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:
Diameter of the bathsphere:
Volume of the bathsphere:
Radius of steel cable:
Cross section area of the cable:
Length of the steel cable:
Tension force on the cable:
Buoyance force
Gravitational Force
M = 13200 kg
D = 4.75 ft
V = 1/6 π D3
R = 1.66cm
A = π R2
L = 923m
T
Fbuoyance
W
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