Blocks on two inclines of a free wedge

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This article is about the following scenario. A freely movable triangular wedge placed on a fixed frictionless horizontal floor has two inclines $I_{1}$ and $I_{2}$ making angles $\alpha _{1}$ and $\alpha _{2}$ with the horizontal. There are blocks of masses $m_{1}$ and $m_{2}$, resting on the two inclines $I_{1}$ and $I_{2}$ respectively. All surfaces of contact are frictionless.

Summary of cases

Case	What happens
$\!m_{1}\sin(2\alpha _{1})=m_{2}\sin(2\alpha _{2})$	The wedge remains stationary and the blocks both slide down their respective inclines.
$\!m_{1}\sin(2\alpha _{1})>m_{2}\sin(2\alpha _{2})$	The horizontal push exerted by $m_{1}$ exceeds that by $m_{2}$, causing the wedge to move in the push direction of $m_{1}$. Both $m_{1}$ and $m_{2}$ slide down.
$\!m_{1}\sin(2\alpha _{1})<m_{2}\sin(2\alpha _{2})$	The horizontal push exerted by $m_{2}$ exceeds that by $m_{1}$, causing the wedge to move in the push direction of $m_{2}$. Both $m_{1}$ and $m_{2}$ slide down.