What illustrative example of the central axis theorm exists? This is not the parallel axis theorem

Question

This is from Ira Freeman's translation of Georg Joos's Theoretical Physics. I've taken a few liberties with the notation.

It may be shown also that there is always an axis such that if any point thereon be chosen as reference point, $\mathfrak{T}$ will have the direction of $\mathfrak{F}$. This axis is called the central axis of the the system of forces. If we denote the moment referred to $\mathcal{\bar{O}}$ by $\bar{\mathfrak{T}}$, The vector $\bar{\mathfrak{T}}$ should be so determined that \begin{align*} \bar{\mathfrak{T}}= & \mathfrak{T}-\mathfrak{d}\times\mathfrak{F}=\gamma\mathfrak{F}, \end{align*} where $\gamma$ is a scalar which is yet undetermined. Scalar multiplication by $\mathfrak{F}$ yields \begin{align*} \mathfrak{F}\cdot\bar{\mathfrak{T}} & =\mathfrak{F}\cdot\mathfrak{T}=\gamma\mathfrak{F}^{\cdot2},\\ \text{that is, }\gamma & =\frac{\mathfrak{T}\cdot\mathfrak{F}}{\mathfrak{F}^{\cdot2}}\\ \text{and, }\mathfrak{d}\times\mathfrak{F} & =\mathfrak{T}-\gamma\mathfrak{F}\\ & =\mathfrak{T}-\frac{\mathfrak{F}\mathfrak{T}\cdot\mathfrak{F}}{\mathfrak{F}^{\cdot2}}\\ & =\frac{\mathfrak{T}\mathfrak{F}^{\cdot2}-\mathfrak{F}\mathfrak{F}\cdot\mathfrak{T}}{\mathfrak{F}^{\cdot2}}\\ & =\frac{\mathfrak{F}\times\left(\mathfrak{T}\times\mathfrak{F}\right)}{\mathfrak{F}^{\cdot2}}.\\ \text{Thus }\mathfrak{d} & =\lambda\mathfrak{F}-\frac{\mathfrak{T}\times\mathfrak{F}}{\mathfrak{F}^{\cdot2}}. \end{align*} where $\lambda$ is an arbitrary variable scalar quantity. This means, however, that the terminus of $\mathfrak{d}$ is on a straight line parallel to $\mathfrak{F}$. If, in particular, $\mathfrak{T}$ is parallel to $\mathfrak{F}$, the central axis passes through $\mathcal{O}$ and is parallel to $\mathfrak{F}$.

I am calling this the central axis theorem. Is there a good illustrative example of this theorem in action? I'm finding it difficult to visualize.

Answer 1

One of the easiest examples to visualize comes from statics.

Consider a general shape under gravity supported at two places. A sliding joint at A and a pin joint at B.

Newton's 3rd law says that the force combination of the supports must equal the applied force (gravity W) not only in magnitude and direction but in the line of action also.

All support forces must converge to the central axis, and the moment about the central axis is going to be zero for a 2D case. But so is the moment of the force of gravity (due to Newton's 2nd law) shall also be zero, which means the central axis is going to be located somewhere along the line of action of gravity

Since the direction of force, A is fixed, the line of action of A is known. Where the two lines of actions A and W converge, has to be the central axis location. The net moment about this point is going to zero for both gravity and the reaction at A.

But also, the line of action of B must pass through the central axis, which designates the direction of B (combination of B_x and B_y) as seen above.

Newton's 3rd law implies that any support load system will have a combination of forces and moments, such that they are equal and opposite of the applied forces and moments, and if the applied loads are of a single force then the support system of forces is equipollent to the applied load and will converge to the central axis.

This sets an equivalency between many forces and one force (the net force) in both magnitude, direction, and location in space (designated by the net moment).

$$\begin{aligned} \boldsymbol{F}_{\rm net} &= \sum_i \left( \boldsymbol{F}_i \right) \\ \boldsymbol{M}_{\rm net} &= \sum_i \left( \boldsymbol{r}_i \times \boldsymbol{F}_i + \boldsymbol{M}_i \right) \end{aligned}$$

The central axis location is always found by

$$ \boldsymbol{r}_{\rm central} = \frac{ \boldsymbol{F}_{\rm net} \times \boldsymbol{M}_{\rm net}}{ \| \boldsymbol{F}_{\rm net} \|^2} $$

and will have direction defined by the net force

$$ \boldsymbol{\hat{e}}_{\rm central} = \frac{ \boldsymbol{F}_{\rm net} }{ \| \boldsymbol{F}_{\rm net} \|} $$

To prove the above you need to prove that

$$ \boldsymbol{M}_{\rm net} = \boldsymbol{r}_{\rm central} \times \boldsymbol{F}_{\rm net} + h\, \boldsymbol{F}_{\rm net}$$

where $h$ is a scalar value, the pitch and it is equal to

$$h = \frac{ \boldsymbol{F}_{\rm net} \cdot \boldsymbol{M}_{\rm net}} { \| \boldsymbol{F}_{\rm net} \|^2} $$

I recommend you read this extensive writeup on the terms twist and wrench since the central axis theorem is essentially describing the geometry of a system of forces. In robotics, a system of forces is called a wrench and the geometry of wrenches is central to Screw Theory of rigid body mechanics.

Answer 2

The aim is to simplify a many force acting on a rigid body situation into a much simpler one.

I will try and show you how this can be done via a series of diagrams.

An assembly of forces acting on a rigid body can be reduced to a single force $\vec F$ acting at an arbitrarily chosen point $O$ together with a couple $\vec \tau$ whose axis passes through $O$.

Diagram 1
$a,\,b$ and $c$ are three orthogonal axes with force $\vec F$ along the $c$-axis and couple $\vec \tau$ passing though point $O$ and is in the $cb$ plane.

Diagram 2
The couple is resolved into two components , one along the $c$-axis, $\tau\cos \theta$, and the other along the $b$-axis, $\tau\sin \theta$.

Diagram 3
Two forces are added (orange) as shown in the diagram and thus the couple $\tau \sin \theta$ is replace by an equal couple $Fd$.

Diagram 4
The only force $\vec F$ is now acting at position $O'$.

Diagram 5
The axis of the couple $\tau\cos\theta$ is transferred to be along axis $c'$, which is parallel to axis $c$ and acting at position $O'$.

Hence any given system of forces acting on a rigid body can be reduced to a single force together with a couple whose axis coincides with the direction of the force. The axis defined by the line $c'O'$ is called the central axis.

Note that if $\theta = 90^\circ$ then the original system of forces can be reduced to just one force acting at position $O'$.
In the derivation given in the book the $\cos \theta$ term comes from the dot product of $\vec \tau$ and $\vec F$, $(\bf M F)$ and the $\sin \theta$ term from the cross product of $\vec \tau$ and $\vec F$, $(\bf [M F] )$.

Although the ideas illustrated here can be used for static equilibrium situations when the forces and couples add up to zero, the reduction of a lot of forces acting on a part of a much larger structure, to a single force acting at a point and the associated couple can be useful when analysing the whole structure.

---

Georg Joos's book is possible not the easiest to follow and you might find one of the more modern Engineering books more useful particularly as they also tend to have lots of examples and exercises?

Answer 3

There are two forces $\mathfrak{f}_1$ and $\mathfrak{f}_2$ acting tangentially on the thin cylindrical disk at opposing points, both lying parallel to the base. They induce the moment $\color{blue}{\mathfrak{T}}.$ The force $\color{magenta}{\mathfrak{F}}$ acts at the center of the disk, which is assumed to be rigid and of uniform density.

The red line running through the vector $\color{green}{\bar{\mathfrak{T}}}$ is the central axis advertised in the theorem. The moment of the system about the central axis is $\color{green}{\bar{\mathfrak{T}}}.$ The small arrow from the origin labeled $\color{green}{\mathfrak{d}}$ is the the relative position vector of a randomly chosen $\color{green}{\mathfrak{d}=\lambda}\color{red}{\mathfrak{F}-\frac{\mathfrak{T}\times\mathfrak{F}}{\mathfrak{F}^{\cdot2}}}$ along the central axis. The green line segment lying in the disk is $\color{green}{\mathfrak{d}=0}\color{red}{\mathfrak{F}-\frac{\mathfrak{T}\times\mathfrak{F}}{\mathfrak{F}^{\cdot2}}}.$