What is the formal mathematical representation of a "force"?

Question

In mechanics, it is usual to represent a force by a 3-vector. When it is necessary to consider the turning effect of a force, the 3-vector is commonly "attached" to a point on its line of action. In this way, calculations are easily done. There is some redundancy in this specification, and only 5 independent parameters are needed to pin down the force. But I don't know of any practical way of representing forces as elements in a 5-dimensional vector space. In particular, I don't know how forces could be combined or how the concept of "couple" could emerge in such a formalism. Has a theory of forces as 5-dimensional objects been developed?

Answer 1

It's still rather unclear to me what you're asking, but it's probably about force-torque vectors in the space of wrenches as defined in the so called screw theory.

Answer 2

In a differential geometrical setting force is represented by a covector field (a differential $1$-form). For example, work is then a path integral of force. In a vector calculus setting force is a vector field. The geometrical setting is IMHO the most useful but the vector calculus setting seems to be (the most) widespread.

Answer 3

It isn't clear in your post why you say a translational force and point's position don't have full degree of freedom. From your comments:

Drop the perpendicular from the origin O to the line of action of the force, meeting it at A. Given A, whose specification requires 3 parameters, the force may be represented by a 2-vector, in the plane through A perpendicular to OA, which points along the line of action.

This is wrong.

Let the vector component of the force be $v$ and let $x$ be the point the force is acting on, so that the force is $(x,v+x)$.

Construct the plane through $A, x(v+x)$, which is the normal plane to $OA$ at $A$. Take a 2-vector in this vector space at $A$ over this plane. What is $x$ and what is $v+x$? The most information you can get is by letting $x$ or $v+x$ be parametrized by the 2-vector, and then using a scalar multiple of that 2-vector to get the other vector, but the 2-vector alone is not enough to describe both vectors.

Answer 4

A force is a screw in 3D with 6 independent parameters. Four parameters for the 3D line of action of the force (position + direction). One parameter for the pitch (torque parallel to the line of action) and one parameter for the magnitude.

Analytically, a force with direction $\vec{e}$, magnitude $F$, position of a point along the force line $\vec{r}$ and pitch $h=\frac{\tau}{F}$ where $\tau$ is the parallel torque is defined by the force moment vectors of

$$ \begin{aligned} \vec{F} &= F \vec{e} \\ \vec{M} & = \vec{r} \times \vec{F} + h \vec{F} \end{aligned} $$

Working backwards, given the above vector pair the following parameters are extracted

• Magnitude $$F = ||\vec{F}||$$
• Direction $$\vec{e} = \frac{ \vec{F}}{F}$$
• Position $$\vec{r} = \frac{\vec{F} \times \vec{M}}{F^2}$$
• Pitch $$h = \frac{ \vec{F} \cdot \vec{M} }{F^2}$$

Just as @RespawnedStuff said, the pair $\hat{f} = (\vec{F},\vec{M})$ is called a wrench in screw theory. It is a screw because geometrically it represents a line with direction and position and a pitch. The magnitude is treated as a homogeneous factor because changing its value does not change the geometry of the force system.

Answer 5

With the aid of other answers, particularly that of @ja72, I have gathered the answer to be no: a theory of forces as 5-dimensional objects has not been developed. The reason is that, although a pure force is 5-dimensional, the set of such forces does not have any natural vector-space structure. Rather, the set of forces is a 5-dimensional submanifold (not a subspace) of the 6-dimensional space of screws. This space accommodates also couples, or torques, forming a 3-dimensional subspace, which arise when forces are combined.

Six-dimensional screw space comprises pairs $(\vec M,\vec F)$ of 3-vectors which combine in the usual way. The basic equations are $$ \begin{aligned} \vec{F} &= F \vec{e}, \\ \vec{M} & = \vec{r} \times \vec{F} + \tau\vec{e}, \end{aligned}$$ where $F$ is the magnitude of the (pure) force, $\vec e$ is a unit vector in the direction of the line of action of the force, $\vec M$ is the moment of the screw about the origin, $\vec r$ is the position vector of any point on the line of action of the force, and $\tau$ represents the torque component of the screw.

Pure forces are represented by screws with $\tau=0$: that is, $$\{(\vec M,\vec F):\vec M=\vec r\times\vec F\;\,\text{for some}\;\,\vec r\}.$$ Pure moments are those screws with zero $F$-component. In this formalism, two pure forces may give rise to a moment. Thus $$(\vec r_1\times\vec F_1,\;\vec F_1)+(\vec r_2\times\vec F_2,\;\vec F_2)=(\vec r_1\times\vec F_1+\vec r_2\times\vec F_2,\;\vec F_1+\vec F_2)$$ may be not of the pure-force form $(\vec r\times\vec F,\;\vec F)$.