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 u and that between the block of mass M and the horizontal surface is u. What is 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 frictional force Acting on the upper block of $m$ should not be more than the Limiting friction $(={\mu }_{1}mg).$

For non-slipping condition

Force in upper block $f\le \mu mg$ (limiting friction)

$ma={\mu }_{1}mg$

$a={\mu }_{1}g.........\left(1\right)$

Friction force on mass $M$, is ${\mu }_2(M+m)g$

Let the system moves with Acceleration $a$ . Then for whole system:

$F-{\mu }_2(M+m)g=(M+m)a.........\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get

$F=(M+m)g({\mu }_1+{\mu }_2)$