# How do we know that $F = ma$, not $F = k \cdot ma$

It seems intuitive that $a\; \propto \frac{F}{m}$, as the greater the force that is applied on an object, the greater its acceleration will be. Inversely, the greater the mass of the object, the slower the acceleration will be.

However, when rewriting proportions as equations, you must introduce a constant of proportionality, and in this case of a direct proportion, if $a \propto \frac{F}{m}$ then when rewriting as an equation you will have $$a = k\cdot\frac{F}{m}$$

In order to get the standard formula $F = ma$ this constant must be $1$. However, how do we know that this is the case? How do we know that the constant isn't $2$ and the formula $F = \frac{1}{2}ma$, for instance?

• Related post physics.stackexchange.com/q/70186 – user26143 Mar 18 '14 at 17:39
• possible duplicate of How did Newton discover his second law? – jinawee Mar 18 '14 at 18:11
• Different, but related: physics.stackexchange.com/q/112959/44126 – rob Jun 24 '14 at 2:56
• It should be understood that Newton's law is not "intuitive", it is highly non-trivial and involved and depends on a bunch of experimental facts that beautifully unite in the form of Newton's law. In particular, there is nothing intuitive about the fact that there exists a force, or that there exists a mass, or that $a\propto\frac{F}{m}$. All of these things depend on crucial experimental facts. See this masterpiece of an answer by tparker: physics.stackexchange.com/a/340890/20427 – Dvij Mankad Feb 26 at 14:38
• F DOES equal kma. In the SI system of units, the unit of force, which is the Newton, has been defined in such a way that k=1. For the English system of units, which is pounds mass, pounds force, and ft/s^2, k is NOT equal to 1. – David White Feb 26 at 18:02

I think it is more intuitive to say that (net) force is proportional to acceleration: $F\propto a$. The proportional constant tells us now how easy it is to accelerate an object with a certain force. This proportional constant is called the (inert) mass of said object. Hence $F=m\cdot a$.

• The comment by @JohnRennie is a clearer answer. – garyp Mar 18 '14 at 23:36
• This answer makes a misleading impression that $m$ of a body is to be determined as the ratio $F/a$, with values of $F$ and $a$ known beforehand. This would be hard to do. In contrast, $m$ is measured very easily by a scale or just counting the number of equal components forming the body.With value of $m$ obtained this way and force measured in independent units (weight or deformation coordinate), the second law reads $F=kma$ where $k$ is some constant. It is a simplifying step to change units of $F$ so that we do not have to maintain $k$ in the equations. Check out John Rennie's comment. – Ján Lalinský Jun 24 '14 at 8:05
• Actually determining the ratio $F/a$ IS the way a scale usually works: The acceleration is known ($g=9.81m/s^2$) and the force is measured by measuring the pressure created by the object upon the scale or by comparing the object to another set of weights with known/defined mass (and therefore known force). Even so, it might be interesting to further read into the (non-)difference between inertial and gravitational mass: wikipedia – Michael Große Jun 26 '14 at 9:59

That's the way the unit of force is defined. One Newton is the force that accelerates a mass of one kilogram by 1 m/sec$^2$. The Newton is chosen to make the constant of proportionality equal to one.

• ok that fixes the value of k as 1, but was there any reason to take it as dimensionless? – user22180 Jun 28 '14 at 13:34
• @user22180 Multiplication by 1 is easy. And 1 is easy to remember. – Navneet Kumar Apr 14 '18 at 17:26

A lot of the other answers kind of point to this, but the best way of looking at Newton's second law is to think of it as a definition of force -- while it encodes the notion of a push or pull, technically, we have to encode this quantitatively. Newton's second law encodes the fact that pulls create acceleration, and it sets the unit of a pull in such a way that you don't need your constant of proportionality.

In the Newton's Second Law, Newton basically defined what Force is. He could have taken that constant as any number that he wanted, he chose 1 for simplicity.

• Not that your answer implies it but it should be made clear that the second law of motion is not just the definition of force. It is a full-fledged law of physics, i.e., it is non-trivial. – Dvij Mankad Feb 26 at 14:33

In Newtonian mechanics, we have the quantity momentum (I'll get to force a bit later):

$$\vec p = m\vec v$$

which is conserved and is thus a quantity of fundamental importance. We can think of the mass as the constant of proportionality between momentum and velocity.

But you might ask "why isn't $\vec p = k \cdot m \vec v$ instead?"

The answer is that, by the appropriate choice of units, $k$ can always be made equal to one.

In other words, we want the following: one unit of momentum equal to the product of one unit of mass and one unit of speed.

For example, in SI units, the unit of momentum is

$$kg \cdot \frac{m}{s}$$

with is the product of one unit of mass and one unit of speed.

Now, suppose that the unit of mass were grams rather than kilograms? Would we write

$$\vec p = 1000 \cdot m \vec v$$

or would the unit of momentum become

$$g \cdot \frac{m}{s}$$

Now, a similar argument could be made for $\vec F = m \vec a$ but we don't really need to because we have

$$\vec F = \frac{d \vec p}{dt} = m \frac{d \vec v}{dt} = m \vec a$$

for $m$ constant.

The equation $F = km.a$ is actually more correct, since this equation might be used to define units, but if the units are pre-defined, then the value of $k$ is to be found. This is the theory that Stroud and Wallot taught to engineers at the turn of last century.

For example, 'pdl' = pound * ft/s/s, pound = 'slug' * ft/s/s, and pound = pound * 'g' are all coherent equations, but pdl = 32.175 * pound * ft/s^2 is likewise valid.

The theory of Wallot and Stroud is quantity analysis, where like on the london underground, distances are measured in km, speeds in mph, and time in minutes. The units are all set, and the object is to find 'k' in D = k.VT.

There is no proportionality constant for the same reason there isn't one in

$$\text{distance} = \text{speed}\times\text{time}$$.

It's from using consistent units. For instance, if distance and time are kilometers and hours, and we express speed in kilometers per hour, then there is no conversion factor.

The $F = ma$ formula defines a force as being mass, times acceleration. The units in which force is measured is derived from the product of the units of acceleration and mass.

In metric units, acceleration is $\displaystyle\frac{\text{m}}{s^2}$, and the product of acceleration and mass in $\text{kg}$ is therefore $\displaystyle\frac{\text{kg}\cdot\text{m}}{s^2}$.

The unit of force, the Newton, is then simply defined as: $$1N = 1\frac{\text{kg}\cdot\text{m}}{s^2}$$ with no additional conversion factor.

If an extra constant were present, it would only be changing the measurement unit of force, creating an annoying inconvenience. For instance if you have mass and acceleration composed of a combination of metric units (meters and seconds), but force is being expressed in pounds, then there will be a constant: the conversion factor between Newtons and pounds.

• I like the basic idea of your answer, except for "The $F=ma$ formula defines a force as being mass, times acceleration." because it makes an impression that force as a concept is defined as $ma$. If this was so, the equation would be exact, not an approximate relation inferred from experience. It is better to say "The $F=ma$ formula is a simplified writing of the second law $F=kma$ allowed by changing the unit of $F$.". – Ján Lalinský Jun 24 '14 at 8:41
• @JánLalinský Would you say that the distance-speed-time formula $d = vt$ is really just a simplified way of writing $d = kvt$ allowe by changing the units of $d$? – Kaz Feb 26 at 16:52
• I wouldn't, because I don't know any preexisting notion of speed that does not depend on the definition of distance per time. However, notice that even $d=vt$ does not define the concept of distance or velocity. It is merely statement that there is some rectilinear motion happening. The situation is completely different for the concept of force, because this was existing before the current concepts of mass and acceleration were established and was measured in units different from mass $\times$ distance $/$ time $/$ time. – Ján Lalinský Feb 27 at 0:39
• @JánLalinský This is true in that the force concept is involved in phenomena such as pressure and stress. A cable not undergoing acceleration can have a tension in it measured in Newtons, and divided by the cross-section area, a stress measurable in Pascals. – Kaz Feb 27 at 0:58
• That is a good example. Force can have various manifestations, not merely acceleration. And so it can be measured in units of that manifestation, such deformation of the body under force or quantity of weights needed to counterbalance that force, etc. – Ján Lalinský Feb 27 at 1:01

A lot of these comments/posts are suggesting that Newton's law is defining force, but I don't think this is a good way of looking at it, otherwise it is trivial and the statement is vacuous. I look at Newton's law as essentially defining mass, ie. $m=\frac{F}{a}$ and the reason Newton's law is then nontrivial is because it says that $m$ is constant.

• That would be even worse than using $F=ma$ to define force. Mass is much more easily measured by scales, counting the number of equal components or by volume. Constancy of mass is also very basic experimental fact. – Ján Lalinský Jun 24 '14 at 8:23
• @JánLalinský Well you haven't proposed an alternate definition of mass...would you prefer it if I said inertial mass? – JLA Jun 24 '14 at 17:16
• It was not my intention to define mass. I do not think the notion is problematic - mass is quantity of matter and can be easily measured by scales or other means. If you want to define technical term "inertial mass" as $F/a$ while $F$ and $a$ are measured in units independent of unit of mass, you may do so, but I do not believe that is what the 2nd law is about. – Ján Lalinský Jun 24 '14 at 19:43
• @JánLalinský I think the notion may be problematic if you don't define it...how do you measure it without knowing what to measure? In any case, I was in fact talking about inertial mass, and I think it does get at what the second law is about. Inertia is a bodies resistance to movement...this makes that statement precise. – JLA Jun 25 '14 at 1:10

Actually $$F=k\cdot ma$$ expression is better, because it helps to make a transition to special relativity. In special relativity you need to take into account of relativistic mass for Newton second law to be fixed. In that case k becomes :

$$k = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$$

Now, it's very easy to understand why object can't move faster than light. Solving for limit of $$k$$ when v approaches c gives us an answer -> $$\infty$$, infinity. This means that any physical object hypothetically moving at speed of light has a infinite relativistic mass.So to be able to give even a slightest acceleration to such object - we need to apply infinite force too. Which is non-sense, thus - no object can't be accelerated faster than light.

• Any reasons for a down-vote ? Is this for a fact, that Newtonian laws breaks in special relativity (in cases when object speed is comparable to the speed of light) ? – Agnius Vasiliauskas Feb 26 at 14:35
• – Kyle Kanos Feb 26 at 14:41
• @KyleKanos Ok, then if you think that term "relativistic mass" is out-dated, then How do you re-write Newton second law F=m*a for v ~ c ? I'm sure you know that it breaks on speeds comparable to speed of light – Agnius Vasiliauskas Feb 26 at 14:56
• $F=\mathrm dp/\mathrm dt$, obviously. In this case, $p=\gamma mv$ so you end up with $F= \gamma^3ma_\parallel+\gamma ma_\perp$, if you do the math right. – Kyle Kanos Feb 26 at 15:03
• @AgniusVasiliauskas See for example the Wikipedia page where it warns against this exact thing. – JMac Feb 26 at 15:12