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Question

In the arrangement shown in the figure, mass of the block B and A is 2m 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 A remains stationary w.r.t. B. Co-efficient of friction between A and B is μ.



A
mμ
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B
2m+1μ+1
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C
3mμ1
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D
6mμ+1
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Solution

The correct option is A 3mμ1
Let the mass of C be M for A to remain stationary,

During this condition, acceleration of system a=MgM+2m+m
A is stationary w.r.t. to B

FBD of A
μN=mg
μ=mgma=g(M+3m)Mg
MμM=3mM=3mμ1

Physics

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