Question about tensions in pulley problems?

[Elmer Correa]

Homework Statement

The scenario I'll use specifically is in the attached file. I can understand why the pulley can have two different tensions, one for each side of the pulley, but what I don't get is why the direction of each individual tension reverses direction? For example, in the free body diagram for mass 2, the direction of tension is in the positive y direction while when drawn from the pulley's perspective, it's in the negative y. And how could tension 1 be in the negative x direction when it's the only force that could be pushing mass 1 to the right?
Thanks for any clarification.

Homework Equations

The Attempt at a Solution

 

[kuruman]

It all depends on what system you choose to draw the FBD for. If your system is the mass on the table, the tension is to the right; if your system is the pulley, then the tension is to the left. Note that the two arrows labeled T₁ in the two FBDs have equal magnitudes and opposite directions. They are action-reaction counterparts as required by Newton's third law.

 

[Elmer Correa]

Note that the two arrows labeled T1 in the two FBDs have equal magnitudes and opposite directions. They are action-reaction counterparts as required by Newton's third law.

Could you elaborate on the force pair at play here? And since this is a force pair, is it even accurate to consider both directions of tension 1 or 2 the same force? If so, is this just the result of naming conventions that make problems like these easier to solve?

 

[kuruman]

You can pull with a rope but you cannot push with a rope. That's why tensions in FBDs are always directed away from the system in question.
Look at the picture of two persons pulling on a rope shown below. If you ask the person on the left "Which way is the rope pulling on you?", he will say "to the right". If you ask the person on the right the same question, he will say "to the left."

The tension in the FBDs for each man will look like this