Answer
$F_2=9.60kN$
$F_1=1.83kN$
Work Step by Step
First we will sum the forces horizontally
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$F_2\sin70^{\circ}+F_1\cos60^{\circ}−5\cos30^{\circ}−4/5(7)=0$
Now we will sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_2\cos70^{\circ}+5\sin30^{\circ}−F_1\sin60^{\circ}−3/5(7)=0$
Solving the system of two equations, we obtain :
$F_2=9.60kN$
$F_1=1.83kN$
