# Buoyancy Problem

1. Sep 24, 2008

### eddieb340

I've got an interesting question about a Buoyancy problem we were given in Lab today. We were given an unknown block and we were asked to find its density without directly measuring its volume.
So we found the mass in air and in water.

Mass in air was 267.65g
Mass in water was 237.76g

Then using the equation T+Bwater=mg we found the volume of the block.
We then found the density using density = m/V.

My desnity came out to be 8.58x103 kg/m3

He then said that because we didn't account for the buoyant force of air that our density will be off a bit. So my question is how would you find the exact mass and volume of the unknown block if they BOTH change with no buoyant force. I am always left with two variables that I cant find no matter how many different ways i try to combine equations. Is there a different approach? Its like trying to find the mass and volume of my unknown block in a vacuum but I don't have one :(

Any help is greatly appreciated!

2. Sep 24, 2008

### cepheid

Staff Emeritus
The block's mass doesn't really change...no matter where it is. Sooo....????

What is that equation? It must have units of force, so B_water must be the buoyant force due to the water. So, what is T?

Exactly what procedure did you follow to make these measurements?

3. Sep 24, 2008

### eddieb340

I understand that the true mass of the block never changes but we cant directly measure the true mass due to the buoyancy force of air.

T + Bwater = mg

In the above equation the mg was recorded by weighing the block in air. The T was recorded by weighing the block in water and the B was the difference of the two.

Bwater = mg - T

So since B also equals pgV we can say

ph2ogVdis = mg - T

you can solve this for V and then plug into p=m/v

So I have the density of the block in air. But not taking into account that the mass and volume that i used to calculate the density was calculated ignoring the Buoyant for in Air

SO...

I decided to set up another equation T + Bair = Mtrueg

Then... T+pairgVdis= true mg

heres the problem. I cant get the true mass of the object without using the true volume of the displaced fluid (volume of the block) becuase both the mass of an object and its volume is affected by buoyant force. (think of the volume change in air compared to the bottom of the ocean)

did i leave anything out?

true density = true mass/true volume
or
density in vacuum = mass in vacuum/volume in vacuum

Last edited: Sep 24, 2008
4. Sep 25, 2008

### cepheid

Staff Emeritus

1. Block is immersed in air only...its apparent weight is the difference between its weight and the buoyant force due to air on it.

2. Block is immersed in BOTH air and water. Its apparent weight is its weight minus the two buoyant forces.

I will use the letter w for weight, B for buoyant force

Situation 1:

$$w_{\textrm{air}} + B_{\textrm{air}} = mg$$

$$B_{\textrm{air}} = mg - w_{\textrm{air}}$$

$$\rho_{\textrm{air}}gV = (m - m_{\textrm{air}})g$$

where m_air is the mass that would be required to have the "apparent weight" that you measured.

$$V = \frac{m - m_{\textrm{air}}}{\rho_{\textrm{air}}}$$

Similarly for situation 2:

$$B_{\textrm{air}} + B_{\textrm{water}} = mg - w_{\textrm{water}}$$

divide both sides by g:

$$(\rho_{\textrm{air}} + \rho_{\textrm{water}})V = m - m_{\textrm{water}}$$

In words, this equation says that the total mass of displaced fluid (left hand side) is what accounts for the difference between the true mass and the apparent mass in water (where the apparent mass is the mass for the object that you would deduce from its weight if you naively assumed that it wasn't immersed in any fluid).

Since you already have an expression for V in terms of m and known constants (from situation 1), you can substitute that expression for V into this latest equation and solve for m.