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 with respect to wedge will be: (where $g$ is acceleration due to gravity)

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

Answer 1

The correct option is B $\frac{m}{M}(M+m)g\tan \theta$

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

It is clear from the diagram horizontal component of force on the block
$F-N\sin \theta =ma\dots \left(i\right)$
It is clear from the diagram vertical component of force on the block
$N\cos \theta =mg\dots \left(ii\right)$
For wedge
$N\sin \theta =Ma\dots \left(iii\right)$
From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$\therefore F\Rightarrow \frac{m}{M}(M+m)g\tan \theta$

Final answer: (c)