Problem Statement
A thin plate is pulled at a velocity \( V \) through a narrow gap of height \( h \). The gap is filled with oil of viscosity \( \mu_1 \) on one side and \( \mu_2 \) on the other side. Derive expressions for:
- The position of the plate where the shear force on both sides is equal.
- The position of the plate where the pull required to drag it is minimum.
Solution
(a) Position where shear forces are equal:
Shear stress on the upper face:
\( \tau_1 = \mu_1 \frac{V}{h-y} \)
Shear stress on the lower face:
\( \tau_2 = \mu_2 \frac{V}{y} \)
Equating \( \tau_1 \) and \( \tau_2 \):
\( \mu_1 \frac{V}{h-y} = \mu_2 \frac{V}{y} \)
Simplify:
\( y = \frac{\mu_2 h}{\mu_1 + \mu_2} \)
(b) Position where the pull required is minimum:
Pull required per unit area:
\( F = \tau_1 + \tau_2 = \mu_1 \frac{V}{h-y} + \mu_2 \frac{V}{y} \)
To minimize \( F \), differentiate with respect to \( y \) and set \( \frac{dF}{dy} = 0 \):
\( \frac{dF}{dy} = \mu_1 \frac{V}{(h-y)^2} – \mu_2 \frac{V}{y^2} = 0 \)
Simplify:
\( \frac{\mu_1}{(h-y)^2} = \frac{\mu_2}{y^2} \)
Cross-multiply:
\( \mu_1 y^2 = \mu_2 (h^2 – 2hy + y^2) \)
Rearrange and solve for \( \frac{h}{y} \):
\( \frac{h}{y} = 1 + \sqrt{\frac{\mu_1}{\mu_2}} \)
Final position of the plate:
\( y = \frac{h}{1 + \sqrt{\frac{\mu_1}{\mu_2}}} \)
Explanation
The derivation follows these principles:
- Equal shear forces: At the position \( y \), the shear stresses \( \tau_1 \) and \( \tau_2 \) are balanced. The formula depends on the relative viscosities (\( \mu_1 \) and \( \mu_2 \)) and the height \( h \).
- Minimizing pull force: To minimize the required pull, the derivative of the total force is set to zero, leading to a quadratic equation in \( y \). The solution gives the optimal position of the plate.
- The derived formulas have practical applications in optimizing drag in systems with fluids of differing viscosities.
