B Why do we take k=1 in the derivation of F=k*ma?

[Herman Trivilino]

Since 1 N = 1 kg m/s^2 by definition it is logically impossible to ever find $f\ne ma$ in SI units.

Huhh?! We currently define the Newton as a kilogram meter per second squared and know that $F \neq ma$. Are you conflating the definition of the Newton with the definition of force?

 

[Dale]

Huhh?! We currently define the Newton as a kilogram meter per second squared and know that $F \neq ma$. Are you conflating the definition of the Newton with the definition of force?

What? I don’t know what you are saying.

 

[Herman Trivilino]

What? I don’t know what you are saying.

$F=ma$ is an approximation, valid only in the limit of slow speeds. The Newton is defined by BIPM to be a kilogram times a meter per second squared.

 

[e-pie]

Good definitions are essential to a good theory!

Since there are 7 most fundamental quantities that cannot be defined, according to you all of Physics is then wrong, because we talk about time, length yet no one can define it.

Newton himself said that we must stop asking "what" and start asking "how".

A lot of "what" s will eventually continue till eternity and nothing productive will come out from a philosophical discussion.

 

[e-pie]

$F=ma$ is an approximation, valid only in the limit of slow speeds. The Newton is defined by BIPM to be a kilogram times a meter per second squared.

We all know this and the discussion has traveled past scientific facts.

 

[Herman Trivilino]

Since there are 7 most fundamental quantities that cannot be defined, according to you all of Physics is then wrong, because we talk about time, length yet no one can define it.

The seven base units are indeed defined. What makes them base units is that they are not derived. But they are very carefully defined. The definitions undergo changes to keep up with the demands of science, technology, and engineering.

 

[Dale]

To arrive at a definition, one must experiment/observe, rationally think and analyze... This is a logical series of thought.

No, to arrive at a definition one need only write the definition down. Observations, experiments, and rational thought then can be used to determine if the definition is useful or not.

Newton never did directly define what a Force is.

Yes, this is a modern view of his theory.

you cannot define "definition" by using itself.

So what? Why should that be at all relevant.

You are just getting silly now. Have you seriously never done a proof in geometry or logic where at some point in the proof you used “by definition” as a justification? If you seriously have not then you should practice doing a few proofs, e.g. prove that a square is a rhombus and a rectangle.

 

[e-pie]

The seven base units are indeed defined. What makes them base units is that they are not derived. But they are very carefully defined. The definitions undergo changes to keep up with the demands of science, technology, and engineering.

If you mean that second is defined in this way

The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.

then strictly speaking that is not definition. That is standardizing.

 

[Dale]

$F=ma$ is an approximation, valid only in the limit of slow speeds.

Oh, I see what you are saying. I missed that at first.

Yes, Newtonian physics is not valid in all domains, but F=ma is valid wherever Newtonian physics is. In relativistic physics the concept of force needs to be redefined.

 

[Dale]

Since there are 7 most fundamental quantities that cannot be defined,

They most certainly can be defined. See chapter 2 of the BIPM brochure on the SI. It clearly identifies the descriptions as definitions of the SI units.

https://www.bipm.org/en/publications/si-brochure/section2-1.html#section2-1-1

 

[Dale]

$F=ma$ is an approximation, valid only in the limit of slow speeds. The Newton is defined by BIPM to be a kilogram times a meter per second squared.

You are right. It may not be a tautology in SI units. I need to give this a little more thought. The whole concept of force is different in relativity, so I need to be careful.

 

[Herman Trivilino]

Oh, I see what you are saying. I missed that at first.

Yes, Newtonian physics is not valid in all domains, but F=ma is valid wherever Newtonian physics is. In relativistic physics the concept of force needs to be redefined.

Yes. But my point is that the definition of the Newton doesn't need to be refined. It is valid even in the realm where $F=ma$ is not.

 

[Dale]

Yes. But my point is that the definition of the Newton doesn't need to be refined. It is valid even in the realm where $F=ma$ is not.

Yes, I got that belatedly.

 

[Dale]

Yes. But my point is that the definition of the Newton doesn't need to be refined. It is valid even in the realm where $F=ma$ is not.

So here is my thinking. In modern Newtonian physics F=ma (in the classical domain) is considered a definition of force. So currently any classical test would be tautological. In order to make a non tautological test you would need to establish an independent definition of force.

Scientifically, that would be done through an operational definition. But the SI system does not admit an independent operational definition of the Newton.

So you would have to do two things to non-tautologically test Newton’s 2nd law in the classical domain. One would be to get another unit system, and the other would be to get a non-2nd-law definition of force. The two would probably go together, but for the reason you point out they are not the same thing.

I was incorrect to associate it entirely with the unit system. That is only part of the picture. I need to refine this in my own mind a bit, so I may still have some mistakes remaining. I will consider it again tomorrow.

 

[gmax137]

I'm still confused

Imagine you don't know what "force" is.
Zz.

But, I do have a pretty good idea of what "force" is -- and I learned it long before I heard of Newton. I learned it as a kid, pulling my brother in his red wagon; carrying shingles to the roof; pushing the pickup out of a ditch; sliding books across the table; getting crushed by playmates at the bottom of the pig pile... I don't see the need for the abstraction.

Well, then how would you quantitatively define “force” without referencing Newton’s 2nd law?

Newton’s first law can be seen as a definition of an inertial frame and the second law as a definition of force. If you don’t use Newton’s laws to define them then you need to find another definition.

There's a difference between "defining" a concept and quantifying an instance. At least, there is to me.

Also, I can quantify a force with a spring and a ruler. So, does F=-kx "define" what force "is"?

 

[DrStupid]

Newton never did directly define what a Force is.

I think he did it in Definition IV. It is not complete (the quantitative properties are specified in the laws of motion) but it is sufficient to understand what Newton means with force.

In relativistic physics the concept of force needs to be redefined.

That's one possibility. Another possibility is not to redefine mass. That is not as convenient but it works.

 

[Dale]

I can quantify a force with a spring and a ruler. So, does F=-kx "define" what force "is"?

Yes, that would be an alternative method to define force. Typically it isn’t defined that way, but you could indeed use such a definition in a self consistent manner. In that case Hooke’s law would be a tautology, and Newton’s 2nd law would be an experimentally testable hypothesis.

The reason Newton’s law is usually used as the definition of force instead of Hooke’s law is practical. It is much easier to reproduce a mass standard than a stiffness standard.

 

[f95toli]

This problem has been discussed for a very long time
See Mach's "The Science of Mechanics"
I read it some 20 year ago and -as far as I remember- he spends many (many. many,many) pages discussing how to "define" force and mass.

 

[girishr]

Please explain further. Maybe I am getting a wrong interpretation.

Are you suggesting the F=kma approach where k=1 because F, m and a are unit value?

Suppose you invent a new thing and name it as Xxxxxxx. That's what happened here. These values were started from thereafter only.

 

[girishr]

Since 1 N = 1 kg m/s^2 by definition it is logically impossible to ever find $f\ne ma$ in SI units.

This unit of force started thence only. Do they could put any value for this. For ex

 

[russ_watters]

So here is my thinking...

That is only part of the picture. I need to refine this in my own mind a bit, so I may still have some mistakes remaining. I will consider it again tomorrow.

I feel like maybe you're struggling to address a circularity problem between the name (definition) given to a phenomena and the definition of a mathematical relationship that describes how it works. I think if they are kept separate, the problem goes away.

Consider a non-mathematical example: evolution. Evolution is both the name given to an observed phenomena and the name of the theory describing how it works. This is sometimes confusing when it comes to evolution (it is discussed often as a creationist misunderstanding), but usually isn't when talking about force because the link between the phenomena and the mathematical definition is exact/specific. I think that's why you are struggling to separate them/view them as separate. Still, f=ma or f=dp/dt can be separated from the name of the phenomena if we want: force is what you feel when you push on something. Inexact, sure (pressure...?), but useful.

A less complicated mathematical example that we nevertheless often get confused questions about is energy. Why? Because energy isn't really a phenomena or property - it isn't a "thing" - so it doesn't have phenomena/property-type definition, only a mathematical one. It's just a useful relationship between properties that was found to be conserved in most cases. Because of this, the verbal definition is much more tautological - and therefore unsatisfying - than for force: energy is the capacity to do work. But unlike f=ma, you can't as easily see or feel w=fd (PE=mgh). It's just a useful mathematical definition.

 

[gmax137]

Just to extend Russ's line of thought, from force to energy, now go to entropy. How many threads are there asking "yes, but what is it?"

 

[Herman Trivilino]

Yes, that would be an alternative method to define force. Typically it isn’t defined that way, but you could indeed use such a definition in a self consistent manner. In that case Hooke’s law would be a tautology, and Newton’s 2nd law would be an experimentally testable hypothesis.

And those experiments would quickly show that $F \neq ma$. The reason is because real springs only approximately obey Hooke's Law, which by the way is the assertion that the $k$ in $F=-kx$ is a constant for a given spring.

I don't know if the following is an example of what Russ said, but I thought of it immediately after I read what he wrote. Suppose we define the Newton as the amount of force required to make an object of mass 1 kg accelerate at 1 m/s² and build a force-measuring device that's so calibrated. What we can then do, for example, is to use that force-measuring device to apply that same amount of force to an object of mass 0.5 kg. If it results in the object having an acceleration of 2 m/s² that's a verification of $F=ma$. And if it doesn't then it's a refutation of $F=ma$.

 

[David Lewis]

Force is defined as the change in momentum over time. No sense putting in messy proportionality constants when it's simply a definition.

Good point now that you bring it up. F=ma only relates physical quantities. It doesn't say anything about units of measure.

 

[Dale]

I feel like maybe you're struggling to address a circularity problem between the name (definition) given to a phenomena and the definition of a mathematical relationship that describes how it works.

No, I am not worried about circularity too much in general. You can always make some operational definitions in science to remove circularity.

 

[Dale]

And those experiments would quickly show that F≠maF≠maF \neq ma. The reason is because real springs only approximately obey Hooke's Law

That isn’t a problem in principle. You can simply limit the definition to small x. For small x and small v you could experimentally show f=ma. Such limitations are common.

 

[Herman Trivilino]

That isn’t a problem in principle. You can simply limit the definition to small x. For small x and small v you could experimentally show f=ma. Such limitations are common.

That's a good point. So I suppose it comes down to an issue of precision. The definition that minimizes these limitations is the better definition.

I don't think, though, that adopting this definition in any way introduces a testability of $F=ma$ that's lacking in a definition that makes use of $F=ma$. Rather, it simply constitutes a less precise definition.

 

[Dale]

The definition that minimizes these limitations is the better definition.

Yes, which is why Newton’s law is used instead of Hooke’s law.

I don't think, though, that adopting this definition in any way introduces a testability of F=maF=maF=ma that's lacking in a definition that makes use of F=maF=maF=ma. Rather, it simply constitutes a less precise definition

You could be right. I haven’t finished thinking through this yet. But my gut feeling is still that it does introduce testability that was previously lacking, but I am mentally stuck on mass and stiffness now.

 

[Edeff]

So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.

 

[Dale]

If I am struck by a car moving at constant velocity,

If the car strikes you then neither you nor the car will move at constant velocity

 

[A.T.]

So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.

The F in F=ma stands for the vector sum of all forces acting on the object, not for some individual force.

 

[CWatters]

So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.

When the car hits you it accelerates you and you go flying down the road. The force you feel is due to your own acceleration.

 

[Dadface]

Experiments show that a is proportional to F/M. We can write:

a = kF/M

If we had agreed units for a and M we can use the equation to define a unit for F. I might suggest that we use a unit of force called a turnip (T) defined as:

One Turnip is the resultant force that gives 12Kg an acceleration of 4.5m/s²

It's unlikely that the Turnip would be widely adopted as a unit because it would make k equal to 54 and amongst other things that can be annoying. The neatest definition would make k equal to one and hence the Newton is adopted because it does just that:

One Newton is the resultant force that gives 1Kg an acceleration of 1 m/s².

 

[bobob]

I think the point of the original question might be related to the equivalence principle. If you take the weight of an object to be be proportional to it's mass and that proportion is W=F=mg = m(GM/r^2) then it might be natural to ask if that is equivalent to a mass x acceleration in general. The underlying question is then, given a gravitational mass m and an inertial mass km, is k=1? In that case, the question is obviously quite reasonable given the number of experiments done look for departures from that equivalence.

 

[jbriggs444]

The underlying question is then, given a gravitational mass m and an inertial mass km, is k=1?

That question might better be posed as "is k a fixed constant".

The formula for [passive] gravitational mass is $F=\frac{G\ m_{passive}\ m_{active}}{r^2}$.

That constant G in there means that any fixed constant k in $m_{passive}=\frac{m_{inertial}}{k}$ is indistinguishable from an adjustment to G.

 

[Gamertag]

You can and sometimes you need to. For example if you measure f in lbf, m in kg, and a in AU/day^2 then k would be 4.5 lbf/(kg*AU/day^2). We can only set it to 1 if you are using SI units or other unit systems that were designed that way, which are called consistent units.

As it says "The unit of force is so chosen that, k = 1, when m = 1 and a = 1." (emphasis added). We can do it because we defined the SI units that way.

So why do we not consider value of G as 1 in F= (GMm)/r^2 ?

 

[etotheipi]

So why do we not consider value of G as 1 in F= (GMm)/r^2 ?

You can (with a caveat) use Planck units where $G = c = \hbar = k_B = 1$, in which case the law of Gravitation becomes$$\tilde{F} = \frac{\tilde{m}_1 \tilde{m}_2}{\tilde{r}^2}$$The caveat is that the $\tilde{F}$, $\tilde{m}_1$, $\tilde{m}_2$ and $\tilde{r}$ are now the dimensionless values of those quantities when expressed in the Planck units, i.e. $\tilde{F} = F/F_P$, $\tilde{m}_1 = m_1/m_P$, $\tilde{m}_2 = m_2/m_P$ and $\tilde{r} = r/l_P$. So you have to be a little careful. (More details here: https://en.wikipedia.org/wiki/Planck_units#Introduction)

As far as I am aware, $G$ is usually never set to $1$, and we mostly choose units where $G$ is explicitly kept in the equations.

 

[russ_watters]

So why do we not consider value of G as 1 in F= (GMm)/r^2 ?

G is different from the "k" the OP was referring to in that it has units. The units of the parts of the gravitational force equation (under most systems) don't reconcile on their own, so the constant is needed to make the statement mathematically complete/consistent.

 

[Dale]

So why do we not consider value of G as 1 in F= (GMm)/r^2 ?

You can choose to do that by choosing a system of natural units. You can also choose units where c=1 or h or any other fundamental physical constant. The SI system chose not to do that for historical reasons and also so that the base units would be "human" scale. I.e. the height of a human is a couple of meters, etc.

 

[pbuk]

Oh dear, it's necroposting season again

 

[Dale]

Oh dear, it's necroposting season again

How prescient of you!

 

[Physics guy]

So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.

Interesting question that even I took a long to get an answer. Fortunately now I have an answer. This can be understood by accepting Newtons laws of motion to be true (even though no one can prove them to be true). Second law of motion say that force is directly proportional to the rate of change of momentum, which gave is the famous equation that F = ma. So this equation tells us that a net force can only act if there is a change in momentum of a body ie, the body undergoes acceleration. During collision what we can expect is the same when a car moving with constant velocity hits a person, during the collision there is a reduction in the velocity of the car for a small instant, this causes a change in momentum and thus a force is applied to the person (may god save him). This is especially so in the case of vehicles having larger mass in which the change in momentum will be larger

 

[pbuk]

So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.

Interesting question that even I took a long to get an answer. Fortunately now I have an answer.

Unfortunately you have missed the same thing as @Edeff missed: the force you feel is the product of your mass and your acceleration. Any (negative) acceleration of the car is only relevant to the car and its occupants.

@Edeff is assuming that the mass of the car is infinite and therefore it does not lose any momentum in the collision; again this is only relevant to the occupants of the car (who will not feel anything). In the limit as the mass of the car tends to infinity the change in velocity of the person that is hit tends to instantaneous and so they suffer the effect of a force tending to infinity.

 

[Physics guy]

Unfortunately you have missed the same thing as @Edeff missed: the force you feel is the product of your mass and your acceleration. Any (negative) acceleration of the car is only relevant to the car and its occupants.

@Edeff is assuming that the mass of the car is infinite and therefore it does not lose any momentum in the collision; again this is only relevant to the occupants of the car (who will not feel anything). In the limit as the mass of the car tends to infinity the change in velocity of the person that is hit tends to instantaneous and so they suffer the effect of a force tending to infinity.

Well replied

 

[jbriggs444]

@Edeff is assuming that the mass of the car is infinite and therefore it does not lose any momentum in the collision;

Be careful. A correct conclusion is that the lost vehicle velocity is zero (or infinitesimal). The lost momentum is non-zero.

 

[pbuk]

Be careful. A correct conclusion is that the lost vehicle velocity is zero (or infinitesimal). The lost momentum is non-zero.

I agree that considering momentum is problematic here, that's why I avoided it.

In the limit as the mass of the car tends to infinity the change in velocity of the person that is hit tends to instantaneous and so they suffer the effect of a force tending to infinity.

 

[berkeman]

The OP's question has been answered, and after some member requests and a Mentor discussion, this thread is now closed. Thank you everyone.