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Upward Acceleration of Hanging Mass Connected Pulley

  1. Nov 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A 27.1 kg block (m1) is on a horizontal surface, connected to a 6.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.097 m and a moment of inertia I=0.070 kgm2. A force F = 213.7 N acts on m1 at an angle theta = 34.5°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

    2. Relevant equations
    Torque = F x R
    Torque = I x alpha
    a = R x alpha
    I = 1/2 m x R^2

    3. The attempt at a solution
    Block 1 : Fcos(theta) - T1 = m1a T1 = Fcos(theta) - m1a (opposite direction of accereration)
    Block 2 : T2 - m2g = m2a T2 = m2a + m2g (in the direction of accleration)

    Torque = (T2-T1)R = I x (a/R)
    (m2a + m2g - Fcos(theta) + m1a)R = 1/2 m x R^2 x (a/R)
    m2a + m2g - Fcos(theta) + m1a = (m x R^2 x a)/(2 x R^2) R^2 cancel out
    m2a + m2g - Fcos(theta) + m1a = (m x a)/2
    m2a + m1a - (m x a)/2 = Fcos(theta) - m2g
    a (m2 + m1 - m/2) = Fcos(theta) - m2g
    a = (Fcos(theta) - m2g)/(m2 + m1 - m/2) -----------------> A
    a = (213.7cos(34.5) - 6.7*9.81)/ (6.7 + 27.1 - 17.88/2)
    a = (110.39/26.36)
    a= 4.187m/s^2

    I know that the answer is 2.68m/s^2 and that the equation A should be

    a = (Fcos(theta) - m2g)/(m2 + m1 + m/2)

    But i dont know how to get that...
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    If m2 is supposed to accelerate upwards then m1 must be accelerating leftwards, right? That means the pulley should be accelerating counter clockwise. That would make T1 > T2 if you want to maintain the assumption of m2 accelerating upwards. So, rearrange your T1 and T2 in the above and redo from there.
    Strange, that's not the value that I get. Let's see how you do and then we can compare notes :)
     
  4. Nov 21, 2014 #3
    When I did

    Torque = (T1-T2)R = I x (a/R)

    It worked out perfectly :) thank you
     
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