$R_x = \Sigma F_x$

$R_x = 300 \sin 30^\circ - 224(\frac{2}{\sqrt{5}}) + 361(\frac{2}{\sqrt{13}})$

$R_x = 149.895 \, \text{ lb to the right}$

$R_y = \Sigma F_y$

$R_y = 300 \cos 30^\circ + 224(\frac{1}{\sqrt{5}}) - 361(\frac{3}{\sqrt{13}})$

$R_y = 59.613 \, \text{ lb upward}$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

$R = \sqrt{149.895^2 + 59.613^2}$

$R = 161.314 \, \text{ lb}$

$\tan \theta_x = \dfrac{R_y}{R_x}$

$\tan \theta_x = \dfrac{59.613}{149.895}$

$\theta_x = 21.69^\circ$

$M_O = \Sigma M$

$M_O = -(300 \sin 30^\circ)(2) + 224(\frac{1}{\sqrt{5}})(2) + 361(\frac{2}{\sqrt{13}})(1)$

$M_O = 100.598 \, \text{ lb}\cdot\text{ft counterclockwise}$

$R_ya = M_O$

$59.613a = 100.598$

$a = 1.688 \, \text{ ft to the right of the origin}$

$R_xb = M_O$

$149.895b = 100.598$

$b = 0.671 \, \text{ ft below the origin}$

Thus, R = 161.314 lb upward to the right at θ_{x} = 21.69° and intercepts at (1.668, 0) and (0, -0.671).

## Comments

## How is it possible that the

How is it possible that the sign conventions on the moments are contradicting the appropriate signs for clockwise and counter clockwise