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Finding Power of an Elevator

  1. Feb 13, 2016 #1

    CGI

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    1. The problem statement, all variables and given/known data
    media%2Fe4c%2Fe4c463dd-a209-4ffa-8091-5bfcc0fffbc0%2Fphp1kc8uI.png


    2. Relevant equations
    So I've been thinking about this one for a while and I've tried to read up about Power in general.
    I think Power = Work/ Time = (Force x Distance)/Time or Force x Speed.

    3. The attempt at a solution
    F = ma, so if velocity is constant, a = 0. So that would mean that the force is zero, so would that make the motor have zero power? That doesn't really sound right. If someone could help me out, I would really appreciate it. Thanks in advance!
     
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  3. Feb 13, 2016 #2
    F = ma, but you're forgetting about gravity working more on one side than the other.

    Gravitational acceleration works on both the elevator and counterweight.. if they're the same mass, it cancels out, so it won't accelerate.. If they're a different mass, the motor will have to do (or absorb) work.

    Draw out free body diagrams for it all to start with.
     
  4. Feb 13, 2016 #3

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    Alright. I got for the counter weight, that the sum of forces in the y direction with up as positive, is T - W = 0 or T = W for the FBD of the counter weight.
    For the Elevator, I have that 3T - W = 0 or that 3T = W for the FBD at the elevator.

    Would these be correct?
     
  5. Feb 13, 2016 #4
    I'm a little lost.. What is T and W here, and why 3T?

    If it's not accelerating, the motor has to do work on the difference of the two weights... so moving the elevator 5000 lb elevator up is moving the 2200 lb counterweight down, The net weight you're lifting will be 3300 lbs @ 15ft/s
     
  6. Feb 13, 2016 #5

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    Oh wow, I'm sorry. T in my FBD stands for tension in the cables and W stands for weight. In the elevator, I thought that it would be 3T - W = 0 because there were three cables lifting that elevator up. As for my counterweight, I thought it would T - W = 0 since there is only one cable so the tension would have to be equal to the weight.

    But okay, if the net force of gravity on the system is 3300lbs, could I just multiply 3300 by 15 and have the Power generated by the motor?
     
  7. Feb 13, 2016 #6
    Forget about the tension in the cables.. simplify it... if you look closely at the diagram, you'll notice it's a 1:1 movement ratio for the elevator : counterweight

    The tension on the cable to the motor is half the weight of the elevator minus counterweight, but the tension in the cable is irrelevant to work.. the motor needs to move twice as far at half the load which means the work stays the same.

    Yes, the motor is making 3300lbs * 15ft/s of work :)
     
  8. Feb 13, 2016 #7

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    Oh okay that makes things so simple! So when it says that there is acceleration upward, how would that change the power from the motor?
     
  9. Feb 13, 2016 #8
    The motor will have to do more work.. and it'll need to apply more force to accelerate both masses.

    Just as an exercise (and it'll give you a hint), how fast would the elevator accelerate if the cable to the motor broke?
     
  10. Feb 13, 2016 #9

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    It would just fall so wouldn't it just be the acceleration due to gravity?
     
  11. Feb 13, 2016 #10
    Nope... there's gravity acting on both weights, so the net force will be the difference of gravitational forces, but both weights need to accelerate at the same speed (one up, one down evidently)
     
  12. Feb 13, 2016 #11
    Well, yeah, OK, the acceleration will be due to gravity, but the acceleration will not be 32ft/s^2
     
  13. Feb 13, 2016 #12

    haruspex

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    3300?
     
  14. Feb 13, 2016 #13

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    Oh that's right, the net weight would be 2800lbs.

    And I don't think I'm understanding the way the acceleration would affect the elevator. I am so sorry :(
     
  15. Feb 13, 2016 #14

    haruspex

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    You cannot simply add an extra power term to handle the acceleration. The power required for an acceleration depends on the current speed.
    Easiest is to use power = force x velocity. What force would be required from the motor to achieve the desired acceleration?
     
    Last edited: Feb 13, 2016
  16. Feb 14, 2016 #15

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    If I used F = ma with (a) as 3, would the force needed just be 3 x 5000?
     
  17. Feb 14, 2016 #16

    haruspex

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    In the equation F=ma, it is important to remember exactly what F represents.
     
  18. Feb 14, 2016 #17

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    F in this case is just the applied amount of force on the elevator, no?
     
  19. Feb 14, 2016 #18

    haruspex

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  20. Feb 14, 2016 #19

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    Okay so F stands for the vector sum of forces on an object. So if I used F = ma, I would have to add up all the forces acting upon the elevator, which would not just be the force of gravity. Is that headed somewhere in the right direction?
     
  21. Feb 14, 2016 #20

    haruspex

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    Yes. F=ma will tell you the net force, and from that you can deduce the force provided by the motor.
     
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