Step 2: find the corresponding physical models Model 1: Elastic rod for the steel cable:

1. The American naturalist Charles William Beebe (1877-1962) set a world record in 1934 when he made a dive to a depth of 923 m below the surface of the ocean. The dive was made in a device known as the bathysphere, which was basically a steel sphere 4.75 ft in diameter. Suppose the bathysphere and its occupants had a combined mass of 13200 kg and that they were lowered into the ocean on a steel cable with a radius of 1.66 cm. How much did the cable stretch when the bathysphere was at the depth of 923 m? (Neglect the weight of the cable itself, but include the effects of buoyancy.) [Answer] Step 1: write down variables Mass of the bathsphere: M = 13200 kg Diameter of the bathsphere: D = 4.75 ft Volume of the bathsphere: V = 1/6 π D3 Radius of steel cable: R = 1.66cm Cross section area of the cable: A = π R2 Length of the steel cable: L = 923m Tension force on the cable: T Buoyance force Fbuoyance Gravitational Force W

2. Step 2: find the corresponding physical models Model 1: Elastic rod for the steel cable: Tensile test: Young’s Modulus; Y = P / ( ΔL/L ) = (F * L) / ( A*ΔL ) Model 2: Object in liquid (with gravity field): Archimedes’ Principle: Fbuoyance = ρwater g V Model 3: State of equilibrium: (no motion) Newton’s second law: ΣF = T + Fbuoyance – W = 0 Step 3: write down applicable equations based on the models Ysteel = T * L / ( A * ΔL ) (1) Fbuoyance= ρwater g V (2) T + Fbuoyance – W = 0 (3) V = 1/6 π D3 (4) A = π R2 (5) Plugging in the numbers from step 1 and finding the Young’s modulus of steel from your textbook, we obtain, ΔL = 60.7cm