Q. In the arrangement shown in the figure,mass of the blocks B and A is 2m and m respectively.Surface between B and floor is smooth.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 with respect to B.Coefficient of friction between A and B is `mu.` A) `m/mu` B)`(2m + 1)/(mu + 1)` C)`(3m)/(mu - 1)` D)`(6m)/(mu + 1)`

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In the system of the three blocks shown in the figure, the block C is suspended by a string. The coefficient of friction between mass A and B is mu. If the mass of block C is M, there is a force equal `M*g` that pulls it downwards. This translates to the block B being accelerated by `(M*g)/(M+2m+m)` . The normal force acting on the block A, pushing it against block B `m*(M*g)/(M+3m)` . The force of friction is `mu*(m*(M*g)/(M+3m))` If the block does not move down, the force of friction is equal to the force with which block B is being pulled down that is `m*g`

Equating the two `mu*(m*(M*g)/(M+3m)) = m*g`

=> `mu*(M/(M+3m)) = 1`

=> `(M + 3m)/M = mu`

=> `1 + (3m)/M = mu`

=> `(3m)/M = mu - 1`

=> `M = (3m)/(mu - 1)`

The required mass of block C to stop block A from sliding down block B is `(3*m)/(mu-1)` . The correct answer is option C.

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