"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow

We learn it very early in the course of our study that Force is vector; But, if I were the physicist defining the the Newton's second law (experimentally) and analysing the result F=ma, how would I determine whether Force is vector or scalar(especially in 3-D).

Actually, when I read the aforementioned sentences from the book, I wanted to know why do the authors expect it to be natural for us to think that in 3-D "Force" behaves like a vector. I know a(acceleration ) is vector and mass a scalar and scalar times vector gives a new vector but is there another explanation for this?

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    $\begingroup$ I think the first evidence of force behaving like vectors is the Stevin law of the triangle of forces, published in De Beghinselen der Weeghconst (1586; “Statics and Hydrostatics”), based on an experiment with three dynamometers. $\endgroup$ – JPG Dec 5 '17 at 10:58

12 Answers 12


Uhm ... you start with an object at rest and notice that if you push on it in different directions it moves in different directions? Then notice that you can arrange more than two (three for planar geometries and four for full 3D geometries) non-colinear forces to cancel each other out (hopefully you did a force-table exercise in your class and have done this yourself).

The demonstration on an object already in motion is slightly less obvious but you can take the ideas here and generalize them.

In a sense this is so obvious that it's hard to answer because almost anything you do with forces makes use of their vector nature.

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    $\begingroup$ It is only obvious to people who are used to vectors. After a while you get so used to it you forget that it was confusing to learn. You forget what you did and didn't know at the time. This makes it hard to explain things well to beginners. E.G. safeshere's comment is correct. But someone who is wondering why force is a vector will also wonder why momentum is. I remember being confused that kinetic energy has an obvious direction, but it isn't a vector. $\endgroup$ – mmesser314 Dec 4 '17 at 3:21
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    $\begingroup$ Kinetic energy does not have a direction. The momentum of an object has a direction. An 500 g object moving at 2 m/s in the positive x direction does not have the same momentum as a 500 g object moving at 2 m/s in the negative x direction, but they both have the same kinetic energy. $\endgroup$ – Bill N Dec 4 '17 at 17:23
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    $\begingroup$ @BillN mmesser314 is aware of that, but it is a common enough misunderstanding among intro students (especially the more thoughtful ones). He criticizing the notion that "look this has a direction" is a good enough tool to give students for distinguishing vectors from non-vectors. I disagree because I'd rather deal with the question of kinetic energy than try to give introductory students a more abstract definition of 'vector', but it is a point worth considering. $\endgroup$ – dmckee --- ex-moderator kitten Dec 4 '17 at 17:28
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    $\begingroup$ @dmckee Yeah, I was hand-waving my way through Biot-Savart today trying to explain why the current,$I$, isn't a vector, but $d\vec{\ell}$ is. I almost choked while mumbling. :) That's still a non-satisfying vector for me, but I hold my nose and move on. $\endgroup$ – Bill N Dec 4 '17 at 17:43
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    $\begingroup$ @BillN I think that your KE example is a good example of why this can be tricky few newcomers to physics. I find it's not necessarily obvious that KE lacks a direction component until you've done a few experiments that show that there is a scalar "energy" worth paying attention to. $\endgroup$ – Cort Ammon Dec 4 '17 at 19:18

Vectors are things that add like little arrows. Arrows add tip to tail.

Number of rocks is not a vector. 2 rocks + 2 rocks = 4 rocks.

Displacement is a vector. If you move 2 feet left and 2 feet left again, you have moved 4 feet. Two arrows 2 feet long pointing left added tip to tail are equivalent to one arrow 4 feet long pointing left.

If you move 2 feet left and 2 feet right, you have moved back to the start. This is the same a not moving at all. You can't add rocks this way.

Force adds like this. Two small forces to the left are equivalent to a big force to the left. Equal forces left and right are equivalent to no force. This is why force is a vector.

Edit - The comments raise a point that I glossed over. This point is usually not raised when introducing vectors.

Mathematicians define a vector as things that behave like little arrows when added together and multiplied by scalars. Physicists add another requirement. Vectors must be invariant under coordinate system transformations.

A little arrow exists independently of how you look at it. A little arrow does not change when you turn so it is now facing forward. Equivalently, little arrows do not change if you rotate the arrow so that it faces forward.

This is because space is homogeneous and isotropic. There are no special places or directions in space that would change you or an arrow if moved to a new location or orientation. (If you move away from Earth gravity is different. If this matters, you must move Earth too.)

By contrast, a scalar is a single number that does not change under coordinate system transformations. Number of rocks is a scalar.

The coordinates that describe a vector change when the coordinate system is changed. The left component of a vector is not a scalar.

There is a 1-D mathematical vector space parallel to the left coordinate of a vector. If you rotate the coordinate system, it may be parallel to what has become the forward component. A physicist would not say it is a vector space.

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    $\begingroup$ What you explained, also matches a signed scalar. You should have included a "forward" or "up" movement to make it clearer. $\endgroup$ – Ralf Kleberhoff Dec 4 '17 at 17:24
  • $\begingroup$ @RalfKleberhoff - True. You raise a good point. $\endgroup$ – mmesser314 Dec 5 '17 at 14:53
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    $\begingroup$ @RalfKleberhoff How is a signed scalar not a vector in a single dimension? Really. This always confused me. It seems to have much, much more in common with vectors than scalars. $\endgroup$ – jpmc26 Dec 5 '17 at 23:35
  • $\begingroup$ @jpmc26 physics.stackexchange.com/questions/35562/… $\endgroup$ – Nico Dec 6 '17 at 12:41
  • $\begingroup$ @jpmc26 - Good question. I updated my answer to address it. $\endgroup$ – mmesser314 Dec 6 '17 at 14:27

A minor nitpick: force is not a vector. Like momentum, it is a covector or one-form, and covariant. You can see this in several ways:

  • from the principle of virtual work: force is a linear function mapping infinitesimal displacements $\delta\mathbf{x}$ (a vector) to infinitesimal changes in energy $F\delta\mathbf{x}$ (a scalar) and hence a covector by definition.
  • Newton's second law $F=ma$: acceleration is a vector, which is "index-lowered" by the mass to give force.
  • conservative forces arise from the differential of potential energy, $F = -dV$, and the differential of a function is a one-form (covariant).

The difference between a vector and covector may not make sense if you're just starting to learn about physics, and for now, knowing that forces can be "added tip to tail" like vectors may be enough for practical calculations. But it is something you should start to pay attention to as your understanding matures: like dimensional analysis, carefully keeping track of what your physical objects are, mathematically, is helpful both for building deeper understanding, and catching errors.

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    $\begingroup$ I think this is a useful comment because it illustrates that "this is the most natural way to think about force" is in fact not necessarily true. Covectors are quite natural things and you can imagine a curriculum that worked with them as much as with vectors. It is a tradition of our education system that we do not (at least explicitly). $\endgroup$ – Francis Davey Dec 4 '17 at 21:29
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    $\begingroup$ @FrancisDavey I would rather say that the tradition is that we do not make the distinction between vectors and convectors until way too late, and just call them all vectors. (I didn't learn the distinction explicitly until I took general relativity, or possibly quantum mechanics with bras and kets. It should've been explicit in the first linear algebra course, where they appeared as column vectors and row vectors, but it wasn't explicit.) $\endgroup$ – Arthur Dec 5 '17 at 6:30
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    $\begingroup$ Not worth a downvote, but most definitely not worth an upvote. I'm not thrilled by this "how things transform" definition of what constitutes a "vector". The mathematical definition of a vector is much simpler: Vectors are members of a vector space -- a space endowed with two operations, that obey eight simple axioms. By this definition, forces (in Newtonian mechanics) are vectors. $\endgroup$ – David Hammen Dec 5 '17 at 11:44
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    $\begingroup$ @DavidHammen A "vector" can mean either 1) a tangent vector, i.e. an element of the tangent bundle (or more generally, the (0,1)-tensors of a tensor algebra) or 2) an element of some general vector space. Usually in physics when we say "vector" we mean "(tangent) vector": we wouldn't call scalars, functions, 2-tensors, or indeed, covectors, "vectors" even though technically all are elements of a vector space. Notice that by definition #2, even the OP's "is force a vector or scalar" is a meaningless question! $\endgroup$ – user2617 Dec 5 '17 at 16:22
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    $\begingroup$ All of those things are genuine vectors. We don't typically call them vectors because that's not typically a useful feature. If you're using a different definition of "vector" it should be spelled out. $\endgroup$ – Era Dec 5 '17 at 20:05

Acceleration transforms like a 3-vector under rotations (group O(3)).

Acceleration transforms like a 4-vector under rotations and boosts (Lorentz group O(3,1)).

Acceleration may well be part of a larger structure (eg: 2 index tensor) under a larger group of transformations including rotations, boosts, strains, and translations.

My point being, when you say acceleration (or force) is a 3-vector (or something else), you have to specify for which group of transformations. For example, "acceleration transforms like a 3-vector under rotations", and that is why we call it a 3-vector.

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    $\begingroup$ This question was clearly about Newtonian physics, which the author doesn't fully understand. You're barging in with stipulations from much more complicated areas of physics (which the author may not even need). It's the equivalent of someone asking about Bernoulli's law and you asking them to specify if the fluid is viscous. Please explain the terms you use and match the level of technicality to the question. $\endgroup$ – Cody P Dec 4 '17 at 17:13
  • $\begingroup$ @CodyP Not barging in at all! Well, maybe group theory is a little higher than needed here, but ... The definition of a vector is intimately tied to how the quantity behaves under the rotation of coordinates. The fact that we simplify that idea to "magnitude and direction" doesn't remove the importance of understanding rotation of coordinate systems and what's invariant and what's not. That may be advanced, but that's essential to answering the OP. At the level of Kleppner and Kalenkow, the person should be introduced to a broader definition of vectors and coordinate rotations. $\endgroup$ – Bill N Dec 4 '17 at 17:32
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    $\begingroup$ @CodyP Questions on Stack Exchange sites aren't just for the OP. They are also a durable resource for later visitors on. Answers of varied level are a desirable thing, though Gary is unlikely to get the OP's acceptance. $\endgroup$ – dmckee --- ex-moderator kitten Dec 4 '17 at 17:34
  • $\begingroup$ True, but it's still valuable to understand your target audience, and define terms like boosts, tensor, or even "group of transformations". You can, for an analogy, talk about effects of viscosity in a question about Bernoulli's law, but doing so without care is more likely to sound pedantic and confusing than helpful and clear. $\endgroup$ – Cody P Dec 4 '17 at 22:03
  • $\begingroup$ @CodyP true, but maybe one day OP revisits their questions and understand this $\endgroup$ – Ooker Dec 5 '17 at 15:11

The real answer in my opinion isn't some underlying philosophical arguments about what a force is. The real answer is that thinking of force as a vector gives you a model that satisfies the single most important criterium for any model: it agrees with experiment. It is also nice and simple, which is an added bonus.

Thinking of forces as vectors will allow you to come up with predictions of what happens when you do experiments, specifically experiments where you apply several forces at once. For instance, put a crate on ice, and pull on it using ropes with spring scales embedded in them to measure the magnitude of all forces involved. Measure and write down all the forces and their directions, think of forces as vectors, and calculate the resultatnat force acting on the crate, which ought to give you a prediction of its acceleration. Then measure its actual acceleration. The two should agree, to within some error.

People have done experiments like this, both more and less sophisticated, for a long time, and so far we haven't found anything to indicate that thinking of forces as vectors gives the wrong result. Thus thinking of forces as vectors will most likely give accurate results the next time we need to calculate a prediction as well.

So we learn to think of forces as vectors because it works. And then philosophers can argue about why it works, usually by putting it into the context of a bigger picture, which has also withstood the test of experiments.

That being said, there are natural ways to come up with the idea of even considering that force is a vector. Specifically, each force has a direction and a magnitude. As pointed out in other comments, this doesnt necessarily mean that is must be a vector (kinetic energy clearly has a direction and a magnitude, but isn't usually thought of as a vector). But it is enough to ask whether it could possibly be a vector, and to start designing experiments around that hypothesis.

  • $\begingroup$ Changes in kinetic energy are scalar. There is no absolute kinetic energy; if an absolute kinetic energy is given as a vector, it is understood to be relative to a frame of reference, and basically indicates the amount of energy that would be converted if the given object were to stop moving relative to that frame. It cannot be treated simply as a vector; for instance two equal masses moving in opposite directions, at the same speed relative to the reference frame, do not add to zero kinetic energy. $\endgroup$ – Kaz Dec 4 '17 at 22:49
  • $\begingroup$ @Kaz Your "no absolute" comment also applies to momentum, though, so that's not a good reason as momentum has proven to be useful to think about as a vector. Also, "two equal masses moving in opposite directions, at the same speed relative to the reference frame, do not add to zero kinetic energy" I don't see the problem. The kinetic energy becomes internal energy if you consider the two objects as one system. The problem appears when you change to a moving reference frame, in which case the sum kinetic energy vector would become non-zero. That is not a good vector transformation property. $\endgroup$ – Arthur Dec 5 '17 at 6:24
  • $\begingroup$ (Of course it becomes non-zero. In just tired. The real problem is that which non-zero vector it becomes depends on the internal properties of the system. Are the two objects the same size and moving at the same velocity, or is one object bigger and slower? This affects the transformed energy "vector".) $\endgroup$ – Arthur Dec 5 '17 at 6:34

I had this question previously too and spent a good 5 hours on it. In the end, the explanation for this is just that the displacement acts like a vector. And acceleration being the double derivative of it also acts like one. Why does displacement acts like a vector?? Well, it follow the rules of trigonometry and displacements in one direction is independent of the displacement perpendicular to it. Hence, we define vector concepts to encompass this behavior. Why does displacement follow the rules of trigonometry?? Well, this has been more or less found by observing rather than deriving. The most fundamental basis of everything in maths is also observation and logic after all.


To get the droll bit out of the way: you know force is a vector from its definition.

To demonstrate that it really is, you would perform experiments: start by attaching three spring scales (like the ones fishermen use to weigh fish) to each other at the same point, and pull the other ends of the scales horizontally at 120 degree angles with equal non-zero force F. The configuration is in the beautiful ascii graphic below, and you can tell the forces are equal by looking at the readings on each scale.

 F ----- o

You'll also notice that the point of attachment in the middle stays stationary, that is, the net force is zero.

If F was a scalar, it would be impossible to add or subtract exactly 3 non-zero Fs in whichever order, and get 0 as a result.

Now that you know that force is not a scalar, you would then try to figure out a way to get the three Fs to add up to zero, and you notice that if you pair the direction of each spring to each F, you can get exactly that:

 F-----F   if you consider the direction each
  \   /    spring was pulled, you can rearrange
   \ /     the forces so that they form a loop,
    F      that is, they add to zero.

You'd then conduct further experiments, in various set-ups, and find that in each case, treating force as a scalar paired with a direction gives the correct result, at which point you would feel justified in saying: for the purposes of calculation, force has both a magnitude and a direction.

A vector, on the other hand, is nothing more than a magnitude paired with a direction, so you have experimentally showed that within the limits of measurement, force is a vector.


It depends on the nature of your approach, and on your interpretation of the word "vector". Conceptually, a spatial vector is a mathematical object used to encapsulate quantities that have both a magnitude and direction. When you apply a force to something, the net result on that object's motion depends not only on how hard you are pushing it, but also the direction you are pushing it, so it's necessary to model forces in a way that takes the direction component into consideration. This is just as true in three dimensions as it is in one. That's the simplest way to think about it.

From a mathematical perspective, as you've already mentioned, it's implicit in the definition.


"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow.

Newton himself made the vectorial nature of forces the first and second corollaries of his three laws of motion:

Corollary I:
A body by two forces conjoined will describe the diagonal of a parallelogram, in the same time that it would describe the sides, by those forces apart.

Corollary II:
And hence is explained the composition of any one direct force AD, out of any two oblique forces AC and CD; and, on the contrary, the resolution of any one direct force AD into two oblique forces AC and CD: which composition and resolution are abundantly confirmed from mechanics.

In short, forces are Cartesian vectors, in the mathematical sense of what constitutes a vector.

The derivation of those corollaries in the Principia is rather suspect. Newton's second law addresses the net force on object while Newton's third law addresses how individual forces come in pairs. But how to relate those individual forces to the net force? Unlike Kleppner and Kolenkow, other texts do a better job, stating that forces are vectors is in effect Newton's fourth law of motion.

A handwave response (e.g., Kleppner and Kolenkow) is to claim that forces obviously act as vectors, and then move on. A non-handwave response is to axiomatically claim that forces are vectors, and then move on. There's subtle but significant difference between these two responses. The handwave response leaves students confused. The axiomatic claim invites students to question the axiom. The next step is of course to test whether the axiom applies in a laboratory setting.


Actually, a physical force is not a vector. It is a line in 3D. A line with a magnitude. A physical force contains the following properties

  • Direction, $\mathbf{e}$
  • A point anywhere along the line, $\mathbf{r}$
  • Magnitude, $F$

To describe a physical force with a vector you combine the magnitude and the direction into $\mathbf{F} = F\, \mathbf{e}$ a single vector. But this still lacks the information needed to describe a physical force.

You also need a location (the point of application, or the line of action as it is called). Here you have a choice between an actual point $\mathbf{r}$, or the equipollent moment about the origin $\mathbf{M} = \mathbf{r} \times \mathbf{F}$. If you choose the latter, you can recover the point with $\mathbf{r} = \frac{\mathbf{F} \times \mathbf{M}}{\| \mathbf{F} \|^2}$.

The force vector that you are familiar with is commonly used because it obeys the vector algebra rules

  • Addition is done by component $$\mathbf{F}_1 + \mathbf{F}_2 = \pmatrix{{Fx}_1+{Fx}_2 \\ {Fy}_1+{Fy}_2 \\ {Fz}_1+{Fz}_2 }$$
  • Scaling is done by component $$\lambda\, \mathbf{F} = \pmatrix{\lambda\,{Fx} \\ \lambda\,{Fy} \\ \lambda\,{Fz} }$$
  • But the locations of two foces do not add up like vetors.

To represents physical forces with vectors you need 6 component quantities called screws $$\hat{f} =\left[ \matrix{ \mathbf{F} \\ \mathbf{r}\times \mathbf{F} } \right]$$ which do follow the rules of linear algebra and carry the positional information inside of them, producing the correct geometric and algebraic results.

  • $\begingroup$ Is this the n-th definition of a force "vector"? $\endgroup$ – jjack Dec 6 '17 at 0:21
  • $\begingroup$ Read this post for the definition of a screw vector. $\endgroup$ – ja72 Dec 6 '17 at 3:41

Let's think about what would happen if force were not a vector.

First, note that:

The laws of physics are invariant in space. An object behaves the same way when acted upon by a force whether it is in Paris or in Beijing.

Furthermore, we note:

The laws of physics are invariant under spatial rotation. Kicking a soccer ball will make it go away from you regardless of whether you are facing West or East.

Now imagine we applied a force to a ball resting on a table. Let's say we observe that:

The ball starts rolling east at a speed of 1 m/s.

Wait. Where did "east" come from? Why isn't the ball rolling west? Thus, we naturally conclude:

There must be some additional information contained in the force we applied to the ball.

That additional information is direction.


According to Newton's 2nd Law of Motion, the force acting on a body is proportional to the rate of change of momentum and is in the direction in which the force is applied. Now from the statement you can see that force has a magnitude and a direction. Hence it is a vector. You can even see it as the dot product of mass(scalar) and acceleration(vector) which will give you a vector.


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