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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?

Qmechanic
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HotDogCannon
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2 Answers2

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  • The engineering answer

Every force $[F_x, F_y, F_z]$ acting on the same line of action will produce the same $[M_x, M_y, M_z]$ moments.

  • The mathematical answer

As you said the singularity of the matrix is telling you that there is not a distinct solution.

If the coefficient matrix has a null determinant the system can be either impossible or have solutions, depending on the determinant of the complete matrix (Rouché-Capelli Theorem).

In this case the system has more than one solution: if you compute the determinant on every $3\times 3$ minor of the complete matrix you'll find out that it is zero, too$\color{red}{^1}$.

Since both the coefficient and the complete matrix are rank $2$ and there are $3$ equations, there are $\infty^1$ solutions to the problem.

$\color{red}{^1}$ This is because $\mathbf F \cdot \mathbf M=0$, that is the force and the moment vectors are perpendicular by definition, since $\mathbf M = \mathbf r \times \mathbf F \Rightarrow \mathbf F \cdot \mathbf r \times \mathbf F=0$ (triple product with two elements being the same).

What I would do to solve the problem is just to pin one of the $\mathbf r$ vector components.

For instance in your case if you pick $r_y$ you end up with

$$r_x=\frac{F_x r_y + M_z}{F_y}$$ $$r_z=\frac{F_z r_y - M_x}{F_y}$$

where $r_y$ is a value of your choice. The third equation will give $F_x M_x + F_y M_y + F_z M_z=0$ being satisfied since $\color{red}{^1}$.

At that point having a point on the line and by knowing the force components you can find any other point on the same line by defining a curvilinear abscissa $t$

$$p_x=r_x+t \cdot F_x$$ $$p_y=r_y+t \cdot F_y$$ $$p_z=r_z+t \cdot F_z$$

so that you can choose the point at which the line of action meets the object (it could also not meet it at all for some loading conditions).

Please correct me if I'm wrong!

Damiano Natali
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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).

John Alexiou
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