# Buoyancy force on a steel boat

1. Dec 1, 2008

### aa1515

1. The problem statement, all variables and given/known data

The bottom of a steel "boat" is a 7.00 m x 8.00 m x 5.00 cm piece of steel (rho-steel=7900 kg/m^3.) The sides are made of 0.540 cm-thick steel.

What minimum height must the sides have for this boat to float in perfectly calm water

3. The attempt at a solution

I figured the water displaced must equal the weight of the boat. Then the volume of that water displaced will equal the volume of the boat. But for whatever reason I am still not getting the correct answer. Any thoughts?

2. Dec 1, 2008

3. Dec 2, 2008

### bokonon

I'm having trouble with this one too.

I figure, the weight of the boat is equal to the weight of the bottom plus the weight of the sides. However, one thing that is unclear is if the sides are added to the very end of the bottom, or if they are placed on top. In the later case the height of the sides, h, would really be h+thickness of the bottom? Anywho, lets say that:

Weight of boat = Vbottom*rho*g + Vsides*rho*g = 7*8*.05*7900*9.8 + (2h*7*.0054+2h*8*.0054)*7900*9.8

Weight of displaced water = Vboat*rho*g = 7*8*(h+.05)*1000*9.8

where h is the height we are looking for. Setting these two equations equal to each other and solving for h, I get h = .398. Is this correct?