# Tension of rope over frictionless edge obstructing block

Consider the diagram below. All corners are assumed to have no friction. Is it true that the tension throughout the rope is equal in all sections? The section over the edge of the building, the section between the edge of the building and the back right corner of the block where the rope turns around it, and finally the section between the back right corner of the block and the eyebolt anchor point.

Maybe another way I could reframe my question is, can the angles around these edges influence the tension in the different sections? For example, the angle between the rope and the block is not the same on the load-side as the angle between the rope and the block on the anchor-side. Does that matter?

• If you don't consider friction edge angle don't affect tension. It will be uniform throughout its length.
– Fire
Commented Jul 14 at 2:49
• Thankyou for your help @Fire Commented Jul 14 at 3:15
• As long as all surfaces are considered frictionless and the rope is considered non extensible, the tension would theoretically be constant throughout. IMO this is an example of “pushing the envelope”. Normally, at least the surfaces are rounded. Commented Jul 14 at 6:50
• All corners are assumed to have no friction. This is quite an assumption! Commented Jul 14 at 10:57
• @Bucephalus The direction of a normal to an ideal corner is an interesting concept. With you two bits of rope, either side of the corner, you can always choose the normal to be one which bisects the angle. Commented Jul 15 at 12:37

I would consider each bent part a point of interest. Starting at the last bend before the hanging mass, let's fix the rope on the edge and calculate the tension for this segment. Call it $$|T_2|=mg$$.
Next, the kink at the block. Pretend the rope is fixed at this point. This time, let's simplify by replacing the edge and hanging mass with a constant force of $$T_2$$. See where this is going? This tension is just $$T_1 = T_2$$.