| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
18 Aug 2005 01:56:11 AM |
| Object: |
bouyancy |
hi,
How can the force of bouyancy be explained starting from
newton's three laws??
ganesh
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| User: "Herman Trivilino" |
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| Title: Re: bouyancy |
18 Aug 2005 07:13:56 AM |
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<gans1973@rediffmail.com> wrote ...
How can the force of bouyancy be explained starting from
newton's three laws??
Start with an object submerged in a static fluid. When released from rest
it wil laccelerate either up or down, depending on the direction of the net
force, in accordance with Law II. (Experimentation with different objects
submerged in different fluids reveals that it's possible to produce a net
upward force, a net zero force, or a net downward force, depending on the
relative magnitudes of the upward buoyant force and the downward weight
force.)
Now imagine removing the submerged object in such a way as to leave the
surrounding fluid undisturbed. There is now a perfect vacuum in the region
where the object used to be. Next, imagine filling that region with some
additional fluid of the same type that surrounds it to create a submerged
"object" that is exactly the same size and shape as the original. The net
"force" on this object is zero. Hence, the upward buoyant force equals the
weight of the "object".
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| User: "Uncle Al" |
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| Title: Re: bouyancy |
18 Aug 2005 01:23:55 PM |
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wrote:
hi,
How can the force of bouyancy be explained starting from
newton's three laws??
ganesh
http://www.theonion.com/news/index.php?issue=4133&n=2
Except the other way.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "PD" |
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| Title: Re: bouyancy |
18 Aug 2005 04:30:25 PM |
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wrote:
hi,
How can the force of bouyancy be explained starting from
newton's three laws??
ganesh
1. Start with a tank of still water.
2. Imagine an imaginary Campbell's Soup can's worth of water in the
middle of this tank and note that it is not moving up or down
(otherwise the water would stir and mix constantly), hence it is in
equilibrium.
3. The water outside this imaginary boundary is pushing in on this blob
of water from all sides, though not necessarily equally.
4. Indeed, because it is in equilibrium, the imaginary can's worth of
water must have its weight (the pull of gravity on that blob of water)
canceled out by another force. The only other thing that is in contact
with the blob of water is the water outside that imaginary can, so that
water pressure must be pushing up.
5. The net force up due to the pressure of the water outside the
imaginary can must be equal to the weight of the water.
6. Now imagine the imaginary can is painted black so you can't see
inside the imaginary can. Nothing else changes.
7. Now imagine replacing the contents of the imaginary can with
Campbell's Cream of Buckshot Soup. The contents inside the can are now
substantially heavier than they were before, but the water outside
doesn't know that (it's painted black), so it pushes up the same as it
always did. But the downward force (the weight of the soup) is heavier,
so it sinks.
8. Now imagine replacing the contents of the imaginary can with
Campbell's Cream of Feathers Soup. The contents inside the can are now
substantially lighter than they were before, but the water outside
doesn't know that (it's painted black), so it pushes up the same as it
always did. But the downward force (the weight of the soup) is less, so
the can rises.
8. Quantitatively, say the can is a cylinder with end-area A and height
h. The pressure of the outside water on the sides of the can obviously
cancel, even though the pressure is probably steadily increasing the
further down along the side of the can you go. The force due to the
pressure of the water on the top is downward and is P1*A, where P1 is
the pressure on top. The force due to the pressure of the water on the
top is upward and is P2*A, where P2 is the pressure on bottom. The net
force upward is P2*A-P1*A=(P2-P1)*A. We know from step 5 that this must
be equal to the weight of the water, which is
m*g = d*V*g = d*(A*h)*g, where d is the water density, V is the volume
of the imaginary can.
Therefore (P2-P1)*A = d*A*h*g
or P2 = P1 + d*h*g
and you've just verified that the pressure increases with depth.
PD
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