Problem type: Block of mass (m) stationary on plane inclined at (\alpha). Find the least value of (\mu) to keep it at rest.
At limiting equilibrium (on the point of sliding down):
So:
The least value of (\mu) is (\tan\alpha). For the block to stay at rest, (\mu \geq \tan\alpha).
Do not assume the system moves in the direction of the heavier particle! Consider the angle too.
If (m_A \sin\alpha > m_B \sin\beta) and they're connected, A tends to move down.
If (m_B \sin\beta > m_A \sin\alpha), B tends to move down.
Two particles on slopes. Compare (m_A g \sin 30°) vs (m_B g \sin 60°) to determine direction.
Block mass (m) on plane at 25° to horizontal. Find least (\mu).
At limiting equilibrium (about to slide down):
Least (\mu = \tan 25°).