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Elevator Problems

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A 56.0 kg girl weighs herself by standing on a scale in an elevator. What is the force exerted by the scale when the elevator is descending at a constant speed of 10 m/s?

    What is the force exerted by the scale if the elevator is accelerating downward with an acceleration of 2.4 m/s2?

    If the elevator's descending speed is measured at 10 m/s at a given point, but its speed is decreasing by 2.4 m/s2, what is the force exerted by the scale?



    2. Relevant equations
    Fnet=ma

    How do we find acceleration without having a time?
     
  2. jcsd
  3. Sep 16, 2007 #2

    Doc Al

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    In all cases you are given the acceleration (or all the information needed to figure it out). You don't need the time.
     
  4. Sep 16, 2007 #3

    nrqed

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    Draw a free body diagram and then apply [tex] \sum F_y = m a_y [/tex]. That is all there is to it, really. Just be careful to include the correct sign of [tex] a_y [/tex].
     
  5. Sep 16, 2007 #4
    Okay so I have F=(m*-a)+w for the second part
     
  6. Sep 16, 2007 #5
    Now for the first part I am not sure what to do with the speed??
     
  7. Sep 16, 2007 #6

    Doc Al

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    All you care about is acceleration.
     
  8. Sep 16, 2007 #7
    I don't understand
     
  9. Sep 16, 2007 #8

    Doc Al

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    Reread the statement of the first problem. How is the velocity changing?
     
  10. Sep 16, 2007 #9
    it's descending at 10 m/s
     
  11. Sep 16, 2007 #10

    Doc Al

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    Acceleration is the rate of change of velocity. How is the velocity changing? (Hint: Read it carefully. :wink:)
     
  12. Sep 16, 2007 #11
    it's not its constant
     
  13. Sep 16, 2007 #12

    Doc Al

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    Exactly! So what is the acceleration?
     
  14. Sep 16, 2007 #13
    0 m/s^2
     
  15. Sep 16, 2007 #14

    Doc Al

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    You got it.
     
  16. Sep 16, 2007 #15
    ohhh so the only force acting is the weight.
     
  17. Sep 16, 2007 #16

    Doc Al

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    No. If the only force acting on her was her weight, the girl would be in free fall.
     
  18. Sep 16, 2007 #17
    so how is the third part any different from the second if the acceleration is the same?
     
  19. Sep 16, 2007 #18

    Doc Al

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    The accelerations are not the same. Direction counts!
     
  20. Sep 16, 2007 #19
    but they are both going down?
     
  21. Sep 16, 2007 #20

    Doc Al

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    No. The second one has acceleration going down (it tells you that), but the third one you have to figure out the direction of acceleration by reading carefully.
     
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