Forces as vectors in Newtonian mechanics

Question

I seem to be confused about the nature of forces as vectors, in the basic Newtonian mechanics framework.

I know what a vector is as a mathematical object, an element of $R^3$. I understand that if a vector is drawn in a usual physical way as an arrow in space, it can be seen as a mathematical vector by translating it to begin at 0 and seeing where the arrow tip ends up. Generally it seems the word "vector" is used in such a way that a vector remains the same vector if it's translated arbitrarily in space (always corresponding to the same mathematical vector).

But now let's say I have a solid object, maybe a metal cube, with some forces acting on it: I push at it with a stick in the center of one facet, it's held by a rope in a different corner, etc. To specify each force that is acting on the cube it doesn't seem enough to specify the vector: I also need to specify the place of application. The cube behaves differently if I push it in the center as opposed in the corner etc.

I'm reading through J.P.Den Hartog's Mechanics that teaches me how to find the resultant force on the cube. I need to sum forces one by one using the parallelogram law, but I should always be careful to slide each force along its line of application, until two forces meet. I could just translate them all to start at the same point and add, but then I won't find the right resultant force, only its direction and magnitude; I will still need to find its line of application (maybe using moments etc.)

So let's say I'm calculating the resultant force "the right way": by sliding arrows along their lines until tails meet, adding, repeating. What am I doing mathematically? (it's not vector addition, that would correspond to just translating them all to 0 and adding) What mathematical objects am I working with? They seem to be specified with 4 free parameters: 3 for direction/magnitude of the vector and 1 more to displace it to the correct line of application; the location at the line of application seems irrelevant according to the laws of statics.

Answer 1

In 3 dimensions, sliding the forces on the line of application until the forces meet does not always work because the two lines of application may be skew.

To capture the line of application it is good to work with pairs of forces and torques. (You can transform the line of application of a force by adding a torque acting on the rigid body. This way you can find a representant whose line of application goes through the origin.) There is an algebra of line vectors that works for such pairs of forces and torques.

You find a description of this stuff in:

R. Featherstone: The calculation of Robot Dynamics Using Articulated-Body Inertias. The international Journal of Robotics Research, Vol. 2, No. 1, Spring 1983.

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In the comments below I mentioned that the generalized forces (force and torque) on the rigid body are a consequence of the rigid body constraints and that line forces are useful but not required for Newtonian mechanics. In the following I sketch how the rigid body constraints lead to the generalized forces. (Note again, that this is just a simplified sketch.)

Rigid body constraints can be stated as the requirement that the placement of the rigid body into (Euclidian) space is an orientation-preserving isometry: \begin{align} \def\l{\left}\def\r{\right}\def\rmL{{\rm L}}\def\rmS{{\rm S}}\def\ph{{\varphi}}\def\SO{{\rm SO}} \def\nR{{\mathbb{R}}} r^\rmL\mapsto r(r^\rmL) = r^\rmS + R\cdot r^\rmL \end{align} where $r^\rmL$ are local coordinates, $r^\rmS$ is a shift vector (3d) and $R$ is a rotation matrix (3x3, orthogonal). The corresponding virtual displacement is constrained by \begin{align} \delta r &= \delta r^\rmS + (\delta R)\cdot r^\rmL\\ &=\delta r^\rmS + \delta\ph \times (R\cdot r^\rmL) \end{align} where $\delta r^\rmS$ is the virtual displacement of the rigid body and $\delta\ph$ is its virtual angular displacement. Now, suppose there is a force (volume) density $f$ applied to the rigid body. The generalized forces $F$ and $T$ for the generalized coordinates $r^\rmS$ and $R$ (with the dimension of the manifold $\SO(\nR^3)$ equal to three) result from the equation \begin{align} \int_{r\in r^\rmS+R\cdot (B)} \delta r \cdot f d V &= \int_{r^\rmL\in B} \l(\delta r^\rmS\r) \cdot f dV + \int_{r^\rmL\in B} \l(\tilde\ph \times (R\cdot r^\rmL)\r)\cdot f dV\\ &= \l(\delta r^\rmS\r) \cdot \int_{r^\rmL\in B} fdV + \delta\ph \cdot \int_{r^\rmL\in B} (R\cdot r^\rmL)\times f dV \end{align} The coefficient \begin{align} F := \int_{r^\rmL\in B} fdV \end{align} of the virtual displacement $\delta r^\rmS$ is the overall force applied to the rigid body and the coefficent \begin{align} T := \int_{r^\rmL\in B} (R\cdot r^\rmL)\times f dV \end{align} of the virtual angular displacement $\delta\ph$ is the overall torque acting on the rigid body.

Answer 2

When you translate-and-add, you do get the correct vector. The point of application is something else, not part of the definition of the vector. Additional information that must be supplied. The addition you refer to in your last paragraph is not "something else". It is vector addition.

I'm surprised that more people don't have the confusion that you have.

Related: There's nothing in the mathematical definition of a vector that allow it to be moved. That's a slight-of-hand that physicists pull to help simplify the analysis. It is more mathematically sound to consider each point in space to be the origin of its own vector space, but that adds way too much extra analysis and obscures the physics. But, the trick does work very well in Euclidean space. I think perhaps you are sensing that something is not quite right in the way the vectors are used in physics. Modeling real space as a vector space works, but it has drawbacks. For example, it assigns special status to the origin whereas there is no physical special status to any point in real space.

Answer 3

I think what you are arriving at, will be rotational mechanics.

As long as objects just translate, you can just sum up the forced, and calculate the results for on the object, as you understand.

When an object is free to rotate, the story becomes a lot more complex and the point of exertion of a force becomes a lot more important. You can easily test that with the closest. Try to close it by pushing near the hinge, and then further away.

I can type a whole story here that explains these concepts, but Wikipedia has some interesting articles for you

• Torque - also moment of force. Takes into account this distance.
• Angular momentum - like linear momentum, but then for rotation.

I hope this gives you some answers on your questions.

Answer 4

tl;dr: you can't superimpose forces when they're acting on different things.

An object is not the same as its center of mass. Under certain conditions (rigid body, force directly in line with the center of mass, etc.) you can assume the object is a point particle at the location of its center of mass and thus you can add up the force vectors because they're acting on the same particle. But when that's not the case, you can't just add up the forces -- they're acting on different particles. That can result in torque, elasticity effects, etc. which you now have to take that into account because the system can no longer be reduced to a simple point particle.

Answer 5

Forces are not vectors. Forces are 3D lines, with a magnitude and a pitch.

• A force acts through a line in space. From geometry you need 4 quantities to fully define a 3D line. This is why you can slide a force vector along its direction line.
• A magnitude described how much force along this line and you need 1 quantity to describe this (duh!) for a total of 5 quantities up to this point.
• For a pure force, 5 quantities is sufficient, but it is possible to have a parallel torque to the line of action. The ratio of torque magnitude to force magnitude is called pitch and it describes a 3D screw in space (see aforementioned Roy Featherstone book). So a force screw (also called a wrench) needs 6 parameters to be fully defined.

These usually are specified as the force and torque vectors at some point A in space. Here is how you can take 3 quantities of a force $\vec{F}$ and the 3 quantities of a torque at $\vec{\tau}_A$ and deduce the screw properties of this equipollent system.

1. Magnitude (scalar) $$F = |\vec{F}|$$
2. Direction (vector) $$\vec{e} = \frac{ \vec{F}}{F}$$
3. Location on line closest to point A (vector) $$ \vec{r} = \vec{r}_A + \frac{ \vec{F} \times \vec{\tau}_A }{ F^2 }$$
4. Pitch (scalar) $$ h = \frac{\vec{F}\cdot\vec{\tau}_A}{F^2}$$

See this answer for more details.