>> We've looked at fluid statics to figure out
what the pressure is at different locations.
And one of the big reasons we're
interested in knowing what the pressure is,
is so that we can know what
the forces are that are acting
on things that are surrounding the fluid.
Now, if we know the pressure at this location,
we also know that pressure acts normal
to a surface and it acts on the surface.
So, if I had a pressure acting on this
surface, it would be pushing in that direction
with a pressure P. The resulting
force would depend on the size
of the area that it was pushing on.
So if I look at just a little small area there,
dA, then this will be the pressure times dA.
So pressure times area gives me a force.
And finally, if I want to get the direction
correct, then I'll need a vector that accounts
for the direction of the surface, and the
usual way we define that with surfaces is
with a unit normal vector that
points outwards from the surface.
So this n vector is a unit normal.
That means it's one unit in magnitude
and it is at a right angle to the surface
and we can define that unit
normal vector for every surface
as pointing outwards from the surface.
In that case, the force from the
pressure is going to be negative.
It's in the opposite direction of that unit
normal pointing outwards from the surface,
negative n times the pressure times the
small amount of area that it's acting over.
Now, if I want to get the total force over the
entire area, then I'm going to have to integrate
over this whole surface S, and I'll get that
the total vector force is equal to the integral.
It's a double integral because
it's over area, over the surface S,
the whole surface that I'm interested in
of the pressure times the incremental area.
That's the magnitude of the force.
And the normal vector is going to be
involved and this normal vector may change
if this area isn't perfectly flat.
So we need to make sure we keep that
normal vector inside the integral signs.
And there's a negative coming
from this reversal of direction.
This is just a sign convention that says
this unit normal always points outwards
so the pressure force is always
pointing in the opposite direction.
So this is how we could get the total
vector force acting on the surface.
Or if we were interested only
in components of that force,
we could make our lives a little simpler by
not worrying so much about the vector stuff.
We could say that the force component in
the x direction will still have the integral
over the surface area and we'll take
the pressure and we'll multiply it
by the area projected in the x direction, so
that f of x is the x component of the force,
and a of x is the projected
area in the x direction.
Now, if we're going to take that approach, we
need to be careful to look after that sign there
and make sure that we get our projected area
pointed in the direction that we're interested
in so that we get this integral right.
But it's going to get a lot simpler if we've
got simpler situations on our arbitrary surface.
And let's look at how we can break this down.
If I had a sloping surface like this
with a pressure acting on it like that,
it's going to be exactly like this one up here.
I'll have a normal vector outwards.
I've got the pressure acting
in the opposite direction.
I can break that down into two
parts, an x and a y component,
with one of them being pressure times
the area projected in the x direction
and the other one being pressure times
the area projected in the y direction
where those are each projected areas of dA.
I can look at it this way or
I can look at it this way.
I can say that that's going to be equal
to the component of the pressure force
over the entire area so I could take P in
the x direction over the entire area plus P
in the y direction, the component of P
in the y direction over the entire area.
And that would give me the same thing.
I'm taking the component either of the area
of projection or I'm projecting the pressure
but not projected pressure
times the projected area
because that way I would be taking the
projection twice, I get my geometry wrong.
So the simplest way to manage
this is look at it in this sense.
I got this area.
I can consider it as made
up of two projected areas,
one in the x direction and
one in the y direction.
I will have the pressure force acting
over each of these projected areas
and I can integrate those separately
to get me the components Fx
or similarly to get me the Fy component.
