$\Sigma M_A = 0$
$6R_D = 2(100) + 4(120)$
$R_D = 113.33 \, \text{ kN}$
At joint F
$F_{EF} = 0$
At joint D
$\Sigma F_V = 0$
$\frac{3}{\sqrt{13}}F_{DE} = R_D$
$\frac{3}{\sqrt{13}}F_{DE} = 113.33$
$F_{DE} = 136.21 \, \text{ kN compression}$
$\Sigma F_H = 0$
$F_{CD} = \frac{2}{\sqrt{13}}F_{DE}$
$F_{CD} = \frac{2}{\sqrt{13}}(136.21)$
$F_{CD} = 75.56 \, \text{ kN tension}$
At joint E
$\Sigma F_V = 0$
$F_{CE} = \frac{3}{\sqrt{13}}F_{DE}$
$F_{CE} = \frac{3}{\sqrt{13}}(136.21)$
$F_{CE} = 113.33 \, \text{ kN tension}$
At joint C
$\Sigma F_V = 0$
$\frac{3}{\sqrt{13}}F_{CG} + F_{CE} = 120$
$\frac{3}{\sqrt{13}}F_{CG} + 113.33 = 120$
$F_{CG} = 8.02 \, \text{ kN tension}$ answer
Comments
Sir, bakit wala pong
Sir, bakit wala pong horizontal component na sinolve sa joint E and C. Sa joint E po, aren't both joint G and E carrying 100kn load? Please do correct me po sir/ma'am. Thank you! God blesss