2 Caclulate the net torque (magnitude and direction) on the beam in
3 Figure~P11.5 about \Part{a} an axis through $O$ perpendicular to the
4 page and \Part{b} an axis through $C$ perpendicular to the page.
10 \sum \tau = 0\U{m}\cdot30\U{N}
11 + 2.0\U{m}\cdot25\U{N}\sin(90\dg-30\dg)
12 + 4.0\U{m}\cdot10\U{N}\sin(-20\dg)
15 Because we used the usual right handed cross product, the positive
16 torque will cause counter clockwise rotation.
20 \sum \tau = 2\U{m}\cdot30\U{N}\sin(45\dg)
22 + 2.0\U{m}\cdot10\U{N}\sin(-20\dg)
25 As in \Part{a}, positive torques are counter clockwise.