A particle P of mass 6 kg lies on a rough plane inclined at an acute angle θ to the horizontal. A horizontal force of magnitude 12N acts on P as shown in the diagram. The line of action of this horizontal force lies in a vertical plane which contains the line of greatest slope of the plane that passes through P. The coefficient of friction between P and the plane is μ. P is in equilibrium and on the point of sliding down the plane. [Figure 7.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] For resolving in any direction to form an equation. Correct number of relevant dimensionally correct terms; allow sin/cos mix; allow sign errors. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved.
2. [1] Perpendicular to plane: R=6gcosθ+12sinθ Or 12+Fcosθ=Rsinθ (resolving horizontally).
3. [1] Parallel to plane: F+12cosθ=6gsinθ or 6gsinθ-12cosθ = μ(...). Or Rcosθ+ F sinθ=6g (resolving vertically).
4. [1] For use of F = μR to get an equation in μ and θ only. From dimensionally correct equations with the correct number of relevant terms (forces that need resolving should be resolved, forces that do not need resolving should not).
5. [1] AG-CWO must give some indication of how the equation in sin/cos leads to the given answer e.g. 6g sinθ-12cosθ then ‘divide through by cos' gives 6g cosθ+12sinθ scores A1 (so therefore mentioning 'dividing by cos'). Seeing μ = (5tanθ-1)/(5+tanθ) only is A0.
6. [1] μ > 0 ⇒ tan θ > 1/5. Set the numerator of μ either =, > or >= to 0 and obtain tank where □ is =, > or >= and k≠0 - ignore any consideration of an upper bound (e.g. saying that μ <1).
7. [1] 11.3 < θ (<90) AWRT 11.3 (11.30993...). Must be strict inequality. Allow θ > 0.197(39...) - if anything other than < 90 then A0.