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

$F = 1 230 N$ $\theta = 6.08°$
We are given both the direction and magnitude of both F1 and F2 and are asked for the magnitude and direction of F (the sum of both forces). By projecting the forces on the x axis $F_{x} = 750 \times \sin(45°) + 800 \times \cos(30°) = 1 223.15 N$ By projecting the forces on the y axis $F_{y} = 750 \times \cos(45°) - 800 \times \sin(30°) = 130.33 N$ The magnitude of F is obtained $F = \sqrt (F_{x}^{2} + F_{y}^{2}) = 1230 N$ The angle from the x axis is obtained $\theta = \tan^{-1} (\frac{F_{x}}{F_{y}}) = 6.08°$