A crane lifts a steel submarine of density $7800 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ and mass 20,000 kg. What is the tension in the lifting cable (a) when the submarine is submerged in water of density $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ and (b) when it is entirely out of the water?

Question

A crane lifts a steel submarine of density $7800 \mathrm {~kg} / \mathrm { m } ^ { 3 }$  and mass  20,000 kg. What is the tension in the lifting cable (a) when the submarine is submerged in water of density $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ and (b) when it is entirely out of the water?

Quizplus · Accepted Answer

When the submarine is submerged, the tension in the cable is reduced by the buoyant force, which is equal to the weight of the displaced water. The buoyant force is $V \rho_{\text{water}} g$, where $V$ is the volume of the submarine, $\rho_{\text{water}}$ is the density of water, and $g$ is the acceleration due to gravity. Since the mass of the submarine is $20,000 \, \text{kg}$ and its density is $7800 \, \text{kg/m}^3$, its volume $V = \frac{m}{\rho_{\text{sub}}} = \frac{20,000}{7800}$. The weight of the submarine is $mg = 20,000 \times 9.8$. The buoyant force is $V \times 1000 \times 9.8$. The tension in the cable when submerged is the weight of the submarine minus the buoyant force. When the submarine is out of the water, the tension is simply its weight, $20,000 \times 9.8$.

A Crane Lifts a Steel Submarine of Density 7800 kg/m37800 \mathrm {~kg} / \mathrm { m } ^ { 3 }7800 kg/m3

A Crane Lifts a Steel Submarine of Density $7800 \mathrm {~kg} / \mathrm { m } ^ { 3 }$