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We all know that "Newton's second law" states that any particle or a system of particles with a constant total mass $m$ under influence of $n$ external forces will move with a constant acceleration $\vec{a}$ proportional to the sum of all forces such that we can write: $$\sum_{i=1}^{n} \vec{F_i}=m\vec{a}$$ Now my question is if the individual external forces $\vec{F_1}$, $\vec{F_2}$, $\vec{F_3}$, ..., $\vec{F_n}$ generate the corresponding accelerations $\vec{a_1}$, $\vec{a_2}$, $\vec{a_3}$, ..., $\vec{a_n}$ independently, can we write the equation for an "Individual external force" as follows? $$\vec{F_i}=m\vec{a_i}$$ I mean can we deduce the equation for each individual external force from 2nd law?

(Notice that the total acceleration should be: $\vec{a}=\sum \vec{a_i}$)

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Well, yes of course. But bear in mind that the single $a_i$ are not observable, so is more of a mathematical/logical tool.

Then the equation becomes, in one dimension (the same applies to vector calculus):

$$\sum_i F_i = ma = m\sum_i a_i = m\sum_i {F_i\over m} = \sum_i F_i $$

it is an identity, you can use any of the forms above!

That is the strength of linear formulas: you can solve everything separately and then add everything up (linear superposition principle)

EDIT: Let's put it another way. If each force was acting without the others you would have $F_i=ma_i$ thus ${F_i\over m}=a_i$.

Then now take: $$\sum_i F_i=ma$$ and divide evrything by $m$. You get:

$$\sum_i {F_i\over m} = a$$

Then, using the formula above: $$\sum_i a_i = a$$

This proves that the decomposition in single forces leading to single accelerations make sense. You could have done another decomposition, such as, supposing you only have $F_1$ and $F_2$, $$F_1=m {a\over 3}$$ and $$F_2=m{2a\over 3}$$. Then you would still have

$${F_1\over m}+{F_2\over m}={a\over 3}+{2a\over 3}=a$$

and as long as the two forces are acting together it is a fine decomposition. Yet it has no physical sense, as if the forces were acting singularly the equations above would be wrong.

So to say, since you can decompose a sum of force anyway you want you just choose the only decomposition which would make sense if the forces were acting alone.

This kind of superposition has physical sense only if the single forces are real (e.g. one is pulling and one is pushing). This may also help you to solve problems in which several forces are acting: you can solve each one separately and then sum the results (make sure you sum the vectors in the right way..!)

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  • $\begingroup$ I'm sorry but I think you confused premise and conclusion when you wrote $m\sum_i a_i = m\sum_i {F_i\over m}$. How do we know $a_i=\frac{F_i}{m}$ before deducing it? $\endgroup$ – Hamed Begloo Nov 5 '16 at 12:58
  • $\begingroup$ Not really. It is you that inverted the line of reasoning! That is Newton second law: each force $F_i$ acting on an object of mass $m$ generates and accelaration $a_i$. Then, following my derivation, you find out that you can also consider the total sum of the forces to compute the total acceleration! Or rather, since the equation are linear, you can do whatever you want separately and the sum up. $\endgroup$ – JalfredP Nov 5 '16 at 13:02
  • $\begingroup$ Agreed. If "2nd law" was just $F_i=ma_i$ then we could easily say $\sum \vec{F_i}=ma$. But in most textbooks says that "Newton's second law" just states that the NET external force(sum of all forces) is equal to the mass times net acceleration. Even wikipedia: en.wikipedia.org/wiki/Newton%27s_laws_of_motion $\endgroup$ – Hamed Begloo Nov 5 '16 at 13:09
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    $\begingroup$ Ok, well then you can refer to here. The principle tells you that you decompose things anyway you want. You could also link each force to whatever acceleration provided that the sum still is $a$. You can decompose each force in sub-forces and solve everything separately and then sum everything up. Yet it makes sense to suppose that each force is acting as if it were the only one (then $F_i=ma_i$) and then sum up. Let's put it this way: since you can decompose the sum anyway you want, you simply do it in the most "right" way. $\endgroup$ – JalfredP Nov 5 '16 at 13:13
  • $\begingroup$ That principle is really strange :). I heard some things about linear superposition in "Electromagnetism" but never heard of it in "Classical mechanics". Anyway if you don't mind I prefer to wait for more answers, and then decide to accept which one. Thank you for your help. $\endgroup$ – Hamed Begloo Nov 5 '16 at 13:19
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My answer to this question is no.

Maybe this is no issue mathematically, as another answer points out, but the splitted portions of "acceleration" have no physical meaning. It hurts my eyes to see - it takes us away from the simple and intuitive and physical Newton's law to something completely fictitious. And I don't see any advantage in ease of work or so. This so often causes so much confusion, so let's not go this way.

There is not a small "portion" of acceleration contributing from each. There is a force from each, and the only thing Newton's 2nd law says is that they together result in one acceleration:

$$\sum \vec F=m\vec a$$

That the sum symbol $\sum$ is only on the force vector and not the acceleration vector is no coincidence or lazy writing. It is on purpose, because this formula show how the world works.

An object can experience many forces simultaneously. But it only has one acceleration at a time.

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    $\begingroup$ Sorry but I do not agree. While it is true that the acceleration is one, that acceleration is the vector sum of all contributions. If you have to perpendicular forces, each one will be acting along their axis providing their own acceleration, and then you sum them. It is the same thing as computing the vector sum of the force and then computing the final acceleration, which maybe is the only really physical answer, but it is often simpler to decompose the effects..! It is a mental picture: as long as you can handle it right it is the same thing..! $\endgroup$ – JalfredP Nov 5 '16 at 13:25
  • $\begingroup$ @JalfredP Acceleration is not the vector sum of all contributions, but it can be a vector sum of several components. But such components are also not single accelerations. Just like the gravitational foce can be split into several components, that doesn't mean that several forces are being exerted; it is still only one gravitational force being exerted. Yes, splitting into components is often a very useful tool, but let's not confuse that with what actually exists. There can be many forces on an object, but only one acceleration. $\endgroup$ – Steeven Nov 5 '16 at 13:34
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    $\begingroup$ What? Of course you can split accelerations too. Why would you be able to split forces and not accelerations? Of course what you observe is the net acceleration and the single $a_i$ are not observable, but yet the total $a$ is the vector sum of all $a_i$. It really depends on what picture you choose to see. The single $a_i$ are unphysical, but can provide insight. $\endgroup$ – JalfredP Nov 5 '16 at 13:39
  • $\begingroup$ @JalfredP When an object is at rest on the ground, is it then being accelerated downwards by gravity while also being accelerated upwards by the normal force from the ground...? $\endgroup$ – Steeven Nov 5 '16 at 13:43
  • $\begingroup$ But that is happening actually, or rather, you could see it as that. Now, choosing normal forces is tricky as they "depend" on gravity. But what if you have a spring and gravity: could not you say "the spring is accelerating upwards, gravity downwards, the net effect is...". If you instantaneously removed the spring the acceleration would be downwards... Again, I agree: you do not observe any $a_i$ but mathematically it is the same. Anyways I edited my answer so that is clear that I am answering on a mathematical basis only. $\endgroup$ – JalfredP Nov 5 '16 at 13:47
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Given the definitions $\vec a_i = \vec F_i/m$ and $\vec F = \sum_i \vec F_i$, and Newton's 2nd law formulated as $\sum_i \vec F_i = m \vec a$, we can prove that all of the following expressions are equivalent:

$$\vec F = \sum_i F_i = \sum_i m \vec a_i = m \sum_i \vec a_i = m \vec a.$$

These definitions together also imply the additivity of acceleration: $$ \vec a = \sum_i \vec a_i.$$

What we cannot do is uniquely derive the definition of $\vec a_i$ (or, for that matter, $\vec F$) from the statement of the 2nd law alone — without knowing what we mean by $\vec a_i$, there's no way to figure that out just from $\sum_i \vec F_i = m \vec a$, since no $\vec a_i$ appears in it.

However, if we take it as granted that accelerations are additive and proportional to forces, i.e. that $\vec a = \sum_i \vec a_i$ and $\vec a_i \propto \vec F_i$, then we can indeed derive from $\sum_i \vec F_i = m \vec a$ that the constant of proportionality between $\vec a_i$ and $\vec F_i$ must, naturally enough, be $1/m$.


Indeed, a perfectly reasonable way to derive Newton's first and second laws would be to start from the following definitions / postulates:

  1. Acceleration is the change in velocity over time: $\displaystyle \vec a = \frac{{\rm d} \vec v}{{\rm d}t}$.
  2. When more than one effect accelerates a body, they combine additively: $\displaystyle \vec a = \sum_i \vec a_i$.
  3. A force acting on a body accelerates it in proportion to its magnitude and direction: $\displaystyle \vec a_i \propto \vec F_i$.
  4. The ratio of the force and the resulting acceleration equals the mass of the body: $\displaystyle \frac{\vec F_i}{\vec a_i} = m$.

These postulates are almost sufficient to derive $\sum_i \vec F_i = m \vec a$, and indeed all the other equivalent forms given above, except for the fact that postulate 3 implicitly leaves open the possibility that there might be accelerations that do not result from forces.

If we insist on assigning to each component acceleration $\vec a_i$ in our system a corresponding force $\vec F_i = m \vec a_i$ (which we can always mathematically do, as long as $m > 0$), then we will indeed recover Newton's 2nd law as conventionally stated. However, I would argue that there are, in fact, sound conceptual and pedagogical reasons not to do this, but to instead leave open the possibility of there being sources of acceleration without a corresponding force.

In particular, such "anomalous accelerations" arise naturally in non-inertial coordinate systems, where they represent changes in the apparent motion of an object that actually result from the movement of the coordinate system we're working in. The standard way of handling such pseudo-accelerations involves artificially multiplying them by the mass of the object, and calling the result a fictitious force, but they aren't really proper forces at all — for one thing, they don't generally obey Newton's third law.

I would argue that, conceptually, it makes a lot more sense to treat these effects as (apparent) accelerations not arising from any force, than to artificially invent imaginary pseudo-forces to attribute them to. Not only is this approach simpler (in terms of not postulating unnecessary entities), but it works better for numerical calculations (where avoiding the needless multiplication and subsequent division by $m$ simplifies the arithmetic and reduces rounding errors) and allows the equations of motion to be naturally applied even to inertially moving objects of unknown or zero mass ("tracer particles"), without having to take awkward limits as $m \to 0$.

A notable special case is gravity, which is a real force in Newton's physics, but a pseudo-force in general relativity. Even when working in Newtonian physics, it can be numerically convenient to treat gravity as a pure forceless acceleration, rather than as a force proportional to the mass of the object it acts on. Indeed, I expect that there's quite a lot of naïvely written physics code out there whose speed and numerical accuracy would be improved (marginally, at least) if programmers were taught to treat gravity as an acceleration and not as a force, and thus avoid unnecessary multiplication and division.


Ps. For a historical perspective, it might be interesting to look more closely at Newton's original formulation of his second law (via Wikipedia):

"Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur."

"Law II: The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which that force is impress'd." (Motte 1729)

Taken literally, this form of the second law may be written mathematically as $\frac{{\rm d}\vec v}{{\rm d}t} \propto \vec F$, or, to use Newton's own notation for time derivatives, $\dot{\vec v} \propto \vec F$. (An even more literal reading might skip the derivatives altogether, and interpret Newton's statement here merely as $\Delta \vec v \propto \vec F$.)

It's worth noting that Newton here merely talks about a single "motive force"; while the following commentary notes that subsequent changes to the velocity are additive, this statement of the second law makes no explicit mention of how multiple non-parallel forces acting simultaneously should combine.

Nonetheless, it is remarkable how closely this original form of Newton's second law resembles what I stated as my postulate 3 above (given that I hadn't actually looked it up before writing this addendum). Indeed, by giving a name — "acceleration" — to the "alteration of motion" over time, and noting that multiple accelerations combine the same way whether they are applied consecutively or simultaneously, we essentially obtain the postulates 1–3 above; all that remains is the explicit constant of proportionality given as postulate 4 above.

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    $\begingroup$ "However, I would argue that there are, in fact, sound conceptual and pedagogical reasons not to do this, but to instead leave open the possibility of there being sources of acceleration without a corresponding force." I think Newton's first law alone is provided as a proof for existence of inertial frames and also 2nd law is considered to be valid if the frame of reference is inertial. Therefore the fictitious accelerations we build are not related to 2nd law and is merely a tool for handling non-inertial frames. $\endgroup$ – Hamed Begloo Nov 5 '16 at 20:57
  • $\begingroup$ Besides I don't know why the additivity of acceleration has so much doubt for everyone. Isn't it really obvious? $\endgroup$ – Hamed Begloo Nov 5 '16 at 20:58
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Well, yes but no. What does that mean to write $\vec{F_i}=m\vec{a_i}$ if all the forces are actually acting simultaneously?

If you want to mean that what you mean by that equation is the relation between force and acceleration if only $\vec{F_i}$ were acting and all the rest were absent then yes, you are perfectly correct and have probably got the essence of the definition of force, mass and the laws of motion in Newtonian Mechanics.

But if you wish to interpret it as the relation between the part of acceleration produced by $\vec{F_i}$ even when the rest of the forces are acting then I would have to disagree with you. There is no meaning to the part of acceleration produced when all the forces are acting simultaneously- at least in the standard way of thinking about classical mechanics that I know. And within the scope of classical mechanics, such a proposal seems metaphysical. If, imaginarily, there were some mechanism that could definitively tell us what part of acceleration is produced by what force component then only such an interpretation can hold any physical significance.

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  • $\begingroup$ I believe there is confusion between understanding whether the dynamics is applied to one single particle (with many forces) or to a system of particles. If the latter, then of course you do have many single one accelerations that contribute to the acceleration of the centre of mass. $\endgroup$ – gented Nov 5 '16 at 14:34
  • $\begingroup$ There is no mention of constituents of the system by OP. Thus, I presumed the context to be regarding a single mass particle. $\endgroup$ – Dvij Mankad Nov 5 '16 at 14:40
  • $\begingroup$ "We all know that "Newton's second law" states that any particle or a system of particles..." it seems there is indeed such mention. $\endgroup$ – gented Nov 5 '16 at 14:48
  • $\begingroup$ No. It mentions the system but the way OP has broken down $\vec{F}$ into $\vec{F_i}$s doesn't mention any reference that indicates that the breakdown is on a constituent basis. Indeed if this were the case then it would have been written that $\vec{F_i}=m_i\vec{a_i}$ not $\vec{F_i}=m\vec{a_i}$- especially when the OP has mentioned that $m$ is the total mass. $\endgroup$ – Dvij Mankad Nov 5 '16 at 14:53
  • $\begingroup$ I believe "system of particles" means "system of particles", independent of any interpretation that follows afterwards. But even in the other case there would be no $a$, rather $a_{cm}$, since every system that is broken down, whether originally one or not, fulfills Newton's law with $a_{cm}$, as there is no general "global" acceleration. $\endgroup$ – gented Nov 5 '16 at 15:03
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Your premises are slightly ambiguous. The result and its interpretation depend on what the actual system is:

  • if you're dealing with one point particle then the statement is clear: Newton's law is what it is.

  • if you are dealing with a system of point particles then you start writing down Newton's law for each single particle; summing them over and taking into account the fact that internal forces cancel out mutually you end up with an expression of the form: $$ \sum_{i=1}^N \mathbf{F}_i^{\textrm{(external)}} = M {(\mathbf{a}_{cm})} $$

In particular one has, for each particle $j$ $$ \mathbf{F}_j = m_j\mathbf{a}_j,\qquad j=1,\ldots, N $$ where the force acting on each particle (left hand side of the above) can be separated into the contributions coming from internal interactions and the contributions coming from external ones; thus we have $$ \mathbf{F}_j = \mathbf{F}_j^{\textrm{(external)}} + \sum_{k<j}^N\mathbf{F}_{jk} = m_j\mathbf{a}_j,\qquad j=1,\ldots, N $$ with $\mathbf{F}_{jk} = - \mathbf{F}_{kj}$ due to Newton's third law, for each pair $(j,k)$. Summing all the equations of the form as above one obtains: $$ \sum_{j=1}^N \mathbf{F}_j^{\textrm{(external)}} + \sum_{j=1}^N \sum_{k<j}^N\mathbf{F}_{jk} = \sum_{j=1}^N m_j \mathbf{a}_j $$ that can be re-arranged as $$ \sum_{j=1}^N \mathbf{F}_j^{\textrm{(external)}} + \sum_{k\neq j}^N(\mathbf{F}_{jk} + \mathbf{F}_{kj}) = \sum_{i=1}^N m_i\,\frac{\sum_{j=1}^N m_j \mathbf{a}_j}{\sum_{i=1}^N m_i} $$ where the second contribution in the left hand side vanishes. We thus have: $$ \sum_{i=1}^N \mathbf{F}_i^{\textrm{(external)}} = M {(\mathbf{a}_{cm})}. $$ There is no "general acceleration": there is instead the acceleration of the centre of mass system (that fulfills Newton's law with mass being the total mass, subject to external forces only) and the single accelerations of each single point particle (that are what they are). The same result, however, does not hold for other quantities like the energy, for example: the total kinetic energy of a system of particles cannot be addressed to the centre of mass only (König's theorem).

  • if you are dealing with a continuous body the reasoning goes along the same lines as above: paying attention to integrate over the corresponding surface and mass elements, one ends up with a similar result, mutatis mutandis, with the resultant of external forces being thought as applied to the centre of mass of the system, so that all other pieces in the integrations cancel out with each other. The corresponding König's theorem holds again too.

The point is that one constructs the laws of dynamics in the general case summing all the single contributions for point particles: therefore the starting point is Newton's law holding true for single particles and hence Newton's law holding true (in the centre of mass system) for the general discrete or continuous body.

What is instead equivalent is the above and Newton's third law for point particles: namely the above holds true if and only if two point particles interacting with each other exercise the same and opposite forces, as vectors.

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  • $\begingroup$ Sorry but I think you misunderstood my question. I'm not talking about internal forces or the 2nd law for each particle inside the system. I'm talking about decomposing the net external force itself and relating it to a corresponding acceleration. $\endgroup$ – Hamed Begloo Nov 5 '16 at 20:24
  • $\begingroup$ Then in that case there is no question at all: for one particle only there is one acceleration. Any other system in the universe is however composition of particles, hence the above applies. $\endgroup$ – gented Nov 6 '16 at 0:28

protected by Qmechanic Nov 5 '16 at 20:38

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