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

[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?"

 

[Mister T]

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.

 

[Mister T]

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.