Pulley system on a double inclined plane

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This article is about the following scenario. A fixed triangular wedge has two inclines $I_{1}$ and $I_{2}$ making angles $\alpha _{1}$ and $\alpha _{2}$ with the horizontal, thus making it a double inclined plane. A pulley is affixed to the top vertex of the triangle. A string through the pulley has attached at its two ends blocks of masses $m_{1}$ and $m_{2}$, resting on the two inclines $I_{1}$ and $I_{2}$ respectively. The string is inextensible. The coefficients of static and kinetic friction between $m_{1}$ and $I_{1}$ are $\mu _{s1}$ and $\mu _{k1}$ respectively. The coefficients of static and kinetic friction between $m_{2}$ and $I_{2}$ are $\mu _{s2}$ and $\mu _{k2}$ respectively. Assume that $\mu _{k1}\leq \mu _{s1}$ and $\mu _{k2}\leq \mu _{s2}$.

We assume the pulley to be massless so that its moment of inertia can be ignored for the information below.

Summary of cases starting from rest

Case	What happens qualitatively	Magnitude of accelerations
$\!m_{1}\sin \alpha _{1}-m_{2}\sin \alpha _{2}>\mu _{s1}m_{1}\cos \alpha _{1}+\mu _{s2}m_{2}\cos \alpha _{2}$	$m_{1}$ slides downward and $m_{2}$ slides upward, with the same magnitude of acceleration	$\!a=g(m_{1}\sin \alpha _{1}-m_{2}\sin \alpha _{2}-\mu _{k1}m_{1}\sin \alpha _{1}-\mu _{k2}m_{2}\sin \alpha _{2})/(m_{1}+m_{2})$.
$\!m_{2}\sin \alpha _{2}-m_{1}\sin \alpha _{1}>\mu _{s1}m_{1}\cos \alpha _{1}+\mu _{s2}m_{2}\cos \alpha _{2}$	$m_{2}$ slides downward and $m_{1}$ slides upward, with the same magnitude of acceleration	$\!a=g(m_{2}\sin \alpha _{2}-m_{1}\sin \alpha _{1}-\mu _{k1}m_{1}\sin \alpha _{1}-\mu _{k2}m_{2}\sin \alpha _{2})/(m_{1}+m_{2})$.
$\!|m_{1}\sin \alpha _{1}-m_{2}\sin \alpha _{2}|\leq |\mu _{s1}m_{1}\cos \alpha _{1}+\mu _{s2}m_{2}\cos \alpha _{2}|$	The system remains at rest	$0$