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

Question

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?

Answer 1

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

Answer 2

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 $F=m\cdot a$.

Answer 3

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

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

instead?

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

Ever wondered why the value of $G$ in Newton's law of gravitation is not 1 just like in this case you stated. Keep reading. I will come to the answer a bit later.

This is Newton's law of Gravitation:

$F =\frac{Gm_1m_2}{R²}$

In this equation, there are four "defined" physical quantities $m_1$,$m_2$, $R$ and $F$ . 1kg of mass is defined as the mass of platinum-iridium aloy kept in at the International Bureau of Weights and Measures at Serves near Paris, similarly length (R) is defined as the length of a platinum-iridium rod and F is defined by Newton's first law of motion: $F=ma$(have patience).

So knowing the rest of the four quantity in the equation the value of unknown $G$(the fifth quantity) can be determined easily(using a torsion balance).

But what to do if there are two unknowns as in equation of Newton's first law: $F=kma$ Here $k$ and $F$ both are unknowns. When Newton discovered the law, the unit of force was not defined. No one thought of measuring forces. So $F$ is also an unknown value in the equation. But it does contains $F$, so it can be used to define $F$ instead! So Newton got an equation which he could use to define the first ever unit of force. But what to do of that constant $k$? Why was it chosen as $1$?

In fact, the value of k can be any real number but IUPAC took it as $1$. Why? Because it was the easiest way. Taking $k=1$,the equation becomes $F=ma$. It is easy to say that $1N$ of force is the force that can produce an acceleration of 1 m/s² to a body of mass $1kg$. But we take $k$ as 4.72675(any arbitrary number) we will have to include it in the definition. Also it will become measurements and calculations more difficult. So taking $k$ as $1$ is easier for us.

Now you can understand that had Newton discovered his law of gravitation first, he would have used it instead to define the unit of force. In that case he would had taken $G$ as $1$. And then the value of $k$ in his first law would been something else ( may be it would had been a complex value just as $G$). Now you can see that it's one because it makes it easy.