Answer
$F_1\approx57.84\;N$,
$F_R\approx379.8\;N$
Work Step by Step
First we resolve each force into its $x$ and $y$ components using the parallelogram law,
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force in which. Thus we have
$F_{1x}=F_1\sin30^\circ\;N=0.5F_1\;N$
$F_{1y}=F_1\cos30^\circ\;N=0.866F_1\;N$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=200\;N$. Thus, we have
$F_{2x}=200\cos0^\circ\;N=200\;N$
$F_{2y}=200\sin0^\circ\;N=0\;N$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=260\;N$. Using proportional parts of similar triangles and considering the directions we have
$\frac{F_{3x}}{260\;lb}=\frac{5}{13}$
or, $F_{3x}=260\;N\Big(\frac{5}{13}\Big)=100\;N$
Similarly,
$\frac{F_{3y}}{260\;N}=-\frac{12}{13}$
or, $F_{3y}=-260\;N\Big(\frac{12}{13}\Big)=-240\;N$
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, $(F_R)_x=(0.5F_1+200+100+)\;N=(0.5F_1+300)\;N$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(0.866F_1+0-240)\;N=(0.866F_1-240)\;N$
$\therefore$ The magnitude of the resultant force is given by
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(0.5F_1+300)^2+(0.866F_1-240)^2}\;N$
or, $F_R\approx\sqrt {{F_1}^2-115.68F_1+147600}\;N$
or, $F_R=\sqrt {{F_1}^2-2\times F_1\times57.84+57.84^2-57.84^2+147600}\;N$
$F_R\approx\sqrt {(F_1-57.84)^2+144254}\;N$
For minimum value of $F_R$, the value of $(F_1-57.84)$ has to be zero. Thus
$F_1-57.84=0$
or, $F_1=57.84\;N$
For $F_1=57.84\;N$, the magnitude of the resultant force becomes minimum.
$\therefore$ The minimum magnitude of resultant force is
$F_R=\sqrt {144254}\;N$
or, $F_R\approx 379.8\;N$
