Question

A block of mass $m$ is placed on another block of mass $M$ which itself is lying on a horizontal surface. The coefficient of friction between two blocks is ${\mu }_1$ and that between the block of mass $M$ and horizontal surface is ${\mu }_2$. What maximum horizontal force can be applied to the lower block so that the two blocks move without separation?

• A. $(M+m)({\mu }_2-{\mu }_1)g$
• B. $(M-m)({\mu }_2-{\mu }_1)g$
• C. $(M-m)({\mu }_2+{\mu }_1)g$
• D. $(M+m)({\mu }_2+{\mu }_1)g$

Answer 1

The correct option is D $(M+m)({\mu }_2+{\mu }_1)g$

Here, the force applied should be such that force acting on the upper block of $m$ should not be more than the force of friction $(={\mu }_{1}mg)$ acting on it. Let the system moves with acceleration $a$. Then
$F-{\mu }_2(M+m)g=(M+m)a$ .. (1)
For block of mass $m:$
${\mu }_{1}mg=ma$ or $a={\mu }_{1}g$ ... (2)
From Eqs. (1) and (2), we get;
$F={\mu }_2(M+m)g+(M+m){\mu }_{1}g$
$=(M+m)g({\mu }_1+{\mu }_2)$
Hence, the correct answer is option (d).