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Posted by on Sunday, January 23, 2011 at 2:30pm.

A 19 kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 23.2 degrees the crate begins to slide downward. what is the coefficient of static friction between the crate and the ramp? At what angle does the crate begin to slide it its mass is doubled? I just need to know how to approach this problem because i'm lost.

  • Physics - , Monday, January 24, 2011 at 7:26pm

    Fc = 19kg * 9.8 = 186.2N @ 23.2Deg = Weight of crate.

    The Wt. of crate is broken down into 2
    components:

    Fp = 186.2*sin(23.2) = 73.35N parallel
    to plane downward.

    Fv=186.2*cos(23.2)=71.1N.perpendicular
    to the plane downward.

    Since the crate just beginsw to
    move at 23.2 deg, we can assume that
    Fp is = to the force of friction(uFv):

    uFv = Fp,
    171.1u = 73.35,
    u = 73.35 / 171.1 = 0.429 = coefficient
    of friction.

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