Solution 14.1a

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(New page: Taking moments about the point A: <math>\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 \text...)
Current revision (09:29, 8 March 2011) (edit) (undo)
 
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Taking moments about the point A:
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The weight acts at the middle point of the beam that is distance 0.5 m from <math>A</math>.
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Taking moments about the point <math>A</math>:
<math>\begin{align}
<math>\begin{align}
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\end{align}</math>
\end{align}</math>
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Taking moments about the point A:
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Taking moments about the point <math>B</math>:
<math>\begin{align}
<math>\begin{align}
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\end{align}</math>
\end{align}</math>
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Or
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Or use
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<math>\begin{align}
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<math>{{R}_{A}}+{{R}_{B}}=20\times 9\textrm{.}8\ \text{N}</math>
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& {{R}_{A}}+65 \textrm{.}3=20\times \textrm{.}7\text{ N} \\
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\end{align}</math>
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and only one of the above moment equations.

Current revision

The weight acts at the middle point of the beam that is distance 0.5 m from \displaystyle A.

Taking moments about the point \displaystyle 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 \displaystyle B:

\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 use

\displaystyle {{R}_{A}}+{{R}_{B}}=20\times 9\textrm{.}8\ \text{N}

and only one of the above moment equations.

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