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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?

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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.

One caveat: it gets more complicated for massive strings and friction. Not that much more, though.

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  • $\begingroup$ Thanks a lot, I have guessed something like this for curved shapes. What about around sharp corners though? I'm not even sure along which direction the reaction force of the corner on a string should point. Could I approximate the corner by starting with an arc and letting the radius of curvature tend to 0? $\endgroup$
    – Si Chen
    Commented Jul 25, 2015 at 20:50
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    $\begingroup$ maybe? keep in mind that massless strings and infinitely sharp corners are mathematical idealizations, not real things. you can get an answer by taking the radius to zero but it may not apply to reality. $\endgroup$
    – knzhou
    Commented Jul 26, 2015 at 2:58
  • $\begingroup$ As radius goes to zero the finite size of the string and its stiffness will all come into play (as does the fact that "massless" doesn't really exist). The limiting approach gives useful insights - but as usual there is no good answer to the question "what is 0 divided by 0", only "what value does it approach". $\endgroup$
    – Floris
    Commented Jul 26, 2015 at 13:46
  • $\begingroup$ @Floris, can you give me a concrete demonstration of how the finite size of the string and its stiffness comes into play as radius of curvature -> 0? Thanks! $\endgroup$
    – Si Chen
    Commented Jul 26, 2015 at 15:56
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If we are talking about real strings then the radius of curvature would never go "all the way" to zero anyway. Therefore they aren't a special case, but the general case you have already described .

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  • $\begingroup$ Very true, but as I already mentioned the string being massless and the peg frictionless... you can see it's just idealisations here, a real life situation taken to some convenient limit. I'm just not sure which limit is right. $\endgroup$
    – Si Chen
    Commented Jul 26, 2015 at 2:58

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