Confusion about force analysis in 3D space

Question

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?

Answer 1

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 \begin{equation} \mathbf{R}=\sum\limits_{\jmath} \mathbf{F}_{\jmath} \ne \boldsymbol{0} \tag{01} \end{equation} acting on point $\,A\,$ and resultant moment with respect to the origin $\,O\,$ \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} This case includes the possibility the two terms to be zero simultaneously \begin{equation} \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} \end{equation} 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 $\,\mathbf{R}\,$ \begin{equation}\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} \end{equation} In this case there exists at least one vector $\,\mathbf{b}\,$ such that \begin{equation} \mathbf{M}=\mathbf{b}\boldsymbol{\times}\mathbf{R} \tag{06} \end{equation} 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 $\,\mathbf{b}\,$.

Answer 2

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

Let's see the following concepts might help you understand the geometry of rigid body loadings:

• 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.

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