= Solution
For a Newtonian fluid,
$$
\sigma_{ij}=-p\delta_{ij}
+\mu\left(\frac{\partial u_i}{\partial x_j}
+\frac{\partial u_j}{\partial x_i}\right).
$$
On $r=a$, with $\mathbf n=\mathbf x/a$,
$$
A'=-\frac{3}{2a},
\qquad B=0,
\qquad B'=\frac{3}{2a^3},
\qquad
p=\frac{3\mu}{2a}(\mathbf U\cdot\mathbf n).
$$
The surface velocity gradient is therefore
$$
\frac{\partial u_j}{\partial x_i}
=\frac{3}{2a}n_i\bigl((\mathbf U\cdot\mathbf n)n_j-U_j\bigr).
$$
Contracting the stress with $n_j$, the normal part of the viscous stress cancels the pressure contribution, leaving the uniform traction
$$
\boxed{(\boldsymbol\sigma\cdot\mathbf n)_i
=-\frac{3\mu}{2a}U_i.}
$$
Back to article page
