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Elevator Problem

  1. Apr 15, 2004 #1
    Hey all..here is another stumper (for me).

    "A 1000 kg elevator has a maximum load of 800 kg. If constant friction retards the elevator at 4000 N, to lift at 2 m/s, what does the motor's hp(horsepower) have to be?"

    Now, I was trying to do a lot of things with this one. I thought it was an obstacle that we had a speed and not a distance. I tried getting around this by simply acknowledging that since we went to go 2 m/s, and since a watt is 1 J/s, that I could jsut sort of throw the /s off each of them and leave it at 2 m. Prolly totally off.

    Any Help?
    (My answer was 5.92 hp)
     
  2. jcsd
  3. Apr 15, 2004 #2
    Use energies. From point A to a higher point B, the equation looks like this:

    [tex]W_{motor} + W_{friction} = \Delta E_m = \Delta E_p + \Delta E_k [/tex]

    [tex]Pt - f\Delta h = mg\Delta h + (0)[/tex]

    Where P is the power of the motor, t is the time the eleavtor takes to travel the distance from A to B, f is the force of friction and Δh is the distance from A to B. We need to get rid of the time there, so we express it as t = Δh/v, where v is the elevator's constant speed:

    [tex]P\frac{\Delta h}{v} + f\Delta h = mg\Delta h[/tex]

    Now rearrange the equation, cancel Δh and you get:

    [tex]P = v(mg - f)[/tex]

    And voila. :smile: I get P = 36.6hp.
     
    Last edited: Apr 15, 2004
  4. Apr 15, 2004 #3
    I love you Chen.
     
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