# Fluid Mechanics and Archimedes Principle

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1. Nov 21, 2016

### a1234

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

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?

2. Nov 21, 2016

### haruspex

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

3. Nov 21, 2016

### a1234

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

4. Nov 21, 2016

### haruspex

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

5. Nov 21, 2016

### 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?

6. Nov 21, 2016

### Cutter Ketch

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

7. Nov 21, 2016

### 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)?

8. Nov 21, 2016

### haruspex

Tell us what you think, and why.

9. Nov 21, 2016

### a1234

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

10. Nov 21, 2016

### ehild

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

11. Nov 22, 2016

### a1234

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

12. Nov 22, 2016

### ehild

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

13. Nov 22, 2016

### 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 ?)

14. Nov 22, 2016

### haruspex

You need to keep track of units.

15. Nov 22, 2016

### a1234

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

Last edited: Nov 22, 2016
16. Nov 22, 2016

### haruspex

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

17. Nov 22, 2016

### 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

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.

18. Nov 22, 2016

### haruspex

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.

19. Nov 22, 2016

### a1234

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

Thank you for all the help!