Change friction direction on uphill and downhill motion
Use different Newton equations for the two sliding directions on a rough slope and test whether the particle restarts after reaching rest.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Resolve weight on an incline.
- Distinguish sliding friction from adjustable static friction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle slides up a rough slope. Which way does sliding friction act?
What you will learn
- Derive separate accelerations for uphill and downhill sliding.
- Check static equilibrium at a turning point before assuming a return.
Friction reverses with sliding direction
Use down the plane as positive throughout. With no other applied forces, R = mg cos θ. During uphill sliding, friction points down with gravity, so ma = mg sin θ + μmg cos θ. The downward acceleration is g(sin θ + μ cos θ). During downhill sliding, friction points up, giving a = g(sin θ − μ cos θ). These are separate force states; the same acceleration cannot be carried through a reversal on a rough plane.
For m = 5 kg, sin θ = 3/5, cos θ = 4/5 and μ = 1/4, reaction is 40 N and sliding friction is 10 N. During uphill motion the downward resultant is 30 + 10 = 40 N, giving downward a = 8. During downhill motion it is 30 − 10 = 20 N, giving downward a = 4. The reaction is unchanged, but friction’s direction changes.
Momentary rest needs a static test
In the same example, a particle projected up at speed 8 m s⁻¹ stops after 1 s and travels 4 m up the plane. At rest, the downslope weight component 30 N exceeds the maximum static friction 10 N, so it starts sliding down. Use the new downward acceleration 4 for the return: travelling 4 m from rest takes √2 s and reaches speed 4√2 m s⁻¹. Uphill and downhill travel times differ.
If μ cos θ is at least sin θ, static friction can balance gravity at rest. A 3 kg particle with the same slope and μ = 1 has R = 24 N and downslope weight 18 N, so after an uphill slide it can remain at rest with friction 18 N up the plane. Do not put maximum friction 24 N into a resting equation and claim it accelerates up spontaneously. The supplied static and sliding rules use the same μ here; real surfaces can require a more detailed model.
Pause and explain
At rest, if mg sin θ is less than μmg cos θ and no other force acts, what must be checked?
Worked example
A 5 kg particle is projected up a rough slope at 8 m s⁻¹. Take sin θ = 3/5, cos θ = 4/5 and μ = 1/4 for both static limit and sliding friction. Find the rise, test restart and find the return speed at its starting point.
Show the worked solution
- R = 40 N and friction is 10 N; uphill downward acceleration is (30 + 10)/5 = 8. Rise to rest is 8²/16 = 4 m.
- At rest, gravity’s 30 N component exceeds the 10 N static limit, so downhill sliding starts with a = (30 − 10)/5 = 4.
- Over the 4 m return, v² = 2 × 4 × 4 = 32, so return speed is 4√2 m s⁻¹.
Answer Rise 4 m; rest cannot persist; downhill return speed 4√2 m s⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 2 kg particle slides on a rough slope with sin θ = 3/5, cos θ = 4/5 and μ = 1/4. Find reaction and downward acceleration separately during uphill and downhill motion.
Give me a hint
Keep down positive, but change the friction sign between the two sliding directions.
Compare my reasoning
- R = 20(4/5) = 16 N and friction magnitude is 4 N; downslope weight is 12 N.
- During uphill sliding, 12 + 4 = 2a, so downward a = 8 m s⁻².
- During downhill sliding, 12 − 4 = 2a, so downward a = 4 m s⁻².
R = 16 N; downward acceleration 8 uphill and 4 downhill, in m s⁻².
Look for these in your work
- Resolve weight correctly.
- Reverse friction, retaining one common positive axis.
A 4 kg particle is projected up at 10 m s⁻¹ on a long rough slope with sin θ = 3/5, cos θ = 4/5 and μ = 1/2. Use the same coefficient for the static limit. Find its rise and its speed when it returns to the start, if it restarts.
Give me a hint
Test the rest state before using the downhill acceleration.
Compare my reasoning
- R = 32 N and friction is 16 N. Uphill downward acceleration is (24 + 16)/4 = 10, giving rise 100/20 = 5 m.
- At rest, 24 N downslope exceeds the 16 N static limit, so downhill motion begins with acceleration (24 − 16)/4 = 2.
- Return speed satisfies v² = 2 × 2 × 5 = 20, so it is 2√5 m s⁻¹.
Rise 5 m; restart occurs; downhill return speed 2√5 m s⁻¹.
Look for these in your work
- Use the two different accelerations.
- Confirm restart from a static inequality before computing return motion.
A 3 kg particle is projected up at 7 m s⁻¹ on a rough plane with sin θ = 3/5, cos θ = 4/5 and μ = 1 for both sliding and the static limit. Find its stopping distance and decide whether it must slide back.
Give me a hint
The friction force while moving differs from the friction required at rest.
Compare my reasoning
- R = 24 N, sliding friction is 24 N and downslope weight is 18 N. Uphill downward a = (18 + 24)/3 = 14.
- Time to rest is 7/14 = 0.5 s and rise is 49/28 = 1.75 m.
- At rest, friction 18 N up the plane balances gravity and lies below the 24 N limit, so it can remain at rest rather than slide back.
Stops after 0.5 s and 1.75 m; rest is possible with friction 18 N up the plane.
Look for these in your work
- Use sliding friction only until zero velocity.
- Use the smaller required static friction to check whether rest can persist.
Common mistakes
- Friction reverses with the motion direction.
- Do not continue the uphill acceleration on the downward leg.
- Test static balance at rest before assuming restart.
What can you explain now?
Must a particle slide down after stopping on any rough incline?
Compare with the explanation
No; static friction may be sufficient to hold it at rest.
Compare mg sin θ with the available limit μmg cos θ at the zero-velocity event.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Rebuild the force equation when motion reverses, and test the static state at zero velocity before continuing.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.