On a slope, resolve along and perpendicular to the plane. The weight splits into \(mg\sin\alpha\) down the slope and \(mg\cos\alpha\) into it, and on a rough plane friction always points against the motion, so going up and coming down give different accelerations.
Along the plane: resultant force \(= ma\). Perpendicular: \(R = mg\cos\alpha\) (plus or minus any other perpendicular components). Moving up a rough plane, friction acts down it, so \(mg\sin\alpha + \mu R\) slows the particle; moving down, friction acts up, so the driving force is \(mg\sin\alpha - \mu R\).
Fig. 1: a 2 kg particle moving UP a rough plane with sin α = 0.6 (α = 36.9°, drawn to scale). Friction acts down the plane, against the motion
Section 01
Smooth slopes.
On a smooth plane the only force along the slope is the weight component \(mg\sin\alpha\), so \(a = g\sin\alpha\) down the plane whichever way the particle is moving. The mass cancels.
Cambridge often gives the angle as \(\sin\alpha = 0.6\) (so \(\cos\alpha = 0.8\), \(\tan\alpha = 0.75\)) to keep the arithmetic exact. Use these values rather than finding \(\alpha\).
Resolve parallel and perpendicular to the plane: the acceleration is along the plane, so there is no acceleration perpendicular to it.
Worked example Released on a smooth plane
A particle is released from rest on a smooth plane inclined at \(30^\circ\) to the horizontal. Find its speed after it has slid 4 m and the time taken.
Down the plane: \(mg\sin 30^\circ = ma\), so \(a = 10 \times 0.5 = 5\ \text{m s}^{-2}\).
\(a = g\sin\alpha\), not \(g\cos\alpha\). Check with a limiting case: on a vertical "slope" (\(\alpha = 90^\circ\)) the particle falls freely with \(a = g\), and \(\sin 90^\circ = 1\).
Section 02
Rough slopes: up and down differ.
Friction always opposes the motion. A particle sliding down has friction up the plane; a particle moving up has friction down the plane, adding to the weight component. That is why the syllabus expects different accelerations for the two directions.
Worked example Sliding down
A particle of mass 2 kg slides down a rough plane inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.6\). The coefficient of friction is 0.25. Find its acceleration.
Perpendicular: \(R = 20\cos\alpha = 20 \times 0.8 = 16\) N; friction \(= 0.25 \times 16 = 4\) N up the plane.
Down the plane: \(20\sin\alpha - 4 = 2a\).
\(12 - 4 = 8 = 2a\).
Answer: \(a = 4\ \text{m s}^{-2}\) down the plane.
Worked example Projected up the plane (Fig. 1)
The same particle is projected up the plane from point A at \(10\ \text{m s}^{-1}\). Find how far up the plane it travels, and its speed when it returns to A.
Answer: 6.25 m; it returns at \(7.07\ \text{m s}^{-1}\), slower than it set off because friction did work both ways.
Your turn
A particle slides from rest down a rough plane inclined at \(25^\circ\) and reaches \(3\ \text{m s}^{-1}\) after travelling 3 m. Find the coefficient of friction.
\(9 = 2a(3)\), so \(a = 1.5\). Per unit mass: \(10\sin 25^\circ - 10\mu\cos 25^\circ = 1.5\), so \(4.226 - 9.063\mu = 1.5\) and \(\mu = 0.301\).
Examiner trap
After a particle stops at the top, do not assume it slides back. Compare the weight component \(mg\sin\alpha\) with the maximum friction \(\mu mg\cos\alpha\): it stays put if \(\mu \ge \tan\alpha\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
A particle slides down a smooth plane inclined at \(30^\circ\). What is its acceleration?
Quick check+10 XP
A 4 kg particle rests on a plane inclined at \(30^\circ\), with no other forces perpendicular to the plane. Find the normal reaction.
Quick check+10 XP
A particle is projected up a rough plane. Which way does friction act while it moves up?
Quick check+10 XP
A particle comes to rest on a rough plane inclined at \(\alpha\). When will it stay at rest?
Section 03
Extra forces on the slope.
With a pulling force, resolve it too. A force parallel to the plane leaves \(R = mg\cos\alpha\); a force at an angle to the plane changes \(R\), and so changes the friction.
Worked example Pulled up the plane
A particle of mass 5 kg is pulled up a rough plane inclined at \(20^\circ\) by a force of 50 N parallel to the plane. The coefficient of friction is 0.3. Find the acceleration.
\(R = 50\cos 20^\circ = 46.98\) N, so friction \(= 0.3 \times 46.98 = 14.10\) N down the plane.
Up the plane: \(50 - 50\sin 20^\circ - 14.10 = 5a\).
\(50 - 17.10 - 14.10 = 18.80 = 5a\).
Answer: \(a = 3.76\ \text{m s}^{-2}\)
Worked example A horizontal push
A particle of mass 3 kg is pushed up a rough plane inclined at \(30^\circ\) by a horizontal force of 40 N. The coefficient of friction is 0.2. Find its acceleration.
The horizontal force makes \(30^\circ\) with the plane and pushes into it.
Up the plane: \(40\cos 30^\circ - 30\sin 30^\circ - 9.196 = 3a\).
\(34.64 - 15 - 9.196 = 10.44 = 3a\).
Answer: \(a = 3.48\ \text{m s}^{-2}\)
Examiner trap
Using \(R = mg\cos\alpha\) when the applied force has a component perpendicular to the plane. In the second example that gives friction 5.20 N instead of 9.20 N.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
A 2 kg particle slides down a rough plane with \(\sin\alpha = 0.6\) and \(\mu = 0.25\). Find its acceleration.
Exam style+20 XP
The particle is instead projected up the same plane at \(10\ \text{m s}^{-1}\). How far up the plane does it travel?
Exam style+20 XP
It then slides back down. Find its speed when it returns to its starting point.
Exam style+20 XP
A 5 kg particle is pulled up a rough plane inclined at \(20^\circ\) (\(\mu = 0.3\)) by a force of 50 N parallel to the plane. Find the acceleration.
Skill unlocked
Slopes conquered.
This skill is now recorded on your profile. Keep the chain going.