Question

A smooth inlclined plane, of mass m₁ , has a mass m₂ kept on it, as shown. The inclined plane is kept on a smooth horizontal surface. A light inextensible string and a light smooth pulley are used to connect it to a mass M hanging vertically. It is observed that when the mass M is ‘let-go’, the mass m₂ does not slip over the inclined plane. The relation between M, m₁ and m₂ , is, then

(1) \(M =\frac{m_1+m_2}{(cot\theta-1)}\)

(2) \(M=\frac{m_1+m_2}{(tan\theta-1)}\)

(3) \(M=\frac{m_1+m_2sin\theta}{(cot\theta-1)}\)

(4) M = \(\frac{m_1+m_2sin\theta}{(tan\theta-1)}\)

Answer 1

 (1) \(M =\frac{m_1+m_2}{(cot\theta-1)}\)

Let ‘a’ be the common acceleration of the mass M and the (staying intact) ‘combination’ of the inclined plane and the mass m₂ . The mass m₂ , by itself, can slide down the inclined plane with an acceleration g sin θ. It would not slip over the inclined plane if the component of the horizontal acceleration, a, of the ‘combination’, equals g sin θ . We thus need to have

a cosθ = g sin θ.....(1)

To find ‘a’ we look at the free body diagrams of (i) mass M and (ii) the ‘combination’ of masses m₁ and m₂ . We then have,

Putting this value of ‘a’, in the above equation (1), we get