Question

A block of mass $m$ lies on wedge of mass $M$, which lies on fixed horizontal surface. The wedge is free to move on the horizontal surface. A horizontal force of magnitude F is applied on block as shown, neglecting friction at all surfaces, the value of force F such that block has no relative motion w.e.t wedge will be: (where g is acceleration due to gravity)

• A. $(M+m) g\,tan\,\theta$
• B. $(M+m)g\,cot\,\theta$
• C. $\frac{m}{M}(M+m)g\,tan\,\theta$
• D. $\frac{m}{M}(M+m)g\,cot\,\theta$

Answer 1

Answer: (C)

For no relative motion between wedge and block, let the acceleration of both block and wedge be ‘a ’ towards left

From FBD of block ; $N\,cos\,\theta=mg \ldots(i)$
and $F-N \,sin\,\theta=ma \ldots(ii)$
From FBD of wedge; $N\,sin\,\theta=Ma \ldots(iii)$
Solving equation (i), (ii) and (iii), we get
$F=\frac{m}{M}(M+m)g\,tan\,\theta$