Tuesday
March 28, 2017

Post a New Question

Posted by on .

A mass of m1 = 12 kg is placed on an incline with an inclination angle of 50 degrees, a coefficient of kinetic friction of 0.16, and a coefficient of static friction of 0.89. m1 is tied to a rope which runs parallel to the incline and runs over a pulley to a hanging mass of m2. In order to move m1 up the incline, what is the minimum mass which m2 needs to be in kg?

please help me solve this problem :]

  • Physics. - ,

    W1 = mg = 12kig * 9.8N/kg = 117.6N. =
    Wt. of m1.

    F1 = {117.6N,50deg). = Force of mi.
    Fp = 117.6sin50 = 90.09N = Force parallel to incline.
    Fv = 117.6cos50 = 75.59N. = Force perpendicular to incline.

    Fs = u*Fv = 0.89 * 75.59 = 67.28N. =
    Force of static friction.
    Fk = 0.16 * 75.59 = 12.09N. = Force of
    kinetic friction.

    Fn = Fap - Fp -Fs = 0,
    Fap - 90.09 - 67.28 = 0,
    Fap - 157.37 = 0,
    Fap = 157.37N. = Min applied force

    Fap = mg = 157.37,
    m2 * 9.8 = 157.37,
    m2 = 16.1kg,min.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question