Buoyancy and Archimedes's Principle Based Question

1. Jul 4, 2012

shinnsohai

I've this to be handed in by tomorrow

I think that my working is completely wrong !!!

1. The problem statement, all variables and given/known data
An empty cylinder bucket, 30cm in diameter and 50cm long, whose wall thickness and weight can be considered negligible is forced, open end first, into water untill its lower edge is 4m below the surface as shown in the picture. Calculate the force F which is required to hold t he bucket in this position assuming the trapped air will remain at constant temperature during the entire operation.

2. Relevant equations
P=ρ x G x H

3. The attempt at a solution

P1V1=P2V2
(4-h)ρg(∏(0.15^2))(4-h) = 3.5ρg (∏ (0.15^2))(h-3.5)
(4-h)^2(0.0707)= 3.5(0.0707)(h-3.5)
(16-8h+h^2)(0.0707)= 0.24745h-0.866075
0.0707h^2-0.565h+1.1312= 0.24745h-0.866075
0.0707h^2-0.81305h+1.997275= 0
(3.5ρg+101.3k)∏(0.15^2)(h-3.5)
(4-H)^2ρg(∏(0.15^2))= 3.5ρg(∏0.15^2)(h-3.5)

0.5=h-3.5+4-h
h-3.5=0.5
h=3.5/0.5
h=7

Last edited: Jul 4, 2012
2. Jul 4, 2012

LawrenceC

To solve this problem you will need to determine the volume of the trapped air in the pail. You have the equation P1*V1=P2*V2 so you can relate the final volume to the intial volume for the air in the pail. In turn, this can be related to depth at the water-air interface inside the pail (depth times density being pressure, P2). Doing this will result in a quadratic equation for the height (h) of the cylinder of air in the pail if you represent the total depth as 3.5+h. Knowing the height of the air, you can easily determine the volume of air. From that you determine, via Archimedes Principle, the force necessary to hold the can under water at the specified depth.

Last edited: Jul 4, 2012
3. Jul 4, 2012

shinnsohai

At Least A Hope !
Gonna Re-Solve This!

4. Jul 4, 2012

LawrenceC

When you set up your equation relating pressure and volume, pressures must be absolute. Also notice the slight change I made in my previous post. The 4 meters should have been 3.5 meters making the interface depth 3.5+h.

5. Jul 4, 2012

shinnsohai

Erms Lawrance..
I Think that I do mistake all the time , but hoping no this time :D

Maybe the 2nd photo having the correct ans
Where small h=3.75
Therefore Trapped Air Height is 0.25m

what's the next step to find force(Brain Stuck!)
It's so tiring ~~
2am~
i shall continue tomorrow , hoping you could give me the answer ASAP

Thx for helping anyway!

6. Jul 4, 2012

LawrenceC

Here is some more help:

P1V1=P2V2 where P2 is trapped air pressure and V2 is volume of trapped air.

Let d = depth to top of can (3.5 m)
h2= depth of compressed air in top of can
Pa= atmospheric pressure

So P2=rho(d+h2)+Pa

V2=Ah2=P1V1/P2=P1V1/(rho(d+h2)+Pa)
A=area of can cross section
Solve above quadraticfor h2. It is written as
Ah2=P1V1/(rho(d+h2)+Pa)
and should be put in standard quadratic form so you can use formula.

Knowing h2 you can get the force from Archimedes Principle or by evaluating pressure at air-water interface and multiplying it by the area followed by subtracting the pressure force pushing down on the top of the pail. The difference is the force needed.

If you work out the equations in symbols it is less messy and therefore less error prone. Plug in the numbers at the end. Make sure your units agree.

7. Jul 4, 2012

1h193

may I know why are you using Boyle's Law ..? I don't get the concept ..

*I was quite in shocked when I looked at this ..
0.5=h-3.5+4-h
h-3.5=0.5
h=3.5/0.5
h=7

8. Jul 4, 2012

LawrenceC

PV^n=constant

For an isothermal process, n=1
For an isentropic process, n=k
etc...