Gaining and losing weight in an elevator

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    Elevator Weight
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Homework Help Overview

The problem involves a person standing on a scale in an elevator and determining the weight displayed under various conditions of motion, including constant velocity and acceleration. The subject area includes concepts from mechanics, specifically relating to forces and weight under different acceleration scenarios.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for weight under different conditions, questioning the validity of certain assumptions and results. There is an exploration of the effects of constant velocity versus acceleration on the perceived weight.

Discussion Status

Some participants have provided feedback on the calculations, indicating that certain interpretations may not align with physical expectations. There is a recognition of correct calculations in specific scenarios, while others are encouraged to reconsider their reasoning regarding weight perception during steady motion.

Contextual Notes

Participants are navigating through the implications of acceleration and constant velocity on weight measurement, with some expressing confusion over the results and seeking clarification on the underlying principles.

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Homework Statement



Person with mass m = 80 kg stands on a scale during a ride in an elevator. How much weight shows on the scale, when the elevator is being moved evenly upwards (downwards) with velocity (speed) v = 1 m/s. How much weight shows on the scale when the elevator starts to move upwards (downwards) or when moving upwards (downwards) starts to stop; if we suppose that at this time the absolute value of acceleration or deceleration is equal to 1 m/s²?

Homework Equations



W= mg

The Attempt at a Solution



Moving with constant velocity of 1 m/s:

Elevator is stopped: a= 0, g= 9.8, W= mg= 80 kg * 9.8 m/s²= 784 N
Elevator is moving up: a= g, W= 2mg= 2 * 80 kg * 9.8 m/s²= 1568 N
Elevator in moving down: a= -g, W= 0

Moving with an acceleration of 1 m/s²:

Elevator is moving up: W= m (g + a)= 864 N
Elevator is moving down: W= m(g - a)= 704 N
Elevator is going down and stopping: W= m(g- (-a))= 864 N
Elevator is going up and stopping: W= m( g+ (-a))= 704 N

Are my conclusions correct?
 
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Elevator is moving up: a= g, W= 2mg= 2 * 80 kg * 9.8 m/s²= 1568 N
Elevator in moving down: a= -g, W= 0
These don't make sense! You don't feel double your weight when going steadily upward and you don't float around like an astronaut when going steadily downward. Use the same formula you had for the last bet W= m (g + a) with a=0 and you'll get the correct answer.

The 864 and 704 look good!
 
So, calculation when elevator is stopped: a= 0, g= 9.8, W= mg= 80 kg * 9.8 m/s²= 784 N is correct?
Elevator is moving up: W= m(g + a)= 80 kg (9.8 m/s² + 0)= 784 N
Elevator in moving down: W= m(g - a)= 784 N.
So, in this cases I get the same result, or one has to positive and the other negative?
 
Last edited:
Those look good. When the acceleration is zero, you always get 784.
You can't tell by your weight whether you are stopped or moving any direction at constant speed. Only acceleration changes your weight.

Unless you leave the Earth and go somewhere where gravity varies.
 
Thank you for explaining!
I always wondered how life would be with no gravity!
 
Most welcome.
No gravity - like being in an elevator when the cable breaks. Not for me.
 
Well, it's seem like no gravity spells disaster! So, no, neither for me!
 

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