Question

The torque of a force $\underset{F}{\to }$ about a point is defined as $\underset{\mathrm{\Gamma }}{\to }=\underset{r}{\to }\times \underset{F}{\to }$. Suppose $\underset{r}{\to },\underset{F}{\to }$ and $\underset{\mathrm{\Gamma }}{\to }$ are all nonzero. Is $r\times \underset{\mathrm{\Gamma }}{\to }\left|\right|\underset{F}{\to }$ always true? Is it ever true?

Answer 1

$$\begin{gathered} \mathrm{No},\underset{\mathrm{r}}{\to }\times \underset{\tau }{\to }\left|\right|\underset{\mathrm{\Gamma }}{\to }\mathrm{is}\mathrm{not}\mathrm{true}. \\ \mathrm{In}\mathrm{fact},\mathrm{it}\mathrm{is}\mathrm{never}\mathrm{true}.\mathrm{This}\mathrm{is}\mathrm{because}: \\ \underset{\mathrm{r}}{\to }\times \underset{\tau }{\to } \\ =\underset{\mathrm{r}}{\to }\times \left(\underset{\mathrm{r}}{\to }\times \underset{\mathrm{F}}{\to }\right) \\ \mathrm{Applying}\mathrm{vector}\mathrm{triple}\mathrm{product},\mathrm{we}\mathrm{get}: \\ \underset{\mathrm{r}}{\to }\times \left(\underset{\mathrm{r}}{\to }\times \underset{\mathrm{F}}{\to }\right) \\ =\left(\underset{\mathrm{r}}{\to }.\underset{\mathrm{F}}{\to }\right)\underset{\mathrm{r}}{\to }-\left(\underset{\mathrm{r}}{\to }.\underset{\mathrm{r}}{\to }\right)\underset{\mathrm{F}}{\to } \\ \because \underset{\mathrm{r}}{\to }.\underset{\mathrm{r}}{\to }={\mathrm{r}}^2 \\ =\left(\underset{\mathrm{r}}{\to }.\underset{\mathrm{F}}{\to }\right)\underset{\mathrm{r}}{\to }-{\mathrm{r}}^2\underset{\mathrm{F}}{\to } \\ \mathrm{If}\underset{\mathrm{r}}{\to }.\underset{\mathrm{F}}{\to }=0;\mathrm{that}\mathrm{is},\underset{r}{\to }\perp \underset{\mathrm{F}}{\to },\mathrm{then}: \\ \underset{\mathrm{r}}{\to }\times \underset{\mathrm{\Gamma }}{\to }=-{\mathrm{r}}^2\underset{\mathrm{F}}{\to } \\ \mathrm{We}\mathrm{know}\mathrm{that}{\mathrm{r}}^2\mathrm{is}\mathrm{never}\mathrm{negative}\mathrm{and}\underset{\mathrm{r}}{\to }\times \underset{\mathrm{\Gamma }}{\to }=-{\mathrm{r}}^2\underset{\mathrm{F}}{\to .} \\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{both}\mathrm{vectors}\mathrm{may}\mathrm{be}\mathrm{antiparallel}\mathrm{to}\mathrm{each}\mathrm{other}\mathrm{but}\mathrm{not}\mathrm{parallel}. \end{gathered}$$