OxfordAQA International AS and A-level Mathematics: Statics: forces, friction and equilibrium of a particle (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 the first part of section M2.3 (Statics and forces) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2. The specification states this content is part of International A-level Unit M2 (Mechanics). Use g = 9.8 m s⁻² and round to 3 significant figures unless the question says otherwise. Study first with the study guide or the revision notes. Course pages: hub and checklist.

Questions

1. A 20 kg crate rests on a rough slope inclined at 30° to the horizontal. A rope attached to the crate runs parallel to the slope and pulls up it. The crate is on the point of sliding down. List every force on the crate, give the direction of each, and state how the friction direction is decided. [3]

2. Three forces act on a particle: F₁ = (7i − 3j) N, F₂ = (−2i + 9j) N and F₃ = (−10i + 2j) N.

(a) Find the resultant force. [2] (b) Find its magnitude. [1] (c) Write down a fourth force that would keep the particle in equilibrium. [1]

3. A box of mass 15 kg is on a rough horizontal floor with coefficient of friction 0.6. A horizontal force P acts on it.

(a) Find the greatest friction the floor can provide. [2] (b) P = 60 N. State the friction force and explain why it is not the answer to part (a). [2]

4. A particle of mass 12 kg rests on a smooth plane inclined at θ to the horizontal, where tan θ = 5/12. A horizontal force H, in the vertical plane containing the line of greatest slope, holds it in equilibrium.

(a) Find H. [3] (b) Find the normal reaction. [2]

5. A 5 kg particle lies on a rough plane whose angle of inclination is 35°. Take the coefficient of friction to be 0.4.

(a) Show that the particle cannot rest on the plane with no other force. [2] (b) A force P acts up the line of greatest slope. Find the range of P for which the particle stays at rest. [4]

6. A particle of mass 4.5 kg is held at rest by two light strings fixed to a horizontal ceiling. One string makes 40° with the ceiling and the other makes 65° with the ceiling, on opposite sides of the particle. Find the two tensions. [6]

7. Four forces act on a particle and keep it at rest: F₁ = (2i + 5j) N, F₂ = (pi − 3j) N, F₃ = (−4i + qj) N and F₄ = (3i − 4j) N.

(a) Find p and q. [3] (b) After F₂ is taken away, find the magnitude of the resultant of the other three forces. [2]

8. An 8 kg block rests on rough horizontal ground, μ = 0.5. A string pulls it at angle α above the horizontal, with tension T and tan α = 3/4. A horizontal force of 40 N pulls the block the other way. Find the range of T for which the block stays at rest. [7]

9. A particle of mass m is at rest on a rough plane inclined at α to the horizontal, with coefficient of friction μ and no other force.

(a) Show that μ ≥ tan α. [3] (b) For α = 24°, find the least possible value of μ. [1]

10. A particle P of mass 6 kg lies on a rough horizontal table. P is joined by a light inextensible string to a 2.5 kg particle that hangs freely beyond the table edge, the string running over a smooth pulley. The system is at rest.

(a) Find the tension in the string, with a reason. [2] (b) State the friction force on P. [1] (c) Find the least possible coefficient of friction. [2] (d) The hanging particle is replaced by one of mass M, and with μ = 0.55, P is on the point of slipping. Find M. [2]

11. A 3 kg particle sits on a rough plane at 30° to the horizontal, where μ = 0.3. A horizontal force X, acting in the vertical plane containing the line of greatest slope and pushing the particle towards the plane, holds the particle at rest. Find the range of X. [8]

Answers

1. Weight 196 N vertically downwards [1]; normal reaction perpendicular to the slope, away from it, and tension up the slope along the rope [1]; friction along the slope, acting up the slope, because it opposes the tendency to slide down [1]. [3] Examiner insight: Name every force and give its direction. A list that says “friction” with no direction, or leaves out the normal reaction, is incomplete.

2. (a) x: 7 − 2 − 10 = −5 [1]; y: −3 + 9 + 2 = 8 [1]. Resultant (−5i + 8j) N. (b) √(25 + 64) = √89 = 9.43 N [1]. (c) The negative of the resultant: (5i − 8j) N [1]. [4] Examiner insight: Add the i-parts and j-parts separately and keep the signs. The force for equilibrium is the negative of the resultant, not the resultant itself.

3. (a) R = 15 × 9.8 = 147 N [1]; greatest friction = 0.6 × 147 = 88.2 N [1]. (b) The block is at rest, so friction balances P: F = 60 N [1]. Friction equals μR only when the block is about to slip, and 60 < 88.2, so it is below the limit [1]. [4] Examiner insight: Say why friction is not μR. Stating F = μR without a reason in a non-limiting case loses the explanation.

4. (a) Along the slope: H cos θ = mg sin θ [1]. H = mg tan θ [1] = 12 × 9.8 × 5/12 = 49 N [1]. (b) Perpendicular: R = mg cos θ + H sin θ [1]. With cos θ = 12/13 and sin θ = 5/13, R = 117.6 × 12/13 + 49 × 5/13 = 127.4 N [1]. [5] Examiner insight: H presses the particle into the slope as well, so R is more than mg cos θ. Check R with the vertical resolution, R cos θ = mg.

5. (a) Without other forces, F = mg sin 35° = 28.1 N and R = mg cos 35° = 40.1 N [1]. Greatest friction = 0.4 × 40.1 = 16.1 N, which is less than 28.1 N, so it cannot rest [1]. (b) Lower limit, friction up the slope at its greatest: P + 16.06 = 28.11 [1], so P = 12.05 [1]. Upper limit, friction down the slope: P = 28.11 + 16.06 = 44.16 [1]. 12.0 N ≤ P ≤ 44.2 N [1]. [6] Examiner insight: A range question needs both limits, with friction in opposite directions. A single value earns less than the full range.

6. T₁ at 40° to the horizontal and T₂ at 65°, weight 4.5 × 9.8 = 44.1 N [1]. Horizontal: T₁ cos 40° = T₂ cos 65° [1]. Vertical: T₁ sin 40° + T₂ sin 65° = 44.1 [1]. T₂ = 1.8126 T₁ [1]. T₁ (0.6428 + 1.8126 × 0.9063) = 44.1, so T₁ = 44.1/2.2856 = 19.3 N [1]. T₂ = 35.0 N [1]. [6] Examiner insight: The angles here are to the horizontal, so cosines go with horizontal components. A quick check is that the vertical components add to the weight.

7. (a) i: 2 + p − 4 + 3 = 0, so p = −1 [1]. j: 5 − 3 + q − 4 = 0, so q = 2 [1]. Both equations written down [1]. (b) The rest sum to −F₂ = (i + 3j) [1]. Magnitude √10 = 3.16 N [1]. [5] Examiner insight: Show the i-equation and the j-equation separately. In part (b), removing a force leaves a resultant equal and opposite to that force.

8. R = 78.4 − T sin α = 78.4 − 0.6T [1]. Let F be friction towards the right, with T cos α = 0.8T to the right: 0.8T + F = 40, so F = 40 − 0.8T [1]. Equilibrium needs |F| ≤ μR = 39.2 − 0.3T. If friction is to the right: 40 − 0.8T ≤ 39.2 − 0.3T [1], so T ≥ 1.6 [1]. If friction is to the left: 0.8T − 40 ≤ 39.2 − 0.3T [1], so T ≤ 72 [1]. At T = 72, R = 35.2 > 0, so the block stays on the floor. 1.6 N ≤ T ≤ 72 N [1]. [7] Examiner insight: A pull upwards lowers R, so friction available falls as T rises. Check R stays positive at the upper end.

9. (a) Perpendicular: R = mg cos α [1]. Along the slope: F = mg sin α [1]. F ≤ μR gives mg sin α ≤ μ mg cos α, and dividing by mg cos α (positive) gives μ ≥ tan α [1]. (b) μ ≥ tan 24° = 0.445 [1]. [4] Examiner insight: A “show that” answer needs each step in order. Say that you divide by a positive quantity so the inequality does not reverse.

10. (a) The hanging particle is at rest: T = 2.5 × 9.8 [1] = 24.5 N; the string is light and the pulley smooth, so the tension is the same on P [1]. (b) P is at rest, so F = 24.5 N [1]. (c) R = 6 × 9.8 = 58.8 N [1]; F ≤ μR gives μ ≥ 24.5/58.8 = 0.417 [1]. (d) Limiting: Mg = 0.55 × 58.8 [1], so M = 3.3 kg [1]. [7] Examiner insight: Give the reason for the tension, not just the number. Limiting equilibrium is where you use equals; at rest and below it use an inequality.

11. R = mg cos 30° + X sin 30° = 25.46 + 0.5X, with mg = 29.4 and mg sin 30° = 14.7 [1]. Lower limit, about to slide down, friction up the slope: X cos 30° + μR = 14.7 [1]. Then 0.866X + 0.3(25.46 + 0.5X) = 14.7 [1], so 1.016X = 7.06 and X = 6.95 [1]. Upper limit, about to slide up, friction down the slope: X cos 30° = 14.7 + μR [1]. Then 0.866X − 0.15X = 14.7 + 7.64 [1], so X = 22.34/0.716 = 31.2 [1]. Range: 6.95 N ≤ X ≤ 31.2 N [1]. [8] Examiner insight: Because X also presses the particle into the plane, R changes between the two cases of the same equation. Keep R in terms of X rather than substituting a number.

Where marks are usually lost

• Taking R = mg when a rope, push or slope changes the normal reaction.
• Using F = μR in a case where the body is not about to slip.
• Giving one limit of a range and not the other.
• Friction drawn the wrong way round and not corrected.
• Swapping sin and cos because the angle is to the other axis.
• Dropping a negative sign when finding the force that holds a particle in equilibrium.
• Rounding early, so that the third significant figure is wrong.
• Leaving out units or a reason where the question says to explain.

Next steps

• Condense with the revision notes.
• Re-read the study guide.
• Course hub and checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section M2.3 Statics and forces, published by OxfordAQA.