Question

In the arrangement shown in the figure, mass of the block B and A is $2$m 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 block A remains stationary w.r.t. B. Coefficient 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

Answer: (C)

$\frac{3m}{\mu -1}$

Let for acceleration $a$ block A is at rest w.r.t to block B

From FBD of block A you can conclude normal force will provide acdelration $a$ and in vertical direction it is in equilibrium

so $N=ma,f=mg$

$\mu N=mg$

$\mu ma=mg$

$a=g/\mu$

As block A and B both are moving with same acceleration so you can consider a single mass system

From FBD of block C and combined mass A and B you will get

$T=3ma,Mg-T=Ma$

$Mg-3ma=Ma$

$M=\frac{3m}{\mu -1}$