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Compound pulley tension question

  1. Feb 11, 2008 #1
    [SOLVED] Compound pulley tension question

    1. The problem statement, all variables and given/known data
    An object of mass M is held in place by an applied force and a pulley system as shown in Figure P5.55. The pulleys are massless and frictionless.
    [​IMG]
    (a) Find the tension in each section of rope, T1, T2, T3, T4, andT5. (Answer in terms of M and g.)
    (b) Find the magnitude of F.

    3. The attempt at a solution
    The only thing I know is that T5=Mg. I'm trying to draw the a FBD for each pulley but I'm getting confused with relating T1, T2, T3, and T4 with Mg and F.
     
  2. jcsd
  3. Feb 11, 2008 #2

    Doc Al

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    Staff: Mentor

    It's easier than it looks. The FBD for each pulley should be easy since the only forces on the pulleys are the string tensions.
    Hint 1: The tension is the same throughout a continuous piece of rope (since the pulleys are massless and frictionless).
    Hint 2: How does F relate to T1?
     
  4. Feb 11, 2008 #3
    F-T1=0 so F=T1?

    When I'm drawing the FBD for the top pulley, what would the downward force be? F?
     
  5. Feb 11, 2008 #4

    Doc Al

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    Right!

    The downward force will equal the three tensions that pull on it.
     
  6. Feb 11, 2008 #5
    So the downward force would be T1+T2+T3? Meaning T4=T1+T2+T3? And how do I relate these tensions to M and g?
     
  7. Feb 11, 2008 #6

    Doc Al

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    Right.
    By combining that with other equations you can come up with. What about the second pulley? And, especially, what about my hint #1? :wink:
     
  8. Feb 12, 2008 #7
    Even if the rope goes through a pulley? So that means T1=T2=T3?

    Ahh, just solved it. Thanks for all of your help. T1=T2=T3=Mg/2 and since T4=T1+T2+T3, T4=(3Mg)/2 and since F=T1, F=Mg/2
     
    Last edited: Feb 12, 2008
  9. Feb 21, 2008 #8
    Hey guys I have a problem I got wrong similar to this. The pulley set up is exactly the same, but the outside rope is pulled at an angle (60) and I was curious to see how the force changes with the angle. Don't get me wrong here, I know that to find the x component of the force it would be FCos(60) and the find the y component you would use FSin(60), but do you use 1/2MG as F even though its pulled at an angle? And furthermore, do you do you square Fx and Fy and then take the root of it to find the final F?
     
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