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Finding the tension in terms of mg

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Attached the question.

    2. Relevant equations

    (total) tension = mg

    3. The attempt at a solution

    I'm not sure how the tension gets divided here. I know the tension on the right hand side of the turnbuckle is the same tension on the left hand side since they are symmetrical. But when does the tension get divided in the pullies, and by how much, I don't know. Help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 13, 2014 #2

    Doc Al

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    Start labeling the tensions and analyzing each pulley. You know there must be equilibrium, so use that.
     
  4. Apr 13, 2014 #3
    I'm not sure how you analyze each pulley separately. I know each of the two cables directl attached to the turnbuckle will have a tension of T/2. I don't know how to divide it from there. How do the cables attached to the ceiling or ground affect T?
     
  5. Apr 13, 2014 #4

    Doc Al

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    What must be the net force on each?

    Good. Keep going. Now that you know that those cables have a tension of T/2, what else can you deduce? Follow the cable.
     
  6. Apr 13, 2014 #5
    I'm a bit confused. Since T = mg, and mg acts downwards, that would mean the cables with the T/2 force would have to act upwards, right?

    Also, does the tension of a pulley always act away from the pulley? If I did this, the first cable attached to the ceiling from either side would have a tension of T (Tx = T/2 + T/2). The first cable attached to the ground would have a tension of -T (T/2 + T/2, but acting downwards). The 2nd cable attached to the ceiling would have a tension of (T/2 - T/2) = 0.

    And the 2nd cable attached to the ground would have a tension of (T/2 + T/2) = T..

    But this gives 2(T - T + T) = 2T = mg, T = mg/2 which is clearly wrong.. what is my mistake?

    Hopefully I am making sense.
     
  7. Apr 13, 2014 #6

    Doc Al

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    No, not right. Why do you think T = mg? (The problem asks to solve for T.)

    Let's do things systematically. Label the 8 pulleys 1 through 8 going from left to right (that way we can refer to them unambiguously). So, what can you say about the tensions in the cables attaching to pulley #3?
     
  8. Apr 13, 2014 #7
    Oh right. Makes sense.


    OK. Here's my deduction:

    A tension T/2 exists in the cable attached to the left hand side of the turnbuckle. So a tension of T/2 would be exerted by pulley 4 on both cables. Both acting away from the pulley. Since the cable on the left hand side of pulley 4 is connected to a cable of pulley 3, the tension exerted by the cable on pulley 3 would also be T/2, and therefore the tensions exerted by pulley 3 on the cables would also be T/2, just in the opposite directions.

    So pulley 3 would exert a tension of T/2 on both cables. Both tensions would act upwards. Since the net upwards force would have to equal the net downwards force of each pulley, the cable attached to the ground of pulley #3 would be -(T/2 + T/2) = -T..

    The problem I have is that I'm not sure about my directions.. and how the tension of the cables attached to the ceiling or ground would affect things.
     
  9. Apr 13, 2014 #8

    Doc Al

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    Good. You correctly deduced that the ground cable attached to pulley #3 pulls down on pulley #3 with a force of T. Thus the tension in that cable is T. (And, by symmetry, you know the tension in the bottom cable attached to pulley #6.)

    Keep going! Next: Analyze the forces on pulley #2.

    You're getting there. So far you have expressions for the tensions in 3 of the 5 cables attached to the mass.
     
  10. Apr 13, 2014 #9
    OK after some calculations I got the ceiling cable tension of pulley #2 (and therefore #7) as 0 (T/2 + ceiling tension = T/2). And the ceiling cable of pulley #1 (and therefore #8) as T (upwards). However I assume we are only needed in the ground cables, and the grounded cable of pulley #1 and pulley #8 should have a tension of T/2 each..

    I take it I'm wrong here, and that the tension of T1 and T8 are actually T/4 since that's whats mentioned in the solution manual. I'm not sure what I did wrong though.
     
  11. Apr 13, 2014 #10

    Doc Al

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    I don't understand your reasoning here. You have a tension of T/2 acting downward on pulley #2, so what must be the tension in the cables pulling up on that pulley?
     
  12. Apr 13, 2014 #11
    Oh right, T/4..

    With this logic the grounded cables on T1 and T8 would be T/4? And that would give the total tension on the mass as 3T + T/4 + T/4 = 3.5T?

    So 3.5T = mg?

    Is this correct?
     
  13. Apr 13, 2014 #12

    Doc Al

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    Right!

    So what's T in terms of mg? (Express it as a fraction.)
     
  14. Apr 13, 2014 #13
    T = (2/7) mg.

    Thank you so much for the help. It feels great to solve this with your help.

    Appreciated. :)
     
  15. Apr 13, 2014 #14

    Doc Al

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    Good work! :thumbs:
     
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