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

$F_{1}x = 141N$ $F_{1}y = 141N$ $F_{2}y = -130N$ $F_{2}y = 75N$
We determine the force components: $F_{1}x =200\times \sin 45 ° = 141N$ $F_{1}y = 200 \times \cos 45 ° =141N$ $F_{2}y =-150 \times \cos 30 ° = -130N$ $F_{2}y = 150 \times \sin 30 ° = 75N$