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

[live4physics]

Hi,

Anyone knows the derivation for Newton´s formula F = ma ?

Any help is importante for me.

Thanks

 

[gutti]

One Newton is the force required to accelerate a one kilogram mass at a rate of one meter per second squared.

The formula works when force is measured in Newtons because of what the Newton is defined as.

 

[Nabeshin]

Mmm... Good question.

At the first level, there is no derivation of F=ma. In a sense, F=ma is a starting point. You assume F=ma to be the case (with experimental evidence leading you to this conclusion) and work out mechanics from there. This is the way it is taught in terms of a first year mechanics class.

At a second level, you can say F=ma is derived from the extremization of a certain action, defined to be the integral of a lagrangian with respect to time. But at some level you bump up against something fundamental, which is that the lagrangian be written as the difference of kinetic and potential energies. If none of this makes sense to you, the first answer is the one you want. If it does make sense, you can easily work out F=ma.

 

[Eric McClean]

It is Newton’s law where force (F) =mass (m)*acceleration (a). It is also called Newton's laws of motion.

 

[mashrukh]

Hi,

Anyone knows the derivation for Newton´s formula F = ma ?

Any help is important for me.

Thanks

I can do that.

See, if a force of f1 acts on a body of mass m then the body starts accelerating and when the mass of the body is increased then the force of f1 is not enough to accelerate the body.

So we can say that force is proportional to mass.

Then if a force of f1 is acted on the same body then then its acceleration is 'a' then when the force of f2 is applied on the body then the body's acceleration 'a2'.

So we can say that force is proportional to acceleration.

So we can say that F = ma.

If u liked it then please reply.

 

[Fightfish]

If u liked it then please reply.

Unfortunately, I didn't like it, but I am still replying anyway.

See, if a force of f1 acts on a body of mass m then the body starts accelerating and when the mass of the body is increased then the force of f1 is not enough to accelerate the body.

If F₁ is the only force acting on the body, and it is non-zero, then the body will experience acceleration no matter how large its mass is. How can the force "not be enough to accelerate the body"?

So we can say that force is proportional to mass.

So we can say that force is proportional to acceleration.

Dependent on and proportional to are very different things. In the rudimentary "examples" that you gave, without experimental values, we can only conclude that a depends on F - our law of motion could very well be F = ma² - this will still give a different a when F is varied.

Furthermore, your "derivation-of-sorts" appears to ignore the possibility of other variables besides m and a coming into play.

 

[harrylin]

Hi,

Anyone knows the derivation for Newton´s formula F = ma ?

Any help is importante for me.

Thanks

His axiom was:

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

This should be taken together with the definition:

"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly."

In formula: F = dp/dt = d(mv)/dt

Assuming that m = constant, we then obtain F = m a.

An axiom is a "self-evident truth". In this case, probably it was based on experience (experiments).

 

[Claude Bile]

Newtons formula is constructed rather than derived.

Newton defined a quantity, Force, which caused objects to accelerate, with the Force being proportional to the acceleration experienced by said object.

So Newton defined Force to be proportional to acceleration.

To turn a proportionality into equality, you need to introduce a constant. This "Constant of proportionality" Newton called mass.

Claude.

 

[chrisbaird]

Nabeshin's answer is the most correct and informative.

 

[jishitha]

From second law of motion, dP/dt propotional to F. i.e,m dV/dt propotional to F. To make this an equation,multiply with a constant. i.e,F=k m dV/dt. Where , k=1, then F=m dV/dt.
We know that, dV/dt=a. There fore, F=ma.

 

[mashrukh]

I think I have got the derivation of f = ma

According to Newton's second law we know that mv-mu/t is proportional to force applied.

so we can write that m(v-u)/t = f

but (v-u)/t = a

We can write that ma = f i.e f = ma

If i am wrong can anyone correct me.Please!

 

[jimgram]

1. Everything has a fixed position and resists a change in that position.

2. A unit of resistance to change in position = mass

3. A change in position can only occur when two things collide. Let’s call thing #1 ‘mass1’ and thing #2 ‘mass2’.

4. When the datum of mass1 is ascribed to be datum of a second mass2, then the relative position between the two things is described as motion (change between the two things might equal zero, and therefore the distance between them will not change).

5. The magnitude of motion is velocity and the change in motion is acceleration.

6. Acceleration is a mathematical relationship that describes the effect of collision, which is the only means that any mass can alter its position, and only by equally and oppositely altering the position of the mass it collides with.

7. If: position of mass1 minus position of mass2 = s (distance), then velocity is s as a function of time: v = ds/dt.

8. Acceleration is v as a function of time: a = dv/dt.

9. So, the position of mass1 minus the position of mass2 = a * t2.

10. Mankind has developed a concept of ‘force’ to deal with the observation of collisions. Since a thing has mass (resistance to change in position) and only a collision with another mass can overcome this resistance and can cause a change, what is really meant by the term ‘force’ is mass1 in a collision.

11. The motion equations then lead to mass1/mass2 = a or v/t.

12. Substituting the term ‘force’ for mass1, we arrive at a = F/mass2, or F = m*a.

 

[harrylin]

His axiom was:

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

This should be taken together with the definition:

"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly."

In formula: F = dp/dt = d(mv)/dt

Assuming that m = constant, we then obtain F = m a.

An axiom is a "self-evident truth". In this case, probably it was based on experience (experiments).

PS Little correction (as already indicated by jigarbageha): I should have written F ~ dp/dt = d(mv)/dt = m a.

Contrary to what some people think, m in classical mechanics is not a proportionality constant but it stands for the amount of matter. In the SI, the units are conveniently chosen such that the proportionality constant is equal to 1.

Note also that Newton did not define force to be proportional to acceleration: instead it is an easily falsifiable axiom (and as a matter of fact, it has been falsified with SR).

 

[Bhargava2011]

Suppose you take a spring which has an extension say l, then the spring would pull any object attached to it with a constant force say F. When you'll perform this experiment for different objects of different mass, you'll notice that the product m*a is always constant. And there you have it, what is important is m*a.
Now as far definition of force is concerned, remember its a definition and it could be any Bijective function of m*a.
eg. you define Force to be g(m*a) = F. where F is the force and g is a bijective function. you can take any bijective function g and you'll able to describe the universe as beautifully as with F=m*a. All the laws involving "Force" would work as fine as with Newton's definition.
Now the simplest form g can take is g(x)=x i.e. F=ma.
And that is precisely Newton's definition of force.
Thanks.

 

[harrylin]

Suppose you take a spring which has an extension say l, then the spring would pull any object attached to it with a constant force say F. When you'll perform this experiment for different objects of different mass, you'll notice that the product m*a is always constant. And there you have it, what is important is m*a.

Quite so - perhaps an elaboration is useful. I'd do this as follows: test for different masses what acceleration is required to achieve the same extension and thus, by common definition, the same "power to influence" (force). You will find that to very good approximation, F ~ m a. This may also reveal Hooke's law, but it doesn't depend on it.

Now as far definition of force is concerned, remember its a definition and it could be any Bijective function of m*a.
eg. you define Force to be g(m*a) = F. where F is the force and g is a bijective function. you can take any bijective function g and you'll able to describe the universe as beautifully as with F=m*a. All the laws involving "Force" would work as fine as with Newton's definition.
Now the simplest form g can take is g(x)=x i.e. F=ma.
And that is precisely Newton's definition of force.
Thanks.

Not really: a variation of the test above with for example two springs in parallel reveals that F ~ d(mv)/dt. Thus when Newton called this an axiom or a law, he was more precise than when he included part of it in his definitions: he was not free to define it according to his wishes.Harald

 

[BruceW]

Nabeshin's answer is the most correct and informative.

Agreed. The extremisation of the action is more fundamental than Newton's laws, and they lead to Newton's laws. And I also agree that you still bump up against something fundamental, even using the Lagrangian formulation.

This 'bump' is that the form of the action, and why it should be extremised is not self-evident (as far as I'm aware).
In other words, there are some physical laws that we have always found to hold true. And for this reason, we use them to make predictions.
In fact, inside a black hole, the laws of general relativity break down, so I would assume that inside a black hole, Newton's laws wouldn't hold either. But that's a whole other topic.

 

[Bhargava2011]

Quite so - perhaps an elaboration is useful. I'd do this as follows: test for different masses what acceleration is required to achieve the same extension and thus, by common definition, the same "power to influence" (force). You will find that to very good approximation, F ~ m a. This may also reveal Hooke's law, but it doesn't depend on it.

What is required is the understanding of the "Force". In the scope of classical mechanics, the law F=ma is absolutely precise and I think not an approximation. Instead of trying to find out what acceleration is required for a particular extension, why don't we have an extension of the spring i.e. everything in the experiment is kept as it is but the mass. This is more easy to perform.
So, the experiment is like this, a spring with a particular extension and a mass. Every part of the experiment is kept constant except the mass and what we observe is, m*a is constant.
Now, this is just an observation about our experiment and I haven't used the word force yet.
So, one thing we observe about our experimental setup is this, irrespective of mass the product m*a is always constant for a given extension if the spring. And now its up to us what we have to do with this m*a. Define it as a "force" or "gorce" or whatever, its just for our convenience.

Not really: a variation of the test above with for example two springs in parallel reveals that F ~ d(mv)/dt. Thus when Newton called this an axiom or a law, he was more precise than when he included part of it in his definitions: he was not free to define it according to his wishes.

I didn't get your example. and the law F ~ d(mv)/dt is same as f=ma. Definition is up to us. We can define anything according to our convenience in describing the nature. That's my understanding of a "definition" whether in physics or in maths. However its better to have logical and useful definitions.

 

[harrylin]

[..] its just for our convenience [...]
I didn't get your example. and the law F ~ d(mv)/dt is same as f=ma. Definition is up to us. We can define anything according to our convenience in describing the nature. That's my understanding of a "definition" whether in physics or in maths. However its better to have logical and useful definitions.

It's not just for our convenience: Newton did not introduce a new concept such as "blurp", instead, he related to already existing concepts such as "force". Our definitions must be consistent too: by common definition, two identical springs push with twice the force as one spring. That double force does not match an arbitrary function of m*a; instead it matches to very good approximation F ~ m a, as can easily be verified with double mass (same acceleration) and single mass (double acceleration). Defining it differently leads to inconsistency (just imagine two springs next to each other, either connected together or not). Note also that ~ is not identical to = : a law may be about proportionality while definitions are about identities.

 

[Bhargava2011]

It's not just for our convenience: Newton did not introduce a new concept such as "blurp", instead, he related to already existing concepts such as "force". Our definitions must be consistent too: by common definition, two identical springs push with twice the force as one spring. That double force does not match an arbitrary function of m*a; instead it matches to very good approximation F ~ m a, as can easily be verified with double mass (same acceleration) and single mass (double acceleration). Defining it differently leads to inconsistency (just imagine two springs next to each other, either connected together or not). Note also that ~ is not identical to = : a law may be about proportionality while definitions are about identities.

I'm getting your idea. Its just our understanding of "Force", a push or pull. But, the universe has got not nothing to do with our understanding. So, by common sense yes, if you double the force m*a gets doubled. But, its not at all necessary to have a linear relation between m*a and force. I think one can very well understand the universe with a concept called "gorce" which doesn't have a linear relation with m*a. But, yes of course it would get very complicated without doing any good. So, better to stick with the simple way F=m*a.

 

[DrStupid]

But, its not at all necessary to have a linear relation between m*a and force.

Indeed, in SRT the relation between Force and m*a is

{\rm F = }\left( {{\rm I + }\frac{{{\rm v} \cdot {\rm v}^{\rm T} }}{{{\rm c}^{\rm 2} - v^2 }}} \right) \cdot m \cdot a<br />

with I = Identity matrix and

m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

 

[harrylin]

I'm getting your idea. Its just our understanding of "Force", a push or pull. But, the universe has got not nothing to do with our understanding. So, by common sense yes, if you double the force m*a gets doubled. But, its not at all necessary to have a linear relation between m*a and force. I think one can very well understand the universe with a concept called "gorce" which doesn't have a linear relation with m*a. But, yes of course it would get very complicated without doing any good. So, better to stick with the simple way F=m*a.

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.

 

[harrylin]

Indeed, in SRT the relation between Force and m*a is

{\rm F = }\left( {{\rm I + }\frac{{{\rm v} \cdot {\rm v}^{\rm T} }}{{{\rm c}^{\rm 2} - v^2 }}} \right) \cdot m \cdot a<br />

with I = Identity matrix and

m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

Indeed, that's the reason why I wrote "to very good approximation".
Note that with that (non-classical) mass convention, Newton's second law,
F ~ dp/dt
is conserved in SR (however, relativity is off-topic for this forum).

 

[DrStupid]

(however, relativity is off-topic for this forum).

You are right, it is. But nature doesn't care about our theories. If you check F=m*a experimentally you will see that it is wrong - even if you do not know anything about SR. In fact there were such experiments years before Einstein published SR. For example Walter Kaufmann showed in his paper "Die magnetische und electrische Ablenkbarkeit der Bequerelstrahlen und die scheinbare Masse der Elektronen" [Göttinger Nachrichten, 1901, (2): 143–168] that the inertial mass of electrons increases at high velocities. But he wasn't aware that this is a falsification of F=m*a because after centuries of Galilei transformation nobody noticed that this formula results from Newtons F=dp/dt only for dm/dt=0. And this is definitely on-topic for this forum.

 

[BruceW]

'the inertial mass of the electrons increases at high velocities' is a relativistic effect, and as such is off-topic.

 

[harrylin]

Y [..] Walter Kaufmann showed in his paper "Die magnetische und electrische Ablenkbarkeit der Bequerelstrahlen und die scheinbare Masse der Elektronen" [Göttinger Nachrichten, 1901, (2): 143–168] that the inertial mass of electrons increases at high velocities. But he wasn't aware that this is a falsification of F=m*a because after centuries of Galilei transformation nobody noticed that this formula results from Newtons F=dp/dt only for dm/dt=0. And this is definitely on-topic for this forum.

That was one of the fundamental experiments that led to SR; IMHO such falsifications of classical mechanics and the introduction of SR equations are definitely off-topic for this forum, and on-topic for the relativity forum.

 

[DrStupid]

'the inertial mass of the electrons increases at high velocities' is a relativistic effect

First of all it's an experimental observation.

such falsifications of classical mechanics and the introduction of SR equations are definitely off-topic for this forum

introduction of SR equations: YES

You are right, it would have been better not to post the equations but references to experimental observations only. Such observations are independent from the theory.

falsifications of classical mechanics: NO

Falsification of F=m*a requires the knowledge of its origin and this is not only on-topic for this forum. That's the topic of this thread.

By the way: According to you understanding of the topic of this forum Bhargava2011's statement

But, its not at all necessary to have a linear relation between m*a and force.

would be off-topic too because in classical mechanic this linear relation is necessary. But I wouldn't take such a narrow view of it.

 

[BruceW]

In the classical limit, the laws of Einstein's relativity and the laws of classical mechanics are the same.

And as far as I am aware, the derivation for Einstein's relativity is no more fundamental than that for classical physics (Apart from the fact that classical physics doesn't work outside the limit of classical phenomena).

So there is no reason to bring in relativity to this discussion apart from to say that relativity works when great relative velocities occur, etc.

So introducing relativity doesn't bring anything useful to the discussion of the classical laws in the classical limit.

 

[harrylin]

[..] According to you understanding of the topic of this forum Bhargava2011's statement .. would be off-topic too because in classical mechanic this linear relation is necessary. But I wouldn't take such a narrow view of it.

The discussion of how the linear relationship was deduced for classical mechanics is of course on topic here. The discussion of how the deviation of that linear relationship for relativistic mechanics was deduced is commonly regarded as the topic of SR.

 

[DrStupid]

The discussion of how the deviation of that linear relationship for relativistic mechanics was deduced is commonly regarded as the topic of SR.

That is true, but I didn't do that. I just posted the formula without discussing its origin. As it describes experimental observations it doesn't matter where it comes from. With sufficient experimental data the same formula could have been found within classical mechanics and as long as nobody shows that it is valid for different frames of reference there would be no need for a new transformation.

 

[harrylin]

That is true, but I didn't do that. I just posted the formula without discussing its origin. As it describes experimental observations it doesn't matter where it comes from. With sufficient experimental data the same formula could have been found within classical mechanics and as long as nobody shows that it is valid for different frames of reference there would be no need for a new transformation.

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?

 

[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.