Fig. $6\mathrm{.}10$

From equilibrium at point $O,$ the sum of force components in both $x-$ and y $-$directions is zero:

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=F_{OA}cos60-F_{OB}cos30+4cos45=0$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=F_{OA}sin60+F_{OB}sin30-4sin45=0$

Solving the equations simultaneously yields:

$F_{OA}=1\mathrm{.}04kN$ and $F_{OB}=3\mathrm{.}86$kNFig. $6\mathrm{.}10$

From equilibrium at point $O,$ the sum of force components in both $x-$ and y $-$directions is zero:

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=F_{OA}cos60-F_{OB}cos30+4cos45=0$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=F_{OA}sin60+F_{OB}sin30-4sin45=0$

Solving the equations simultaneously yields:

$F_{OA}=1\mathrm{.}04kN$ and $F_{OB}=3\mathrm{.}86kN$