Weight of a person in a elevator going down

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    Elevator Weight
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SUMMARY

The discussion focuses on calculating the apparent weight of a passenger in an elevator descending with constant acceleration. For a passenger with a mass of 72.2 kg, the scale reads 477.242 N when the elevator descends at 3.20 m/s². If the elevator descends at 9.8 m/s², the scale would read zero, indicating free fall. The effective acceleration on the passenger decreases as the elevator accelerates downward, resulting in a lower normal force, which is reflected in the scale reading.

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  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of forces and normal force
  • Familiarity with the concept of apparent weight
  • Ability to perform calculations involving mass and acceleration
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces in non-inertial frames, particularly in elevator systems.

stealthcataf
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Im new to physics and I am trying to do some problems here, I hope someone can help me, here it goes :

If a passenger with a mass of 72.2kg stands on a weight scale in a elevator. What is the reading on the scale if the elevator goes down with a constant acceleration of 3.20 m/s^2.
 
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If the elevator was descending at constant acceleration of 9.8m/s^2, what would the scale read? Why?

What does this tell you about the effective acceleration on the passenger?
 
Ive done some calcs, correct me if I am wrong

m=72.2kg
a=-3.20 m/s^2

SumFx=0
SumFy=Fn-mg=ma

ma=72.2(-3.20)=-231.04n

Fn=mg+ma=72.2(-9.8)+72.2(-3.20)=-938.6n
Is this the apparent weight?
 
If the elevator is going down, then is he pressing MORE on the scale or less?
It is the normal force that gives the reading on the scale. How is the normal force affected when the elevator is falling?
 
Weight of person=72.2*9.81=708.282
upward lift on person=72.2*3.2=231.04
Therefore net 'weight' of person=708.282-231.04=477.242
 

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