forcesfree-body-diagramnewtonian-mechanicsstring
Suppose, me and my brother are playing tug of war. If we both pull the rope to ourselves at 50N force, won't the tension force be zero as the net force is zero? However, practically, this doesn't happen: if the tension force becomes zero, the rope should slack over due to gravity, but this doesn't happen; it remains tight. So, does the tension force become zero or not?
It is always best to draw a diagram to convince yourself of things in a case like this.
This is intended to represent a steady state situation: nobody is moving / winning. As you can see, there are two horizontal forces on A: the floor (pushing with 100N) and the rope (pulling with 100N). There will be two vertical forces (gravity pulling down on center of mass, and ground pulling up) to balance the torques - I did not show them because they are not relevant to the answer.
Now I drew a dotted line between A and B. Consider this a curtain. A cannot see whether the rope is attached to B (an opponent) or a wall. A can measure the tension in the rope by looking (for instance) at the speed at which a wave travels along the rope - or by including a spring gage.
Now ask yourself this question: if A feels a tension of 100N in the rope (this is the definition of the force on A), and can confirm (by looking at the gage) that the tension is 100 N, but he cannot see whether the rope is attached to a ring or to an opponent, then how can the tension be 200N? If I pull on a gage with a force of 100N, it will read 100N - it cannot read anything else (in a static situation, and where the gage is massless, ... )
I think I understand the source of your confusion based on the earlier q/a that you referenced - so let me draw another diagram:
In this diagram, I have move the point of attachment of the rope with which A pulls B away from B's hands, to his waist. Similarly, the rope with which B pulls on A is moved to A's waist.
What happens? Now there are two distinct points where A experiences a force of 100 N: one, his hands (where he is pulling on the rope attached to B's waits); and another where the rope that B is pulling on is tied around his waist.
The results is that there are two ropes with a tension of 100N each, that together result in a force of 200N on A (two ropes) offset by a force of 200N from the floor, etc.
This is NOT the same thing as the first diagram, where the point on which B's rope is attached is the hands of A - there is only a single line connecting A and B with a tension of 100 N in that case.
As was pointed out in comments, you can put a spring gauge in series with your rope to measure the tension in it; and now the difference between "a single person pulling on a rope attached to a ring at the wall (taken to be the dotted line) and two people pulling across a curtain (so they cannot see what they are doing) is that in one case, a single spring (with spring constant $k$) expands by a length $l$, while in the second case you find a spring that's twice as long, with constant $k/2$), expanding by $2l$.
These are all different ways to look at the same thing.
If the pulley is not accelerating, or sliding against the rope, then the rope exerts a purely radial normal force on it at each point of contact. The resultant force 2T is the vector sum of these forces around the arc of contact.
In the diagram on the left, when the tension in the rope $T$ changes direction by angle $d\theta$ there is a resultant force $dF=2T\sin(\frac{d\theta}{2})\approx Td\theta$. This force acts on the pulley, and has components $dF_x=T\cos\theta d\theta, dF_y=T\sin\theta d\theta$.
In the diagram on the right, we see the rope continually changing direction. Over the element of arc length shown, it changes direction by $d\theta$, which is the same as the angle subtended from the centre. The force $dF$ from each element of the rope of length $ds$ has to be integrated along the arc of contact :
$F_x= \int T\cos\theta d\theta = T\sin\theta$
$F_y= \int T\sin\theta d\theta = T(1-\cos\theta)$
taking $\theta=0$ at the starting point.
If we are integrating round a semicircle then $\theta=\pi$ so $F_x=0, F_y=2T$. If we are integrating round a quarter circle then $\theta=\frac12 \pi$ so $F_x=T, F_y=T$.
These results are obtained far more simply by looking at the pulley and the rope in contact with it as a single Free Body and considering the external forces acting on it, as in your diagrams.
Best Answer
If you and your brother are pulling there has to be tension in the rope of 50 N. But if you are both holding your ground it means that the net external force acting on the combination of you as a system is zero. See the free body diagrams on each of you below.
The external forces would be the static friction forces that are exerted on each of your feet in opposition to the tension. Those static friction forces would be equal and opposite for a net force of zero. However, there is a maximum possible static friction force on each of you that equals your weight, $Mg$, times the coefficient of static friction between your feet and the ground before slipping occurs. If one of you is able to pull hard enough so that the maximum static friction force on the other is exceeded, the other will slip and lose the tug of war.
One other point. In order that there is no net torque one each of your causing you to fall, each of you would normally be tilted backward (not as shown) so that your weight provides a torque about your feet to counter the torque due to the rope.
Hope this helps.