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Question

# A block placed on a horizontal surface is being pushed by a force F making an angle θ with the vertical. The coefficient of friction between block and surface is μ. The force required to slide the block with uniform velocity on the floor is

A
μmg(sinθμcosθ)
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B
(sinθμcosθ)μmg
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C
μmg
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D
None of these
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Solution

## The correct option is A μmg(sinθ−μcosθ)The given condition can be reframed as shown below in the figure For vertical equilibrium of the block N=Fcosθ+mg......(1) As block will move with uniform velocity, the acceleration a will be zero. Hence, for horizontal motion Fsinθ−μN=0 ⇒ Fsinθ=μN....(2) From Eqs. (1) and (2) we get Fsinθ=μ(Fcosθ+mg) i.e., F=μmg(sinθ−μcosθ)

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