Integral of Rope Tension on Sphere

A segment of the chain between $\phi$ and $\phi+d \phi$ has a mass of $\rho d \phi$ , where $\rho=3m/\pi$ is a constant. The tangential tension force pulling this segment backwards is $F$ , the tension pulling forward is $F+d F$ and the tangential component of the gravitational force is $\rho g \sin\phi d \phi$ . Newton's law for this segment is

$(\rho g \sin\phi )d \phi - F+ F+dF= (a \rho) d \phi$

where $a$ is the magnitude of the acceleration that is independent of the angle $\phi$ . This leads to

$\frac{1}{\rho g}\frac{dF}{d \phi}=\frac{a}{g}- \sin \phi$

We solve this differential equation with the boundary condition that the tension is zero at $\phi=0$ and at $\phi=\pi/3$ . The solution is

$F=\rho g\frac{a}{g}\phi +cos \phi -1$

where $\frac{a}{g}=\frac{3}{2\pi}$ , otherwise the boundary condition is not satisfied. Inserting $\rho=3m/\pi$ and performing the integration yields

$\int_0^{\pi/3} F d\phi=mg \frac{3}{\pi}\left(\frac{3}{4\pi}\phi^2 +\sin \phi -\phi\right)|_0^{\pi/3}=mg \left(\frac{3}{\pi}\sin\frac{\pi}{3}-\frac{3}{4}\right)= 0.7699$