Find the equilibrium range on a rough inclined plane
Resolve along a slope, test whether friction can prevent sliding and find force intervals with opposite friction directions at the two limiting endpoints.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.1. Paper 1 algebra and trigonometry are assumed. Bodies are modelled as particles: forces act at one point and rotational effects are not considered. Resolve numerically; a scale drawing is not a substitute for calculation. Force diagrams are qualitative unless stated otherwise. Where mass is supplied in this sequence, use the stated g = 10 m s⁻². Check non-negative tension and contact reaction and use F ≤ μR except at limiting friction. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Resolve weight in plane directions.
- Solve an absolute-value force-capacity inequality.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With no other normal force, what is R for weight W on an incline θ?
What you will learn
- Test a rough incline’s equilibrium using the required friction and μR.
- Find both limiting applied forces and explain the change in friction direction.
Without a parallel support force, friction must act uphill
A weight W on a plane inclined at θ has reaction R = Wcos θ when no other force has a normal component. Its downslope component is Wsin θ, so rest requires that much friction uphill. The capacity test is Wsin θ ≤ μWcos θ, or μ ≥ tan θ for 0 ≤ θ < 90°. Equality describes impending downward slip. A steeper angle increases the required ratio; the mass cancels from this specific particle model.
For W = 80 N, θ = 30° and μ = 0.75, required friction is 40 N uphill, while capacity is 0.75·40√3 = 30√3 N ≈ 52.0 N. Equilibrium is below the limit. The minimum coefficient would be tan30° = 1/√3. Do not replace the actual 40 N by its larger limiting value when balancing the particle.
An uphill force can require friction in either direction
Let P ≥ 0 act parallel uphill and take signed friction f positive uphill. Parallel balance gives f = Wsin θ − P, with |f| ≤ μR. Hence Wsin θ − μR ≤ P ≤ Wsin θ + μR, intersected with P ≥ 0. At the lower endpoint friction is uphill, opposing impending downward slip; at the upper it is downhill, opposing impending upward slip. At P = Wsin θ friction is zero.
For W = 60 N, θ = 30° and μ = 0.4, R = 30√3 N and capacity is 12√3 N. The range is 30 − 12√3 ≤ P ≤ 30 + 12√3 N, approximately 9.22 to 50.8 N. Zero applied force is outside this interval, so friction alone cannot hold the particle. Both endpoints are limiting but have opposite directions; assuming uphill friction everywhere would lose the upper threshold.
Pause and explain
At the greatest uphill P allowing equilibrium, which way does friction act?
Worked example
A 60 N particle rests on a rough 30° incline with μ = 0.4 under a non-negative parallel uphill force P. Find the equilibrium interval and friction directions at its endpoints.
Show the worked solution
- Normal balance gives R = 60cos30° = 30√3 N, so μR = 12√3 N.
- With uphill friction positive, f = 30 − P; require |30 − P| ≤ 12√3.
- Thus 30 − 12√3 ≤ P ≤ 30 + 12√3; friction is uphill at the lower and downhill at the upper endpoint.
Answer 30 − 12√3 ≤ P ≤ 30 + 12√3 N (about 9.22 to 50.8 N); uphill friction at lower, downhill at upper limit.
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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.
An 80 N particle rests without another force on a rough 30° incline, μ = 0.75. Find R, required friction, capacity and the minimum coefficient for rest.
Give me a hint
Balance the downslope component and compare it with the normal-based limit.
Compare my reasoning
- R = 80cos30° = 40√3 N; required friction is 40 N uphill.
- Capacity is 0.75R = 30√3 N ≈ 52.0 N, so the particle can rest below the limit.
- The minimum coefficient is (40)/(40√3) = 1/√3.
R = 40√3 N; actual F = 40 N uphill; capacity 30√3 N; minimum μ = 1/√3.
Look for these in your work
- Use actual friction rather than its capacity in balance.
- Check the minimum coefficient from the required force ratio.
A 100 N particle on a rough 30° incline, μ = 0.2, is supported by a non-negative parallel uphill tension T. Find its equilibrium range and the value for zero friction.
Give me a hint
Normal reaction is independent of a force parallel to the plane.
Compare my reasoning
- R = 50√3 N and μR = 10√3 N.
- Tangential balance needs signed friction f = 50 − T, so |50 − T| ≤ 10√3.
- Thus 50 − 10√3 ≤ T ≤ 50 + 10√3 N; friction is zero at T = 50 N.
50 − 10√3 ≤ T ≤ 50 + 10√3 N; zero friction at T = 50 N.
Look for these in your work
- Retain both limiting endpoints.
- Identify where the friction direction changes.
A 60 N particle is on a rough 45° incline with μ = 0.5. An uphill force P acts parallel to the plane. Find the equilibrium range and identify the impending motion and friction at both limits.
Give me a hint
The weight’s equal plane components are 30√2 N.
Compare my reasoning
- R = 30√2 N and capacity is 15√2 N.
- Use |30√2 − P| ≤ 15√2 to obtain 15√2 ≤ P ≤ 45√2 N.
- At the lower limit it is about to slip down with friction uphill; at the upper it is about to slip up with friction downhill.
15√2 ≤ P ≤ 45√2 N; lower: impending down, friction up; upper: impending up, friction down.
Look for these in your work
- Calculate the normal-dependent capacity.
- Explain both limiting directions from the tendency to slide.
Common mistakes
- Only use R = Wcos θ when other forces have no normal component.
- Static friction may act uphill, downhill or be zero depending on the applied force.
- Intersect a mathematical force interval with the permitted non-negative applied-force range.
What can you explain now?
At P = Wsin θ for a force parallel uphill, what friction is required?
Compare with the explanation
Zero.
The applied force already balances the downslope weight component.
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Make it stick
Use signed friction on a rough incline, derive the full equilibrium interval and explain its two limiting directions and zero-friction point.
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