Gaining and losing weight in an elevator

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AI Thread Summary
When a person with a mass of 80 kg stands on a scale in an elevator, the weight displayed varies based on the elevator's movement. At a constant velocity of 1 m/s, the scale shows 784 N regardless of the elevator's direction. When the elevator accelerates upwards at 1 m/s², the weight increases to 864 N, while it decreases to 704 N when accelerating downwards. The calculations confirm that only acceleration affects the weight reading, not constant velocity. Overall, the discussion emphasizes that weight changes only during acceleration, not during steady movement.
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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?
 
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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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