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+F2mg−F2sinθ
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C
F1+F2cosθmg+F2sinθ
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D
F1sinθ−F2mg−F2cosθ
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Solution
The correct option is AF1+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]