Distance Traveled by an object on a pulley

[jk494]

Hi everybody, I've really been struggling with this basic idea, I have drawn it out about half a dozen times, watched numerous videos, read descriptions, played with applets but I still can't see it.

Lets just start with a simple atwood machine with a massless rope and frictionless pulley. There are two masses A and B at heights y1 and y2 from the 0 point at the center of the pulley. You pull the rope through a distance d. because the rope does not stretch, the distance between points on the rope does not change(this is why the acceleration is the same for both sides of the atwood machine). I have marked points P, Q, and R in the attached diagram. As I said before, the relative distance between these points are constant. Therefore, when P moves distance d, so should Q and R since this is just equivalent to sliding the rope along a numberline (also drawn). The final y coordinates of the masses are Ya = y1 + d and Yb = y2 - d. The correct answer is stated as Ya = y1+d/2 and Yb = y2 - d/2.
What am I missing?

 

[PeterDonis]

Hi @jk494 and welcome to PF!

Unfortunately, it looks like the images you posted aren't visible in your post. To make them visible you will need to upload them to PF (there should be an "UPLOAD" button at the lower right of the edit window when you are editing a post).

 

[Dullard]

Your answer appears correct, for the problem that you state.

 

[jk494]

Wow, I did not expect to see that. I really thought I was going crazy. I did an experiment with measuring tape and got the same result as my calculations.
For some additional context, here is the problem I was working on and its solution. Is there an error in this?

 

[PeterDonis]

Your answer appears correct, for the problem that you state.

This is ambiguous; the problem that he states in the OP is not precisely the same as the problem in the images given in post #4. See below.

Is there an error in this?

No, it's correct. But note carefully what it says: if a length $L$ of rope passes through the person's hands, then both the person and the block rise a distance $L / 2$. Do you see the key distinction that is being made?

 

[Dullard]

No ambiguity. The preposition was included in my response precisely because I suspected that the stated problem (OP) and the actual problem were different. Not sure how I could (in post #3) respond precisely to what was divulged in post #4.

 

[jk494]

Please forgive me for the confusion, I should make a separate thread addressing the problem I just posted in my reply.

 

[PeterDonis]

The preposition was included in my response precisely because I suspected that the stated problem (OP) and the actual problem were different.

Then you should say so instead of obliquely hinting at it.

Not sure how I could (in post #3) respond precisely to what was divulged in post #4.

You couldn't. But you could be clear about what you suspect.

 

[jk494]

I included the diagrams in the other thread to show you how I was visualizing the problem.

I see the distinction but if all points on the rope move a distance L then the block also accelerates with acceleration a to move up a distance L. When you are climbing the rope, you are accelerating relative to the rope with a as well. because there is no stretching the rope is accelerating in the same direction as the block which is -a in the direction opposite you. relative to the ground your acceleration is a-a = 0 so you are moving along the rope but staying in place as the rope moves down just as much as you climb.

What am I missing here?

 

[PeterDonis]

all points on the rope move a distance L

That's not what the solution says. It says that a length $L$ of rope passes through the person's hands. Think carefully about what that means.

 

[jk494]

Does that mean the person moves a distance L along the rope? I can only picture myself climbing a rope and I moved a distance L along it. Because the length of the rope is constant, the L I pulled down had to come from the other end of the rope, so the point of the rope that was formerly distance L away from me is in my hand when I'm done climbing?

 

[PeterDonis]

Does that mean the person moves a distance L along the rope?

Yes.

the point of the rope that was formerly distance L away from me is in my hand when I'm done climbing?

Yes.

 

[jk494]

Ok so doesn't that mean that the point 2L away is now only L away? So for any number c the distance away after climbing is cL - L? That sounds to me like every point on the rope has to move L.

 

[PeterDonis]

doesn't that mean that the point 2L away is now only L away?

Yes. Away from your hands.

That sounds to me like every point on the rope has to move L.

Move L relative to what? See above. And again, think carefully about what that means.

 

[jk494]

Every point is moving L relative to the person on the rope. That can happen 2 ways. If the block stays in place and the person climbs a distance L up the rope. Or, the person stays in place relative to the ground and pulls a length L of the rope down through their hands. My solution says that the block is accelerating upwards relative to the ground at a m/s^2. The tension is constant in the rope so the entire block rope system also accelerates at a. that means the rope on the person's side of the pulley is accelerating -a m/s^2 relative to the ground. the person is accelerating a m/s^2 relative to the rope. Therefore the person is accelerating a- a = 0 relative to the ground. Therefore the only possibility is that the person is pulling the rope while staying in place relative to the ground.

 

[PeterDonis]

Every point is moving L relative to the person on the rope. That can happen 2 ways. If the block stays in place and the person climbs a distance L up the rope. Or, the person stays in place relative to the ground and pulls a length L of the rope down through their hands.

No, neither of these are physically possible. Think carefully about what happens as the person moves the rope through their hands.

My solution says that the block is accelerating upwards relative to the ground at a m/s^2.

The key to the problem is not acceleration relative to the ground. It's the force balance. Consider: in terms of distance along the rope, the block is a length $L$ closer to the person if a length $L$ of rope passes through the person's hands. How is that length $L$ divided between the distance the person rises relative to the ground, and the distance the block rises relative to the ground?

 

[jk494]

No, neither of these are physically possible. Think carefully about what happens as the person moves the rope through their hands.

Why aren't they possible? I really have no idea how to go about visualizing this properly.

The key to the problem is not acceleration relative to the ground. It's the force balance. Consider: in terms of distance along the rope, the block is a length $L$ closer to the person if a length $L$ of rope passes through the person's hands. How is that length $L$ divided between the distance the person rises relative to the ground, and the distance the block rises relative to the ground?

What does "force balance" mean? Why would the person rise relative to the ground if the rope is moving down as they are pulling it? I have watched videos of atwood machines and when one side goes up a distance the other side goes down the same distance. However the masses remain at the ends of the rope which are fixed along its length. The person climbing would be equivalent to a block with a hole in it which is small enough to hang on the rope through friction but loose enough to pull a length of rope through it. You can hold the block with the hole in place relative to the ground and pull a length L through the hole. Because the masses are balanced, the block will remain at the same height but be moved along the length of the string.

I don't know how to imagine this situation any other way. I am quite literally stuck in my mindset. This happens to me frequently when doing physics. How do I get unstuck?

 

[PeterDonis]

Why aren't they possible?

Because one object can't stay in the same place while the other changes height.

What does "force balance" mean?

.
The forces on the person, the rope, and the block. The solution you linked to describes them.

I have watched videos of atwood machines

This isn't the same as an atwood machine. In an atwood machine, both sides have a block that is fixed to the rope, so the length of the rope between them is constant. In the problem you're considering, the person moves relative to the rope, so the length of the rope between them shortens. That makes a difference.