Using Archimedes' principle in engineering applications

[Emmanuel]

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!

 

[berkeman]

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.

 

[haruspex]

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?

 

[scottdave]

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.

 

[Delta2]

if there is some sort of figure or picture that comes along with this assignment then be kind enough to upload it here.

 

[Emmanuel]

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

 

[haruspex]

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.

 

[Emmanuel]

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

 

[Emmanuel]

So volume of iron plate is 20.096 sorry

 

[scottdave]

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.

 

[Emmanuel]

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

 

[Emmanuel]

Volume would then equal 60.288

 

[haruspex]

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.

 

[Emmanuel]

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?

 

[haruspex]

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.

 

[Emmanuel]

Ok so the surface area is 6.03

So i times by 0.01m rather than 10mm

Which is 0.0603 ?

 

[haruspex]

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?

 

[Emmanuel]

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.

 

[haruspex]

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.

 

[Emmanuel]

I've solved this now, covered Archimedes principle in college so I've managed to do it but thanks for the input