# Need help with a buoyancy problem

1. Jun 4, 2014

### AlphaLima

1. The problem statement, all variables and given/known data
A cube of side length 3 cm floats in water (density = 1g/cm^3) with 1 cm floating above the water. What is the density of this cube?

2. Relevant equations
Fbouyant = ρfluid*g*Vsubmerged
Fg= mg

3. The attempt at a solution

I am having problems figuring out what the submerged volume of the cube is to plug into the formula. I have a formula that states that volume submerged for an object = Az, where A = the area of the object and z = the depth that the object is submerged so:

V=Az = 6(3cm^2) * 2cm (that is submerged) = 108cm^3 It seems that this is wrong though since the volume should be 18cm^3 to give me the right answer.

I then use the formula

Fb = Fg for an object that is floating
ρfluid*g*Vsubmerged = ρobject*g* Vtotal of object

cancelling out the g from both sides gives a ratio that allows me to calculate for the density of the object. The answer is supposed to be 2/3 g/cm^3 which I am not getting.

Any guidance would be appreciated :)

2. Jun 4, 2014

### Nathanael

One side of the cube is $9cm^2$ and the part that is submerged is $2cm$ in height.

So what's the volume of a rectangular-cube (probably not the right name for it) with a base of $9cm^2$ and a height of $2cm$?

3. Jun 4, 2014

### TSny

Hi AlphaLima. Welcome to PF!

In the formula V = Az, A is not the total surface area of the cube. See this link .

4. Jun 4, 2014

### AlphaLima

L * W * H = 3*3*2 = 18cm^3. Thank you guys! This makes a lot more sense now :)