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Need help, mechanics problem with pulleys and tensions

  1. May 12, 2013 #1
    I've already tried this problem but yielded an impossible answer:

    1. The problem statement, all variables and given/known data

    We have a pulley with a mass-less rope attached to two masses hanging off each side of the pulley:
    -m1 (4kg), suspended 1m off the ground
    -m2 (2kg), on the ground

    The mass (M) of the pulley is 2.6 kg
    The rotational friction of the pulley is given as .51 Nm
    The radius of the pulley is .085

    How long will it take for the 4kg mass to hit the ground once the masses are released?

    2. Relevant equations

    Moment of inertia (I) = (1/2)(mass)(radius)2
    Torque (τ) = (moment of inertia)(angular acceleration)
    Angular acceleration (α) = acceleration of point on outside edge of pulley / radius

    3. The attempt at a solution

    I defined moving the pulley so that m1 falls as "positive" rotation and angular acceleration.

    Regarding mass one,
    Fnet = m1g + T1 = ma
    T1 = m1a - m1g

    For mass two,
    Fnet = - m2g + T2 = ma
    T2 = m2a + m2g

    τnet = I α
    τ(from tension) - τ(from friction) = ((1/2) M r2) (a / r)
    T1r - T2r - .51 = (1/2)(2.6)(r)(a)
    T1 - T2 - .51/r = (1/2)(2.6)(a)
    m1a - m1g - m2a - m2g = 1.3a
    a(m1 - m2 - 1.3) = m1g + m2g
    a(4 - 2 - 1.3) = 39.2 + 19.6
    a = 84

    An acceleration greater than the acceleration due to gravity would be impossible for mass one.

    Would any please check what I did wrong? Thank you!
     
  2. jcsd
  3. May 12, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Check the sign of the tension.
     
  4. May 12, 2013 #3
    -m1a + m1g - m2a - m2g = 1.3a
    a(-m1 - m2 - 1.3) = -m1g + m2g
    a(-4 - 2 - 1.3) = -39.2 + 19.6
    a = 2.68

    That was the problem. Thanks!
     
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