A block of mass 12kg is placed on a plane which is inclined at an angle of 24° to the horizontal. A light string, making an angle of 36° above a line of greatest slope, is attached to the block. The tension in the string is 65N (see diagram). The coefficient of friction between the block and plane is μ. The block is in limiting equilibrium and is on the point of sliding up the plane. [Figure 5.1]

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [*M1] Attempt at resolving parallel to the plane; 3 terms. Allow sign errors, sin/cos mix. Allow g missing, otherwise dimensionally correct.
2. [A1] 65 cos 36 = 12g × sin 24 + F F=3.777707......
3. [*M1] Attempt at resolving perpendicular to the plane; 3 terms. Allow sign errors, sin/cos mix. Allow g missing, otherwise dimensionally correct.
4. [A1] 12g x cos 24 = R + 65 sin 36 R=71.419......
5. [DM1] Use F = µR; To get an equation in μ only. Dependent on two previous M marks. Allow g missing
6. [A1] μ = (65 cos 36-12g x sin 24) / (12g x cos 24-65 sin 36) = 52.586-48.808 / 109.625-38.206 = 3.777... / 71.419...; μ = 0.0529; Allow AWRT 0.053; Do not accept fractional equivalent.