Floating 9cm Post: Solving Submersion Depth

[tennisgirl92]

Homework Statement

A 9 cm square wooden post (density=420kg/m³) is 1.3m long and floats in sweater (density=1029 kg/m³. How deep is the post submerged in the water? The picture shows a 9cm x 9cm cross section of the post.

Homework Equations

F_b=density of liquid x volume of liquid submerged x gravity
Fmg=density of post x volume of post x gravity
P=density x gravity x height
P=Force/area

The Attempt at a Solution

I first found the force mg.
420 x (.09x.09x1.3) x 9.8=.103194 N
Then I found pressure
P=.103194/ (.09x.09)=12.74

I don't think this is right, and I am unsure where to next. Can anyone help me understand how I should be doing this?

 

[kuruman]

Use Archimedes's principle.

 

[tennisgirl92]

Ok, Archimedes's principle states that the magnitude of the buoyant force=weight of the fluid displaced by object

Fb=density of fluid x volume of object submerged x g
=1029 x (.09 x .09 x 1.3) x 9.8
=106.187 N

I don't see where to go next.

 

[kuruman]

Fb=density of fluid x volume of object submerged x g

Is the entire volume of the wood under water? The answer is probably not because the problem is asking you to find how deep into the water the wood is when it floats. What you calculated is the maximum buoyant force that you can have as in if you pushed the wood entirely under water.

I also noticed this

... and floats in sweater (density=1029 kg/m3

I am sure you meant to say that the post floats in water not sweater.

 

[tennisgirl92]

ha! oh yes-I did mean water. Sorry.

Ok, no the wood is not entirely submerged. But the problem does not given us the percent it is submerged-how do we find that?

 

[kuruman]

Can you get a number for the buoyant force? If yes, then that is equal to the weight of the displaced water. Hint: The post floats.

 

[tennisgirl92]

If the post is floating, wouldn't the buoyant force have to be greater? Would we use the density of the wood on the water to find it instead of the density of the water?

 

[kuruman]

If the post is floating, wouldn't the buoyant force have to be greater?

Greater than what? If you place the post on a table, how big is the table force that supports the post? How different from the table force is the buoyant force that supports the post?

 

[tennisgirl92]

Greater than the gravitational force or m x g. But on the table. the force acting on the post is equal to that of the post acting on the table. In other words, the gravitational force is equal to the normal force. Am I wrong to say that the buoyant force must be greater than the gravitational force? Is it equal?

 

[kuruman]

Am I wrong to say that the buoyant force must be greater than the gravitational force? Is it equal?

You can answer that yourself. What would happen to the net force on the post if the buoyant force were greater than the gravitational force? How big is the gravitational force anyway?

 

[tennisgirl92]

I believe the buoyant force is equal to the gravitational force, as the object is at rest.

 

[kuruman]

Very good. Start with $BF=mg$. Re-express each side of the equation in terms of volumes and densities. Remember Archimedes's principle.

 

[tennisgirl92]

Ok, it can either be

density_waterx Volume_post x g=mass_postxg

or

density_waterx Volume_post x g=density_post x Volume_water x g

does this look right?

 

[kuruman]

does this look right?

No. Left side (BF) first. Archimedes says the buoyant force is the weight of the displaced water.
W_dw = m_dw g. What is the mass of the displaced water m_dw? Answer: It is the density of water ρ_w, times the volume of the displaced water. You don't know what the volume of the displaced water is, so call it a name and write an expression for the buoyant force. Then we will work on the left side.

 

[tennisgirl92]

Fb=density_w x Volume_dw x g
=1029 x V_dw x 9.8

would we set this equal to the density of the post x Volume of post x g?

 

[haruspex]

Fb=density_w x Volume_dw x g
=1029 x V_dw x 9.8

would we set this equal to the density of the post x Volume of post x g?

Yes.

 

[tennisgirl92]

So setting it equal...

1029 x Vdw x 9.8=420 x (.09 x .09 x 1.3) x 9.8
Vdw=.0043 m³

How do we find the height from this? Do we subtract it from the volume of the post?

 

[kuruman]

How do we find the height from this? Do we subtract it from the volume of the post?

What does does the submerged volume look like? It has a square 0.09 cm x 0.09 cm base and unknown height h that you are looking for.

 

[tennisgirl92]

I know Volume would equal .09x.09xh, but wouldn't this height be referring to be side of the 1.3m? aren't we looking for the part of the .09 that is submerged? the post seems to be laying horizontally in the water, not straight up.

Also, when I reduce and isolate the height, I obtain .531m or 5.31 cm. The correct answer is 3.67cm

 

[kuruman]

OK, then. You have a rectangle 0.09 m x 1.3 m base and unknown height h that you are looking for. If you don't get the correct answer, you need to show what you did in detail to reduce and isolate the height.

 

[tennisgirl92]

Got it! THANK you! You have been so helpful