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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...)
Current revision (09:29, 8 March 2011) (edit) (undo)
 
(One intermediate revision not shown.)
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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}
Line 6: Line 8:
\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}
Line 13: Line 15:
\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 A.

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.