Question

In the arrangement shown in figure. There is a fricion force between the block of masses $m$ and $2m$. Block of mass $2m$ is kept on a smooth horizontal plane. The mass of suspended block is $m$. Block $A$ is stationary with respect to block of mass $2m$. The minimum value of coefficient of fricion betwen $A$ and $B$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{4}$

The FBDs of the block can be shown as below



Given, Block $A$ and $B$ are moving together and the surface is smooth.
Let coefficient of friction between $A$ and $B$ be $\mu$
Then, acceleration of the system is
$a_{\text{system}}=\frac{\left(\text{Supporting force-Opposing force}\right)}{\text{Total mass}}$
$\Rightarrow \frac{mg-0}{4m}=\frac{g}{4}$

For block $A$ not to slip on $B$, we have
$\Rightarrow f=m\left(a_{\text{system}}\right)$
$\Rightarrow \mu mg=m\left(\frac{g}{4}\right)$
$\Rightarrow \mu =\frac{1}{4}$