# Equilibrium of double Atwood machine via lagrangian

I am training to solve the following physical system via lagrangian equations for equilibrium Points: I have got the following boundary condition: $\frac{1}{2}(y_1+y_2) = l_2-y_3$ $\Rightarrow$ $y_3 = l_2 - \frac{1}{2} (y_1 - y_2)$. $y_1, y_2, y_3$ are the vertical positions of $m_1, m_2, m_3$, and $l_1$ and $l_2$ are the lengths of the ropes on the small and the big wheel. I get the following equations for the kinetic energy $$T = \frac{1}{2} \left( m_1\dot{y_1} + m_2 \dot{y_2} + m_3 \dot{y_3} \right) = \frac{1}{2} \left( \left( m_1 + \frac{m_3}{4}\right)\dot{y_1} + \left(m_2 + \frac{m_3}{4} \right) \dot{y_2} - \frac{m_3}{2} \dot{y_1}\dot{y_2}\right)$$ and the potential Energy $$V = mg(h-y_1) + m_2g(h-y_2) + m_3g(h-y_3) = \left( \frac{m_3}{2} - m_1\right)y_1g + \left( \frac{m_3}{2} - m_2\right) y_2g + const.$$

Which leads to the following 2 lagrangian equations: $$(i) \quad \frac{d}{dt} \frac{\partial L}{\partial \dot{y_1}} - \frac{\partial L}{\partial y_1} = \left( m_1 + \frac{m_3}{4}\right)\ddot{y_1} - \frac{m_3}{4}\ddot{y_2} + \left(\frac{m_3}{2} - m_1\right) g = 0$$ and $$(ii) \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{y_2}} - \frac{\partial L}{\partial y_2} = \left( m_2 + \frac{m_3}{4} \right) \ddot{y_2} - \frac{m_3}{4}\ddot{y_1} + \left( \frac{m_3}{2} - m_2 \right)g = 0.$$

Now for one of the equibilirium points i guessed the following condition: $m_1 = m_2 \Rightarrow$

$$(i) - (ii) = m_1\ddot{y_1} - m_2\ddot{y_2} + \frac{m_3}{4} \left( \ddot{y_1} - \ddot{y_2}\right) + (m_2 - m_1) g = 0$$ with $m_1 = m_2 = m$ $$\Rightarrow m \ddot{y_1} - m \ddot{y_2} + \frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2}) = 0\\ \Rightarrow m(\ddot{y_1} - \ddot{y_2}) = -\frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2})$$ $$\Rightarrow m = -\frac{m_3}{4}$$ Where is my error? Please help me. D:

• "equibilirium" $\to$ "equilibrium" Sep 10, 2017 at 3:55
• write the lenghts and the vertical positions on the picture. Sep 10, 2017 at 14:12
• Your error $y_3 = l_2 - \frac{1}{2} (y_1 -y_2)$ must be $y_3 = l_2 - \frac{1}{2} (y_1 +y_2)$
– Eli
Jan 18 at 7:48

Okay i think i just figured it out myself, and the answer is rather obvious. From the condition: $$m(\ddot{y_1} - \ddot{y_2}) = -\frac{m_3}{4}(\ddot{y_1} - \ddot{y_2})$$ it follows that $$\ddot{y_1} - \ddot{y_2} = 0.$$ Which is the expected result, because this means, that $m_1$ doesn't experience any acceleration relative to $m_2$.
• actually you must find $\ddot y_1=-\ddot y_2$. Sep 10, 2017 at 18:24
• Hm, can you explain why. I thought this is the case for the equilibrium $m_1+m_2=m_3$? Sep 10, 2017 at 19:29