buoyancy

Chocolaty

Hi, i'm not sure how to prove this
http://img402.imageshack.us/img402/5...oyancyq0wn.jpg

I tried developing a bit but i'm stuck, can anyone point me in the right direction plz

thanks

Chocolaty

See i've understand that
F=1000*9.8*3*15
F=441000N

But how do I prove that this is the difference between the upward and downward force?

Fermat

Well, this is the difference between the upward and the downward force.
You have to show that this is the same as the buoyant force.

Hmm. What is your definition of buoyant force ?

Chocolaty

the weight of the liquid displaced

Fermat

That's right, but from your attachment you wrote,

[tex]\Delta P = \rho g \Delta h[/tex]

which looks like you are caculating the force due to the pressure difference.

Have you actually calculated this force ?

Chocolaty

I'm not exactly sure which force you're talking about. I found this formula in my textbook. I scanned the formulas we have for buoyancy, I got it from the one on the right of this page
http://img474.imageshack.us/img474/641/buoyancy1rq.jpg

We have another formula as well which is not on that page.
Fb = F2 - F1
F2 = upward force applied on the lower surface
F1 = downward force applied to the upper surface

But that's all i've got :s

Fermat

OK.
I can't see the bit on the right very well.

I'll post a few bits to clear things up.
However I'm half-way through my lunch at the moment and my wife wants my attention, so I may be a few minutes gettting back.

Ciao.

Fermat

The buoyant force, call it BF, is equal to the weight of the fluid displaced.

[tex]BF = \rho gV[/tex]

where V is the volume of fluid displaced.

When a body is completely immersed in a fluid, as shown in your original problem, then the difference in pressure between the top and bottom faces is given as follows,

[tex]P_0 = \mbox{ pressure on upper surface}[/tex]
[tex]P_1 = \mbox{ pressure on lower surface}[/tex]

[tex]P_0 = \rho g h_0[/tex]
[tex]P_1 = \rho g h_1[/tex]

[tex]\Delta P = \rho g \Delta h[/tex]

[tex]\mbox{where } \Delta h \mbox{ is the difference in height between the top and bottom surfaces.}[/tex]

The upward force due to this difference in pressure is,

[tex]F = \Delta P A[/tex]
[tex]F = \rho g \Delta h A[/tex]
[tex]F = \rho gV[/tex]

[tex]\mbox{since } \Delta h A \mbox{ is just the volume of displaced fluid.}[/tex]

So, the force, F, due to the difference in pressures, is equal to the buoyant force, giving

[tex]BF = \rho gV = F = \rho g \Delta h A[/tex]
=====================

Fermat

I managed to see that bit on the right hand side of the page.

That formula,

ΔP = ρgΔh

is for the difference in pressure and gives the upward force as,

F = ΔP*A = ρgΔh*A

Now, this is equal to the buoyant force, but the buoyant force is defined as the weight of the fluid displaced. i.e.

BF = ρgV - by definition
=======

Chocolaty

Thanks Fermat, forgive the weird question, i've got a book full of them. Tip to anyone who reads this: "never choose to take distant courses when you can take them at school"