# Buoyant Force Problem

1. Feb 8, 2010

### ACE_99

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

Suppose a buoy is made of a sealed steel tube of mass 5 kg with a diameter D = 7 cm and a length of 6 meters. At the end of the buoy is a spherical weight of galvanized steel (specific gravity=7.85). If the buoy floats in fresh water, what must be the mass of the steel M at the bottom to make the distance h=195 cm?

2. Relevant equations

FB = W
F = $$\rho$$gV

3. The attempt at a solution

I know that in order for this object to float the buoyant force must equal the mass of the submerged object. So

FB = Wcyl + Wsph = Wwater which is also

Vcyl$$\rho$$sg +Vsph$$\rho$$sg = Vwater$$\rho$$g

This is where I get confused. In order to find the mass of the sphere I need to find its volume since I have the density, but how do I determine the volume of water displaced if I dont know the volume of the sphere. Hopefully my reasoning is correct. Any help would be great! I've also attached a copy of the picture provided.

http://i429.photobucket.com/albums/qq12/ACE_99_photo/ps-222-1-q6-1.jpg" [Broken]

Last edited by a moderator: May 4, 2017
2. Feb 8, 2010

### rl.bhat

Mass of tube is given.
Mass of the sphere = ρs*V.
weight of the displaced liquid = (Volume of the immersed tube + Volume of the sphere)*ρw
Volume of the immersed tube = π*D^2/4*(L-h)
From these information find the volume of the sphere and then mass of the sphere.

3. Feb 8, 2010

### ACE_99

Based on what rl.bhat stated I managed to figure out the following.

mtube + $$\rho$$sVsph = [Vcyl sub + Vsph]$$\rho$$w

isolate for Vsph to get

Vsph = Vcyl$$\rho$$w - 5 kg / $$\rho$$w + $$\rho$$w

Solving for Vsphere I get V = 0.011962 therefore making the mass 9.39 kg.

4. Feb 8, 2010

### rl.bhat

The equation should be
Vs = (Vc*ρw - 5 kg)/(ρw + ρs).
Μay be typo.

Last edited: Feb 8, 2010
5. Feb 8, 2010

### ACE_99

Ya that was just a typo. Thanks for your help