 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/38c/e/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The leading shear stress exerted by the fluid on the lower surface is

τ (x) = μ u_{y} (x, 0) = - \frac{μU}{h} - \frac{h}{2} p_{x} .

The horizontal force per unit span is

\int_{0}^{L} τ d x

. Substituting the pressure gradient from part (c) gives

\frac{1}{μL} \int_{0}^{L} τ d x = - 4 U ⟨ h^{- 1} ⟩ + 6 V ⟨ x h^{- 2} ⟩ + 6 C ⟨ h^{- 2} ⟩ .

Using the value of

C

and setting this force to zero yields

U (4 ⟨ h^{- 1} ⟩ - 3 \frac{⟨ h ^{- 2} ⟩ ^{2}}{⟨ h ^{- 3} ⟩}) = 6 V (⟨ x h^{- 2} ⟩ - \frac{⟨ h ^{- 2} ⟩ ⟨ x h ^{- 3} ⟩}{⟨ h ^{- 3} ⟩}) .

U = 6 V \frac{⟨ x h ^{- 2} ⟩ - ⟨ h ^{- 2} ⟩ ⟨ x h ^{- 3} ⟩ / ⟨ h ^{- 3} ⟩}{4 ⟨ h ^{- 1} ⟩ - 3 ⟨ h ^{- 2} ⟩ ^{2} / ⟨ h ^{- 3} ⟩} .

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