A-LevelMathematicsMechanics9709/41 · May/June 2017 · Q712 Marks

As shown in the diagram, a particle A of mass 0.8 kg lies on a plane inclined at an angle of 30° to the horizontal and a particle B of mass 1.2 kg lies on a plane inclined at an angle of 60° 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 parallel to lines of greatest slope of the respective planes. The particles are released from rest with both parts of the string taut. [Figure 7.1]

(i)Given that both planes are smooth, find the acceleration of A and the tension in the string.[6]
(ii)It is given instead that both planes are rough, with the same coefficient of friction, μ, for both particles. Find the value of μ for which the system is in limiting equilibrium.[6]

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[T – 0.8g sin 30 = 0.8a; 1.2g sin 60 – T = 1.2a; 1.2g sin 60 – 0.8g sin 30 = 2a]
For A: T – 4 = 0.8a
For B: 6√3 – T = 1.2a
System equation is 6√3 – 4 = 6.4 = 2a
Solve for a or T
a = 3√3 – 2 = 3.20 ms⁻²
T = (12/5)(1 + √3) = 6.56 N
RA = 0.8g cos 30 = 4√3
RB = 1.2g cos 60 = 6
FA = 4√3 μ and FB = 6μ
12 sin 60 – 6μ – T = 0 or T – 8 sin 30 – 4√3 μ = 0
System equation is 12 sin 60 – 8 sin 30 – 6μ – 4√3 μ = 0
Eliminate T and/or find μ
μ = (6√3 – 4) / (6 + 4√3) = 0.494

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