# Homework Help: Weight Change in an Elevator

1. Oct 14, 2008

### keemosabi

1. The problem statement, all variables and given/known data
A woman stands on a scale in a moving elevator. Her mass is 58.5 kg, and the combined mass of the elevator and scale is an additional 815 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of 9510 N. What does the scale read during the acceleration?

2. Relevant equations
Fnet = ma

3. The attempt at a solution
I found the acceleration of the entire elevator by doing 9510 - 815(9.8) = 1523. I divided this by 815 to get 1.8687. It is in the negative direction, so it is -1.8687 If this is the acceleration of the elevator, it must also be the acceleration of the person. As a result, I did Fnet = ma, Fnet = (58.5)(-1.8697) and got Fnet = 682.31895. I put 682.31895 as the what the scale reads but it says I got the answer wrong. What did I do incorrectly?

2. Oct 14, 2008

### D H

Staff Emeritus
How many scales have you stepped on that give a reading in Newtons? When you are in an elevator that starts going up, do you feel momentarily heavier or lighter?

3. Oct 14, 2008

### keemosabi

The space to fill in the answer requests that it be given in Newtons.
I believe you feel momentarily heavier.

4. Oct 14, 2008

### D H

Staff Emeritus
You forgot about the mass of the woman.

5. Oct 14, 2008

### keemosabi

Is that because it says that the elevator and the scale provide an additional mass?

6. Oct 14, 2008

### D H

Staff Emeritus
Yes. That's why I highlighted the word additional.

7. Oct 14, 2008

### keemosabi

Code (Text):
9510 - 815(9.8) = 1523. I divided this by 815 to get 1.8687.

So basically just add 58.5 to each of those calculations where I used 815?

8. Oct 14, 2008

### D H

Staff Emeritus
What does that get you?

9. Oct 14, 2008

### keemosabi

It accounts for the weight of the woman in all of my calculations.

10. Oct 15, 2008