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Forces and Newton's Laws

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    To hoist himself into a tree, a 72 kg ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 358 N . Neglect amy friction between the rope and the branch, and determine the man's upward acceleration.

    2. Relevant equations
    Newton's laws of motion.

    3. The attempt at a solution
    image.jpg

    The answer must be a =0.14m/s2
     
  2. jcsd
  3. Feb 11, 2015 #2

    Svein

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    How do you get from -2.414m/s2 (handwritten) to 0.14m/s2 (your answer)?
    Suppose I changed the problem to: What force must he pull with in order to stay in place (a=0)?
     
  4. Feb 11, 2015 #3
    That's easy:
    F = mg = 72 kg * 9.8 m/s2 = 705 N
     
  5. Feb 11, 2015 #4
    Firstly:
    WMAN = mg = 72x9.8 = 705.6N
    So force required to lift the man is 705.6N.

    Surely if he pulls on the rope with a force of 358N, he won't be lifted off of the ground at all?
     
  6. Feb 11, 2015 #5
    Don't worry, I'm not this stupid, I was mimicking the other commenter hahaha!
     
  7. Feb 11, 2015 #6
    I know, but the book gives the answer that a = 0.14 m/s2 . Are you sure it's a mistake?
     
  8. Feb 11, 2015 #7
    I believe that because there is a pulley, the question isn't as simple as "he weighs more than the force he exerts so therefore he must not be lifted"; have you learned about the physics of pulleys? I think that is the area of physics this question might be about...
     
  9. Feb 11, 2015 #8
    We're dealing with a fixed pulley, so it doesn't change anything.
     
  10. Feb 11, 2015 #9
    I thought the lifting force required is still reduced
     
  11. Feb 11, 2015 #10

    Nathanael

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    What about newton's 'equal and opposite law'? When the man pulls down on the rope, does the rope not pull back up on him?
     
  12. Feb 11, 2015 #11

    Svein

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    Yes - but I think the problem lies in the fact that when the man pulls down on the other rope, he does so by transferring some of his "weight" to the other rope. From where else would he get any force?
     
  13. Feb 11, 2015 #12

    Nathanael

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    You can say it like that if you'd like, but that's saying exactly the same thing. The man pulls down on the rope and the rope pulls up on the man. The rope supports part of the weight of the man, and the weight of the man supported by the rope pulls down on the rope. Etc.

    The reason I mention this is because the OP did not include it in his free body diagram.


    To the OP:
    That's incorrect. Fix your FBD, fix your equations, and try again.

    Also, in your equations you wrote, "T-F=-ma" but that "m" is not the same "m" as in the equation above it, is it?
     
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