# Given Newton's third law, why are things capable of moving?

Given Newton's third law, why is there motion at all? Should not all forces even themselves out, so nothing moves at all?

When I push a table using my finger, the table applies the same force onto my finger like my finger does on the table just with an opposing direction, nothing happens except that I feel the opposing force.

But why can I push a box on a table by applying force ($$F=ma$$) on one side, obviously outbalancing the force the box has on my finger and at the same time outbalancing the friction the box has on the table?

I obviously have the greater mass and acceleration as for example the matchbox on the table and thusly I can move it, but shouldn't the third law prevent that from even happening? Shouldn't the matchbox just accommodate to said force and applying same force to me in opposing direction?

• There are excellent answers below. I wanted to add that on the system scale (i.e. all objects together) the forces DO cancel out---that's why momentum is conserved. – DilithiumMatrix Dec 2 '12 at 2:49
• Duplicate of ? physics.stackexchange.com/q/2095 – Hobo Dec 2 '12 at 14:23
• Here's a point of view that helped me to "get" this question: If the matchbox didn't push back on your finger with equal force, your finger would go right through it as if it were a ghost! – wim Dec 3 '12 at 0:45
• Note that the acceleration of the object (i.e. matchbox) depends on its mass and the net sum of forces acting upon it. Crucially, it does not depend on forces which the object exerts upon other things (i.e. finger). – wim Dec 3 '12 at 0:47
• There must be hundreds of questions similar to this one, all as a result of physics teachers forgetting to insert the words "..acting on different bodies" when explaining the 3rd law. – ja72 Dec 20 '13 at 14:29

I think it's a great question, and enjoyed it very much when I grappled with it myself.

Here's a picture of some of the forces in this scenario.$^\dagger$ The ones that are the same colour as each other are pairs of equal magnitude, opposite direction forces from Newton's third law. (W and R are of equal magnitude in opposite directions, but they're acting on the same object - that's Newton's first law in action.)

While $F_{matchbox}$ does press back on my finger with an equal magnitude to $F_{finger}$, it's no match for $F_{muscles}$ (even though I've not been to the gym in years).

At the matchbox, the forward force from my finger overcomes the friction force from the table. Each object has an imbalance of forces giving rise to acceleration leftwards.

The point of the diagram is to make clear that the third law makes matched pairs of forces that act on different objects. Equilibrium from Newton's first or second law is about the resultant force at a single object.

$\dagger$ (Sorry that the finger doesn't actually touch the matchbox in the diagram. If it had, I wouldn't have had space for the important safety notice on the matches. I wouldn't want any children to be harmed because of a misplaced force arrow. Come to think of it, the dagger on this footnote looks a bit sharp.)

• This answer is completely awesome. My question to you is how on earth did you decide it that it was better to write in an entire warning label instead of removing the word "match"? – Steven Lu Dec 2 '12 at 5:41
• Just a little thing: the reaction force from the table shouldn't be paired with the matchbox's weight. – Javier Dec 2 '12 at 7:54
• @StevenLu because I found it funny, particularly "May cause fire.". – AndrewC Dec 2 '12 at 9:28
• @JavierBadia Fixed now. Thanks for pointing out my silly (and ironic) but key mistake. My answer is better now because of your comment. – AndrewC Dec 2 '12 at 21:20
• A nice exercise is to draw the table, matchbox, person and earth and find as many third law matched pairs you can (remember to make sure they're acting on different objects). There's an answer to be found in the revision history of my answer (click the link after the word edited), but I hid it because I feel it distracts from the main part of the answer. – AndrewC Dec 2 '12 at 21:32

I had similar problem in understanding the 3rd law. I found the answer myself while sitting in my study chair which has wheels!

sitting in the chair, I folded my legs up so that they are not in touch with ground. Now I pushed the wall with my hands. Of course, wall didn't move but my chair and I moved backward! why? because wall pushed me back and wheels could overcome the friction.

I was mixing up things earlier : trying to cancel the forces where one cannot.

Movement of the matchbox is due to the force which you apply on it. period.

Now why you didn't move when matchbox applied the equal force on you is because of the friction. If you reduce the friction like I did sitting in the chair, you would also move in opposite direction.

Equilibrium can only establish itself when the forces are on the same object..

Alas, I am free from this confusion.. such a relief

• @user1062760 Agreed. Moreover, the accepted answer confuses the reader by the sentence "While Fmatchbox does press back on my finger with an equal magnitude to Ffinger, it's no match for Fmuscles (even though I've not been to the gym in years). ". One should rather speak about friction, as in this answer. – wondering Apr 9 '18 at 20:19
• @wondering Maybe, but to the beginner, Fmuscles clearly act on my hand (which is in the picture), whereas the friction is on my feet or rear end (which are not). Also, since my hand and my body move independently of each other in this scenario, I feel that considering my entire body as a single particle in this instance is invalid. A force acts on my hand to move it forwards. It is not friction; friction always opposes motion rather than producing it. For all these reasons I conclude that calling it friction would cause more confusion for a beginner, not less. – AndrewC 19 hours ago

Good! This question implies that you're thinking hard and questioning the laws. It turns out that you are misunderstanding Newton's 2nd Law though. Motion of a body is due to an external force. F1 acts on your box, but not F2. An object can never act on itself.

• Think you mean "motion of a body". – Eugene Seidel Dec 1 '12 at 23:48

Forces related to Newton's third law apply to different bodies, therefore they cannot cancel each other out.

For example, the reaction to Earth's gravitational pull on the Moon is the Moon's pull on Earth. That force won't have any relevance to the Moon.

In any financial transaction the money given is equal to the money received. (If I give you \$10 I am \$ 10 poorer and you are \$10 better off.) So how does anyone get rich? • This fantastic answer goes directly to the heart of the question. If we sum +\$10 and -\$10 we get \$0, but the mistake is in considering those two numbers as applying to the same person, whereas in reality they apply to different people. This is epically clear and deserves more prominence. – AndrewC 18 hours ago

While considering 3rd law, forces act on different bodies , and not on same bodies. So the body which is hit is under the influence of applied external force only. The force which the hit body applies back to the hitting object is acting on the hitting object, so no point of cancelling of forces as they are acting on different objects.

I too used to think that way. Try this experiment : Ask your friend to stand in front of you and both of you try to push each other with approximately same strength, see what happens. Try this with friends of different masses.

There is a common misconception about newtons 3rd law because of the words"equal and opposite" and many of us think that net force is zero. But these forces act on two different bodies and hence the bodies accelerate. If you have a table in space with zero gravity and if you pushed it with your fingers , then the table would move in the direction of force and you would move in the opposite direction. If you consider the table and yourself as one system , then the net force on that system is zero.

If I could only change one thing about physics education, it would be the phrasing of Newton's 3rd law. According to my copy of Magnificent Principia (by Colin Pask, Prometheus Books, 2013) the "To every action there is always opposed an equal reaction..." phrasing is Newton's. And it's been causing confusion ever since.

To get a sense of what Newton really meant, consider the universal gravitation equation: $$F=G\frac{m_1m_2}{r^2}$$

Notice there are two masses specified, but there is no "source" mass and no "target" mass. And there is only one force produced by this equation. Now, you can look at it as two different forces: $m_1$ attracting $m_2$ and $m_2$ attracting $m_1$. But that is misleading. It gives the impression that the forces somehow have independent existences. But they don't. They are completely, inextricably linked. So much so, that I think it makes much more sense to this of this as one attractive force between two masses.

Coulombs law follows the same format:

$$F=k_c\frac{q_12_2}{r^2}$$

Again, you can think of this as two different forces. But I think the equation really hints at a single attractive force (different charge signs) or a single repulsive force (identical charge signs) between two charges.

That is what Newton meant by his third law. It's not possible for $m_1$ to attract $m_2$ without $m_1$ being caught up in the very same force of attraction between the two particles. And it's not possible for $q_1$ to attract or repel $q_2$ without $q_1$ being caught up in the very same force.

This is harder to see with contact forces. Part of the problem is that human muscles must constantly expend energy at a molecular level in order to stay contracted. So it's easy to confuse force exertion with expenditure of energy. And humans have cognition and agency. So to say, "The person pushes on the matchbox and the matchbox pushes on the person" feels wrong because the person is expending energy; the matchbox is not. The person has agency and initiates the push; the matchbox is inanimate.

To get a better feel for Newton's third law, consider yourself in a deep swimming pool where your feet are off the bottom. You're next to the wall. Now push on the wall. What happens? You push yourself away from the wall. The traditional explanation is that you push on the wall, and "the wall pushes back on you." And while that is technically true, it doesn't make intuitive sense because you know darn well that you're the one doing the pushing.

What's really happening is that you create a repulsive force between the wall and yourself. The wall is fixed to the earth and the earth is mighty big and hard to move. So the repulsive force manifests itself in you pushing yourself away from the wall.

When you "push the matchbox," you're really setting up a repulsive force between your finger and the matchbox. (At a molecular level, this is just the Coulomb repulsion, of course.) But you're much more massive than the matchbox. Your weight and the friction between your shoes and the floor essentially fix you to the floor and make you immovable. So the repulsive force manifests itself as the matchbox moving.

Finally, when dealing with forces where one mass (or one charge) is so much larger than the other (such as an apple falling towards the earth) it's very common to ignore that fact that the masses are attracting each other, and to phrase the interaction as if it were just the earth attracting the apple and nothing more. That is an oversimplification. But it's justified by the fact that the attractive force between the two masses is overwhelmingly manifested in the motion of the apple.

In fact, Newton phrased that part well in The Principia,

"The changes made by these actions are equal . . . if the bodies are not hindered by any other impediments . . . the changes of velocities made towards common parts are reciprocally proportional to the bodies [the masses]."

• Very good answer. – Dude156 Feb 19 '19 at 6:04

You are using one law (third) that is true, to try to invalidate another unrelated law (second).

Using your own examples, the reason you are able to move the box, is because you apply a force larger that the force produced by friction of the box against the table. If you glue the box on the table, it will take a much larger force to move it! The equal but opposite force that the box exerts against your finger, can only be as large as the friction force (or the glue force), if you exceed it, the box will have to move.

Similarly, the table you mention, can only exert a force against your hand equal to the friction exerted by the table legs on the floor. If you exceed it, the table will definitively move! Just to make this clear, if you put rollers on the table legs, it will take little force to move it, but if you nail the legs to the floor, you might break the legs or nails before it moves. If the force is less that the required amount, nothing (no movement) happens.

If you really got interest in it then let's understand it with an example:

By law of gravitation you know that earth is attracting a freely falling body by a force GMm/r^2,and the body is attracting the earth by a force -GMm/r^2(negetive sign indicate the opposite direction).

The misconception people have is that Net force= GMm/r^2+(-GMm/r^2)=0,and the question they ask is why don't the falling body hang in air(as there is no force on it).

let's use our mind,what can you say about the forces acting on falling body,i assume the answer is force acting on body is gravitational attraction towards earth(nothing else) that's why body is moving towards earth,we don't have to consider -GMm/r^2 as it is acting on earth not on body. For the system of body + earth you can say that GMm/r^2+(-GMm/r^2)=0,but for individual bodies there is only one force(no counter of it)

One of my books tells me how to overcome this. You must always specify the system. Which block is considered? By the way, Newton's third law would be: "The force exerted by A on B, is equal and opposite to force exerted by B on A. You must specify which block is in consideration. If you suppose consider both blocks as a system, the forces would become internal, and should be left out.

I have added a few additional forces to the diagram produced by AndrewC to show 5 groups of forces which are Newton's third law pairs and made the hand massless to simplify the diagram.

The Newton's third law pairs are colour coded and labelled.
These pairs of forces:

• are equal in magnitude and opposite in direction
• act on different objects
• are of the same type ie both contact, both gravitational etc.

For example $R_{\rm be}$ is the reaction on the box due to the earth and its Newton's third law pair is $R_{\rm eb}$ the reaction on the earth due to the box, $W_{\rm be}$ is the gravitational attraction on the box due to the earth and $W_{\rm eb}$ is the gravitational attraction on the earth due to the box.

The $F$ forces are the frictional forces between the box and the earth, the $X$ forces are the contact forces between the box and the hand, and the $Y$ forces are the forces on the hand and the person and the earth as a result of the action of muscles in the arm.

If the system is assumed to be the box, the hand and the person & the earth then the net external force on that system is zero and the centre of mass of the system does not undergo an acceleration.

Looking at vertical, y-direction, forces acting on the box alone system and applying Newton's second law gives $R_{\rm be} - W_{\rm be} = 0$ and the equivalent equation for vertical forces acting on the person & earth system is $R_{\rm eb} - W_{\rm eb} = 0$ so the box and the earth do not accelerate in the vertical, y-direction.

Now consider the forces acting on the box in the x-direction and apply Newton's second law $F_{\rm be} - X_{\rm bh} = m_{\rm b}a_{\rm b}$ where $m_{\rm b}$ is the mass of the box and $a_{\rm b}$ is its acceleration.
Now if the left hand side of this equation is zero the box could be at rest or moving at constant velocity.
If the left hand side of the equation is no zero then the box will accelerate and if the force on the box due to the hand has a larger magnitude than the frictional force on the box due to the earth then the box will accelerate to the left.
So even though you have all these Newton's third law pairs supposedly cancelling each other out they do not because the act on different bodies.

For the hand system the equation of motion is $X_{\rm hb} - Y_{\rm he}=0$ which means that the magnitude of the force on the box due to the hand $X_{\rm bh}$ is equal to the magnitude of the force on the person & earth due to the hand $Y_{\rm eh}$.

And of course even though you would notice the effect because the earth is so massive, the person and earth system of mass $m_{\rm e}$ would suffer an acceleration $a_{\rm e}$ in a direction opposite to that of the box given by the equation $Y_{\rm eh} - F_{\rm eb} = m_{\rm e}a_{\rm e}$.

Note that the magnitude of the force the person & earth system is exactly the equal to the magnitude of the force on the box system.

Forces can only cancel themselves out when they act on the same object. All action-reaction pairs identified by Newton's 3rd Law act reciprocally, meaning that if one of the forces acts from object A onto object B, then the reaction force acts from object B onto object A, which cannot cancel since they act on different objects.

• How is this different from the other 15 answers? Your answer is basically a less detailed version of the already existing answers, so I'm not sure what's the point of posting a new one... – AccidentalFourierTransform Sep 30 '18 at 21:31
• Brevity can be of value in understanding. – Trevor Kafka Oct 1 '18 at 3:13

Think of the "a" in F=ma as the instantaneous rate of change in velocity - meaning, how quickly velocity changes at an instant. In calculus terms, a is the derivative of v(t), where t represents time and v(t) = at.

The moment you start moving that box, you are creating a force, because the velocity then is changing instantaneously. At any point, you can reduce the force to be equal to the opposing forces, at which time the "net force", the sum of the aforementioned forces, becomes zero.

So, if you are pushing that box, at some point you must have caused that box to accelerate. The acceleration may have been unnoticeable, but it must have been there, otherwise there would be no change in the velocity.

When you push an object, it is true the the object pushes you back with the same force. However, this does not mean that the force that you are exerting on the body has got cancelled. The object will continue to to experience the push and so would you from the object being pushed. In a way they are 2 separate forces, each acting on separate object.

To understand the concept better imagine yourself pushing your friend and your friend pushing you back with the same force. Just because the forces are equal and opposite in magnitude, does not mean that you feel at ease. You feel the pain in your muscles because there is a force acting on your arms that stresses your muscles.

Scientifically put, you need to see the body being pushed in isolation. The force you are putting on the body is “tangible” and makes it move once the force of friction has been overcome.

This is the reason problems around Newton’s laws of motion are solved by using “free body diagrams”. This essentially requires you to label all forces that act on a body and then find the “Net force”, using vector algebra. This net force is then equated with the product of mass and the acceleration this net force is creating, to find the unknown in the equation. This also Newton’s 2nd Law of Motion that is used to answer problems like this (Net Force = ma)

I have created these 2 video that will bring more clarity to you-

Newton's 2nd law of motion

Newton's 3rd law of motion

When you say ,i apply force ,on match box,and match box apply force on me, so forces cancel out ,these.forces are on two different bodies,they have different acceleration .for match box to remain at rest,forces on him should be cancel out,you can think this using newton formulas, suppose match box has mass of 5kg and you apply force of 5N produces of a=1m/s^2 now to produce the same acceleration ,to you (let say your weight is 60kg),force should need to be,60N hence , you are in rest position .this is the best possible way to explain it .thank you

This is a really valid doubt and most of us have this on our Mind while trying to understand Newton's third law. Now yes, $$\vec{F_1}=-\vec{F_2}$$ is valid and the forces here are an action reaction pair acting in opposite direction with the same magnitude.

So why doesn't a body remain in equilibrium?

These forces(the action reaction pair) act on different bodies and not on the same body.A body is said to be in equilibrium if two forces acting on the same body cancel out each other but that is not the case here. Therefore when we represent Newton's third law we write $$\vec{F_{12}}=-\vec{F_{21}}$$ which means force on body $$1$$ due to body $$2$$ is equal to the negative of force on body $$2$$ due to body $$1$$

• The boldface and italics here don't really make this more readable. – Chris Feb 16 at 5:26

I don't see how this is considered a good question given how obvious the answer is. Newton's 3rd law is in fact one of the reasons so many things move at all. If it wasn't for it you would be able to push something without having to move yourself. Some things are moved(from being stationary) more or less easily depending on their inertia..

action and reaction Depends on the frame eg: if you push matchbox with your fingers kept on a table, from matchbox's frame, we have to see for only those force acting ON the matchbox, not those forces that matchbox applies i.e the reaction force to your fingers So, from matchbox's frame, forces acting on matchbox are: your push, downward mg and normal reaction from table, that is why it moves

• Try to avoid abbreviations ("ur") and run-on sentences. It makes your answer hard to read. – Javier Dec 2 '12 at 7:55
• Aside from that, you can "see" all forces from a reference frame, so this answer is wrong. The frames don't matter, it is just that an object moves only due to the forces acting on it. – Manishearth Dec 2 '12 at 14:16