Solving the Elevator Problem: Acceleration in Downward Direction?

[Balsam]

Homework Statement

You are standing on a scale in the elevator. You weigh 500N. What would happen to the scale reading if you slow down, going upwards?

Homework Equations

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The Attempt at a Solution

My answer: Acceleration would occur in the downwards direction because if you decelerate in one direction, you are accelerating in the opposite direction(not sure if this is true). So, FN(the reading on the scale) must be less than Fg(500N) in order to have a net force in the direction of acceleration(down).

Is my answer correct?

 

[Chestermiller]

How about demonstrating this with free body diagrams and equations?

 

[Balsam]

How about demonstrating this with free body diagrams and equations?

The question just asks for the action and reaction forces between the scale and the person and what the reading on the scale would be. My teacher wrote something different. She basically said that the scale does not have to push up on the person with as much force in this case

 

[Chestermiller]

The question just asks for the action and reaction forces between the scale and the person and what the reading on the scale would be. My teacher wrote something different. She basically said that the scale does not have to push up on the person with as much force in this case

Is this a high school Physics course?

 

[Balsam]

Is this a high school Physics course?

yes

 

[Chestermiller]

yes

For a high school course, I guess her "arm waving" explanation is OK, but I liked you own attempt to explain it in terms of forces and accelerations much better. Here's what I had in mind: There are two forces acting on you, the upward force of the scale N and the downward force of gravity mg (aka your actual weight). So a force balance on you in the upward direction gives $N-mg=ma$, where a is your upward acceleration. If we solve this equation for the upward force that the scale exerts on you, we get $N=m(g+a)$. Since the elevator is slowing down, it's upward acceleration is negative, and the upward force of the scale is less than mg. This is basically what you were saying in words, and is a better explanation than your teacher's.

 

[Balsam]

For a high school course, I guess her "arm waving" explanation is OK, but I liked you own attempt to explain it in terms of forces and accelerations much better. Here's what I had in mind: There are two forces acting on you, the upward force of the scale N and the downward force of gravity mg (aka your actual weight). So a force balance on you in the upward direction gives $N-mg=ma$, where a is your upward acceleration. If we solve this equation for the upward force that the scale exerts on you, we get $N=m(g+a)$. Since the elevator is slowing down, it's upward acceleration is negative, and the upward force of the scale is less than mg. This is basically what you were saying in words, and is a better explanation than your teacher's.

What if the elevator sped up, going downwards. How would you explain it then?

 

[Chestermiller]

What if the elevator sped up, going downwards. How would you explain it then?

If a were negative enough to be equal to -g, the elevator would be in free fall, and it would as if there were no gravity in the elevator. It would be the same as if you cut the cable. You would float. If a were even more negative, it would be as if you were firing rockets upward from the roof of the elevator, and you would get pinned against the ceiling.

 

[Balsam]

If a were negative enough to be equal to -g, the elevator would be in free fall, and it would as if there were no gravity in the elevator. It would be the same as if you cut the cable. You would float. If a were even more negative, it would be as if you were firing rockets upward from the roof of the elevator, and you would get pinned against the ceiling.

Is there a way to explain the first case/question using apparent weight

 

[Chestermiller]

Is there a way to explain the first case/question using apparent weight

Yes. The apparent weight is N = m(g + a) < mg