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

$F_{2u}=6.00kN$ $F_{2v}=3.11kN$
We are asked to resolve the force $F_2$ into components acting along the u and v axes and determine the magnitudes of the components. We can use the law of sines to find the magnitudes of the components of $F_2$. $\sin75^{\circ}/6kN=\sin75^{\circ}/F_{2u}$ $F_{2u}=\sin75^{\circ} *6kN/\sin75^{\circ}$ $F_{2u}=6.00kN$ $\sin75^{\circ}/6kN=\sin30^{\circ}/F_{2v}$ $F_{2v}=\sin30^{\circ} *6kN/\sin75^{\circ}$ $F_{2v}=3.11kN$