# Newton's third law's intuition

This is a dumb question but I don't seem to understand Newton's third law of motion. If an object is at rest on a table, it exerts a force of magnitude $$F=mg$$ in the downward direction. Similarly, the table exerts equal and opposite force to the object. But what is the use of the second pair of forces in this process? What would happen if there were no force acting by the table to the object. I mean what is the usefulness of the action-reaction pair in this system. Please help, I'm a high school student with a very low understanding of physics.

• If the object exerted force on the table but the table did not exert force on the object, then the object would just sink through the table. Commented Dec 29, 2016 at 5:30
• One piece of advice: when dealing with forces in Physics, if you have a body that is experiencing a force, make sure that you know what body is exerting that force. Commented Jun 22, 2021 at 12:54
• The object exerts a downward force of mg on the table and the table exerts an upward force of mg on the object. That's an action reaction pair per Newton's 3rd law. So, what do you mean by "the second pair of forces"? Commented Jun 22, 2021 at 20:36

Let's start with a statement of Newton's third law (N3L): If body A exerts a force on body B, then B exerts an equal and opposite force on A.

[It might – or might not – help to think of there being an interaction between A and B, with symmetrical consequences in terms of forces on A and B.]

Now suppose that you place a book on a table.

The Earth exerts a downward (gravitational) pull, $$W$$, on the book. According to N3L the book exerts an equal upward pull, $$-W$$ on the Earth, though you don't notice the latter.

The downward pull makes the book accelerate downwards, slightly distorting the table. The more the table is distorted the greater the normal contact (electromagnetic) force between book and table: the book pushing down on the table and the table pushing up on the book: another N3L pair of forces.

So the book is acted upon by two forces, the downward pull of the Earth and the upward force from the table. These are NOT an N3L pair: they originate from different interactions and they act on the same object! But when the book reaches equilibrium (which is usually very soon indeed after placing it on the table) these forces are equal and opposite. The table has been distorted enough to produce an upward force on the book equal to the downward pull of the Earth.

You have not understood Newton's third law.

The reaction to the action force on object due to gravitational attraction of the Earth is the force on the Earth due to gravitational attraction of b.

The reaction to the action force on object due to table is the force on table due to object.

It so happens that when the object is resting on a table the force on object due to gravitational attraction of the Earth is equal in magnitude but opposite in direction to the force on the object due to the table.
Note that these two forces both act on the object whereas a N3L pair of forces must act on different objects.
Further confirmation that they are not N3L pairs is imagine removing the table.
The gravitational attraction still acts but the force due to the table has vanished, so where is the N3L pair?

In this example, I think it could be useful to consider a balance of forces on the center of mass of the object.

You have an object standing still on a table. You correctly wrote that the gravitational force acts on the object: $$F=mg$$ with $$m$$ mass of the object and $$g$$ gravitational acceleration.

Suppose for a moment that this is the only force acting on the object (i.e. that there is no force from the table, but that the object still somehow exerts a force on the table.) Therefore, there would be an acceleration, in this case $$g$$, acting on the object; this is a consequence of Newton's second law. But an acceleration means a change in speed, therefore your object couldn't remain still on the table, and that would contradict what we stated above. So there must be a force acting on the object that exactly cancels the force of gravity, and we can say this force is caused by the table because with the table absent, the object would fall normally. Also, if the object could somehow exert a force on the table that wasn't reciprocated, it would have an effect on the measured weight, which we can check in several ways (you might consider disassembling the table and measuring the weights of each piece [and then reassembling it], or seeing if the table accelerates faster than expected in free fall, or trying to bring the table into orbit.)

Now, suppose that Newton's third law is right, and on the object is acting a force, exerted by the table, equal and opposite to the one that the object is exerting on it. This means that on the object are acting two forces, one directed downwards and the other upwards, both with the same magnitude: $$F=mg$$ By adding them as vectors, you can see that the resulting net force on the body is equal to zero. This means, from Newton's second law, that acceleration is zero and therefore the object has always the same speed: if it was still at the beginning of the example, it will always be still.

The same reasoning can be applied to the table.

This how they usually explain Newton's third law in high school examples.

• But why doesn't this apply if the table is suddenly removed?Where does the balance reaction pair go? Commented Dec 29, 2016 at 14:02
• @SurazBasnet what do you mean? If you remove the table, you have just the gravitational force that is acting on the object, and it falls. Commented Dec 29, 2016 at 14:08
• Thanks,now I think I understood it a little bit.But what when we push a stone across the road?According to this law,stone must also exert equal and opposite force to us when we do but we anyways push the stone.How does this happen? Commented Dec 29, 2016 at 14:12
• @SurazBasnet in this case, I think you can say that when you push the stone, you are experiencing a force pushing you, a sort of resistance by the stone to be pushed. Commented Dec 29, 2016 at 14:50
• but shouldn't these two forces cancel each other out and stone still continue to be in the rest like the object in the table did? Commented Dec 29, 2016 at 15:45

But what is the use of the second pair of forces in this process? What would happen if there were no force acting by the table to the object.

Gravity exerts a downward force of $$mg$$ on the object. If the table did not exert an upward force of $$mg$$ on the object, then there would be a net downward force of $$mg$$ acting on the object. Then, applying Newton's 2nd law to the object ($$F_{net}=ma$$), it would experience a downward acceleration through the table of

$$a=\frac{F_{net}}{m}=\frac{mg}{m}=g$$

But we know the object is not accelerating downward, so the table must be exerting an upward force of $$mg$$ on the object in order for the net force acting on the object to be zero.

Now consider the table. The object exerts a downward force of $$mg$$ on the table. The force of gravity also exerts a downward force on the table of $$Mg$$ where $$M$$ is now the mass of the table. So the total downward force acting on the table is

$$F=(M+m)g$$

The floor supporting the table exerts an upward force on the table. Since the table is not accelerating, we know that the upward force exerted by the floor on the table has to equal $$F=(M+m)g$$ in order for the net force acting on the table to be zero.

In this example, the action-reaction pairs of forces on the object and the table cancel out because there is no acceleration of either the object or the table. This sometimes leads those new to physics to wonder why anything accelerates since it seems the action-reaction forces always cancel each other out. But an object doesn't accelerate only because the net force acting on it is zero. And the net force consists of the action (or reaction) force on the object, plus any other external force acting on the object.

As a high school student, perhaps the most important thing you need to know is that Newton's 3rd law alone doesn't determine what happens to each object involved in the action-reaction pair. You need to apply Newton's 2nd law to each object individually to determine that. And that means you need to know what other forces, if any, are acting on each object.

Hope this helps.