Questions on friction and inclined planes

Question 1

State the friction law F ≤ μR, explaining when the inequality is an equality.

Accepted Answer

Friction matches whatever force is trying to slide the surfaces past each other, up to a ceiling of μR. Equality holds only at that ceiling, which happens in limiting equilibrium, on the point of slipping, or while the object is actually sliding. Below the ceiling friction is exactly as large as it needs to be and no larger. B1 for the inequality, B1 for the conditions giving equality.

Question 2

A 6 kg box rests on a rough horizontal floor, μ = 0.4. A horizontal force of 20 N is applied. Show that the box does not move, and state the friction force acting.

Accepted Answer

Vertically, R = mg = 58.8 N, so friction can supply up to μR = 0.4 × 58.8 = 23.52 N. The applied 20 N is below that ceiling, so the box stays put and friction takes the value 20 N, exactly enough to balance. A 'show that' demands the comparison in writing, so state that 20 < 23.52 before concluding. M1 for R = mg, A1 for μR = 23.52 N, M1 for the comparison, A1 for the friction being 20 N.

Question 3

A block is pushed up a rough slope by a force acting along the slope. The force is then removed and, after coming to rest, the block slides back down. State the direction in which friction acts on the block during each stage, and explain why.

Accepted Answer

While the block moves up the slope, friction acts down the slope; while it slides back down, friction acts up the slope. B1 B1, one for each direction with its reason: friction opposes the sliding that is happening, so its direction is fixed by the block's motion relative to the surface and reverses when that motion reverses. Fixing the direction before writing an equation is what sets the sign of the friction term, which is why the deceleration going up and the acceleration coming down have different magnitudes.

Question 4

A block of mass 4 kg rests in equilibrium on a rough plane inclined at 25° to the horizontal. Taking g = 9.8 m s⁻², find the normal reaction and the friction force, and hence find the magnitude of the total contact force exerted by the plane on the block.

Accepted Answer

Perpendicular to the plane, R = 4g cos 25° = 35.5 N. Along the plane, friction balances the weight component: F = 4g sin 25° = 16.6 N, acting up the slope. The contact force is the resultant of these two perpendicular components: √(35.5² + 16.6²) = 39.2 N, which is exactly the weight 4g. M1 for resolving perpendicular and along the plane, A1 for R = 35.5 N and F = 16.6 N, A1 for 39.2 N. That it equals the weight is no coincidence: the only two forces on the block are its weight and the contact force, and in equilibrium they must be equal and opposite.

Question 5

A 6 kg box rests on a rough horizontal floor, μ = 0.4, and the horizontal force applied to it is increased to 30 N. Find the acceleration.

Accepted Answer

30 N beats the ceiling of 23.52 N, so the box slides and friction locks at its maximum value. The resultant is 30 − 23.52 = 6.48 N, so a = 6.48/6 = 1.08 m s⁻². M1 for the resultant, A1 for 6.48 N, A1 for the acceleration.

Question 6

A block is released on a smooth slope inclined at 30°. Find its acceleration, and explain why the mass does not matter.

Accepted Answer

Resolving along the slope, the only force component is mg sin 30°, so ma = mg sin 30° and a = g sin 30° = 4.9 m s⁻². The mass multiplies both the driving force and the inertia, so it cancels. M1 for resolving along the slope, A1 for the acceleration, B1 for the mass cancelling.

Question 7

A 3 kg block sits on a rough slope at 20°, μ = 0.5. Show that it remains at rest.

Accepted Answer

Down the slope the pull is mg sin 20° = 3 × 9.8 × sin 20° = 10.1 N. Perpendicular to the slope, R = mg cos 20° = 27.6 N, so friction can supply up to μR = 0.5 × 27.6 = 13.8 N. The ceiling of 13.8 N beats the pull of 10.1 N, so equilibrium is possible and the block stays at rest with friction actually providing 10.1 N up the slope. On a slope R = mg cos θ and never mg, so writing R = 29.4 N carries a wrong normal reaction into everything after it. M1 for mg sin 20°, A1 for the friction ceiling of 13.8 N, M1 for the comparison, A1 for the conclusion.

Question 8

A block of mass 10 kg rests on a rough horizontal floor, μ = 0.35. A force of magnitude P N is applied to the block in a direction 25° above the horizontal. Find the value of P for which the block is on the point of sliding.

Accepted Answer

The applied force has a vertical component, so the normal reaction is no longer the weight. Resolving vertically, R + P sin 25° = 10g, so R = 98 − P sin 25°. The friction ceiling is μR = 0.35(98 − P sin 25°), and it falls as P rises. On the point of sliding the friction is at that ceiling and the block is still in equilibrium, so resolving horizontally, P cos 25° = 0.35(98 − P sin 25°). Collecting the terms in P: P(cos 25° + 0.35 sin 25°) = 34.3, so P = 34.3/1.0542 = 32.5 N. M1 for resolving vertically with the vertical component of the applied force, A1 for R = 98 − P sin 25°, M1 for resolving horizontally with friction at its maximum, dM1 for collecting the terms in P and solving, A1 for 32.5 N. Pulled horizontally the block would need the full 34.3 N, so tilting the pull upwards helps: it lifts some of the weight off the floor and lightens the friction, and the gain outweighs the loss of horizontal component at this angle. Writing R = 98 N ignores the vertical component of P, in the same way that R = mg ignores the slope.

Question 9

A block rests on a rough plane whose inclination to the horizontal is slowly increased from zero. The block first begins to slide when the angle reaches 21°. Find the coefficient of friction between the block and the plane, and hence find the acceleration of the block when the plane is held at 30°.

Accepted Answer

At the instant it begins to slide the block is in limiting equilibrium, so friction is at its ceiling. Resolving along and perpendicular to the plane, mg sin 21° = μR and R = mg cos 21°, so mg sin 21° = μ mg cos 21° and μ = tan 21° = 0.384. The mass cancels, which is why the question can be answered without it: the angle at which any block begins to slide depends only on μ. That angle is the angle of friction, and tan of it is the coefficient. At 30° the plane is steeper than the angle of friction, so the block slides, friction acts up the plane at its maximum, and ma = mg sin 30° − μ mg cos 30°. Then a = 9.8(sin 30° − 0.384 cos 30°) = 9.8(0.5 − 0.3324) = 1.64 m s⁻². M1 for limiting equilibrium at the critical angle with friction at μR, A1 for μ = tan 21°, A1 for 0.384, M1 for Newton's second law along the plane at 30°, A1 for 1.64 m s⁻². A negative answer at the second stage would mean the plane was not steep enough to start the block, so the sign is a check on the first stage as well as an answer.

Question 10

A block of mass 10 kg rests on a rough horizontal floor, coefficient of friction 0.3. A force of 60 N is applied to the block at 30° below the horizontal, pushing it along the floor. Taking g = 9.8 m s⁻², find the normal reaction and the acceleration of the block.

Accepted Answer

Fix the direction of friction first: the block moves in the direction of the push, so friction acts against it. Vertically the push has a downward component: R = 10g + 60 sin 30° = 98 + 30 = 128 N. Friction is then 0.3 × 128 = 38.4 N. Horizontally: 60 cos 30° − 38.4 = 51.96 − 38.4 = 13.56 N, so a = 13.56/10 = 1.36 m s⁻². M1 for resolving vertically with the downward component of the push, A1 for R = 128 N, A1 for friction 38.4 N, M1 for the equation of motion along the floor, A1 for 1.36. Pushing downwards increases R and therefore the friction, the opposite effect to pulling upwards; the same 60 N applied at 30° above the horizontal would give R = 68 N and friction 20.4 N.

Question 11

A particle A of mass 3 kg lies on a rough plane inclined at 30° to the horizontal, coefficient of friction 0.2. A light inextensible string attached to A runs up the line of greatest slope, over a smooth pulley at the top of the plane, to a particle B of mass 4 kg hanging freely. The system is released from rest. Taking g = 9.8 m s⁻², find the acceleration of the system and the tension in the string.

Accepted Answer

Decide the direction first. B is pulled down by 4g = 39.2 N. A is pulled down the slope by 3g sin 30° = 14.7 N, and even with the largest friction, 0.2 × 3g cos 30° = 0.2 × 25.46 = 5.09 N, added to that, 19.8 N is less than 39.2 N. So B descends, A moves up the slope, and friction acts down the slope on A. For B, down positive: 39.2 − T = 4a. For A, up the slope positive: T − 14.7 − 5.09 = 3a. Adding: 39.2 − 14.7 − 5.09 = 7a, so a = 19.41/7 = 2.77 m s⁻². Then T = 4(9.8 − 2.77) = 28.1 N. B1 for the direction of motion, with friction placed down the slope, M1 for R = 3g cos 30° and friction 5.09 N, M1 for the two equations of motion, A1 for 2.77, A1 for 28.1 N. Check from A: 3 × 2.77 + 14.7 + 5.09 = 28.1 N. The tension lies between the down-slope pull on A and B's weight, as it must while A accelerates up and B accelerates down.

Question 12

A block of mass 2 kg is projected up a rough slope inclined at 25° with an initial speed of 6 m s⁻¹. The coefficient of friction between block and slope is 0.3. Find the distance the block travels up the slope, and determine whether it then slides back down.

Accepted Answer

Going up, gravity and friction both act down the slope, so ma = mg sin 25° + μmg cos 25° and a = 9.8(sin 25° + 0.3 cos 25°) = 6.81 m s⁻² of deceleration. The mass cancels. Then 0 = 6² − 2 × 6.81 × s gives s = 2.64 m. At the top the block is at rest, so ask whether the down-slope pull can beat the friction ceiling: mg sin 25° = 8.28 N against μmg cos 25° = 5.33 N. It can, so the block slides back. The same test written as an angle is tan 25° = 0.466 > 0.3 = μ, which is the standard criterion for slipping on a rough slope. Coming down, friction reverses and acts up the slope, so the return acceleration is 9.8(sin 25° − 0.3 cos 25°) = 1.48 m s⁻², much gentler than the climb. M1 A1 for the deceleration up the slope, M1 A1 for the distance, M1 A1 for the comparison that settles the slide-back, B1 for the return acceleration.

Question 13

A block of mass 5 kg rests on a rough plane inclined at 30° to the horizontal, the coefficient of friction between block and plane being 0.25. A force of magnitude P N acts on the block up a line of greatest slope. Find the range of values of P for which the block remains in equilibrium.

Accepted Answer

Perpendicular to the plane nothing moves and the applied force has no component there, so R = 5g cos 30° = 42.4 N and the friction ceiling is μR = 0.25 × 42.4 = 10.6 N. Down the plane the weight pulls with 5g sin 30° = 24.5 N. Since 24.5 beats 10.6, the block cannot be left to itself, so P has a least value as well as a greatest one, and friction points a different way at each end of the range.
At the bottom of the range the block is on the point of sliding down, so friction acts up the plane at its full 10.6 N and helps P: P + 10.6 = 24.5, giving P = 13.9 N.
At the top of the range the block is on the point of being dragged up, so friction reverses and acts down the plane against P: P = 24.5 + 10.6 = 35.1 N.
So the block stays in equilibrium for 13.9 ≤ P ≤ 35.1.
B1 for R = 5g cos 30°, M1 for the friction ceiling, A1 for 10.6 N, M1 for the limiting case with the block about to slide down and friction acting up the plane, A1 for 13.9 N, M1 for the limiting case with the block about to be dragged up and friction reversed, A1 for 35.1 N, B1 for presenting the answer as a range.

Question 14

A crate rests on the flat horizontal floor of a lorry. The coefficient of friction between the crate and the floor is 0.4. The lorry accelerates along a straight level road. Explain which way friction acts on the crate while it does not slip, and find the greatest acceleration of the lorry for which the crate does not slip, taking g = 9.8 m s⁻². State why the answer does not depend on the mass of the crate.

Accepted Answer

The crate accelerates forwards with the lorry, so it needs a forward resultant, and the only horizontal force on it is friction from the floor: friction acts forwards on the crate, in the direction of the lorry's acceleration, because it opposes the sliding that would otherwise happen, the crate slipping backwards along the floor. Vertically R = mg. Along the road F = ma, and F ≤ μR, so ma ≤ 0.4mg and a ≤ 0.4 × 9.8 = 3.92 m s⁻². B1 for friction acting forwards with the reason, M1 for R = mg with F = ma along the road, M1 for F ≤ μR, A1 for 3.92, B1 for the mass cancelling from ma ≤ μmg, so that any crate slips at the same acceleration.

Question 15

A particle A of mass 5 kg lies on a rough plane inclined at 30° to the horizontal, coefficient of friction 0.3. A light inextensible string attached to A runs up the line of greatest slope, over a smooth pulley at the top of the plane, to a particle B of mass m kg hanging freely. Taking g = 9.8 m s⁻², find the range of values of m for which the system can remain at rest.

Accepted Answer

Perpendicular to the plane, R = 5g cos 30° = 42.4 N, so the friction limit is 0.3 × 42.4 = 12.7 N. Down the slope the weight component of A is 5g sin 30° = 24.5 N, and while the system is at rest the string pulls A up the slope with T = mg. Friction can act either way, so there are two limiting cases. A about to slide down, friction up the slope: mg + 12.7 = 24.5, so m = 11.8/9.8 = 1.20. A about to slide up, friction down the slope: mg = 24.5 + 12.7 = 37.2, so m = 3.80. Between the two, friction is below its limit and takes whatever value balances A, so the system can rest for 1.20 ≤ m ≤ 3.80. M1 for R and the friction limit, A1 for 12.7 N, M1 for the slide-down case with friction up the slope, A1 for m = 1.20, M1 for the slide-up case with friction reversed, A1 for m = 3.80 with the range. In one line the condition is |mg − 5g sin 30°| ≤ 0.3 × 5g cos 30°, and the modulus is the two cases. Testing only whether mg exceeds 24.5 + 12.7 misses the lower bound, where A slides down because B is too light to hold it.

Question 16

The diagram shows a particle P held in equilibrium on a rough inclined plane by a light inextensible string. The four forces acting on P are its weight, the normal reaction R, the friction F and the tension T in the string, and the string lies in the vertical plane containing the line of greatest slope. The coefficient of friction between P and the plane is 0.3, and P is on the point of moving up the plane. Taking g = 9.8 m s⁻², use the diagram to find the tension in the string.

Accepted Answer

Read the diagram first: the plane is at 30° to the horizontal, the string is at 20° to the plane, and the particle has mass 5 kg, so its weight is 5g = 49 N. Fix the direction of the friction before writing anything: P is on the point of moving up the plane, so the friction acts down the plane, and because the equilibrium is limiting it is at its ceiling, F = 0.3R.
The string is not along the line of greatest slope, so it has a component perpendicular to the plane as well as one along it, and that perpendicular component lifts part of the weight off the plane. Resolving perpendicular to the plane: R + T sin 20° = 5g cos 30° = 42.4 N. M1 for resolving perpendicular to the plane with the perpendicular component of T included, A1 for R = 42.4 − T sin 20°.
Resolving along the plane, with the friction down the slope beside the weight component: T cos 20° = 5g sin 30° + 0.3R = 24.5 + 0.3(42.4 − T sin 20°). M1 for resolving along the plane with the friction at 0.3R down the slope, A1 for the equation in T alone.
Collecting the terms in T: T(cos 20° + 0.3 sin 20°) = 24.5 + 12.73, that is 1.0423T = 37.23. dM1 for collecting the terms in T and solving. So T = 37.23/1.0423 = 35.7 N. A1 for 35.7 N.
Back-substituting, R = 42.44 − 35.72 sin 20° = 30.2 N and the friction is 0.3 × 30.2 = 9.07 N, and along the plane 35.72 cos 20° = 33.6 N against 24.5 + 9.07 = 33.6 N, which checks. Writing R = 5g cos 30° and stopping there ignores the string's perpendicular pull and makes the friction too large. The tilt of the string matters: a string pulling straight up the line of greatest slope would need 24.5 + 0.3 × 42.4 = 37.2 N, more than the 35.7 N found here, because lightening the normal reaction saves more friction than the tilt costs in useful component.