Homework Help: Archimedes Principle Problem: Floating Object

1. Apr 2, 2014

AMacias2008

1. The problem statement, all variables and given/known data
A cylindrical log of uniform density and radius R=30.0cm floats so that the vertical distance from the water line to the top of the log is d = 12.0cm. What is the density of the log?

2. Relevant equations
Fbouyant=Wwaterdisplaced

ρwater * V displaced water = ρlog * Vlog

3. The attempt at a solution

I started off by making a FBD. Buoyant force going up, mg coming down. No acceleration so I ended up getting to the second equation up above.

Solving for ρlog, I get

ρlog = ρwater * (Vdisplacedwater/Vlog) where Vdisplacedwater = ∏R2

I'm having trouble figuring out the volume of the displaced water. Since only part of the log is submerged, would the volume of the displaced water be the volume of the log that is submerged? And if so, how I would go about determining the ratio of the volumes?

I tried several things already to no avail. I tried taking the ratio of the radius of the log to the distance to the top of level (.12/.30) and used the answer of .4 to use into the R of Vdisplacewater but didn't get the answer.

I tried switching it and still got nothing. Some guidance in the right direction would be greatly appreciated. The answer is 858kg/m3.

2. Apr 2, 2014

SteamKing

Staff Emeritus
Obviously, yes.

You must use some subtle tricks called geometry to calculate the volume of displacement of the log. It always helps to draw a picture of the problem first.

http://en.wikipedia.org/wiki/Circular_segment

I assume you had no problem calculating the total area of the cross section of the log?

3. Apr 2, 2014

AMacias2008

Thank you for your response SteamKing. Yeah I did.

Ah I see. Geometry is a cool little I should use more haha. Thank you for the help and I used that equation to find the area, took the answer minus the entire cross sectional area, did some more work with another helpful tool called algebra and got the correct answer. Thank you for the help and sorry for the silly question.