Answer
The net torque about $O$ at $P$ is
total ${\bf{\tau}} = \left( {204.79 - 2940\cos \theta } \right){\bf{k}}$ N-m
Work Step by Step
Since the angle between the position vector ${\bf{r}}$ and the force ${{\bf{F}}_g}$ is not given, we call it $\theta$. From the Figure 22(B) we have then
${{\bf{F}}_g} = - 9.8m{\bf{j}}$
${\bf{r}} = 10\cos \theta {\bf{i}} + 10\sin \theta {\bf{j}}$
Thus, the torque due to the force ${{\bf{F}}_g}$ is
${{\bf{\tau}} _g} = {\bf{r}} \times {{\bf{F}}_g} = \left( {10\cos \theta {\bf{i}} + 10\sin \theta {\bf{j}}} \right) \times \left( { - 9.8m{\bf{j}}} \right)$
${{\bf{\tau}} _g} = {\bf{r}} \times {{\bf{F}}_g} = - 98m\cos \theta \left( {{\bf{i}} \times {\bf{j}}} \right) - 98m\sin \theta \left( {{\bf{j}} \times {\bf{j}}} \right)$
Since ${\bf{i}} \times {\bf{j}} = {\bf{k}}$, ${\ }$ ${\bf{j}} \times {\bf{j}} = {\bf{0}}$ and $m=30$ kg, we get
${{\bf{\tau}} _g} = {\bf{r}} \times {{\bf{F}}_g} = - 2940\cos \theta {\bf{k}}$ N-m
Recall from the result in Exercise 63, the torque due to ${\bf{F}}$ is
${\bf{\tau}} = 204.79{\bf{k}}$ N-m
Thus, the net torque about $O$ at $P$ is
total ${\bf{\tau}} = 204.79{\bf{k}} - 2940\cos \theta {\bf{k}}$ N-m
total ${\bf{\tau}} = \left( {204.79 - 2940\cos \theta } \right){\bf{k}}$ N-m
