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Archimedes's principle

by sadhu
Tags: archimedes, principle
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sadhu
#1
Jan16-08, 04:08 AM
P: 158
can anyone prove Archimedes's principle for a sphere and a cylindrical vessel of same volume
, and prove that the forces are same in both cases.



I mean buoyant forces..................
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Doc Al
#2
Jan16-08, 04:56 AM
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Well, you could just crank it out--integrate the fluid pressure over the surface area. Sounds like a good exercise.

But you can also use an argument that requires no integration and works for any shape. Imagine any object submerged in the fluid. Now replace that object by an equal volume of fluid. Since the fluid is in hydrostatic equilibrium, the net force of the surrounding fluid on that imaginary "object" must equal the weight of the fluid contained within its boundary. That's Archimedes's principle. Done!
sadhu
#3
Jan16-08, 10:10 AM
P: 158
I prefer to use that integration idea

but on calculation it is getting to complicated to solve , however somehow when i solved it
, I just got it wrong........i think

can any one show me its integration...........pleazzzz

Shooting Star
#4
Jan16-08, 01:11 PM
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P: 1,979
Archimedes's principle

P = dgh = dg(-y), where I’m writing d for the density of the fluid, and choosing +y in the upward direction, and g is a scalar, and g=-gj.

Buoyant force B = total force on submerged body = Surface Integral Pda = Volume integral [(grad P)dV].

Now, grad P = grad (-dgy) = -dgj

So, B = the vol integral = -dgVj= -dV*gj = weight of fluid displaced, acting upward.

p.s. Archimedes didn't know calculus. Anyway, now you can try your hand on calculating the buoyant force on a really big object floating in the ocean by calculus. I mean, in a non-uniform g field.


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