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 µ and pulley is ideal)
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FBD of A
If the acceleration of ‘C’ is a
For block ‘A’ N = 8 ma....
8 mg – µN = 0....
and acceleration a can be written by the equation of system (A + B + C)
m 1 g = (10 m + m 1 ) a....
8 mg = 
10 m + m 1 = μ m 1
10 m = ( μ – 1) m 1 ⇒ m 1 = 
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