin 50° and state the sense of the resultant. A bare number leaves the answer incomplete.

2. Σm = 10 [1] Σ(m x) = 3(0) + 2(1.0) + 5(2.4) = 14 [1] x̄ = 14/10 = 1.4 m from A [1] [3] Examiner insight: the rod is light, so it adds nothing to either sum.

3. (a) Σm = 10 [1] Σ(m x) = 2 + 15 − 15 = 2, so x̄ = 0.2 [1] Σ(m y) = 6 − 6 + 20 = 20, so ȳ = 2; the centre of mass is (0.2, 2) [1] (b) total mass 20; x: 2 + 10p = 20 × 1 [1], so p = 1.8 [1] y: 20 + 10q = 20 × 3, so q = 4; coordinates (1.8, 4) [1] [6] Examiner insight: in (b) use total mass 20 on the right-hand side.

4. (a) whole area 36π, hole area 4π [1] take O as origin and the hole’s side as positive: 36π(0) − 4π(3) = 32π x̄ [1] x̄ = −12/32 = −3/8 [1] the centre of mass is 0.375 cm from O, on the side away from the hole [1] (b) G is on the line through O and the hole, so angle BOG = 90°, and G is vertically below B: tan θ = OG/OB = 0.375/6 [1] θ = 3.58° [1] [6] Examiner insight: show the hole as a negative area, and say which side of O the centre of mass lies on.

5. (a) areas: rectangle 24, triangle 9 [1] heights of centres of mass: rectangle 2; triangle 4 + 3/3 = 5 [1] ȳ = (24 × 2 + 9 × 5)/33 [1] = 93/33 = 31/11 = 2.82 cm [1] (b) M is the midpoint of AB; AG is vertical and triangle AMG is right-angled at M [1] tan(GAB) = MG/AM = (31/11)/3 = 31/33 [1] AB makes 43.2° with the vertical [1] [7] Examiner insight: the symmetry axis fixes the horizontal coordinate, so one calculation is enough. Keep 31/11 exact until the last step.

6. distance OG = 4(0.9)/(3π) = 1.2/π = 0.382 m, where O is the midpoint of AB [1] G is on the symmetry axis, perpendicular to AB at O, and vertically below A [1] in right-angled triangle AOG, tan θ = OG/AO = (1.2/π)/0.9 = 4/(3π) [1] θ = 23.0° [1] [4] Examiner insight: the radius cancels, so keep r symbolic.

7. (a) moments about A; the weight acts at the centre, 0.6 m to the right of A: 6g(0.6) [1] P acts at C, 0.8 m below A: moment P(0.8) [1] 0.8P = 3.6g, so P = 44.1 N [1] (b) horizontal component of the axis force X = P = 44.1 N [1] vertical component Y = 6g = 58.8 N up [1] magnitude = √(44.1² + 58.8²) = 73.5 N [1] direction: tan⁻¹(58.8/44.1) = 53.1° above the horizontal, pointing upwards and towards the side of AD [1] [7] Examiner insight: moments about A remove the unknown axis force. Name the line your angle is measured from.

8. (a) G is 0.2 m above AB and 0.35 m from the lowest edge horizontally [1] topples when G is above that edge: tan α = 0.35/0.2 = 1.75 [1] α = 60.3° [1] (b) slipping occurs when tan α = μ = 0.45, giving α = 24.2° [1] 24.2° < 60.3°, so it slips first [1] [5] Examiner insight: in a “show that”, end with both angles and a sentence saying which is smaller.

9. (a) moments about C: 2R_D = 30g(1.5) + 5g(4) [1] R_D = 32.5g = 318.5 N [1] vertically: R_C + R_D = 35g [1] R_C = 2.5g = 24.5 N [1] (b) tilting about D means R_C = 0 [1] moments about D: m g(2) = 30g(0.5), so m = 7.5 kg [1] [6] Examiner insight: say which reaction is zero before writing the equation.

10. (a) moments about A: S(4 sin 55°) = 12g(2 cos 55°) + 52g(1.5 cos 55°) [1] 4 sin 55° = 3.277 and the right-hand side is 102g cos 55° = 573.3 [1] S = 573.3/3.277 = 175 N (174.98) [1] (b) vertically R = 64g = 627.2 N [1] friction needed = S = 175.0 N, and μR = 0.45 × 627.2 = 282.2 N; 175.0 < 282.2, so no slip [1] (c) limiting: S = μR = 282.24 N [1] 282.24(4 sin 55°) = 12g(2 cos 55°) + 52g(x cos 55°), which is 924.8 = 134.9 + 292.3x [1] x = 2.70 m [1] (d) the man is a particle, so his weight acts at one point [1] [9] Examiner insight: in (b) quote the friction needed and the maximum available, then compare. In (c) keep S unrounded.

11. (a) horizontal: S = 0.5R, where S is the wall reaction and R the ground reaction [1] wall friction 0.25S acts upwards [1] vertical: R + 0.25S = W [1] R + 0.125R = W, so R = 8W/9 (and S = 4W/9) [1] (b) moments about A, with ladder length L and angle α: S L sin α + 0.25S L cos α [1] = W(L/2) cos α [1] substituting S = 4W/9: (4/9)(tan α + 0.25) = 1/2 [1] tan α = 9/8 − 1/4 = 7/8 [1] α = 41.2° [1] [9] Examiner insight: draw both friction forces first. At the wall friction acts upwards, and it appears in the vertical equation and in the moments.

Where marks are usually lost

• Using the length along a rod where the perpendicular distance is needed.
• Forgetting to make a hole negative, or using lengths rather than areas.
• Stating that the beam tilts without saying which reaction is zero.
• Using F = μR when the ladder is not on the point of slipping.
• Dropping the friction at a rough wall from the vertical resolution.

Next steps

• Revision notes and study guide.
• Course hub: /boards/edexcel/a-level/mathematics/. Checklist: /checklists/edexcel/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, unit M2: sections 2 (centres of mass) and 5 (statics of rigid bodies).