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Tension on a Cable

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data
    An elevator in a tall building is allowed to reach a maximum speed of 3.89 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.70 m if the elevator has a mass of 1320 kg including occupants?



    2. Relevant equations
    Vfy^2 = Viy^2 + 2ay(Yf-Yi)

    F = ma (Newton's Second Law in the y-direction)


    3. The attempt at a solution
    Vi = 3.89
    T = ?
    Yf = 2.70 m
    Yi = 0
    m = 1320 kg
    a = ?

    Using the first equation: 0 = (3.89)^2 + 2(ay) (2.70)
    ay = -2.80 m/s^2

    Applying Newton's Second Law, I got: T - mg = ma

    T = mg + ma = (1320)(9.81) + (1320)(-2.80) = 9250.24 N

    However, the system says it's incorrect. Any help would be appreciated!
     
  2. jcsd
  3. Feb 22, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    Think over the signs. Considering the upward direction positive, the displacement of the elevator is negative, -2.7 m. The acceleration points upward, it is positive.

    ehild
     
    Last edited: Feb 22, 2012
  4. Feb 22, 2012 #3
    Oh, okay. Thank you, Ehild!
     
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