Question

In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is $\mu .$ Find the minimum horizontal force F required to keep the block stationary with respect to wedge.

• A. $\frac{(M+m)g}{\mu }$
• B. $\frac{Mg}{\mu }$
• C. $\frac{(M+m)g}{\mu +1}$
• D. None of these

Answer 1

The correct option is A $\frac{(M+m)g}{\mu }$

Let acceleration of the whole system be $a$.

Friction force acts on the block of mass $m$ in upward direction, wedge of mass $M$ applies a normal force $N$ on mass $m$ as shown in the figure.

Acceleration of block is $a$.

Thus normal force acting on it $N=ma$

Friction force balances the weight of the block of mass $m$ to keep it stationary w.r.t. wedge. So $f=mg$

Using $f=\mu N$ and $N=ma$ we get $\mu ma=mg$

$⟹$ $a=\frac{g}{\mu }$

Thus minimum horizontal force required $F=(M+m)a$

$⟹$ $F=\frac{(M+m)g}{\mu }$