Mathematical connection between systems of particles and rigid bodies

Question

This question goes right back to the first principles of mechanics, so to speak. I'm confused as to the nature of forces applied to rigid bodies. Mathematically, a force acts on a particle of a known mass, but I don't see the mathematical connection that allows a rigid body of the same mass to behave similarly under the influence of the same force.

I understand that a rigid body can be obtained by extending the idea of a system with only a finite number of particles to a system with its number of particles tending towards infinity. Thus each particle in a rigid body will have a differential mass.

Therefore, what I don't understand is how a force applied at a particular location on a rigid body doesn't lead to an infinite acceleration due to the infinitesimal mass of the point at which the force is applied. Obviously this is not the case in real life, however this is what occurs when I try to take the definitions that surround systems of particles and try to build up rigid body mechanics around it.

So, does a force applied to a rigid body simply imply that this force is spread over the body? This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point), right? What am I missing?

Answer 1

Suppose you have body with mass M experiencing a force F. If you model the force as being applied to the whole body, then this gives an acceleration F/M. Now suppose you cut this body into n equal parts. The part that the force is directly applied to will experience force F over mass M/n. So if you model the force as being applied just to one part, that appears to give an acceleration of nF/M. Contradiction? No. We're modeling this as a rigid body. So if one part accelerates, it must take the rest of the body with it. The rest of the body has mass M(n-1)/n. If this mass is accelerating at F/M, then it is experiencing a force of (F/M)(M(n-1)/n) = F(n-1)/n. So if we're modeling the force as applying only to one part, the rest of the body isn't being accelerated by this force, at least not directly. Instead, it must be being accelerated by the part that has the force applied to it. So that part is applying a force of F(n-1)/n to the rest of the body, and by Newton's Third Law, the rest of the body is applying a force of F(n-1)/n to that part. So the total resultant force for the part with an external force is the external force F minus the force F(n-1)/n from the rest of the body, which gives F/n. A force of F/n applied to a mass of M/n gives an acceleration of F/M.

If you want, you can break this analysis down even further. Suppose you have a object of n sections, and you apply a force to a section. Then that section will experience the exterior force plus forces from the adjacent sections. Those sections will experience a force from that original section, plus forces their adjacent sections. And so on. You can then set up a system of equations involving all these forces, relate the acceleration of each part to the net force it experiences, apply the constraint that the acceleration on each part is equal to the acceleration of each other part, and solve that system of equations. Intuitively, we should find that the acceleration for each part is F/M, regardless of how we divide the body into "parts".

Or we can simply recognize that we are dealing with a model to begin with, and different aspects of the model apply to different situations. We can treat the body as being made up of differential masses for such purposes as calculating moments of inertia, but a single mass for calculating acceleration from force. We can treat the body as being a point mass for purposes of calculating what gravitational force it creates, and yet treat forces as being displaced from the center of mass for purposes of calculating torque.

This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point)

The definition of force is the same. What model we use depends on the context. The term "rigid body" does not refer to any actual physical object. Forces are transmitted through an object at the speed of sound in that object, which cannot be higher than c. No object is truly "rigid". When we talk about a "rigid object", that is simply shorthand for "an object that, in the scenario currently under consideration, deviates so little from the theoretical ideal of a rigid body that it can be modeled as a rigid body to the degree of accuracy required". This is similar to how we talk about applying the ideal gas law to gases, even though there is no such thing as an ideal gas in the real world. So when we model something as a rigid body, we treat all forces applied to it as being applied to its whole mass. When we model something as a system of particles, we model each force as being applied to a specific particle. The actual physical laws are the same in both cases, we're simply applying different models.

Note that treating a force as being applied to a particular point is itself a simplification used in particular models. A full quantum mechanical analysis would have the particle not localized to a point, so the force would not be localized to a point. In fact, in quantum mechanics, forces are themselves carried by particles, and those particles have some positional spread.

Answer 2

You are right to some degree. Two clarifications though, there is no such thing as a rigid object, nor is there a real zero area point to carry the force.

In real life, a force applied to an object at a certain point means applied through a sharp enough object such as a tooltip, small sharp hard object which has a surface area, start to accelerate and move that molecules at that point, however, due to the existing bond between the molecules of the object that point will immediately radiate the stress and start a dynamic shock wave which will distribute the stress throughout the part. Ther are many ways of measuring the propagation of stress such as imaging or dying the surface with special paints. One of the causes of fatigue in metals is this.

In solids: A force applied to a point in a solid object if strong enough and the impact is fast enough so that the stress cannot be spread fast will, compress a small part of the impacted area and push it into an indentation hole similar to a bullet hole. Or the stress wave will propagate through the part and may crash it along the way. Modern car frames take advantage of this as crumple zones.

If the part is brittle like rock or glass it will break. These are addressed in studies on the continuum medium.

In liquids: The same force applied faster than the viscosity and elasticity of the liquid can accommodate it will cause turbulence and create a shock wave opening into concentric circles or shooting out some of the liquid as a projectile.

In gases: A sharp point load such as the tip of a wing will create vortices or shock waves that can be detrimental to the wing or propeller.

Answer 3

The answer for most commonly used definition of rigid bodies is quite banal: It does not include any such thing. Instead an assumption (simplification) is made that the body is fully rigid, as in no part can move in relation to other parts. Hence the name rigid body.

From this the investigation goes further asking how such a element would be minimally defined. This eventually leads to a definition where the entire body is reduced to 6 degrees of freedom (3 in 2D), and they come up with moment of inertia to reduce the rotational energy to a higher order particle etc etc. So rigid bodies work like particles with different formulas governing their motion. Mainly handling rotation is added.