# Infinite Atwood's machine

(I'm not allowed to post the entire problem here since I'm a new user.)

Here are my equations:
$$T - mg = ma$$ (equation 1)
$$Mg - T = Ma$$ (equation 2)
(Upper case "M" used for infinite series of masses on the right arm of the first pulley. Lower case "m" for the small, single mass on the left arm.)

algebra is as follows:
$$Mg - (ma + mg) = Ma$$
$$g (M - m) = a (M + m)$$
$$a = g (M - m) / (M + m)$$

Here is my reasoning to get final answer: The limit of $a$ for large $M$ goes to $g$. Therefore, acceleration of the top mass is $g$. However, the correct answer is $g/2$. Why does my reasoning get the wrong answer? Any ideas?

(Please excuse the formatting, I'm a newbie here).

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–  John Rennie Mar 4 '12 at 10:44

Your second equation is wrong. You can't consider a nonrigid system as a single blob of mass $M=\Sigma m$. IIRC, this problem is done via a recursive relation.

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No, you can! "M" here will be the effective mass and this IS the way to solve the problem (still using a recursive relation).... all that is left to do is find "M", and DUE to it being nonrigid, it won't be infinite! –  Chris Gerig Mar 4 '12 at 20:34
@ChrisGerig Yeah that works as well. But the OPs method of taking $M=\sum m$ is wrong. When I solved it a few years ago I recursive'd the acceleration. –  Manishearth Mar 5 '12 at 0:34
Right, the sum of the two masses on the pulley is not the mass of the (single) atwood machine... but proving that the mass is $4m_1m_2/(m_1+m_2)$ is a good task, and that ultimately leads to the solution of this problem, without appealing to relative accelerations (which are usually beyond the scope of a first course in mechanics). –  Chris Gerig Mar 5 '12 at 6:51

You are making a mistake because you are considering $M$ to be large assuming infinitely many masses are attached to the system but, the masses attached are free to slide and the net force due to them does not add up to infinity.

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## protected by Qmechanic♦Apr 2 '14 at 17:33

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