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Max and min forces on an elevator.

  1. Oct 28, 2005 #1
    " A elevator with a mass of 4850 kg is to be designed so that the maximum acceleration is 0.0600 g's. What are the max and min forces the motor should exert on the cable?"
    I converted the .0600 gs to .558 m/s/s, but when iplugged it in to the F=ma formula i got the wrong answer? am i supposed to use another formula? :bugeye:
     
  2. jcsd
  3. Oct 28, 2005 #2
    Yes.

    I think you use Fnet = Ft - Fg

    Where Ft is the forces the motor exerts on cable.

    ^
    |
    | Ft
    ^ |
    | ______
    Fnet | |
    _______
    |
    | Fg
    V


    For the min Fnet = 0 so Fg = Ft so Ft = 4850 x 9.8

    and the max Fnet = Ft - Fg
    would be (4850)(0.06)(9.8) + (9.8)(4850) = Ft

    I could be wrong because I'm crap at Physics but, eh, at least I moved your question to the top again.
     
  4. Oct 28, 2005 #3

    daniel_i_l

    User Avatar
    Gold Member

    You used the right equation but with the wrong variables. Think, on the way down gravity makes the elevater to accelerate at 9.8m/s/s so to cancel that out the moter needs to pull enough to accelerate the elevator upwards with the acceleration of 9.8 + .588 m/s/s. (this is what kirby said but I didn't see him untill I posted)
     
    Last edited: Oct 28, 2005
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