Calculate limiting pulls and pushes when reaction changes
Keep the applied force in the normal equation, solve the limiting friction balance and check contact and model restrictions.
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 angled forces on horizontal surfaces.
- Solve linear equations and inequalities with force-dependent reaction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is R under a rightward pull P at θ above the horizontal, with weight W?
What you will learn
- Solve coupled tangential and normal equations for an angled limiting pull or push.
- Check contact feasibility and recognise when an ideal model gives no finite force bound.
An upward pull reduces both reaction and friction capacity
A particle of weight W on a rough horizontal plane is pulled right at magnitude P and angle θ above the horizontal. For contact and vertical balance, R = W − Psin θ ≥ 0. Rest requires horizontal friction F = Pcos θ ≤ μR. At impending rightward slip, solve Pcos θ = μ(W − Psin θ), giving P = μW/(cos θ + μsin θ). The reaction must be calculated at that same P, not at zero pull.
For W = 100 N, μ = 0.5 and θ = 30°, P = 200/(2√3 + 1) ≈ 44.8 N. Then R = 100 − P/2 ≈ 77.6 N and F = μR ≈ 38.8 N, equal to Pcos30°. Contact loss would require P = 200 N for this angle, so impending slip occurs first. A computed threshold is valid only if it respects non-negative reaction and all stated force assumptions.
A downward push increases the capacity
For a rightward push at θ below the horizontal, R = W + Psin θ and the equilibrium condition becomes P(cos θ − μsin θ) ≤ μW. If cos θ − μsin θ > 0, the limiting value is P = μW/(cos θ − μsin θ). With the same W = 100 N, μ = 0.5 and θ = 30°, it is 200/(2√3 − 1) ≈ 81.2 N, greater than the upward-pull limit because the push increases reaction.
Check the coefficient before dividing an inequality. If cos θ − μsin θ ≤ 0, the stated ideal particle-contact model permits the balance for every non-negative P, rather than producing a positive finite slipping threshold. For W = 60 N, μ = 0.8 and θ = 60°, required friction is P/2 while capacity is 48 + (0.4√3)P, whose P coefficient exceeds 1/2. This is a consequence of the supplied model, not a claim that a real object or material tolerates unlimited loads; deformation and tipping are omitted.
Pause and explain
For a downward push, what must be checked before dividing by cos θ − μsin θ to find a force bound?
Worked example
A 100 N particle is on a rough horizontal plane with μ = 0.5. Find the limiting rightward pull at 30° above the horizontal and check R.
Show the worked solution
- Use R = 100 − P/2 and F = P√3/2.
- At the limit P√3/2 = (1/2)(100 − P/2), giving P = 200/(2√3 + 1).
- Thus P ≈ 44.8 N, R ≈ 77.6 N > 0 and F ≈ 38.8 N; the contact assumption is valid at impending slip.
Answer P = 200/(2√3 + 1) ≈ 44.8 N; R ≈ 77.6 N; F ≈ 38.8 N.
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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 on a rough horizontal plane has μ = 0.25. Find its limiting rightward pull P at 30° above the horizontal and the reaction at that value.
Give me a hint
Use the upward-pull reaction in the limiting equation.
Compare my reasoning
- R = 80 − P/2 and F = P√3/2.
- Set P√3/2 = (1/4)(80 − P/2), giving P = 160/(4√3 + 1).
- Thus P ≈ 20.2 N and R = 80 − P/2 ≈ 69.9 N, which is positive.
P = 160/(4√3 + 1) ≈ 20.2 N; R ≈ 69.9 N.
Look for these in your work
- Keep P in both component equations.
- Check the threshold’s reaction remains non-negative.
A 100 N particle on a rough horizontal plane has μ = 0.5. Find the limiting rightward push P at 30° below the horizontal and its reaction.
Give me a hint
The downward component adds to the weight in the normal equation.
Compare my reasoning
- R = 100 + P/2 and F = P√3/2.
- At the limit P√3/2 = (1/2)(100 + P/2), so P = 200/(2√3 − 1).
- Thus P ≈ 81.2 N and R ≈ 140.6 N; the larger reaction supports more friction than an upward pull.
P = 200/(2√3 − 1) ≈ 81.2 N; R ≈ 140.6 N.
Look for these in your work
- Use the push sign in the reaction equation.
- Verify the same P in both limiting component balances.
A 60 N particle is on a rough horizontal plane, μ = 0.8. A rightward push P ≥ 0 acts 60° below the horizontal. Does this ideal particle-contact model give a finite upper P for equilibrium? Show the inequality and state the modelling limit.
Give me a hint
Compare the coefficient of P in the required friction with the coefficient in its capacity.
Compare my reasoning
- R = 60 + (√3/2)P > 0, while required F = P/2.
- Capacity is μR = 48 + (0.4√3)P; because 0.4√3 > 0.5, F ≤ μR for every P ≥ 0.
- There is no finite upper P from this model; this does not establish unlimited real-load capacity because deformation and rotational effects are omitted.
No finite upper P within the stated model: P/2 ≤ 48 + (0.4√3)P for every P ≥ 0; real-load limits are not established.
Look for these in your work
- Check the inequality coefficient without dividing by a negative value incorrectly.
- State the ideal model’s conclusion and its physical limits separately.
Common mistakes
- Use reaction at the threshold, not reaction before the force is applied.
- Distinguish an upward pull from a downward push in the normal equation.
- Check the inequality coefficient and contact restriction before reporting a force bound.
What can you explain now?
Why does an angled downward push usually permit a greater limiting horizontal drive than an upward pull?
Compare with the explanation
It increases R and therefore μR, whereas the upward pull decreases them.
The friction capacity changes together with the applied force’s vertical component.
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Make it stick
Resolve an angled pull or push, solve its reaction-dependent friction threshold and check coefficient signs, contact and model limitations.
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.