Internal force disintegrating a solid body?

Question

Let $M$ be a block on a frictionless surface. Now let us mentally divide (not physically) the block into 2:1 ration (i.e $1/3$ of the left be called $M_1$ and $2/3$ right be called $M_2$). So $M_1$ applies force $F_1$ on $M_2$ and $M_2$ applies force $F_2$ on $M_1$ and by 3rd law they are equal. Hence acceleration of $M_1$ would be $2a$ and that of $M_2$ would be $a$. Shouldn't this deform the block?

Answer 1

If the body is isolated, then $a=0$ and the internal forces are $0$. This is a trivial case where there is no longer a contradiction or possibility of the body being torn apart. Therefore, let's consider the scenario of a body that has external forces acting on it.

You are forgetting about external forces being applied to the object. In other words, the net force acting on section $i$ in general is not going to just be $F_i$. You need to specify what external forces are acting on the entire object (and where these forces are being applied). Once you do this, you will find that treating the system as a whole or as many parts should still give you consistent results.

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Let's take a simple example of a rigid cube of mass $M$ on a frictionless surface, and we push on the left side of this cube to the right with a force of magnitude $F$. Then by Newton's second law the acceleration of the cube is $a=F/M$.

Now let's hypothetically partition our cube into two pieces like you describe, with $1/3$ of the cube as one part on the left and $2/3$ of the cube as the other part on the right. Let's call the interaction force between the two parts $F_{\text{int}}$. Applying Newton's second law to each partition with acceleration $a=F/M$ we have $$F-F_{\text{int}}=\frac M3\cdot\frac FM$$ $$F_{\text{int}}=\frac{2M}{3}\cdot\frac FM$$ Both equations show that $F_{\text{int}}=2F/3$. Therefore, we don't have any issues here, and this shows the mistake you were making. Each partition has an acceleration of $a=F/M$, and the partition with mass $2M/3$ mass has twice the net force to the right in order for it to have the same acceleration as the partition with mass $M/3$. No deformation occurs due to how we hypothetically partition the system.

In fact, you can generalize the above example for the case of partitioning the left side to have mass of $xM$ and the right side to have mass $(1-x)M$ for $0\leq x\leq1$. You can use the same above analysis to show that the interaction force has a magnitude of $(1-x)F$. This makes sense: the father you are from where the external force is applied, the weaker the interaction force becomes. It even drops to $0$ once we consider the entire block as a single partition ($x=1$), and it is equal to $F$ right at the point of application ($x=0$).

Answer 2

Most of your intuition is correct. The piece that you're missing is the constitutive relationship that describes how the force $F$ between the pieces depends on their relative position and velocity.

A decent model for most materials is that the constitutive forces act like some combination of a "spring" holding the pieces together, and a "friction damper" that makes their relative motion slow down (you can think of this combination like the mechanism on an automatic door closer).

The spring-damper forces are at equilibrium (zero force) when the block is undeformed and not changing shape, so that the block is moving like the rigid object you expect ($F=0$ gives $a=0$ in your model).

If something did disturb the pieces of the block (i.e. deform the whole block), then the spring-damper constitutive forces would pull it back into shape. The strength of the friction in the damper would determine whether it settles quickly back into shape (for large friction) or "rings" by having the parts of the block oscillate outward and inward (for small friction).

Answer 3

Actually you haven't truly addressed as to what kind of material we are dealing with over here, so let me dissect it into two types:

1. Totally rigid

2. Jelly like

Totally Rigid Bodies

For objects made up of materials behaving as such as soon as an internal force tries to deform the object an opposite restoring force is generated which balances it. So here you were missing a restoring force in your calculation.

Jelly Like Bodies

In these when an internal force acts on the particle then a deformation is caused which leads to greater or lesser volume than originally it would have.

Also the forces aren't unidirectional and hence the deformation occurs in all direction.

$$\underline {\text {Reality}}$$

No real body is a perfect example of both the given case and hence there always is some kind of deformation.

A significant effect of this can be seen in stars (main internal force being gravity).

[Note: It must be noted that since internal forces cancel each other therefore the center of mass doesn't accelerate even a bit(unless an external unbalanced force acts). ]

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Suggested Materials:

1. Does a particle exert force on itself?

2. Neutron Stars

Answer 4

I believe you are making a mistake in the way you are imagining force is being applied on the masses. The scenario where $M_1$ and $M_2$ move with acceleration $2a$ and $a$ respectively is when $F$ is applied on both the bodies separately, imagining both are rigid bodies.

The idea of deformation in your mind, what I perceive is, that if $F$ is applied at either body when they are kept adjacent to each other, difference in acceleration will mean that they should deform. BUT, like Aaron clearly showed you, this is not equivalent to application of $F$ on both bodies independently.

Here the force will be redistributed in such a way, that both the bodies will have same acceleration. The whole idea works only when assuming rigid body condition of $F_{12} = F_{21}$. This is the key. You can read more about $\textbf{strong}$ and $\textbf{weak}$ laws of action and reaction.