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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- | Taking moments about the point A: | + | Taking moments about the point <math>A</math>: |
<math>\begin{align} | <math>\begin{align} | ||
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\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} | ||
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\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 \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.