Solution 14.1a - Revision history

Ian at 09:29, 8 March 2011

Ian — Tue, 08 Mar 2011 09:29:52 GMT

←Older revision		Revision as of 09:29, 8 March 2011
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		+	The weight acts at the middle point of the beam that is distance 0.5 m from <math>A</math>.
		+
	Taking moments about the point <math>A</math>:		Taking moments about the point <math>A</math>:

Ian at 09:19, 8 March 2011

Ian — Tue, 08 Mar 2011 09:19:39 GMT

←Older revision		Revision as of 09:19, 8 March 2011
Line 1:		Line 1:
-	Taking moments about the point A:	+	Taking moments about the point <math>A</math>:

	<math>\begin{align}		<math>\begin{align}
Line 6:		Line 6:
	\end{align}</math>		\end{align}</math>

-	Taking moments about the point A:	+	Taking moments about the point <math>B</math>:

	<math>\begin{align}		<math>\begin{align}
Line 13:		Line 13:
	\end{align}</math>		\end{align}</math>

-	Or	+	Or use

-	<math>~~\begin{align}~~	+	<math>{{R}_{A}}+{{R}_{B}}=20\times 9\textrm{.}8\ \text{N}</math>
-	& {{R}_{A}}+~~65 \textrm~~{.}3=20\times \textrm{.}7\text{ N~~} \\~~	+
-	~~\end{align~~}</math>	+	and only one of the above moment equations.

Ian: New page: Taking moments about the point A: $\begin{align} & 1 \textrm{.}5\times {{R}_{B}}=0 \textrm{.}5\times 20\times 9 \textrm{.}8 \\ & {{R}_{B}}=\frac{ 0\textrm{.}5\times 20\times 9 \text...$

Ian — Wed, 01 Sep 2010 12:53:05 GMT

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...

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 \textrm{.}8}{1 \textrm{.}5}=65 \textrm{.}3\text{ N} \\
\end{align}</math>

Taking moments about the point A:

<math>\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}</math>

Or

<math>\begin{align}
& {{R}_{A}}+65 \textrm{.}3=20\times \textrm{.}7\text{ N} \\
\end{align}</math>

Solution 14.1a - Revision history

Ian at 09:29, 8 March 2011

Ian at 09:19, 8 March 2011

Ian: New page: Taking moments about the point A: \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...

Ian: New page: Taking moments about the point A: $\begin{align} & 1 \textrm{.}5\times {{R}_{B}}=0 \textrm{.}5\times 20\times 9 \textrm{.}8 \\ & {{R}_{B}}=\frac{ 0\textrm{.}5\times 20\times 9 \text...$