According to the third law of Newton, every action has an equal and opposite reaction. This would (to me) suggest that every surface should put a force exactly opposite the weight of the body (imagine the experiment in vacuum). If so, why does compression occur? Say you keep a weight on a sponge, then the sponge compresses, why is this?

Does the sponge not give an equal reaction to the weight on it?

Edit: If Newton's third law is true, why can we sink in sand? This question is different from this one because there the question is mainly about which bodies the forces act on according to third law while the question I asked is mainly about compression and how much force a body applies against a weight put placed on it.


5 Answers 5


Imagine you are in a crowd of people. A huge crowd, with everyone, almost squeezed together. Imagine you run into the crowd. In that case, you will have an almost rigid body, because, you apply a force on him, he applies a force back on you, you stop, but he will have a force, which will be balanced by the next person and next and so on till you reach a wall. enter image description here

But, if the crowd has a little less number of people, or more like people standing in rows holding hands to form a chain, it will be different. If you run into the crowd from outside, you will push a few people, and they will accelerate (inelastic collision) and you and the other person moves with the same velocity as you, but you are not falling into the person. He exerts an equal force, thus preventing you from sinking into him. But as there is no other person to provide an equal force on him, he starts falling in till he reaches another person because of the unbalanced forces on him

enter image description here But that does not mean that he is not moving, because he is, and if you were large enough it would be similar to the case you talk about. Several people moving in a localized region that would 'appear' as if the forces are not equal. But if you go a bit more closer, you will see that it is not violated.

The sponge is not one body, rather it is like the crowd. The particles can move, to some extent, independent of the other particles. So, when you push a sponge, you are actually making some sponge mass to move, like crashing into the crowd.

So, Newton's third law is not violated

  • $\begingroup$ Some formula references in the answer would be nice and like what things you are referring to irl when you use the analogy( more highlighted like) $\endgroup$
    – Babu
    Commented Apr 12, 2020 at 15:29
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    $\begingroup$ What you have in your (very nice, BTW) picture is really more of a closed-cell foam rather than an open-cell sponge. In the foam, the air is compressed until it reaches a pressure equal to the weight on it (approximately, of course, there's also material stiffness &c). In the sponge, the air is pushed out so it becomes like your top picture. $\endgroup$
    – jamesqf
    Commented Apr 12, 2020 at 17:38
  • $\begingroup$ @jamesqf I realize that, because an open-cell sponge cannot exist like the one I'd drawn, because it would be practically impossible to balance the weight of the middle particle in the topmost layer unless I consider the air locked. Though I think I can still make a sponge with air gaps, by balancing normal forces between particles (creating gentle curves) without making it look exactly like the first one (a bit less rigid). $\endgroup$
    – Elendil
    Commented Apr 12, 2020 at 17:57
  • $\begingroup$ @jamesqf thanks! $\endgroup$
    – Elendil
    Commented Apr 13, 2020 at 2:49

Your mistake is in assuming that N3L describes the relationship between the force of gravity on the weight and the force the sponge applies to the weight. This is not the case. N3L relates the gravity on the weight from the Earth to the gravity on the Earth from the weight, and it relates the force the sponge exerts on the weight to the force the weight exerts on the sponge. N3L does not guarantee the force the sponge exerts on the weight is equal to the weight of the weight. All it says is that the force the weight exerts on the sponge is equal and opposite to the force the sponge exerts on the weight.

As an explicit example, if you put a $10\, \rm N$ weight on the sponge, as it compresses the sponge is not exerting $10\,\rm N$ of force on the weight. N3L doesn't relate the $10\,\rm N$ force of gravity on the weight to the force the sponge exerts on the weight. These forces will be equal when compression ends at the weight is at rest, but to conclude this you need to bring in N2L, not N3L.

Therefore, this statement

This would suggest (to me) that every surface should put a force exactly opposite the weight of the body.

is false. As a simple counterexample, in an elevator the normal force acting on you by the floor when the elevator starts moving upwards has to be larger than your weight in magnitude in order for you to accelerate you upwards.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Apr 12, 2020 at 17:42
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    $\begingroup$ Call the object an "object" instead of a "weight". Then you would have, for example, "the weight of the object" instead of "the weight of the weight". Much clearer. Also say "Newton's third (second) law" instead of "N3L (N2L)". $\endgroup$ Commented Nov 25, 2020 at 14:28

Compression occurs because of Newton's second and third laws.

The crux of the issue is what these laws really mean (and a little bit treating the sponge like a whole, single object). You know the laws, but we have to dig into them to make the answer clear.

Newton's second law:

The first problem is that this law is often abbreviated to "force equals mass times acceleration." It is critically important to note, however, that

$\overrightarrow{F} = m\overrightarrow{a}$

describes the net force acting on an object - the sum of forces on that object.

Newton's third law:

The second problem is that this law is given to us in terms of "actions" and "reactions," which are mathematically meaningless. Instead, think of actions and reactions as two halves of a pair of forces that always occur whenever anything happens.

Before you put the weight on the sponge...

...the net force on both objects is zero.

Both are subject to Earth's gravity. Both are sitting on the table (or whatever), which is, itself, pushing on the Earth, all equally pushing back on each other. Their net forces are all zero, therefore their accelerations are all zero. They aren't going anywhere.

As you put the weight on the sponge...

...the sponge moves.

You might be thinking, no it doesn't.

In Newtonian mechanics, we tend to treat objects such as sponges like whole, discrete things, and this is incredibly convenient. It is important to remember, however, that nothing is truly this way. Sponges, like everything else, are made of molecules.

So what's happening?

Zooming in, the molecules which constitute the sponge are in equilibrium with each other before the experiment as well. They are some natural distance from each other due to intermolecular, electromagnetic forces, which result from whatever property makes sponges the way they are. This is a rabbit hole of explanation beyond the scope of the question.

As the weight initially pushes down on the sponge, the sponge does push back with some force. However, the top layer of molecules also begins to accelerate. This is because the forces which keep the sponge's molecules that initial distance from each other are weaker than the forces experienced by the top layer of molecules.

In this moment, Newton's third law is not violated. The sponge is moving and simply isn't experiencing the full weight of the object compressing it.

We do know, however, that the force exerted by the sponge on the weight in return reduces the weight's net force and therefore reduces its acceleration.

Following through (keep this in mind when you get to the section about Hooke's Law):

As the sponge's molecules get more and more compressed, they also become less and less able to compress. This is to say that the magnitudes of the intermolecular electromagnetic forces become greater.

Eventually, these forces overcome the downward net force of the weight and it begins to slow down. During this time, the weight is still falling because it still has velocity. The force exerted by the sponge on the weight is momentarily greater than the weight of the object.

This phenomenon is the same reason that objects seem heavier when they are falling (e.g. it might be easy to hold a stack of books, but difficult to catch a stack of books).

The result is that the weight bounces upward. This effect will vary depending on whether or not you carefully place the weight down or drop it (how much kinetic energy the weight has).


...the molecules of the sponge will all be compressed according to the new sum of force acting on them (due to gravity on themselves, the weight of all the sponge molecules above them, and the object sitting on the sponge).

This new net force is once again zero, but each action/reaction pair of forces is greater in magnitude.

Making all of that easier with Hooke's Law:

In the 17th century, Robert Hooke simplified this whole process with a law which looks like

$F_s = kx$

where $F_s$ is the force required to deform a spring by length $x$.

$k$ is a value called the spring constant and is different for every type of spring-like object. This is the value one would use to take those aforementioned intermolecular electromagnetic forces and reduce them to a nice, easy constant. Someone else could then reference that value and predict the compression of a similar sponge in the future.

Does that answer your question?

  • $\begingroup$ Nice answer ( well - written) but what I was missing was that the block is not applying it's full weight on the sponge. $\endgroup$
    – Babu
    Commented Apr 13, 2020 at 9:30
  • $\begingroup$ Thanks, hope it helped! $\endgroup$ Commented Apr 13, 2020 at 15:26

It might be more informative to go through an entire process, rather than just skipping to the end. (For the sake of making the numbers easier to understand, I'm going to round Earth's gravitational acceleration to 10 m/s/s.)

Let's say the sponge is 1kg (it's a big sponge). If I just drop the sponge in the middle of a room, gravity pulls it down with a force of 10N. So, by Newton's third law, something else must be pulled up with 10N of force.

That "something" is the Earth itself. If the Earth pulls something with 10N of force, that something pulls Earth in the opposite direction with 10N of force. So as the sponge falls toward the Earth, the Earth falls toward the sponge. (Of course, since the Earth is much more massive than the sponge, that force is going to have proportionally less effect.)

Eventually, of course, the sponge will hit the floor. (Or does the floor hit the sponge?) Since the sponge decelerates back to zero far faster than it accelerated during its brief fall, we know that there's a much larger upward force produced by the chemical bonds between the molecules of the floor not allowing the sponge to pass through. But even here, the forces still balance: however much the floor is pushing up on the sponge, the sponge is pushing down with that same force.

Once the sponge is at rest, we now have two different forces opposing each other on the sponge. The whole Earth is pulling on the whole sponge with 10N of force (balanced by the sponge pulling on the Earth just as hard), and we have the physical interaction of the floor pushing up on the sponge with 10N of force (balanced by the sponge pushing down on the floor just as hard).

Now let's set a 10kg jug of water on top of the sponge. Similarly to the sponge, the jug will have 100N of downward force on it from gravity, balanced by 100N of upward force on the Earth. The sponge will push up on the jug, balanced by the jug pushing on the sponge. However, because the sponge is flexible, it will push up much less than 100N (again balanced by the same amount of downward force on the sponge).

This is when the compression happens. Newton's third law doesn't specify a balance between the sponge/jug force and the jug/Earth force. It says that the sponge/jug force must be equal to the jug/sponge force. If the jug is pushing down on the sponge with 5N of force, then the sponge is pushing up on the jug with 5N of force, leaving 95N to accelerate the jug down and the Earth up. However, as the jug falls, the amount of upward force produced by the sponge will increase because of its elasticity, eventually reaching 100N. There will be some bouncing around this point because of the inertia of the jug, but eventually it will settle down to an equilibrium state where all forces balance, with the sponge compressed under the jug.

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    $\begingroup$ How do we find exactly how much force the weight puts on the sponge? Is this related to pressure by any chance? $\endgroup$
    – Babu
    Commented Apr 12, 2020 at 16:59
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    $\begingroup$ @DDD4C4U The amount of force at any given instant depends on how far the jug has deformed the sponge, and the elasticity of the sponge itself. At the precise moment of contact, there's zero deformation and thus zero force. As the deformation increases, so does the force. (Unless, of course, the force is great enough to tear or otherwise disrupt the structure of the sponge, in which case, the force is anyone's guess.) $\endgroup$ Commented Apr 12, 2020 at 17:14

A surface only provides an equal but opposite reaction when the object is at rest. When the object is at rest the net force is zero, so the normal force must be equal to the weight. The forces are in equilibrium. The case of the sponge is a good example how this equilibrium comes about.

Let's drop an object when it is just hovering above the sponge. In the beginning the sponge is not deformed at all, but the sponge also isn't exerting any force. The object starts accelerating and as the object moves into the sponge it deforms. When the sponge deforms it exerts a force upwards because, in a crude approximation, it is like a spring: when you push a spring down it exerts a force upwards. The force that the sponge exerts increases with the deformation until the force matches the weight of the object. At that point the object is in equilibrium. It is possible for the object to bounce a bit before it reaches equilibrium but in the end the forces have to match.

When you rest an object on a table the same process happens but the deformations are just very, very small.

  • $\begingroup$ shouldnt third law be active always hence sponge should put equal force for the objects weight? $\endgroup$
    – Babu
    Commented Apr 11, 2020 at 22:28
  • $\begingroup$ @DDD4C4U I get what you mean but when the object just starts falling it is exerting zero weight. The third law still holds. If you've ever been on an analog scale you probably know that you can change your weight by bobbing your upper body up and down. When an object is not completely at rest the force is not $m\cdot g$ but could be more or less. $\endgroup$ Commented Apr 11, 2020 at 22:40
  • $\begingroup$ but to fall for a second the forces should be unequal, I'm saying the instant at which the force of the mass is applied then instantly the sponge applies an opposite force and hence it should never fall. $\endgroup$
    – Babu
    Commented Apr 11, 2020 at 22:53
  • $\begingroup$ the instant at which the force of the mass is applied then instantly the sponge applies an opposite force and hence it should never fall. The thing is that it doesn't. You never see this happening because it happens so fast, but it takes time for the force to build up. Imagine you drop the object on a spring. The spring gets compressed a little so the object must have fallen a little. This means for a short amount of time there was an unbalanced force pulling the object down. $\endgroup$ Commented Apr 11, 2020 at 23:20
  • $\begingroup$ and when I say fast I mean extremely fast $\endgroup$ Commented Apr 11, 2020 at 23:20

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