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Maximum tension a cable can withstand.

  1. Feb 22, 2009 #1


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    1. The problem statement, all variables and given/known data
    If the maximum tension that a cable can safely withstand is 9000N and the maximum upward acceleration of the elevator is 1.00 ms2, how many 80kg passengers may be safely carried in the elevator?

    (The mass of the cable is negligible)

    2. Relevant equations

    3. The attempt at a solution
    If I let ay=1.00 ms2 and F=9000N then I have:
    9000N = m * ay and m = 9000. The number of 80kg people the elevator can withstand is 9000 / 80 = 112 people.

    Is this correct?
  2. jcsd
  3. Feb 22, 2009 #2


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    No. You first need to know the mass of the elevator. Is it given? Then note that weight and mass are not the same, and that net force = ma. You need to identify all the forces acting on the elevator/people system. Hint: don't forget the weight acting down.
  4. Feb 22, 2009 #3


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    The mass of the elevator is 520 kg. Sorry I forgot to include this. Correct me here: There are two forces acting on the elevator, FG and FN.
    FG = mg = 520 kg * (-10) = -5200N.

  5. Feb 22, 2009 #4
    I think the 2 forces you want to analyze in the Y direction are Fw and Ft, which oppose each other. Fw is acting in the negative Y direction and Ft is acting in the positive Y direction. Ft of course being the force of tension

    So id look at it as the max force of tension it can hold is 9000N=Ft,

    Ft-Fw=ma ----> Ft=ma+(Fw)
    Last edited: Feb 22, 2009
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