The minimum value of F such that the block is at rest is (tanθ>μ)
A
mg(cosθ−μsinθ)sinθ+μcosθ
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B
mg(cosθ+μsinθ)sinθ−μcosθ
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C
mg(sinθ+μcosθ)cosθ−μsinθ
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D
mg(sinθ−μcosθ)cosθ+μsinθ
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Solution
The correct option is Cmg(sinθ+μcosθ)cosθ−μsinθ N=Fsinθ+mgcosθ....(1)Fcosθ=f+mgsinθ=μ[fsinθ+mgcosθ]+mgsinθ=F[cosθ−μsinθ]=mg[μcosθ+sinθ]F=mg[μcosθ+sinθ]cosθ−μsinθHence,optionCiscorrectanswer.