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प्रश्न
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.
उत्तर
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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