# Object floating on water with oil poured in

1. Sep 8, 2014

### happysmiles36

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

A block of wood floats in a bucket filled with water. You pour some oil into the bucket. The oil floats on top of the water and that a part of the block is now in water, part is in oil, and part is in air. What happened to the volume of the block that is in air when you put in the oil?

a) Stayed the same
b) Decreased
c) Increased

2. Relevant equations

None

3. The attempt at a solution

I don't think it increased because oil is less dense than water. I am having trouble figuring out whether the extra buoyant force from the oil is enough to cause the volume of wood in air stay the same. I think that it isn't enough and that the volume of wood in air would decrease. I also don't think staying the same would make sense because if the volume is the same, then the wood would keep rising so it can't be covered with oil and I don't think this happens because most pictures show the wood still being in water when oil is covering it.

Last edited: Sep 8, 2014
2. Sep 8, 2014

### Tanya Sharma

I would suggest to work things out mathematically .

Before oil is put - Let height of the block in air and in water be h1 and h2 respectively .What is the buoyant force in this case?

After oil is put - Let height of the block in air , oil and water be x1 , x2 , x3 respectively .What is the buoyant force in this case ?

Compare the two forces .

3. Sep 8, 2014

### happysmiles36

F(buoyancy) = g*(density)*volume(in fluid)

The l*w*(h2)* 9.81 would be the buoyant force since that is the weight of water displaced

After oil is put in, the buoyant force is (l*w*(x3)*9.81) + (l*w*(x2)*9.81*(density of oil))

I am kind of confused at this part because I'm not sure what I am looking for between these two. Without any numbers to go off of, its hard for me to visualize it in a mathematical way.

EDIT:
Bit more stuff:
If the area of the base of the block is (imagining) 10 and the density of oil is (0.8) then:

(10*9.81*(h2))= 98.1(h2)

and

(10*9.81*0.8*(x2))= 78.48(x2)
(10*9.81*(x3))= 98.1(x3)
Total force = 78.48(x2) + 98.1(x3)

Again I don't see how these two separate things can be related since x and h are different.

Last edited: Sep 8, 2014
4. Sep 8, 2014

### Tanya Sharma

What is 'l' ? What is 'w' ? Which variable represents density of water ? Which variable represents density of oil ?

Please rewrite your expressions using ρw as density of water and ρo as density of oil .Instead of 9.81 use 'g' for the sake of clarity .

5. Sep 8, 2014

### happysmiles36

ρw=density of water
ρo=density of oil
ρ=density
l= length
w= width
g=9.81
h2= height in water
x2=height in oil
x3=height in water

The l*w*(h2)*g*ρw would be the buoyant force since that is the weight of water displaced

After oil is put in, the buoyant force is (l*w*(x3)*g*ρw) + (l*w*(x2)*g*ρo)

I am kind of confused at this part because I'm not sure what I am looking for between these two. Without any numbers to go off of, its hard for me to visualize it in a mathematical way.

EDIT:
Bit more stuff:
If the area of the base (l*w) of the block is (imagining) 10 and the density of oil is (0.8) then:

Total buoyant force = (10*g*(h2))= 98.1(h2)

and

(10*g*0.8*(x2))= 78.48(x2)
(10*g*(x3))= 98.1(x3)

Total buoyant force = 78.48(x2) + 98.1(x3)

Again I don't see how these two separate things can be related since x and h are different.

Last edited: Sep 8, 2014
6. Sep 8, 2014

### Tanya Sharma

Ok .

So, we have F1 = ρwgh2(lw) , where F1 is buoyant force before oil is put

and F2 = ρ0gx2(lw)+ρwgx3(lw) , where F2 is buoyant force after oil is put .

Now , what is the relation between F1 and F2 ? Is F1>F2 or F1<F2 or F1=F2?

Don't look at the expressions on the right side . Just think ,how the two buoyant forces are related.(Hint:The block is in equilibrium in both the cases)

7. Sep 8, 2014

### happysmiles36

F1=F2

If the block is in equilibrium then in the first one, F1=Fg and all the displaced fluid is water
and in the second one, F2=Fg but the fluid is in water and oil.
So F1=F2

If the buoyant force is the same, then the volume of air surrounding the block decreases because there is no change in buoyant force meaning it can't rise with the height of the added oil?

Edit:
When i think of the block floating in oil and water, I see the buoyant force greater in water than it will be in the oil, so the block moves up slightly (relative to water) because the block is displacing oil. But since the oil is less density, the effectiveness of the buoyant force is less meaning it moves down (relative to air). Does that make sense? or am I misunderstanding something.

Last edited: Sep 8, 2014
8. Sep 8, 2014

### Tanya Sharma

Yes , F1=F2 .

So, ρwgh2(lw) = ρ0gx2(lw)+ρwgx3(lw)

Cancel the common factors on both the sides . What do you get ?

9. Sep 8, 2014

### happysmiles36

ρwgh2(lw) = ρ0gx2(lw)+ρwgx3(lw)
g and (lw) cancel
so, ρwh2 = ρ0x2+ρwx3

The height of the block in water will decrease since there is oil and oil is less dense than air. But how does this tie back to the block being in air? It seems to me that I am only comparing the height in water because I don't know how much the buoyant force is causing the block to rise relative to the increasing height of oil.

Edit:
I thought of something else. Since the oil is on top of the water, the block has to first displace oil before water. Since the density of oil is lower than water, more fluid has to be displaced to counter act this (same volume in oil = less buoyant force than in water). The wood is in water as well as oil so the buoyant force of the wood in oil isn't enough and that means the water has to make up for this to allow it to float. Therefore more volume is submerged and less is in air?

Last edited: Sep 8, 2014
10. Sep 8, 2014

### Tanya Sharma

Now divide by ρw on both the sides and express h2 in terms x2 and x3 .This expression will give you the answer.

From the above expression can you deduce whether h2 < (x2+x3) or h2 > (x2+x3) )

Last edited: Sep 8, 2014
11. Sep 8, 2014

### happysmiles36

Can't you only express h2 in terms of x2 and x3 if the heights submerged are the same? Like saying h2=(x2+x3) would be assuming that the height of the block in water is the same as in oil+water.

po/pw will be less than 1 since p0 is smaller than pw.

Edit
if h2 is = to x2+x3

then

x2+x3 > (#<1)x2+x3

but the original equation was equal, so this would be wrong right?

Last edited: Sep 8, 2014
12. Sep 8, 2014

### Tanya Sharma

The height submerged are not the same .Nobody is assuming h2=(x2+x3) . I simple asked you to find whether h2>(x2+x3) or h2<(x2+x3) from the final expression we have got.

Right.

13. Sep 8, 2014

### happysmiles36

Sorry I misunderstand what you meant by expressing h2 in terms of x2 and x3.

Anyways, h2<(x2+x3) since the p0/pw is not here.

because x2+x3 is larger, than more of the volume is submerged and less is in air when compared to h2.

Thanks I got it now, sorry for so many questions.

14. Sep 8, 2014

### Tanya Sharma

You are welcome