Using Archimedes' principle in engineering applications

In summary, a cylindrical buoy made of iron plate 10mm thick with a diameter of 0.8m and a height of 2m is floating in sea water with its axis vertical, submerged three-fourths. Using the relative density of iron (7.8) and sea water (1.03), the task is to calculate the mass of iron chain securing the buoy. To begin, the weight of the buoy itself needs to be calculated, which can be done by finding the volume of the iron plate used in its construction. The surface area of the iron plate can be calculated by multiplying π by the radius squared, and then multiplying that by 2 to account for both ends of the cylinder. The volume of the iron
  • #1
Emmanuel
10
1

Homework Statement


A cylindrical buoy floats in sea water with its axis vertical so that it's three-fourths submerged. The buoy is 0.8m in diameter and 2m in height. Its fabricated from iron plate 10mm thick. Calculate the mass of iron chain securing the buoy.

The relative density of iron is 7.8 and of sea water is 1.03

Homework Equations


Not sure of this as I haven't been taught it at college, yet it is on my assignment, eek!

The Attempt at a Solution


Really don't know where to start, I would be really grateful if somebody could help me through this. Thanks in advance!
 
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  • #2
Emmanuel said:

Homework Statement


A cylindrical buoy floats in sea water with its axis vertical so that it's three-fourths submerged. The buoy is 0.8m in diameter and 2m in height. Its fabricated from iron plate 10mm thick. Calculate the mass of iron chain securing the buoy.

The relative density of iron is 7.8 and of sea water is 1.03

Homework Equations


Not sure of this as I haven't been taught it at college, yet it is on my assignment, eek!

The Attempt at a Solution


Really don't know where to start, I would be really grateful if somebody could help me through this. Thanks in advance!
Welcome to the PF.

Re-read your textbook and reference materials on how Archimedes' Principle is applied to calculate buoyancy of floating objects, and post the Relevant Equations and make a sketch of the problem. That's the best way to get started... Use the Upload button in the lower left to Upload a JPEG image of your sketch please.
 
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  • #3
Emmanuel said:
A cylindrical buoy floats in sea water with its axis vertical so that it's three-fourths submerged. The buoy is 0.8m in diameter and 2m in height. Its fabricated from iron plate 10mm thick. Calculate the mass of iron chain securing the buoy.
There's not enough information to solve the problem as given.
If there is a diagram it will indicate whether the chain is taut, but it still is not enough either way. If it is taut then there is an unknown force from the riverbed; if it is slack there is an unknown mass of chain lying on the riverbed.
So I would assume that a) the chain is slack and b) that they only want the mass of the hanging part.

First up, what can you calculate about the buoy that will be useful?
 
  • #4
First, you need to calculate the weight of a buoy with the given construction specs. Do you know what you need to be able to do that? Hint: calculate the volume of iron plate used, first.
 
  • #5
if there is some sort of figure or picture that comes along with this assignment then be kind enough to upload it here.
 
  • #6
There is no figure that comes with it and I have provided all the information I was given.

I think to calculate the volume of the buoy it would be 4/3•π•r^3 but I'm not sure how I would do this with the iron plate or put the two together

Please forgive me I have only just started this course, I have no background in these subjects, I'm a true beginner and this subject hasn't been taught
 
  • #7
Emmanuel said:
the volume of the buoy it would be 4/3•π•r^3
Yes. To get the volume of iron in the shell, multiply the surface area by the thickness.
 
  • #8
So to find the surface area it is 4•π•r^2

4 × 3.14 × 0.4^2= 2.0096

And the volume of the buoy

4/3 × 3.14 × 0.4^3= 0.267

Do I just add these together to find total volume
 
  • #9
So volume of iron plate is 20.096 sorry
 
  • #10
Your volume of iron is incorrect. Surface area of a cylinder is area of each end plus area of the side. The side can "unroll" into a rectangle.
 
  • #11
Sorry for some reason I had a sphere in my head

So it would be 6.0288?

(2•π•r^2 + π•d•h)

2 × 3.14 × 0.4^2 + 3.14 × 0.8 × 2
 
  • #12
Volume would then equal 60.288
 
  • #13
Emmanuel said:
Sorry for some reason I had a sphere in my head
And when I saw your 4πr3/3 I assumed it was a sphere and did not go back to check... Whoops.
 
  • #14
Bloody hell I've already made a mess of it ha

So the volume of the buoy alone would be 1.0048
And the volume of the iron is 60.288

Where do I go from there?
 
  • #15
Emmanuel said:
Bloody hell I've already made a mess of it ha

So the volume of the buoy alone would be 1.0048
And the volume of the iron is 60.288

Where do I go from there?
There cannot be more iron by volume than the whole buoy. Check your units.
 
  • #16
Ok so the surface area is 6.03

So i times by 0.01m rather than 10mm

Which is 0.0603 ?
 
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  • #17
Emmanuel said:
Ok so the surface area is 6.03

So i times by 0.01m rather than 10mm

Which is 0.0603 ?
Looks ok, but you should always state the units.
So what is the mass of the iron?
 
  • #18
Is mass: volume × density?

If so the mass of iron would be 0.06 × 7800= 468

To be completely honest I'm not sure if the units

Do I just add the volume of iron and the buoy together to get the total volume? Seems too simple to be that.
 
  • #19
Emmanuel said:
I'm not sure if the units
If you do not keep track of units then you have no idea what your answer means.
What were the units of the volume you calculated? If not sure, do the same calculation with the units that you did with the numbers. E.g. if you are told that a rectangle is 2m by 30mm then after converting that to 2m by 0.03m multiplying them together gives 0.06 m2.

For the total volume, it is not clear whether the given diameter of the buoy is internal or external. Assume external.
 
  • #20
I've solved this now, covered Archimedes principle in college so I've managed to do it but thanks for the input
 

FAQ: Using Archimedes' principle in engineering applications

How is Archimedes' principle used in engineering applications?

Archimedes' principle is used in engineering applications to determine the buoyant force on an object submerged in a fluid. This information is crucial in designing ships, submarines, and other watercrafts. It is also used in designing and calculating the stability of offshore structures such as oil rigs and platforms.

What is the equation for Archimedes' principle?

The equation for Archimedes' principle is FB = ρf * V * g, where FB is the buoyant force, ρf is the density of the fluid, V is the volume of the submerged object, and g is the acceleration due to gravity.

How does Archimedes' principle relate to the weight of an object?

Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. This means that if the weight of the fluid displaced is greater than the weight of the object, the object will float. If the weight of the fluid displaced is less than the weight of the object, the object will sink.

Can Archimedes' principle be applied to gases?

Yes, Archimedes' principle can be applied to gases as well as liquids. In this case, the density of the gas is compared to the density of the surrounding air. If the density of the gas is greater, it will rise, and if the density is less, it will sink. This is how hot air balloons work.

What are some real-life examples of Archimedes' principle in engineering?

Some real-life examples include designing submarines to control their buoyancy, designing boats and ships to ensure they can float and remain stable, and designing swimming pools and water parks to ensure proper water displacement. It is also used in designing pipelines and underwater structures to calculate the buoyant force and prevent them from floating to the surface.

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