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Acceleration of elevator and force in cable

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-12-17_20-25-21.png

    Mass of elevator = 500kg
    Counterweight = 134kg
    Force in cable C1 is 1.2kn

    Determine the acceleration of the elevator and the force in cable C2

    2. Relevant equations


    3. The attempt at a solution

    a = f / m

    a = (C2 - C1) / m2

    a = ((m2 * g) - C1) / m2

    a = ((134 * 9.81) - 1200) / 134 = 0.86m/s2

    Not sure if I need to be taking the mass of the elevator into account somewhere?
     
  2. jcsd
  3. Dec 17, 2015 #2

    gneill

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    Staff: Mentor

    Ignoring the elevator and motor for the moment, draw a free body diagram for the 134 kg counterweight alone. What forces are acting on it? Which ones do you know? What can you conclude about the motion of the counterweight?
     
  4. Dec 17, 2015 #3
    The counterweight has a force of 1.314kN acting downwards and 1.2 kN acting upwards. Therefore the counterweight is going down and pulling the elevator upwards?
     
  5. Dec 17, 2015 #4

    jbriggs444

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    That appears to be what OP has done already. That takes care of the motion of the counterweight.

    The next step would be to see what that implies about the motion of the elevator.
     
  6. Dec 17, 2015 #5

    gneill

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    I saw the C2 in the acceleration equation (second line) and had my doubts as to what the OP had in mind.
     
  7. Dec 17, 2015 #6

    gneill

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    Yes. And what's more, you can find the acceleration of the counterweight, which you actually did when you replaced the "C2" term in your equation with the weight of the counterweight. I couldn't tell if this was a fortuitous mistake or intentional... ... which is why I asked you to concentrate on the counterweight alone.
     
  8. Dec 17, 2015 #7
    I am confused now as to whether or not I have used C2 in the right application?
     
  9. Dec 17, 2015 #8

    gneill

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    If you concentrate on the counterweight alone, C2 is nowhere in the picture. What appears in the FBD for the counterweight?

    We'll get to C2 later when we look at the elevator.
     
  10. Dec 17, 2015 #9
    Disregarding C2. The acceleration on the counterweight alone is:

    F=ma
    a = F/ m
    a = (1314N - 1200N) / 134
    a = 0.85 ??
     
  11. Dec 17, 2015 #10

    gneill

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    That's right. Be sure to include units on results, and indicate the direction of the acceleration.

    Now, what does that tell you about the direction of motion of the elevator car and its acceleration?
     
  12. Dec 17, 2015 #11
    Units are in m/s^2 for acceleration? The elevator car will be moving upwards as the counterweight is moving down, I'm not sure what this suggests about the acceleration..
     
  13. Dec 17, 2015 #12

    gneill

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    Yes, units of acceleration are m/s2.

    How are the counterweight and elevator car connected? Can they have different velocities or accelerations?
     
  14. Dec 17, 2015 #13
    They are connected through the same pulley therefore will have the same velocity and acceleration. Which would mean the acceleration of the elevator is the same as the counter block... Is that the logic you were trying to make me realise?
     
  15. Dec 17, 2015 #14

    gneill

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    It is indeed :smile:

    So with that fact in mind, can you use the known forces operating on the elevator and its acceleration to find the "missing" force C2? Think Newton's 2nd law. An FBD for the isolated elevator car is in order.
     
  16. Dec 17, 2015 #15
    Is it as simple as doing F=ma, f=500*0.85, giving me an answer of 0.425kN. Or do I need to take the 1.2 kN and weight of the elevator into account as well?
     
  17. Dec 17, 2015 #16

    gneill

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    Yup. Hence the FBD for the elevator car. F = MA relates the mass, acceleration, and NET force acting.
     
  18. Dec 17, 2015 #17
    (m1*a) + (m1*g) - C1 = C2
    (500 * 0.85) + (500 * 9.81) - 1200 = 4.13kN

    Is this correct taking all variables into account?
     
  19. Dec 17, 2015 #18

    gneill

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    Looks good to me! :oldbiggrin:
     
  20. Dec 17, 2015 #19
    Thanks for your help! Got there in the end...
     
  21. Dec 17, 2015 #20
    So far i believe this is correct, however I'm
    Well Done! o_O
     
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