Question

In the arrangement shown in the figure mass of the block B and A is 2m and m respectively surface between B and the floor is smooth.The block 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 Cso that A remains stationary w.r.t. B. Coefficients of friction between A and B is emu

Answer 1

Given that,

Mass of block B = 2m

Mass of block A = 8m

We need to calculate the friction force on the block A should be sufficient to counter the weight of the block A

Using formula of frictional force

$F=\mu N$

$mg=\mu N$

Where, $\mu$ = coefficient of friction

N = normal force

So, The normal force will be

$N=\dfrac{\mu}{mg}$....(I)

We need to calculate the acceleration of block A

Using formula of normal force

$N=m_{a}a$

$a=\dfrac{N}{m_{a}}$

So, The acceleration of whole system is

$a=\dfrac{N}{m_{a}}$

The requires force will be

$N=(m_{a}+m_{b}+m_{c})a_{requires}$

We need to calculate the minimum value of mass of block C

Using formula of force

$F=(m_{a}+m_{b}+m_{c})a$

Put the value of a

$m_{c}g=(m_{c}+10m)\times\dfrac{N}{m_{a}}$

Put the value of N from equation (I)

$m_{c}g=(m_{c}+10m)\times\dfrac{2m\times g}{\mu}\times\dfrac{1}{2m}$

$m_{c}=\dfrac{10m}{\mu-1}$

Hence, The minimum value of mass of block C is $\dfrac{10m}{\mu-1}$