Interpreting force components in statics problems involving inclined planes or similar

Question

I'm having troubles to understand how to interpret the components of forces on inclined plane in quite complicated situations. I will make an example of one of the cases where I get confused.

Consider the system in the picture, find the minumum coefficient of static friction between $B$ and the floor and $A$ and the floor, such that there is static equilibrium

So on $C$ there is weight acting of course (first picture), and if $C$ must not move, then there is a force equal and opposite to $m_cg$. Let's focus on the interactions between $C$ and $B$. I interpreted this as a situation similar to an inclined plane: if $C$ must not slip then, besides the normal force $N$ (in orange) there must be a force of friction $F$, in green, both represented in the second picture.

If all of this is correct then the resultant of the two forces $F$ and $N$ is the red $R$ in the third picture, which is vertical, equal and opposite to $\frac{m_c}{2} g$ as it should be.

But then a force $-R$ acts on $B$ and the question lose sense because $R$ is vertical and there is no need of friction to keep $B$ in static equilibrium.

Am I missing something important in interpreting the free body diagrams?

Answer 1

Free body diagram of C is as below:

Am I missing something important in interpreting the free body diagrams?

Yes. You have assumed that $\vec N+\vec F+\large{\frac 12 m_C\vec g}=\vec 0$. But, this is not correct. The correct equation is: $$(\vec N)_y+(\vec F)_y-\large{\frac 12}m_Cg=0$$ or $$N\sin(\large{\frac \alpha 2})+F\cos(\large{\frac \alpha 2})=\large{\frac 12}m_Cg$$