From Mechanics

The diagram shows the forces acting on the beam

In this case \displaystyle {{R}_{A}}=0. Taking moments about the right hand support:

\displaystyle \begin{align} & 0\textrm{.}4\times m\times 9\textrm{.}8=0\textrm{.}6\times 58.8+1\textrm{.}6\times 11\textrm{.}76 \\ & m=\frac{0\textrm{.}6\times 58\textrm{.}8+1\textrm{.}6\times 11\textrm{.}76}{0\textrm{.}4\times 9\textrm{.}8}=13\textrm{.}8\text{ kg} \\ \end{align}