How Thick Should the Aluminum Sheet Be for the Box to Float as Described?

[Menisto]

A closed cubical box made of aluminum sheet floats in water with (1/4) of the volume above water. Determine the thickness of the sheet.

I first found out what the density would need to be for this orientation:

75% below water or

X % = density of box/ density of water

.75 = p/(1000)

p = 750 kg/m^3 --> effective density.

p = M(sheet) / Volume(box)

Mass of the sheet is then = 6p(s^2)x

where s is the length of a side of the cube, x is the thickness of the sheet and p is the density of aluminum (2700 kg/m^3)

M = 16200 x(s^2)

Therefore: 750 = 16200 x(s^2) / s^3

.0463 = x/s

The problem is, there seems to be not enough information.

 

[Menisto]

Am I correct in assuming this?

 

[OlderDan]

The problem is, there seems to be not enough information.

You are correct. Surface area is not proportional to volume. Even if you did a more precise calculation of the volume of aluminum as the difference between outer volume and inner volume, you can still go from essentially a solid cube of aluminum to a huge cube with a very small ratio of aluminum volume to air volume (low density) using the same thickness of walls. There has to be more information to solve the problem.

 

[Menisto]

Ok, thank you. I thought I was going crazy there for a second, too many hours of homework. It doens't help when the professor makes his own course packet, and has to correct problems and answers for every homework set. It gets a little frustrating.