Accelerate a particle on a rough horizontal surface
Resolve an angled pull, calculate its effect on the normal reaction and use the stated sliding-friction model in the horizontal Newton equation.
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 forces using sine and cosine.
- Use the smooth and rough contact models and check motion direction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which normal reaction applies to an upward angled pull on a horizontal particle in contact?
What you will learn
- Use the perpendicular balance to find reaction before friction.
- Calculate horizontal acceleration with a parallel or angled drive.
The reaction can change when the drive is angled
For a particle constrained to a horizontal surface, vertical acceleration is zero while contact persists. A pull P at angle θ above the surface gives R + P sin θ − mg = 0. Thus R = mg − P sin θ; using R = mg would miss the lifting component. A downward angled push instead increases R. In either case check that the resulting normal force is non-negative before using the surface-contact model.
In our stated sliding model, friction has magnitude μR and opposes the current motion. For rightward motion under an upward-right pull, the horizontal equation is P cos θ − μR = ma. A 5 kg particle with P = 25 N, sin θ = 3/5, cos θ = 4/5 and μ = 1/5 has R = 50 − 15 = 35 N, friction 7 N and a = (20 − 7)/5 = 2.6 m s⁻² rightwards.
Use the force law appropriate to the current state
A simple horizontal drive with no vertical component leaves R = mg. For a moving 4 kg particle with forward force 18 N and μ = 0.2, friction is 8 N and a = (18 − 8)/4 = 2.5. A light rod can provide either a pulling or a pushing horizontal force; its direction must appear in the diagram rather than being decided from its name.
For a particle initially at rest, friction adjusts up to its limiting value instead of automatically equalling μR. A drive of 5 N on that 4 kg particle can be balanced by static friction 5 N, below the 8 N limit; the particle need not start moving. A negative acceleration during rightward sliding is valid, but after it stops, recheck static balance before extending the same sliding equation into leftward motion.
Pause and explain
A particle slides right while slowing. Which way does its sliding friction act?
Worked example
A 5 kg particle slides right under a 25 N pull with sin θ = 3/5 and cos θ = 4/5. The sliding coefficient is 1/5. Find reaction, friction and acceleration while contact and rightward motion persist.
Show the worked solution
- Vertical balance gives R = 50 − 25(3/5) = 35 N.
- Friction is (1/5) × 35 = 7 N leftwards; the pull’s horizontal component is 20 N.
- Therefore 20 − 7 = 5a, so a = 2.6 m s⁻² rightwards.
Answer R = 35 N; friction 7 N leftwards; a = +2.6 m s⁻².
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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 4 kg particle slides right on a rough horizontal surface under a horizontal 18 N drive. With μ = 0.2, find friction and acceleration.
Give me a hint
There is no vertical component in the drive, so first use R = mg.
Compare my reasoning
- Normal reaction is 40 N.
- Friction is 0.2 × 40 = 8 N leftwards.
- Horizontal balance gives 18 − 8 = 4a, so a = 2.5 m s⁻² rightwards.
Friction 8 N; a = +2.5 m s⁻².
Look for these in your work
- Use the perpendicular balance before the friction law.
- Subtract friction from the driving force.
A 5 kg particle slides right with an upward-right pull 25 N, sin θ = 3/5, cos θ = 4/5 and μ = 1/5. Calculate R and acceleration.
Give me a hint
Use the vertical component to reduce R, then resolve along the surface.
Compare my reasoning
- R = 50 − 25(3/5) = 35 N.
- Friction is 7 N leftwards and horizontal pull is 25(4/5) = 20 N.
- Hence 5a = 13, so a = 2.6 m s⁻² rightwards.
R = 35 N; a = +2.6 m s⁻².
Look for these in your work
- Do not assume reaction equals weight under this angled pull.
- Use μR with the calculated reaction.
A 4 kg particle is initially at rest on a horizontal surface with static-friction limit μR, μ = 0.2. A horizontal 5 N pull is applied. Decide whether it must move and give the friction and acceleration in a possible resting state.
Give me a hint
Compare the friction needed for rest with its maximum available value.
Compare my reasoning
- R = 40 N, so the friction limit is 8 N.
- Rest requires friction 5 N opposing the pull, which is below 8 N and is admissible.
- The resting state has resultant zero and a = 0; motion is not forced by these data.
Rest is possible: friction 5 N opposing the pull; a = 0.
Look for these in your work
- Use the static inequality rather than setting friction to its maximum.
- Test the claimed motion before choosing a sliding branch.
Common mistakes
- An angled force changes reaction and therefore friction.
- Friction opposes motion, not necessarily acceleration.
- Use a static-friction inequality for a resting particle.
What can you explain now?
Must a resting particle have friction equal to μR?
Compare with the explanation
No; static friction supplies the needed balance up to μR.
Equality describes the limiting state, not every resting contact.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Resolve perpendicular first, choose static or sliding friction, then use the along-surface resultant in ma.
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.