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

$54.2N; 43.5^{\circ}$
We know that $F_1=50(\frac{3}{5})i+50(\frac{4}{5})j$ $F_2=80 cos\space 255i+80 sin\space 255j$ and $F_3=30i+0j$ Now we find the horizontal and vertical resultant forces as $R_x=50(\frac{3}{5})+80cos 255+30=39.3N$ and $R_y=50(\frac{4}{5})+80sin\space 255+0=-37.3N$ Now the magnitude of the resultant force is $F_R=\sqrt{R_x^2+R_y^2}$ We plug in the known values to obtain: $F_R=\sqrt{(39.3)^2+(-37.3)^2}=54.2N$ Now the direction can be determined as $\theta=tan^{-1}(\frac{R_y}{R_x})$ We plug in the known values to obtain: $\theta=tan^{-1}(\frac{37.3}{39.3})=43.5^{\circ}$