Suppose we have a supply of devices to exert one unit of force, such as a set of identical springs stretched to a specified length. We can exert two units of force by putting two such springs in parallel. In this way one can create any desired force that is a multiple of the unit force, but that can be worked around by taking the unit force to be small. Suppose we have a mass and we exert several forces on it in different directions. It is not a priori clear to me why it should be the case that the mass will not accelerate iff the vector sum of the forces is zero. It is clear to me for 2 forces, but not for 3 forces, though I can see why it's true for more than 3 forces if it is true for 3 forces. Is this purely an experimental fact, or can this be (partially) explained on a logical basis?

  • $\begingroup$ I suspect you can produce a compelling argument from symmetry: space is translationally-symmetric, therefore by Noether's theorem momentum is conserved. Forces represent the transfer of momentum between parts of a system, and now, mumble, space looks like $\mathbb{R}^3$ so things have to sum as vectors in $\mathbb{R}^3$. Yes this isn't coherent, that's why it's a comment not an answer! I think it could be made coherent. (Of course the real answer is 'because experimentally it is true'.) $\endgroup$ – tfb Sep 6 '16 at 11:21
  • $\begingroup$ Thanks, that makes sense. I guess that leaves the question of why macroscopic springs act as (independent) momentum transfer devices, and the ultimate answer to that will be something about the microscopic behaviour of particles in the springs. In terms of discovery it's putting the cart before the horse, as Newton didn't know all that, so I'm wondering how he figured it out...maybe he actually did play around with springs and simply observed it. $\endgroup$ – Jules Sep 6 '16 at 14:42

I looked at Newton's Principia Mathematica, and it appears that he handled the question of adding oblique forces right after his famous three laws, and he did it by pure reasoning. In Corollary I, he reasons that two forces applied in sequence have the same result as the diagonal of a parallelogram formed by those two forces. Then, in Corollary II, he deduces that the same construction can be used if the forces are applied at the same time. The extension to more than three forces is just a matter of adding one force at a time. If, at the end of all these constructions, you're back where you started, then the result is the same as applying no force. And that is covered by his first law.

Interestingly, the concept of vectors had not yet been invented, which is one of many reasons that it's really hard to read the Principia. But the idea of parallelograms of velocities had been around since Greek times. See Michael J. Crowe, "A History of Vector Analysis: The Evolution of the Idea of a Vectorial System", Courier Corporation, 1967.


It is not a priori clear to me why it should be the case that the mass will not accelerate iff the vector sum of the forces is zero.

Newton's 2nd law: $$\sum \vec F=m\vec a$$

If the vector sum is zero, then what is $\vec a$? Zero as well.

Think of it like this:

  • If two forces pull equally in an object in opposite directions, they cancel out and it doesn't move.
  • If three forces pull, then maybe one pulls left, and two right. If the left one pulls just as much as the sum of the two right ones, then their vector sum is zero. And since the total pull left and the total pull right are equal, the object still doesn't move.

The point is that forces add up, so we can consider all forces in a certain direction as one by summing them. It doesn't matter if one force pulls right or if two forces of half the magnitude pull right - the result is the same, and if that equals the one pulling left, they cancel out.

Now if a force points at an angle instead of straight to the right, then remember the superposition principle: you can break it into components. That force namely pulls a bit up and a also a bit right, so you can replace that one force with two forces in each of these directions. And then you can again do the summing up of force pointing left and right, and also up and down and see, if they cancel out.

Yes, Newton's law is an empirical, experimental law that is never really proven, but simply always showed to be correct (or rather: it has never been disproven with a counterexample).

  • $\begingroup$ I understand the situation where one force pulls left and two pull to the right. My question is about the more general case where all 3 point in a different direction. If it is purely empirical, does that mean that Newton came up with his law by doing experiments? Or did he have a logical reason for it? $\endgroup$ – Jules Sep 6 '16 at 10:14
  • $\begingroup$ Yes, the law is made from making experiments. When all three or more point in different directions, remember the superposition principle. You can split them up in components. So if a force points at an angle, then that means that it pulls a bit up and a bit to the side. So you can draw this force as two other such forces (which are it's components). And then you can still look at how much pulls left and right, and also up and down, and you can see if they cancel out. $\endgroup$ – Steeven Sep 6 '16 at 10:31
  • $\begingroup$ The superposition principle is equivalent to saying that forces add like a vector. I am familiar with the mathematics; what I am interested in is the connection to physical reality. Do you have a reference for Newton's experiments with which he figured out that forces add like vectors? $\endgroup$ – Jules Sep 6 '16 at 14:31
  • $\begingroup$ @Jules, you said: "I understand the situation where one force pulls left and two pull to the right". This is also vector addition. The two pulling right add up to correspond to one bigger force. And this bigger force then balances the leftwards force so they cancel out. Is it the superposition principle itself that is your question? That is, is it the act of breaking a force into it's components, that is your question? $\endgroup$ – Steeven Sep 6 '16 at 14:35
  • $\begingroup$ I am not good at explaining my question. Maybe somebody who can decipher it can explain it better. $\endgroup$ – Jules Sep 6 '16 at 14:37

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