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.
Section 4.1 · Forces and equilibrium
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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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.
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.