Apparent weight loss 2nd Newtons Law elevator problem

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Homework Help Overview

The problem involves a scenario where a woman is descending in an elevator that accelerates downward. The focus is on understanding the apparent weight as measured by a scale during this acceleration, specifically in the context of Newton's second law of motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the woman and the reading on the scale. There is confusion regarding why the scale reads a lower mass value during the downward acceleration and the role of gravitational acceleration in this context.

Discussion Status

Some participants have attempted to clarify the reasoning behind the scale's reading and the forces involved, but there remains uncertainty about the elimination of gravitational acceleration in the scale's reading. The discussion is ongoing with multiple interpretations being explored.

Contextual Notes

Participants note that the scale measures force but displays a mass reading, which adds to the confusion regarding the relationship between force and mass in this scenario.

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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