3
$\begingroup$

Assuming a constant mass, which would be best way of expressing Newton's second law? $F=ma$ or $a=\frac{F}{m}$? The most common way is the first one, but I feel more comfortable explaining it the second way. It seems more natural to talk about the consequence (acceleration $a$) of applying the cause (force $F$) on an object (of mass $m$). How is the second law stated in each case?

Edit: What I mean is that a scientific law is more than the formula. I want to know of the law itself would be closer to Newton's interpretation if phrased as "the acceleration of an object is proportional to the net force applied on it and inversely proportional to the mass" or as "the net force needed to accelerate an object is proportional to the acceleration and the mass". I usually I see it explained using the first, but accompanied by $F=ma$, which I find weird (and confusing for students). I am aware of the relationship between force and change of momentum, but I am working with 10th graders.

$\endgroup$

closed as primarily opinion-based by John Rennie, Kyle Kanos, Jon Custer, Daniel Griscom, JMac Oct 24 at 13:24

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Aha, it seems that you are looking for a comparison with Newton's actual original words on this. I have made an answer that includes a quote with his own words from his publication of his laws in the 17th century. I hope this clears it out. $\endgroup$ – Steeven Oct 17 at 15:32
  • 2
    $\begingroup$ Neither, I would use $\mathrm{d}/\mathrm{d}t(\mathbf{p})=\mathbf{F}$. That is: the change in momentum is equal to the (total) applied external force. The version you post is a special case when the mass is constant, but there are cases where that doesn't hold, like a rocket that uses up fuel as it flies. $\endgroup$ – S V Oct 17 at 15:39
  • $\begingroup$ @SV Newton's laws are not valid for variable mass systems, so in Newtonian mechanics the two expressions are equivalent. In relativistic mechanics, only $F=dp/dt$ is valid. $\endgroup$ – garyp Oct 17 at 16:26
  • $\begingroup$ In pedagogical settings I prefer the first way. System on the right, resulting motion on the left. I also prefer Ohm's Law: $I=V/R$ for similar reasons. $\endgroup$ – garyp Oct 17 at 16:28
  • $\begingroup$ @garyp fluids are variable mass systems and I see no Lorentz factors there. You simply generalize "contact forces" to stress and body forces acting pointwise. No relativity there. Are you saying that the whole field of fluid mechanics is wrong? $\endgroup$ – S V Oct 17 at 16:30
5
$\begingroup$

There are a number equivalent mathematical forms of Newton's second law. The choice is a matter of convenience for solving different problems or understanding different concepts.

There is, of course

$$F=ma$$

Which is useful in understanding Newton's first law relationship to the inertial property of mass, i.e., the larger the mass the larger the force needed to effect a change in velocity (acceleration).

And what you prefer

$$a=\frac{F}{m}$$

When one is interested in the acceleration of a mass due to a net force.

Then there is

$$F=m\frac{dv}{dt}$$

which can be re-written as

$$F=\frac{d(mv)}{dt}=\frac{dp}{dt}$$

which demonstrates that a net force results in the change in momentum of an object.

I want to know if it would be closer to Newton's interpretation if phrased as "the acceleration of an object is proportional to the force applied on it and inversely proportional to the mass" or as "the force needed to accelerate an object is proportional to the acceleration and the mass". I usually I see it explained using the first, but accompanied by 𝐹=𝑚𝑎, which I find confusing for students.

I usually see it phrased the first way (except it's missing a key word, as noted below), together with the formula $F=ma$. I agree with you the equation expressed in terms of $a$ better fits the phrasing. And I also agree it can be confusing to students. The problem is, that is the way they will see it if they continue their education in physics. But if I were teaching it, I would present it both ways like you.

Equally, if not more, important is your phrasing is missing the key adjective "net" in front of force. You can have a force without acceleration, like when you push against a wall. I would really stress the phrase "net force" with students. If that were more often done, perhaps students would not get so confused when the learn the third law and wonder why forces don't always cancel each other and no acceleration occur.

Hope this helps.

$\endgroup$
  • $\begingroup$ The last way of representing it (Force is equal to rate of change of momentum) might seem an odd phrasing at first but it is the one that can be carried over to problems in relativity and quantum physics most easily. It also is true even for photons. Photons have zero mass - so cannot use the other equations, but do carry momentum and exert force in exactly the way of Bob D's last equation. $\endgroup$ – Dast Oct 16 at 12:30
  • $\begingroup$ What I mean is that a scientific law is more than the formula. I want to know if it would be closer to Newton's interpretation if phrased as "the acceleration of an object is proportional to the force applied on it and inversely proportional to the mass" or as "the force needed to accelerate an object is proportional to the acceleration and the mass". I usually I see it explained using the first, but accompanied by $F=ma$, which I find confusing for students. I am aware of the relationship between force and change of momentum, but I am working with 10th graders. Sorry if it was unclear. $\endgroup$ – jrglez Oct 17 at 15:13
  • $\begingroup$ @jrglez I have incorporated your comment and my response in a revision to my post. Hope it helps. $\endgroup$ – Bob D Oct 17 at 16:15
  • $\begingroup$ It does, thank you! $\endgroup$ – jrglez Oct 19 at 9:28
2
$\begingroup$

The two expressions are mathematically equivalent, so you can use either depending on the context. Your preference might be influenced depending on the factor your are trying to calculate. For example, if you are given F and m it would be more natural to think of the relationship as a = F/m.

$\endgroup$
2
$\begingroup$

In my opinion $F = ma$ is not a good way to describe Newton's 2nd Law as one will see in rocket equations, Impulses, Variable mass systems, etc.

According to me the best way to describe 2nd law is $\vec F = \cfrac {d \vec p}{dt}$ As it shows how Force induces a change in momentum of the particle.

$\endgroup$
  • 1
    $\begingroup$ $F=dp/dt$ yields the rocket equation only in special cases, and then almost by accident. Either formulation of Newton's second law is not valid for variable mass systems. $\endgroup$ – garyp Oct 17 at 16:42
  • $\begingroup$ As @garyp says, your first paragraph is quite misleading. Please have a look at this answer and this other one of mine for a more detailed discussion. $\endgroup$ – Massimo Ortolano Oct 19 at 10:10
  • $\begingroup$ I know its not fully answered that why some must use f=dp/dt I was just writing it short thanks for your comment @MassimoOrtolano $\endgroup$ – Dr_Paradox Oct 20 at 4:50
1
$\begingroup$

It seems from your comments that you are interested in the relation between our usual formula $F=ma$ and with the actual original phrasing by sir Isaac Newton himself.

Newton's own phrasing was actually entirely different. He didn't mention acceleration and arguably also not even force. Instead he described the proportionality between the total impulse $J$ (which he called "impressed force" or "impressed motive force") and motion change $\Delta v$: $$J\propto \Delta v$$

Newton didn't really write out his law as an expression like this. In his original Principia publication in the 1680s he wrote it in words like this (an exact quote):*

"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" - Sir Isaac Newton

To make use of his law, we can rewrite this proportionality into an equation with a proportionality constant that we call mass $m$:

$$J=m\Delta v$$

Often his words are interpreted as not "change in motion" but "change in amount of motion", which might mean "change in momentum". So, you will often see the interpretation of his words instead written as:

$$J\propto \Delta p$$

which translates to $J=\Delta p$ with a proportionality constant of 1, and is equivalent to the motion-interpretation since $m\Delta v=\Delta (mv)=\Delta p$. This latter version turns out to be more accurate, since it doesn't have to assume constant mass.

The background of and Newton's own thoughts on and doubts about his three laws of motion is an interesting read that you can dig deeper into here.

To finalise, impulse relates to force: $$J=F\Delta t$$ and after a rearrangement, momentum change over time relates to acceleration: $$F\Delta t=\Delta p\quad\Leftrightarrow \\F=\frac{\Delta p}{\Delta t}=\frac{\Delta (mv)}{\Delta t}=m\frac{\Delta v}{\Delta t}\quad \rightarrow \quad F=m\frac{d v}{d t}=ma$$

So, the different formulations are equivalent (assuming constant mass $m$).


* $\quad$‘Philosophiæ Naturalis Principia Mathematica’, Isaac Newton, 1st ed., vol. 1, 1687, (English translation published 1728)

$\endgroup$
1
$\begingroup$

I favour your choice, $a = F/m$ and I would explain it to pupils as follows. That the acceleration of a body is directly proportional to the net force on it and is inversely proportional to its mass. Thus, $a \propto F/m$. We choose units of physical quantities in such a way that the constant of proportionality is zero. I would also emphasize that although $F = ma$ and $a = F/m$ are mathematically identical, the physics of the situation comes out clearer in the latter.

$F = ma$ has led several pupils astray into believing that force is equal to mass into acceleration and therefore concluding that 'there is force because the body is accelerating'.

$\endgroup$
1
$\begingroup$

I think all the answers are ignoring the fact that we are talking about a law . What are laws in physics? They are the corresponding axioms that define which solutions of the general mathematical set up for a physics theory fit the data, and can make the theory predictive.

Axioms mean they cannot be proven. As in mathematics, a theorem can replace an axiom and then the axiom becomes a theorem, the same with laws of physics, so it is OK to use different forms .In physics though, Occam's razor is used: trying to give to the laws the simplest form. $F=ma$ to me seems clearest, using the known measurable in the lab quantities of mass and acceleration, to define axiomatically what a force is.

$\endgroup$
0
$\begingroup$

There is no one form suitable for all. Concrete law should be expressed in a way most suitable to context needed.

We can even express Newton second law as: $$ \large{ F = m \cdot \begin{bmatrix} \partial ^{2}x \over \partial t^{2} \\ \partial ^{2}y \over \partial t^{2} \\ \partial ^{2}z \over \partial t^{2} \end{bmatrix} } $$

Or

$$\vec{F} = m \left(x''(t)\space\hat{\textbf{i}} + y''(t)\space\hat{\textbf{j}} + z''(t)\space\hat{\textbf{k}} \right) $$

So really, everything depends on context & notation.

$\endgroup$
0
$\begingroup$

I would say it actually depends on what you want to do with the formula. If you want an interpretation on relations between mass, force and acceleration, I would agree with you. I think it is good to see the consequence in terms of the cause (and a property of the object).

I think the second version is more useful for problem solving. If you solve a force or torque-related problem in Newtonian mechanics, you use force diagrams and calculate the net force in each direction, then apply Newton's Second Law. It is useful to have the geometric-dynamic part on one side and the rest on the other one.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.