Imagine a pulley with a massless and frictionless rope. On both sides of the pulley (and both sides of the rope), we attach a mass ($m_1$ and $m_2$). These masses are connected with two other masses ($m_3$ an $m_4$) by means of two ideal springs (spring constants $k_1$ and $k_2$). The rope has a mass ofs(kg/m).
We hold the two upper masses at equal height after which we let the system go its way. How do we have to model the system mathematically?
For two single masses at each end the situation is simple. The heavier mass gets an acceleration of $g-a$, the lighter mass $g+a$ and it's easy to calculate $a$, if the masses are known. But what happens in this (non-equilibrium) case? Can it be that it's an unsolvable four-body problem?
For sure there are four bodies (the four masses). But this is not a four body problem because the masses are not moving in each other's force field. Two pairs of masses are connected by a string and both these pairs are subjected to the force of gravity.
After some thought, the problem is easier than I thought. For the acceleration, a, for both sets of masses after letting them go, we solve the following formula (if $(m_1+m_2)>(m_2+m_4)$):
$(m_1+m_2)(g-a)=(m_3+m_4)(g+a)$
After setting $g=10$ and some rearranging we get:
$a=\frac{10(M-m)} {(M+m)}$, where $M=m_1+m_2$, and $m=m_3+m_4$
So the two spring connected masses
When knowing a, it's easy to find how much the springs have lengthened or shortened.
But what if we take into account the (equally) changing mass on either side of the pulley because of the mass of the rope?
