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Derivation for F = ma

  1. Mar 2, 2010 #1

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

    Any help is importante for me.

  2. jcsd
  3. Mar 2, 2010 #2
    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.
  4. Mar 2, 2010 #3


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    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.
  5. Mar 3, 2010 #4
    It is Newton’s law where force (F) =mass (m)*acceleration (a). It is also called Newton's laws of motion.
  6. Aug 24, 2011 #5
    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.
  7. Aug 24, 2011 #6
    Unfortunately, I didn't like it, but I am still replying anyway.
    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"?
    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.
  8. Aug 24, 2011 #7
    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).
  9. Aug 26, 2011 #8

    Claude Bile

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    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.

  10. Aug 26, 2011 #9
    Nabeshin's answer is the most correct and informative.
  11. Aug 29, 2011 #10
    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.
  12. Sep 1, 2011 #11
    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!
  13. Sep 1, 2011 #12
    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.
  14. Sep 12, 2011 #13
    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).
    Last edited: Sep 12, 2011
  15. Sep 12, 2011 #14
    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.
  16. Sep 14, 2011 #15
    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.
    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.

  17. Sep 14, 2011 #16


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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.

    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.
    Last edited: Sep 14, 2011
  18. Sep 14, 2011 #17
    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.

    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.
  19. Sep 14, 2011 #18
    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.
    Last edited: Sep 14, 2011
  20. Sep 14, 2011 #19
    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.
  21. Sep 14, 2011 #20
    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

    with I = Identity matrix and

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