Solution 14.3

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(New page: The diagram shows the forces acting. Image:14.3.gif Taking moments about the left hand support: <math>\begin{align} & {{R}_{B}}\times 3=49\times 1+196\times 1\textrm{.}5 \\ & ...)
Current revision (17:03, 10 March 2011) (edit) (undo)
 
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[[Image:14.3.gif]]
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<math>\begin{align}
<math>\begin{align}
& {{R}_{A}}+114\textrm{.}3=49+196 \\
& {{R}_{A}}+114\textrm{.}3=49+196 \\
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& {{R}_{A}}=49+196-114\textrm{.}3=130.7\text{ N} \\
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& {{R}_{A}}=49+196-114\textrm{.}3=130\textrm{.}7\text{ N} \\
\end{align}</math>
\end{align}</math>

Current revision

The diagram shows the forces acting.


Image:14.3a.gif


Taking moments about the left hand support:


\displaystyle \begin{align} & {{R}_{B}}\times 3=49\times 1+196\times 1\textrm{.}5 \\ & {{R}_{B}}=\frac{49\times 1+196\times 1\textrm{.}5}{3}=114\textrm{.}3\text{ N} \\ \end{align}


Taking moments about the right hand support:


\displaystyle \begin{align} & {{R}_{A}}\times 3=49\times 2+196\times 1\textrm{.}5 \\ & {{R}_{B}}=\frac{49\times 2+196\times 1\textrm{.}5}{3}=130\textrm{.}7\text{ N} \\ \end{align}


Or:

\displaystyle \begin{align} & {{R}_{A}}+114\textrm{.}3=49+196 \\ & {{R}_{A}}=49+196-114\textrm{.}3=130\textrm{.}7\text{ N} \\ \end{align}