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physics

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A 22 kg box rests on the back of a truck. The
coefficient of static friction between the box
and the truck bed is 0.369.
The acceleration of gravity is 9.81 m/s2 .
What maximum acceleration can the truck
have before the box slides backward

  • physics - ,

    Wb = mg = 22kg * 9.8N/kg = 215.6N. =
    Weight of box.

    Fb = (215.6N,0 deg).

    Fp = Fh = 215.6sin(0)= 0 = Force paral-
    lel to plane = Hor. force.

    Fv = 215.6cos(0) = 215.6N. = Force perpendicular to plane = Normal.

    Ff = u*Fv = 0.369 * 215.6 = 79.56N. =
    Force due to friction.

    Fn = Fp - Ff = 22a,
    0 - 79.56 = 22a,
    a = -3.62m/s^2.

  • physics - ,

    Post it.

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