# Tension on a string while an elevator is moving

1. Feb 25, 2012

### iJamJL

1. The problem statement, all variables and given/known data
A girl in an elevator is holding one end of a string, and a weight hangs from the other end. In which of the following cases is the magnitude of the tension in the string greatest?

(a) while the elevator moving downward and it is slowing down
(b) while the elevator moves upward at constant speed
(c) while the elevator is moving upward at it is slowing down
(d) while the elevator moves downward with constant speed
(e) while the elevator is at rest

2. Relevant equations
F=ma
F=mg

3. The attempt at a solution
This is an online homework problem. I thought the answer was simple, but I actually got it wrong. I'm trying to figure out why, and here's what I thought as I solved the problem.

When the elevator is at constant speed, that means that there is no acceleration, and gravity is the only resulting force that pushes the weight down. This is the same for when the elevator is at rest. Therefore, choices B, D, and E are all of equal value. In choice A, the elevator is moving downward, and slowing down. That means the hanging weight is being exposed to a changing velocity, or acceleration, in the downward direction. The acceleration causes the weight to become heavier as it slows down. In choice C, the elevator is moving upward and slowing down. The acceleration is already causing the weight to be heavy, and it is becoming lighter.

The way I thought of it was that choices B, D, and E are all incorrect because they have the same values. Choice A means that the tension has some value but has not reached its heaviest yet, whereas choice C means that the tension is decreasing FROM a heavier weight. I thought choice C was the correct answer.

2. Feb 25, 2012

### tiny-tim

hi iJamJL!
i don't see the sense of your own reasoning …

you say that (with C) "it is becoming lighter", and "the tension is decreasing" …

then why do you say that (with C) the tension is greatest??

anyway, that's a confusing way to do it

use F = ma … there are only two forces that make up F, and one of them (the weight) is constant …

so if a (measured upward) increases, that can only be because the other force (tension) increases …

T = ma + mg

so just find the one with the greatest acceleration (upward), and forget about "heavier" and "lighter"​

3. Feb 25, 2012

### iJamJL

Hi again, tiny-tim

Thanks for your reply. Now, I understand how to solve it mathematically. However, when we think about it in real life, it doesn't seem so true..

If I'm in an elevator, and the elevator is already in motion going up toward the 5th floor, I will feel like I am getting lighter as the elevator stops. Wouldn't that mean tension has gone from a stronger force to a weaker force? The same should apply when going down from the 5th floor to the main floor: the second the elevator begins to accelerate downward, I feel as though I am extremely light. As the elevator begins to stop, I feel as though my body is becoming heavier, thus the tension on my body is increasing.

That is what I was thinking when I first read this problem, and that's what I was trying to explain in my original post. However, mathematically, the values are this:

Accelerating down, where g is negative.
F = ma + mg
F = m(-a) + m(-g)
F = -m*(a+g)

Accelerating up:
F = ma + mg
F = ma + m(-g)
F = m*(a-g)

Thank you for helping with the solution. :tongue:

4. Feb 25, 2012

### tiny-tim

hi iJamJL!
yes, all this is correct

(case C: the tension is less)
ah, stop there!

that's where your're going wrong …

mg is not an acceleration, it's a force!!

it has to go on the left:

T - mg = ma

then you can shove mg over to the right …

T = ma + mg

(btw, can't you see that writing these two lines …
F = ma + mg
F = ma + m(-g)​
… is asking for trouble? )