Question

In the arrangement shown in the figure mass of the block $B$ and $A$ are $2m,8m$ respectively. Surface between $B$ and floor is smooth. The block $B$ is connected to block $C$ by means of a pulley. If the whole system is released then the minimum value of mass of the block $C$ so that the block $A$ remains stationary with respect to $B$ is (Co-efficient of friction between $A$ and $B$ is $\mu$)

• A. $\frac{m}{\mu }$
• B. $\frac{2m}{\mu +1}$
• C. $\frac{10m}{1-\mu }$
• D. $\frac{10m}{\mu -1}$

Answer 1

The correct option is D $\frac{10m}{\mu -1}$


Let mass of block $C$ be $m_c$
$a=\frac{m_{c}g}{8m+2m+m_c}$
$a=\frac{m_{c}g}{10m+m_c}...........\left(1\right)$
FBD of $A$ in the rest frame of $B$ (non-inertial)

Since $A$ should remain stationary in this frame,
$8mg=f$
$N=8ma$
Minimum value of $m_c$ implies friction $f$ is limiting.
Thus, $f=f_L=\mu N$
$\therefore 8mg=\mu \left(8ma\right)$
$a=\frac{g}{\mu }$
From $\left(1\right)$

$\frac{m_{c}g}{10m+m_c}=\frac{g}{\mu }$

$\Rightarrow m_c=\frac{10m}{\mu -1}$