\begin{problem*}{11.5}
Caclulate the net torque (magnitude and direction) on the beam in
Figure~P11.5 about \Part{a} an axis through $O$ perpendicular to the
page and \Part{b} an axis through $C$ perpendicular to the page.
\end{problem*}

\begin{solution}
\Part{a}
\begin{equation}
  \sum \tau =   0\U{m}\cdot30\U{N}
              + 2.0\U{m}\cdot25\U{N}\sin(90\dg-30\dg)
              + 4.0\U{m}\cdot10\U{N}\sin(-20\dg)
    = \ans{29.6\U{J}} \;.
\end{equation}
Because we used the usual right handed cross product, the positive
torque will cause counter clockwise rotation.

\Part{b}
\begin{equation}
  \sum \tau =   2\U{m}\cdot30\U{N}\sin(45\dg)
              + 0\U{m}\cdot25\U{N}
              + 2.0\U{m}\cdot10\U{N}\sin(-20\dg)
    = \ans{35.6\U{J}} \;.
\end{equation}
As in \Part{a}, positive torques are counter clockwise.
\end{solution}
