Question

In the arrangement shown in the figure mass of the block $B$ and $A$ are $2 \,m$, $8 \,m$ 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: (Coefficient of friction between A and B is $\mu$ and pulley is ideal)

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

Answer 1

Answer: (D)

$FBD$ of $A$

If the acceleration of ‘C’ is a
For block ‘A’ N = 8ma \ldots(1)
$8mg -\mu\,N=\mu\,8ma \ldots(2)$
and acceleration a can be written by the equation of system $(A+B + C)$
$m_{1}g=(10\,m+m_{1})a \ldots(3)$
$8\,mg=\mu \,8m\left(\frac{m_{1}g}{10m+m_{1}}\right)$
$10m+m_{1}=\mu\,m_{1}$
$10m=(\mu-1)m_{1}$
$\Rightarrow m_{1}=\frac{10m}{\mu-1}$