$\Sigma M_D = 0$

$6R_A = 5(12) + 3(20)$

$R_A = 20 \, \text{ kN}$

$\Sigma M_A = 0$

$6R_D = 1(12) + 3(20)$

$R_D = 12 \, \text{ kN}$

**At joint A**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{AG} = R_A$

$\frac{\sqrt{21}}{5}F_{AG} = 20$

$F_{AG} = 21.82 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{AB} = \frac{2}{5}F_{AG}$

$F_{AB} = \frac{2}{5}(21.82)$

$F_{AB} = 8.73 \, \text{ kN tension}$

**At joint G**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{BG} + 12 = \frac{\sqrt{21}}{5}F_{AG}$

$\frac{\sqrt{21}}{5}F_{BG} + 1 2 = \frac{\sqrt{21}}{5}(21.82)$

$F_{BG} = 8.73 \, \text{ kN tension}$

$\Sigma F_H = 0$
$F_{FG} = \frac{2}{5}F_{AG} + \frac{2}{5}F_{BG}$

$F_{FG} = \frac{2}{5}(21.82) + \frac{2}{5}(8.73)$

$F_{FG} = 12.22 \, \text{ kN compression}$

**At joint B**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{BF} = \frac{\sqrt{21}}{5}F_{BG}$

$F_{BF} = F_{BG}$

$F_{BF} = 8.73 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{BC} = F_{AB} + \frac{2}{5}F_{BG} + \frac{2}{5}F_{BF}$

$F_{BC} = 8.73 + \frac{2}{5}(8.73) + \frac{2}{5}(8.73)$

$F_{BC} = 15.71 \, \text{ kN tension}$

**At joint F**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}F_{BF} = 20$

$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}(8.73) = 20$

$F_{CF} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{EF} + \frac{2}{5}F_{CF} = \frac{2}{5}F_{BF} + F_{FG}$

$F_{EF} + \frac{2}{5}(13.09) = \frac{2}{5}(8.73) + 12.22$

$F_{EF} = 10.48 \, \text{ kN compression}$

**At joint C**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{CE} = \frac{\sqrt{21}}{5}F_{CF}$

$F_{CE} = F_{CF}$

$F_{CE} = 13.09 \, \text{ kN tension}$

$\Sigma F_H = 0$
$F_{CD} + \frac{2}{5}F_{CE} + \frac{2}{5}F_{CF} = F_{BC}$

$F_{CD} + \frac{2}{5}(13.09) + \frac{2}{5}(13.09) = 15.71$

$F_{CD} = 5.24 \, \text{ kN tension}$

**At joint E**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{DE} = \frac{\sqrt{21}}{5}F_{CE}$

$F_{DE} = F_{CE}$

$F_{DE} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{EF} = \frac{2}{5}F_{CE} + \frac{2}{5}F_{DE}$

$10.48 = \frac{2}{5}(13.09) + \frac{2}{5}(13.09)$

$10.5 = 10.5$ *check*

**At joint D**

$\Sigma F_V = 0$
$R_D = \frac{\sqrt{21}}{5}F_{DE}$

$12 = \frac{\sqrt{21}}{5}(13.09)$

$12 = 12$ *check*

$\Sigma F_{H} = 0$
$F_{CD} = \frac{2}{5}F_{DE}$

$5.24 = \frac{2}{5}(13.09)$

$5.24 = 5.24$ *check*

**Summary**

*F*_{AB} = 8.73 kN tension

*F*_{AG} = 21.82 kN compression

*F*_{BC} = 15.71 kN tension

*F*_{BF} = 8.73 kN compression

*F*_{BG} = 8.73 kN tension

*F*_{CD} = 5.24 kN tension

*F*_{CE} = 13.09 kN tension

*F*_{CF} = 13.09 kN compression

*F*_{DE} = 13.09 kN compression

*F*_{EF} = 10.48 kN compression

*F*_{FG} = 12.22 kN compression
## Comments

## How come that the hypotenuse

How come that the hypotenuse in the solution became 5m where it is 2.5m?

## What is shown in the solution

What is shown in the solution is slope. 1/2.5 = 2/5. If you are not used in solving with slope, go with angles.

## Got it. Thank you

Got it. Thank you

## I dont get how you used the

Great

## how to find slope triangle

how to find slope triangle please help

## Use your scientific

Use your scientific calculator that is capable of displaying fraction. If you input 1/2.5 the calculator will output 2/5.