Apparent weight loss 2nd Newtons Law elevator problem

AI Thread Summary
A 65-kg woman in an elevator accelerating downward at 0.20g experiences a change in her apparent weight as measured by a scale. The net force equation shows that the normal force (Fn) is 0.80mg, leading to a scale reading of 52 kg instead of her actual weight of 640 N. The confusion arises from the scale reading in mass (kg) rather than force (N), as it measures the force exerted by the woman against it. The gravitational constant (g) is effectively dropped in the scale's reading because it converts the force back to mass. Understanding this relationship clarifies why the scale reflects a lower weight during the elevator's acceleration.
Frankenstein19
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Homework Statement


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?

Homework Equations


ƩF = ma

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?
 
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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
 
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?
 
Biker said:
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

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?
 
Frankenstein19 said:
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?
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.
 
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