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Question

A block of mass m on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block, the coefficient of friction between block and surface is

A
F1+F2sinθmg+F2cosθ
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B
F1cosθ+F2mgF2sinθ
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C
F1+F2cosθmg+F2sinθ
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D
F1sinθF2mgF2cosθ
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Solution

The correct option is A F1+F2sinθmg+F2cosθ
By the free diagram of the block,
and resolving the force F2 along the horizontal and the vertical components.
Here the reaction force R will be ,
R=mg+F2cosθ, f=μR, hence the frictional force we get as f=μ[mg+F2cosθ]

Also, for the equilibrium condition f=F1+F2sinθ

Equating the above equations we get,

μ[mg+F2cosθ]=F1+F2sinθ or μ=F1+F2sinθmg+F2cosθ

Hence option A is correct answer.

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