A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start the crate movi

Question

A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start the crate moving with the weakest possible applied force, in what direction should the force be applied ?

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Thanh Thu 3 weeks 2021-08-26T18:43:49+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-26T18:44:54+00:00

    Answer:

    Explanation:

    The direction of force will be in upward direction making an angle of θ with the vertical .

    Reaction force R = mg – F cosθ

    Friction force = μR

    = .36 (mg – F cosθ )

    Horizontal component of applied force

    = F sinθ

    For equilibrium

    F sinθ = .36 (mg – F cosθ)

    F sinθ + .36 F cosθ =.36 mg

    F (sinθ + .36 cosθ) = .36 mg

    F R( cosδsinθ +sinδ cosθ) = .36 mg ( Rcosδ = 1 . Rsinδ= .36 )

    F R sin( θ+δ )  = . 36 mg

    F = .36 mg / Rsin( θ+δ )

    For minimum F , sin( θ+δ ) should be maximum

    sin( θ+ δ ) = sin 90

    θ+ δ  = 90

    Rsinδ / Rcosδ  = .36

    δ = 20⁰

    θ = 70⁰ Ans

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