A crane lifts the 18,000 kg steel hull ship out of the water.

Question

A crane lifts an 18,000 kg steel hull ship out of the water. The density of steel is 7.8 x 10^3 kg/m^3, and the density of water is 1,000 kg/m^3. 
A) What is the volume of water displaced when the ship is fully submerged? I am unsure how to approach this part.
B) What is the tension in the cable when the ship is fully submerged? I calculated it as follows: 
B = Density of water × Volume of steel × g 
= 1,000 kg/m^3 × (18,000 kg / 7,800 kg/m^3) × 10 m/s² 
= 23,077 N.
C) What is the tension in the cable when the ship is out of the water? The tension can be calculated using the formula: 
tension = mg, where g = 10 m/s². 
Thus, tension = 18,000 kg × 10 m/s² = 180,000 N.

Anonymous · Accepted Answer

A)
Volume of water displaced = Volume of submerged ship
Vw = (18,000 kg) / (7.8e^3 kg/m^3)
Vw = 2.3077 m^3
Weight of water displaced = (density of water)(volume of water)(acceleration due to gravity)
Ww = (1000 kg/m^3)(2.3077 m^3)(9.81 m/s^2)
Ww = 2.2638e4 N
B)
By Archimedes’ principle, the buoyant force on the ship equals the amount of fluid displaced.
(Tension of cable) + (Weight of displaced water) = Weight of ship
T + 2.2638e4 N = (18,000 kg)(9.81 m/s^2)
T = 1.539e5 N
C)
Your approach to this one is correct.

Anonymous

A crane lifts the 18,000 kg steel hull ship out of the water.

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