Stick the circle mechanics

Considering first the forces on the circle, taking moment about the centre $O$ tells us that the frictional forces at the two points of contact are the same. Resolving forces horizontally gives $F + F \cos60^\circ \; = \; R_2\cos30^\circ$ and hence $R_2 = F\sqrt{3}$ .

Now considering forces on the rod, taking moments about $X$ tells us that $R_2\times L \; = \; W_1 \times \tfrac12L \cos60^\circ$ and hence $R_2 = \tfrac14W_1$ , and hence $F = \tfrac{W_1}{4\sqrt3}$ , where $L = 2\sqrt{3}$ is the length of the rod, and $W_1 = QLg$ is the weight of the rod. Thus $F = \tfrac12Qg = \boxed{5Q}$ .

Interestingly, resolving forces on the rod horizontally tells us that $F_3 = F$ as well.