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

$F_{R} = 983N$ $\theta = 21.8°$
Projecting both forces on the x axis: $F_{R}x = 400 \times \cos(30°) + 800 \times \sin (45°) = 912.10N$ Projecting both forces on the y axis: $F_{R}y = 400 \times \sin(30°) + 800 \times \cos (45°) = 365.69N$ The magnitude is obtained: $F_{R} = \sqrt {912.1^{2} + 365.69^{2}} = 983N$ The direction about the x axis is: $\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 21.84° \approx 21.8°$