# Homework Help: Apparent weight loss 2nd Newtons Law elevator problem

1. Mar 4, 2016

### Frankenstein19

1. The problem statement, all variables and given/known data
A 65-kg woman descends in an elevator that briefly acclerates at 0.20g downward when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read?

2. Relevant equations
ƩF = ma

3. The attempt at a solution
I understand that mg - Fn = m(0.20g) happens and then solving for Fn = 0.80mg.

I also understand that F'n is -0.80mg.

What I don't understand is why the "scale only needs to exert a force of only 0.80mg and that it will only give a reading of 0.80m"; why did the g get eliminated?

2. Mar 4, 2016

### Biker

Because you want the net force to be downward so it can actually accelerate downward and n is just the force that the scale will push you up with. It also equal to the apparent weight because the springs in the scale gets compressed for example. Draw a FBD and try to figure what I am saying

3. Mar 4, 2016

### Frankenstein19

I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?

4. Mar 4, 2016

### Frankenstein19

I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?

5. Mar 4, 2016

### haruspex

The scale reads in kg, not Newtons. But the spring inside it is sensitive to force, not mass. So the scale interprets a force of mg as a mass m, and displays it as such.