How Is Newton's F=ma Formula Derived and Understood in Physics?

[DrStupid]

If you merely wanted to remind us of the fact that F = ma turned out to be inaccurate for high speed electrons, what does that have to do with the topic here?

It was a reply to Bhargava2011 (see above for details).

 

[harrylin]

It was a reply to Bhargava2011 (see above for details).

Ok. What Bhargava2011 apparently meant was that we may freely define the relationship as any function that we wish - which is rather different from our opinion.

 

[Bhargava2011]

I'm afraid that you still didn't get my proof; let me elaborate. Your "gorce" may be, for example:
F = (m*a)²
Two masses next to each other must have the same relationship; the total "gorce" of the two springs is thus F+F=2F. Now put a thin piece of wood (with negligible mass) on top of both springs, so that they act together on a total mass of 2m. Now the gorce of these springs must be 4F according to the formula. It must be a magic piece of wood to double the gorce.

Let all the vectors be represented with capital letters and their magnitude by lowercase letters.
Let the "gorce" be G
and let G = [(m*A)^2]*[(m*A)/(m*a)] (the latter term is the unit vector in the direction of A)
Now, to find the net gorce of two gorces G#1 and G#2 we have to use a vector operation, let it be called "vector addition X" with symbol $
So, for two vectors A and B
let K = {A/[(a)^(1/2)] + B/[(b)^(1/2)]}
then A$B = (K^2)*(K/k)

So, to find the net gorce of two gorces G#1 and G#2, a "vector addition X" has to be performed on the two gorces and we'll get the net gorce :)

PS: I don't know how to properly write the symbols here so, I've used notations which looks a bit confusing.

 

[Bhargava2011]

It was a reply to Bhargava2011 (see above for details).

No, I'm not at all going outside of simple Classical Mechanics. I'm just trying to explain the original poster about the derivation of F=m*A. When going by Newton's second law we'll get a relationship between F and m*A of the form F=K*m*A (where k is a constant). Take any value of K and define force :) but, for simplicity k is taken to be 1. As, I said the simplest bijective relation is f(x)=x.

 

[epenguin]

Not read all thread so this may have been said. The point is often fudged in teaching.

The point stems from Newton's realisation that, contrary to the naive idea that things left to themselves would stay still, and that if they are moving you have to invoke a cause of the movement, things left to themselves will just keep moving at constant velocity. (examples of this are pretty hard to find, and it is a pity courses never mention them, even if they did come along long after Newton).

So after that you say that for a change of this constant rectilinear velocity you have to seek a cause. And we do find and identify them. We call them a force. A force causes an acceleration.

Which is one term in F = ma. You can measure a, but you cannot in the first place measure F and m independently. So this is not an empirical law, more of a definition. In fact I think it would have little relation with experiment or observation at all if there were not in Nature some masses that are near enough constant. The real experimental observation is these constant masses. So you can get standards and comparisons of masses from interactions between them. For instance you can make sense of elastic collisions (note definition creeping in again, risk of circularity, but there are collisions and other interactions of that kind). Fortunately we can make sense of them without knowledge of forces. So from things like that we can I think put masses on an empirical basis without the circularity which I think is what rightly bothers you.

A rather incomplete answer but I hope OK. Oh by the way, I don't think the alternative mathematical formulations of Newton's laws that have been mentioned change any of this. If there is any problem with F = ma, the others have the same problem. Nor would I admit that any formulation is epistemologically better or more basic than any other. If all can be derived from each other, then all have equal status, at least epistemologically.

 

[DrStupid]

No, I'm not at all going outside of simple Classical Mechanics.

That depends on the type of systems your statement ("its not at all necessary to have a linear relation between m*a and force") was related to. For open systems you are right but for closed systems you already are outside classical mechanics.

When going by Newton's second law we'll get a relationship between F and m*A of the form F=K*m*A (where k is a constant).

For closed systems this is basically correct but in addition to Newtons second law you also need the third law, definition II, Galilei transformation, isotropy and maybe even more.

 

[harrylin]

Not read all thread so this may have been said. The point is often fudged in teaching.
[..] You can measure a, but you cannot in the first place measure F and m independently. So this is not an empirical law, more of a definition. [..]

As mentioned before: The equality sign F=ma is a later simplification (k=1) that came with the introduction of standard unit systems; it is due to a free choice of units.

For deriving F~ma Newton could compare identical "impressed forces" by observing an equal amount of impression of a spring, and he could also compare identical "masses" by means of equilibrium of a balance. Neither Newton nor Hooke were completely free to define "Force" as they liked.