How to know the rotation of moment vector in 3D?

Question

For a single position and force vector, it is easy to just use the curling hand, as shown in the below picture where we expand our hand in the direction of position vector and curling it in the direction of force vector.

However, it is not lucid if we have a plethora of position and force vectors as shown below.

The only notion I can think of is to construct the resultant force vector but of course I need some sort of resultant position vector but I have no idea how to construct such resultant position vector if it exists altogether.

Answer 1

Each force $\vec {F_i}$ exerts a moment $\vec {M_i}$ about $O$. You can add the moments of the individual forces just as if they were vectors (technically they are pseudovectors but they can still be added like vectors). So the total moment $\vec{M_{R_0}}$ is just the vector such of the individual moments i.e.

$\displaystyle \vec{M_{R_0}} = \sum_i \vec {M_i}$

Answer 2

Use the individual contributions of each force, to sum up the combined moment.

$$ \begin{aligned} \vec{F}_{R} & = \sum_i \vec{F}_i \\ \vec{M}_{R_O} & = \sum_i \vec{r}_i \times \vec{F}_i \end{aligned} \tag{1}$$

Now you can where the resultant force's line of action is in 3D.

$$ \vec{r}_{R_O} = \frac{ \vec{F}_{R} \times \vec{M}_{R_O} }{ \| \vec{F}_{R} \|^2} \tag{2}$$

Now you can write

$$ \vec{M}_{R_O} = \vec{r}_{R_O} \times \vec{F}_{R} $$

Well almost, because there is also a component of the moment parallel to the net force. That component is found with

$$ \vec{M}_{{R_O}_\parallel} = \underbrace{ \left( \frac{ \vec{F}_{R} \cdot \vec{M}_{R_O} }{ \| \vec{F}_{R} \|^2} \right)}_{\text{pitch }h_R} \vec{F}_{R} \tag{3}$$

So the decomposition of moments is

$$ \boxed{ \vec{M}_{R_O} = \vec{r}_{R_O} \times \vec{F}_{R} + h_R\, \vec{F}_{R} } \tag{4}$$

with $\vec{r}_{R_O}$ as defined in (2) and $h_R$ as defined in (3).

Proof using the vector triple product identity $a\times(b\times c) = b(a\cdot c) - c (a \cdot b)$

$$\begin{aligned} \vec{M}_{R_O} & = \vec{r}_{R_O} \times \vec{F}_{R} + h_R\, \vec{F}_{R} \\ & = \frac{ \vec{F}_{R} \times \vec{M}_{R_O} }{ \| \vec{F}_{R} \|^2} \times \vec{F}_{R} + \frac{ \vec{F}_{R} \cdot \vec{M}_{R_O} }{ \| \vec{F}_{R} \|^2} \vec{F}_{R} \\ & = \frac{-\vec{F}_{R} \times (\vec{F}_{R} \times \vec{M}_{R_O} ) + \vec{F}_{R} (\vec{F}_{R} \cdot \vec{M}_{R_O} ) }{\| \vec{F}_{R} \|^2} \\ & = \frac{-\vec{F}_{R} (\vec{F}_{R} \cdot \vec{M}_{R_O}) + \vec{M}_{R_O} (\vec{F}_{R} \cdot \vec{F}_{R} ) + \vec{F}_{R} (\vec{F}_{R} \cdot \vec{M}_{R_O} )}{\| \vec{F}_{R} \|^2} \\ & = \frac{\vec{M}_{R_O} \| \vec{F}_{R} \|^2} {\| \vec{F}_{R} \|^2} = \vec{M}_{R_O} \;\; \checkmark \end{aligned}$$

Answer 3

Lets write the torque $\vec{M}_{Ro}$ like this:

$$\vec{M}_{Ro}={M}_{Ro}\,\vec{\hat{n}}$$

where ${M}_{Ro}$ is the magnitude of the torque and $\vec{\hat{n}}$ is the axes (normalized to one) about which the torque is applied.

if you take now the sum of the torques $~\vec{\tau}~$ about point o ,from all forces you obtain.

$$\vec{\tau}=\sum_i \vec{r}_i\times\vec{F}_i$$

project $\vec{\tau}$ towards the axes $\vec{\hat{n}}$

$$\vec{\tau}=\tau\,\vec{\hat{n}}$$

thus:

$${M}_{Ro}=\tau=\sqrt{\vec{\tau}\cdot\vec{\tau}}$$

and

$$\vec{M}_{Ro}=\sqrt{\vec{\tau}\cdot\vec{\tau}}\,\vec{\hat{n}}$$