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Solution 14.1a

From Mechanics

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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...)
Line 1: Line 1:
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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>
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Or
+
Or use
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<math>\begin{align}
+
<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>
+
and only one of the above moment equations.

Revision as of 09:19, 8 March 2011

Taking moments about the point A:

1.5RB=0.5209.8RB=1.50.5209.8=65.3 N

Taking moments about the point B:

1.5RA=1209.8RA=1.51209.8=130.7 N

Or use

RA+RB=209.8 N

and only one of the above moment equations.