# Why do tensions cancel out in some places but not others?

I was answering a question regarding two rotating particles $$P$$ and $$Q$$ with $$P$$ attached to a fixed point $$O$$ by a light inextensible string and $$Q$$ attached to $$P$$ by a light inextensible string with $$P$$ and $$Q$$ rotating about the vertical through $$O$$ in the same vertical plane and at constant angles to the vertical:

If I consider the whole system and consider the forces acting on the vertical plane I believe I get something like this:

Even if this is slightly wrong I know that I consider the tension $$T_{1}$$ acting in the direction $$\vec{PO}$$ but not in the direction $$\vec{PQ}$$ because it is cancelled out by the tension in the direction $$\vec{QP}$$ which is equal and opposite. My question is therefore why is there not an equal and opposite tension in the direction $$\vec{OP}$$ that cancels out $$T_{1}$$. Is it the fact that $$O$$ is a fixed point rather than a particle? If so why does this make a difference?

If you are considering the whole system to be both particle $$P$$ and $$Q$$, then this makes sense. You are only drawing external forces in your second picture. The tension between $$P$$ and $$Q$$ is an internal force, so you have not drawn it. However, the tension between $$P$$ and $$O$$ is an external force, so you have drawn it.
If you wanted to "get rid of" the tension between $$P$$ and $$O$$ you would need to include point $$O$$ (whatever it is physically) in your system. However, you would then also need to consider all of the other forces acting on $$O$$ that keeps in fixed in space. So you would get rid of the tension between $$P$$ and $$O$$ from your diagram, but you would then include other forces acting on $$O$$.