# Confusion about force analysis in 3D space

A book I'm using states that it is possible for a system of forces to produce a net translational force with no turning moment if the line of action of the net force does not pass through the origin if the resultant vector moment of the system is perpendicular to the resultant force. However, I'm having trouble understanding this statement. How is it possible for the net force to produce no torque if it does not act through the origin?

• Do you have a reference for the book and the location of the statements within it? – Farcher Jul 23 '17 at 13:15
• The statement of the book would be: "If it reduces to a single force R then either the line of action of R passes through the origin O or the line of action of R does not pass through O and the resultant moment of the system about O acts in the plane containing R and O, i.e. the resultant vector moment of the system about O is perpendicular to the resultant force (i.e. $(\sum{r_{i}\times F_{i}})(\cdot \sum{F_i})=0$)" – EigenFunction Jul 23 '17 at 14:09
• The book is an M5 Edexcel textbook , page 18. – EigenFunction Jul 24 '17 at 12:14

Suppose that in a frame $\,Oxyz\,$ a system of forces $\,\mathrm{S}\,$ consists of forces $\,\mathbf{F}_{1},\mathbf{F}_{2},\cdots,\mathbf{F}_{n}\,$ acting through points with position vectors $\,\mathbf{r}_{1},\mathbf{r}_{2},\cdots,\mathbf{r}_{n}\,$ respectively. The question is under what conditions this system $\,\mathrm{S}\,$ is equivalent to a single force in case of nonzero resultant.

So, let a point $\,A\,$ with position vector $\,\mathbf{a}$. Every force $\,\mathbf{F}_{\jmath}\,$ is parallel transported so that its action point to be the point $\,A$. Then the system of forces is transformed to an equivalent one with resultant force $$\mathbf{R}=\sum\limits_{\jmath} \mathbf{F}_{\jmath} \ne \boldsymbol{0} \tag{01}$$ acting on point $\,A\,$ and resultant moment with respect to the origin $\,O\,$ $$\overline{\mathbf{M}}=\sum\limits_{\jmath} \left(\mathbf{r}_{\jmath}-\mathbf{a}\right)\boldsymbol{\times}\mathbf{F}_{\jmath}= \left(\sum\limits_{\jmath} \mathbf{r}_{\jmath}\boldsymbol{\times}\mathbf{F}_{\jmath}\right)-\left(\mathbf{a}\boldsymbol{\times}\mathbf{R}\right)=\mathbf{M}-\mathbf{a}\boldsymbol{\times}\mathbf{R} \tag{02}$$ Now, this last equivalent system may be in turn equivalent to a single force in the following two cases :

$\underline{\textbf{CASE 1 :}}$

The resultant moment $\,\overline{\mathbf{M}}\,$ is zero $$\overline{\mathbf{M}}=\boldsymbol{0} \quad \Longrightarrow \quad \sum\limits_{\jmath} \mathbf{r}_{\jmath}\boldsymbol{\times}\mathbf{F}_{\jmath}=\mathbf{a}\boldsymbol{\times}\mathbf{R} \qquad \text{that is} \qquad \mathbf{M}=\mathbf{a}\boldsymbol{\times}\mathbf{R} \tag{03}$$ This case includes the possibility the two terms to be zero simultaneously $$\mathbf{M}=\sum\limits_{\jmath} \mathbf{r}_{\jmath}\boldsymbol{\times}\mathbf{F}_{\jmath}=\boldsymbol{0} \qquad \text{and} \qquad \mathbf{a}\boldsymbol{\times}\mathbf{R}=\boldsymbol{0} \tag{04}$$ in which case the line of action of the resultant force $\,\mathbf{R}\,$ passes through the origin $\,O$.

$\underline{\textbf{CASE 2 :}}$

The line of action of the resultant force $\,\mathbf{R}\,$ doesn't pass through the origin $\,O\,$ and the resultant moment $\,\mathbf{M}\ne \boldsymbol{0}\,$ is normal to the resultant force $\,\mathbf{R}\,$ $$\mathbf{a}\boldsymbol{\times}\mathbf{R}\ne \boldsymbol{0} \qquad \text{and} \qquad \mathbf{M}\boldsymbol{\cdot}\mathbf{R}=\left(\sum\limits_{\jmath} \mathbf{r}_{\jmath}\boldsymbol{\times}\mathbf{F}_{\jmath}\right)\boldsymbol{\cdot}\left(\sum\limits_{\imath} \mathbf{F}_{\imath}\right)=0 \tag{05}$$ In this case there exists at least one vector $\,\mathbf{b}\,$ such that $$\mathbf{M}=\mathbf{b}\boldsymbol{\times}\mathbf{R} \tag{06}$$ a moment which could be balanced by a parallel transport of the resultant force $\,\mathbf{R}\,$ by the vector $\,\left(\mathbf{b}-\mathbf{a}\right)\,$, giving a final equivalent system of a single force : that of $\,\mathbf{R}\,$ acting on a point $\,B\,$ with position vector $\,\mathbf{b}\,$.

• I'm sorry for asking such a stupid question, but if you take $\overline{M}$ to be $\sum{(r_{j}-a) \times F_{j}}$, why is this not the resultant moment about the point A instead of the origin O? – EigenFunction Jul 25 '17 at 11:04
• You don't ask stupid questions. To the contrary. So, let a force $\mathbf{F}$ acting on a point $\mathrm{P}$ with position vector $\mathbf{r}$. This is as single force, that is except its own moment $\mathbf{M}=\mathbf{r}\times\mathbf{F}$ it's not accompanied by any other moment. Now, if you parallel transport this force to a point of action $\mathrm{A}$ with position vector $\mathbf{a}$ then in order to have an equivalent system you must compensate the change of its moment...(*) – Frobenius Jul 25 '17 at 14:52
• (*)...The new equivalent system now is a (not single any more) force $\mathbf{F}$ acting on point $\mathrm{A}$ PLUS A MOMENT WITH RESPECT TO THE ORIGIN $\,O\,$ equal to $\overline{\mathbf{M}}=\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{F}$. – Frobenius Jul 25 '17 at 14:53

The book is talking about a force couple that results in a pure torque (perpendicular to the forces). Imagine two equal and opposite forces ${\bf F}_1$ and ${\bf F}_2$ not along the same force line. If their relative position is described by the vector ${\bf r} ={\bf r}_1-{\bf r}_2$ then the resultant force is $${\bf R}={\bf F}_1+{\bf F}_2 = {\bf 0}$$ and the resultant moment $${\bf M} = {\bf r}_1 \times {\bf F}_1 +{\bf r}_2 \times {\bf F}_2 = ({\bf r}_2+{\bf r}) \times {\bf F}_1 -{\bf r}_2 \times {\bf F}_1 = {\bf r} \times {\bf F}_1$$

The moment is perpendicular to the forces and the relative position of the force lines.

## More Theory

• All complex loading cases (multiple forces ${\bf F}_i$ located at ${\bf r}_i$) reduce to a single resultant force and couple that may be zero or not

$${\bf R}=\sum_i {\bf F}_i \\ {\bf M}=\sum_i {\bf r}_i \times {\bf F}_i$$

• The combined loading represents a line in space (the load line), a force magnitude $f$ and a scalar pitch value $h$. This is called a force screw. The pitch value is a geometrical value describing how much moment the system has along the force per unit force.

• The general state of loading from a system of forces is a force in space plus a parallel moment. Any ${\bf R}$ and ${\bf M}$ values are decomposed into a force (magnitude and direction) located at a position with a parallel moment described by the pitch value.

Here is how you get the above 4 screw quantities that decompose a system of forces:

• The force magnitude is $$f = \| {\bf R} \|$$
• The force line has direction $${\bf e} = \frac{ {\bf R} }{ f }$$
• The point on the force line closest to the origin is $${\bf r} = \frac{ {\bf R} \times {\bf M}}{f^2}$$
• The scalar pitch value is $$h = \frac{ {\bf R} \cdot {\bf M}}{f^2}$$

Using the four above quantities you can fully reconstruct the combined loading (proof in the Appendix)

\begin{align} {\bf R} &= f\, {\bf e} \\ {\bf M} & = {\bf r} \times {\bf R} + h {\bf R} \end{align}

In the above expression, $f$ is the quantity of force, and ${\bf e}$, ${\bf r}$ and $h$ is the geometry of force. The quality of the force system is described by these quantities. You have to allow for not only zero values, but also infinities for the math to work out. Here are some special cases

1. Force magnitude is zero and pitch is finite --> no net force or moment \begin{align} {\bf R} &= 0\, {\bf e} = {\bf 0} \\ {\bf M} & = {\bf r} \times {\bf 0} + h {\bf 0} = {\bf 0} \end{align}

2. Location and pitch are zero --> force line through the origin \begin{align} {\bf R} &= f \,{\bf e} \\ {\bf M} & ={\bf 0} \times {\bf R} + 0 {\bf R} = {\bf 0} \end{align}

3. Location is zero, but pitch isn't --> force through the origin, with parallel moment. \begin{align} {\bf R} &= f\, {\bf e} \\ {\bf M} & ={\bf 0} \times {\bf R} + h {\bf R} = h {\bf R} \end{align} This happens with 3 froces. A force couple causing a net moment along a direction and a third parallel force to the moment.

4. Force magnitude is zero and pitch is infinite --> pure torque of magnitude $m$ \begin{align} {\bf R} &= 0\, {\bf e} = {\bf 0} \\ {\bf M} & ={\bf r} \times {\bf 0} + \infty {\bf 0} = m {\bf e} \end{align} This is just a force couple. The magnitude $m$ is not encoded into the 4 screw values and it is considered an arbitrary value. Here is why keeping things in the $({\bf R},{\bf M})$ system is advategneous to using the $(f,{\bf e},{\bf r},h)$ system. Nevertheless the screw properties help us understand the geometry of the loading.

## Apendix

You prove the loading compisition by expanding ${\bf R} \cdot {\bf M}$ and ${\bf R} \times {\bf M}$ with vector identities.

$$\require{cancel} \begin{cases} {\bf R} \cdot {\bf M} = {\bf R} \cdot \left( {\bf r}\times {\bf R} + h {\bf R} \right) = h ({\bf R}\cdot{\bf R}) = h f^2 \\ {\bf R} \times {\bf M} = {\bf R} \times \left( {\bf r}\times {\bf R} + h {\bf R} \right) = {\bf R} \times ({\bf r}\times {\bf R}) = {\bf r} ({\bf R}\cdot{\bf R}) - {\bf R} (\cancel{{\bf R}\cdot{\bf r}}) \end{cases}$$