As far as my knowledge is concerned, a vector quantity should possess magnitude and direction & more over it should also obey the laws of vector addition.

As we all know that the vector sum of 3 newtons in the x direction and 4 newtons in the y direction will acting at a point will produce a resultant of 5 newtons. Where did the remaining 2 newtons go?

I mean we have applied a total of 7 newtons of force on a point sized particle but the output is only 5 newtons, so it appears as if a 2 newton force is disappearing here. In which other form does it reappear rather than the resultant? Or is it something like the remaining 2 newtons of force just vanishes and it doesn't appears in any other form?

Please correct me if I am going wrong on this issue.

  • 1
    $\begingroup$ Agha, I've correct what I think is a typo in your question i.e. "product" should be "sum". If I've got this wrong please roll back my changes. $\endgroup$ Aug 9, 2014 at 18:49
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    $\begingroup$ Since the OP's first paragraph refers to vectors having both size and direction, why don't the vectors in the second paragraph have their directions defined? $\endgroup$
    – DJohnM
    Aug 9, 2014 at 18:50
  • $\begingroup$ @JohnRennie thank you for that edit and sorry it was a typo , am i misconcepting this really $\endgroup$ Aug 9, 2014 at 19:02
  • 8
    $\begingroup$ Probably easier to consider 3 N of force in the pos x-dir and 3 N of force in the neg x-direction. Where did the 6 go? $\endgroup$
    – BMS
    Aug 10, 2014 at 0:59
  • 2
    $\begingroup$ Agreed with @BMS; easiest way to think about it. Now change the direction a little bit, bit by bit, and it won't seem so counter-intuitive that at 90°, you could still say that the forces kind of cancel out (though, more correctly, don't fully add up). $\endgroup$ Aug 10, 2014 at 9:58

7 Answers 7


There are some quantities that only come with one sign, for example mass/energy. So in any process if you start with some mass/energy $m$ then whatever happens you can't end up with less than $m$.

But other quantities come in both positive and negative magnitudes, and these will cancel. If you add a velocity of 10 m/s North to a velocity of 10 m/sec South then obviously they will cancel and sum is zero. You don't ask "hey, what happened to the 20 m/s I started with" because we all know equal and opposite velocities cancel. Likewise accelerations. If you add an acceleration of 9.81 m/s$^2$ up to an acceleration of 9.81 m/s$^2$ down then the total acceleration is zero. Again you wouldn't be puzzled as to where the 19.62 m/s$^2$ you started with have gone.

But remember that force is just mass times acceleration (it's one of Newton's laws though I always forget which). So if accelerations cancel it shouldn't be a surprise that forces cancel as well.

To see why you're getting cancellation in your example consider the simpler example of adding two 1N forces at right angles to produce a force of $\sqrt{2}$ N:


To see what is going on first rotate the diagram 45º:


and then split the vectors $F_a$ and $F_b$ that you're adding together into $x$ and $y$ components:


Now it should be obvious what is going on. The total $y$ component of $F_a + F_b$ is $F_{ay} + F_{by}$, and these components point in the same direction so they add together. The total $x$ component of $F_a + F_b$ is $F_{ax} + F_{bx}$, but these components point in opposite directions so they cancel and sum to zero.

And that's how two 1N forces can sum to less than 2N. It's because the two forces have a components with the same sign that sum and components with opposite signs that cancel.

  • $\begingroup$ i think i am very near to vanish away confusion now but what i am thinking now is we have done some work to produce a total force of 7 newton but the result only has 5 newtons so now in which form this force reappears as i have done some work to produce that force and i am getting the resultant less than that $\endgroup$ Aug 9, 2014 at 19:21
  • $\begingroup$ okay now i got you, you mean that as mass remains same and the magnitude of force is decreased so obviously the acceleration must be increased in the new resultant $\endgroup$ Aug 9, 2014 at 19:25
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    $\begingroup$ @agharehanabbas: I've extended my answer to try and make it clearer. $\endgroup$ Aug 9, 2014 at 19:29
  • $\begingroup$ thanks a lot for your explanation, now i got my doubt cleared $\endgroup$ Aug 10, 2014 at 6:36
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    $\begingroup$ Splitting vectors like this requires understanding of vector addition. $\endgroup$
    – gronostaj
    Aug 10, 2014 at 16:30

Fear not! You are not alone in your confusion. While it is common these days to teach that forces are vectors and follow vector addition (as the other answers beautifully present it), historically, you have hit upon a quandary that lay at the center of a dispute throughout much of the 19th century, involving such characters as Newton, Lagrange, Heaviside, Poisson, Maxwell, Bernoulli, De Morgan, Young, Earnshaw, to name a few you might recognize.

For a full account of all of the back and forths that transpired and a full analysis of the weights of the arguments presented, please see A Tale of Two Vectors by Marc Lange (2009) [doi][pdf], which served as my introduction to the controversy, and reference for preparing what follows.

Just as you hint in your answer, that forces ought to add as vectors is entirely intuitive, and so it is not well known who first came up with the idea. Throughout time, everyone agrees on what the answer should be when you add two forces (as you can readily test in everyday life), but the dispute lay in exactly why it should take the form that it does. There are hints that it appeared in a lost work of Aristotle (384 - 322 B.C.E), and it definitely appeared in Heron's Mechanics (first century A.D.) [cite]

Newton and dynamics

But it was in Newton's Principia (1687) [wikisource] that we see the first proof, right at the top in Corollary I:

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.

Plate I Figure I from Principia

If a body in a given time, by the force $M$ impressed apart in the place $A$ should with a uniform motion be carried from $A$ to $B$; and by the force $N$ impressed apart in the same place, should be carried from $A$ to $C$; complete the parallelogram $ABCD$, and by both forces acting together, it will in the same time be carried in the diagonal from $A$ to $D$. For since the force $N$ acts in the direction of the line $AC$, parallel to $BD$, this force (by the second law) will not at all alter the velocity generated by the other force $M$, by which the body is carried towards the line $BD$. The body therefore will arrive at the line $BD$ in the same time, whether the force $N$ be impressed or not; and therefore at the end of that time it will be found somewhere in the line $BD$. By the same argument, at the end of the same time it will be found somewhere in the line $CD$. Therefore it will be found in the point $D$, where both lines meet. But it will move in a right line from $A$ to $D$, by Law I

So, here Newton proves that force addition should take the form of vector addition (vectors won't be invented for a good 150 years), and gives as his proof a dynamical proof, one founded on the ideas of dynamics. He uses the motion that would be created by the forces on some hypothetical mass $m$, by his law of motion $\vec F = m \vec a$ to prove how forces should add given we know how the paths add.

This would give birth to the controversy, between the those in the dynamic camp, and those in the static camp.

I think Bernoulli gave the most compelling damnation of the dynamical proof, in which he points out that even if were to live in a universe where Newton's law didn't hold, but instead some other law $\vec F = m \vec v$, say, forces would still add by the parallelogram law, so the ultimate reason that forces add in this way must be independent of dynamics itself.

†: Here Daniel Bernoulli, the one of the many famous Bernoullis responsible for Bernoulli's principle

Duchayla and statics

As an alternative to the dynamical explanation of Newton, the static proof usually given is ultimately traced back to Charles Dominique Marie Blanquet Duchayla (a name if I ever saw one) in 1804. In which his proof proceeds by induction, with the inductive step:

If forces $P$ and $Q$, acting together at a point, result in a force directed along the diagonal of the parallelogram representing the two forces, and likewise for forces $P$ and $R$ acting together at the same point, with $R$ acting in $Q$'s direction, then likewise for $P$ together with the resultant of $Q$ and $R$.

Duchayla diagram

And proof of the inductive step:

Let $P$ be represented by segment $AB$. Grant that the resultant of $Q$ and $R$ is in their common direction and equal in magnitude to the sum of their magnitudes; let it be represented by segment $AE$, with $Q$ represented by $AC$, so that segment $CE$ is the proper length and direction to represent $R$ except that $R$ is actually applied at $A$ rather than at $C$. Nevertheless, when a force acts on a body, the result is the same whatever the point, rigidly connected to the body, at which it is applied, provided that the line through that point and the force's actual point of application lies along the force's direction. So although $R$ is applied at $A$, its effect is the same if it is applied at $C$, since $AC$ is in the forces' direction. Continuing to treat the parallelograms in figure as a rigid body, we can move the three forces' points of application to other points along the forces' lines of action without changing their resultant. We cannot move $P$'s point of application directly to $C$, since $AC$ does not lie along $P$'s direction. But by hypothesis, the resultant of $P$ and $Q$ acts along diagonal $AD$, so the resultant can be applied at $D$. It can then be resolved into $P$ and $Q$, now acting at $D$. $Q$'s direction lies along $DG$, so $Q$ can be transferred to $G$. $P$'s direction lies along $CD$, so $P$ can be transferred to $C$, where it meets $R$. By hypothesis, their resultant acts along diagonal $CG$, so it can be transferred to $G$, where it meets $Q$. By the converse of the force transmissibility principle, $AG$ must lie along the line of action of the force resulting from $P$ composed with resultant of $Q$ and $R$...

Quotation and figure reproduced from Lange's paper

The proof goes on from here to establish that you can also demonstrate that it gets both the directions and magnitudes right, for a full reproduction, see Lange's paper.

This proof is a bit hard to read, and reviews were mixed, some thought it was the best thing since sliced bread, "very simple and beautiful", such as Mitchell, Young, Imray, Earnshaw and Pratt, other's, not so much:

forced and unnatural... a considerable waste of time

Besant 1883, Lock 1888

the proof of our youth... now voted cumbrous and antiquated, and only retained as a searching test of logical power

A.G.G. 1890

brainwasting... elaborate and painstaking, though benumbing

Heaviside 1893

certainly convincing...but...essentially artificial...cunning rather than honest argument


Poisson and symmetry

But not all is lost. Poisson offered an alternative proof of the static case in 1811, which is based on symmetry arguments, dimensional considerations and a uniqueness constraint. I quite like this version. You start by assuming you have two forces of equal magnitude $P$ but different directions, separated by an angle $2\theta$ (The red forces in the diagram). Invoking rotational invariance and symmetry, the resultant $R$ (black) must bisect the two in direction, and its magnitude must be a dimensionally consistent formula of the magnitude of $P$ and the angle $\theta$ alone, or of the form: $$ R = P g(\theta) $$ with some unknown function $g(\theta)$. To figure out the function, Poisson then considers two new problems:

Poisson's Explanation

At the top and bottom, we've set up two new copies of the same task, now adding pairs of forces $Q$ (blue) to create the $P$s, and can see that $$ P = Q g(\phi) \implies R = Q g(\theta) g(\phi) $$ Meanwhile the two inner $Q$ forces (slightly darker blue) and two outer $Q$ (slightly lighter blue) forces set up the same scenario, all four of them adding to the resultant $R$, and each pair a part, and since we assume that forces in the same direction just add in magnitude, we have also $$ R = Q g(\theta + \phi) + Q g(\theta - \phi) $$ together: $$ g(\theta) g(\phi) = g(\theta+\phi) + g(\theta-\phi) $$ the only solution to this functional equation are solutions of the form: $$ g(\theta) = 2 \cos (\alpha \theta) $$ with an unknown $\alpha$, but we can fix $\alpha = 1$ by requiring that equal forces in opposite directions cancel. So finally we recover the force addition formula for equal magnitude forces: $$ R = 2 P \cos \theta $$ cast in polar form. From here it is easy to generalize to the full vector addition results by decomposing forces into various pieces.

I'm quite fond of this proof as it relies only on symmetry and dimensional considerations, and is completely free from any discussion of dynamics. But, some people equate this to a deficit:

Many have been puzzled by finding that the thing which, by its very definition tends to produce motion, is reasoned on... under a compact that any introduction of the idea of motion would be out of place. The statical proofs... seem to be all geometry and no physics

Augustus De Morgan (1859)

But others find it quite elegant:

The proof which Poisson gives of the "parallelogram of forces" is applicable to the composition of any quantities such that turning them end for end is equivalent to a reversal of their sign

Maxwell in 1873

Onto the philosophical

Nowadays, people continue to argue over the force addition law's origins, but mostly in philosophy journals. But In the second half of Lange's paper, he summarizes these various philosophy arguments and comes to the rough conclusion that in the end, whether you want to believe a dynamical or static origin is sort of question of interpretation, similar to whether you want to believe Newtonian Mechanics of Lagrangian mechanics. It's up to you to decide for yourself. The historic lineup of heavyweights is roughly:

In the dynamic camp:

  • Newton
  • Lagrange
  • Heaviside
  • De Morgan

In the static camp:

  • Poisson
  • Maxwell
  • Bernoulli
  • Young
  • Earnshaw

Where will you stand?

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    $\begingroup$ +1, for the historical background, this by itself, apart the physics, is worth alot $\endgroup$
    – Nikos M.
    Aug 9, 2014 at 23:59
  • $\begingroup$ thanks a lot for this elaborate and rhetoric answer, and yes now my doubt is cleared $\endgroup$ Aug 10, 2014 at 6:43
  • $\begingroup$ i have another thing in my mind, to apply some force some work must be done probably what about the work done in producing 7 newtons, is it totally conserved or in whcih other form it reappears $\endgroup$ Aug 10, 2014 at 13:45
  • $\begingroup$ 1) could you please explain what AB and AC stand for in the Newton picture? the whole post is not comprehensible if it is not specified what arrows stand for in various sketches, is it space, v or a? 2) you are interpreting Newton diagram with current F= ma, which is not Newton's, his second law was simply $F \propto v$ $\endgroup$
    – user59485
    Nov 9, 2014 at 6:06
  • $\begingroup$ @Alba 1) The figure and accompanying text is not mine, it's Newton's. He uses the parallelogram to describe both motions, and forces. 2) The included text is Newton's own proof of the addition law for forces, I don't really have to interpret it much at all. The point was just that his proof is a dynamical one, wherein he shows how forces would compose by showing how the motions induced by those forces would compose. $\endgroup$
    – alemi
    Nov 9, 2014 at 15:37

If I walk 3 kilometers East and 4 kilometers North, how far have I gone from my original position? Only 5 kilometers. Where have the other 2 kilometers gone?

If I walk 7 kilometers East and then 7 kilometers West, how far have I gone from my original position? Nowhere. Where have all 14 kilometers gone? This situation avoid vectors entirely. But it goes to show you can break up the sum ($0$) arbitrarily ($7+(-7)$, $-3+2+1$, etc.), but that doesn't change the sum.

The fact is, the displacement from the initial position, calculated after all is said and done, is not obliged to match the length of the path walked. With forces, it is the same - we want the sum of all forces that acted on the system, when all is said and done.

In math notation, we are interested in the quantity $\lvert \vec{F}_1 + \vec{F}_2 + \cdots\, \rvert$, not $\lvert \vec{F}_1 \rvert + \lvert \vec{F}_2 \rvert + \cdots$.

  • $\begingroup$ thanks for you explanation,it totally cleared my doubt on displacement vectors, but i think i am again in confusion, i mean to produce a force of 7newtons (say) we have to do some work on the system which is at rest i.e. point sized particle in this case but we did a work with 7 newton and now we are getting the result with 5 newtons only how does this happen, $\endgroup$ Aug 9, 2014 at 19:17
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    $\begingroup$ @agharehanabbas, depending on the details of the problem, you will either recover the energy you spent or lose it to friction. E.g. applying a forward force to a car initially at rest, force·distance is positive, but applying the same force backward to bring it to a stop again, force·distance is negative ("regenerative braking"). In the idealized case, these will be equal and opposite and you'll recover everything. In the case of 3N and 4N forces at right angles, you are doing less work than you think because a force perpendicular to the velocity does no work. $\endgroup$
    – benrg
    Aug 9, 2014 at 20:37

I think others have answered the question with regards to the vector addition of forces very well and alemi gave a fascinating insight into the historical "controversy" surrounding the question of a dynamic vs. a static "proof". Much of his exposition was new to me and I thank him for the valuable historical insight.

I would like to add an important comment to this: there are no mathematical proofs in physics, at all! The one and only instance that "proves" anything in science is nature, by answering scientific experiments. One can therefor not "prove" geometrically or analytically that force addition follows a vector addition law. That has to be done experimentally by applying multiple forces to an actual test mass and by measuring its actual acceleration. As long as such experiments, within the confines of the validity of Newtonian mechanics and the experimental error bars confirm, that forces do add as vectors, we can use the vector addition law. Once we find an experiment that does not, we will know, that we have identified a (potentially new) sector of nature, for which Newtonian mechanics does not hold.

  • $\begingroup$ This is a useful addition to this thread. But, while this is part of the story we tell ourselves, you have to admit: for there being no proofs in physics, our books and papers sure are full of a lot of them. The practice of physics seems more than just being constrained by experimental observations, we also are unsatisfied unless we have some kind of reason why we should expect those facts, some form of "proof" by means of a consistent mathematical theory. $\endgroup$
    – alemi
    Aug 10, 2014 at 14:19
  • $\begingroup$ I agree, that many physics theory textbooks look like as if they were full of mathematical proofs. In reality, of course, those are not proofs in the sense of axiomatic mathematics, but merely derivative calculations. A well known example of this can be found in Landau and Lifshitz "Course on theoretical Physics, Volume 1: Mechanics", where the authors start by introducing generalized coordinates and the least action principle, after which all of Newtonian mechanics becomes a more or less straight forward derivation. $\endgroup$
    – CuriousOne
    Aug 10, 2014 at 15:00
  • $\begingroup$ I'd go further: do forces exist? Then it comes down to a philosophical question. BTW your last sentence refers to Popper's definition of science. $\endgroup$
    – anderstood
    Aug 17, 2014 at 7:32

(add my comment as answer)

Remaing newtons cancelled out since vectors have directions.

Remember also forces that do not do work when being vertical to direction of motion.

Lets say we have 3 newtons in 1 direction and 2 newtons in opposite direction, result is 1 newton where did all the other newtons gone? Cancelled out


I know this topic is old, but this is a great question and as an Engineering student, I am struggling with same confusion, questions and I hope my answer may help you:

As we all know that the vector sum of 3 newtons in the x direction and 4 newtons in the y direction will acting at a point will produce a resultant of 5 newtons. Where did the remaining 2 newtons go?

It did not go anywhere. I think, the question you are really asking and the objective of calculations you are actually making are two different concepts.

Imagine there is any object laying on a table with some constant friction term, pushed by 3N force in X direction and 4N force in the y direction, at the same time for some given amount of time. The final destination of an object would be the same/EQUIVALENT, as if you pushed by 5N in the 52 degrees (arctan4/3) direction from x axis. Is the Force needed to move object to destination point the same in 2 cases? No, it is not. Obviously in one case you would need 7 newtons of force and in second just 5 newtons. But net effect is the same. So the GEOMETRY MATTERS and the geometry is one of the pillars of physics & mechanics.

Also remember, adding up or cancelling out does not mean that force is not there (so it does not vanishes). It just means that the net effect of forces which act on a body, are added up to a single, resultant vector, because vector addition works and NET EFFECT on a body by resulting force is the same as NET EFFECT on a body by several different forces. But obviously, In 1 case you only have one resultant force with F newtons force, in second case there may be dozens of forces acting on various directions, but the NET EFFECT on a body would still be the same.

Can the body resist? That's different question. Definitely, if total of 5000 newton force vectors are acting on an egg with resultant force vector = 0 (i.e. all vectors are cancelling out), still an egg would crack. But the NET EFFECT of a 5000 newton forces acting on egg is the same as 0 newtons, as an egg does not move F=ma=0.

I hope this can help anyone confused.


***the 2N force was cancelled because of the force acting horizontally which i call frictional Force............but when the same situation is carried in a interstellar space free from gravity and other forces the mass will experience a total of 7N force because there is no friction to cancel the resultant forces.

  • $\begingroup$ The question has nothing to do with friction. It has to do with forces partially canceling because they're meeting at an angle and partially working against each other (vector addition). $\endgroup$ Sep 21, 2015 at 6:07
  • $\begingroup$ i hope u r not clear with forces..........please refer principia mathematica $\endgroup$
    – user93364
    Oct 11, 2015 at 17:54

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