# Problem regarding a pulley and a chain

1. Jun 26, 2013

### DorelXD

Hello everybody! First of all,I would like to apologize for my English because I'm not a native speaker. I'm 16 and I have a pretty good knowledge on physics but I need help with solving certain type of problems. In my country we now have a summer vacation so I decided to look for a forum where I could post my problems and get help with their solutions. Also, I would like to help others. Here is my first problem, in field of mechanics: "Over a fixed pulley with a very small radius and negligable mass passes a chain of mass m and length l. The chain begins to slide. Let x be the distance between its ends. find in term of x: a)the acceleration of the chain and b) the force that the chain exerts on the pulley". I've managed to solve point a: a=g*x/l. But i'm struggling on point b. My first thought was that the force exerted by the chain on the pulley must be equal to the weight of the chain. But this can't be true, because as the chain slides down, the forces exerted by each piece of chain (the one on the left ogfthe pulley and the one on its right) don't remain the same so that must change something on the resultant force acting on the pulley, I think....Please, help solve this problem! I'll wait for your thoughts, suggestions, ideas. And if you don't understand something I wrote here, please tell me so I can reformulate. As I said, I am not a native speaker. Thanks a lot!

2. Jun 26, 2013

### BruceW

Hi, welcome to physicsforums. Your English is very good :) For part b, think about the total forces on the chain in the vertical direction, and the resultant acceleration. You should be able to work out what the force from the pulley must be, by considering the relation between total forces and resultant acceleration.

edit: ah, there is also the problem that the 'resultant acceleration' is different on parts of the chain either side of the pulley. You need to consider both parts of the chain individually, then work out the combined resultant force acting on them.

Last edited: Jun 26, 2013
3. Jun 26, 2013

### DorelXD

Thank you very much for welcoming me and for your answer! Are there any other vertical forces acting on the pulley besides the tensions due to the partial weight of the chain? So far I found I've found only two forces: a tension on the right side of the pulley and another one on the left. I can't see anything else. But I guess there is, isn't it?

4. Jun 26, 2013

### PhanthomJay

How did you solve for the acceleration? Is this a chain on a frictionless table that overhangs one end of the table over a pulley and starts to fall down? And if you solved for the chain tension forces at the pulley, the force of the chains on the pulley must be the vector sum of those tensions. Nice English by the way.

5. Jun 27, 2013

### DorelXD

@PhantomJay, @BruceW, thank you for your answers! I've managed to solve point b, too!

Here's how. First, I drew the following sketch:
[URL=http://s1168.photobucket.com/user/DorelXD/media/IMG_20130627_182856_0_zps01a510c3.jpg.html][PLAIN]http://i1168.photobucket.com/albums/r496/DorelXD/IMG_20130627_182856_0_zps01a510c3.jpg[/URL][/PLAIN]

x is the distance between the two ends of the chain. Y is the distance between the pulley and the first end. Here's a little comment here: because the radius of the pulley is very small (from the statement of the problem) we don't need to worry about the fact that the chain warps around the pulley i.e. we don't need to calculate how much of the length of the circumference of the pulley is covered by the chain. (Hope you guys understood what i meant). The other distance is obviously y-x.

The next step would be to see how much of the total mass of the chain hangs on each side. This is basic math: "If at a length L the chain has a mass m, at a length y what will be the mass of the chain?". There is a (I don't know exactly how to say it in English) a relation of proportionality between those two:

Great, so now we've found out how much of the chain hangs on each side. The next step I took was to make a little idealisation (for the moment). I pictured the original machine like this: an ideal rope that passes over a pulley and two masses m1 and m2 that hang on each side.
I wrote down the second laws of mechanics for each of the masses. This problem is easy. The acceleration is the same for both objects due to the fact that the distance between them stays the same (we had supposed that the rope is ideal i.e. inextensible). The tension in wire is also the same. I obtained the following system and I solved for a:
[URL=http://s1168.photobucket.com/user/DorelXD/media/IMG_20130627_184834_0_zpse4759415.jpg.html][PLAIN]http://i1168.photobucket.com/albums/r496/DorelXD/IMG_20130627_184834_0_zpse4759415.jpg[/URL][/PLAIN]

Ok, so we've found the acceleration for the chain, great.

Here's how I solved point b.

First we take away the idealization we made earlier. Let's consider the initial picture. We write down the same set of equations as earlier, BUT, we keep in mind that now the tension is not the same. So we'll have a tension T1 and a tension T2. We now know the acceleration. So, we solve the system for T1+T2, because that is the force acting on the pulley. The only forces acting on the pulley are the two tensions I've just mentioned. So, we've found out the force acting on the pulley in terms of m,g,l,x which we know all.
[URL=http://s1168.photobucket.com/user/DorelXD/media/IMG_20130627_192433_0_zps9d6cab12.jpg.html][PLAIN]http://i1168.photobucket.com/albums/r496/DorelXD/IMG_20130627_192433_0_zps9d6cab12.jpg[/URL][/PLAIN]

I chose to explain in detali because I thought that maybe there are other people intrested in the soulution of these problems. Please, give some suggestion or critics about this solution. I have a question, though. I've checked the solution and my answer is good (the solution has only the final formula for the acceleration and force, not the complete steps in solving the problem). But, I'll have to admit that I don't know if I can explain clearly why I did that idealization to find the acceleration? I think I've done it instinctively. While visualising how the system moves, I said to my brain: "freeze the image!". And after that it became clear, that if the system is freezed, the situation can be idealized as I did earlier. But that kind of explanation simply is not enough for me. There must be another one, more logical. Can somebody explain why it was ok to idealize? Or the idealization was wrong from the very begging and I found the good result by accident.

I'll wait four your opinions and feedback! Thanks again for answering me!

6. Jun 27, 2013

### PhanthomJay

I will review your work and questions later when I get some free time, but in the meantime, physics, engineering, math, or doctorate college professor seems to be your calling. You express yourself extremely well, and your work is of very high quality for a person of your age. I wish you well. I or others will get back to you soon.

7. Jun 27, 2013

### DorelXD

Thank you sir, I am really flattered. Nobody has ever told me something like that. I do it with passion, and I try really hard to understand completely every statement I make when I talk about physics. Also, when I learn something new I try to get it right, because once you got it you never forget it, you simply can't. Right now I'm trying to find a better way to browse this forum on Android because I'm not satisfied with how Opera manages its content, and I'd really want to start helping others around, beacause It's a very very beautiful forum. Do you have any solution for my problem regarding viewing this forum on Android? I'll wait for you or for others to take a more detailed look at my solution, as you said. Thank you again, for everything! Oh, and my English is improving constantly because I learn every day. After I finish this summer I hope I will be at a new level.

8. Jun 27, 2013

### PhanthomJay

I don't, but try asking this question in the "Forum Feedback" sub-forum at the bottom of the home page. Some IT expert can probably help.
Incredibly brilliant work! In regard to the tension in the chain on either side of the rope, in the ideal situation with the chain being inextensible, and with the pulley massless, and with no axle friction in the pulley, then the tensions in the chain on either side of the pulley ARE the same. It is OK to make this idealization since the problem assumes the ideal case. There can be no net torque on the pulley if its mass and friction are negligible, so the tensions must be the same. You said in part b that you removed the idealization, but you really did not. You solved for T1 +T2, but you will find that T1 = T2. I do want to point out that T1 and T2 are the maximum tensions in the chain. The tension in the chain is variable, varying from 0 at the free ends to T1max or T2max (T1max=T2max) at the pulley. When you solved for the acceleration, you took the entire weight of the chain on either side, implying in your free body diagram you were solving for the maximum tension where the rope enters or exits the pulley. I hope this explanation is clear to you, but you could probably say it better. The solution to this problem to solve for other unknowns, like the time it takes for the chain to fall completely off the pulley, is quite complex.
Amazing! Keep up the good work!

9. Jun 29, 2013

### DorelXD

Thank you for your review!!!! It helped a lot. Indeed, the two tensions are equal, one can easily verifiy it. I put the pen down and I verified it myself. It becomes clerarer to me as the time passes that my assumption was ok. But I'm not 100% percent convinced yet. Also I read carefully your review and I totally agree that my idealization was for the whole problem not just for point a. I also understood from you that the two tensions have a maximum value at the contact points with the pulley and I totally agree. Thank you again! I hope I'll understand completely in a shor amount of time the solution that I found using the idealization.

10. Jun 29, 2013

### PhanthomJay

OK, great.

If you haven't found this yet
try