Buoyancy of ship, volume displaced, and tension of crane (in/out) of water

[orbits]

A crane lifts the 18,000kg steel hull of a ship out of the water.
The density of steel is known to be 7.8 x 10³ kg/m³, while that of water is 1000 kg/m³.


A) While the steel hull is fully submerged in the water, what is the volume of water displaced by the hull? I really have no idea how to work this problem out.


Buoyancy = m* density * V displaced * G
m=pv





B) What is the tension in the crane's cable when the hull is fully submerged?
I believe this one is related to A

Density * Volume * G = F m=(density) Volume
1000* 18000/10 *10

f=1.8 x 10⁷


C) What is the tension in the crane's cable when the hull is out of the water?

Ft= mg
Ft = 18,000 * 10 (class standard)
Ft= 180,000N

 

[glueball8]

If you have the mass of the material and its density, can't you find its volume? That volume displace the same volume of water.

 

[DaveC426913]

If you have the mass of the material and its density, can't you find its volume? That volume displace the same volume of water.

Only if the hull fills with water. If it does not, it displaces far more water than its own volume.

The question seems a bit tricky. It talks about a fully-submerged the hull, which, to me, strongly suggests that the hull has filled with water (unless it's a watertight submarine hull I suppose ).

So, I guess you're right. They're simply talking about the volume of the steel itself, not any kind of buoyant hull-shape.