Answer
$F_{1v}=2.93kN$
$F_{1u}=2.07kN$
Work Step by Step
We are asked to resolve the force $F_1$ 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 $F$.
$\sin105^{\circ}/4kN=\sin45^{\circ}/F_{1v}$
$F_{1v}=\sin45^{\circ} *4kN/\sin105^{\circ}$
$F_{1v}=2.93kN$
$\sin105^{\circ}/4kN=\sin30^{\circ}/F_{1u}$
$F_{1u}=\sin30^{\circ} *4kN/\sin105^{\circ}$
$F_{1u}=2.07kN$
