# Buoyant Force

1. Apr 30, 2017

### tennisgirl92

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Fb=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

3. 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?

2. Apr 30, 2017

### kuruman

Use Archimedes's principle.

3. Apr 30, 2017

### 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.

4. Apr 30, 2017

### kuruman

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

5. Apr 30, 2017

### 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?

6. Apr 30, 2017

### 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.

7. Apr 30, 2017

### 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?

8. Apr 30, 2017

### kuruman

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?

9. Apr 30, 2017

### 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?

10. Apr 30, 2017

### kuruman

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?

11. Apr 30, 2017

### tennisgirl92

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

12. Apr 30, 2017

### kuruman

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

13. Apr 30, 2017

### tennisgirl92

Ok, it can either be

densitywaterx Volumepost x g=masspostxg

or

densitywaterx Volumepost x g=densitypost x Volumewater x g

does this look right?

14. Apr 30, 2017

### kuruman

No. Left side (BF) first. Archimedes says the buoyant force is the weight of the displaced water.
Wdw = mdw g. What is the mass of the displaced water mdw? Answer: It is the density of water ρw, 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.

15. Apr 30, 2017

### tennisgirl92

Fb=densityw x Volumedw x g
=1029 x Vdw x 9.8

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

16. May 1, 2017

### haruspex

Yes.

17. May 1, 2017

### 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?

18. May 1, 2017

### kuruman

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.

19. May 1, 2017

### 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

20. May 1, 2017

### 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.