Question

In the arrangement shown mass of the block B and A are $2$m and $8$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:(Co-efficient 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

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

to keep block A at rest wrt to block B, friction force on the block A should be sufficient to counter the weight of the block A.

i.e. $m_{a}g=\mu .N$

where $\mu$ is the coefficient of friction and $N$ is normal force between block A and block B.

this gives $N_{required}=\frac{m_{a}g}{\mu }.$

for this much $N$, acceleration of block A in rightward direction is required to be $=\frac{N_{required}}{m_a}$

for this acceleration of whole system is to be $=\frac{N_{required}}{m_a}$

for this force required is $(m_a+m_b+m_c)a_{required}$

putting values of mass of block B and block A we get, $m_c.g=(m_c+10m)\frac{2m.g}{\mu }\times \frac{1}{2m}$

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

so answer to the question is D.