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a tapered bar

https://classes.mst.edu/civeng2210/concepts/05/elongation/index.html

If the bar is separated in to segment AC and segment CB, to maintain its equilibrium, Shouldn't the N(x)s be directed in opposite?

In Static, I learned that if we apply two equal, opposite forces at the ends of a member and cut it at the middle, there should be internal forces equal and opposite to each forces at the ends of the segments like this.

a member cut into two pieces

PS:I should've been clearer. What I'm asking is why all the forces(P and N(x))of the portion AC in the first pic are pointing right, whereas F and F' of the portion AC in the second pic are in different directions. Why is that? I'm totally confused.

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From both images you have the internal forces ($N(x)$) are opposite for facing cross-sections. enter image description here

The portion denoted as C, is a very thin cylindrical disk with two opposite facing surfaces.

PS: This is not a real answer. I need to show the image. If the question updates, I will probably update this post to something more meaningful.


UPDATE

PS:I should've been clearer. What I'm asking is why all the forces(P and N(x))of the portion AC in the first pic are pointing right, whereas F and F' of the portion AC in the second pic are in different directions. Why is that? I'm totally confused.

I assume you are referring to the green encircled arrows in the image below.

enter image description here

The reason is that the author of that image wants to point out that when P is applied at the end, then at any point between P and the support there is a tensile force.

Essentially that the beam "carries" along the force between the point of application (B), and the support (A). In my native language, a more generic term to describe the beam (or any system that "carries" the force) is literally translated as "carrier" (I've not seen it in English though).

The point is that when you take a sliver (a thin disk of material) along AB , at any point you would get C the same behavior as the sliver C. Actually, it doesn't have to be a thin disk of length dx. A portion of any thickness between A and B would exhibit the same behavior (such as the bottom figure with FF' at the original post).

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  • $\begingroup$ Sorry, It's not my first language so there was a mistake. $\endgroup$ Jan 28 at 10:27
  • $\begingroup$ I should've said "reversed" instead of "directed in opposite". Or "inversed"? Anyway, thanks a lot. $\endgroup$ Jan 28 at 10:34

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