Solution 14.4a
From Mechanics
(Difference between revisions)
Ian (Talk | contribs)
(New page: Taking moments about the left hand support: <math>\begin{align} & 3\times {{R}_{2}}=18\times 9\textrm{.}8\times 2 \\ & {{R}_{2}}=\frac{18\times 9\textrm{.}8\times 2}{3}=117\textrm{.}6\t...)
Next diff →
Current revision
Taking moments about the left hand support:
\displaystyle \begin{align}
& 3\times {{R}_{2}}=18\times 9\textrm{.}8\times 2 \\
& {{R}_{2}}=\frac{18\times 9\textrm{.}8\times 2}{3}=117\textrm{.}6\text{ N} \\
\end{align}
Taking moments about the right hand support:
\displaystyle \begin{align}
& 3\times {{R}_{1}}=18\times 9\textrm{.}8\times 1 \\
& {{R}_{1}}=\frac{18\times 9\textrm{.}8\times 1}{3}=58\textrm{.}8\text{ N} \\
\end{align}
Or:
\displaystyle \begin{align} & {{R}_{1}}+117\textrm{.}6=18\times 9\textrm{.}8 \\ & {{R}_{1}}=18\times 9\textrm{.}8-117.6=58\textrm{.}8\text{ N} \\ \end{align}