How Does the Capstan Equation Derive Rope Tension?

[Charles Link]

But why are $T_B$ and $T_A$ the bounds of integration that correspond to the lower and upper limit of $\theta$?

It is only between these points that your differential equation applies. From the sail to the point of contact of the rope the tension is constant at $T=T_A$ and from the point where the rope leaves the cylinder to go to the sailor's hand it is constant at $T_B$. Between these two points, the tension decreases steadily as it wraps around the pole because of the frictional force. You have your equation $\Delta T=-\mu T \Delta \theta$. Perhaps if you would go over its derivation in more detail, you would see how the tension $T$ is changing from the frictional force. Meanwhile $\theta$ has limits which we call $\theta=0$ and $\theta$. The upper limit we could have called $\theta_B$, but in the final formula, $\theta$ is sufficient. As previously mentioned, this problem can be computed using length $ds$ instead of angle, but I think you can recognize that $ds=r d \theta$. e.g. you could find how the tension in the rope changes with distance around the pole.

 

[Mr Davis 97]

So does $\displaystyle \frac{dT}{d \theta} = - \mu T$ represent how to tension changes over the contact area of the rope and capstan?

 

[Charles Link]

So does $\displaystyle \frac{dT}{d \theta} = - \mu T$ represent how to tension changes over the contact area of the rope and capstan?

The capstan is a cylinder. The rope is wrapped around the cylinder as many times as necessary to reduce the force needed to keep the rope (sail end) still. Suggestion: Try it with a friend with a lamppost=have them hold a rope and pull on it and start wrapping it around the pole. You wrap it enough times and you won't really need to pull on it at all...Even one loop and they won't be able to win a tug-o-war... You won't be able to pull them either though=the same equations apply going the other way=i.e. if they stay fixed with a light steady tension and you tug... And yes, the differential equation only computes $T$ for the region of contact with the pole. Outside of this region $T$ is constant. (On one side it is equal to $T_A$, and on the other side $T_B$. If it is 10 ft. from the sail to the pole, $T=T_A$ stays constant for 10 ft.)