= Solution
Set $G=\Delta p/L>0$, so $dp/dz=-G$. For the unidirectional velocity $\mathbf u=u(r)\mathbf e_z$, the axial inertialess <Cauchy momentum equation> is
$$
-G=\frac1r\frac{d}{dr}(r\tau_{rz}).
$$
The free surface is shear-free, $\tau_{rz}(b)=0$, and integration gives
$$
\boxed{\tau_{rz}(r)=\frac G2\left(\frac{b^2}{r}-r\right).}
$$
This result follows from force balance alone, so it is independent of the <constitutive equation> and of $\sigma_y$.
The axial shear force exerted by the liquid on the cylindrical fibre is therefore
$$
F_z=2\pi aL\tau_{rz}(a)
=\pi aLG\left(\frac{b^2}{a}-a\right)
=\boxed{\pi\Delta p\,(b^2-a^2)}.
$$
Its direction is downstream; the reaction on the fluid is upstream. In particular, the yield stress changes the velocity field but not the total force required by the imposed pressure drop.
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