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This question has been asked many a times on physics stack exchange but I can't understand this in simple terms. This is a definition I got online:

The third law states that all forces between two objects exist in equal magnitude and opposite direction: if one object $A$ exerts a force $F_A$ on a second object $B$, then $B$ simultaneously exerts a force $F_B$ on $A$, and the two forces are equal in magnitude and opposite in direction: $F_A = −F_B$.

My question for this is that if a force is applied to body $B$ by body $A$, and the magnitude is $10$N, body $B$ is applying the same magnitude of force against body $A$. So, the forces should cancel out and motion won't be possible in that case as Newton's first law states that there should be an unbalanced force for bringing a body from rest to motion.

One thing people are answering is that they are on different body but they are also in different direction so force should cancel out as it is simultaneously exerted on the two bodies.

jng224
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    Indeed you are correct: the center of mass of the system of bodies is always at rest or moving with constant velocity due to the vector sum of all forces in the system vanishing. – Prof. Legolasov Mar 24 '21 at 11:26

5 Answers5

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The equal and opposite forces of magnitude 10N are applied to different objects, not the same, so forces can't cancel out.

Nihar Karve
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Karim Chahine
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  • but the forces are applied simultaneously and the two forces are also in different directions. – Ram Kaushal Mar 24 '21 at 08:49
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    In order for forces to cancel out they need to be applied to the same object. Newton's second law, $\boldsymbol{F}=m\boldsymbol{a}$ refers to the sum of forces acting on one object. If you took both objects a and b to be one system then yes, the forces would cancel, but as long as they are separated objects you have to treat them as such. – Karim Chahine Mar 24 '21 at 08:51
  • I totally understood your point but when I apply the force forwards, it applies a forces backward the very same moment so if I move ten steps forward simultaneously move ten steps backwards, I did not move at all even though I moved in two different directions – Ram Kaushal Mar 24 '21 at 08:55
  • In order to move the object, you would have to account for the reaction force, you don't move because the force pushing you back is balanced by something else. – Karim Chahine Mar 24 '21 at 09:08
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    You say "I totally understood your point". I say "you don't". Read this answer carefully and think about Newton's second law and third law. Note that Newton's second law applies to a single object, while the third law applies to two objects. – garyp Mar 24 '21 at 11:28
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I suggest you might like to think about a simple example, and really think it through. The idea is that you use what you have been told about Newton's third law, but don't regard it as a sort of mysterious utterance which you have to accept. Rather, regard it in the first instance as a hint or pointer; it is saying "look, notice this about the forces, this is a valuable thing to notice ..." and then by applying it to some simple examples you eventually come to understand it, and that is the moment when you can also accept it.

The simplest example is perhaps a collision when two objects A and B first approach one another, then interact, and then move away from one another. During the interaction, each is exerting a force on the other. Let $\bf F$ be the total force on A. Never mind for a moment what caused that force. Let $\bf G$ be the total force on B. Again, never mind for a moment what caused that force (I use the letter $\bf G$ just to have a different letter from $\bf F$). The equation of motion of each body is (Newton's 2nd law) $$ {\bf F} = m_A {\bf a}_A $$ and $$ {\bf G} = m_B {\bf a}_B $$ where I am assuming the masses are constant, and the accelerations are ${\bf a}_A$, ${\bf a}_B$ respectively. (I am not sure if you are familiar yet with writing Newton's second law as rate of change of momentum, which is in fact a better way to write it, so that is why I chose to talk about acceleration).

Now let's think about how those forces come about. The idea is that a force always comes about as a result of an interaction between two or more things. In the present example, if there is a third thing nearby, say a charged metal plate or a planet or something like that, then it can contribute to both the forces $\bf F$ and $\bf G$. But if there is nothing nearby, then the forces $\bf F$ and $\bf G$ are each caused only by the single interaction between A and B. And in this case the third law predicts $$ {\bf G} = - {\bf F}. $$

If we now put this fact into the two equations above, then we see that $$ {\bf a}_A = \frac{1}{m_A} {\bf F}_A $$ $$ {\bf a}_B = -\frac{1}{m_B} {\bf F}_A $$ so we find that the two bodies must be accelerating in opposite directions: either away from each other (when they repel) or towards one another (when they attract). This is of course exactly what is observed, because Newton's third law is in fact a correct observation about the nature of the physical world.

In this example there is something which is not accelerating: the location of the centre of mass of the system. So your intuition that "it all balances out and nothing accelerates" is not completely untrue, it is simply that it applies to the overall centre of mass, not the individual parts A and B.

Finally, let's briefly look at a case where nothing is accelerating, such as a brick sitting on a table on planet Earth. In this case there is a gravitational force downwards on the brick and upwards on the Earth, pulling the two together. So that is one pair of forces: $$ \begin{array}{rcl} \mbox{gravitational force on brick } &=& -m g \\ \mbox{gravitational force on Earth } &=& m g \end{array} $$ where to write these equations in 1 dimension, I am taking the upwards direction as positive. If these were the only forces acting, then the brick and the Earth would accelerate towards one another. But in the scenario under discussion, we we also have an electromagnetic force where the brick meets the table, pushing them away from each other: $$ \begin{array}{rcl} \mbox{normal reaction force on brick } &=& N \\ \mbox{normal reaction force on table } &=& -N. \end{array} $$ These are again equal and opposite, as Newton's third law says.

When the brick first lands on the table, it may wobble or bounce, because $N$ might not be equal to $mg$. But when it settles down and comes to rest, then $N$ is equal to $mg$ so the total force on the brick is then $N - mg = 0$. Notice that in this final result the balance is between two different types of force both acting on the same object. This balance is not an example of Newton's third law! But the physics behind it does involve two applications of Newton's third law, as shown above.

Andrew Steane
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The forces don't cancel because, as you say, they act on different bodies. Consider the graphic below, where two bodies have come into "contact" (really, they don't necessarily truly touch each other since the electromagnetic force prevents them from doing so - they get very close to each other so we say they touch). Here $F_a = - F_b$. The forces add up to zero, yes; if they didn't, the universe would not conserve momentum. But, each body reacts separately to forces acting upon it.

enter image description here

Consider the implications if we should cancel all forces on all bodies in the universe. Altering the state of motion of any object would not be possible.

jpf
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As you correctly said, the bodies exert a force on each other. So body one starts to move due to the force exerted by the other body and vice versa. As simple as that.

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Fa is applied on object B by object A. Fb is applied on object A by object B. Two forces are applied on different objects. When we consider both objects together those forces should cancel out. But when we consider them individually they do not cancel out.