An inner force within solid objects. Is it tension? Does it even exist or am not understanding something?

There is a famous problem in which you are given two masses separated by a rope (with mass). Those two masses are pulled up by some force, and you are asked to find the tension at some point in the rope.

The way to do this is to look at the system in parts (2 masses and rope into two parts) that being a lower and upper part and to apply Newton's second law on each part.

I like to over complicate things because I feel by doing so I can reveal misunderstandings I have and correct them, as well as strengthen what I already know. Here is the problem I will like to over complicate.

Suppose there is a uniformly distributed block of mass $m$ being pushed with a force $F$ to the right, calculate the acceleration.

Idea: Look at the block in two parts (in half to make things simpler)

On the left half with have a mass of $m/2$. That tells me that the net force on the left half is not $F$ since I already know the acceleration is $\frac{F}{m}$. So I conclude there must be some force force $T$ from the right half pushing the left half back. I also conclude this force pushes the right half forward (right) and is the only force acting on the right half. If this force actually exists my math tells me $T=(1/2)F$ so that the net force on the left half works out:

$$\sum F_{\text{left}}=m_{\text{left}}a_{\text{left}}$$

$$F-T=(\frac{m}{2})a$$

$$F-(1/2)F=(\frac{m}{2})(\frac{F}{m})$$

My Question: Does this force $T$ actually exist or am I using Newton's second law incorrectly. What is this force if it exists? Where is it located?

• Actually, it is a compressive force, not a tensile force. The left half exerts a compressive force to the right on the right half, and the right half exerts an equal and opposite compressive force to the left on the left half. Nov 12, 2017 at 4:23

In a sense, yes. This is how internal forces work. If you divide the block into four equal-size pieces, then the total force on each piece is one-fourth of the total external force owing to the internal forces. In general, any arbitrary piece (mass $m$) of a solid object (mass $M$) experiences a total force of $f = mA = mF/M,$ where $A$ is the acceleration of the whole object and $F$ is the total external force on the object. This is a simple application of Newton's law. The difference between the external force and the force felt by the piece is due to internal forces that keep the solid object the same shape.
This is part of the reason why you can survive getting hit by a car when you are in another car. The total acceleration you feel is equal to $A = F_{impact}/(m_{car} + m_{you}).$ Thus, the force you feel is $F_{you} = m_{you}A = F_{impact}m_{you}/(m_{car} + m_{you}),$ which is much smaller than $F_{impact}$.