Answer
$F_{2u}=6.00kN$
$F_{2v}=3.11kN$
Work Step by Step
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 $F_2$.
$\sin75^{\circ}/6kN=\sin75^{\circ}/F_{2u}$
$F_{2u}=\sin75^{\circ} *6kN/\sin75^{\circ}$
$F_{2u}=6.00kN$
$\sin75^{\circ}/6kN=\sin30^{\circ}/F_{2v}$
$F_{2v}=\sin30^{\circ} *6kN/\sin75^{\circ}$
$F_{2v}=3.11kN$
