- #1
Roger Wilco
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Homework Statement
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 [tex]\rho_c=.300\frac{g}{cm^3}[/tex], and 2.00 cm of its height is above the water. What is the radius of the ball?
Homework Equations
[tex]\sum F=0[/tex] [tex]F_b=\rho*V*g[/tex]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??
The Attempt at a Solution
Subscript c is cylinder, b is the ball, and V' is the portion of the cylinder under water.
[tex]\sum F=0[/tex]
[tex]\Rightarrow W_c+W_b-F_{bouyant}=0[/tex]
[tex]\Rightarrow m_cg+m_bg-\rho_cV_c'g=0[/tex]
[tex]\Rightarrow \rho_cV_c+\rho_bV_b=\rho_cV_c'[/tex]
[tex]\Rightarrow V_b=\frac{\rho_c(V_c'-V_c)}{\rho_b}[/tex]
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?
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