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 $\mu _{k2}\leq \mu _{s2}$.

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	$0$