In this lesson, we will discuss how to determine
the hydrostatic force on a curved surface.
Here we have a rectangular tank of liquid
with specific weight gamma.
The side walls of the tank consist of flat
plane surfaces.
Let’s examine the pressure force along the
right wall.
Although the pressure increases with depth
along the wall, notice that it always points
in the same direction – toward the right.
This is because pressure force acts normal
to the wall and the walls does not change
its orientation relative to the free surface.
The resultant force on any planar surface,
FR, is found by first determining the small
net force at each small piece of area dA along
the wall, then adding all these small forces
through integration.
The small net force at a given area dA is
equal to the specific weight gamma, times
the distance y from the free surface to the
area, times the sine of the wall orientation
angle theta, times dA.
We are able to pull gamma sin(theta) out of
the integral because they are constant, and
the integral of y dA is the y-coordinate of
the centroid, yC, times the wall area.
For curved surfaces, integration is usually
much more difficult because the pressure force
does not have a constant orientation angle.
On the right we have a fish bowl filled with
the same liquid.
Notice that theta changes along the entire
surface.
This means we cannot pull sin(theta) out of
the integral and cannot develop a general
equation to calculate the resultant force.
However, it is still possible to find the
resultant force on a curved surface using
the following procedure.
First, we need to specify the surface we want
to examine.
The drawing on the right shows a cross section
of a tank with a liquid of specific weight
gamma.
The side of the pool is flat and vertical
to a depth d, then becomes curved.
The curved surface is highlighted in purple
and has a height hS and length lS.
The surface also has a width wS out of the
screen.
Next, we isolate the body of fluid that is
adjacent to the curved surface.
We do this by creating planar surfaces.
Here is what the body of liquid looks like
when rotated slightly.
The volume of the isolated body of liquid
will be denoted as Vf.
Next, we draw a free body diagram of the isolated
fluid body.
The weight of the isolated fluid body acts
at its center of mass and points downward.
We will call this force W.
There is a pressure force caused by the weight
of the fluid above which acts at the top planar
surface.
We will call this force F1.
There also is a pressure force acting on the
left planar surface, which we will call F2.
For the last force, Newton’s 3rd law tells
us if the fluid exerts a force FR on the wall,
then the wall exerts an equal and opposite
force on the fluid.
This normal force will be denoted as FN.
So if we can find FN, we will know the resultant
force FR.
We apply Newton’s second law on the isolated
fluid body to determine the components of
FN.
The sum of the forces acting on the isolated
fluid body is equal to its mass times its
acceleration.
Since the body fluid is not moving, its acceleration
is zero and the right side of the equation
becomes zero.
The normal force vector FN can be broken into
a horizontal component, which we label FH,
and a vertical component, which we label FV.
The sum of forces in the x-direction is equal
to zero.
So F2 minus FH is equal to zero.
Rearranging the equation, we find that the
component in the horizontal direction FH is
equal to F2.
F2 is the resultant force on the left planar
surface and can be calculated from the equation
gamma times hC times the area on the left
side of the isolated fluid body.
hC is the vertical distance from the free
surface to the centroid of the left planar
surface, and is d plus one-half hS.
The area of the left planar surface is wS
times hS.
The sum of the forces in the y-direction is
also equal to zero.
So FV minus F1 minus W is equal to 0.
Rearranging the equation, we find that FV
is equal to F1 plus W.
F1 is the pressure force due to the weight
of the fluid directly above the isolated fluid
body.
This weight is equal to the mass of the fluid
directly above the top surface times the gravitational
acceleration g.
The mass of the fluid above is equal to its
density rho times volume it occupies.
We can combine rho and g into gamma, and the
volume of the fluid above is the height d
times the area wS lS.
We also could have calculated F1 from the
pressure at the top of the isolated body P1,
which is gamma times the depth d, times the
area of the planar surface wS lS.
The weight of the isolated fluid body is the
mass mf times g.
Mass is replaced by density rho times the
volume of the fluid body Vf.
Rho and g can be replaced by gamma, and we
find that the weight of the isolated fluid
body is gamma times Vf.
At this point we would need to know the volume
of the isolated fluid body to continue.
For simple shapes, it may be easy to determine
the volume.
However, if the surface is irregular then
it may difficult to determine the volume.
The magnitude of normal force is the square
root of the sum of the squares of the components,
and the orientation angle is equal to the
inverse tangent of FV divided by FH.
The resultant force FR is equal in magnitude
and opposite in direction to FN.
If the curved surface is a gate hinged about
some point O, we may need to determine the
moments of all forces about the hinge.
This requires knowledge of the moment arms
of all the forces involved in the problem.
If the fluid is static and the gate is not
moving, there must be no net moment on the
gate or the fluid body.
That is, F1, F2, the weight, and the components
of FN must produce no net moment.
F2 and FH both act only in the x-direction,
while F1, W, and FV act only in the y-direction.
The means that F2 and FH must have the same
line of action and FH is at the same depth
as F2.
F2 acts at the center of pressure of the left
planar surface
The distance from the free surface to F2 is
yR
yR can be calculation from the equation yC
plus Ixx,C divided by yC A.
yC is the y-coordinate of the centroid of
the left planar surface, and is equal to d
plus one-half hS.
A is the area of the left planar surface and
is equal to wS hS,
Ixx,C is the second moment of area about the
centroid of the left planar surface.
For rectangles, this is equal to one-twelfth
times the height to the third power times
the width.
If we take the moment about point O, the moment
arms of F2 and FH are equal to the distance
between point O and point B, where point B
is the location where F2 acts.
We will call this distance yH.
The length yH is equal to the distance from
the free surface to point O, which is d plus
hS, minus the distance from the free surface
to point B, which is yR.
F1 acts at the center of the top surface of
the fluid body, so the moment arm is lS divided
by 2.
The moment arm of the weight is equal to the
distance between the centroid of the fluid
body and the left planar surface.
You can look up this distance in tables for
simple shapes.
Notice that the moment of arm FV, which we
will call xV, is the same whether we take
the moment about point O or about point B.
Additionally, since the gate is not moving,
the sum of the moments about any point must
be zero.
To determine xV, we will calculate the sum
of the moments about point B and set it equal
to zero.
By taking the moment about point B, the moment
of F2 and FH are zero.
If we take the counter clockwise direction
to be positive, the sum of the moments is
FV times its moment arm xV, minus F1 times
lS divided by 2, minus W times xC.
We can rearrange the equation to solve for
xV.
