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Strain on a slowing elevator

  1. Nov 7, 2004 #1
    A 1260kg freight elevator is supported by a steel cable of diameter 34.9mm. It is loaded with a crowd of people collectively having a mass of 2850kg and it is descending.
    What is the strain in the cable when it is brought to a stop in 0.600s?

    I'm not really sure how to figure in the fact that the elevator is slowing in this equation? a little push in the right direction would be GREATLY GREATLY appreciated. thanks!
     
  2. jcsd
  3. Nov 7, 2004 #2
    i believe this is an impluse question, idk what the diameter has to do with anything...

    [tex]\frac{dp}{dt} = F [/tex]
    [tex]-(1260kg + 2850kg)v = F(.6s) [/tex]
    solve for F and that is what i would assume "strain" means, however idk what the v would be is it given? idk...sorry...
     
  4. Nov 7, 2004 #3
    you need the diameter to find the cross-sectional area of the cable. i understand all that, but i need the strain when it is slowing. i got the strain when it was moving at a constant velocity (i think).
    does anyone know how to figure this?
     
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