# Why does $F=ma$? Is there a straightforward reason?

Why is force = mass $\times$ acceleration? I have searched in many sites but didn't actually get at it. Simply I want to know that if a mass in space moves (gains velocity thus further accelerates), how can I think, postulate and further believe that force = multiplication of mass and its acceleration ?

• That's one of Newton's laws, which are to be taken as axioms in Newtonian mechanics. It also seems to describe the world around us pretty well. I don't know what you want. – ACuriousMind Mar 28 '15 at 17:19
• That equation should be the definition of either mass or force. – Jerry Schirmer Mar 28 '15 at 17:26
• It is worth remembering that physics (like all science) is at its heart descriptive. We seek a set of rules that allow us to correctly describe the world. And while we have some very powerful chains of reasoning that allow us to explain why certain complicated phenomena happen in terms of simple rules the foundation that every single one of those chains rests on is "because the world is observed to work this way". The reason for $\vec{a} = (\sum \vec{F})/m$ is "because the world is observed to work this way". – dmckee --- ex-moderator kitten Mar 28 '15 at 17:32
• @JerrySchirmer No. Both mass and force (and acceleration) are defined as independent measurable quantities. A balance measures mass, a spring scale measures force, and a meter stick and clock measure acceleration. Newton's Second Law relates the three quantities. (This leaves open the distinction between inertial mass and gravitational mass, but separate experiments can show that they are numerically the same.) – garyp Mar 28 '15 at 17:58
• @garyp: a spring scale measures force only because there is a known theoretical construct relating force and acceleration. At heart, force is "amount of push", but how do you define this? You can't, without making some appeal to a deeper law. Your statement about inertial and gravitational mass belies this -- gravitational mass isn't a thing without a gravitational law. There is no a priori reason that there should be such a thing as "gravitational mass" – Jerry Schirmer Mar 28 '15 at 18:13

I'm going to give you another approach.

Suppose that you where to perform the following experiments:

1.)

a) You happen to find a ball that is not moving, well its going to stay like that if we are in a place where the ball is not interacting with anything (air, gravity, etc). Like this : b) If You happen to find a ball that is moving with constant velocity, that is it moves in a straight line and it moves equal distances in equal times. Like this What we find is that if we are in a place where the ball is not interacting with anything (air, gravity, etc). Something like this will happen Video of ball over ice. The ball will keep moving forever.

2.) Now suppose that the ball is not moving and you shot this arrow to move it, you have the same conditions as before :  As you can see the ball wasn't moving and now it is moving, so it accelerated. Suppose it has an acceleration $a_1$.

Now you look for a Weight scale and you find the mass of the ball: You find that the mass is $m_1$.

Now you repeat the experiments with two balls of different masses, $m_1$ and $m_2$. That is you shot two arrows exactly the same. As you can see the acceleration $a_2$ of the bigger mass is less that $a_1$ the one of the smaller one. If you multiply the two masses by its corresponding acceleration you find something 1 $S_1$ and something 2 $S_2$ , that is:

$$S_1=a_1\times m_1 \text{ and } S_2=a_2\times m_2$$

If you compare $S_1$ and $S_2$ you find that they are equal (BECAUSE OF EXPERIMENT !!):

$$S_1=S_2$$

This makes you think that the only thing that was the same in your experiment was the fact that you shot the arrows with the exact same conditions, so that thing that was the same is that we call force $F$. And we define that $$F=S_1=S_2$$

We say that the arrows hit the balls with a force $F$ and we find that useful because it is a quantity that let us predict what is going to happen, that is what is going to be the acceleration.

So from that we have experimental evidence that

$$F=m_1\times a_1=m_2 \times a_2$$

If we consider some arbitrary mass and force, we generalize to this with the known equation:

$$F=ma$$

Let's break this down in two pieces.

First - the proportionality of force with mass. Imagine you need a certain force $F$ to accelerate a mass $M$. Now imagine that instead of one mass, you have two - each with mass $\frac{M}{2}$. Attach a piece of string to each. The total force on the two strings must be the same as before, so each string sees half the force for half the mass. This should convince you that $F\propto M$.

Now for the acceleration part. Imagine the same mass as before, but now it is mounted on a train: If the train is accelerating with acceleration $a/2$, you feel a certain force on the string holding the mass which is accelerating with $a/2$ - and the same force on the string pulling the train. If the train is standing still, you can "reel in" the mass, and still need the same force. But once again, you will need to apply that force also on the outside of the train in order to prevent it from moving.

Now if I try to accelerate the mass inside the accelerating train, I have to add these forces together. There is the force I need to move the train+mass with acceleration $a/2$, and the force I need to counter the force of accelerating the mass relative to the train.

From the above it follows that each of these is the same force - so when I double the acceleration, I double the force needed. It follows that force must be proportional to acceleration.

Putting these two together, you find

$$F \propto M\cdot a$$

The above only proves that $F$ is proportional to both mass $M$ and acceleration $a$ - the fact that it's actually $F=M\cdot a$ and not $F\propto M\cdot a$ is a matter of definition.

• I realize the second argument needs to be constructed a bit more carefully. I will be back when I can make a diagram... – Floris Mar 28 '15 at 18:11
• +1. The same approach of yours was followed by A . P. French, last century's MIT physics professor in his book. – user36790 Mar 28 '15 at 18:44
• @user36790 - do you happen to know what book it was? – Floris Mar 28 '15 at 18:55
• I do have that book; this is MIT introductory book on mechanics. You can go here. – user36790 Mar 28 '15 at 19:05

If a particle is left alone in an inertial system, it will travel on a straight line, that is $$\frac{d\vec{v}}{dt} = 0 \tag1$$ it will not change the magnitude and direction of its velocity. This is an empirical fact. So if a particle does change its velocity the right hand side of (1) is not zero. Lets call it $\vec{A}$: $$\frac{d\vec{v}}{dt} =: \vec{A} \tag2$$. This is rather a definition for $\vec{A}$, because one can only measure the change in position and then assign a value for $\vec{A}$. Lets assume we know the cause of $\vec{A}$, e.g. because we applied the action of spring under tension to our particle. Furthermore we observed that $\vec{A} = \textrm{const.}$. If we now apply the same action to a second particle of double the mass what we will empirically observe is $$\frac{d\vec{v} \ '}{dt} = \frac{1}{2} \vec{A}$$. This suggests that $\vec{A}$ is not an intrinsic property of the action (since we changed the particle and not the action) but rather $m\cdot\vec{A}$ is. One can now repeat that process with particles that don't differ in mass but for example in color, volume, etc. What one will find is that really only $\vec{A}$ and $m$ matter. Lets give that intrinsic property of action a name and call it force: $$\vec{F} := m \cdot \vec{A} \ \underbrace{=}_{(2)} \ m \frac{d\vec{v}}{dt} = m \vec{a}$$ with $\vec{a}$ the acceleration of the particle. So the formulation of the equation of motion is the desire to find an expression with quantities that belong to delimitable subsystems only. In this sense $\vec{F}$ is an abstract quantity that is by definition equal to $m\vec{a}$.

The acceleration is the effect on the objects motion due to the cause(s), which is the vector sum/total of the forces acting on the object, i.e. the total force (composed of many forces) is responsible for the acceleration (which is just one vector). The mass is just the constant of proportionality that links total force and acceleration.

Simply put, it's a correlation between a certain action (applying force on a given mass) and reaction (gaining or losing velocity). The mass is a just a factor that was found to give accurate predictions in the world Newton lived in (relativity theory introduced a slight modification).

Both force (over time or distance) and velocity represent energy, that's why there's a direct relation between them.

This relation is held via the multiplication operator because of the type of proportionality between these qunatities, as observed in nature.

Force comes to play when the body is moving with changing velocity.

To quote one author's statement is rather noteworthy:

Force is the agent that is at the forefront for the initiation of the motion & also responsible for change in the velocity.

The famous relation can be deduced intuitively by means of calculus.

If $v$ is the velocity of a body of mass $m$ at an instant $t$, then its momentum at that instant is given by $$p = mv$$. The time rate change of momentum of the body at that instant is $$\dfrac{dp}{dt} = \dfrac{dmv}{dt}$$. Acc. to Newton's second law, $$P \propto \dfrac{dp}{dt} \implies P = k. \dfrac{dp}{dt} = k. \dfrac{dmv}{dt}$$ where $k$ is the constant of proportionality. Since, mass $m$ of the body is constant(it need not be, it can change & this can be given as a function of time), we may write, $$P = k. m\dfrac{dv}{dt}$$. But $$\dfrac{dv}{dt} = \text{acceleration of the body} = f$$. Hence, $P = k. mf$. Now, if we agree that force that produces unit acceleration on unit mass will be called unit force ($P = 1$), then the equation yields: $$1 = k\cdot 1\cdot 1 \implies k = 1$$.

An interesting historical fact, often overlooked, is that Newton's own statement of the basic law of mechanics was not in the form of $P = mf$; it appears nowhere in Principia. Instead, Newton spoke of the change of "motion" by which he meant momentum & related this to the value of the product of force & its time of application. In other words, Newton's version of the second law was essentially $$P\cdot\Delta t = m\Delta v$$.

I would also like to jot down A.P. French's way to prove the law:

A.P. French's deduction of 2nd law:-

Let an object is placed on a horizontal table pierced with holes through which air is blown from below. It is then possible to pull horizontally on the object & make observations as the following:

1. A spring, stretched by a constant amount, causes the velocity of the object to change linearly with time - the acceleration produced in a given object by a given force is constant.

2. If a second spring, identical with the first & stretched by the same amount, is used side by side with the first, the acceleration is doubled. That is, if we take a known multiple of a force, then the acceleration produced in a given object is directly proportional to the total force.

It thus becomes possible to write down simple equations expressing the above relation: $$a = k.F \quad \text{or} \quad F = k' .a$$ where $k \quad , \quad k'$ describe the inertial properties of the particular object. Now, which result is more convenient??

Let's see:

If we place on the first object a second, identical object, it is observed that all accelerations produced by given arrangements of the sprins are reduced to half of what was obtained with one object alone. We can express this most easily by choosing the second equation, so that the inertial property is additive. $$F = (k'_1 + k'_2) . a_{1 + 2} \implies a_{1 + 2} = \dfrac{F}{k'_1 + k'_2}$$. It is by such steps that one can be led to the equation that is universally known as "Newton's (second) law": $$F = ma = m\dfrac{dv}{dt}$$ where $m$ is the inertial mass of the object identical with $k'$ above.

• I think you need a dot over the $P$ in your last equation, right? – Physics_Plasma Mar 28 '15 at 19:32
• Who are you quoting? – innisfree Mar 28 '15 at 19:44
• This is circular reasoning. If one defines $p=mv$ and postulates $dp/dt \propto F$ then sure $F=ma$. So $F=ma$ because $F=ma$ ... – image Mar 28 '15 at 21:17
• @Marcel Köpke: Newton's second law is an established fact. Here, we are not questioning about the validity of second law but rather we have to find the relation between force and acceleration. Now, one can ask whether acceleration is proportional to force or vice versa. So, this is not a circular reasoning . You are not proving the second law but rather using it to find the quantitative picture of the law. – user36790 Mar 29 '15 at 1:17