Solution 14.1a
From Mechanics
Taking moments about the point A:
\displaystyle \begin{align} & 1 \textrm{.}5\times {{R}_{B}}=0 \textrm{.}5\times 20\times 9 \textrm{.}8 \\ & {{R}_{B}}=\frac{ 0\textrm{.}5\times 20\times 9 \textrm{.}8}{1 \textrm{.}5}=65 \textrm{.}3\text{ N} \\ \end{align}
Taking moments about the point A:
\displaystyle \begin{align} & 1 \textrm{.}5\times {{R}_{A}}=1\times 20\times 9 \textrm{.}8 \\ & {{R}_{A}}=\frac{1\times 20\times 9 \textrm{.}8}{1 \textrm{.}5}=130 \textrm{.}7\text{ N} \\ \end{align}
Or
\displaystyle \begin{align} & {{R}_{A}}+65 \textrm{.}3=20\times \textrm{.}7\text{ N} \\ \end{align}