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Nonequilibrium/Newton's Laws

  1. Sep 23, 2004 #1
    I am stuck on this pulley problem:
    To hoist himself into a tree a 72.0kg man ties one end of a nylon rope around his waist and throws the other end over a branch of a tree. He then pulls downward on the free end of the rope with a force of 358N. Neglet any friction between the rope and the branch and determine the man's upward acceleration.

    I understand the basic concept if I have two masses but I only have one here, so I need some guidance!
    Thanks
     
  2. jcsd
  3. Sep 23, 2004 #2

    Pyrrhus

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    What are the forces acting on the man?

    What do you know about the tension when the man is pulling downward?
     
  4. Sep 23, 2004 #3
    I am not sure that I understand. The tension pulling down is negative and the force of gravity is acting on the man?
     
  5. Sep 24, 2004 #4

    Pyrrhus

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    On one end you got weight and tension acting on the man, what you got on the other end?
     
  6. Sep 24, 2004 #5
    Isn't tension acting on the other end and isn't it equal to the other tension?
     
  7. Sep 24, 2004 #6

    Doc Al

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    The tension in the rope is the same throughout.

    As Cyclovenom suggested, start by describing every force acting on the man. Hint: how many ropes attach to the man? What's the tension in the rope? What's the net force acting on the man? Apply Newton's 2nd Law.
     
  8. Sep 25, 2004 #7
    I am still stuck on this I know that F=MA and that the tension on the rope is equal. I guess that there are two ropes attached to the man. But I keep plugging in numbers and not getting any closer to the answer!
     
  9. Sep 25, 2004 #8

    Doc Al

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    The tension is the same throughout the rope.
    Right!
    Stop "plugging in numbers". Instead, answer the questions I asked in my previous post.
     
  10. Sep 25, 2004 #9
    The tension in the rope is 358N plus the man's Weight(mg)?
     
  11. Sep 25, 2004 #10

    Doc Al

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    No. The man pulls on the rope with 358N. And the rope pulls on the man with 358N (Newton's 3rd law!). Thus the tension in the rope is 358N.

    Continue answering those questions.
     
  12. Sep 25, 2004 #11
    I'm sorry, but I give up! I can't answer the questions, that is the problem!
     
  13. Sep 25, 2004 #12

    Pyrrhus

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    You know the tension right?, what are the forces acting on the man, use Newton's 2nd Law, you are almost there!

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]
     
  14. Sep 25, 2004 #13
    the force on the man is his weight(mg)?
     
  15. Sep 25, 2004 #14
    I got it!! Finally, I had to multiply the tension by 2 and then subtract the mans weight! And then divide by his mass!!
     
  16. Sep 25, 2004 #15

    Doc Al

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    Right! The two ropes attached to the man each pull up with a force equal to the tension in the rope (total upward force = 2T); his weight pulls down (total downward force = mg). So, the net force is [itex]F_{net} = 2T - mg[/itex] upwards. To find the acceleration, use Newton's 2nd law: [itex]a = F_{net}/m[/itex] upwards.

    Good job.
     
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