From Mechanics

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Revision as of 12:21, 2 September 2010

The diagram shows the forces acting.

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=130.7\text{ N} \\ \end{align}