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Apparent Weight in an Elevator

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    While stationary on Earth you have a weight of 550N. When in an elevator that accelerates upward your weight temporarily becomes 590N. When descending, your weight temporarily becomes 510N. Find a) the acceleration you experience as the elevator moves up and b) the acceleration you experience as the elevator moves down.

    2. Relevant equations
    [itex]\Sigma[/itex]F=m*a


    3. The attempt at a solution
    There are two forces acting on you in the elevator: the force due to gravity Fg, which acts downward, and the force of the floor on you Ffloor, which acts upward. According to Newton's 2nd law, the sum of these forces will equal m*a, so we can write
    Ffloor - Fg = m*a
    We know that Fg equals m*g, where g = 9.8m*s-2, so we can find the mass by dividing Fg by g, which yields about 56.1kg. Substituting in all of the known values and solving for a, we find that
    for a)
    a = [itex]\frac{(590 - 550)N}{56.1kg}[/itex] = .713m*s-2
    and for b)
    a = [itex]\frac{(510 - 550)N}{56.1kg}[/itex] = -.713m*s-2
    The answer key and the others students got 10.5m*s-2 for (a) and 9.01m*s-2 for (b), which are the answers I got, plus g.
    I'm pretty sure I'm the one who's wrong here, but I don't understand where I messed up. Help would be greatly appreciated.
     
  2. jcsd
  3. Sep 2, 2011 #2

    lewando

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    Gold Member

    When the elevator is not moving, your acceleration is zero. Therefore, I like your answers better.
     
  4. Sep 2, 2011 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    i agree with lewando :smile:
     
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