1. Initial Equilibrium (Figure 1)

In the first case, the half ring hangs freely from the hinge A. The Center of Mass (CM) aligns vertically below the hinge. From the geometry shown in Figure 1, the distance of the CM ($d$) from the geometric center ($O$) is related to the radius ($r$) and the angle of tilt $\theta_0$: $$ \tan \theta_0 = \frac{d}{r} \implies d = r \tan \theta_0 $$

2. Geometry of Case 2

The ring is hinged at A and rests on nail B, which is vertically below A at a distance $r$. Since the distance $AB = r$ and the radius of the ring is $r$, the triangle $\triangle ABO$ (where O is the center of the ring) is equilateral. Therefore, the line $AO$ makes an angle of $60^\circ$ with the vertical wall $AB$.

3. Torque Equilibrium about A

We take torques about the hinge A to solve for the normal force $N_B$.

• Torque from $N_B$: The force acts along the radius $BO$. The lever arm is the perpendicular distance from A to line $BO$. In an equilateral triangle of side $r$, this altitude is $r\frac{\sqrt{3}}{2}$.
• Torque from Gravity ($mg$): The Center of Mass (CM) is located relative to the diameter $AO$ based on $\theta_0$. The angle of the CM vector from the vertical is $(60^\circ – \theta_0)$. Lever arm for gravity: $L_{cm} \sin(60^\circ – \theta_0)$. Expanding this: $L_{cm}(\sin 60^\circ \cos \theta_0 – \cos 60^\circ \sin \theta_0)$. Using $L_{cm} \cos \theta_0 = r$ and $L_{cm} \sin \theta_0 = d = r \tan \theta_0$: $$ \text{Lever arm} = r \frac{\sqrt{3}}{2} – d \frac{1}{2} = \frac{r}{2}(\sqrt{3} – \tan \theta_0) $$

Balancing torques: $$ N_B \left( \frac{r\sqrt{3}}{2} \right) = mg \left[ \frac{r}{2}(\sqrt{3} – \tan \theta_0) \right] $$ $$ N_B \sqrt{3} = mg (\sqrt{3} – \tan \theta_0) \implies N_B = mg \left( 1 – \frac{\tan \theta_0}{\sqrt{3}} \right) $$

4. Reaction Forces at A

The normal force $N_B$ acts along $BO$, which is at $30^\circ$ to the horizontal. $$ \vec{N}_B = N_B \cos 30^\circ \hat{i} + N_B \sin 30^\circ (-\hat{j}) \quad (\text{Note: standard axes}) $$ Let’s use components relative to the diagram (x right, y up): $N_B$ points up-right towards O? No, the nail pushes the ring OUT. Since the ring hangs and presses *in* on the nail, the nail pushes *out* towards the center O. Angle with horizontal is $30^\circ$.
Horizontal Force Balance ($\sum F_x = 0$): $$ R_x + N_B \cos 30^\circ = 0 \implies R_x = – N_B \frac{\sqrt{3}}{2} $$ Magnitude: $\frac{mg\sqrt{3}}{2} \left( 1 – \frac{\tan \theta_0}{\sqrt{3}} \right)$ (Direction: Left)
Vertical Force Balance ($\sum F_y = 0$): $$ R_y + N_B \sin 30^\circ – mg = 0 \implies R_y = mg – \frac{N_B}{2} $$ $$ R_y = mg – \frac{mg}{2} \left( 1 – \frac{\tan \theta_0}{\sqrt{3}} \right) = \frac{mg}{2} \left( 1 + \frac{\tan \theta_0}{\sqrt{3}} \right) $$