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Suppose there are two blocks kept side by side on a frictionless surface . Since they are in contact with one another , they should both , by Newton's 3rd law , exert equal and opposite normal forces on each other. Now there is no friction so if the blocks are looked at individually, they shall get separated with accelerations in opposite directions, right ?

Whenever I have asked this question ,people have replied, "Normal forces get cancelled out . Net force on the system is zero . So they shall continue to be in contact etc etc"

But what is the exact meaning of this answer ? If we look at both the blocks as a system, of course, there is no external force . But if the blocks are individually examined , the forces do not cancel out !

I have displayed my thought process in the above text . Please correct me if I am wrong .

M.B.
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4 Answers4

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In real life the blocks are slightly compressible. If you force the two blocks together with a large force they will slightly compress (deform), resulting in normal forces between the blocks because the blocks act like springs. In physics problems like these we usually don't care about this deformation so the normal force has to be determined from context. Example: a block is resting on a table. Gravity is acting on the block but there is no net force since the block is stationary. The normal force must be equal to the gravitational force.

In your case, if the blocks are resting on the table, the most sane answer would be to say the normal force is zero. If two hands are pushing the blocks together with 10N for each hand then there will be a normal force of 10N experienced by each block. We can also separate the blocks and slide them towards each other, resulting in a collision. During the collision there will be a normal force which will look somewhat like a downwards parabola if you plotted it over time.

AccidentalTaylorExpansion
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The magnitude of the normal force is not fixed, but is whatever is necessary to prevent the objects from passing through each other. If you start with two blocks at relative rest, and there are no other forces, then the normal force is zero, and the blocks remain at relative rest.

people have replied, "Normal forces get cancelled out . Net force on the system is zero . So they shall continue to be in contact etc etc"

That's wrong, and your rebuttal is correct. If it were true, there could be no relative motion, since all forces obey Newton's third law.

benrg
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It is correct that there is no external force on the system, so the two blocks will remain in contact with one another. However, when examined individually, the two blocks do not experience a net force on them, due to the equal and opposite nature of the forces that they are exerting on one another. Therefore, the two blocks will remain in contact with one another and will not move due to the lack of a net force on them. This could be the misunderstanding in your way of thinking.

This is because Newton's Third Law states that for every action, there is an equal and opposite reaction. The two blocks are pushing against each other with equal and opposite forces, so when examining the them individualy the net force on each individul block is 0, meaning that they will stay in contact and not move.

Masterrun80
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It sounds like your question is about the nature of normal forces.

A massive object is subject to gravity. If you drop it, the only force acting on it is gravity, and so it accelerates downward.

The same object sitting on a table is motionless. The acceleration is $0$, so the total force is $0$. The table exerts an upward force just as strong as the downward force of gravity.

If you stack another mass on top, now the downward force on the object is twice as big. The table makes its upward force twice as big to cancel it.

On the other hand, if you pick up the mass, the table does nothing to prevent you. It reduces its force to $0$, but does not reverse it to pull the object downward.

This illustrates the nature of reaction forces. Without the reaction force the mass would penetrate into the table. The reaction force is just strong enough to prevent this.

It is a little more intuitive with a spring. If you compress a spring, the spring pushes back. The harder you push, the more the spring is compressed, and the more it pushes back. If you lift your hand off, the spring does not hold onto you.

The origin of the tabletop and spring forces is the same. Atomic bonds are stiff. If you deform them, the push back to try to keep their shape.

A spring has a shape that is makes it easier to bend than a solid block of steel. The spring is made of thin coils. A small bend in each piece of the coil adds up to a big compression. The block of steel would bend a microscopic amount. And so does the table.

So for your example, two blocks sit side by side with nothing pressing them together. The force of one block pressing against the other is $0$. So the reaction force from the other block is $0$. The blocks do not push each other apart.

Suppose you squeeze them together. Then the blocks push each other apart just hard enough to keep from penetrating into each other. They stay still. If you stop pushing them together, they stop pushing each other apart.

mmesser314
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