String Tension on a Sphere in Current

Equations: drag force F_D = ½·C_D·ρ·A·V², with A the frontal (projected) area of the sphere, A = πD²/4. Submerged weight W′ = mg − ρ_water·g·Vol, with Vol = πD³/6. At equilibrium the string tension balances both, so tan θ = F_D / W′. Values: C_D = 0.47, ρ = 1000 kg/m³ and V = 1.5 m/s are given. D = 0.10 m gives A = π(0.10)²/4 = 0.00785 m² and Vol = π(0.10)³/6 = 5.24 × 10⁻⁴ m³. m = 2.0 kg. Step 1: drag, F_D = 0.5 × 0.47 × 1000 × 0.00785 × 1.5² = 4.15 N (horizontal, downstream). Step 2: weight, mg = 2.0 × 9.81 = 19.62 N; buoyancy, 1000 × 9.81 × 5.24 × 10⁻⁴ = 5.14 N; so W′ = 19.62 − 5.14 = 14.48 N (vertical, down). Step 3: tan θ = 4.15 / 14.48 = 0.287, so θ = 16.0° from the vertical, deflected downstream. Ignoring buoyancy gives tan θ = 4.15/19.62 and 12.0°. Leaving out the ½ doubles the drag and gives 29.8°. 74.0° is the angle measured from the horizontal.