How Is the Mass of the Spherical Weight Calculated for Buoyancy Equilibrium?

[ACE_99]

Homework Statement

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?

Homework Equations

F_B = W
F = $$\rho$$gV

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

F_B = W_cyl + W_sph = W_water which is also

V_cyl$$\rho$$_sg +V_sph$$\rho$$_sg = V_water$$\rho$$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 don't 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"

 

[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.

 

[ACE_99]

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

m_tube + $$\rho$$_sV_sph = [V_{cyl sub} + V_sph]$$\rho$$_w

isolate for V_sph to get

V_sph = V_cyl$$\rho$$_w - 5 kg / $$\rho$$_w + $$\rho$$_w

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

 

[rl.bhat]

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

 

[ACE_99]

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

Ya that was just a typo. Thanks for your help