# Double Double Atwood machine

1. Nov 1, 2016

### Incand

1. The problem statement, all variables and given/known data
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 $m, 4m , 2m, m$.

2. Relevant equations
Lagrange's equation of motion
$\frac{d}{d t} \frac{\partial L }{\partial \dot q^i} -\frac{\partial L}{\partial q^i} = 0$.

3. 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.

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2. Nov 1, 2016

### nrqed

You are thinking of the lower Atwood machines as single masses, which we cannot do.

Let's go back to a simple Atwood machine as an example. Let's say you ask someone to calculate the tension of a string holding the pulley and someone says: obviously, this will simply be (m1+m2)g since there are two masses m1 and m2 hanging from the whole setup. In other words, that person considers the whole Atwood machine as a single mass. But that does not give the correct tension in the string attached to the pulley, right? (unless the two masses are equal, in which case the answer is indeed 2 mg). The reason is the setup contains parts that are accelerating (and have different accelerations). The same explanation applies here, to your double Atwood machine.

Hope this helps.

3. Nov 1, 2016

### Incand

Thanks, that was a really good explanation! And it gave me some practice trying the single Atwood machine example.

Doing some calculations for a single Atwood machine I found that if the tension in the string is $T$ then the acceleration of each mass must be $T/m_1-g = g-T/m_2$. It follows that $T=\frac{2m_1m_2}{m_1+m_2}g$.
The force on the string must then be double this so $F_1 = 16mg/5$ and $F_2 = 8mg/3$ (Using the masses in the original exercise).
But had the Atwood machines instead been point masses they must have had the masses of $m_L = 16/5m$ and $m_R = 8/3m$ and then $\ddot z = g(16/5-8/3)/(16/5+8/3) = \frac{(8/15)}{(88/15)}g = g/11$.