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Quick Static Equilibrium Questions

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A horizontal beam of weight W is supported by a hinge and cable as shown. The force exerted
    on the beam by the hinge has a vertical component that must be:
    2yy4jsg.jpg

    Answer is nonzero and up.

    The pull P is just sufficient to keep the 14-N block and the weightless pulleys in equilibrium
    as shown. The magnitude T of the tension force of the upper cable is:
    v475kw.jpg

    Answer is 16 N.

    2. Relevant equations

    1. τ=r x F
    W=mg
    F=ma
    Tsin(θ)=mg

    2. T=mg+P

    3. The attempt at a solution

    1. Not quite sure why the answer for this one is nonzero and up. Shouldn't the cable already have a tension force in the y direction that cancels out the weight of the beam? Then the only force that the hinge needs to exert is in the positive x direction to counteract the tension in the y-direction. Can someone shed some light on this one?

    2. I'm having a bit of trouble setting up a free-body diagram for this one, but I do know that pulleys reduce the required pulling force by a factor of 2, and since there are three pulleys it reduces the pulling force by a factor of 8. 14/8 is 1.75, which using T=mg+P equals 15.75N for T, or 16 N. Is this logic even correct? And if so, could someone help me set up the FBD so I can prove my logic?
     
  2. jcsd
  3. Apr 4, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Consider torques about the point where the cable attaches to the beam.

    Just work your way through each pulley, starting with the one at the bottom. If the tension in that pulley's rope is P, what must be the tension in the rope above it? And so on. Then you can solve for P and use it to answer the question. (Draw free body diagrams for each pulley and for the block.)
     
  4. Apr 4, 2013 #3
    Oh, understand the first one now, thanks!

    10oog8i.jpg
    I'm not sure if I got it. P+42/3=14N, thus P=2. T=14+P=16N, which is right, but then the top string has two different tensions. What did I do wrong?
     
  5. Apr 4, 2013 #4

    mukundpa

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    Homework Helper

    I think as the pulleys are considered friction-less the tension in the string over it will remains same everywhere. If it is P on the right side part of the string it will be P on the left side as well. If we consider equilibrium of lowest mass less pulley, what should be the tension in the string over the middle pulley? (tensions in the three strings will not be same)
     
  6. Apr 4, 2013 #5

    Doc Al

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    Staff: Mentor

    For one thing, you seem to have assumed that the three ropes attached to the block have the same tension. Do not assume any such thing.

    Start with the bottom pulley. Since the tension is P, the rightmost rope attached to the block must have tension P. Then keep going until you've analyzed all the pulleys and the block.
     
  7. Apr 4, 2013 #6
    Hm, well if that's the case, then the tension in the string over the middle pulley is 2P, and the tension in the string over the top pulley is 4P. So together the tensions equal 4P+2P+P=7P, which has to balance out the 14N weight. So P=2. And since the tension in the top string equals 14+P, the tension is 16N?
     
  8. Apr 4, 2013 #7

    Doc Al

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    Staff: Mentor

    You got it! :approve:

    I would say that the tension in the top string must equal 8P = 16N, since the string through the top pulley has a tension of 4P.
     
  9. Apr 4, 2013 #8
    Awesome! Thanks, both of you!
     
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