# Need Help finding radius of ball using bouyancy

1. Nov 11, 2007

### Roger Wilco

1. The problem statement, all variables and given/known data
An iron ball is suspended by a thread of negligible mass from an upright cylinder that floats partially submerged in water. The cylinder has a height of 6.00cm, a face area A=12.0 cm^2 on the top and bottom, and a density $$\rho_c=.300\frac{g}{cm^3}$$, and 2.00 cm of its height is above the water. What is the radius of the ball?

2. Relevant equations $$\sum F=0$$ $$F_b=\rho*V*g$$
Using Newton's Second and Archimedes' Principle I have used the following method. My concern comes at the point that I get the expression (V'-V)<-- this will yeild a NEGATIVE quantity and r cannot = negative. My problem is that the V'=volume of water displaced and V= volume of the cylinder. I think I need to assume that the CYLINDER IS HOLLOW in order to get a positive quantity. But whay is the volume of a hollow cylinder if I am not given an inner and outer radius??

3. The attempt at a solution

Subscript c is cylinder, b is the ball, and V' is the portion of the cylinder under water.

$$\sum F=0$$

$$\Rightarrow W_c+W_b-F_{bouyant}=0$$

$$\Rightarrow m_cg+m_bg-\rho_cV_c'g=0$$

$$\Rightarrow \rho_cV_c+\rho_bV_b=\rho_cV_c'$$

$$\Rightarrow V_b=\frac{\rho_c(V_c'-V_c)}{\rho_b}$$

I don't find it necessary to move any further than this last step as finding the r is easy enough from there. However it is in this last step that you can see that if I use
V=height*cross-sectional area...I will get a negative number for V'-V.

What should I be using for V? Shoud it be zero? I think that is a bold assumption, or is it?

Thank you,
RW

Edit: After looking at my diagram, I have encountered another problem: Do I need to consider the bouyant force on the ball, too?

I do not not see why I wouldn't.

Okay. I also noticed that for F_bouyant I should have used rho_water NOT of the cylinder.

Guess I need to re-work this. :( So I guess in re-working this my question is still: do I need to consider the bouyant force on the ball, too?

Last edited: Nov 11, 2007
2. Nov 11, 2007

### Kurdt

Staff Emeritus
I think the problem is that you're multiplying V'c by the density of the cylinder and not the density of water which is what you need to do for the buoyant force.

3. Nov 11, 2007

### Roger Wilco

Yeah Kurdt, I just caught that. But tell me, should I be taking into account the Bouyant Force of the ball?

See "Edits" in post #1.

4. Nov 11, 2007

### Kurdt

Staff Emeritus
Thats a good thought, and yes I would do that. Its not that much harder.