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Calculating Tension in Cord in 2-Pulley System

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    I attached a picture of the given figure to this thread. The suspended 2.5kg mass is moving up, the 2.9kg mass on the ramp is sliding down, and the 9kg mass is moving down. It is given that there is friction between the block on the ramp and the ramp itself with a coefficient of friction = 0.11. The pulleys are massless and frictionless.

    What is the tension in the cord connected to the 9kg block?

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    So, I drew three free-body diagrams for the three separate masses. I'll call the furthermost left mass m1, the middle block on the ramp m2, and the rightmost mass m3.

    The forces acting on m1 are the force of gravity and the tension in the far left cord (I'll call this tension T1, and m1's overall acceleration a1). Therefore, since m1 is falling:

    Sum(forces) = m1g - T1 = m1a1

    For m2, in the x direction the forces are the force of friction in opposite direction to the motion, the tension in cord T1, the tension in the other cord T2, and the x-component of gravity, m1gsin(33). In the y direction, the forces are the normal force and the y-component of gravity. Therefore:

    Sum(x forces) = T2 + Ffriction - T1 - m2gsin(33) = m2(a2 - a1)
    Sum(y forces) = n - mgcos(33) = 0

    The free-body diagram for m3 is the same as m1, but the object is moving in opposite direction so:

    Sum(forces) = T2 - m3g = m3a2

    I've reached this point and I believe my equations are right (except possibly the sum of x forces for m2), but I can't seem to figure how to combine these equations to determine T1.

    Any help is greatly appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Sep 13, 2010 #2

    Doc Al

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    Staff: Mentor

    (1) Since m2 is moving down the incline, why not choose that as your positive direction?
    (2) Since the masses are connected by cords, how must their accelerations relate?
    (3) How can you express Ffriction in terms of other quantities?

    Your analysis of m1 and m3 is good.
  4. Sep 13, 2010 #3
    (1) Yeah, that makes a lot more sense.
    (2) I figured there was no related between the acceleration of the two cords since they were not on the same pulley. Are they equal but opposite (a1 = -a2)?
    (3) I actually meant to fill that in. Following from friction force = u x n,

    Ffriction = (0.11)m2gcos(33) = 2.6219

    Therefore, using the equation for the x forces on m2:

    T1 + m2gsin(33) - T2 - 2.6219 = 2(m2)(a)
    T1 - T2 = 5.8a - 12.8567

    Solving the equations for m1 and m3, we find:

    T1 = 88.2 - 9a
    T2 = 24.5 - 2.5a

    T1 - T2 = 63.7 + 11.5a

    Equating this equation with the previous:

    11.5a + 63.7 = 5.8a - 12.8567
    5.7a = 76.5567
    a = 13.431

    Plugging this into the equation for T1,

    T1 = 88.2 - 9(13.431) = 32.679

    But that answer is definitely wrong. I feel kinda lost, can someone give me some guidance on where I need to go with this problem? I been trying for a couple of days hahha.
  5. Sep 14, 2010 #4

    Doc Al

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    Staff: Mentor

    The three masses are connected, so they all have the same speed and the same magnitude of acceleration.

    Redo this calculation.

    Tip: Generally, it's a good idea to hold off from using the calculator as long as possible. Try to solve the equations symbolically until the last step. This way if you make an arithmetic mistake you won't have to redo everything.
    Last edited: Sep 14, 2010
  6. Sep 14, 2010 #5
    Thanks a ton, Doc Al! I finally got the answer right lol.
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