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Question

# Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is:

A
Wtanθμ
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B
μWtanθ
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C
μW1+tanθ
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D
μWsinθ
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Solution

## The correct option is B μWtanθThe block B is under equilibrium by action of tension force on it rightwards(say T′), and force of static friction of it leftwards(f).Hence f=μN=μW=T′Consider the forces acting on the knot..Balancing the forces on knot horizontally,T′=TcosθAlso balancing the forces on knot vertically,Tsinθ=W⟹W=T′cosθsinθ=T′tanθ=μWtanθ

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