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Newton originally wrote his second law as:

"The rate of change of momentum of a body is directly proportional to the resultant force applied to the body, and is in the same direction as the force."

This definition implies,

$F=k\frac{dp}{dt}$ where $k$ is the constant of proportionality, and $p$ is the object's momentum.

From this relationship, why is it that we are able to deduce that $k=1$, and consequently form the equation:

$F=ma$. (mass being constant).

Qmechanic
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J.Gudal
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    That is true in SI system of units, if you change the system, the proportionality will change – Courage Jan 07 '16 at 04:55
  • So the SI units were designed in such a way that $k=1$ ? – J.Gudal Jan 07 '16 at 04:57
  • @TheGhostOfPerdition: Unless this is a duplicate, you should make that the answer, maybe with a short note on the definition of the unit of force in SI? – CuriousOne Jan 07 '16 at 04:59
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    @TheGhostOfPerdition I don't think that's true. An equation such as $F = m a$ is not written in any unit system. Setting $k=1$ is a definition of either $F$ or $p$, or if force and momentum are already defined then $k=1$ is an experimental fact. – DanielSank Jan 07 '16 at 05:04
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    @DanielSank , If we use grams instead of kgs then the proportionality would be equivalent to (1/1000), I think the system of units were chosen so as to call '$1$ kg or mass accelerated at $1m/s^2$' as 1 newton, to get rid of the proportionality constant – Courage Jan 07 '16 at 05:15
  • @TheGhostOfPerdition $F = dp/dt$ means "force is equal to time rate of change of momentum". There is absolutely no reference to any system of units. I can plug values into this equation, and those values will be expressed in a particular system of units, so for example 1 Newton = (1 kg m / s)/(1 s), but the values there have nothing at all to do with whether or not there's a $k$ in Newton's law. – DanielSank Jan 07 '16 at 05:36
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    One should think of $F=\dfrac{dp}{dt}$ as the definition of force just as $v=\dfrac{dx}{dt}$ is the definiton of velocity. The constant of proportionality then by definition is 1. – Omar Nagib Jan 07 '16 at 05:44
  • @DanielSank I agree that the mathematical expression $F=\frac{dp}{dt}$ doesn't depend on the system of units, but when I measure the quantities physically it would define 1 Newton as (1 kg m / s)/(1 s). – Courage Jan 07 '16 at 05:54
  • @TheGhostOfPerdition Yeah, fine, you can define units however you want. That does not mean that Newton's law gets an extra factor in it. – DanielSank Jan 07 '16 at 06:04
  • @DanielSank wouldn't the same argument hold true for the electrostatic force $F=\frac{1}{4 \pi \epsilon}\frac{q_1 q_2}{r^2}$? which as we know holds true only with SI units – Courage Jan 07 '16 at 06:10
  • @TheGhostOfPerdition. I would say those are different things: Newton's 2nd Law is a definition of either force or momentum, whereas Coulomb's Law is a relationship between already-defined physical quantities. – march Jan 07 '16 at 06:16
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    @TheGhostOfPerdition There's an important difference in Coulomb's law: the $q$ you write "in cgs units" has different dimensions that the $q$ you write "in SI units". The two $q$'s are not the same physical thing. – DanielSank Jan 07 '16 at 06:28
  • @DanielSank umm may be I get it now, still not 100% convinced, can you may be add an answer and point out why the newtons and coulomb's law cases are different? thank you – Courage Jan 07 '16 at 06:39
  • @DanielSank: One can measure F in any units one wants. Don't the engineers in the English speaking world sometimes use pound-force? In Crazytown they may also measure momentum in stones-furlong/fortnight, just to make life really miserable. In that case one will need a non-trivial factor. Does it make sense? No. Does it do Newton's genius honor? No. Or, if we want to be literal, maybe it does? – CuriousOne Jan 07 '16 at 08:08
  • @march: What J.Gudal pointed out correctly is that Newton may not have defined the units of force this way. His relationship is more general than the modern version and it is 100% correct, except that it doesn't require the SI convention that forces k to 1 (which works just fine). – CuriousOne Jan 07 '16 at 08:11
  • According to http://plato.stanford.edu/entries/newton-principia/#NewLawMot the literal translation of the Principia yields "A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed.". To me that seems to leave the possibility of a proportionality factor open. BTW: should we close this as belonging into science history? – CuriousOne Jan 07 '16 at 08:17
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    @CuriousOne no, I don't think so. It seems like a perfectly fine science question to me. It may be a duplicate of this, though - if not a duplicate, at least closely related. – David Z Jan 07 '16 at 08:41
  • @DavidZ: As you can see, I didn't vote that way, so I am with you... I also think your answer pretty much kills it. +1. – CuriousOne Jan 07 '16 at 08:50
  • @CuriousOne yeah, I notice that nobody has voted to close, and given how much attention this has gotten, that probably reflects a consensus that this is on topic. The duplicate thing is another matter. I'll mark it as a duplicate if other people also vote accordingly, but I'd rather not do it unilaterally. (FWIW, I don't think moderators can tell who has cast close votes until they are publicly displayed when - if - the question is actually put on hold. Or if we can, it's buried somewhere in a moderator tool, not immediately apparent.) – David Z Jan 07 '16 at 09:30
  • @DavidZ: I didn't know about the duplicate, which I believe this is.. but neither title really indicates the actual content very well, do they? I certainly like the questions and they made me think about the issue a little more, too. When questions become duplicates they get deleted, right? If that's the case I will not vote for duplicate, I rather let it stand. It would be nice to have the link to your old answer stand out more. – CuriousOne Jan 07 '16 at 09:40
  • @CuriousOne When we write physics equations we usually do not write them "in a system of units". You can do that, but we usually don't. For example, $F=ma$ is a perfectly happy equation which is not written "in units". On the other hand, $F_\text{newtons} = 1000 m_\text{grams} \times a_\text{meters / second /second}$ is written "in units" and look! it has a numerical factor now. – DanielSank Jan 07 '16 at 15:50
  • Who remembers $F=\frac{ma}{g_c}$, where F is in $lb_f$, m is in $lb_m$, a is in ft/$sec^2$, and $g_c=32.2\frac{lb_m-ft}{lb_f-sec^2}$? – Chet Miller Jan 07 '16 at 21:25
  • @DanielSank: Yes, we do write with units, hence there are textbooks with cgs, SI, different choices of electromagnetic notation etc.. One of the first things a textbook author does is to explain the unit system. In relativity one has to point out the choice of metric. I have recently seen papers on quantum gravity that had further explanations of the indexing system. One size fits all does not even exist in physics, even though I am with you that it would be nice if it did. That, however, was not even the question, if I am not mistaken. $F=ma$ is only correct for special choices of units. – CuriousOne Jan 07 '16 at 23:36
  • @ChesterMiller: That's a good point and that's why we teach physics usually with SI. Logically, however, Newton was correct and the equation with a proportionality constant to account for strange choices of units has the same logical consequences. It's still a definition of both mass and force and a law of how both relate to acceleration. – CuriousOne Jan 07 '16 at 23:38
  • @CuriousOne The difference between the co-called "SI" and "cgs" versions of the E&M equations is that the word "charge" means two different things; it's a difference in dimensions, not a difference in units. This is an incredibly common and unfortunate misunderstanding which leads countless people to confusion. – DanielSank Jan 08 '16 at 00:37
  • @DanielSank: cgs means centimeters-grams-seconds, i.e you change the preferred scale, not the dimensions. In the end we always define physical measurements as ratios to a preferred scale. That's the price we pay for precision metrology. If we were only interested in the abstract physics (or, the "philosophy of nature"), then we could replace the "=" sign with a "is proportional to" (i.e. Newton's original language) and reproduce all effects qualitatively. If, however, we want to understand scale dependent effects, we have to chose a scale and resort to numerical equivalence. – CuriousOne Jan 08 '16 at 00:59
  • I am 99% sure the statement "i.e. you change the preferred scale, not the dimensions" is false. If I write $F = q_1 q_2 / r^2$, then $q$ has dimensions of $\sqrt{\text{force}} \times \text{length} = \sqrt{\text{mass} \times \text{length}^3 / \text{time}^2}$. If I write $F = q_1 q_2 / (4 \pi \epsilon_0 r^2)$, then I find that $q$ has dimensions which cannot be expressed in terms of mass, length, and time. The two different versions of Coulomb's law imply two different dimensions for $q$. Right? – DanielSank Jan 08 '16 at 01:24
  • Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth Jan 11 '16 at 10:06
  • A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. In the body of the Principia this law is applied both to discrete cases, in which an instantaneous impulse such as from impact is effecting the change in motion, and to continuously acting cases, such as the change in motion in the continuous deceleration of a body moving in a resisting medium. ...the above statement of Newton appears in the Principia so we may consider it to be his stand on the issue. – drvrm Jun 03 '16 at 18:36

1 Answers1

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At the time of Isaac Newton there was no defined unit for force. This gave him the freedom of choosing any unit which was convenient.So he defined a unit of force such that the proportionality constant $k$ becomes 1 which simplified the formula.

Jatin
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  • Agree. I'd even say that at Isaac Newton time there was no force unit named 'Newton' :). He just didn't know that 1 kg m / s2 equals exactly 1 N. (joke, but maybe not joke). – dmafa Jun 03 '16 at 18:43