# Tension in a massless string as it bends around a concave shape

I understand that tension in a massless string must be the same throughout when the string is straight, to prevent any section of it from accelerating at an infinite rate. However, I fail to see how this remains true when the string bends around some general concave outline (e.g. I have a diamond or elliptical peg, and I wrap the string around a corner/a part of the perimeter to hang a picture off it).

I can see how it would be true for a circular peg. Again as we do not want the section of the string wrapped around a peg to accelerate at an infinite rate, the reaction force of the peg on the string must balance the sum of the tensions on each side. As this is a three-force member the lines of action of the forces must meet at one point. As the tensions on each side are tangents of the circle and the reaction force's line of action goes through its centre, the reaction force's line of action must bisect the angle made by the lines of action of the tensions on each side. Thus, to ensure the tangential force on the string is 0, the two tensions must be the same. Can I extend this logic to any convex shape?

Yes, your reasoning is correct. One way to see this is that if the radius of curvature of your shape is at some point is $R$, then it's locally just like a circle of radius $R$, and you already know the result for circles.