Question

In the arrangement shown in the figure, mass of the block $B$ and $A$ is $2m$ and $m$ respectively. Surface between $B$ and floor is smooth. The block $B$ is connected to the block $C$ by means of a string pulley system. If the whole system is released, then find the minimum value of mass of block $C$ so that $A$ remains stationary w.r.t. $B$. Co-efficient of friction between $A$ and $B$ is $\mu$.

• A. $\frac{m}{\mu }$
• B. $\frac{2m+1}{\mu +1}$
• C. $\frac{3m}{\mu -1}$
• D. $\frac{6m}{\mu +1}$

Answer 1

The correct option is C $\frac{3m}{\mu -1}$

Let the mass of ${}^{\prime }{C}^{\prime }$ be $M$ for ${}^{\prime }{A}^{\prime }$ to remain stationary,

During this condition, acceleration of system $a=\frac{Mg}{M+2m+m}$
$A$ is stationary w.r.t. to ${}^{\prime }{B}^{\prime }$

FBD of $A$
$\mu N=mg$
$\therefore \mu =\frac{mg}{ma}=\frac{g(M+3m)}{Mg}$
$\Rightarrow M\mu -M=3m\Rightarrow M=\frac{3m}{\mu -1}$