Viscosity of oil between parallel plates

Use Newton's law of viscosity, $\tau = \mu\, du/dy$, with a linear velocity distribution between the plates.

Given: gap $dy = 5$ mm $= 0.005$ m, $F = 2$ N, $u = 0.8$ m/s, $A = 0.09$ m², $S = 0.9$.

Dynamic viscosity

$$\begin{aligned} \tau &= \frac{F}{A} = \frac{2}{0.09} = 22.22\ \text{N/m}^2 \\ \frac{du}{dy} &= \frac{0.8 - 0}{0.005} = 160\ \text{s}^{-1} \\ \mu &= \frac{\tau}{du/dy} = \frac{22.22}{160} = 0.1389\ \text{N·s/m}^2 \end{aligned}$$

Kinematic viscosity

$$\begin{aligned} \rho &= 0.9 \times 1000 = 900\ \text{kg/m}^3 \\ \nu &= \frac{\mu}{\rho} = \frac{0.1389}{900} = 1.543 \times 10^{-4}\ \text{m}^2/\text{s} \end{aligned}$$

Answer: $\mu = 0.1389$ N·s/m² (= 1.389 poise), $\nu = 1.543 \times 10^{-4}$ m²/s (= 1.543 stokes).