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Person pulling themselves up a double pulley

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data
    A man is pulling himself up to a pulley that consists of two disks welded together as shown(same center). The man is currently pulling straight down on the rope in his hands with a force of magnitude 447.2 N (on the bigger disk). The other rope is also vertical and is attached to the man's waist at his center of mass(this is on the smaller disk). The man's mass is 76.3 kg, the pulley's total moment of inertia is 2.74 kg⋅m2, the radius of the small disk is 0.33 m, and the radius of the big disk is 0.62 m.

    The man is hanging on a rope attached to the inner disk and is pulling on the rope attached to the outer disk. if you need a better description of the picture let me know.
    2. Relevant equations
    a(t) = angular acceleration*R
    n2l for rotation and translation

    3. The attempt at a solution
    R=outer radii r=inner radii
    sys: man Tpm+Tpm2-Mg=Ma
    a=(Tpm+Tpm2-mg)/M
    sys: pulley Tmp*R - Tmp2*r=angular acceleration*I
    Tmp*R - Tmp2*r= a*I/r
    (r*Tmp*R - Tmp2*r*r)/I= a
    so far i have 2 equations and 3 unknowns, how would i solve for a from here?
     
  2. jcsd
  3. Mar 19, 2017 #2

    TSny

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    Can you list the three unknowns? I see only 2.
     
  4. Mar 19, 2017 #3
    the 2 different tension forces and the acceleration
     
  5. Mar 19, 2017 #4

    TSny

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    Aren't you essentially given one of the tension forces?
     
  6. Mar 19, 2017 #5
    yes, yes i am. Thank you I didn't really make that connection
     
  7. Mar 19, 2017 #6

    TSny

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    The statement of the problem does not really state what you are supposed to find.
    Is the following picture representative of the setup?
    upload_2017-3-19_22-2-52.png
     
  8. Mar 19, 2017 #7
    sorry i guess i forgot to include that, i am supposed to find the magnitude of acceleration at the mans waist or CM. And yes that picture is the right setup
     
  9. Mar 19, 2017 #8

    TSny

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    OK. I think you have set up the equations correctly.
     
  10. Mar 19, 2017 #9

    TSny

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    Hold on. Check your signs in your equations to make sure your positive direction for ##a## corresponds to your positive direction for ##\alpha##.
    [EDIT: Never mind, I think everything is OK!]
     
  11. Mar 19, 2017 #10
    So i solved for a=(F*R*r+F-M*g)/(M*r^2+I) and got the wrong answer which is probably a sign problem. If I made a positive clockwise torque and defined positive y to be up would the second tension force be negative in the first equation?
     
  12. Mar 19, 2017 #11

    TSny

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    No, both tensions act upward on the person. So, I think you are OK there.

    What did you get for the ##a##?
     
  13. Mar 19, 2017 #12
    -18.99 m/s^2
     
  14. Mar 19, 2017 #13

    TSny

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    That's now what I get. I get about 5.3 m/s2.
     
  15. Mar 19, 2017 #14

    TSny

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    You can see something's wrong with this equation. The terms in the parentheses do not have the same dimensions.
     
  16. Mar 19, 2017 #15
    5.3m/s^2 worked but how did you get that? Did you just solve for Tmp in terms of a and plug that in to the other equation?
     
  17. Mar 19, 2017 #16

    TSny

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    I plugged your two equations (from your first post) into Mathematica and let it do the grunt work!
     
  18. Mar 19, 2017 #17
    ok i tried again and got the right answer, just an algebra mistake. Thank you so much
     
  19. Mar 19, 2017 #18

    TSny

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    OK, good work. Did you notice that this person is going to go head-over-heels heels-over-head?
     
  20. Mar 19, 2017 #19
    haha yes
     
  21. Mar 19, 2017 #20

    TSny

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    On second thought, I think it is head-over-heels. (Counterclockwise rotation of the person's body). Oh well.
     
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