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

$86.03^{\circ};1.97KN$
We find the force components: $F_x=8cos 30^{\circ}=6.93KN$ and $F_y=8 sin 30^{\circ}=4KN$ Thus $F_x=6+Fsin\theta-4sin 15^{\circ}=6.93$ $\implies Fsin\theta=1.9653$ and $F_y=Fcos\theta+4cos 15^{\circ}=4$ $\implies Fcos\theta=0.1363$ Now $tan\theta=\frac{1.9653}{0.1363}$ $\implies \theta=86.03^{\circ}$ and $F=\frac{1.9653}{sin\theta}$ $\implies F=1.97KN$