Conceptual elevator and atwood pulley problem

In summary: I will try and answer any remaining questions tomorrow morning.In summary, the elevator is accelerating upward.
  • #1
runphysicsrun
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Homework Statement



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?

Homework Equations



[itex]\Sigma[/itex]F=m*a
Y=y0 +v0t+1/2at2

The Attempt at a Solution


I found the [itex]\Sigma[/itex]F on m2 [itex]\Sigma[/itex]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:
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  • #2
runphysicsrun said:

Homework Statement



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?

Homework Equations



[itex]\Sigma[/itex]F=m*a
Y=y0 +v0t+1/2at2

The Attempt at a Solution


I found the [itex]\Sigma[/itex]F on m2 [itex]\Sigma[/itex]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?


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
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.
 
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  • #4
runphysicsrun said:
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?

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
runphysicsrun said:
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.

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.
 

Related to Conceptual elevator and atwood pulley problem

1. What is a conceptual elevator and Atwood pulley problem?

A conceptual elevator and Atwood pulley problem is a physics problem that involves a system of pulleys and masses, typically an elevator or a counterweight. The problem requires the use of Newton's laws of motion to determine the acceleration and tension in the system.

2. How do you approach a conceptual elevator and Atwood pulley problem?

The first step is to draw a free-body diagram of the system, labeling all the forces acting on each mass. Then, use Newton's second law (F=ma) to set up equations for each mass. Finally, solve the equations simultaneously to find the acceleration and tension in the system.

3. What are the common misconceptions when solving a conceptual elevator and Atwood pulley problem?

One common misconception is assuming that the acceleration of the system is equal to the acceleration due to gravity (9.8 m/s^2). In reality, the acceleration depends on the masses involved and the direction of motion. Another misconception is neglecting the tension in the string, which is crucial in determining the acceleration of the system.

4. How does the number of pulleys affect the solution to a conceptual elevator and Atwood pulley problem?

The number of pulleys does not affect the solution to the problem. The key factor is the number of supporting ropes or strings. Each supporting rope adds an additional tension force to the system, which must be taken into account when solving the equations.

5. Can a conceptual elevator and Atwood pulley problem be solved using energy conservation?

Yes, it is possible to solve a conceptual elevator and Atwood pulley problem using energy conservation. However, this method is more complex and requires a good understanding of the conservation of mechanical energy and work-energy theorem.

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