[mechanics] Regardlessness of point of moment calculation about an axis

Moment of a force about an axis through the origin $O$ can be calculated in two ways:

 

(1) ${\overrightarrow{M}}_L=\overrightarrow{r_2}\times \overrightarrow{F_2}$
(2) ${\overrightarrow{M}}_L=\overrightarrow{\lambda }\cdot (\overrightarrow{r}\times \overrightarrow{F})$

 

where $\lambda$ is a unit vector in the direction of $L$, ${\overrightarrow{F}}_1$ is a component of $\overrightarrow{F}$ normal to the plane $P$, ${\overrightarrow{F}}_2$ parallel to the plane $P$, $P$ is a plane perpendicular to $L$. In eq. (2) the moment is calculated about the origin. But it can be shown that the moment of $\overrightarrow{F}$ about the axis $L$ is obtained using any point on the axis $L$. Let us consider a point on $L$, $S$, which satisfies

 

$\overrightarrow{OS}=s\overrightarrow{\lambda }$.

 

Moment of $\overrightarrow{F}$ about $L$ using $S$ is

 

${\overrightarrow{M}}_{L,S}=\overrightarrow{\lambda }\cdot ((-s\overrightarrow{\lambda }+\overrightarrow{r})\times \overrightarrow{F}).$

 

Then,

 

${\overrightarrow{M}}_L-{\overrightarrow{M}}_{L,S}=\overrightarrow{\lambda }\cdot ((-s\overrightarrow{\lambda }+\overrightarrow{r})\times \overrightarrow{F})-\overrightarrow{\lambda }\cdot (\overrightarrow{r}\times \overrightarrow{F}).$

 

Using the property of mixed triple product which is

 

$\overrightarrow{A}\cdot (\overrightarrow{B}\times \overrightarrow{C})=\overrightarrow{B}\cdot (\overrightarrow{C}\times \overrightarrow{A})=\overrightarrow{C}\cdot (\overrightarrow{A}\times \overrightarrow{B})$,

 

difference between two moments is

 

$$\begin{array}{rl}{\overrightarrow{M}}_L-{\overrightarrow{M}}_{L,S}=& (-s\overrightarrow{\lambda }+\overrightarrow{r})\cdot (\overrightarrow{F}\times \overrightarrow{\lambda })-\overrightarrow{r}\cdot (\overrightarrow{F}\times \overrightarrow{\lambda })\\ =& -s\overrightarrow{\lambda }\cdot (\overrightarrow{F}\times \overrightarrow{\lambda })\\ =& -s\overrightarrow{F}\cdot (\overrightarrow{\lambda }\times \overrightarrow{\lambda })\\ =& 0\end{array}$$

 

which means ${\overrightarrow{M}}_{L,S}={\overrightarrow{M}}_L$ for any point $S$ on $L$.

 

 

- lazy engineer