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Man connected to harness standing on plank

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A professor (m= 75 kg )is seated on a light harness connected to a rope and pulley system. The professor's feet touch a uniform plank (mplank = 15kg) which is supported by a hinge at the wall. A bucket (M) is suspended from the right end of the plank.

    m = 75kg
    mplank = 15kg
    l = 2m
    L = 4.9m

    Find: Label all the forces on a FBD
    Find: T if M=0


    oo2vGDh.png
    2. Relevant equations
    ΣMhing = 0

    3. The attempt at a solution
    I took the moment about the hinge and looked at the man by himself. I think I'm drawing my FBD diagram incorrectly.

    KpST7JT.png
     
  2. jcsd
  3. Nov 19, 2015 #2

    BvU

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    Hello JT, :welcome:

    From your drawing it's clear that the rope is attached to the end of the plank. That should be in the problem statement !

    If prof is sitting in the harness, I expect one more force in his FBD. Otherwise I think you have most of it covered. Fry is the only horizontal force I see, so it may well be 0....
     
  4. Nov 19, 2015 #3
    Have you tried writing the equations for the torques about the hinge end of the plank?
    Also, you know that T is constant throughout the length of the rope.
    This greatly would simplify the solution since you don't need to consider the forces
    acting through the hinge.
     
  5. Nov 19, 2015 #4

    haruspex

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    Well, not quite. Since the rope is attached to the end of the plank, the section of rope below the plank will be at a different tension.
     
  6. Nov 19, 2015 #5

    BvU

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    Yes, but the problem statement reads M = 0
     
  7. Nov 19, 2015 #6

    haruspex

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    Which post is that in response to? It does not appear to contradict my post #4.
    Also, that M=0 condition is only for the final part of the question.
     
  8. Nov 19, 2015 #7
    ##T## is the tension in the section of rope above the plank. The section below will have a tension equal to ##Mg## which will generally be different than ##T##, particularly if ##M=0##.
     
  9. Nov 19, 2015 #8

    BvU

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    You are absolutely right :smile: ! (Haru and T2)
     
  10. Nov 21, 2015 #9
    Since the rope is connected to the man, why isn't T=(mass of man)*g
     
  11. Nov 21, 2015 #10

    BvU

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    Part of his weight is taken by the harness !
     
  12. Nov 22, 2015 #11
    I'm still having trouble understanding this question. Can someone draw a FBD ? Is Normal Force suppose to be on the FBD?
     
  13. Nov 22, 2015 #12

    haruspex

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    No, that's why it could have been mg. The reason that T is not mg is that not all of the weight is taken by the harness.
     
  14. Nov 22, 2015 #13

    haruspex

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    An FBD relates to a single rigid body. There are two appropriate FBDs here, one for the man and one for the plank.
    Your diagram depicted both, which is ok except that where those two bodies interact you need to show the equal and opposite forces between them.
     
  15. Nov 22, 2015 #14

    BvU

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    Part of his weight = not all of the weight
     
  16. Nov 22, 2015 #15

    haruspex

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    It's a question of emphasis in English. If you write that the reason that T is not mg is that part of his weight is taken by the harness it implies that the part taken by the harness accounts for the discrepancy. If you mean that the part not taken by the harness accounts for it then you need to write "only part of his weight...".
     
  17. Nov 23, 2015 #16

    BvU

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    Didn't want to give too much emphasis, just get the OP to think it over.

    Non-native english typist... But then again, weren't you in Australia ?
    :smile:
     
  18. Nov 23, 2015 #17

    haruspex

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    I am now.
     
  19. Nov 23, 2015 #18

    BvU

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    Can't beat Oxford English :cool:
     
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