I tried to ask this question in Electrical Engineering Stackexchange but was told I was better asking here.

Newton's Second Law of Motion states that the vector sum of the forces $\mathbf{F}$ on an object is equal to the mass $m$ of that object multiplied by the acceleration vector $\mathbf{a}$ of the object:


By this equation we may associate, among others, the following three properties with any object in newtonian mechanics:

  1. A mass $m$.
  2. The vector sum of forces $\mathbf{F}$ acting on it.
  3. An acceleration vector $\mathbf{a}$.

These properties are related by Newton's second law, as stated earlier. However, they seem to have a different status. We can change the mass of an object (by simply removing parts of the object) independently of the other two properties. We can also change the sum of forces on the object independently of the other two (e.g. by adding a new force). However we are unable to change the acceleration vector of the object directly, we would need to change either $\mathbf{F}$ or $m$ to do that.

This difference seems strange to me, because after all the equation treats all of these characteristics the same (we can choose to solve any of them by substituting for the other two). For example, in the model $V=RI$ of a resistor this doesn't happen. We can choose to alter directly any of the three properties of the resistor (the voltage across it, the current flowing through it or its resistance). Why is $\mathbf{a}$ in $\mathbf{F}=m\mathbf{a}$ different?

I know it's pretty common sense to us that we can't change acceleration directly (at least for me it is), but still the mathematical equation wouldn't suggest that. So is there a physical (theoretical) explanation as to why this third property (acceleration) can only be changed indirectly? Why does it seem to have a different status than the other two (in the sense that we humans can't alter it directly, only indirectly)?

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    $\begingroup$ Your reasoning (very insightful indeed) shows acceleration is not the same type of property as force and mass. It is simply a measure of (change of) movement being actually the result of the other two properties - mass and force. (Gravity is a little different, but then who knows what gravity is?) In the same way - a dent in a body is the result of a force working against the rigidity and the hardness of this body. But you cannot just produce a given dent and expect it will influence the force or the body's properties to match the dent. Certain things come first and others follow ... $\endgroup$ – bright magus Jul 12 '14 at 18:03
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    $\begingroup$ In other words, in real world F=ma is not just some computer-made equation where you can move a slide for one variable and expect the others will follow accordingly. :) $\endgroup$ – bright magus Jul 12 '14 at 18:21
  • $\begingroup$ Well I think even $p = mv$ is similar. You cannot directly change $p$. Just giving another example :) $\endgroup$ – rahulgarg12342 Jul 12 '14 at 18:30
  • $\begingroup$ If you interpret "we" more liberally, one CAN and does consider "applied accelerations" rather than forces. An excellent example : the kinematic boundary conditions applied to structures during earthquake ground motion are acceleration b.c.s $\endgroup$ – user_of_math Jul 13 '14 at 15:01
  • $\begingroup$ I don't see the difference between how you're treating V=IR and F=ma. Sure you can change the resistance of the object but once that is fixed the main way you change current is by changing voltage. $\endgroup$ – Jordan Jul 14 '14 at 21:12

Your observations are spot on.

I usually write Newton's second law this way: $\vec{a} = \vec{F}/m$. This form makes it clear that the law is a relationship between the dynamic variables force and mass, and the kinematic variable, acceleration. $F$ and $m$ describe the situation, $a$ is the result. Cause and effect, if you will. In fact, that's one reason why it's a law that can't be derived from some other principle. It is based on observation: with $\vec{F}$ and $m$ I always observe $\vec{a}$

That's why it seems like it has a different status: because it does have a different status. As @brightmagus points out, you've made a very astute observation.


My phrase "cause and effect" was poorly chosen. But still, force and mass are the setup, and acceleration is what is observed. I'm given a property of the system (mass), and its interaction with the environment is specified (force). What results is a change in the position (via a constant acceleration). I can't cause an acceleration without first creating a force. But mass exists regardless of force or acceleration, and force exists, well, actually gradients of potentials, exist independent of the mass of the test particle, and even the existence of a test particle.

Note also that the equation of motion of a particle, the fundamental question, is found starting with $\vec{a} = \vec{F}/m$ which leads to $\vec{v} = \left(\vec{F}/m\right)t + \vec{v}_0$, etc. (for constant force in this case).

Concerning Ohm's Law, the situation seems similar. I do not think it's true that any of the three parameters can be changed externally and independently. We have a property of the system (resistance), and an interaction with the outside world (a potential or EMF), and the result is the motion of charge carriers. I can create an EMF chemically with a battery, or electromechanically via a generator. But I can't think of a way to generate a current without first creating an EMF. One can model a lumped circuit element as a current source, but under the hood, I think it has to be an EMF source and a resistor. I can't think of any other way of generating a current, but I'm ready to be corrected! Current is what is observed given a voltage and a resistance.

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    $\begingroup$ I disagree with this. $F=ma$ is, from a certain point of view, and equation describing the allowed shapes of world-lines of things in space-time. You really don't have to think about cause and effect. Doing so assumes that time flows in a specific direction which is all kinds of problematic. Just like $V = IR$, Newton's law is simply a constraint equation. $\endgroup$ – DanielSank Jul 12 '14 at 19:44
  • $\begingroup$ @DanielSank Again, we're on different wavelengths. In what sense is Newton's second law a constraint equation? $\endgroup$ – garyp Jul 12 '14 at 20:17
  • $\begingroup$ If an object has mass $m$, Newton's law is an equation of constraint on $F$ and $a$ for that object. For example: if you want to shoot off a rocket and use the minimum amount of fuel, you can write down a cost function for the fuel as a function of the rocket's trajectory. When you do that, the laws of physics (ie Newton's law) appear as equations of constraint on the trajectory :) $\endgroup$ – DanielSank Jul 12 '14 at 22:49
  • $\begingroup$ garyp, man you were also "spot on" my doubt. I even had the same exact thought you had about V=Ri ( can't create current withouth creating voltage first ). Appreciate a lot for the input and sorry for taking so long to acept it. $\endgroup$ – nerdy Jul 16 '14 at 1:58

We should think a bit more carefully what force, mass and acceleration really are:

For simplicity, we consider a classical point particle.

The force $\vec F$ is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, $\vec F = - \nabla V$, but it need not be. In general, it is some function of position (and velocity, if friction is in play, and higher derivatives of position, if you've constructed a particulary weird example) function $F(x,\dot x,\dots)$. (If you want, let it also depend on time.)

The mass $m$ is an intrinsic property of the particle, it does not change, it is simply some real number.

The acceleration is the description of where the particle will go. It is $a = \ddot{x}(t) $. Newton's law is a differential equation that is inherently meant to be solved for $x$ since $a = \ddot{x}$, because if you know $x(t)$, then you know everything you could possibly want to know about the kinematics. Contrary to force, the acceleration is only defined on the trajectory of the particle, not everywhere in space, and thus not external. You cannot directly change acceleration because it is just the measure of the change of velocity $\dot x$, which is just the measure of the change of position $x$, which is just the space coordinate of the particle. And Newton's law tells you exactly how this change is determined.

Now, of course, if you know $a(t)$ and $m$, then you can trivially see what $F$ is, and if you know $a(t)$ and $F(x,\dot x,\dots)$, then you know $m$ after plugging $x(t),\dot{x}(t)$ and so on into the force. But the equation $\vec F = m\vec a$ is really an ODE whose solution determines the kinematics $\vec x(t)$.

(I apologize for inconsistent vector arrows.)

  • $\begingroup$ One could argue that a body moving by means of its own engine is not being exerted an external force on, but that's just an aside. :) $\endgroup$ – bright magus Jul 12 '14 at 19:56

In my view, an expression like V=RI, practically speaking, isn't much different. You have to be concrete with a real-world example. I can connect a resistor across a voltage source, and then I can only alter the voltage or change the resistance value to change the current; I cannot change the current directly in this configuration.

Your statement that $V=RI$ doesn't behave this way is only correct if you bias the circuit two different ways. For example, I can directly control the current if I use a current source in conjunction with the resistor, but then I can't directly change the voltage.

So, which variable can be directly modified is a question of topology.

$F=m a$ is seemingly different because of topology and practical considerations, but it's really the same. For example, if a box is free-falling, I can change the acceleration by dropping the box on Earth or on the Moon or on Mars (not as simple as dialing in a current, but philosophically, it's the same thing). In this topology, I control the acceleration and the mass, and the force is determined for me.

  • $\begingroup$ I'd say that if you drop a box on Mars you've changed the force, not the acceleration. $\endgroup$ – garyp Jul 12 '14 at 19:33
  • $\begingroup$ @garyp: I strongly disagree. If going to Mars were changing only the force, the acceleration of a given falling would depend on that object's mass. The acceleration for any falling object is the same, regardless of mass. $\endgroup$ – DanielSank Jul 12 '14 at 19:38
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    $\begingroup$ @garyp Semantics. The question has to do with controlling $a$ directly, independent of the other two variables. One way to pick $a$, independent of the other two variables, is to do what I said. I read the question as "you can't ever fix $a$ independent of the other two variables so $F=ma$ is different, and I'm providing you with a counterexample. $\endgroup$ – user3814483 Jul 12 '14 at 19:41
  • $\begingroup$ @user3814483: Gravity is a very bad example, because it is unique. And saying that deciding whether to dropp a body on the Moon or on the Earth means controlling the acceleration is simply false. One has no direct control over the particular acceleration of the Earth or the Moon. Something does produce this acceleration, but it is not you. This something is the mass. Therefore, you could have control of the acceleration by removing a part of Moon's mass. Which shows again that acceleration is the result of mass. $\endgroup$ – bright magus Jul 12 '14 at 20:05
  • $\begingroup$ @brightmagus: Whether a particular human person has control over a parameter isn't really a good anchor point for a discussion of Nature. $\endgroup$ – DanielSank Jul 12 '14 at 20:13

A better electrical analogy to Newton's second law might be inductance:

$$ V = L \frac{\mathrm d i}{\mathrm d t} $$

The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not exist. I'd then write Newton's second law as:

$$ F = m \frac{\mathrm d v}{\mathrm d t} $$

Looks similar, right? Feels similar in subjective terms of cognition also: force is like voltage, mass is like inductance, and velocity is like current.


I'm not sure if I'm just misinterpreting your question, but we can most certainly influence the acceleration of an object directly.

To do this we must take into consideration what is actually happening as we accelerate; Acceleration, in physics, is the rate at which the velocity of an object changes over time. While it is a sum of the net forces over the mass, for conceptual cases, this being one, I choose to think of acceleration in the calculus notion.

Mathematically, instantaneous acceleration—acceleration over an infinitesimal interval of time—is the rate of change of velocity over time:

$\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$, i.e., the derivative of the velocity vector as a function of time. (Here and elsewhere, if motion is in a straight line, vector quantities can be substituted by scalars in the equations.)

Average acceleration over a period of time is the change in velocity $( \Delta \mathbf{v})$ divided by the duration of the period $( \Delta t)$

$\boldsymbol{\bar{a}} = \frac{\Delta \mathbf{v}}{\Delta t}$.

So we can think about influencing acceleration by changing velocity over time.

We don't need to simply change the F or m to change the acceleration. However a common reason we think that is because Force ($f=ma$) is often defined as the ability to influence acceleration. And by that language we wouldn't think of changing the acceleration, but rather the mass or force to effect change in acceleration.

  • $\begingroup$ Like garyp said, i still don't think you can change acceleration ( change in velocity over time ) withouth first changing the resulting force on the object. $\endgroup$ – nerdy Jul 16 '14 at 2:00
  • $\begingroup$ @nerdy they are interconnected yes. But what I'm saying is, there's no precedence as to which one you change. You can also change the mass of an object, which would then allow you to change the acceleration without necessarily changing the force. $\endgroup$ – David BasedMathematician Coven Jul 16 '14 at 8:34

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