Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Friction. 5 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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Find the greatest horizontal force that the student can apply without the crate moving.
78.4 N (accept 78 N) N
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A crate of mass 20 kg rests on a rough horizontal floor. The coefficient of friction between the crate and the floor is 0.4. A student pushes the crate with a horizontal force of 50 N, but the crate does not move. The crate is modelled as a particle. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
(a) State the magnitude of the frictional force acting on the crate, giving a reason for your answer.[2]
(b) Find the greatest horizontal force that the student can apply without the crate moving.[2]
(c) Explain the meaning of the inequality \(F \leqslant \mu R\) in this situation.[1]
Show the answer and mark scheme
(a)Answer: 50 N: the crate is in equilibrium, so friction balances the 50 N push. N
B1 for 50 (N)
B1 for the crate is in equilibrium (does not move), so the friction is equal and opposite to the applied force
Worked solution: The crate is at rest, so the resultant force is zero: friction = 50 N, opposing the push.
(b)Answer: 78.4 N (accept 78 N) N
M1 for \(\mu R = 0.4 \times 20g\)
A1 for 78.4 or 78 (N)
Worked solution: \(F_{\max} = \mu R = 0.4 \times 20 \times 9.8 = 78.4\) N
(c)Answer: Friction takes whatever value is needed to prevent motion, up to a maximum (limiting) value of \(\mu R\); it only equals \(\mu R\) when the crate is on the point of moving or is moving.
B1 for friction can take any value up to the limiting value \(\mu R\), which is reached only when the crate is about to move (or moving)
Worked solution: The frictional force adjusts to prevent motion, but can never exceed \(\mu R = 78.4\) N. At 50 N the friction is not limiting.
Question 2Medium8 marks
A box of mass 6 kg rests on a rough plane inclined at \(25^\circ\) to the horizontal. The coefficient of friction between the box and the plane is 0.6. The box is modelled as a particle. A student writes: \(R = 6g\cos 25^\circ = 53.3\) N, so the friction on the box is \(F = \mu R = 0.6 \times 53.3 = 32.0\) N up the plane. Take \(g = 9.8\) m s−2.
(a) Show that the box remains at rest.[3]
(b) Explain the error in the student's statement, and find the actual frictional force on the box.[2]
(c) A force of magnitude \(P\) newtons, acting up the plane along a line of greatest slope, is applied to the box. Find the value of \(P\) for which the box is on the point of moving up the plane.[3]
Show the answer and mark scheme
(a)Answer: Component of weight down the plane \(= 6g\sin 25^\circ = 24.8\) N; maximum friction \(= 0.6 \times 6g\cos 25^\circ = 32.0\) N; \(24.8 \lt 32.0\).
M1 for the component of the weight down the plane: \(6g\sin 25^\circ = 24.8\) N
M1 for the limiting friction \(\mu R = 0.6 \times 6g\cos 25^\circ = 32.0\) N
A1 for comparing: \(24.8 \lt 32.0\), so friction can hold the box at rest
Worked solution: Down the plane, the weight has component \(6g\sin 25^\circ = 24.85\) N. The maximum friction is \(\mu R = 0.6 \times 6g\cos 25^\circ = 31.97\) N. Since \(24.85 \lt 31.97\), friction can balance the weight component, so the box stays at rest.
(b)Answer: 24.8 N up the plane (\(F = \mu R\) only when friction is limiting) N
B1 for \(F = \mu R\) only when the box is on the point of slipping (limiting friction); in general \(F \le \mu R\)
B1 for 24.8 N (accept 24.9 or 25) up the plane
Worked solution: \(\mu R\) is the maximum possible friction, reached only when the box is about to slip. Here the box is in equilibrium with friction less than the maximum, so resolving along the plane, \(F = 6g\sin 25^\circ = 24.8\) N up the plane.
(c)Answer: 56.8 N
M1 for friction now acting down the plane, with \(F = \mu R = 31.97\)
Worked solution: On the point of moving up the plane, friction is limiting and acts down the plane: \(P = 6g\sin 25^\circ + 0.6 \times 6g\cos 25^\circ = 24.85 + 31.97 = 56.8\) N.
Question 3Hard11 marks
A crate of mass 12 kg rests on a rough horizontal floor. The coefficient of friction between the crate and the floor is 0.2. The crate is modelled as a particle. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
(a) A force of magnitude \(P\) newtons, acting at an angle of \(30^\circ\) above the horizontal, pulls the crate, which is on the point of moving. Find the value of \(P\).[5]
[object Object]
(b) Instead, a force of magnitude \(Q\) newtons, acting at an angle of \(30^\circ\) below the horizontal, pushes the crate, which is on the point of moving. Find the value of \(Q\).[4]
[object Object]
(c) Explain why it is easier to move the crate by pulling it than by pushing it at the same angle.[2]
Show the answer and mark scheme
(a)Answer: \(P = 24.3\) (accept 24)
M1 for resolving vertically: \(R = 12g - P\sin 30^\circ\)
M1 for resolving horizontally with limiting friction: \(P\cos 30^\circ = 0.2R\)
(c)Answer: Pulling upwards reduces the normal reaction, and so reduces the maximum friction; pushing downwards increases the normal reaction and the friction, so a larger force is needed.
B1 for the vertical component of the pull reduces \(R\) (the push increases \(R\))
B1 for so the limiting friction \(\mu R\) is smaller when pulling, and a smaller force is needed
Worked solution: The upward component of the pull reduces the normal reaction, and hence the limiting friction \(\mu R\). The downward component of the push increases \(R\) and the friction, so \(Q \gt P\).