$\Sigma M_A = 0$

$24R_B + 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) = 18(\frac{2}{\sqrt{5}})(22.4) + 18(30) + 12(20) + 6(10)$

$R_B = 46.27 \, \text{ kN}$ *answer*

$\Sigma M_B = 0$

$24A_V = 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) + 6(\frac{2}{\sqrt{5}})(22.4) + 6(30) + 12(20) + 18(10)$

$A_V = 33.76 \, \text{ kN}$

$\Sigma F_H = 0$

$A_H = 20 + \frac{1}{\sqrt{5}}(22.4)$

$A_H = 30.02 \, \text{ kN}$

$R_A = \sqrt{{A_H}^2 + {A_V}^2}$

$R_A = \sqrt{30.02^2 + 33.76^2}$

$R_A = 45.18 \, \text{ kN}$

$\tan \theta_{Ax} = \dfrac{A_V}{A_H}$

$\tan \theta_{Ax} = \dfrac{33.76}{30.02}$

$\tan \theta_{Ax} = 48.36^\circ$

Thus, $R_A = 45.18 \, \text{ kN}$ up to the right at $48.36^\circ$ from horizontal. *answer*

## Comments

## how did you get the side of

how did you get the side of the triangle= square root of 5, 2 and 1.in the force 22.4kn?

im just newbie for this. thanks!

## The slope of the top chord is

The slope of the top chord is 1 vertical to 2 horizontal. The 22.4 kN force is perpendicular to the top chord thus the slope is 2 vertical to 1 horizontal.