I would like to demonstrate how calculate the resultant hydrostatic force with an example.
In this particular
example, I've got a reservoir with a gate.
The gate itself is two meters wide,
going into the board, by
1.2 meters tall. That would be this
dimension here. The gate is
three meters below the water surface.
What we're after is is how to
calculate the resultant force. Now
there's two different approaches to do
this. With the first approach what we're
going to do is use the Statics
approach. So I'll go ahead and do
that.  So using the statics approach,
 
this could be a refresher for you
in statics. With the statics approach
what we're looking at is, let me
redraw this,  that's my water surface.
Right down here I have the gate, and make
it larger.
OK so if we look at the pressure
distribution, remember from the previous
lesson, we learned that pressure is a
function of depth. So at the water
surface there's no pressure. As we
increase death we will increase pressure,
so the distribution looks something like
this.  So if we look at this specific piece
 
right here, what we see is we have a
trapezoidal distribution. So here is the
gate itself, and draw that as a gate,
then we have a distribution looks
something like this.
Now the easiest way for us to handle a
trapezoid distribution,  if you recall from
statics, to break it up into two different
objects. We have a rectangle and we
have a triangle. The rectangle will
have a force associated with it,
call that F1,  and the triangle will
have another force associated F2.
OK now for the base of this what we need
to do is we need to calculate the pressure
at this point,  I'll call it P1.
Ok, so as I just mentioned, pressure is a
function of depth and actually what it
is that pressure is equal to the specific
weight times the depth H.
So at the top of this gate here
with three meters down from the
water surface. So we're going to
have specific weight of water at the
depth of 3 meters. For the pressure down
here, it's a similar calculation, but in this
particular case at the bottom of this
surface we have  a depth of 3 plus
1.2. So we have gamma water times a
total depth of 4.2 m. Now what we have
to do is solve for the
magnitude of these forces.
OK, F1 is going to be equal to,
so think back to statics when you
had distributed loads.
The magnitude of the force within a
distributed load is the area of the
distributed load. So the area of
this triangle, sorry this rectangle,
has a base of gamma water times three.
It has a height dimension of 1.2 and then
going into the board it has a dimension
of was the two meters.
OK, so let's see, that gives us F1 is
equal to have 3 times 1.2 which is  3.6, times 2 is
7.2. So Ff1 is equal to 7.2 times gamma
water. For F2 we're going to do a
similar calculation but instead having a
rectangular distribution we have a
triangular distribution. So for F2 we're
going to have this base,  right
her, will the 4.2 gamma minus
the 3 gamma.  First of all we're
going to solve for this piece
here so we have a triangle. For the
area of a triangle we have one half the
base,  this piece here,   will be
gamma water and then you
have the 4.2 minus 3. The height of
this again is when be 1.2,  and then going
into the board as we had earlier,  we have
a width of two meters.
OK so you got the half here,  the two,
so I've got basically 1.2 x 1.2 gamma
water that comes out to be 1.44 gamma
water. So my total force is going to be
the sum of those two forces which is
equal to 8.64 gamma water.  At this
stage let me go and just put in the
values. If I use a value, lets see this
would be 8.64. Technically
this was cubic meters, right? Because
these were all meter dimensions and
then if we use a value of that say 9.81
kN per meter cubed.
So if we calculate the magnitude of
that force we have 9.1 x 8.6 for which
comes out to be 84.76 kN.
As you can see the statics approach was
pretty extensive. We had to go back to
old school:  we had to break it in terms
of rectangle and a triangular
distribution,  and sum of the forces.  Now if
we use the fluids approach, which I'm
gonna go back over here and do because
it's actually very simple approach and
we don't need a lot of space to do it.
So with the fluids approach, it's a
formula that says the resulting force is
the gamma the water times the depth of
the centroid times the area of the plate.
The only calculation involved in this
that they can be a little complex involves
calculating the depth of the
centroid. The depth of the centroid is
this dimension right here.
So literally we're thinking of a
rectangle here that's 1.2 meters tall.
So we want to get the depth of that.
We have gamma water, OK our
depth of centroid is going to be
the three meters plus half of this
dimension which is 1.2. So we have
1.2 over 2, that times the area of the plate, right?
Because this force is acting
normal to that plate. So it's a 1.2 x 2
meter area.
This comes out to be exactly the
same value we got earlier, 8.64 times gamma water.
Or the resultant force comes out to be
84.76 kN.  Obviously this
makes us very happy because we get the
same answer using two different
techniques. I think all of us could agree
that, even though the statics technique is
fundamentally very sound from an
engineering perspective, if you
don't know what to do you do what you
know, this would be a really really
good technique if you can remember.
At the same time if you have this in
your toolbox and remember this
formula the fluids approaches a lot
easier! That concludes our example
