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Quick elevator force problem

  1. Dec 16, 2004 #1
    The question reads "A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only .75 of the person's regular weight. Calculate the acceleration of the elevator, and the direction."

    I know that the direction is going down because..it just seems like common sense, and I know the answer is 2.5m/s^2, but I don't understand how that was reached...any help? Thank you.
     
  2. jcsd
  3. Dec 16, 2004 #2
    Net sum of forces has to be constant. Taking 'up' as the positive direction, we can write:

    N-ma=mg,

    where a is the acceleration and N the normal force. The scale shows the normal force. Thus, N=0.75mg and we can write

    0.75mg-ma=mg => a=-0.25*g=-2.45 m/s^2 (9.8 m/s^2)

    Thus, acceleration is 2.45 in the negative direction, i.e. down.
     
  4. Dec 16, 2004 #3
    Good intuition on the direction. Consider that a reading of the person's weight is a reading of the force the person is exerting on the scale. In other words, 1.00*mg if the person is only under the force of Earth's gravity near sea level. Therefore, the net vertical force on the person right now is 0.75*mg.
     
  5. Dec 16, 2004 #4
    Ah, thank you very much! I guess I was looking at the wrong thing..I was trying to figure out the forces of the elevator, not the person standing on the scale.
     
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