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Riding the elevator

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A 48 kg passenger rides in an elevator that starts from rest on the ground floor of a building at t = 0 and rises to the top floor during a 10 s interval. The acceleration of the elevator as a function of the time is shown in Fig. 3-32, where positive values of the acceleration mean that it is directed upward. Give the magnitude and direction of the following forces.

    Fig. 3-32 (in words)
    -max positive acceleration = 2 m/s^2 at t = 2 s
    -max negative acceleration = -3 m/s^2 at t = 8.5 s
    -0 acceleration from 4 s to 7 s

    (a) the maximum force on the passenger from the floor

    (b) the minimum force on the passenger from the floor

    (c) the maximum force on the floor from the passenger

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    For part a), I thought it would just be F =(48kg)(2), but that was not correct. I know it's in an upward direction, but I can't seem to find the magnitude.

    Part b) I figured that the minimum force should be when the acceleration = 0, so I tried F=(48)(0) = 0. But it just doesn't seem logical to me...

    Part c) I think the maximum force should be when a = -3, but I plugged it into the formula and it was not correct.

    Any help would be appreciated, thanks!
  2. jcsd
  3. Sep 10, 2008 #2


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    Welcome to PF.

    What do you figure is the effect of gravity on the passenger?
  4. Sep 10, 2008 #3
    Thanks :)

    Since the elevator is accelerating upward for a) and b), would you subtract the acceleration from gravity?
  5. Sep 10, 2008 #4


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    No. You would add the forces. But ... one of the forces is negative because 1 of the accelerations is negative. Accelerating up would add the force. Decelerating (slowing at the top) would subtract.

    (Draw from your own experience in an elevator.)
  6. Sep 10, 2008 #5
    Okay I got it now.. so then would a) and c) have the same magnitude because of Newton's 3rd law?

    Thank you very much by the way, much appreciated.
  7. Sep 10, 2008 #6


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    Homework Helper

    Looks like it don't you think?
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