$+F_{Rx}=25+35sin30=42\mathrm{.}5lb$

$+$darr$F_{Ry}=20+35cos30=50\mathrm{.}31lb$

$M_{RA}=35cos30(2)+20(6)-25(3)=105\mathrm{.}6$

$lb*ft$

$M_{RA}=-35sin602-206+253$

$=-105\mathrm{.}6$

$F_R=({42\mathrm{.}5}^2+{50\mathrm{.}31}^2{)}^{\frac{1}{2}}=65\mathrm{.}9lb,$

$=ta{n}^{-1}(\frac{50\mathrm{.}31}{42\mathrm{.}5})=49\mathrm{.}8$

The equivalent single force $F_R$ can be located on the beam $AB$ at a distance $d$ measured from $A.$

$d=\frac{M_{RA}}{F_{Ry}}=\frac{105\mathrm{.}6}{50\mathrm{.}31}=2\mathrm{.}10ft.$