# How does Newton's third law hold up when a block kept on a table breaks the table?

Suppose there's a block of some mass that is kept on a table. But, instead of staying at rest on the table, it breaks through the table. How does Newton's third law hold up here?

If the block is at rest on the table then the push of the block onto the table due to its weight and the normal force by the table on the block are the action-reaction forces. But what about this case?

• Newtons thirds law still applies, but now parts of the table and the block are accelerating downwards. The forces are smaller, but they are still equal. Can you explain in more detail which aspect of the problem you are struggling with? – Andy Newman Jul 21 at 18:25
• I can't understand how does the block break through the table if the force the block and the table apply on each other are equal. – AxSmasher Jul 21 at 19:37

The law still applies, but we have to be careful with the forces. If the book has a mass $$m$$, we cannot simply say that the book applies a force of $$F=mg$$. If the table broke, then there was some maximum force it could withstand, which was less than $$mg$$.

The key to remember is that the sum of the forces on an object equals 0 only if that object is not accelerating. It actually always equals $$ma$$, where $$a$$ is the acceleration on the object. We write this $$\sum F = ma$$, with $$\sum$$ being how we notate summing everything up. It just happens to be that if the object is not accelerating (we say "at rest") $$\sum F = m\cdot 0 = 0$$

So the reality in your scenario is that the table pushes up with some force equal to exactly the amount of force it could apply without breaking. This is less than $$mg$$, so the result is that the book is accelerating. Newton's Third Law holds.

Now very soon this book will start moving closer to the atoms in the table. To understand what happens beyond the breaking point, we can't treat the table as one monolithic entity -- the velocity of the different parts can be different, and will be different during breaking. We can analyze the individual pieces of wood, or we can go to extremes and look at the forces on the atoms. But that all goes beyond your question. The fundamental answer is that the sum of the forces does not have to be zero, unless the object is not accelerating.

And much of these questions can be answered by remembering that there's many operations that we think of as "instantaneous" but are not. If the book falls on the table and WHAM! makes a loud sound, we say it "hit" the table at that time. But, if you look in slow motion, you will see that there was a whole lot of bending and wobbling which let that "hit" work out according to Newton's Laws. For example, in cases like these, the table can't "break" right away. It's actually a long series of smaller motions which, over the course of a few milliseconds, result in what we call "breaking."

• I can't understand that how can the book apply a force of mg and the table only apply its maximum withstandable force. If the table applies a certain amount of force on the book, then the book should also be applying that much force on the table, according to the law. Also, to what you said in the first paragraph, if the book is accelerating it would have to apply a force greater than the max withstandable force to break through, right? Could you please clarify a bit on this? – AxSmasher Jul 21 at 19:24
• @AxSmasher, the book doesn't apply a force of $mg$ to the table. – BowlOfRed Jul 21 at 21:02
• @AxSmasher BowlOfRed is right. The book does not apply a force of $mg$. Ask yourself why you feel it is obliged to. It isn't an immediate consequence of newton's 3rd law. The Force of the table on the book is not the reactionary force to the force of gravity. The force of the table on the book is the recationary force of the book on the table. That is all. – Cort Ammon Jul 21 at 21:31
• The challenge you are having is really common. You get taught that all of the forces equal out. Well, in this case, they absolutely do not. There must be accelerations in the scenario you describe. I think the confusion comes from how we have to teach. We typically teach static situations first -- ones with no motion, and thus no acceleration. In those situations, the forces on any object sum to 0. We then teach Newton's 3rd law, which says that every force has an equal and oppoisite reaction force. – Cort Ammon Jul 21 at 21:33
• I find it always helpful to think about the reality of a situation even though this physics problem postulates a perfectly rigid table. Think of the table as a bunch of big stiff springs (which it is), the book doesn't cause the reactive force to exist, it is instead the bending of the table that generates the force. What happens when you place a book on the table is that while the book is falling it's also bending the table. If the table is strong enough it'll generate a force strong enough to stop the book; if it's not strong enough it keeps bending until it breaks. – csiz Jul 22 at 3:58

There is always some amount of material deformation when two objects are held against each other by a force. In this case material deformation of the table exceeds its structural integrity, so it breaks. Now your action reaction pair is the block falling towards the Earth and the Earth falling towards the block. Since Earth's mass is so much greater than the block's mass the Earth's movement is negligible but not zero.

In the case where the table holds, we have a lot of information:

• The book is not accelerating
• The book has mass $$m$$
• The only vertical forces on the book are gravity and the normal force from the table.

Given the lack of acceleration, we know the total vertical forces on the book must sum to zero and $$F_N = mg$$. The weight of the book is entirely supported by the table. The force of the book on the table and the force of the table on the book are equal in magnitude.

When the book breaks through the table, the first assumption no longer holds. The total forces on the book no longer sum to zero. Gravity is still the same, so the only thing that can have changed is the normal force from the table. Instead of remaining equal to $$mg$$, the normal force is less.

This is not a problem to Newton's third law. The force the book is placing on the table is also less.

It may be helpful to think of an analogy, such as an arm-wrestling or a tug-of-war match.

During much of the match, both combatants may be applying equal force, and there's no motion (in actual practice it's hard to be consistent, so there will be some fluctuation back and forth, but we can simplify this). But eventually one of them will weaken more than the other; when the forces become significantly unbalanced, the stronger player(s) will overcome the weaker ones, and the match will end.

This is what's going on between the block and the table. Gravity is applying a downward force on the block, while the table's rigidity counteracts that, so there's no motion. But the continual force being applied to the table is slowly breaking the bonds that provide its rigidity. If this goes on long enough, there will be enough bonds broken that the table can't provide an equal upward force. Gravity wins the match and the block breaks through.