Why are the traction vectors on each surface independent?

[itamar123]

Hey y'all, my first thread here,
Got a burning question that has been disturbing my serenity.
In all derivations of the stress tensor that I've seen they didn't explain it that much,
So my question is, why do the traction vectors on each surface are independent?
From what I understood, the infinitesimal cube is actually a free body diagram of a small cube taken from a material, and so you put force (or traction) vectors acted on the surfaces that you cut,
But the sum of all the force (or traction) vectors is zero, why can't I just present it as two opposing and equal vectors?

 

[Chestermiller]

Hey y'all, my first thread here,
Got a burning question that has been disturbing my serenity.
In all derivations of the stress tensor that I've seen they didn't explain it that much,
So my question is, why do the traction vectors on each surface are independent?
From what I understood, the infinitesimal cube is actually a free body diagram of a small cube taken from a material, and so you put force (or traction) vectors acted on the surfaces that you cut,
But the sum of all the force (or traction) vectors is zero, why can't I just present it as two opposing and equal vectors?

I don't quite understand your question. Can you please give a specific example with a figure?