Elementary proof of the minimum number or parameters needed to uniquely identify a force-torque (aka wrench) in 2D vs. 3D 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...
 A: 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.
