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Tension problems

  1. Oct 23, 2006 #1

    Im supposed to find T1 and T2 using m1, m2, and g. The whole system is accelerating downwards with magnitude g/2.

    Maybe if I get this I can figure out the more complex ones. Thanks.
  2. jcsd
  3. Oct 23, 2006 #2


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    Draw free body diagrams for both the masses and see if you can work it out. We will help you if you show us your attempt.
  4. Oct 23, 2006 #3
    Its tough to put in paint, but for m1 I have T1 going up, T2 going down, and .5(m1)g going down. On m2 I have T2 up and .5(m2)g down.

    Looking at m2, I have the sum of forces being T2 - .5(m2)g

    I'm not used to having the system accelerate, but right now I can only think of making T2 = .5(m2)g. T1 would be .5(m1 + m2)g then. I just dont think that is correct.
  5. Oct 23, 2006 #4
    Nevermind, it is :)
  6. Oct 23, 2006 #5
    OK heres the one Im really stuck on. Im just not completely sure whats going on. I'm looking for T. The units are kg for mass of the blocks, and Im assuming massless/frictionless pulleys.


    I drew the FBD for each mass, Im just not sure how to connect it all.
  7. Oct 23, 2006 #6
    So far Im thinking about trying to balance the force on each side of that middle pulley. I have the total mass on the right side that its carrying as 7+2.5? (since half of that 5 is attached the the "floor".) For the left side, Im thinking the mass there now is .5 (half of the 1 is attached to the "ceiling", so I need to get another 9 kgs (and 9 x 9.8 = 88.2 N as T).

    Is this completely wrong thinking?
  8. Oct 23, 2006 #7


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    You have two strings and therefore two tensions. You need to know how far each mass moves in relation to the other masses. With that information, your FBDs should get you a solvable system of equations.
  9. Oct 23, 2006 #8
    Still stuck. I think Im missing the effect that the masses have when they are attached to the pulley, not the actual string.
  10. Oct 23, 2006 #9


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    I think I misinterpreted the problem. Nothing is moving here? Is that correct?

    It is so frustrating when this system freezes up like this. I’ve been frozen out for a long time.
    In case I cannot get back in, here is what I have assuming nothing is moving.

    TL is the tension in the upper left string
    TR is the tension in the lower right string.

    T + 1kg*g = 2TL
    TL = 7kg*g + 2TR
    TR = 5kg*g
    Last edited: Oct 24, 2006
  11. Oct 24, 2006 #10
    Thank you. That makes sense now.
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