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 the two blocks is ${\mu }_1$ and that between the block of mass $M$ and the 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 $(M+m)({\mu }_2+{\mu }_1)g$


The FBDs of the upper block and of the system (upper+lower block) are shown above. The force applied should be such that frictional force acting on the upper block $m$ should not be more than the limiting friction
i.e $f_1\le {\mu }_{1}N={\mu }_{1}mg$

Let the system move with a common acceleration $a$, then for the whole system,
$F-f=(M+m)a$ where $f={\mu }_2{N}^{\prime }={\mu }_2(M+m)g$
$\Rightarrow F-{\mu }_2(M+m)g=(M+m)a$
$\Rightarrow F=(M+m)(a+{\mu }_{2}g).....\left(1\right)$

For block of mass $m$,
$f_1=ma\le {\mu }_{1}mg$
$\Rightarrow a\le {\mu }_{1}g......\left(2\right)$
Therefore, from $\left(1\right)$ and $\left(2\right)$, we get
$F\le (M+m)({\mu }_1+{\mu }_2)g$
Option D is correct.