Buoyant force: wooden block in oil and water

[physgirl]

So there's a wooden block floating on water in such way that 78% of the block is submerged in water. Now oil (with density of 700 kg/m^3) is poured on top so that the wooden block is wholly covered by the oil. Now how much of the block is submerged in water? (Hint: do not ignore the buoyant force of air before the oil is submerged)

I'm not really sure where to go with this question :(

 

[theCandyman]

Figure the displacement of water plus the displacement of air is equal to the weight of the block. The density of air should be easy to find, this should help you out a bit.

 

[OlderDan]

So there's a wooden block floating on water in such way that 78% of the block is submerged in water. Now oil (with density of 700 kg/m^3) is poured on top so that the wooden block is wholly covered by the oil. Now how much of the block is submerged in water? (Hint: do not ignore the buoyant force of air before the oil is submerged)

I'm not really sure where to go with this question :(

Every layer of a fluid contributes to the buoyant force. The weights of each fluid that is displaced add together to give the total buoyant force. The buoyant force of the air is almost nothing compared to that of water, but it can be calculated. The 22% of the block that was originally in the air displaces some weight of air, so it gets a little buoyant force from that; most of the buoynat force is from the water. The combined weight of displaced water and air is the weight of the block. When the oil is added, the two fluids contributing a buoyant force are the water and the oil. Their combined displaced weight has to be the weight of the block. You need to figure out what fraction of the block is in water and what fraction is in oil.

 

[physgirl]

but how can you find the buoyant force that a fluid is exerting if you don't have the volume of the fluid? the numbers i gave on the original post are the only numbers that are given in the problem...

 

[tarful]

i was wondering about another issue. do all oils act the same way in this scenario

12.5 lb crude oil = canola oil?

 

[OlderDan]

but how can you find the buoyant force that a fluid is exerting if you don't have the volume of the fluid? the numbers i gave on the original post are the only numbers that are given in the problem...

You know the density of the oil and you can look up the density of the water and air. You can find the density of the block from the information given. Everything is proportional. You can assume any volume you want for the block. The volume will divide out.

 

[physgirl]

wait... I am confused. so let the volume of block be "x"... and the density of the block is 780 kg/m3 (0.78*1000)... and an object floats when the weight of fluid displaced equals the weight of the object... the weight of the object is "(780)(9.8)x"... so that's also the weight of the water displaced... and now what?

 

[OlderDan]

wait... I am confused. so let the volume of block be "x"... and the density of the block is 780 kg/m3 (0.78*1000)... and an object floats when the weight of fluid displaced equals the weight of the object... the weight of the object is "(780)(9.8)x"... so that's also the weight of the water displaced... and now what?

The hint you were given about the air is telling you that you cannot use the 78% of the block submerged in water to conclude the density of the block is 78% of the density of water. Instead, you have to use 78% of the volume of the block displacing water plus 22% of the volume of the block displacing air and add together the weight of the water and the air to equal the weight of the block. It's not a big difference, but the concept is important for doing the rest of the problem. You will apply the same reasoning when the block is surrounded by water and oil instead of water and air.

 

[physgirl]

ok... so if let's say the volume of the block is 100 m^3...

then 78 m^3 of water is displaced. and since W=pVg (where by p, I mean density)... weight of water displaced is: (1000)(78)(9.8)=764400N

and also, 22 m^3 of air is displaced... so weight of air displaced is: (1.29)(22)(9.8)=278.124N

together, the block displaced 764400 + 278.124 = 764678.124N of fluid, and since we assumed its volume to be 100 m^3, its density should be: W/(gV)=764678.124/(9.8*100)=780.2838 kg/m^3

so when oil is added... do you just set up two equations with x=volume of box (when total volume is 100) that's submerged in water and y=volume of box in oil:

x + y = 100
1000*x*9.8 + 761*y*9.8 = 764678.124
[[[[761 is the actual density of oil given in my problem, not 700]]]]

so for x you get 8.069?
I'd be very grateful if someone could double check the number for me... it's an online submission homework and I only have 1 try left for this problem :(

 

[OlderDan]

The method is correct, and the numbers look reasonable. I am interpreting your comment about 761 to mean that the 700 you originally quoted is not what was given. There are lots of different oils. Be sure to use what was given in the problem, and be sure to express your answer as the % submerged. (Your 100m^3 is not a number given in the problem. You have to use a % of the unspecified volume)

 

[merry]

I am confused with why the weight of the oil does not push the block down =S Also, this may sound stupid but since the bottom of the block is not exposed to the oil, why is the oil able to exert an upward force? @.@ Can someone please help me out? THANKS!