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: (Coefficient of friction between A and B is $\mu$ and pulley is ideal)

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

Answer 1

Answer: (D)

$FBD$ of $A$



If the acceleration of ‘C’ is a

For block ‘A’ N = 8ma \ldots(1)

$8mg-\mu N=\mu 8ma\dots (2)$

and acceleration a can be written by the equation of system $(A+B+C)$

$m_{1}g=(10m+m_1)a\dots (3)$

$8mg=\mu 8m\left(\frac{m_{1}g}{10m+m_1}\right)$

$10m+m_1=\mu m_1$

$10m=(\mu -1)m_1$

$\Rightarrow m_1=\frac{10m}{\mu -1}$