Solution 14.1a

From Mechanics

Revision as of 09:19, 8 March 2011 by Ian (Talk | contribs)
Jump to: navigation, search

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.