Hydrostatic Forces on a Plane Surface Video Lecture | Fluid Mechanics for Civil Engineering - Civil Engineering (CE)

In this lesson, we will derive the equation
for calculating the magnitude of the hydrostatic
force on a plane surface.
Here we have a rectangular fish tank filled
with water to a depth d.
In order to determine whether the fish tank
walls can support that much water without
breaking, we would need to know the total
force exerted on the walls.
First we will determine the resultant force
along the bottom wall.
The tank is surrounded on all sides by the
atmosphere, which has a pressure Pa.
On the underside of the bottom wall, which
is exposed to the atmosphere, the pressure
force is Pa times the area of bottom wall,
and this force points upward.
Inside the tank, the pressure force increases
as one descends through the water, reaching
a maximum at the bottom of the tank.
The absolute pressure on the upper part of
bottom wall at depth d is the atmospheric
pressure plus the specific weight of the water
times the depth.
This creates a downward force on the wall
equal to the Pa plus gamma d, times the area
of the bottom wall.
If we examine just the fluid pressure forces
acting along the bottom wall, the resultant
force is Pa times the area of the bottom wall,
minus the quantity Pa plus gamma d, times
the area of the bottom wall.
The atmospheric pressure terms cancel out
and we are left with negative gamma times
d times the area.
This means the resultant force points downward
with a magnitude of gamma times d times the
area.
Finding the resultant pressure force along
the bottom wall was relatively easy because
the pressure is constant along the entire
wall.
This is not the case for the side walls.
The pressure along the exterior of the side
walls is Pa and the related pressure force
is Pa times the area of side wall.
However, on the interior of the side walls,
the pressure increases linearly with depth.
At a depth h, the pressure is the atmospheric
pressure plus gamma h.
Multiplying by the area of the side walls
gives the related pressure force.
We want to determine the magnitude of the
resultant force caused by the fluid pressure
on the side walls.
Here is a different tank.
This tank is open to the atmosphere and partially
filled with a liquid of specific weight gamma.
We are going to examine a section of the flat
wall, highlighted in pink, that has an arbitrary
shape and which is oriented at an arbitrary
angle theta relative to the free surface.
We set up a coordinate system where the x-coordinate
comes out of the screen and is oriented along
the free surface.
The y-coordinate is oriented along the section
of wall we wish to examine.
We will need to examine some important points
on the side wall, so rotate the surface 90
degrees about the y-axis to get a better view.
Point (x,y) is the location of some arbitrary
small area dA on the wall.
dA has an area dx times dy, and is at depth
h from the free surface.
Point C is the centroid of the wall and is
located at coordinates (xC, yC).
The depth of the centroid from the free surface
is hC.
The centroid is the geometric center of the
wall, and the centroid and center of mass
are the same if the density of the wall is
constant.
Point CP is the center of pressure, which
is where the resultant force FR acts.
The coordinates of the center of pressure
are labeled (xR, yR).
The resultant force and center of pressure
location produce an equivalent force and moment
on the wall as the original fluid pressure
field.
In order to find the resultant force over
the entire highlighted surface, we will start
by examining the force at the location of
dA.
The small pressure force on the exterior of
the wall at area dA, labeled dFext, is the
atmospheric pressure Pa times the area dA.
The small pressure force on the interior of
the wall at area dA, which is labeled dFin,
is the pressure at dA times the area dA.
The pressure at that depth is the atmospheric
pressure plus gamma h.
Using trigonometry, h can be rewritten as
y times sin(theta).
The net force due to fluid pressure at dA
is dFinterior minus dFexterior.
After plugging in the expressions for the
two forces, and eliminating Pa times dA, we
obtain gamma times y sin(theta) times dA.
To find the resultant force of the fluid pressure
on the entire wall, we integrate dFnet along
all points of the wall.
Plug in the expression for dFnet, and pull
gamma sin(theta) out of the integral because
they are constant.
The integral of y dA can be simplified if
we recall the definition of the centroid.
The y-coordinate of the centroid, yC, is the
integral of y dA, integrated over the entire
surface, divided by the entire surface area
A.
We now have an expression for the resultant
force FR.
FR is equal to gamma times sin(theta) times
yC times the area.
yC times sin(theta) is equal to the depth
of the centroid, hC, so we can rewrite the
resultant force equation as gamma times hC
times A.
This form of the equation is often more convenient
to use when solving fluid mechanics problems.
Gamma hC is the pressure at the centroid,
PC, so we can further rewrite the resultant
force equation as the pressure at the centroid
PC times the area.
These equations are valid for any flat surface
in which the free surface of the liquid and
the side of the tank are exposed to the same
pressure.
In this case, they are both exposed to the
atmospheric pressure.
We are free to use either yC or hC to find
the resultant force and it is important to
understand the difference between the two
quantities.
hC is the vertical distance from the free
surface to the centroid.
yC is the distance from the free surface to
the centroid as measured along the orientation
of the wall.
If the wall is oriented at an angle less than
90 degrees relative to the free surface, hC
will be shorter than yC.
If the wall is vertical and theta is 90 degrees,
hC and yC will be the same distance.
In both cases, the distance yR will be greater
than yC.
Although not proven in this video, if the
tank is pressurized to a gage pressure of
P0, the equation for the resultant force would
need to be modified to (gamma sin(theta) yC
plus P0) times the area A, which can be rewritten
as (gamma hC plus P0) times A.