A loaded penguin sled weighing 80 N rests on a plane inclined at 휃=20°to the horizontal. Between the sled and the plane, the coefficient of

Question

A loaded penguin sled weighing 80 N rests on a plane inclined at 휃=20°to the horizontal. Between the sled and the plane, the coefficient of static friction is 0.25 and the coefficient of kinetic friction is 0.15.a.What is the least magnitude of force F, parallel to the plane, that will prevent the sled from slipping down theplane?b.What value of F is required to move the sled up the plane at a constant velocity?

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niczorrrr 1 week 2021-07-21T21:11:23+00:00 1 Answers 3 views 0

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    2021-07-21T21:12:58+00:00

    Answer:

    A) F = 8.57N

    B) F = 38.64N

    Explanation:

    A) The formula for least magnitude of force F, parallel to the plane, that will prevent the sled from slipping down the plane is given as;

    F = mgsinθ – μ_s•mgcosθ

    Where;

    mg is weight of Penguin = 80N

    θ is angle of the inclined plane = 20°

    μ_s is coefficient of static friction = 0.25

    Thus,F = mg(sinθ – μ_s•cosθ) = 80(sin20 – 0.25cos20)

    F = 80(0.342 – 0.2349)

    F = 8.57N

    B) The formula to calculate the value of F required to move the sled up the plane at a constant velocity is given as;

    F = mgsinθ + μ_k•mgcosθ

    F = mg(sinθ + μ_k•cosθ)

    Where;

    mg is weight of Penguin = 80N

    θ is angle of the inclined plane = 20°

    μ_k is coefficient of kinetic friction = 0.15

    Thus,

    F = mg(sinθ + μ_k•cosθ) = 80(sin20 – 0.15cos20)

    F = 80(0.342 + 0.141) = 38.64

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