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Laws of Motion problem

  1. May 8, 2007 #1
    « Laws of Motion » problem

    1. The problem statement, all variables and given/known data
    A man of mass = 60 kg is standing on a weighing machine inside an cabin (of mass = 30kg). The cabin is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself.
    a) If the man manages to keep the cabin at rest, what is the weight shown by the machine ?
    b) What force must he exert to get his correct weight ?


    2. What I feel :uhh:
    since the man is holding the rope himself, a tension T is acting on him upwards. His weight is acting downwards. In case a) , there's no acceleration.
    Normal rxn. N is also upwards.
    So, N+T= mg , where m is the mass of the man.
    Also T balances the downward force of the lift and the man.
    so, T= (M+m)g - T
    Is this argument correct ? I am specially confused about the Normal forces . I get the correct answer with this but I have not included the normal force in the second equation...and I am supposed to consider all the forces . Does it cancel out ?

    for b), he must accelerate the cabin upwards to get his correct weight. Let the acc. be a, then, in the cabin's frame of ref. -:
    N+T-mg=ma (ma is pseudo force)
    or, N+T=m(a+g) <-- Here N must be 60g to get correct weight.
    So i get T=60a
    Now in earth's frame,
    T-((M+m)g-T)=(M+m)a <--I have used (M+m)g-T to denote downward force
    Solving, I get a=30, hence T=1800N

    Actually, i have worked backwards from the answers to get these equations, and hence I am in doubt whether they are correct or not. In this case, the equation in earth's frame considers force due to (M+m)g-T but the acceleration is for (M+m) only ! Why ?

    Any help is appeciated

    3. Answers
    a)15 kg
    b) 1800 N
     
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2
    a. It's correct.

    It is more convenient if you consider the man and the cabin as the whole thing. By that, the total force acting on the whole thing is 2T-(m+M)g=0

    The N forces between the man and the cabin are just internal ones.

    b. Considering like a), you only see the whole thing going up with acceleration a, Newton's law 2 now is:

    2T-(m+M)g=(m+M)a

    which is the same as you derived.

    BTW,

    in the cabin's frame of ref. -:
    N+T-mg=ma (ma is pseudo force)


    is not true.

    you are still in the Earth's frame, the right-hand side is the result of the net force in the left hand side.

    If you want to write in the frame sticked with the cabin, it should be:
    N+T-mg-ma=0 zero because the cabin sees you standing rest, and ma now is the pseudo force. Of course, the result must be the same.
     
  4. May 8, 2007 #3
    Thx for the help Weimin, much appreciated :biggrin:
     
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