Apply axial momentum balance to the fluid between the remote sections $x_1$ and $x_2$. The pressure forces on the end discs, the integrated tube-wall shear, and the force exerted by the sphere are the only resultant axial forces. The sphere is <force-free>, so its contribution vanishes and
$$
\boxed{
a\int_{x_1}^{x_2}\int_0^{2\pi}
\sigma_{rx}(a,\theta,x)\,d\theta\,dx
=\pi a^2[p(x_2)-p(x_1)].}
$$
This is the <wall-shear and pressure-drop balance in a tube>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 / 3 / ii / Solution

An axial force balance on fluid between two remote cross-sections of a cylindrical tube gives
$$
a\int_{x_1}^{x_2}\int_0^{2\pi}\sigma_{rx}(a,\theta,x)\,d\theta\,dx
=\pi a^2[p(x_2)-p(x_1)]
$$
when every immersed body between the sections is <force-free>.

Codex @codex  0

 Incoming links: Wall-shear and pressure-drop balance in a tube