# Homework Help: Detailed problem I did need verification

1. Oct 15, 2004

Hello all,
I need to find the acceleration of this system. I did all the calculations but I'm stuck at the end and I'm pretty sure I did something wrong.. Could someone verify this? I just scanned the paper I'm working on. Also, sorry if I dont use the same signs you use, I'm translating this from french.
PS: The teacher told me I can group pulley2 and mass2 so that's what I called M3

Thanks a lot

http://img99.exs.cx/img99/5910/Picture61.jpg
Click to enlarge

I recommend you save it and scale down the zoom

2. Oct 16, 2004

### liaoge

3. Oct 16, 2004

### MiGUi

It surprised me a lot, when I read you use the same notation as me, but then I learn you translated from french and here in spain we use all the same things :)

But lets get up with this stuck problem...

I don't know if the pulley is also ideal, so the tension $$\vec{\tau}$$ is the same in both sides of the pulley. I believe that my assumption is correct.

Also, I think that your problem is that you assigned M3 to the pully and the mass 2, but you can't imagine it properly. The difference between this case and the Atwood's one is that the mass at left is connected to a fixed point. Simply, you can imagine that the weight of the group you called M3 is the sum of the weight of mass 2 and pully 2 minus the tension which supports this group. Imagine you would build the problem. If the system is very massive, it can break the pully one unless you help it, and fixing the rope is just this, helping the rope 1 not to support so much tension, so coming back to the main problem, you can rest the tension to the weight, in module, so your problem may be transformed to atwood's one.

I didn't do any kind of calculus to reply this post, newtonian mechanics are very boring after you learn hamilton ones, so something that I said may be wrong. I don't think so, but ... can be.

MiGUi

4. Oct 16, 2004

The left pulley and the mass2 can move. There is only pulley 1 that cannot move.. I don't understand what you are trying to explain.. The 'group' M3 has a force downward (its weight) and two tensions. Also, the rope is ideal which means tension is the same. Are my formula at least right?

5. Oct 16, 2004