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Boatswans Chair pulley Question

  1. Apr 16, 2017 #1
    1. The problem statement, all variables and given/known data
    The question has been screenshotted, along with the given diagram. Basic stuff I know, but I can't seem to get my head around pulley systems. T
    Screenshot (44).png

    Screenshot (43).png
    2. Relevant equations
    F(net) = m ⋅ a(system)

    3. The attempt at a solution
    I understand the rope to be considered inextensible. Also, I understand that the tension must equal the weight of the man & chair, but beyond that I don't know how to break it down.

    I labelled a reference frame with x being positive down.
     
  2. jcsd
  3. Apr 16, 2017 #2

    gneill

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    Staff: Mentor

    Hi Ryansf98,

    Welcome to Physics Forums!

    What leads you to that conclusion?

    Have you made a free body diagram of the man&chair?
     
  4. Apr 16, 2017 #3
    I just read that over and now realise it doesn't make sense, as equal tension & weight would mean an acceleration of 0m/s^2 wouldn't it? What I meant was that tension must be equal throughout the rope.
     
  5. Apr 16, 2017 #4

    gneill

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    Okay. Draw the resulting free body diagram for the man&chair.
     
  6. Apr 16, 2017 #5
    Thank you by the way! I've used it for sometime looking at existing threads, but I couldn't find an existing problem like this one so thought it'd be a good time to start my own account!
     
  7. Apr 16, 2017 #6
    I done that, labelling F(down) as 200N + mg, => F(down total) = 887N.

    Regarding the tension, I labelled T moving up from the man to the smaller pulley. Would that then split into 1/2T either way up the pulley, or would it remain T?

    I also labelled T pointing up on the rope which the man pulls down on.
     
  8. Apr 16, 2017 #7
    I realise I jumped a gun a little there as you only suggested a free body diagram for the man and chair. For this, I labelled mg down and T up towards the smaller pulley. Would I in fact insert 200 N here also?
     
  9. Apr 16, 2017 #8

    gneill

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    How many ropes attach to the man&chair? If each has a tension T, what's the net force upwards?
     
  10. Apr 16, 2017 #9
    One rope attached directly to the man and chair, therefore T up? Also, would 200N be T, due to Newtons 3rd Law?

    Therefore, Net Force Up = 200N?
     
  11. Apr 16, 2017 #10
    In fact, there are 3 ropes above the man and pulley, therefore F(up) = 3T.

    T = 200N as the man exerts this force down the rope and so the rope exerts this force up on him.

    That means that F(net of the system) = mg - 3T, => 686.7 - (3*200) = 86.7N downwards.

    F = ma, a= 86.7/70, => a = 1.24 m/s^2?
     
  12. Apr 16, 2017 #11

    gneill

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    No, there are two ropes connected to the man and chair: One he holds directly, one connected to the chair.

    The one he holds has a tension T = 200 N. What's the tension in the other rope? (consider a FBD of the small pulley).
     
  13. Apr 16, 2017 #12

    gneill

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    Yes! That's much better!
     
  14. Apr 16, 2017 #13
    So generally no matter what, each rope has a tension T throughout the system, even if the rope reaches a pulley and looks to go into 2 separate ropes?

    Thanks for the help!!
     
  15. Apr 16, 2017 #14

    gneill

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    Staff: Mentor

    If the pulleys are massless and frictionless, then continuous ropes have the same tension. When separate ropes meet at a pulley, their tensions sum (vector sum).
     
  16. Apr 16, 2017 #15
    Makes sense. Again thanks for your help!
     
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