\[ \text{À l’équilibre : } T = P \] \[ k \times ΔL = m \times g \] \[ k = \frac{m \times g}{ΔL} = \frac{0.2 \times 9.81}{0.05} ≈ 39.2\ \text{N/m} \]

\[ \text{Équilibre : } Π = P \] \[ ρ_{eau} \times V_{imm} \times g = ρ_{bois} \times V_{total} \times g \] \[ h_{imm} = \frac{ρ_{bois}}{ρ_{eau}} \times h_{total} = \frac{700}{1000} \times 10 = 7\ \text{cm} \]

\[ \text{Équilibre : } T + Π = P \] \[ Π = P – T = m \times g – T = 0.5 \times 9.81 – 3 ≈ 1.9\ \text{N} \]

\[ \text{Équilibre : } T = F \] \[ k \times Δx = F \] \[ Δx = \frac{F}{k} = \frac{10}{50} = 0.2\ \text{m} = 20\ \text{cm} \]

\[ \text{Équilibre : } Π = P \] \[ ρ_{eau} \times V_{imm} = ρ_{glace} \times V_{total} \] \[ \frac{V_{imm}}{V_{total}} = \frac{ρ_{glace}}{ρ_{eau}} = \frac{920}{1025} ≈ 0.898 \]

Soit ≈ 89.8% du volume immergé

\[ \text{Équilibre : } T + Π = P \] \[ k \times ΔL + ρ_{eau} \times V \times g = m \times g \] \[ \text{Volume du corps : } V ≈ \frac{m}{ρ_{corps}} ≈ 0.07\ \text{m}^3 \] \[ ΔL = \frac{m \times g – ρ_{eau} \times V \times g}{k} ≈ \frac{686 – 686}{2000} = 0\ \text{m} \]

Le plongeur est en flottabilité neutre!

\[ V = \frac{4}{3}πr^3 ≈ 0.0141\ \text{m}^3 \] \[ Π = ρ_{air} \times V \times g ≈ 0.166\ \text{N} \] \[ P = ρ_{He} \times V \times g ≈ 0.025\ \text{N} \] \[ F_{asc} = Π – P ≈ 0.141\ \text{N} \]

\[ P_x = m \times g \times \sin(30°) = 9.81\ \text{N} \] \[ T = k \times Δx = 100 \times 0.1 = 10\ \text{N} \]

À 1.9% près, T ≈ P_x (les frottements expliquent la différence)

\[ \text{Équilibre : } Π = P \] \[ ρ_{liq} \times V_{imm} \times g = m \times g \] \[ \text{Comme il flotte : } V_{imm} < V \] \[ \text{Si complètement immergé : } ρ_{liq} = \frac{m}{V} = 800\ \text{kg/m}^3 \]

Donc ρ_liq > 800 kg/m³

\[ V = 0.1^3 = 0.001\ \text{m}^3 \] \[ P = ρ_{bois} \times V \times g = 5.886\ \text{N} \] \[ Π = ρ_{eau} \times \frac{V}{2} \times g = 4.905\ \text{N} \] \[ T = P – Π = 0.981\ \text{N} \] \[ Δx = \frac{T}{k} ≈ 0.0123\ \text{m} = 1.23\ \text{cm} \]