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Homework Help: Simple newtonian mechanics problem

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    a man with mass m=66kg is standing on top of a platform with mass M=120kg. The man is pulling himself up using a pair of ropes suspended over massless pulleys. he pulls each rope with force of F=600N and is accelarating towards the ceiling at acceleration a. g=9.8 m/sec^2. find the value of a in m/sec^2.



    * - pulley
    2. Relevant equations
    [tex]\sum F = ma[/tex]



    3. The attempt at a solution
    I tried treating the man and the platform as one body but i got negative acceleration, as if the man was accelerating away from the ceiling. I would like to see how you guys solve this problem.
     

    Attached Files:

  2. jcsd
  3. Apr 6, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi ENgez! Welcome to PF! :smile:

    (have a sigma: ∑ and try using the X2 tag just above the Reply box :wink:)
    Show us your full calculations, and then we'll see what went wrong, and we'll know how top help! :smile:
     
  4. Apr 6, 2010 #3
    i summed up the forced on the y axis treating the man and the platform as one body :
    600*2 - (m+M)*g = (m+M)*a
    a=-3.707 m/sec2
     
    Last edited: Apr 6, 2010
  5. Apr 6, 2010 #4

    tiny-tim

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    Hi ENgez! :smile:
    Why 600 times two? :wink:
     
  6. Apr 6, 2010 #5
    The man pulls each rope with a force of 600N, as seen in the attached picture.
     
  7. Apr 6, 2010 #6

    tiny-tim

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    So why two? :wink:
     
  8. Apr 6, 2010 #7
    Is this supposed to be a hint?:confused:. I think it should be two as there are two ropes.
     
  9. Apr 6, 2010 #8

    tiny-tim

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    How do you know it isn't one rope?

    If the rope continued under the platform, so that there was only one rope, would that make any difference? If not, how can the number of ropes matter?
     
  10. Apr 6, 2010 #9
    so you are saying that the equation should be
    600-(m+M)*g=(m+M)*a
    a= -6.574?
     
  11. Apr 6, 2010 #10

    tiny-tim

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    ie 600*1 ?

    No, I'm saying that the number of ropes is irrelevant.

    Hint: what would be the tension in each rope if the platform was stationary? :wink:
     
  12. Apr 6, 2010 #11
    I thinks i see where you are getting here... if the platform was stationary the tension each rope felt was given by: 2T-(m+M)*g=0 => T = (m+M)*g/2. So if the man is supplying 600N of force the equation becomes 2T+600+(m+M)*g/2 = (m+M)*a?
     
    Last edited: Apr 6, 2010
  13. Apr 6, 2010 #12

    tiny-tim

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    Sorry, that equation doesn't make sense …

    if it's meant to be the equation for force on the man-and-platform, how can you include force from the man?
     
  14. Apr 6, 2010 #13
    hmmm. let me try again... the equation for the platform-and-man is:
    2T-(m+M)*g= (m+M)*a
    and the equation for the man:
    T-600-m*g=m*a
    is this correct?
     
  15. Apr 6, 2010 #14

    tiny-tim

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    No.

    Try it for a = 0, ie the man-and-platform is stationary …

    draw all the external forces on the man-and-platform.

    i really have no idea what that's supposed to be :confused:

    (and i'm off to bed :zzz:)​
     
  16. Apr 6, 2010 #15
    thanks alot for your help :) ill keep working on it.
     
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