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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Vân Khánh 5 years 2021-08-23T09:26:33+00:00 1 Answers 11 views 0

Answers ( )

  1. ngochoa
    0
    2021-08-23T09:27:39+00:00
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    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 = μR

    F = μW

    where,

    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 lb

    Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

    F = μ₁W₁

    where,

    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 lb

    μ₁ = 0.15

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