The diagram shows a particle A, of mass 1.2kg, which lies on a plane inclined at an angle of 40° to the horizontal and a particle B, of mass 1.6kg, which lies on a plane inclined at an angle of 50° to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, μ, between the particles and the planes [Figure 5.1]. Find the value of μ for which the system is in limiting equilibrium.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [*M1] Resolving parallel to the slope at A or B to form an equation. Correct number of terms; allow sin/cos mix.
2. [A1] 1.6g sin 50-T - FB = 0. If using the same Fs, then M1A1A0B1 max.
3. [A1] T-FA -1.2g sin 40 = 0. System equation (must be four different terms): 1.6g sin 50 - FB - FA −1.2g sin40=0 only scores M1A1A1. Any sign errors scores M1 only.
4. [*B1] R_A =1.2g cos 40 or R_B =1.6g cos 50. Either correct. Must be explicitly linked to the correct contact (so could be seen on a diagram), or as part of a resolving parallel to the slope equation(s) (so must be combined with μ).
5. [*M1] F_A =1.2gμ cos40 or F_B =1.6gμ cos50. Use of F = µR at either A or B. Must be explicitly linked to the correct contact (so could be seen on a diagram) or as part of a resolving parallel to the slope equation(s). Allow sin/cos mix error only.
6. [DM1] Eliminating T, F_A and F_B to form an equation in μ only. 1.6g sin50-1.6gμcos50=1.2g sin40+1.2gμcos 40
7. [A1] μ = (1.6g sin 50 - 1.2g sin 40) / (1.2g cos 40 + 1.6g cos 50) = 0.23326119... μ = 0.233