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physics

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A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 24° to the horizontal (Fig. 6-28). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.13. (a) What is the minimum magnitude of the force , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?

  • physics -

    Fs = (75N,24deg).
    Fp=75sin24 = 30.51N. = Force parallel to plane acting down the plane.
    Fv = 75cos24 = 68.52N. = Force perpen-
    dicular to plane.

    Ff = u*Fv = 0.25 * 68.52 = 17.13N. =
    Force of static friction.
    Fk = 0.13 * 68.52 = 8.91 = Force of
    kinetic friction.

    a. Fap - Fp - Fk = 0,
    Fap=Fp + Ff = 30.51 + 8.91 = 39.42N =
    Min Force applied.

    b. Fap - Fp - Ff = 0,
    Fap = Fp + Ff = 30.51 + 17.13 = 47.64N.
    = Min force applied.

    c. Fap = Fp + Fk = 30.51 + 8.91 = 39.42N.

  • physics -

    part a is wrong

  • physics -

    pt A: Fp - Ff. in this case, you would do 30.51 - 17.13 = 13.38

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