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Current in elevator's engine

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    An elevator is running upwards at the constant speed v. The magnitude of opposing forces is [tex]F_o[/tex]. The elevator's mass is [tex]m[/tex] and the voltage is [tex]V[/tex]
    Find the magnitude current in the elevator's engine.

    2. The attempt at a solution

    I tried to solve this task by energy equations:
    [tex]VI\Delta t = (mg+F_o)\Delta x[/tex]
    But now I'm stuck. Two variables in one equation and it doesn't seem as if a simultaneous equation could by found.

    Thank you for advice in advance.

    EDIT:
    Actually, I have just noticed that if we re-engineer this equation:
    [tex]I = \frac{(mg+F_o)\Delta x}{V\Delta t}[/tex]
    and [tex]\frac{dx}{dt} = v[/tex]
    then:
    [tex] I = \frac{mg+F_o}{V}v [/tex]
    Is this answer correct?
     
  2. jcsd
  3. Oct 18, 2016 #2

    gneill

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    Staff: Mentor

    It's not clear from the problem statement what is included in ##F_o##, the "magnitude of opposing forces". Is it the net force acting against lifting the elevator (in which case it would include gravity) or is it just the sum of the friction forces (excluding gravity)?
     
  4. Oct 18, 2016 #3
    Gravity is excluded.
     
  5. Oct 18, 2016 #4

    gneill

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    Staff: Mentor

    Then your second attempt should give the correct result. Essentially you've inadvertently used the well known formula P = F*v, the power expended to maintain a constant velocity v is equal to the force required multiplied by the velocity.
     
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