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

$F_{R} = 413N$ $\theta = 24.2°$
Projecting both forces on the x axis: $F_{R}x = 400 \times \sin 30° + 250 \times \cos 45° = 376.78N$ Projecting on the y axis: $F_{R}y = 400 \times \cos 30° - 250 \times \sin 45° = 169.63N$ The magnitude of the force is obtained: $F_{R} = \sqrt {376.78^{2} + 169.63^{2}} = 413N$ The direction is obtained: $\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 24.2°$