[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) $\vec{M}_{L} = \vec{r_2} \times \vec{F_2}$
(2) $\vec{M}_{L} = \vec{\lambda} \cdot \left( \vec{r} \times \vec{F} \right)$

 

where $\lambda$ is a unit vector in the direction of $L$, $\vec{F}_{1}$ is a component of $\vec{F}$ normal to the plane $P$, $\vec{F}_{2}$ 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 $\vec{F}$ about the axis $L$ is obtained using any point on the axis $L$. Let us consider a point on $L$, $S$, which satisfies

 

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

 

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

 

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

 

Then,

 

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

 

Using the property of mixed triple product which is

 

$\vec{A}\cdot \left( \vec{B} \times \vec{C}\right) = \vec{B}\cdot \left( \vec{C} \times \vec{A}\right) = \vec{C}\cdot \left( \vec{A} \times \vec{B}\right) $,

 

difference between two moments is

 

\begin{align*} \vec{M}_{L}-\vec{M}_{L,S} =& \left(-s\vec{\lambda} + \vec{r}\right) \cdot \left( \vec{F} \times \vec{\lambda} \right) - \vec{r} \cdot \left( \vec{F} \times \vec{\lambda} \right) \\ =& -s\vec{\lambda} \cdot \left( \vec{F} \times \vec{\lambda} \right) \\ =& -s\vec{F} \cdot \left( \vec{\lambda} \times \vec{\lambda} \right) \\ =&0 \end{align*}

 

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

 

 

- lazy engineer