What is the Apparent Weight in an Accelerating Elevator?

[Zach Knight]

Homework Statement

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.

Homework Equations

$\Sigma$F=m*a

The Attempt at a Solution

There are two forces acting on you in the elevator: the force due to gravity F_g, which acts downward, and the force of the floor on you F_floor, which acts upward. According to Newton's 2^nd law, the sum of these forces will equal m*a, so we can write
F_floor - F_g = m*a
We know that F_g equals m*g, where g = 9.8m*s^-2, so we can find the mass by dividing F_g by g, which yields about 56.1kg. Substituting in all of the known values and solving for a, we find that
for a)
a = $\frac{(590 - 550)N}{56.1kg}$ = .713m*s^-2
and for b)
a = $\frac{(510 - 550)N}{56.1kg}$ = -.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.

 

[lewando]

When the elevator is not moving, your acceleration is zero. Therefore, I like your answers better.

 

[tiny-tim]

i agree with lewando