(a) The resistance force F of a ship is a function of its length L, velocity V, acceleration due to gravity g and fluid…

(a) Variables: F (MLT⁻²), L (L), V (LT⁻¹), g (LT⁻²), ρ (ML⁻³), μ (ML⁻¹T⁻¹). There are n = 6 variables and m = 3 fundamental dimensions, so by Buckingham's π-theorem there are 6 − 3 = 3 dimensionless groups. Take ρ, V and L as repeating variables (they contain M, L and T and represent fluid, flow and geometry).

• Π₁ = F ρᵃ Vᵇ Lᶜ: M: 1 + a = 0; T: −2 − b = 0; L: 1 − 3a + b + c = 0, so a = −1, b = −2, c = −2 and Π₁ = F/(ρV²L²) (resistance coefficient).
• Π₂ = g ρᵃ Vᵇ Lᶜ: a = 0, b = −2, c = 1, so Π₂ = gL/V², the inverse square of the Froude number Fr = V/√(gL).
• Π₃ = μ ρᵃ Vᵇ Lᶜ: a = −1, b = −1, c = −1, so Π₃ = μ/(ρVL), the inverse of the Reynolds number Re = ρVL/μ.

Dimensionless relationship:

F/(ρV²L²) = φ(V/√(gL), ρVL/μ), i.e. F = ρV²L² φ(Fr, Re).

Interpretation: wave-making resistance depends on Fr and skin friction on Re. Both cannot be matched in a model tested in water, so model tests are run at the prototype Froude number and the frictional part is corrected separately (Froude's method).

(b) With the convention u = ∂ψ/∂y, v = −∂ψ/∂x and ψ = 2xy:

• u = 2x, v = −2y. At P(2, 3): u = 4, v = −6 units.
• Resultant velocity = √(4² + 6²) = √52 = 7.21 units, directed at tan⁻¹(6/4) = 56.3° below the positive x-axis (tangent to the streamline ψ = 12 through P).
• Rotation: ω_z = ∂v/∂x − ∂u/∂y = 0 − 0 = 0, so the flow is irrotational and a velocity potential exists (it is also continuous: ∂u/∂x + ∂v/∂y = 2 − 2 = 0).
• With u = ∂φ/∂x and v = ∂φ/∂y: ∂φ/∂x = 2x gives φ = x² + f(y); then f′(y) = −2y gives f(y) = −y². So φ = x² − y² + C.
• At P(2, 3), taking C = 0: φ = 4 − 9 = −5 units.

With the opposite convention (u = −∂ψ/∂y, v = ∂ψ/∂x, u = −∂φ/∂x), the components change sign (u = −4, v = 6) but the magnitude 7.21 and φ = x² − y² (−5 at P) are unchanged. This is flow into a 90° corner with a stagnation point at the origin; streamlines (rectangular hyperbolas xy = constant) cut the equipotentials x² − y² = constant at right angles.

(c) Undrained cohesion c_u = q_u/2: 30 kN/m² in the top 5 m and 50 kN/m² in the next 10 m. No adhesion factor is given, so α = 1 is assumed. Pile diameter 0.3 m, length 15 m.

Individual action:

• Skin friction per pile: π × 0.3 × (30 × 5 + 50 × 10) = 0.9425 × 650 = 612.6 kN.
• End bearing per pile: N_c c_u A_b = 9 × 50 × (π/4 × 0.3²) = 9 × 50 × 0.0707 = 31.8 kN.
• Capacity of one pile = 644.4 kN; nine piles = 9 × 644.4 = 5800 kN.

Block action (3 × 3 group, 1.00 m centres in a row, rows 0.75 m apart):

• Block size = (2 × 1.00 + 0.3) × (2 × 0.75 + 0.3) = 2.30 m × 1.80 m; perimeter = 8.2 m; base area = 4.14 m².
• Side shear = (30 × 5 + 50 × 10) × 8.2 = 5330 kN.
• Base resistance = N_c c_u A = 9 × 50 × 4.14 = 1863 kN.
• Block capacity = 7193 kN. (Using the gross base pressure c_uN_c + γD and subtracting the weight of the soil block, γ = 16.4 kN/m³ cancels, giving the same net value.)

Governing capacity: the smaller value, 5800 kN (individual action; group efficiency = 7193/5800 = 1.24 > 1, so block failure does not govern).

Safe load on the group = 5800/2.5 ≈ 2320 kN.

(If the end bearing of individual friction piles is neglected, 5513 kN and 2205 kN are obtained; including it is consistent with using N_c for the block base.)