Sliding motion for adjacent blocks along an inclined plane

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This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
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The two blocks are adjacent to each other on the incline.

The scenario here is of two adjacent dry cuboidal blocks as shown in the figure, on an incline of a fixed inclined plane making an angle \theta with the horizontal.

The upper mass is denoted as m_1, and the coefficients of static and kinetic friction between that mass and the incline are denoted \mu_{s1} and \mu_{k1}. The lower mass is denoted as m_2, and the coefficients of static and kinetic friction between that mass and the incline are denoted \mu_{s2} and \mu_{k2}.

The extremes are \theta = 0 (whence, the plane is horizontal) and \theta = \pi/2 (whence, the plane is vertical).

We assume that the block undergoes no rotational motion, i.e., it does not roll or topple. Although the diagram here shows a block with square cross section, this shape assumption is not necessary as long as we assume that the block undergoes no rotational motion.

Behavior assuming system initially at rest

Note that the answer in the third column must be yes if the first two columns are yes and no if the first two columns are no.

In cases where the two masses separate, their kinematics can be studied separately. In cases where the masses stay together, they can be collapsed into a single mass m_1 + m_2 whose kinematics can be studied.

Is \tan \theta < \mu_{s1}? Is \tan \theta < \mu_{s2}? Is \tan \theta < (\mu_{s1}m_1 + \mu_{s2}m_2)/(m_1+m_2)? Is \mu_{k2} < \mu_{k1}? Behavior Do m_1,m_2 stay together?
Doesn't matter Yes Yes Doesn't matter Both blocks remain stationary. Yes
Yes No Doesn't matter Doesn't matter m_2 slides down, m_1 stays stationary. No
No Doesn't matter No Yes Both slide down, but m_2 slides down faster, with accelerations g(\sin \theta - \mu_{k1}\cos \theta) and g(\sin \theta - \mu_{k2}\cos \theta) for m_1,m_2 respectively. No
No Doesn't matter No No Both slide down together, with acceleration a = g\left[\sin \theta - \frac{\mu_{k1}m_1 + \mu_{k2}m_2}{m_1 + m_2} \cos \theta\right]. Yes