I have information about total force and moment on a body for three points, whose coordinates I know. From this information I would like to determine the point at which moment would be zero.
Generally,
$\vec{M} = \vec{r} \times \vec{F}$
…which can be written as …
$\vec{M} = \begin{bmatrix} \vec{i}&\vec{j}&\vec{k} \\ r_x&r_y&r_z \\ F_x&F_y&F_z \end{bmatrix} $
…or expanded out …
$\vec{M} = (r_yF_z-r_zF_y)\vec{i} + (r_zF_x-r_xF_z)\vec{j} + (r_xF_y-r_yF_x)\vec{k}$
Taking a cross product of the two vectors is trivial and even I'm capable of grasping this, but what I would really like to do is backsolve for a point at which $\vec{M}=[0,0,0]$. Is this possible?
Here's what I'm seeing…
$M_x = r_yF_z-r_zF_y = 0\\
M_y = r_zF_x-r_xF_z = 0\\
M_z = r_xF_y-r_yF_x = 0$
…so in my head I'm thinking that if the forces are known and we want the moment to equal zero, this system of equations can be set up and solved for the vector:
$\begin{bmatrix}M_x\\M_y\\M_z\end{bmatrix} = \begin{bmatrix}r_x\\r_y\\r_z\end{bmatrix} \cdot \begin{bmatrix}0&F_z&-F_y\\-F_z&0&F_x\\F_y&-F_x&0\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$
…but obviously this is a singular matrix and there is not a distinct solution.
What am I missing here?
Best Answer
You are looking for the equipollent force (and its location). This is also known as the force axis (line). Given the force vector $\boldsymbol{F}$ and moment $\boldsymbol{M}$ defined at a point, the force axis is located at
$$ \boldsymbol{r} = \frac{\boldsymbol{F} \times \boldsymbol{M}}{\| \boldsymbol{F} \|^2}$$
Note also that you can see if any moments are parallel to the force axis. You do this by calculating the scalar pitch value
$$ h = \frac{ \boldsymbol{F} \cdot \boldsymbol{M}}{\| \boldsymbol{F} \|^2}$$
This gives the parallel moment magnitude ratio to the force magnitude.
NOTE: $\cdot$ the the vector dot product, $\times$ the vector cross product and $\| \mbox{ } \|$ the vector magnitude
Proof
A force $\boldsymbol{F}$ is applied at a location $\boldsymbol{r}$ causing a moment at the origin $\boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F}$. We can get the location (at least the closest point to the force axis) by
$$ \boldsymbol{r} = \frac{\boldsymbol{F} \times \boldsymbol{M}}{\| \boldsymbol{F} \|^2} = \frac{\boldsymbol{F} \times (\boldsymbol{r} \times \boldsymbol{F})}{\| \boldsymbol{F} \|^2}$$
using the vector triple product identity $\boldsymbol{a}\times(\boldsymbol{b}\times\boldsymbol{c}) = \boldsymbol{b} (\boldsymbol{a}\cdot \boldsymbol{c})- \boldsymbol{c} (\boldsymbol{a}\cdot \boldsymbol{b})$ the above is
$$\require{cancel} \boldsymbol{r} = \frac{\boldsymbol{r} (\boldsymbol{F}\cdot \boldsymbol{F})- \boldsymbol{F} (\boldsymbol{F}\cdot \boldsymbol{r})}{\| \boldsymbol{F} \|^2} = \frac{\boldsymbol{r} \|\boldsymbol{F}\|^2 - \boldsymbol{F} \cancel{ (\boldsymbol{F}\cdot \boldsymbol{r})}}{\| \boldsymbol{F} \|^2} = \boldsymbol{r}$$
The cancellation happens because the location vector must be perpendicular to the force (since location along the force do not change the moments).