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Question

Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is μ. Find the acceleration of the system and the force by the rod on one of the blocks.

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Solution


From the free body diagram


R1 = M1g cos θ (1)
R2 = M2g cos θ (2)
T + M1g sin θ − M1a − μR1 = 0 (3)
T − M2g + M2a + μR2 = 0 (4)

From Equation (3),
T + M1g sin θ − M1 a − μM1g cos θ = 0 (5)

From Equation (4),
T − M2 g sin θ + M2 a + μM2 g cos θ = 0 (6)

From Equations (5) and (6),
g sin θ(M1 + M2) − a(M1 + M2) − μg cos θ(M1 + M2)
⇒ a(M1 + M2) = g sin θ(M1 + M2) = μg cos θ(M1 + M2)
⇒ a = g(sin θ − μ cos θ)a − g(sin θ − μ cos θ)
∴ The acceleration of the block (system) = g(sin θ − μcos θ)

The force exerted by the rod on one of the blocks is tension, T.
T = −M1g sin θ + M1a + μM1g cos θ
T = −M1g sin θ + M1(g sin θ − μg cos θ) + μM1g cos θ = 0

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