## Engineering Mechanics: Statics & Dynamics (14th Edition)

$-1.98KN; 184.53^{\circ}$
We know that $F_x=-30sin 30^{\circ}-26(\frac{5}{13})$ $\implies F_x=-25KN$ and $F_y=-30 cos30^{\circ}+26(\frac{12}{13})$ $\implies F_y=-1.98KN$ Now the resultant force can be calculated as $F_R=\sqrt{(F_x)^2+(F_y)^2}$ We plug in the known values to obtain: $F_R=\sqrt{(-25KN)^2+(-1.98KN)^2}$ $\implies F_R=25.08KN$ The orientation can be determined as $\phi=tan^{-1}(\frac{F_y}{F_x})$ $\phi=(\frac{-1.98KN}{-25KN})=4.53^{\circ}$ $\implies \theta=184.53^{\circ}$