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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 F = ma ?
Any help is importante for me.
Thanks
Hi,
Anyone knows the derivation for Newton´s formula F = ma ?
Any help is important for me.
Thanks
Unfortunately, I didn't like it, but I am still replying anyway.If u liked it then please reply.
If F_{1} 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"?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.
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 = ma^{2} - this will still give a different a when F is varied.So we can say that force is proportional to acceleration.
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).
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.
Nabeshin's answer is the most correct and informative.
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.
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.
[..] 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.
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 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.
But, its not at all necessary to have a linear relation between m*a and force.
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.
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]
(however, relativity is off-topic for this 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.
'the inertial mass of the electrons increases at high velocities' is a relativistic effect
such falsifications of classical mechanics and the introduction of SR equations are definitely off-topic for this forum
But, its not at all necessary to have a linear relation between m*a and force.
[..] 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 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.
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).
Let all the vectors be represented with capital letters and their magnitude by lowercase letters.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.
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.It was a reply to Bhargava2011 (see above for details).