Archimedes Principle Problem: Floating Object

In summary, the cylindrical log of density 858kg/m3 floats so that the vertical distance from the water line to the top of the log is 12.0cm.
  • #1
AMacias2008
3
0

Homework Statement


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?


Homework Equations


Fbouyant=Wwaterdisplaced

ρwater * V displaced water = ρlog * Vlog


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.
 
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  • #2
AMacias2008 said:
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?
Obviously, yes.

And if so, how I would go about determining the ratio of the volumes?

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
SteamKing said:
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?

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.
 

Related to Archimedes Principle Problem: Floating Object

1. What is Archimedes' Principle?

Archimedes' Principle states that the upward buoyant force on an object immersed in a fluid is equal to the weight of the fluid that the object displaces.

2. How does Archimedes' Principle apply to floating objects?

When an object is placed in a fluid, it will displace a certain amount of fluid. The weight of the displaced fluid is equal to the buoyant force acting on the object, which allows it to float.

3. What factors affect whether an object will float or sink?

The density of the object and the density of the fluid it is placed in are the main factors that determine whether an object will float or sink. If the object is less dense than the fluid, it will float, and if it is more dense, it will sink.

4. How can Archimedes' Principle be used to calculate the buoyant force?

The buoyant force can be calculated by multiplying the density of the fluid by the volume of the displaced fluid and the acceleration due to gravity (Fb = ρVg).

5. Can Archimedes' Principle be applied to objects in air?

Yes, Archimedes' Principle can be applied to objects in air, as air is considered a fluid. However, the buoyant force in air is usually negligible compared to the weight of the object, so it is not as noticeable as in fluids such as water.

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