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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
    F=ma..


    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

    rock.freak667

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    Homework Helper

    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

    rock.freak667

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

    as

    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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