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Force question on elevator cable

  1. Nov 7, 2007 #1
    A 700 kg elevator suspended by a cable accelerates upwards at 3 m/s^2. The force exerted by the cable on the elevator is?
    (use g=10 m/s^2)

    m=700 kg
    a= 3.0 m/s^2

    Fnet= 700 kg * 3.0 m/s^2
    = 2,100.0 N Upward force

    Fnet= 700 kg * 10 m/s^2
    = 7,000.0 N Downward force

    Fnet= 2,100.0 N + 7,000.0 N = 9,100.0 N

    So is 9,100.0 N the total force exerted on the cable?
     
  2. jcsd
  3. Nov 7, 2007 #2

    Dick

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    Yes, but you didn't express that very well. All of those different numbers can't be Fnet. One of them is Fnet=ma, one of them is gravitational force and one is the cable tension. Can you label them correctly?
     
  4. Nov 7, 2007 #3
    Fnet= 700 kg * 3 m/s^2
    = 2,100.0 N

    Fg= 700 kg * 10 m/s^2
    = 7,000.0 N

    T= 2,100.0 N + 7,000.0 N
    = 9,100.0 N

    What is the variable for tension. I am not sure so I just used T.
    Would this be a more suitable answer?
     
  5. Nov 7, 2007 #4

    Dick

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    Better. Fnet is the sum of all forces. Fnet=T-Fg. So yes, T=Fnet+Fg.
     
  6. Nov 7, 2007 #5
    Thanks a lot. Appreciate your help.
     
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