# Apparent weight (elevator)

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1. Oct 4, 2016

### AlexandraMarie112

1. The problem statement, all variables and given/known data
Hi everyone,I just want to know if my answers are correct.

2. Relevant equations

3. The attempt at a solution
1. For each situation, indicate whether the normal force exerted on you will be gater tha,less than or equal to your true weight.

a) Elevator starts to move down from the top floor n<w
b)Elevator moves down between floors with a constant speed n=w
c)Elevator moving down slows down at bottom floor n>w
d) Elevator starts to move up from the bottom floor n>w
e)Elevator moves up between floors with a constant speed n=w
f)Elevator moving up slows down at the top floor n<w

2. In what direction is the elevators velocity in the case when the normal force acting on you is less than the weight force? Downwards

In what direction is the elevators acceleration in the case when the normal force is less than the weight force? Downwards

2. Oct 4, 2016

### Efrain Garcia

Actually, they would always be equal. Weight is the force of gravity acting on mass, so as the elevator goes down, the normal force decreases, but so does your weight. They will always be in balance. I think what he meant was not weight, but mass. Mass doesn't change, as it is a measure of "stuff" something is made of. In that case, the work seems about right.

3. Oct 4, 2016

### TSny

The force of gravity (weight) acting on a mass is the attractive force which the earth exerts on the mass through gravitational interaction. The force of gravity on the mass does not depend on whether or not the mass is moving. If you throw a baseball upward, the force of gravity acting on the ball is the same whether the ball is moving upward, instantaneously at rest at the top of the flight, moving downward, or sitting on the ground after it lands.

4. Oct 4, 2016

### TSny

Your answers to 1 look good.

Is this answer in agreement with all of your answers to 1?

Yes.

5. Oct 4, 2016

### Efrain Garcia

That can't be correct. If you step on an elevator and stand on a scale, then the elevator goes down, the scale will register LESS weight. And this is where points of reference come in. If you take the point of view of the baseball, and not that of an outside observer, the force of gravity acting upon you IS different. It decreases and increases depending on what point of the arc you are in. Similarly, if you are inside an elevator, that means the only point of reference you have is inside the elevator. That being the case, the force you feel does change.

6. Oct 4, 2016

### AlexandraMarie112

So for my question 2 , it would be upwards and downwards?

7. Oct 5, 2016

### TSny

Yes, but the scale is not measuring the force of gravity acting on you. The scale reads whatever force is pushing on the scale. This force that you are pushing on the scale is called the "normal force". Sometimes the reading of the scale is called the "apparent weight". But this is not the force of gravity.
This is not true. Why would the earth attract the baseball differently while it is moving upward compared to moving downward?
The force you "feel" in an elevator is not the force of gravity. You feel the force that the floor of the elevator pushes upward on you. Again, this is the "normal force".

8. Oct 5, 2016

### TSny

Yes, the direction of the velocity could be upward or downward. Would it also be possible to have n < W when the elevator has zero velocity?

9. Oct 5, 2016

### Efrain Garcia

What is the difference, really? To the observer in the elevator, the force of gravity does change. If you are an observer outside the elevator, then you could say the attraction of gravity is the same, but both would be correct. I guess it just comes down to Newtonian v Einsteinian definitions of "gravity". For example, if you replaced that ball with an elevator, and pretend you are in it, you will experience a greater amount of force, then you would feel a normal amount of force, then you would feel less as the ball dropped back. As your weight increases and decreases, so will the normal force. And if your weight increases and decreases, that must mean that the net forces acting on you are influencing one another. While you could say the force of gravity stays the same, and you would be factually correct, that is not what is experienced inside the elevator. As for the scale, the scale is reading weight, so it is solely influenced by "gravity". The normal force is what counteracts whatever force gravity and other factors are presenting.

10. Oct 5, 2016

### TSny

Efrain,
The usual interpretation, in Newtonian mechanics, of "the force of gravity" on an object near the earth is that it refers to the attractive action-at-a-distance force that the earth exerts on the object. It is equal to mg where m is the mass of the object and g is the acceleration of gravity (which has a fixed value near the surface of the earth). The value of mg does not depend on the motion of the mass m. The force of gravity is also called "weight"; or, sometimes "true weight" when distinguishing it from "apparent weight" (e.g., scale reading in the elevator).

In general relativity, "force of gravity" no longer has a definite meaning. But I believe that the problem stated in post #1 is given in the context of Newtonian mechanics.

11. Oct 5, 2016

### vela

Staff Emeritus
According to your answers to #1, you can also have n>w when the elevator is going up or when it's going down.

So what do you suppose is the point of the questions in #2? What are they trying to get you to realize?

12. Oct 5, 2016

### Orodruin

Staff Emeritus
This is true only in the accelerating phase. When the elevator decelerates in order to come to a halt the scale will register more weight and when the elevator travels at constant velocity it will register the same weight, even if the elevator is going down. The direction of acceleration is what is important, not the direction of motion.

13. Oct 7, 2016

### David Lewis

That's correct. Both definitions are acceptable but may yield different answers. For example, an astronaut in orbit is weightless only under the apparent weight definition. Most problems want you to assume the force of gravity definition unless otherwise specified.

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