Rod on a Hemisphere

Resolving perpendicular to the rod, AB: $S\sin\theta + R = mg\cos\theta$

Resolving parallel to AB: $S\cos\theta - mg\sin\theta = 0$

$\Rightarrow S = \frac{mg\sin\theta}{\cos\theta}$

$\Rightarrow mg\frac{\sin^{2}\theta}{\cos\theta} + R = mg\cos\theta$

$\Rightarrow R = mg\cos\theta - mg\frac{\sin^{2}\theta}{\cos\theta}$

$\Rightarrow R = \frac{mg}{\cos\theta}(\cos^{2}\theta - \sin^{2}\theta) = \frac{mg}{\cos\theta}(\cos2\theta)$

$\Rightarrow R = \frac{mg\cos2\theta}{\cos\theta}$

Taking moments about A: $\frac{l}{2}(mg\cos\theta) = 2a\cos\theta\cdot R$ as $R$ acts through C and length AC $=$ $2a\cos\theta$

$\Rightarrow l = \frac{4a\cdot R}{mg}$

$\Rightarrow l = \boxed{\frac{4a\cos2\theta}{\cos\theta}}$ .

The rod is static and thus there's neither net force nor net torque. From these conditions, we will derive relation between $l$ and $a$ ( $x$ is horizontal and $y$ is vertical axis): $\begin{aligned} &\sum X = 0 \implies F_{C}\sin{\theta} = F_{A}\cos{2\theta} \\ &\sum Y = 0 \implies F_{C}\cos{\theta} + F_{A}\sin{2\theta} = G \\ &\sum T^{(A)} = 0 \implies F_{C}\,2a\cos{\theta} = G\frac{l}{2}\cos{\theta} \end{aligned}$ From the first two equations we get: $G = F_{C}\frac{\cos{\theta}}{\cos{2\theta}}$ And plugging that into the third, one gets: $l = 4a\frac{\cos{2\theta}}{\cos{\theta}}$ Point E is the center of mass of the rod A more elegant way to solve this problem is to note that, in equilibrium, the center of mass of the rod is lowest possible. So, expressing the height of rod's center of mass as a function of $\theta$ and setting derivative to zero gives us desired condition: $\begin{aligned} h &= -a\sin{2\theta} + \frac{l\sin{\theta}}{2} \\ \frac{\partial h}{\partial \theta} &= -2a\cos{2\theta} + \frac{l\cos{\theta}}{2} = 0 \\ &\implies l = \frac{4a\cos{2\theta}}{\cos{\theta}}\end{aligned}$