Distinguish static friction from its limiting value
Find the friction actually required for equilibrium, compare it with μR and use equality only when the particle is about to slip.
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
- Calculate normal reaction and required tangential balance.
- Solve linear inequalities with non-negative force magnitudes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does ‘about to slip’ mean in this model?
What you will learn
- Use the coefficient of friction as limiting friction divided by normal reaction.
- Choose F ≤ μR or F = μR according to the stated rest or limiting condition.
At rest friction adjusts to the required balance
In the supplied dry-friction model, the magnitude of static friction satisfies 0 ≤ F ≤ μR, where μ is the dimensionless coefficient of friction and R is normal reaction. At limiting friction, F = μR; this is the state described as ‘about to slip’ or limiting equilibrium. A particle merely stated to be at rest need not be at that limit. Find the friction needed from tangential balance first, then test it against the capacity.
For weight 50 N on a horizontal rough plane with μ = 0.4 and a 12 N horizontal pull, R = 50 N and the capacity is 20 N. Equilibrium requires F = 12 N opposite the pull, not 20 N. The spare capacity is 8 N. At zero horizontal pull the required friction is zero, despite the same non-zero capacity. Friction is not a constant extra force that always acts at maximum strength.
Use equality only at the limiting state
If a particle at limiting friction has R = 60 N and F = 18 N, then μ = 18/60 = 0.3. The coefficient has no units because it is a ratio of force magnitudes. For that same reaction, a later equilibrium requiring only 12 N friction is below the limit. A measured F/R at a non-limiting rest state does not in general equal μ; it provides only a lower bound on the coefficient needed to support it.
For weight 100 N with μ = 0.35 and a non-negative horizontal pull P, maintained horizontal-surface equilibrium requires P = F ≤ 35 N. Thus 0 ≤ P ≤ 35 N; P = 35 is the limiting case. If P exceeds this, the assumed rest state is impossible under the friction model. Describe the failure of equilibrium here; acceleration and moving-particle laws belong to the later Dynamics section.
Pause and explain
A resting particle needs 12 N friction while μR = 20 N. What is actual F?
Worked example
A particle of weight 50 N rests on a horizontal rough plane with μ = 0.4 under a horizontal pull of 12 N. Find actual friction, its limiting value and spare capacity.
Show the worked solution
- Vertical balance gives R = 50 N.
- Horizontal balance requires friction 12 N opposite the pull.
- The limiting value is μR = 20 N, so the rest state is possible and spare capacity is 8 N.
Answer Actual F = 12 N; limiting friction 20 N; spare capacity 8 N.
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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 particle of weight 80 N rests on a horizontal rough plane, μ = 0.25, under a 14 N horizontal pull. Find R, actual F, maximum F and spare capacity.
Give me a hint
Find the required balance before applying the capacity inequality.
Compare my reasoning
- The reaction is 80 N and actual friction is 14 N against the pull.
- Maximum friction is 0.25 × 80 = 20 N.
- Since 14 ≤ 20 the rest state is supported, with spare capacity 6 N.
R = 80 N; actual F = 14 N; maximum F = 20 N; spare capacity 6 N.
Look for these in your work
- Distinguish required friction and limiting friction.
- Check the inequality and remaining capacity.
A contact is at limiting friction with R = 60 N and F = 18 N. Find μ. For the same R and coefficient, what actual friction acts when a rest state requires 12 N tangential resistance?
Give me a hint
Use the limiting observation to determine μ, then use balance for the later state.
Compare my reasoning
- At the limit, μ = 18/60 = 0.3.
- The capacity at R = 60 remains 18 N.
- The later required 12 N is below capacity, so actual friction is 12 N rather than 18 N.
μ = 0.3; later actual friction 12 N, below the 18 N limit.
Look for these in your work
- Use equality only in the supplied limiting observation.
- Let actual static friction adjust in the later equilibrium.
A 100 N particle rests on a horizontal rough plane with μ = 0.35. A horizontal rightward pull P increases from zero. Find the full non-negative range permitting equilibrium and describe its endpoints.
Give me a hint
Reaction is unchanged by a purely horizontal pull.
Compare my reasoning
- R = 100 N, so friction capacity is 35 N.
- Equilibrium needs leftward F = P, giving 0 ≤ P ≤ 35 N.
- At P = 0 friction is zero; at P = 35 N it is limiting; above 35 the rest model fails.
0 ≤ P ≤ 35 N; zero friction at 0, limiting friction at 35 N.
Look for these in your work
- Use a force inequality for the whole range.
- Explain both endpoints without assuming all rest states are limiting.
Common mistakes
- Do not set F = μR merely because the particle is at rest.
- Use the normal reaction, which need not equal weight.
- A coefficient is dimensionless; a below-limit F/R gives a bound rather than its exact value.
What can you explain now?
At rest with no tangential applied force on a horizontal plane, must friction be μR?
Compare with the explanation
No: required friction is zero.
The surface’s capacity does not force friction to act when no tangential balance requires it.
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Make it stick
Find required friction, compare with μR and reserve equality for a stated or calculated limiting condition.
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