AS & A Level · Practice questions · Friction and limiting equilibrium

Up to the limit.

Six original Cambridge-style questions on friction and limiting equilibrium, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find R first, then decide about F.

Diagram: weight, normal reaction, friction and any applied forces.
Perpendicular to the surface: find \(R\), including components of applied forces.
Direction of friction: opposite to the way the particle moves or is about to move.
Limiting? If moving or about to slip, \(F = \mu R\); otherwise \(F \le \mu R\).
Parallel to the surface: resolve and solve.

Recall

[3 marks]

A crate of mass 12 kg rests on rough horizontal ground, \(\mu = 0.35\). A horizontal force of 30 N acts on it. Determine whether the crate moves, and state the friction force.

\(R = 120\) N, so maximum friction \(= 0.35 \times 120 = 42\) N. ✓
\(30 < 42\), so the crate stays at rest. ✓
Friction \(= 30\) N, opposing the applied force. ✓

Calculation

[2 marks]

A particle of mass 0.8 kg is about to slip down a rough plane inclined at \(28^\circ\) to the horizontal. No other force acts. Find \(\mu\).

Limiting with no other forces: \(mg\sin 28^\circ = \mu mg\cos 28^\circ\). ✓
\(\mu = \tan 28^\circ = 0.532\). (The mass does not matter.) ✓

Calculation

[4 marks]

A block of mass 6 kg rests on rough horizontal ground. A force of 40 N pushes it at \(30^\circ\) below the horizontal, and the block is on the point of moving. Find the coefficient of friction.

Vertically: \(R = 60 + 40\sin 30^\circ = 80\) N. ✓
Horizontally: \(F = 40\cos 30^\circ = 34.64\) N. ✓
\(\mu = \dfrac{34.64}{80} = 0.433\). ✓

Analysis

[5 marks]

A particle of mass 5 kg rests on a rough plane inclined at \(30^\circ\), \(\mu = 0.4\). A force of \(P\) N acts on it, parallel to a line of greatest slope and up the plane. Find the set of values of \(P\) for which the particle stays at rest.

\(R = 50\cos 30^\circ = 43.30\) N, so \(\mu R = 17.32\) N. Weight component down the plane \(= 50\sin 30^\circ = 25\) N. ✓
About to slip down (friction up): \(P = 25 - 17.32 = 7.68\). ✓
About to slip up (friction down): \(P = 25 + 17.32 = 42.3\). ✓
\(7.68 \le P \le 42.3\). ✓

Analysis

[4 marks]

A block of mass 4 kg on rough horizontal ground is about to move under a horizontal force of 18 N. Find \(\mu\), the magnitude of the total contact force, and the angle the contact force makes with the vertical.

\(R = 40\) N and \(F = 18\) N, so \(\mu = 0.45\). ✓
Contact force \(= \sqrt{18^2 + 40^2} = 43.9\) N. ✓
Angle with the vertical \(= \tan^{-1}\dfrac{18}{40} = 24.2^\circ\) (in limiting equilibrium this angle is \(\tan^{-1}\mu\)). ✓

Analysis

[6 marks]

A sledge of mass 15 kg rests on a slope inclined at \(10^\circ\) to the horizontal. The coefficient of friction between sledge and snow is 0.1. A rope pulls the sledge with tension \(T\) N at \(20^\circ\) above the line of greatest slope, up the slope. The sledge is about to move up the slope. Find \(T\).

Perpendicular: \(R = 150\cos 10^\circ - T\sin 20^\circ\). ✓
Parallel (friction down the slope): \(T\cos 20^\circ = 150\sin 10^\circ + 0.1R\). ✓
\(T\cos 20^\circ = 150\sin 10^\circ + 15\cos 10^\circ - 0.1T\sin 20^\circ\). ✓
\(T(\cos 20^\circ + 0.1\sin 20^\circ) = 26.05 + 14.77\), so \(0.9739T = 40.82\). ✓
\(T = 41.9\) N (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.