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Tension on chain on pulley

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A worker lifts a weight w by pulling down a rope with a force F. The upper pulley is attached to a ceiling by a chain, and the lower pulley is attached to the weight by another chain. The weight is lifted at constant speed. Assume the chain, rope, and pulley all have negligible weights. In terms of w, find the magnitude of force F if the weight is lifted at constant speed.


    2. Relevant equations
    F=ma where a=0


    3. The attempt at a solution
    Okay, so I made a fbd again. This time I'm confused about the tensions, I'm not sure if it is pointed up or down, for both of the chains. So far, I have the weight pointed down, F is down, and both tensions are up (I assume that tension is up...but I'm not sure). With that, my equation is

    F = -w+2T where I substitute T with w.

    Can someone pls explain to me about tension? Many thanks in advance.
     
  2. jcsd
  3. Feb 4, 2007 #2

    PhanthomJay

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    I'm not sure how to visualize your problem, but in terms of tension forces, tension forces external to an object in a FBD always pull away from that object in your FBD, never towards its.
     
  4. Feb 4, 2007 #3
    Sorry, I have the picture right here. I figured out that both of the tensions, in upper and lower chains are w, but putting the whole problem together to find the magnitude of the force, I'm bit stuck, keep getting the wrong answer. Is both tensions not pointed up?
     

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  5. Feb 5, 2007 #4

    PhanthomJay

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    There are a few tension forces to deal with here, so be careful. You have the tension in each of the chains, and the tension in the rope. For massless, frictionless pulleys, the tension in the rope is always the same on each side of the pulley it is wrapped around.

    Since the top pulley does not move, and the lower pulley moves at constant speed, there are no net external forces acting on the system. Newton 1 applies.

    Draw an FBD of the weight first. That will give you the tension in the lower chain....w, acting up on the block, pulling away from it, which you have correctly identified. Now draw an FBD of the lower pulley....isolate it with a cloud around it. You have the tension in the chain, w, acting down away from the pulley. What can you now say about the tension in the rope and the force F??
     
  6. Feb 5, 2007 #5
    The tension is pointed downward right? So if the tension is the same on both side, then counting all the forces together, there's two tensions (in the chains) pointed up and two tensions in the ropes, pointing downward. Doesn't that cancel out all the tensions? Unless I'm wrong about the direction of the tensions...?
     
  7. Feb 5, 2007 #6

    PhanthomJay

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    I don't think you are taking your FBD's properly. Let me start one of them off for you, or else we'll be talking forever about what acts up and what acts down.

    Draw a circle or cloud around the lower pulley. The circle will cut through the lower chain and through the 2 ropes immediately above that lower pulley. The tension in the chain acts down on the pulley, pulling away from it, and equals w. The tension , T, in 'each' of the ropes above the pulley act up on the pulley, pulling away from it . The tension in 'each' of the ropes are equal. (I put 'each' in quotes because really it's the same rope, you know). From Newton 1,
    [tex] F_{net} = 0 [/tex]
    [tex]T + T - w= 0 [/tex]
    [tex]2T = w[/tex]
    [tex]T = w/2[/tex]
    NOW, try solving for F.
    Solve
     
  8. Feb 5, 2007 #7
    Yeah, I don't understand about the direction of the tension....I'm not sure if it goes up or down when attached to an object.

    Um, since the tension of the rope is same on both side of the pulley it's wrapped around, does that mean that the force F = T = w/2?
     
  9. Feb 5, 2007 #8

    PhanthomJay

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    Yes! This particular setup has a 'mechanical advantage' of 2....you only have to apply a force equal to half the weight in order to lift up that weight at constant speed. Remember again, when you are looking at tension forces acting on objects in a FBD of that object, that the force could act up or down or at an angle, but what really matters is that it must always pull away from that object.....the tail of the tension force vector originates at the object, and the arrow points away from it... so when you look at the weight in a FBD, the tension in the chain acts up....when you look at a FBD of the lower pulley. the chain tension acts down....when you look at a FBD of the upper pulley, the top chain tension acts up on it...etc.
     
  10. Feb 5, 2007 #9
    Ohh. I get it now. Tension is always away from the object...thanks so much for the help!
     
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