Problem 264 | Resultant of Non-Concurrent Force System

$R_x = \Sigma F_x$

$R_x = 141.4(\frac{1}{\sqrt{2}}) + 300 \sin 60^\circ + 260(\frac{12}{13}) - 240 \sin 30^\circ$

$R_x = 479.79 \, \text{ N to the right}$
 

 

$R_y = \Sigma F_y$

$R_y = 141.4(\frac{1}{\sqrt{2}}) - 300 \cos 60^\circ + 260(\frac{5}{13}) + 240 \cos 30^\circ$

$R_y = 257.83 \, \text{ N upward}$
 

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

$R = \sqrt{479.79^2 + 257.83^2}$

$R = 544.68 \, \text{ N}$
 

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

$\tan \theta_x = \dfrac{257.83}{479.79}$

$\theta_x = 28.25^\circ$
 

$M_O = \Sigma Fd$

$M_O = -141.4(\frac{1}{\sqrt{2}})(3) - (300 \sin 60^\circ)(4) - (300 \cos 60^\circ)(4) - 260(\frac{12}{13})(1) + 260(\frac{5}{13})(4)$

$M_O = -1779.18 \, \text{ N}\cdot\text{m}$

$M_O = 1779.18 \, \text{ N}\cdot\text{m clockwise}$
 

 

$R_xb = M_O$

$479.79b = 1779.18$

$b = 3.71 \, \text{ m above point O}$
 

$R_ya = M_O$

$257.83a = 1779.18$

$a = 6.9 \, \text{ m to the left of point O}$
 

Thus, R = 544.68 N upward to the right at θ_x = 28.25°. The intercepts of R are (-6.9, 0) and (0, 3.71).