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

• AlphaLima
In summary, the conversation discusses the problem of finding the density of a cube that floats in water with part of it submerged. The formula for finding the submerged volume of an object is introduced, but the calculation is incorrect due to a misunderstanding of the formula. After clarification and using the correct formula, the correct answer is found.
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

Fbouyant = ρfluid*g*Vsubmerged
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
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 :)

AlphaLima said:
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 $2cm$?

Hi AlphaLima. Welcome to PF!

AlphaLima said:
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 .

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

It seems like you are on the right track with using the buoyancy equation and the ratio of the densities. However, the issue may lie in your calculation of the submerged volume. Remember, the entire cube is not submerged, only 1 cm of it is above the water. So the volume submerged would actually be (3cm x 3cm x 1cm) = 9cm^3. Plugging this into the buoyancy equation along with the known values for the fluid density and the object's total volume, you should be able to solve for the density of the cube. Keep in mind that the density of the cube will be in g/cm^3, not cm^3. I hope this helps!

## 1. What is buoyancy?

Buoyancy is the upward force exerted by a fluid on an object placed in it. It is the reason why objects float in water.

## 2. How do I calculate buoyancy?

Buoyancy can be calculated using the formula Fb = pVg, where Fb is the buoyant force, p is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

## 3. What affects buoyancy?

The buoyancy of an object is affected by its density, the density of the fluid, and the volume of fluid displaced by the object.

## 4. Can an object sink in water?

Yes, an object can sink in water if its density is greater than the density of water. In this case, the object will experience a downward force greater than the buoyant force, causing it to sink.

## 5. How can I increase the buoyancy of an object?

The buoyancy of an object can be increased by increasing its volume or decreasing its density. This can be achieved by adding air pockets or using materials that are less dense than water.

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