A crane lifts a steel submarine of density 7800 kg/m³ and mass 20,000 kg.What is the tension force that the lifting cable exerts on the submarine (a)when the submarine is submerged in water of density 1000 kg/m³,and (b)when it is entirely out of the water?

Question

Quizplus · Accepted Answer

(a) When the submarine is submerged in water, the buoyant force (FB) acts upward, given by the equation FB = ρVg, where ρ is the density of water, V is the volume of the submarine, and g is the acceleration due to gravity. We can find the volume of the submarine by dividing its mass by its density: V = m/ρ = 20000/7800 = 2.564 m^3. Thus, FB = 1000 x 2.564 x 9.81 = 25123.6 N. This buoyant force acts against the weight of the submarine (W = mg = 20000 x 9.81 = 196200 N). Therefore, the tension force in the cable is T = W - FB = 196200 - 25123.6 = 171076.4 N, which is approximately 1.7 x 10^5 N.
(b) When the submarine is entirely out of the water, the tension force in the cable is simply equal to its weight, which is T = mg = 20000 x 9.81 = 196200 N, which is approximately 2.0 x 10^5 N.

A Crane Lifts a Steel Submarine of Density 7800 Kg/m3

A Crane Lifts a Steel Submarine of Density 7800 Kg/m³