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In the solution to a problem, the author considers the normal force provided by an arc length of string with a differential subtended angle size, $\textrm{d}\theta$. The author reasons that this normal force is provided by the nonzero sum of the tension vectors on both ends of this small arc length. These tension vectors nearly cancel, but they do not entirely cancel since their directions are almost but entirely antiparallel to each other. The resultant sum is the normal force.

What I cannot follow are the author's geometric arguments. Here is the figure provided: enter image description here

The author reasons that if the subtended angle is $\textrm{d}\theta$, then the radial component of both tension vectors is $T\sin{\textrm{d}\theta/2}$. I do not follow how the subtended angle of $\textrm{d}\theta$ leads to an angle of $\textrm{d}\theta/2$ in that small right triangle.

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Imagine a vector $\mathbf{T}$ tangent to a circle. If you move the vector on an arc subtending angle $\theta$, then the vector at its final position rotates by the same angle, and it's projection on the axis to its initial position is $|\mathbf{T}|\sin\theta$. Here the tensions at each end have equal magnitude (necessary condition for lack of stretching), but as the arc opens by an infinitesimal amount $d\theta$ you can imagine the two vectors rotating in opposite directions, but they will go only half the way. Thus both project on the vertical axis with component $|\mathbf{T}|\sin(d\theta/2)$ as pointed out by the author.

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  • $\begingroup$ I understand your argument up to the part where you discuss angles of $\textrm{d}\theta$. Why must I view one vector as rotating $\textrm{d}\theta/2$ in one direction while the translated vector rotates the same amount in the other direction? Why does the original vector rotate at all considering we're only viewing the change in angle from the original vector to the translated vector? $\endgroup$ – Andrew Paul Jun 5 at 20:54
  • $\begingroup$ The tensions are tangent to the ends of the string arc, they are not just arbitrary vectors, that's why the argument holds true. Start rotating them along the arc backwards and they will meet at the middle of it, each of them having subtended an angle of $d\theta/2$. If this doesn't satisfy you, then you might as well do some geometry to see it. :) $\endgroup$ – DinosaurEgg Jun 5 at 20:58
  • $\begingroup$ Why doesn't the same argument work for non-differential angles $\theta$? $\endgroup$ – Andrew Paul Jun 6 at 0:46
  • $\begingroup$ the argument is fine for all angles, big or small. $\endgroup$ – DinosaurEgg Jun 6 at 0:55

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