@PhantomJay, @BruceW, thank you for your answers! I've managed to solve point b, too!
How did you solve for the acceleration?
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:
[URL=http://s1168.photobucket.com/user/DorelXD/media/IMG_20130627_183955_0_zpsf1698ad4.jpg.html][PLAIN]http://i1168.photobucket.com/albums/r496/DorelXD/IMG_20130627_183955_0_zpsf1698ad4.jpg[/URL][/PLAIN]
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 interested 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!
