Answer
$F\approx2.035\;kN$
$F_R\approx7.86\;kN$
Work Step by Step
First, we resolve each force into its $x$ and $y$ components by using the parallelogram law:
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force, in which $F_1=8\;kN$. Thus we have
$F_{1x}=8\cos0^\circ\;kN=8\;kN$
$F_{1y}=8\sin0^\circ\;kN=0\;kN$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=F\;kN$. Thus we have
$F_{2x}=-F\cos45^\circ\;kN=-\frac{F}{\sqrt 2}\;kN\approx-0.707F\;kN$
$F_{2y}=-F\sin45^\circ\;kN=-\frac{F}{\sqrt 2}\;kN\approx-0.707F\;kN$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=14\;kN$. Thus we have
$F_{3x}=-14\cos30^\circ\;kN=-7\sqrt 3\;kN\approx-12.12\;kN$
$F_{3y}=14\sin30^\circ\;kN=7\;kN$
Second, using scalar notation and indicating the positive directions of components along the $x$ and $y$ axes with symbolic arrows, we sum these components algebraically:
Summing the x components, we have
$\xrightarrow{+} (F_R)_x=\sum F_x$
or, $(F_R)_x=(8-0.707F-12.12)\;kN=(-0.707F-4.12)\;kN$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(0-0.707F+7)\;kN=(-0.707F+7)\;kN$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {(0.707F+4.12)^2+(-0.707F+7)^2}$
or, $F_R\approx\sqrt {F^2-4.07F+65.97}\;kN$
or, $F_R=\sqrt {F^2-2\times F\times2.035+2.035^2-2.035^2+65.97}\;kN$
or, $F_R=\sqrt {(F-2.035)^2-2.035^2+65.97}\;kN$
or, $F_R=\sqrt {(F-2.035)^2-2.035^2+65.97}\;kN$
or, $F_R=\sqrt {(F-2.035)^2+61.83}\;kN$
Now, for non-zero minimum of $F_R$, the value of $(F-2.035)$ has to be zero. Thus
$F-2.035=0$
or, $F=2.035\;kN$
For $F=2.035\;kN$, the magnitude of resultant force becomes
$F_R=\sqrt {61.83}\;kN\approx7.86\;kN$
