$R_x = \Sigma F_x$
$R_x = (1120 + 2240 + 1120)(\frac{1}{\sqrt{5}}) + 2000$
$R_x = 4003.52 \, \text{ N to the right}$
$R_y = \Sigma F_y$
$R_y = (1120 + 2240 + 1120)(\frac{2}{\sqrt{5}}) + 3000 + 2000 + 1000$
$R_y = 10\,007.03 \, \text{ N downward}$
$R = \sqrt{{R_x}^2 + {R_y}^2}$
$R = \sqrt{4003.52^2 + 10007.03^2}$
$R = 10\,778.16 \, \text{ N}$
$\tan \theta_x = \dfrac{R_y}{R_x}$
$\tan \theta_x = \dfrac{10007.03}{4003.52}$
$\theta_x = 68.2^\circ$
$M_A = \Sigma Fd$
$M_A = 2240(3.354) + 1120(3.354)(2) + 2000(1.5) + 3000(3) + 2000(6) + 1000(9)$
$M_A = 48\,026.37 \, \text{ N}\cdot\text{m clockwise}$
$R_yx = M_A$
$10\,007.03x = 48\,026.37$
$x = 4.8 \, \text{ m to the right of A}$
Thus, R = 10 778.16 N downward to the right at θx = 68.2° passing 4.8 m to the right of A.
Why did you use (1/SQ root 5…
Why did you use (1/SQ root 5) in Rx & use (2/SQ root 5) in Ry?
$\cos \theta = \dfrac{2}{…
In reply to Why did you use (1/SQ root 5… by rjvilla
$\cos \theta = \dfrac{2}{\sqrt{5}}$ and $\sin \theta = \dfrac{1}{\sqrt{5}}$