# Finding possible forces that could have caused the motion of a system of particles

1, Assuming we have a system of particles and we know the position of each particle at any time, what are the possible forces acting on each particle at any time? Is there anything "interesting" we can say about this problem, and how would we state this sort of problem more formally, in a rigorous way?

Assuming there is only one particle, or non the particles do not interact in any way and aren't constrained in any way, the solution for a certain particle is to simply take the second derivative and multiply it by the mass of the particle.

2, Even assuming additional constraints on the system (which are satisfied at any time), like some of the particles being connected by massless rods etc, taking the second derivative and multiplying will also yield a solution, right?

But how would I go about proving that? How do I show that in some sense, maybe a force acting on some particle A connected to another particle B by a massless rod won't cause some additional force to act on B? Perhaps I should be more exact about what a the "massless rod"/constraint will do ( it won't do anything as long as the particles are the proper distance apart - a massless rod is really just a tool that doesn't formally make sense). I could also instead think of some band/stick that will contract when stretched, and push out when compressed, so it won't create any force as the system goes through the motions.

Still I don't feel completely convinced that it couldn't happen, that maybe the derived forces might cause some sort of weird movement. Ideally, I'd like a proof that if the derived forces act on the system and we add constraints that are satisfied at any time, then no force will be transferred through a rod in some sense or anything like that. But maybe the analogy with a band is as good of a argument as possible. I'm sorry if this is too vague.

• Hm, your question is quite unclear. You talk about derivatives - could you try and write out the math you're referring to? Are you talking about Newton's second law? It would also help if you clearly separated the questions out into separate points. – Codename 47 Jul 31 '19 at 16:14

1, Assuming we have a system of particles and we know the position of each particle at any time, what are the possible forces acting on each particle at any time? Is there anything "interesting" we can say about this problem, and how would we state this sort of problem more formally, in a rigorous way?

Assuming that you're viewing the system from an inertial (i.e. non-accelerating) reference frame, you can determine the total force on each particle based on its acceleration, using Newton's Second Law $$\vec{F}(t)=m\vec{a}(t)$$. This total force is the sum of all the forces exerted on this particle by every other particle in the system, as well as any forces coming from outside the system, including constraint forces*.

2, Even assuming additional constraints on the system (which are satisfied at any time), like some of the particles being connected by massless rods etc, taking the second derivative and multiplying will also yield a solution, right?

It depends on what you mean by "a solution", as it's not clear what specifically you're trying to solve for. That said, taking the second derivative of the position of each particle and multiplying by the mass will yield the total force on each particle, which includes the constraint forces*. In order to study the dynamics of the system to any greater extent using forces**, you must know what the constraint forces are. Usually, these are not explicitly known, so there's not much more than can be said without more information.

But how would I go about proving that? How do I show that in some sense, maybe a force acting on some particle A connected to another particle B by a massless rod won't cause some additional force to act on B?

If the massless rod is rigid, then there will be a force acting on B. We know this because the center of mass of the system must accelerate, and B must be kept a fixed distance from the center of mass, so B must also accelerate.

I could also instead think of some band/stick that will contract when stretched, and push out when compressed, so it won't create any force as the system goes through the motions.

If there is a finite speed of sound in the elastic rod, then you can exert force on A without exerting force on B, but only for a short time (namely, on the order of $$\frac{L}{v}$$ for a rod with length $$L$$ and sound speed $$v$$). This is because, when you exert force on A, the compression in the rod must be transmitted from one end of the rod to the other, and propagates at the speed of sound; in the meantime, elastic potential energy is stored in the propagating disturbance in the rod. Once the disturbance reaches the end of the rod, B accelerates, and the elastic potential energy is turned into kinetic energy, pulling A along with B***. You can't do this in the steady state, though, because a time-indepenently stretched or compressed rod has to have force exerted from both sides, which means you're exerting force on both A and B.

*In order for a constraint to constrain the motion of a particle, it must exert a force on that particle. The forces which implement the constraint are called constraint forces.

**There is a way to get around not knowing constraint forces: don't use forces to study dynamics. Instead, use Lagrangian mechanics or Hamiltonian mechanics, which are based on energy rather than forces, and deal with constraints without having to actually compute what the corresponding forces are.

***In reality, the elastic potential energy is never perfectly and entirely transmitted to B. Some of it gets turned into kinetic energy, and the rest reflects back down the rod towards A, where A picks up some kinetic energy, and so on, and both masses end up oscillating while the whole system moves with the remaining kinetic energy.

• Thank you for the answer. By solution, I meant a specific combination of force vectors that add up to the movement of the system. In the context of no constraints, that seems fairly simple (basically the sum of the force vectors acting on the particle has to be the total force). In constrained systems, say two particles connected by a rigid massless rod, a solution would be some external forces acting on the system that cause the wanted movement (e.g. a solution for translation of the particles A,B in (B-A) direction could a force acting just on A in (B-A) direction. – John P Aug 5 '19 at 23:58
• I'm interested in figuring out these "solutions" (external forces that cause the movement of the system) of constrained systems, but essentially ignoring the constraint forces. I would expect that if I know the movement of the system, then one trivial solution would an external force acting on each particle determined by $F=ma$ - such external force would create the movement of the system (and the constraint forces here would be zero). But I'm not really sure how to prove this. – John P Aug 6 '19 at 0:05
• I'll study Lagrangian mechanics, but I thought I'd first like to learn how to solve these types of problems without it. Similarly to determining the movement of 2 particles connected by a rod by the acceleration at the center of mass and the net external torque, I'd expect that maybe you can use some reasoning for more complicated systems, or at least figure out a good approximation of the movement. Say, a double pendulum, apply short impact on the lower object - surely we can figure out the acceleration of both the lower and upper object without Lagrangian mechanics? – John P Aug 6 '19 at 0:13

About all the "mass-less rod", there is a way to solve questions with particles that are connected in some kind of way, you need to use torque and angular momentum. They replace force and momentum in the world that things aren't volume-less. To calculate the torque acting on an object, take the cross product between the pointing vector to the force vector If you are looking on a way to prove this thing, it's like newton's laws. There aren't real proofs for these formulas.