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Massless Pulley Question

  1. Sep 23, 2008 #1
    A small man is sitting in a bosun chair that dangles from a massless rope, which runs over a massless frictionless pulley and back down to the mans hand. The combined mass of the man and the chair is 95kg. with what force magnitude must the man pull on the rope if he is to rise?

    Heres a link to a drawing
    http://img438.imageshack.us/img438/8349/pulley27xt.png

    I'm having troubles drawing the FBD for the man on the chair.
    The following is my attempted drawing of FBD of the man.

    W down
    Fn up, by the chair, equivalent in magnitude of W.

    F down, man pulling the rope
    Fn up, equal and opposite reaction by the chair

    and T up

    Someone help me out here pls?
     
  2. jcsd
  3. Sep 23, 2008 #2

    PhanthomJay

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    Instead of drawing a FBD of the man, try drawing a FBD of the chair and ropes (a FBD that encircles the chair and cuts through both ropes). Oh, welcome to PF!
     
  4. Sep 23, 2008 #3
    ok so the chair would have

    weight downward

    and the rope would have
    force down and T up

    which doesnt makes sense

    i'm not understanding this...
    help me out
     
  5. Sep 23, 2008 #4

    PhanthomJay

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    yes, correct
    No. When you draw a FBD, draw a 'cloud' or circle around the portion of the system you wish to analyze. In this case, draw that circle starting around the chair, then encircling beneath the bottom lower rope (hanging beneath the guy's hands), and then cutting through the upper 2 ropes, and closing back to where you started. Wherever that circle cuts through a 'contact' point, there is generally a force applied at that point. And always the non-contact gravity force acts. Note the rope tensions on either side of a massless frictionless pulley are always equal. Then use newton 1 to solve for T, which is also his pulling force.
     
  6. Sep 23, 2008 #5
    you lost me at when you said drawing a cloud. Is there any logical way to do it without learning new concepts.

    so

    weight down

    force down

    2T up, left and right side of the rope to balance the two downward forces to move in a constant velocity?
     
  7. Sep 23, 2008 #6

    PhanthomJay

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    I always draw a squiggly circle around the piece I'm analyzing. You can draw a smooth circle if you like. Or forget the circle/cloud concept if it confuses you. Personally, I find it difficult to find forces acting on the free body without it, though.
    yes
    this is where you are in error, what force might this be?
    yes
    what's the other one, please explain?
    I think you may be assuming a downward tension force from the man puling on the rope? This is not correct when you draw a FBD of the man-chair system. You are looking for the forces acting on the man-chair system. Tension forces always pull away from the objects they act on. There are 2 tension forces acting up on them, and the weight of the man and chair acting down. Solve for T.
     
  8. Sep 23, 2008 #7
    Good hint PhantomJay. I was trying to solve the OP's question as well and got hung until I read this.
     
  9. Sep 23, 2008 #8
    i don't understand how there could be 2 Tensions upward.
    Why is it 2T= mg?
    where is the extra T coming from

    can you explain this in detail

    i appreciate your help
     
  10. Sep 23, 2008 #9

    PhanthomJay

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    I know you don't like the cloud, so let me try to explain without it. We are looking at a free body diagram of the man and chair together. That's the portion we are focusing on. One of the ropes is pulling up on the man (the one he is holding), the other is pulling up on the chair. That's 2T pulling up on the man-chair system. The weight acts down. In FBD's you must look at the forces acting only on the portion you are isolating.
     
  11. Sep 23, 2008 #10
    oh ok thx

    if you were to explain in terms of clouds and circles
    how would u explain this behavior?
     
  12. Sep 23, 2008 #11

    PhanthomJay

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    Draw the circle completely around the man-chair and ropes, such that it's circumference 'slices' or cuts through the two top ropes. This is the FBD you are examining for the forces acting on it. In a FBD, you identify all external forces acting on it (such as gravity and any other externally applied forces), and you also identify the internal contact forces acting on it, at the points where the circle 'cuts' through those forces. In this problem, the circle cuts through the 2 top ropes, so you label these unknown forces each as "T" (actually, it could be T1 and T2, respectively, but T1=T2 when the cord passes over a frictionless, massless pulley, so call them each "T".) It doesn't cut through anything else, so the remaining forces are the external ones, in this case, gravity only (the weight of the man and chair). So its 2T up, mg down, or the man need only pull with half his weight to move at constant velocity. This problem is a level of difficulty up from the more simple case of a man just hanging from a rope attached to a ceiling, where for that case, T is simply his weight, mg.
     
  13. Sep 23, 2008 #12
    great explanation :cool:

    When we are assigning T's at the two cut points,
    if you don't mind, can you also explain why both the T is upwards as opposed to pointing opposite direction? (one up, one down)
     
  14. Sep 24, 2008 #13

    PhanthomJay

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    Remember again, when you are looking at tension forces acting on objects in a FBD, that tension forces 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. Since you are looking at the FBD of the tension forces acting on the man and chair, both those forces must pull away from them, that is, they act UP.
    I think you are incorrectly envisioning the right rope tension acting down, but visualize this: The man's hands are pulling down on that rope, that should be obvious; thus by Newton's 3rd law, the rope must be pulling up on the man. Same with the rope acting on the chair, it's pulling the chair UP, therefore, it points up, which is a bit more obvious.
     
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