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## The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction

Question

The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.

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Physics
3 years
2021-08-23T09:26:33+00:00
2021-08-23T09:26:33+00:00 1 Answers
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## Answers ( )

Answer:F = 22.75 lb

μ₁ = 0.15

Explanation:The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,

F = μRF = μWwhere,

F = Smallest force needed to move dresser = ?

μ = coefficient of static friction = 0.25

W = Weight of dresser = 91 lb

Therefore,

F = (0.25)(91 lb)F = 22.75 lbNow, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

F = μ₁W₁where,

μ₁ = coefficient of static friction between shoes and floor

W₁ = Weight of man = 151 lb

Therefore,

22.75 lb = μ₁ (151 lb)μ₁ = 22.75 lb/151 lbμ₁ = 0.15