# Homework Help: Calculating Volume Using Buoyancy

1. Oct 30, 2007

### BayernBlues

1. The problem statement, all variables and given/known data

http://img524.imageshack.us/img524/1743/buobu1.png [Broken]

2. Relevant equations

Vsphere= pi/6(d^3)
W = mg

3. The attempt at a solution

Wc + Wo' = 0.650 N
Wc' + Wo' = 0.550 N

I'm unsure of how to do this because it asks to find the volume ONLY BY USING THE two weight measurements. The simpler way to do it would be to measure how much fluid is displaced but that's not allowed apparently so I'm stuck.

Last edited by a moderator: May 3, 2017
2. Oct 30, 2007

### learningphysics

So the volume and mass of the metal sphere are unknown?

3. Oct 30, 2007

### BayernBlues

I'm supposed to find the volume of the wooden sphere using only the measurements for Wc'+Wo' and Wc+Wo'. I think that what I have to do is find the amount of volume displaced by the wooden and metal spheres in water and subtract it by the volume of just the metal sphere in water.

I'm guessing to find the volume for the metal sphere, I'd need to see the apparent weight of the metal sphere in water (it's 0.650N), find the mass from that (m=W/g), use 1g/cm^3 for the density of water, and find the volume by V=m/D.

Then I'd do the same for the other one and subtract the two volumes.

But the thing is, I'm not sure about what weight value to use for finding the volume of just the metal sphere in Wc+Wo' because this would involve the weight of the cork sphere in the air plus the metal sphere in water and I just want the weight of the metal sphere in water.

4. Oct 30, 2007

### BayernBlues

Here's what I've done so far, I know it's wrong because the volume of the cork should be more than the volume of the metal sphere:

In order to find the volume of the cork, the volume of fluid displaced by the metal sphere in water must be found. This amount can be found by using Wc + Wo’.

Wc’ + Wo’ = (0.650 ± 0.010 N)/9.81 m/s² = 0.0662 kg = 66.26 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vcork+sphere = 66.26 g/1g/cm³ = 66.26 ± 0.10 cm³

A similar procedure must be used to find the volume of water displaced by the cork and the metal sphere:

Wc + Wo’ = (0.550 ± 0.010 N)/9.81 m/s² = 0.056 kg = 56.07 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vsphere = 56.07 g/1g/cm³ = 56.07 ± 0.10 cm³

To find the volume of the cork, the value for Vsphere must be subtracted from Vcork+sphere.

Vcork = 66.26 ± 0.10 cm³ - 56.07 ± 0.10 cm³ = 10.19 ± 0.10 cm³

5. Oct 30, 2007

### learningphysics

yeah, don't think that's right. That measurement isn't the weight of the displaced water...

I believe the equations should be:

m1*9.8 + m2*9.8 - (v1 + v2)1000*9.8 = 0.550N

m1*9.8 + m2*9.8 - v2*1000*9.8 = 0.650N

where m1 is the mass of the cork, and m2 is the mass of the metal sphere. v1 is the volume of the cork... v2 is the volume of the metal sphere...

you can solve for v1 here using these two equations... but I don't see how to get m1...

Last edited: Oct 30, 2007