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Homework Help: Quick question about Equilibrium w/ Stress

  1. Nov 28, 2005 #1
    Quick question about Equilibrium w/ Stress :D

    Hello guys!

    I am new here but I plan to stick around for a long time :D

    I have a quick question, and I was wondering if you guys could help me...

    1) A steel cable with cross sectional area 3.0 cm^2 has an elastic limit of 2.4*10^8Pa. Find the maximum upward acceleration that can be given a 1200-kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.
    Answer is 10.2 m/s2.

    This is what I did :D,
    1/3 of 2.4*10^8 is 80000000Pa

    Tensile Stress=F/A

    80000000Pa=F/.03m
    F=2400000

    F=Ma
    2400000Pa/1200kg=a
    a=2000m/s^2

    Anyone? It's late so I gave it my best shot :D
     
  2. jcsd
  3. Nov 28, 2005 #2

    Fermat

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    Homework Helper

    1cm² = (0.01)² m² = 0.0001 m²
     
  4. Nov 28, 2005 #3
    With that I get 2m/s^2(Which you are right, I didn't see that initially)

    Using N2L, T-W=Ma
    So, T=F from above?
    F-mg=ma
    24000N-11760N=1200kg*A
    A=10.2m/s^2

    Thanks a bunch L:D
     
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