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Homework Statement
This is actually a problem I solved but I'm having trouble reconcile the answer with my intuition
(Check the end of the post for my actual question).
The problem:
Find the acceleration of the masses for the double double Atwood machine (see figure)
The masses are from left to right ##m, 4m , 2m, m##.
Homework Equations
Lagrange's equation of motion
##\frac{d}{d t} \frac{\partial L }{\partial \dot q^i} -\frac{\partial L}{\partial q^i} = 0##.
The Attempt at a Solution
Here's the short version of my solution:
##m_1: \; \; T =\frac{m}{2}(\dot x- \dot z)^2, \; \; V = mg(x-z)##
##m_2:\; \; T = 2m(\dot x+ \dot z)^2, \; \; V = -4mg(x+z)##
##m_3: \; \; T =m(\dot z- \dot y)^2, \; \; V = 2mg(z-y)##
##m_4: \; \; T = \frac{m}{2}(\dot y+ \dot z)^2, \; \; V = mg(y+z)##
Inputting this into L.E. we end up with an equation system
##\begin{bmatrix}5 &0 & 3\\ 0 & 3 & -1 \\ 3 & -1 & 8\end{bmatrix}\begin{bmatrix}\ddot x\\\ddot y\\ \ddot z\end{bmatrix} = \begin{bmatrix}3g\\g\\2g\end{bmatrix}##
This gives ##\ddot x = 6/11g##, ##\ddot y = 4/11g##, ##\ddot z =1/11g##. Which gives the acceleration of ##a_{m_1} = 5/11g, a_{m_2} = -7/11g, a_{m_3} = -3/11g, a_{m_4} = 5/11g##. This is the correct answer according to the book.
My question is about why ##\ddot z = g/11##. Why isn't this the same acceleration as that of a single Atwood machine with masses of ##1m+4m=5m## and ##2m+1m = 3m## i.e. ##a=g\frac{5-3}{5+3} = g/4##. I can't see how the motion of the two other Atwood machine should change anything. When drawing a free body diagram I still see these forces but this doesn't agree with my other solution.