Statics Exercise

Since the rod is not attached at the wall or floor, the wall and floor can supply forces but cannot supply moments. Therefore, we have three unknowns: the magnitude of the reaction force on the semi-circle, and the horizontal and vertical forces at the floor / wall junction. We can write three equilibrium equations (x, y, rotational) to find these quantities.

First, some geometry:

Simple right triangle geometry gives us the necessary angles for the problem. The diagram also shows the forces.

Horizontal equilibrium equation:

$N cos(60^\circ) = F_x \\ \boxed{N = 2 F_x}$

Vertical equilibrium equation:

$N sin(60^\circ) + F_y = W \\ (2 F_x) \frac{\sqrt{3}}{2} + F_y = W \\ \boxed{\sqrt{3} F_x + F_y = W}$

Rotational equilibrium equation (with wall / floor junction as the reference point):

$W \frac{L}{2} cos(30^\circ) = N \sqrt{3} \\ \frac{\sqrt{3}}{2} W = 2 \sqrt{3} F_x \\ \boxed{F_x = \frac{W}{4}} \\ \implies \boxed{F_y = W(1 - \frac{\sqrt{3}}{4})}$

The ratio therefore works out to approximately $\boxed{0.6197}$