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Since the term force-torque (aka wrench vector) is probably more common in Robotics than in Physics, let's try to start with a definition of what is sought: a force-torque is a parsimonious set (well, actually vector) of parameters that are sufficient to deduce all the implied effects (translational and rotational) of a force acting on a rigid body. For illustration, we want to be able to tell apart from our parameters the case of pushing along a line passing through the center of mass of box [with a given force] versus the case of pushing with an equal (in the sense of usual 2D or 3D-vector) force on a different line; the first scenario will cause only a translation but the second will cause a compound movement of translation and rotation see diagram. Please contrast the 2D and the 3D cases.

I actually know the answer to this question, what I'm crowd-sourcing here is a request for a nice elementary proof. I hope such a proof might be useful addition to this site because people confused by this issue often aren't experts in Lie algebras (e.g. saying that in the 2D case we're talking about elements of the Lie algebra se(2)* and that in the 3D case of se(3)* and that the dimension of the former is three but of the latter is six, will not be illuminating for many). This question here is motivated by a somewhat nebulous claim made in a question on math.SE which I interpreted as claiming that force-torque in 3D is 5-dimensional [note that the tile of the question there does not reflect the actual content of the question; you need to read all of its body.] I suspect that those of you who teach physics might have had to explain this before to someone...

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    $\begingroup$ I think the person in the math question you linked to is thinking about a the force and torque resulting from a force acting on a single pont. This requires only five parameters because the component along the force of where the force is being applied doesn't matter. For example if my force is in the $\hat{x}$ direction, it doesn't matter if I apply it at $(1,1,0)$ or $(0,1,0)$. $\endgroup$
    – Brian Moths
    Commented Feb 20, 2015 at 18:47
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs: what do you mean by "a force acting on a single pont"? He is clearly talking about rigid bodies, see his example dated "Jan 12 at 17:22". $\endgroup$
    – got trolled too much this week
    Commented Feb 20, 2015 at 23:41
  • $\begingroup$ By "force acting on a single point" I mean consider the case of a rigid body with one point charge in it. Now suppose this body is put in a uniform electric field. Then the body will only feel a force at the position of the point charge. Contrast this to the case of a massive body in a gravitational field. In this case, every little piece of mass in the body feels a gravitational force. $\endgroup$
    – Brian Moths
    Commented Feb 21, 2015 at 11:40
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs: I still don't see how this scenario (single charged particle in a rigid body) causes a dimensional reduction of precisely 1 in the 3D force-torque, i.e. how the force-torque is 5-dimensional in this case. $\endgroup$
    – got trolled too much this week
    Commented Feb 21, 2015 at 11:58
  • $\begingroup$ Both wrenches and twists span 3 DOF space in 2D and 6 DOF space in 3D. These are 2 force components and one torque component, or two velocity components and one rotational speed component. Both are elements of the projective geometry with the line at infinite being a pure torque and the points at infinity being unit translations. $\endgroup$
    – John Alexiou
    Commented Nov 29, 2016 at 18:40

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Three parameters are needed for 2D force (as opposed to 6 for 3D, see https://math.stackexchange.com/a/1157906/3301).

Composition

A force with magnitude $F$ along a direction $\vec{e}=(e_x,e_y)$ going through a point $\vec{r} = (r_x,r_y)$ is described by the three parameters

$$ f =(a,b,c)= ( F e_x , F e_y , F (e_y r_x - e_x r_y) ) $$

Decomposition

Given a force $f=(a,b,c)$ find the parameters

  • Magnitude $$F = \sqrt{a^2+b^2}$$
  • Direction $$\vec{e} = (\frac{a}{F}, \frac{b}{F})$$
  • Distance of line from Origin $$d = \frac{c}{F}$$
  • Position $$ \vec{r} = (\frac{b d}{F}, -\frac{a d}{F}) = (d e_y , -d e_x)$$

Together the above make the force $$f=(F_x,F_y,d F)$$ which contains the equipollent torque of the force at a distance in the last parameter.

Note that if a moving planar rigid body has 3 motion components (aka twist) $v=(v_x,v_y,\omega)$ at the origin, then a force with components $f=(F_x,F_y,d F)$ is applied then the power produced/required is $$P = f \cdot v $$ where $\cdot$ is the inner product.

Summary

Forces, momenta and motions in 3D are all screws with 6 components. Their planar projections (planar screws) have 3 components. These force a set of homogeneous coordinates on the plane where motions are points and forces/momenta are lines. The points represents the instant center of rotation, and the lines the line of action. When these form a pole-polar pair the line of action is called the axis of percussion.

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  • $\begingroup$ Well, this is a decent proof that, in 2D, 3 parameters suffice to reconstruct the force and its application point, but it's not a proof that you can't get away with just two parameters (somehow). Also, there's nothing in this proof about why you need 6 parameters (not 5) to describe the force-torque in 3D. I play a nitpicking undergrad here... as you probably noticed; this after I tried to write a similar proof in my comments on math.SE, but realized how many gaps such an elementary proof has (to deal with). $\endgroup$
    – got trolled too much this week
    Commented Feb 21, 2015 at 12:13
  • $\begingroup$ A force located in 2D space is characterized by the 2D vector (direction+magnitude) and the minimum distance of the line of action to the origin (equipollent torque). Those are the 3 parameters, and they are identical to the 3 parameters needed to describe a line in 2D with homogeneous coordinates. These parameters are not unique though, as an alternate set exists, where line of action normal vector is used instead of along, the same way planes are represented in 2D with homogeneous coordinates. $\endgroup$
    – John Alexiou
    Commented Feb 21, 2015 at 14:38

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