I'm dealing with a double Atwood's machine problem, and so far I got these $F=ma$ equations:
$T - m_2g = m_2 a_2$
$T - m_3g = m_3 a_3$
$2T - m_1g = m_1 a_1$
$a_1 = -\frac{(a_2 + a_3)}{2}$ (relating acceleration of mass 1 to the others).
However, now I somehow have to solve for $a_1, a_2$ and $a_3$, and get the following results:
\begin{align*} a_1 = g \frac{4m_2 m_3 - m_1(m_2 + m_3)}{4m_2 m_3 + m_1 (m_2 + m_3)} \\ a_2 = -g \frac{4m_2 m_3 + m_1 (m_2 - 3m_3)}{4m_2 m_3 + m_1 (m_2 + m_3)} \end{align*} and \begin{align*} a_3 = -g \frac{4m_2 m_3 + m_1 (m_3 - 3m_2)}{4m_2 m_3 + m_1 (m_2 + m_3)}. \end{align*}
I've tried adding, substracting, substituting forever...nothing seems to work. Can someone explain me how one derives these equations? Would mean a lot! Please don't delete/move this post, since this is a 'mathematical' problem (I've done the physics part).
We have four equations $$T - m_2g = m_2 a_2 \text{ (1)}$$ $$T - m_3g = m_3 a_3 \text{ (2)}$$ $$2T - m_1g = m_1 a_1 \text{ (3)}$$ $$a_1 = -\frac{(a_2 + a_3)}{2}\text{ (4)}$$ Add $m_3\times(1)$ to $m_2\times(2)$ to obtain $$\begin{align} T(m_2+m_3)-2m_2m_3g&=m_2m_3(a_2+a_3) \\\Rightarrow T(m_2+m_3)-2m_2m_3g&=-2(m_2m_3)a_1 \\\Rightarrow 2T(m_2+m_3)-4m_2m_3g&=-4m_2m_3a_1\\\Rightarrow (m_1g+m_1a_1)(m_2+m_3)-4m_2m_3g&=-4m_2m_3a_1\\\Rightarrow a_1=g\frac{4m_2m_3-m_1(m_2+m_3)}{4m_2m_3+m_1(m_2+m_3)} \end{align}$$ Having found $a_1$, using equation $(3)$ we can express $T$ as $$T=\frac{m_1(a_1+g)}{2}\Rightarrow T=g\frac{4m_2m_3}{4m_2m_3+m_1(m_2+m_3)}$$ and substitute $T$ into equation $(1)$ to find $a_2$ as follows:- $$g\frac{4m_2m_3}{4m_2m_3+m_1(m_2+m_3)}-m_2g=m_2a_2\\\Rightarrow a_2=-g\frac{4m_2m_3+m_1(m_2-3m_3)}{4m_2m_3+m_1(m_2+m_3)}$$ Having found $a_1$ and $a_2$, we can evaluate $a_3$ using equation $(4)$ as follows $$a_3=-(2a_1+a_2)=-g\frac{4m_2m_3+m_1(m_3-3m_2)}{4m_2m_3+m_1(m_2+m_3)}$$