Answer
$F_1=1.83 kN$
$F_2=9.60 kN$
Work Step by Step
First we will sum the forces horizontally
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$F_2 \sin 70^{\circ} + F_1 \cos 60^{\circ} - 5 \cos 30^{\circ} - 4/5 (7) = 0$
Now we will sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_2 \cos 70^{\circ} + 5 \sin 30^{\circ} - F_1 \sin 60^{\circ} - 3/5 (7) = 0$
Solving the system of two equations, we obtain :
$F_2=9.60 kN$
$F_1=1.83 kN$
