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Homework Help: Magnitude of force (elevator question)

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Pat Summit (60 kg) is standing in an elevator which has a
    velocity of 10.0 m/s downward and an acceleration of 5.00
    m/s^2 upward. What is the magnitude of the force exerted by
    the floor of the elevator on her? Is the elevator speeding up
    or slowing down?

    2. Relevant equations

    3. The attempt at a solution
    I really am not sure where to start here..
    I drew a free body diagram but, does acceleration count as a force?
    How does velocity come into play?
  2. jcsd
  3. Oct 1, 2009 #2


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    On the free body diagram, there is normal reaction and weight. So what is the resultant force 'ma' equal to?
  4. Oct 1, 2009 #3
    I got the answer :)
    using V*m + W - F = m*a_y... so..
    F = (60*9.81) - (-5m/s^2)*m + (60kg * 10m/s)
    thanks for the help.
  5. Oct 1, 2009 #4


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    While I am glad to see that you got your answer, your equation doesn't seem to 'add' up


    60*9.81 = N

    5 m/s2 * 60kg = N

    60kg*10m/s = kgm/s

    which means you are adding force+force+ momentum and getting force. :confused:
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