Proof: Consider a rope being pulled with tension T at an angle ##\theta## to the horizontal. To simplify the situation, I split the rope up into a finite number of segments. I pull on segment 1, which then pulls on segment 2, etc all the way down to the last segment which pulls on the sled. When I exert a force T on segment 1, let the force that segment 1 exerts on segment 2 be T'. Then segment 2 exerts an equal and opposite force T' on segment 1. Hence segment 1 is being pulling simultaneously to the left and right. Also, segment 1 has a weight mg acting downwards. The component of this weight in the direction parallel to the rope (ie "down the rope") is ##mg\sin \theta##. The tensions and weight are the only forces acting on segment 1, hence ##mg\sin \theta+T'=T##, giving ##T-T'=mg\sin \theta##. Now because m is negligible this is roughly ##T-T'\approx 0## ie ##T\approx T'##. Since this proof can be applied to any segment of rope, it follows that the tension is the same throughout the rope. ##\square##

If the proof is correct, I am still wondering if a special proof is needed for each situation in which tension is applied to ropes. Is there a general proof that deals with all cases? For example, one can give a separate proof dealing with the case of a rope hanging from a ceiling in order to show that the tension is the same throughout if the rope is considered massless, but this doesn't prove the rule in general: if a string is massless then the tension is the same throughout.

Thanks