Using fluid mechanics to show that force is directly proportional to velocity So I am writing a paper about viscous dampers in harmonic oscilators, however I was looking at some old fluid mechanic notes and I thought I had come across what I needed although I have gotten stuck. So I have that $\tau=\mu\frac{\delta u}{\delta y}$ this yields that $F=A*\mu\frac{\delta u}{\delta y}$. I want to get $F\propto\frac{\delta u}{\delta t}$ somehow but I am getting stuck, I'm not even sure its possible. Although I would like some guidance. Also this is assuming that we do not know that drag force is -$kmv$.
 A: At low Reynolds number, as in a creeping flow, one can ignore the advective acceleration terms in the Navier-Stokes equation. If we also assume a steady state, the equation becomes
\begin{equation}
0 = -\nabla p + \mu\nabla^2\vec{v},
\end{equation}
where $p$ is the hydrostatic pressure, $\mu$ is the viscosity of the fluid and $\vec{v}$ is the flow velocity.
Let us consider the motion of a small, spherical object in a fluid. An example could be motion of an oil droplet in Millikan's experiment. The moving object induces a flow in the fluid, which in turn, drags the object. If we solve this equation for a sphere of radius $R$, we first get the viscous force per unit area on the sphere to be 
\begin{equation}
\vec{f} = \frac{3\mu\vec{v}}{2R}
\end{equation}
We can then compute the drag force by integrating $\vec{f}$ over the entire surface. The result is 
\begin{equation}
\vec{F} = 6\pi\mu R\vec{v}
\end{equation}
A: If you solve stokes flow (time-independent navier-stokes equation without the non-linear term) for a sphere, you will find a linear dependence between the drag and the velocity. 
