Problem 002-ms | Method of Sections

$\Sigma M_A = 0$

$8R_H = 2(55) + 4(90) + 6(45)$

$R_H = 92.5 \, \text{ kN}$
 

From section to the right of a-a
$\dfrac{x + 2}{1.5} = \dfrac{x + 4}{2.5}$

$2.5x + 5 = 1.5x + 6$

$x = 1 \, \text{ m}$
 

$\Sigma M_O = 0$

$(x + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + xR_H = (x + 2)(45)$

$(1 + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + 1(92.5) = (1 + 2)(45)$

$\frac{15}{\sqrt{41}}F_{DG} + 92.5 = 135$

$\frac{15}{\sqrt{41}}F_{DG} = 42.5$

$F_{DG} = 18.14 \, \text{ kN tension}$           answer