Physics

A block P of mass 2.75 kg is lying on a rough inclined plane of angle è=0.87 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically. The coefficient of static friction between the block P and the inclined plane is ì=0.39.

Calculate the value of the mass of block Q, in kg, that is just sufficient to initiate the sliding motion of block P up the inclined plane.

Thanks!

  1. 2
asked by Anonymous
  1. (360Deg/6.28rad) * 0.87rad = 49.9 Deg.

    Wb = mg = 2.75kg * 9.8N/kg = 26.95 N. =
    Wt. of block P.

    Fb = 26.95N @ 49.9 Deg.
    Fp = 26.95*sin49.9 = 20.61 N. = Force
    parallel to incline.
    Fv = 26.95*cos49.9 = 17.36 N. = Force
    perpendicular to incline.

    Fn = Fb2 - Fp - Fs = 0.
    Fb2 - 20.61 - 0.39*17.36 = 0.
    Fb2 - 27.38 = 0.
    Fb2 = 27.38 N. = Force of block Q.

    mg = Fb2 = 27.38.
    m = 27.38/g = 27.38/9.8 = 2.79 kg = Mass 0f block Q.

    posted by Henry

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