(i) $\displaystyle \sum \tau = 0 \implies W \cdot L\sin{\theta} = F \cdot 2L\cos{\theta} \implies F = \dfrac{W\tan{\theta}}{2} = \dfrac{W}{6}$
(ii) horizontally ... $H_x = F = \dfrac{W}{6}$
vertically ... $H_y = W$
(iii) $F = \dfrac{W}{6} \implies 2F = \dfrac{W}{3}$ ...
$W \cdot L\sin{\theta} = \dfrac{W}{3} \cdot 2L\cos{\theta} \implies \tan{\theta} = \dfrac{2}{3}$
Thanks. Yes your answers are correct.
How did you know the direction of Rc and that it was going up vertically. It does not say the direction of it in the question.
What are you doing in your second equation.? Are you taking moments about a point and if so where ? Why do you not include the Rc in the second equation.?
both the 100N and 50N weights are in the downward direction ... what direction for $R_c$ is necessary for equilibrium in the y-direction?
Moments are about point $c$ ... $R_c$ does not contribute to the net torque about point $c$ since its moment arm is zero.What are you doing in your second equation.? Are you taking moments about a point and if so where ? Why do you not include the Rc in the second equation.?