$R_x = \Sigma F_x$

$R_x = (1120 + 2240 + 1120)(\frac{1}{\sqrt{5}}) + 2000$

$R_x = 4003.52 \, \text{ N to the right}$

$R_y = \Sigma F_y$

$R_y = (1120 + 2240 + 1120)(\frac{2}{\sqrt{5}}) + 3000 + 2000 + 1000$

$R_y = 10\,007.03 \, \text{ N downward}$

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

$R = \sqrt{4003.52^2 + 10007.03^2}$

$R = 10\,778.16 \, \text{ N}$

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

$\tan \theta_x = \dfrac{10007.03}{4003.52}$

$\theta_x = 68.2^\circ$

$M_A = \Sigma Fd$

$M_A = 2240(3.354) + 1120(3.354)(2) + 2000(1.5) + 3000(3) + 2000(6) + 1000(9)$

$M_A = 48\,026.37 \, \text{ N}\cdot\text{m clockwise}$

$R_yx = M_A$

$10\,007.03x = 48\,026.37$

$x = 4.8 \, \text{ m to the right of A}$

Thus, R = 10 778.16 N downward to the right at θ_{x} = 68.2° passing 4.8 m to the right of A.

Why did u use 1/sq5 instead of 2/sq5 in Rx

Use sin θ to find the horizontal component of inclined loads.

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