# How can forces be added?

I know this sounds elementary; that is why it has taken me this long to ask this question.

I understand how forces can be added this way (above).

But I don't see how it can be added in this way (second picture). Why is the resultant force (vector) equivalent to the line between those two forces (vector a and vector b) ?

• What do you not understand exactly? Please consider editing the question and adding some more details. Sep 29, 2022 at 15:28
• It is more about equivalence. Think about pushing an object with such forces (4N in the y-direction and 3N in the x-direction). The resultant motion would be exactly the same if you pushed the object with a force of 5N with an angle equal to $$\tan^{-1}(\frac{4}{3})$$. Sep 29, 2022 at 15:38
• Experiments are just means to prove existing theoretical work. You can see that such addition of vectors is intuitive because you would see it in daily life. Physicists began to show the mathematical relations underlying such concepts. Therefore, you can think of it as a daily situation that is proved mathematically as I showed. You would only need trigonometry to prove vector resolution, and hence, addition, subtraction, etc. Sep 29, 2022 at 15:49
• Make up a couple of vectors, A and B. Calculate their sum S. Then, try plotting A and B, tip-to-tail on graph paper—see where it gets you. Try plotting B first, then A, also tip-to-tail. See where that gets you. (Hint, it should get you to S, either way.) The tip-to-tail method is just a graphical representation of vector addition. OTOH, if your question really is, "why can forces be added?" then maybe you should edit your question to clarify. Sep 29, 2022 at 15:51
• Since you seem to be essentially asking why force is a vector, physics.stackexchange.com/q/372380/50583 and its linked questions are potential duplicates Sep 29, 2022 at 16:35

It is self-evident that successive displacements add by the tail-to-head arrow method. Velocities, being rates of change of displacement, add in the same way, as must momenta, and therefore forces, being rates of change of momentum.

This is a hand-waving 'justification', but it worked for me as a better alternative to the parallelogram rule which, as a high school student, I had been presented with as if it were almost self-evident – but which seemed to me quite arbitrary.

Addendum: One has to realise, of course, that with the tail-to-head arrow method, the positions of the arrow-tails have nothing to do with the point(s) in space at which the forces act. The diagram performs a mathematical operation that could equally well be carried out on a calculator on the moon.

• This is the correct answer. I also remember feeling that this was left unexplained and not at all justified in high school physics. Sep 30, 2022 at 4:17
• @tobi_s Thank you. I'd also say that justifications in terms of components for adding forces seem to me circular, because the whole concept of a component surely derives from knowing how forces add ! Sep 30, 2022 at 7:39
• @tobi_s Be careful. This is not the correct answer if you are a mathematician, at least not without the scaffolding of the concept of the manifold. A mathematician can't simply move a vector. And applying this logic to a force has a problem: the forces being added add at a point. I don't apply the first force at one point and the second force at a different point. The value of the parallelogram picture is that it keeps all the vectors at one point. Of course, for most physicists, the mathematical niceties can be ignored. Still, "the correct answer" is a bit too strong. Sep 30, 2022 at 12:10
• @garyp But isn't that conflating "moving" the vector in a diagram with physically moving the application point of the force? The force vectors themselves are not attached to any point in space. For that you bring in the position vector indicating where the force is being applied. Sep 30, 2022 at 15:27
• @garyp I'd also add that from a physicist point of view, I'd consider this answer to correctly capture the intuition that we are trying to model mathematically. It turns out that a vector space is the mathematical object that captures the intuition in a precise way. I definitely agree that from the point of view of linear algebra or differential geometry, this answer is not rigorous, but physics isn't pure math, and if this intuition turned out to correspond to some other object than a vector space, we'd be teaching that object instead of vector spaces. Sep 30, 2022 at 15:38

Here is one way of looking at it that might help you. To add vector A to vector B in the diagrams you provided, take each vector and draw its horizontal and vertical component like in this image.

It actually doesn't matter which directions you take as horizontal and vertical, so do whatever is most convenient (as long as your are consistent for both vector A and B).

Then add the vertical components of vectors A and B. They are pointing in the same direction, so this is what you said you are comfortable with. Do the same for the horizontal components. Now you have the vertical and horizontal components of the resultant vector, and you can draw the resultant vector itself.

Mathematically:

$$\vec{A} + \vec{B} = \begin{pmatrix} A_x \\ A_y\end{pmatrix} + \begin{pmatrix} B_x \\ B_y\end{pmatrix} = \begin{pmatrix} A_x + B_x \\ A_y + B_y\end{pmatrix}$$

This shows to add two vectors, you add their components in each dimension. Which is what you are comfortable with. Using triangles (or parallelograms) are useful and ultimately equivalent alternatives.

• Edited, thank you. Oct 1, 2022 at 19:58

The idea of vectors addition, subtraction, resolution, etc. is found everywhere in life. Using basic trigonometry, one could show that some force $$F$$ that makes an angle $$\theta$$ with the horizontal would be equivalent to two forces. One force equal to $$F\sin(\theta)$$ in the y-direction and $$F\cos(\theta)$$ in the x-direction. This is reversible, which leads to the idea of vector additions and subtraction in diagrams (i.e. the tail-to-head method).

Hope that helps!

Let's rearrange this last situation as you are regarding force instead of displacement. The above diagram actually more likely represents displacement. The forces seems to be on seperate points. This is the diagram that should be.

Now, the forces (vectors) are at the same point and the resultant force have become the diagonal of the parallelogram. As we know that, the direction of force applied on a body is the same direction of the change of momentum of the body (acc.to Newton's 2nd law). But in this case, there are two different forces with different directions acting on a single body. So, in which direction the momentum of the body should change? Is it in the direction of vector $$a$$ or is it vector $$b$$? If it is vector $$a$$ then why not $$b$$ or the reverse?

The answer should be in both of them which is the direction of the diagonal. But why? It is true that both of the forces will do some work and because they are not opposite and so they will not sum up to zero. The object will go along the every direction of every non zero forces. But a condition can come in mind that the body will seperate in two parts and each of them will go to the direction of the different forces with different directions. Yes, it is possible and it happens in nature. Suppose the body is a ball of sponge and when you apply too much force from two different directions on it, it will break into two parts and each of them goes to the direction of applied forces. It is because the ball of sponge isn't rigid. But when the object is rigid it doesn't break and goes along the diagonal. So that it goes in the direction of both forces. Now we can consider the object as a undividable point.

The mathematical form of this paragraph is considered in the answer of electronpusher.

$$\vec{A} + \vec{B} = \begin{pmatrix} A_x \\ A_y\end{pmatrix} + \begin{pmatrix} B_x \\ B_y\end{pmatrix} = \begin{pmatrix} A_x + B_x \\ A_y + B_y\end{pmatrix}$$

The magnitude of resultant force can be calculated by the relation below.

$$R = \sqrt {P^2{} + Q^{2} + 2PQcos\theta}$$ Where, $$R$$ is the magnitude of resultant force, $$P$$ and $$Q$$ are the magnitude of applied forces (or magnitude of vector $$a$$ and $$b$$) and $$\theta$$ is the angle between force vectors.

N. B. It doesn't mean that there will always be only two forces. The number of forces may be any whole number.

And what happens for the first case? The both directions are same. So, the object goes along only one direction.

Both of the additions of force vectors mathematically are the laws of vector algebra. Both of the additions are useful in their manner. Both incidents are different from each other. So, we have to use different processes to get the resultant force vector.

Hope this helps.

You are right that this is a non-trivial fact about forces, and it can only be justified by appealing to experiment. There is some discussion of this in Chapter 7 of Spivak's Physics for Mathematicians, where Spivak points to some purported proofs of the parallelogram rule and describes their limitations. As far as I know, the only justification for the parallelogram rule is that it is an assumption of Newton's laws (one that we nowadays treat as completely obvious, but that Newton was aware was not at all obvious, and which he attempted to prove independently of the other laws). We can therefore appeal to the tremendous empirical successes of Newton's laws and use the parallelogram rule without concern.

Just vector algebra. Force are vector physical quantities. You can sum them. When you sum them, you need to apply vector sum

• I think the question could be re-phrased as "why can we treat forces as vectors" ;-). Oct 1, 2022 at 13:13

Consider the forces acting consecutively on an object of unit mass. There is a push (acceleration) and then another push. Assume a coordinate system with its y-axis aligned with the first push such that $$v_y =0$$ at $$y=0$$ and $$v_y =1$$ at $$y=1$$. Assume the second push is along the x-axis with $$v_x =0$$ at $$x=0$$ and $$v_x=1$$ at $$x=1$$. The total acceleration must be given by $$(1,1)$$. If, instead, the force in the x direction brings the object to velocity 10 then the acceleration vector would be $$(10,1)$$. You said that you understand addition of forces along the same direction. So, that's it. Forces add like vectors.

It may help if you consider forces as infinitesimal momentum changes. I was recently convinced (also read the comments) that they are equivalent: $$\vec F = d\vec p/dt.$$

This may help because momentum is, for a given mass, just velocity; it is much more intuitive that velocity is a vector and can be added that way. For example, one can have a platform moving at constant velocity, and on that platform a vehicle moving in a different direction: It is clear that the resulting velocity relative to a "resting" observer is the addition of the two velocity vectors.1 Any change to the momentum vector — equivalent to a force — is, obviously, a vector as well.

Thinking about it, it is possible to experimentally (and intuitively) show that forces are also vectors which can be added as such, e.g. with two springs pulling at an angle at the same object, which will then be accelerated according to the vector-addition of the two forces. And, of course, the change in momentum is, as we said, equivalent to the acting force anyway :-) — it is the force.

1 Nobody except Einstein would think it could be anything else.

• It should be velocity not speed. Speed is a scalar quantity. Oct 1, 2022 at 15:14
• @DebanjanBiswas Thanks. In German there is only one word and I thought it's the other way around in English. Oct 1, 2022 at 15:53

There is a simple answer. Forces add this way because they are vectors, and vectors add this way. It's just how their math works.

Of course, this is a bit of a cop-out. It's not quite satisfying because it starts from the assumption that forces are vectors. If they are vectors, it is easy to show they add this way, because they do. But why can we assume forces are vectors in the first place?

If we look at the mathematics, vectors have the following properties:

• $$a + (b + c) = (a + b) + c$$ -- The "associative property"
• $$a + b = b + a$$ -- The "commutative property"
• $$0 + a = a$$ -- There is a "zero" vector
• $$a + -a = 0$$ -- Every vector has an inverse
• $$1\cdot a = a$$ -- There is a "scalar identity" (multiplying by 1 doesn't change the vector)
• $$k(a + b) = ka + kb$$ -- multiplying by a scalar "k" distributes
• $$(j+k)a = ja + ka$$ -- multiplying scalars by a vector commutes.

These are the properties of vector spaces. They're just a set of properties that showed up often enough in mathematics that we gave it a name and studied them a bit. That's all.

So the question for us is "do we believe" that forces have all of these properties? We can go down the list and convince ourselves that they do. For example, the first one (the associative property) can be shown that if Alice joins Bob and Charlie in pushing on an object, the forces they exert are the same as if Alice and Bob join Charlie. The difference between those two is so small that we intuitively might not even realize they're different. The fact that those situations are equivalent lets us know that forces follow this associative property.

Likewise we can find that it doesn't matter if we consider the force by Alice first, then Bob, or if we consider the force by Bob and then consider Alice, the result doesn't change. That's why it is commutative.

Its reasonably easy to think through the zero force and the idea that every force can have an inverse. In fact, we see that inverse quite often in the concept of "an equal and opposite reaction."

Multiplying by scalars is pretty easy to work through too. If Alice and Bob are applying forces, and then they apply 5x as much force, that is identical to Alice applying 5x more force and Bob applying 5x more force (again, so natural that one might not even realize those are different). And we can see that if Alice applies 2+5 pounds of force to something, that is equivalent to Alice applying 2 pounds and Bob applying 5 pounds in the same direction as Alice.

If you accept all of these things as true statements about forces, then you can conclude that forces are indeed vectors. And after concluding that, you can use all of the imagery we see in this question/answer to handle addition of vectors, such as the parallelogram image. Why do they add this way? If you sit down and hammer on the math long enough, you will realize that that is the only way two vectors could add which satisfies the above 7 properties. Any other addition will cause one of them to break down.

So that's really how forces can be added. You determine that forces have the properties of vectors, and then you know they must add in the only possible way vectors can add.

Probably the most important part of this addition approach is that addition of vectors is commutative. $$a + b = b + a$$. In most visualizations, you will notice that there are two paths you can take through the addition. One starts with $$a$$ and adds $$b$$ to it. The other starts with $$b$$ and adds $$a$$ to it. By the rules of vectors, those additions must commute, meaning they must give you the same value. You are free to try to find another way to add vectors, but most of the intuitive ones you may think of will not have this important property.

It is a question of an equivalence relation: equipollence.

https://math.stackexchange.com/questions/3350499/equipollent-couple-of-points-to-define-vectors