# Homework Help: Conceptual elevator and atwood pulley problem

1. Sep 25, 2011

### runphysicsrun

1. The problem statement, all variables and given/known data

A simple Atwood machine composed of a single pulley and two masses, m1 and m2 is on an elevator. When m1= 44.7kg and m2=45.3kg, it takes 5.00s for mass m2 to descend exactly one meter from rest relative to the elevator. What is the elevator's motion? (That is, is it moving with constant velocity or accelerating up or down?

2. Relevant equations

$\Sigma$F=m*a
Y=y0 +v0t+1/2at2

3. The attempt at a solution
I found the $\Sigma$F on m2 $\Sigma$F on m1 and solved for a, which I took to be the acceleration of mass 2 when the pulley wasn't on the elevator (which I got to be -0.06) Then, I used the kinematic equation to calculate the actual acceleration of mass 2 (which I got to be -0.08 ) Comparing the two, I reasoned that mass 2 was accelerating faster downward on the elevator, which meant some downward force had been applied. Thus, I came to the conclusion that the elevator must have been accelerating downward. I was wrong. It's accelerating upward. Can anyone explain this to me?

Last edited: Sep 25, 2011
2. Sep 25, 2011

### PeterO

You correctly found the mass accelerated down at a greater rate than in a stationary lift.
The same thing would have happened in a stationary lift on a planet where gravity was a little stronger than here on Earth.
In a stronger gravity field, you would feel a little bit heavier.

OK. When do you feel heavier in a lift? When it is accelerating up? When it is accelerating down?

3. Sep 26, 2011

### runphysicsrun

When it's accelerating up? Because sense of weight comes from Normal Force and the normal force is greater in a lift accelerating up. But I still don't get how this gives mass 2 a more negative accel?

When you're going up, if you feel heavier, mass 2 would be heavier as well. But wouldn't the whole pulley system be heavier? The acceleration is changing by the same amount for both objects. So, really I have no idea how to tell that the elevator is going up or down.

Last edited: Sep 26, 2011
4. Sep 26, 2011

### PeterO

You accurately calculated an acceleration of -0.06 using a g value of 9.8 for a stationary lift.

Try re-calculating with a g value of 10.8 and see what you get.

5. Sep 26, 2011

### PeterO

The net force on the two mass system is m2.g - m1.g

so (m2 - m1).g

m2 - m1 doesn't change, so the only way to get a larger acceleration is to have a higher value of g - an environment in which your weight is higher.

One way to do that is put the masses in a lift that is accelerating up. You know the effective g is higher in there, as you are quite familiar with feeling heavier in such a lift.

EDIT: A pendulum even swings with a smaller period in a lift that is accelerating up, due to the artificially high g value.

EDIT 2: 1:00 am here, I am off to bed.