Fluid Mechanics and Archimedes Principle

[a1234]

Homework Statement

[/B]
A rectangular object has a width of 40 meters, height of 15 meters, and length of 2 meters. It floats consistently when 3 meters of its height is below the surface of the water.

1. Find the volume of the displaced water.
2. How much is the buoyant force on the object?
3. What is the object's weight?
4. What is the mass density of the block?
5. How would the amount of the object under the surface change if it were floating in vegetable oil?


2. The attempt at a solution

1. This is 40*15*2, or 1200 m^3.
2. This is 1000*240*9.8, or 2352000 N.
3. Same as part 2.
4. Since weight/g = mass...

2352000/9.8 = 240000

Since density=mass/volume...

240000/1200 = 200

I'm not sure of the units to use on these two.

5. Am I supposed to use the density of vegetable oil instead of water in step 2?

 

[haruspex]

This is 40*15*2,

That is the volume of the whole block.
Draw a diagram.

 

[a1234]

The volume of the displaced part of the object is 40*3*2, or 240 m^3.

 

[haruspex]

The volume of the displaced part of the object is 40*3*2, or 240 m^3.

Right (well, volume of the displaced water). Retry the other parts.

 

[a1234]

I think the original answer for part 2 stays 2352000 N, based on the formula density of fluid*displaced volume*gravity.

Is this also the object's weight, or is there another way to obtain the answer for the third part?

 

[Cutter Ketch]

I think the original answer for part 2 stays 2352000 N, based on the formula density of fluid*displaced volume*gravity.

Is this also the object's weight, or is there another way to obtain the answer for the third part?

The title of your post is "...Archimedes principle". You should probably look that up.

 

[a1234]

So for part 3, I calculated the following using Archimedes Principle:

1 cm^3 of water has a mass of 1 g.
240 m^3 = 240,000,000 cm^3, and that many grams of water displaced
240,000,000 g = 240,000 kg
weight = mass * force of g
w = 240,000*9.8
w = 2352000 N, which is the weight of the displaced water and is also the buoyant force.

But how do I find the weight of the entire object (if that is what they are asking for)?

 

[haruspex]

Is this also the object's weight

Tell us what you think, and why.

 

[a1234]

I think so, because the buoyant force on a submerged object is equal to the weight of the fluid displaced.

 

[ehild]

I think so, because the buoyant force on a submerged object is equal to the weight of the fluid displaced.

The object floats in water, is in equilibrium. What is the net force acting on it?

 

[a1234]

If the weight of the object and buoyant force are equal, I'd say the net force is 0.

 

[ehild]

If the weight of the object and buoyant force are equal, I'd say the net force is 0.

Yes. And the net force is zero in equilibrium. The floating object is in equilibrium. So its weight is equal to the buoyant force

 

[a1234]

Thanks! For part 4, I did the following:

weight = mass * gravity
2352000 = 9.8m
m = 240000

density = mass/volume
d = 240000/1200
(1200 cm^3 is the volume of the entire object.)
d = 200
(200 g/cm^3 ?)

 

[haruspex]

Thanks! For part 4, I did the following:

weight = mass * gravity
2352000 = 9.8m
m = 240000

density = mass/volume
d = 240000/1200
(1200 cm^3 is the volume of the entire object.)
d = 200
(200 g/cm^3 ?)

You need to keep track of units.

 

[a1234]

m = 240000 kg
1200 m^3 is the volume
d = 240,000/1200
d = 200 kg/m^3

 

[haruspex]

m = 240000 kg
1200 m^3 is the volume
d = 240,000/1200
d = 200 kg/m^3

Right. Note that this differs from your answer in post #13 by a factor of 1000.

 

[a1234]

Okay. I mixed up the kg and m the first time.

For part 5, I think the first step is to figure out the density of vegetable oil, which is about 0.91 g/cm^3, or 910 kg/m^3.
Density of object/density of fluid = portion of object underwater

200/910 is about 0.22
The total volume is 1200 m^3
22% of that is 264 m^3, which is the portion underwater
So the height then below the surface of the oil is about 3.3 meters.

 

[haruspex]

Okay. I mixed up the kg and m the first time.

For part 5, I think the first step is to figure out the density of vegetable oil, which is about 0.91 g/cm^3, or 910 kg/m^3.
Density of object/density of fluid = portion of object underwater

200/910 is about 0.22
The total volume is 1200 m^3
22% of that is 264 m^3, which is the portion underwater
So the height then below the surface of the oil is about 3.3 meters.

Yes, but I suspect they just wanted a qualitative answer: does it increase, decrease or stay the same? For that, you only needed to know vegetable oil is less dense than water.

 

[a1234]

Right, because as the denominator gets smaller while the numerator stays the same, the quotient gets bigger.

Thank you for all the help!