>> Now, we're going to have a look
at buoyancy and why things float.
We all know that if something is lighter
than water, it'll float on top of the water.
And we all saw at school that Archimedes
with his eureka moment told us that the force
of buoyancy was going to be equal
to the weight of water displaced.
So, let's unpack that a little bit.
The weight of water displaced,
well, if we look at the amount
of this body here that's
actually below the waterline,
then we can find that there's a
volume of water that's been displaced.
So if we're unpacking, that'll
be the integral over the volume.
Well, let's simplify.
Let's start just by multiplying by the volume.
It'll be density times gravity times
the total volume of water displaced.
And in this case, make sure that you notice that
this density, this is going to be the density
of the fluid that's being displaced.
So in this case, if it's something floating in
the water, this is the density of the water.
Now, if we wanted to figure out what that was,
we could do an integral over the entire volume,
this whole region here of rho g times dv.
Or, if we take rho and g as
constants, that's just the integral
over the volume of the incremental volume.
So, that's not telling us too
much that we don't already know.
Let's look at this in a little more detail.
If we put a coordinate system on here,
and we'll call the vertical dimension zed,
then we've got x and y defining
the plan form projection.
This is the horizontal plane, is the xy plane.
If we've got our body sitting here
and it's underwater to some extent,
then everywhere over this body,
we can define what the height is,
the distance that this surface
is under the water.
So, h is going to be some function of
x and y. And in the vertical forces,
we would have to integrate
over the entire plan form area.
Now, horizontal forces-- well, let's
see where the pressure is acting.
There's pressure forces acting upwards on that
flat bottom, there's a pressure force over here,
it's a little smaller magnitude of pressure
because we're not as far under the water,
and we've got pressure forces
acting all the way around this body.
And the deeper under the water we
are, the higher the pressure is.
Now, if we consider horizontal forces
only, only the horizontal components,
then the force pushing this way on this
side and the force pushing that way
on that side, they're all going to cancel out.
However, the forces pushing upwards vertically
don't have any corresponding forces pushing
downwards if we were to engage pressure.
So the horizontal components, they cancel out.
Vertical components, we will have to integrate
over the projected area in the horizontal plane.
So, each little elemental area, dA,
in the horizontal plan form plane.
And our h is a function of x and y, and it's
what defines the shape of this floating body.
So, our force to the buoyancy is going to be
equal to the integral over that projected area,
the area under the body here, of rho gh.
That's going to be the pressure at any one
of these points defined by h of x and y,
how far under the water we
are times the projected area,
this area in the horizontal plan form plane or
just equal to integral over the area of PdA.
So, this buoyancy force is
really a pressure force.
It's the same kind of pressure force that
we've been calculating all over the place.
And whether we start from
Archimedes' principle or start
from integrating the pressure
force, we'll wind up back here.
And let's follow this one
through from Archimedes.
There is the buoyancy force, is the
integral over the volume of rho gdV
or just the integral rho g-- or sorry,
the integral over the volume dV times rho g.
That'll be equal to, well, rho g times dV.
Well, if we look at a region in here, then
the elemental volume that we're looking
at will be h high and dA plan form area
in projected area in the plan form.
So, that would be integral over the area hdA
or since rho g times h is the pressure locally,
that'll be equal to the integral
over the area PdA.
And those are exactly the same thing whether
we start from thinking about what's going
on with Archimedes or whether
we start from thinking
about a distributed pressure
force acting over an area
which is supporting this thing
that's floating in the water.
Either approach is valid, either
approach will tell you what the answer is,
but I like this approach better in
terms of understanding pressure forces
and how they're distributed over the body.
And it's going to tell us more
information later on if we keep thinking
about these distributed forces in
terms of the way the body here floats
and where that force is more concentrated.
Next, we'll move on and we'll look at what
happens when we have submerged bodies.
