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Hanging a weight on a pulley

  1. Mar 16, 2015 #1
    Suppose we have a weight D (1 Kg) hanging from a pulley on a rope connected to a wall A

    uPHmi.png

    suppose also the distance AB from the wall is 1 m.

    If I hung a weight C =1 kg as a counterweight and the pulley were not connected to the wall, the length of the rope BD (1) would equal the length of the rope BC.

    If I hang a weight C = 0 on the rope AB this will not sag (h = 0) what will the distance of C from the line AB if I hang a weight of 1 Kg? My guess is that h should be sin=cos 45, since a imagine that half the weight is carried by the wall at A.

    In general, is there a simple formula to determine the distance h wrt to the weight? I know this is all about forces and tension http://en.wikipedia.org/wiki/Tension_(physics), where can I find an article that explains all that with examples?

    Any help is appreciated! Thanks
     
  2. jcsd
  3. Mar 16, 2015 #2

    jbriggs444

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    As I understand this, you are allowing weights C and D (both 1 kg) to hang from the pulley. The wall is not involved. You assert that, in this situation, rope length BD must be equal to rope length BC.

    For a massless rope, this is not the only possible solution. Rope length BD could be greater, equal or less than rope length BC and the weights could still be in equilibrium.

    Now you reconnect the rope to the wall and...

    As I understand this, you are saying that rope segment AB will not sag until you hang weight C from it. After you hang it, weight C will be a distance h below the previous position of rope segment AB. You may be assuming a frictionless rope so that weight C is free to slide.

    If the tension in the rope is 1 kg-force, what is the vertical component of that force when the rope is at a 45 degree angle?
     
  4. Mar 16, 2015 #3
    I already said I think the components are equal to sin=cos 45°

    If you look at the sketch the angle at B is roughly 30°, does the distance h from C to the line AB correspond to the weight of P1? or is it the length of BC? what I do not know is what part of the weght of P1 is discharged on the wall.

    Can anybody tell me if it is at all possible to determine a simple relation between h (or BC) and the weight of P1/P2?
    Thanks
     
    Last edited: Mar 16, 2015
  5. Mar 16, 2015 #4

    jbriggs444

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    Your justification for that was incorrect.

    Reading angles from drawings in textbook exercises is an inappropriate way to solve problems. Was this sketch drawn from real life or from a textbook exercise?

    Yes, it is possible to determine a simple relation. I am working to help you do so.
     
    Last edited: Mar 16, 2015
  6. Mar 18, 2015 #5
    You are not really helping me, you did not give me a single clue or a link. The sketch is just a quick way to visualize the problem, it is not an exercise. I want to learnthe general procedure. I think that if P2 is 1 the tension on the rope (both if it goes to rhe wall and if it goes to P1) is 1 and that P1 must be 1/cos30 and h 1*sin30*cos30 = .433
     
  7. Mar 18, 2015 #6

    jbriggs444

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    Is the weight P1 free to slip on the rope or is it firmly tied to the rope? Are you trying to compute the weights based on the angles in the picture or the angles in the picture based on the weights?

    If the diagrram is intended as a visualization aid then you cannot have correctly determined the 30 degree sag angle by looking at the diagram.

    If the sag angle is 30 degrees, how did did you conclude that the total supporting force on weight P1 was 1/cos(30 degrees)?
     
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