A classical mechanics problem by Aly Ahmed

Let the length of the rod be $a\sqrt 3$ , the reaction of horizontal support at the lower end of the rod be $N_1$ and that of the inclined plane at the upper end of the rod be $N_2$ . Then, force balance equations yield

$2N_1+N_2=2mg$

$\mu N_1=\dfrac{\sqrt 3}{2}N_2$ .

The moment balance equation about the lower end of the rod yields

$N_2\cos 60\degree \times \dfrac{3a}{2}+N_2\sin 60\degree \times \dfrac{a\sqrt 3}{2}=mg\times \dfrac{3a}{4}\implies N_2=\dfrac{mg}{2}=\boxed {120} \implies N_1=\dfrac{mg\sqrt 3}{4\mu}$ , both are in Newton units. So,

$\dfrac{mg}{2}(1+\frac{\sqrt 3}{\mu})=2mg\implies \mu =\dfrac{1}{\sqrt 3}\approx \boxed {0.57735}, N_1=\dfrac{3mg}{4}=\boxed {180}$ Newton.