Fluid Mechanics and Archimedes Principle

In summary, a rectangular object with dimensions of 40 meters by 15 meters by 2 meters floats consistently when 3 meters of its height is below the surface of the water. The volume of the displaced water is 1200 m^3 and the buoyant force on the object is 2352000 N. The weight of the object is also 2352000 N based on Archimedes' principle. The mass density of the block is 200 kg/m^3. If the object were floating in vegetable oil, the portion of the object under the surface would increase to about 22%.
  • #1
a1234
78
6

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?
 
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  • #2
a1234 said:
This is 40*15*2,
That is the volume of the whole block.
Draw a diagram.
 
  • #3
The volume of the displaced part of the object is 40*3*2, or 240 m^3.
 
  • #4
a1234 said:
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.
 
  • #5
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
a1234 said:
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.
 
  • #7
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
a1234 said:
Is this also the object's weight
Tell us what you think, and why.
 
  • #9
I think so, because the buoyant force on a submerged object is equal to the weight of the fluid displaced.
 
  • #10
a1234 said:
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?
 
  • #11
If the weight of the object and buoyant force are equal, I'd say the net force is 0.
 
  • #12
a1234 said:
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:smile:
 
  • #13
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
a1234 said:
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.
 
  • #15
m = 240000 kg
1200 m^3 is the volume
d = 240,000/1200
d = 200 kg/m^3
 
Last edited:
  • #16
a1234 said:
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.
 
  • #17
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.
 
  • #18
a1234 said:
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.
 
  • #19
Right, because as the denominator gets smaller while the numerator stays the same, the quotient gets bigger.

Thank you for all the help!
 

Related to Fluid Mechanics and Archimedes Principle

1. What is fluid mechanics?

Fluid mechanics is the study of how fluids (liquids and gases) behave under various conditions, such as flow, pressure, and viscosity.

2. Who was Archimedes and what is his principle?

Archimedes was a Greek mathematician, physicist, engineer, inventor, and astronomer who lived in the 3rd century BC. His principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the displaced fluid.

3. How does Archimedes principle relate to buoyancy?

According to Archimedes principle, an object submerged in a fluid will experience an upward force (buoyant force) equal to the weight of the fluid it displaces. This is why objects float if they are less dense than the fluid they are in.

4. What is the significance of Archimedes principle in daily life?

Archimedes principle has many applications in our daily lives, such as in shipbuilding, understanding the behavior of hot air balloons, and determining the volume of irregularly shaped objects. It also helps in understanding why some objects float while others sink.

5. How is fluid mechanics used in engineering?

Fluid mechanics plays a crucial role in engineering, particularly in the design of aircraft, ships, and cars. It is also used in the design of pipes, pumps, and turbines for various industrial processes. Understanding fluid mechanics is essential for engineers to optimize the performance and efficiency of these systems.

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