# Why do you not double calculated tension when there are two tension forces? [duplicate]

So there is a weight acting at the centre of a rope causing two equal tension forces either side of it. I’m having difficulty working out why you do not double the tension calculated at the end. For example:

There is a person walking on a rope between two buildings. The weight of the person is 800N and they are directly over the centre of the rope. The rope sags by 5 degrees as a result.

When working out the tension in the rope I did sin5 = 400 / T and therefore T = 400 / sin5 = 4600N and then doubled it to get the total tension as my thinking was that the same tension applied to the left and the right of the person. I’m not sure if my problem is that the question wants the tension only in one part - the exact wording was “tension in the rope” - or if I am misunderstanding something conceptual about tension. I couldn’t find any similar examples in my textbook.

In short can somebody explain to me why you do not double the calculated tension for problems with two tension forces.

Edit: thanks for the link just what I needed couldn’t find that one when I looked earlier.