How Do You Calculate the Density of a Partially Submerged Cube?

[AlphaLima]

Homework Statement

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?

Homework Equations

F_buoyant = ρ_fluid*g*V_submerged
Fg= mg

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
ρ_fluid*g*V_submerged = ρ_object*g* V_{total 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 :)

 

[Nathanael]

I am having problems figuring out what the submerged volume of the cube is

...

the volume should be 18cm^3 to give me the right answer.

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 $2cm$?

 

[TSny]

Hi AlphaLima. Welcome to PF!

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.

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

 

[AlphaLima]

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