Processing Math: Done
Solution 14.1a
From Mechanics
(Difference between revisions)
(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...) |
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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 | + | 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> | + | <math>{{R}_{A}}+{{R}_{B}}=20\times 9\textrm{.}8\ \text{N}</math> |
- | + | ||
- | + | and only one of the above moment equations. |
Revision as of 09:19, 8 March 2011
Taking moments about the point
RB=0.5
20
9.8RB=1.50.5
20
9.8=65.3 N
Taking moments about the point
RA=1
20
9.8RA=1.51
20
9.8=130.7 N
Or use
9.8 N
and only one of the above moment equations.