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Weighing yourself in the elevator

  1. Jul 5, 2009 #1
    Need help:Weighing yourself in the elevator-acceleration problem

    1. The problem statement, all variables and given/known data

    If you stand on a spring scale in your bathroom at home, it reads 600N, which means your mass is 60kg. If instead, you stand on the scale while accelerating at 2 m/s^2 upward in an elevator, what would the scale read?

    a. 120N
    b. 480N
    c. 600N
    d. 720N

    2. Relevant equations


    3. The attempt at a solution

    So I asked my professor this question and he came up with some complicated answer that I didn't really understand. My own logic was that W=mg, W=60kg * (2m/s^2+9.8m/s^2), W=720N which is the correct answer. However, this is not the same steps the professor used which was something like F=N-W=ma and he did some subtraction work and somehow got 720N.

    So can somebody explain to me the correct way to solve this problem?
    Last edited: Jul 5, 2009
  2. jcsd
  3. Jul 5, 2009 #2


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    Re: Need help:Weighing yourself in the elevator-acceleration problem

    That works too of course.

    The 60 kg student is accelerating upward at 2 m/s2. The Normal force of the student less the gravitational weight must yield an m*a of the same acceleration as the floor.
    N is what the spring will register and rearranging that normal force = m*a + W = m*(g + a)

    Your way works fine as well.
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