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Kinematics Newton's Laws Question

  1. Sep 16, 2009 #1
    Suppose the cord in Fig. 4-22 is a heavy rope of mass 1.0 kg. Calculate the acceleration of each box and the tension at each end of the cord, using the free-body diagrams shown in Fig. 4-46. Assume the cord doesn't sag. (m1 = 10.5 kg , m2 = 12.5 kg and FP = 35.0 N)

    m1 stands for the mass of the first box and m2 stands for the mass of the second fp is the force that the man is pulling on the rope which is connected to m1 and anther rope connecting m1 to m2 which is 1 kg.
    [m2]t2---1kg rope---t1[m1]- Fp

    above is a small illustration of how it looks like im trying to solve for t1 and t2 then tensions.

    2. Relevant equations
    so their are 2 boxes one being pulled by anther by a 1 kg string i found the acceleration of both boxes to be 1.458 m/s^2 by doing f=ma mass was 24kg and force is 35 35/24 = acceleration i had no idea where to start looking for the tension of the two boxes.


    3. The attempt at a solution

    have to idea i think it would be 35N because the tension would stay the same through out but doesn't seem to be right. Any Ideas?
     
  2. jcsd
  3. Sep 17, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi davidkis ! Welcome to PF! :smile:
    Because the rope is heavy, the "front" end has to pull more mass than the "back" end …

    it's like a series of weights hanging from the ceiling, each joined to the one above by a (massless) string …

    if you think about it, the lowest connecting string has to support far less weight than the highest one, and so its tension will be far less. :wink:
     
  4. Sep 17, 2009 #3
    i get where your going and tried to imagine it hanging down ur right that t2 would have less tension but i still dont completly understan might go to my teacher for this one.
     
  5. Sep 17, 2009 #4
    thanks for the help tho
     
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