- #1

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Hi,

Anyone knows the derivation for Newton´s formula

Any help is importante for me.

Thanks

Anyone knows the derivation for Newton´s formula

**F = ma**?Any help is importante for me.

Thanks

- Thread starter live4physics
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- #1

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Hi,

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

Any help is importante for me.

Thanks

Anyone knows the derivation for Newton´s formula

Any help is importante for me.

Thanks

- #2

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The formula works when force is measured in newtons because of what the newton is defined as.

- #3

Nabeshin

Science Advisor

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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.

- #4

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- #5

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I can do that.Hi,

Anyone knows the derivation for Newton´s formulaF = ma?

Any help is important for me.

Thanks

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.

- #6

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Unfortunately, I didn't like it, but I am still replying anyway.If u liked it then please reply.

If FSee, 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.

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 = maSo we can say that force is proportional to acceleration.

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

- #7

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His axiom was:Hi,

Anyone knows the derivation for Newton´s formulaF = ma?

Any help is importante for me.

Thanks

This should be taken together with the definition:

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).

- #8

Claude Bile

Science Advisor

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

So Newton

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

Claude.

- #9

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Nabeshin's answer is the most correct and informative.

- #10

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We know that, dV/dt=a. There fore, F=ma.

- #11

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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!

- #12

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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.

- #13

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

Contrary to what some people think, m in classical mechanics is

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).

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- #14

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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.

- #15

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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.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.

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.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.

Best,

Harald

- #16

BruceW

Homework Helper

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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.Nabeshin's answer is the most correct and informative.

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.

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- #17

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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.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.

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.

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.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.

- #18

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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[..] 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.

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- #19

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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.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 doesnotmatch 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.

- #20

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Indeed, in SRT the relation between Force and m*a isBut, its not at all necessary to have a linear relation between m*a and force.

[itex]{\rm F = }\left( {{\rm I + }\frac{{{\rm v} \cdot {\rm v}^{\rm T} }}{{{\rm c}^{\rm 2} - v^2 }}} \right) \cdot m \cdot a

[/itex]

with I = Identity matrix and

[itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex]

- #21

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I'm afraid that you still didn't get my proof; let me elaborate. Your "gorce" may be, for example: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.

F = (m*a)

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.

- #22

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Indeed, that's the reason why I wrote "to very good approximation".Indeed, in SRT the relation between Force and m*a is

[itex]{\rm F = }\left( {{\rm I + }\frac{{{\rm v} \cdot {\rm v}^{\rm T} }}{{{\rm c}^{\rm 2} - v^2 }}} \right) \cdot m \cdot a

[/itex]

with I = Identity matrix and

[itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex]

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).

- #23

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

- #24

BruceW

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- #25

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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.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.

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