# Distribution of friction forces

This question is based on this one.

Assume the following initial state, where no external forces are initially applied (no force of 20N) and the initial positions are stable.

Now, the external force of 20N is applied.

Will total friction done by first block be equal to the one done by second block ? That is, will friction be constant in all contact surface or will friction in surface of first block be bigger than in last block?

("first" and "last" block from the point of view of near to the point of the external application, left block is "first", right block is "second" or "last").

In the original question I understand there are two groups of answers:

• One group (including accepted answer) supports that total friction done by first block will be bigger than the one done by second block, in fact, that second block will not do any force because the maximal friction by first one (around 25N) is enough to null the external force.
• the other group believes (for this scenario where initial state was free of forces and stable) forces will be equal, friction per unit of area uniform in all contact surface.

In the comments to the question and answers, you can find also references to situations with more than two blocks or references to consider them as a continuous of differential blocks (vertical splits with a differential width).

• Are you talking about static or dynamic friction ? Jan 6, 2021 at 7:42
• @AgniusVasiliauskas: any one applicable to the experiment described Jan 6, 2021 at 11:58
• Outcomes may be different for different friction types. For example, while first block experiences static friction,- second block - does not experience any. Second block starts to experience static friction only when first block starts to experience dynamic friction. Jan 6, 2021 at 12:33
• @AgniusVasiliauskas " while first block experiences static friction,- second block - does not experience any" Are you so kind of develop this statement as an answer ? Jan 7, 2021 at 18:33

You have an indeterminate situation for this problem. With the friction coefficient given, and by drawing a free body diagram, it is immediately obvious that only one block will provide a friction force of approximately 25 Newtons, meaning that there is not enough force to overcome the friction force from only one block.

Having said that, note that this situation is dealing with static friction, not kinetic friction. The static friction coefficient is only defined at the maximum amount of static friction that can be achieved before a block slips and kinetic friction becomes involved. This means that the static friction force varies depending on the amount of force that is applied to the blocks. That static friction force will keep increasing until there is enough force to overcome the static friction of both blocks, which means that it's reasonable to assume that the static friction force is equally divided between each block. Of course, it's also reasonable to assume that the first block must start moving before the static friction force of the second block becomes involved. This naturally means that your answer is indeterminate.

• "The total friction will be divided equally between both blocks" has the hidden assumption in it that each block is experiencing the same amount of force on it at all times as the force on the two blocks is increasing to the stated value of 20N. What is the basis for this assumption? Also, how can this assumption be experimentally measured and verified? Jan 8, 2021 at 1:27
• Hi David. Not sure I understand the statement "it is immediately obvious that only one block will provide a friction force of approximately 25 Newtons" . Neither block can provide a friction force of 25 N. The static friction force is a variable that opposes a net external applied force up until that external force equals the maximum possible static friction force. In this case, the maximum possible static friction force is 25 N. That can't be reached because the maximum possible net external force is the 20 N applied force (assumes no force between the blocks, which would make it even less). Jan 9, 2021 at 21:21

The total friction will be divided equally between both blocks, provided they are touching. This is because if they are touching, they can in fact be regarded as a continuum. In order to elaborate, consider what you mean by 'one' block. Even when you say 'one' block it can be divided into discrete particles i.e. atoms or molecules which have space in-between. When you apply a force on one side of a block, it is first being applied on the first layer of atoms on the side, and they attempt to move in the direction of the force. But because of the Pauli exclusion principle and electromagnetic repulsion, they can't pass through each other, and the force is carried over to the next layer of atoms/molecules, and so on until all atoms/molecules in the block experience the force. And because of intermolecular/atomic attraction, the block does not deform, and so the atoms/molecules at the top cannot just slide over the ones at the bottom. This causes the friction at the bottom to affect the atoms/molecules at the top, and it gets divided uniformly across all the atoms/molecules. So when the two blocks are touching, a similar mechanism will apply to the parts where they are touching and they will behave as if they are one block. The only difference would be that there would be no significant intermolecular/atomic attraction between the atoms/molecules of one block and another, but it will still be there within each block, so there can be no deformation, and the same result would happen. The exception to this would be that if they are pulled apart instead, these principles wouldn't apply and in that case they would behave like two separate blocks with separate friction

• "This is because if they are touching, they can in fact be regarded as a continuum". I have a hard time agreeing with this, but I am open to an argument you can give to the contrary. If there were no friction on the supporting surface, the blocks could be separated with no external applied force. But as one continuous rigid body, they could not be so separated without an applied tensile load. In that respect, how can the two individual blocks be considered the equivalent of one rigid body. Jan 10, 2021 at 3:01

I agree with the idea that the first block provides total friction in that case.

Here is an imaginary experiment, assume there are "two" equal forces pushing in obverse directions on the two blocks.

In the beginning, the two blocks are "separated", and they move to each other until they are in "contact".

Now, as they are in contact, would you expect that the "normal force" between them is F? Cause if you regard it as "one" block, the strain force in the center is the compression force.

Or, due to the friction force individually, the normal force btw them is smaller than F.

Which statement do you prefer?

Imagine that the blocks are not actually touching. They are as close as you may want (or very far away if you prefer, it does not make any difference).

What happens when you apply that force in the first block? Nothing, it will not move, since the static friction is more than enough to overcome the force being applied. As @David White, the friction is equal to the force applied, up to the limit given by the product of the weight with the static friction coefficient.

So, the first block does not move. If you now bring the block closer, are you expecting any force to act on it? Imagine that instead of the block, you just pressed your hand against the first block. What are you expecting? You expect nothing to happen.

If there is any friction in the second block, then there must be a force applied on it, which implies that, when you touch the first block, your hand is pushed away, which is not very reasonable.

I think that the subtlety of this problem is in the difference between a rigid body and two rigid body's in contact. When you push a rigid body, you "magically" pushing all "atoms" in the rigid body at the same time. The second box, although it is in contact with the first it is not actually a part of this "magic bound" that constitutes the rigid body. In summary, although they are in contact, they are not a rigid body. I think this is why people say that the two experience friction, because they are assuming that the two block, by being in contact with each other, can be treated as a single rigid body (this was also my first thought).

The answer is actually a little more complex than it would seem. It can be either, depending on how exactly they are touching and the atomic structure of the blocks. In order to understand the answer, we have to consider a few things:

First, why does friction happen? Friction is essentially like collisions between the uneven protruding edges between two rough surfaces, like the image below:

So if there were no intersecting edges at the smaller scales, there would be no friction.

Secondly, considering an example where there is only one block, how is the friction getting distributed in the whole block? It is only the bottom surface where the friction is actually happening, but the whole block seems to be experiencing it. Also the initial force is only being applied in one area, then how are the other areas also experiencing the force? This is because the block is a rigid body. In order to understand this, we have to look at it at the molecular level, which brings us to the third point.

What do you mean by a block, or two blocks? A block can be divided into atoms(or molecules if it is a compound, but I'll just use atoms for simplicity), which are held together by forces of attraction, have small amounts of space in-between, and repel when pushed against each other. So even one block is made up of many tiny blocks i.e. atoms which have gaps between them. So what happens when you apply a force on one area of the block? The atoms only in that particular area experience the force initially, and so they should be on the verge of being displaced, but then 3 things happen:

1. The atoms are about to be displaced but because there is a force of repulsion when atoms are squeezed against each other, they can't pass through each other, and a minimum gap is maintained. So the force is carried over to the next atoms, and then to the next and so on, in a line along the force. This is how the force gets spread horizontally, parallel to the initial force.

2. The atoms that are about to be displaced pull on the other atoms surrounding it, attempting to pull them in the direction of the force. Those atoms then pull on the other atoms surrounding them, and so on until atoms at both the bottom and the top experience the initial force. This is how the force gets spread vertically and also diagonally.

3. Finally when initial force is carried over to the atoms at the bottom, they are about to be displaced, but they experience a frictional force, which prevents them from being displaced. Then just like the 2nd point except this time starting from the bottom, the frictional force also carries over to the surrounding atoms, and this keeps on happening, until all atoms experience the frictional force, and so none of them get displaced.

That was the detailed mechanism of how forces get distributed within one block. Now we can take the same principles and see how it applies to two separate blocks that are 'touching'. As we have already seen, at the atomic scale, the atoms of the two block will have a repulsive force between them and a minimum gap will be maintained. If two blocks are 'touching' such that the gap between the atoms of one block to another are already at a minimum then the forces will get distributed between both blocks. Otherwise not, and only the friction of the first block will apply

The practical answer, corresponding to real rigid blocks, is that there is no friction force at the second block.

The reasons are:

1. the macroscopic contact between blocks doesn't mean a full microscopic contact as happens inside each of the blocks.

2. The original question assumes no deformation, that is: rigid bodies. It translates in real life for blocks where the elastic deformation is neglible compared with the microdistances between the blocks.

3. There is a level arm between the applied force and the reaction friction force. But I assume that the microdetails of the contact area of the first block can generate enough couple to prevent any twist.