Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif In this case ${{R}_{A}}=0$ . Taking moments about the right hand support: $\begin{align} & 0\textrm{...$

Ian — Thu, 02 Sep 2010 14:18:52 GMT

New page: The diagram shows the forces acting on the beam Image:14.5a.gif In this case <math>{{R}_{A}}=0</math>. Taking moments about the right hand support: <math>\begin{align} & 0\textrm{...

New page

The diagram shows the forces acting on the beam

[[Image:14.5a.gif]]

In this case
<math>{{R}_{A}}=0</math>. Taking moments about the right hand support:

<math>\begin{align}
& 0\textrm{.}4\times m\times 9\textrm{.}8=0\textrm{.}6\times 58.8+1\textrm{.}6\times 11\textrm{.}76 \\
& m=\frac{0\textrm{.}6\times 58\textrm{.}8+1\textrm{.}6\times 11\textrm{.}76}{0\textrm{.}4\times 9\textrm{.}8}=13\textrm{.}8\text{ kg} \\
\end{align}</math>

Solution 14.5b - Revision history

Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif In this case {{R}_{A}}=0. Taking moments about the right hand support: \begin{align} & 0\textrm{...

Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif In this case ${{R}_{A}}=0$ . Taking moments about the right hand support: $\begin{align} & 0\textrm{...$