Problem 435 - Transmission Tower by Method of Sections

$\Sigma M_G = 0$

$2(4.5 + 2)\left( \frac{3}{\sqrt{10}}F_{BC} \right) = 12(7.8)$

$F_{BC} = \frac{12}{5}\sqrt{10} ~ \text{kN tension}$

$\Sigma M_F = 0$

$9\left( \frac{4}{5} \right)F_{CJ} + 9\left( \frac{3}{\sqrt{10}} \right)\left( \frac{12}{5}\sqrt{10} \right) = 6(7.8)$

$F_{CJ} = -2.5 ~ \text{kN} = 2.5 ~ \text{kN compression}$       answer