Fluid Statics
Lecture -3
Force on a Plane Area




No tangential force can exist within a fluid at rest. All forces
are normal to the surface.
If pressure is uniformly distributed over an area, the force is
equal to pressure times the area and the point of application of
force is at the centroid of the area.
For submerged horizontal areas, the pressure is uniform.
In case of gases, the pressure variation with vertical distance
is very small due to low specific weight. Therefore when we
compute the static fluid force exerted by a gas, we usually
treat p as a constant. Thus for such cases:
F
 pdA  p  dA  pA
(3.14)
Force on a Plane Area


In the case of liquids the
distribution of pressure is
generally not uniform, so
further analysis is necessary.
Let us consider a vertical
plane whose upper edge lies in
the free surface of a liquid
(Fig). Let this plane be
perpendicular to the plane of
the figure, so that MN is its
edge.
Fig-3.15
Force on a Plane Area


The gage pressure will vary from zero at M to NK at
N. The total force on one side of the plane is the sum
of the products of the elementary areas and the
pressure upon them.
From the pressure distribution, we can see that the
resultant of this system of parallel forces must act at
a point below the centroid of the area, since the
centroid of an area is the point where the resultant of
a system of uniform parallel forces would act.
Force on a Plane Area


If we lower the plane to
position M'N', the change of
pressure from M' to N' is less
than it was from M to N.
Hence the resultant pressure
force will act nearer to the
centroid of the plane surface.
The deeper we submerge the
plane, the smaller the pressure
variation becomes, and the
closer the resultant moves to
the centroid.
Fig-3.15
Force on a Plane Area (Sloping)
Fig-3.16

In Fig. let MN be the edge of a plane area making an angle q
with the horizontal. To the right we see the projection of this
area onto a vertical plane.
Force on a Plane Area (Sloping)



The pressure distribution over the sloping area forms a
pressure prism (MNKJ times width in Fig.), whose volume is
equal to the total force F acting on the area.
If the width x is constant then we can easily compute the
volume of the pressure prism, using a mean pressure = 0.5
(MJ+NK), and so obtain F.
If x varies, we must integrate to find F. Let h be the variable
depth to any point and let y be the corresponding distance
from OX, the intersection of the plane containing the area and
the free surface.
Force on a Plane Area (Sloping)

Choose an element of area so that the pressure over it is
uniform. Such an element is a horizontal strip, of width x, so
dA = xdy. As p = gh and h = y sin q, the force dF on the
horizontal strip is:
dF  pdA  ghdA  gy sinqdA
Integrating



F  dF  g sinq ydA  g sinq y c A
(3.15)
Where yc is, by definition, the distance from OX along the
sloping plane to the centroid C of the area A.
Force on a Plane Area (Sloping)

If hc is the vertical depth to the centroid, then,
hc  yc sinq

And in general we have:
F  g hc A (3.16)

Thus we find the total force on any plane area submerged in a
liquid by multiplying the specific weight of the liquid by the
product of the area and the depth of its centroid. The value of
F is independent of the angle of inclination of the plane so
long as the depth of its centroid is unchanged.
Since ghc is the pressure at the centroid, we can also say that
the total force on any plane area submerged in a liquid is the
product of the area and the pressure at its centroid.

Center of Pressure


The point of application of the resultant pressure force on a
submerged area is called the center of pressure. We need to
know its location whenever we wish to work with the moment
of this force.
The most general way of looking at the problem of forces on a
submerged plane area is through the use of the recently
discussed pressure prism concept (Sec. 3.6 and Fig. 3.16). The
line of action of the resultant pressure force must pass through
the centroid of the pressure prism (volume). As noted earlier,
this concept is very convenient to apply for simple areas such
as rectangles. For example, if the submerged area in Fig. 3.15
is of constant width then we know that the centroid of the
pressure prism on area MN is 2/3 MN below M.
Center of Pressure
Fig-3.16
Center of Pressure

If the shape of the area is not so regular, i.e., if the width x in
Fig. 3.16 varies, then we must take moments and integrate.
Taking OX in Fig. 3.16 as an axis of moments, the moment of
an elementary force dF = gy sin q dA is:
ydF  g y 2 sinq dA

and if yp denotes the distance to the center of pressure, using
the basic concept that the moment of the resultant force equals
the sum of the moments of the component forces,
y p F  g sinq


y 2 dA g sinq I o
Where we recognize that Io is the moment of inertia of the
plane area about axis OX.
Center of Pressure

If we divide this last expression by the value of F given by Eq.
(3.15), we obtain
g sinqI o
Io
yp 

(3.17)
g sinqyc A yc A

The product yc A is the static moment of area A about OX.
Therefore Eq. (3.17) tells us that we can obtain the distance
from the center of pressure to the axis where the plane
(extended) intersects the liquid surface by dividing the
moment of inertia of the area A about the surface axis by its
static moment about the same axis.
Center of Pressure

We may also express this in another form, by noting from the
parallel axis theorem that:
I o  Ayc 2  I c

Where Ic is the moment of Inertia of an area about its
centroidal axis. By substituting for Io into Eq. 3.17.
Ayc 2  I c
yp 
yc A
So,
Ic
y p  yc 
yc A
(3.18)
Center of Pressure

From this equation we see that the location of center of pressure P is
independent of the angel q that is, we can rotate the plane area about axis
OX without affecting the location of P. Also, we see that P is always below
the centroid C and that as the depth of immersion is increased, yc increases
and therefore P approaches C.

We can find the lateral position of the centre of pressure P by considering
that the area is made up of a series of elemental horizontal strips. The
centre of pressure for each strip is at the midpoint of the strip. Since the
moment of the resultant force F must be equal to the moment of the
distributed force system about an axis, say, the y-axis.

X P F  x p pdA

(3.19)
Where Xp is the lateral distance from the selected y axis to the center of
pressure P of the resultant force F, and xp is the lateral distance to the
center of any elemental horizontal strip of area dA on which the pressure is
p.
Problem
The cubic tank shown in Fig. is half full of water. Find (a)
the pressure on the bottom of the tank, (b) the force
exerted by the fluids on a tank wall, and (c) the location of
the center of pressure on a wall.
Solution:
(a)
p bott  pair  g water hwater  8  9.81x1
 17.81kN/m 2  17.81kPa
(b) The forceaction on the tank end is divided into two com ponents,
labeled A and B on the pressure distributi on sketch. Com ponet A
has a uniform pressuredistributi on, due to the pressureof the
confined air, which acts throughout the water :
FA  pair Aair  8 x 4  32kN
For com ponentB i.e., the vary ingwater pressuredistributi on
on the lower half of the tank wall, the centroid C of the area
of application is at
h c  yc  0.5 x1  0.5m below the water topsurface,
FB  g water hc Awater  9.81x0.5 x 2  9.81kN
Total Force  FA  FB  32  9.81  41.8kN
(c) The location of centersof pressure of the com ponentforces,
as distance y p below the water topsurface,are
(yp ) A  0m
below the water topsurface,to the centroid of the 2m square
area for the uniform air pressure.
2
2
(yp ) B  hwater  (1)  0.667 m
3
3
below the water topsurfacefor the vary ingpressure on the
rectangular wetted wall area.
We can also find by using this equation :
Ic
y p  yc 
yc A
y c  0.5m,
bh3
Ic 
 2 x13 / 12  0.1667 m 4 A  bh  2m 2
12
y p  0.667 m
Taking m om ents:
F(y p )  FA (yp ) A  FB (yp ) B
y p  0.1565m below the water topsurface.
Problem
Water and oil in an open storage tank are in contact with
the end wall as shown in Fig. S3.7. (a) Find the pressure at
the bottom (lowest point) of the tank caused by the liquids.
Also find (b) the total force exerted on the end wall by the
liquids, and (c) the depth of its center of pressure.
Solution:
(a) P bott = g oil h oil + g water h water
= (0.8x62.4Ib/ft3) (1.5ft) + (62.4Ib/ft3) (1.0 ft)
= 137.3 lb/ft3 = 0.953psi
(b) The force acting on the end consists of three components, labeled
A, B and D, on the pressure distribution sketch. Note that
component B has a uniform pressure distribution, due to the oil
(A) above, which acts throughout the liquid below.
As a preliminary, we note for the semicircular end area (r = 1 ft)
that
(i) A = pr2/2 = p12/2 = 1.571 ft2;
(ii) from Appendix A, Table A.7, the centroid is 4r/3p= 0.424 ft
from the center of the circle, i.e., below the water top surface.
Solution:
For component A, i.e., the varying oil pressure distribution on the l.5-ft
height of the end wall, the centroid C of the area of application is at:
hc= yc = 0.5(1.5 ft) = 0.75 ft below the free oil surface,
so, from Eq. (3.16),
FA = goil hc Aoil = (0.8 x 62.4)0.75(1.5 x 2) = 112.3Ib
For component B, the force FB on the water-wetted area of the end wall
due to the uniform pressure produced by the 1.5-ft depth of oil above is
FB = pA = g h A= (0.8 x 62.4 )1.5(p12/2) = 117.6 lb
For component D, i.e., the varying pressure distribution due to the water
(only) on the water-wetted area of the end wall, the centroid C is at
hc = yc = 0.424 ft below the water top surface,
so:
FD= g hc A = 62.4(0.424) p12/2 = 41.61b
Solution:
The total force F on the end of the tank is therefore
F = FA + FB + FD = 272 lb
(c) As a preliminary to locating the center of pressure, we note
that for the semicircular end area with D = 2 ft,
(i) from Table A.7: I about the center of the circle, is I =
pD4/128 = p24/128 = 0.393 ft4, and
(ii) by the parallel axis theorem: Ic about the centroid, distance
0.424 ft below the center of the circle, is Ic = I + Ad2 = 0.393
+ (p12/2) (0.424) 2 = 0.1098 ft4.
Solution:
The locations of the centers of pressure, below the free oil surface,
of the component forces are:
(yp)A = 2/3(1.5 ft) = 1.00 ft for the varying oil pressure on the oilwetted area, and
(yp)B = yc = 1.5 + 0.424 = 1.924 ft to the centroid of the waterwetted semicircular area, for the uniform pressure on this area due
to 1.5 ft of oil above the water; and
Eq. (3.18):
( y p ) D  yc 
Ic
0.1098
 0.424
 0.589 ft
2
yc A
0.424(p 1 / 2)
below the water top surface.
Solution:
for the varying water pressure on the water-wetted semicircular
area,
=1.5 + 0.589 = 2.09 ft below the free oil surface
Finally,
F(yp) =FA(yp)A +FB (yp)B + FD(yp)D
yp = 1.567 ft
Exercise Problems:
1.
2.
3.
A circular area of diameter d is vertical and submerged in a
liquid. Its upper edge is coincident with the liquid surface.
Derive an expression for the depth to its center of pressure.
If a triangle of height d and base b is vertical and submerged
in a liquid with its base at the liquid surface, derive an
expression for the depth to its center of pressure.
If a triangle of height d and base b is vertical and submerged
in liquid with its vertex at the liquid surface, derive an
expression for the depth to its center of pressure.
Force on a Curved Surface

On any curved surface such as MN in Fig (a), the force on the
various elementary areas that make up the curved surface are
different in direction and magnitude, so algebraic summation is
impossible.

Hence we can apply Eq 3.16 (in previous section) only to plane
area. But for non-planer areas, we can find the component forces
in certain directions.
a. Horizontal Force on a Curved Surface:


We may project (extend) any irregular curved area MN (Fig a)
onto a vertical plane whose trace is M’N’ (Fig b).
The projecting elements (M’MNN’), which are all horizontal,
enclose a volume whose ends are the vertical plane M’N’ and
the irregular area MN. This volume of liquid is in static
equilibrium.
a. Horizontal Force on a Curved Surface:



A force F’ acts on the projected vertical area M’N’. The
horizontal force component Fx’ acts on the irregular end area
MN and is equal and opposite to the Fx of Fig (a).
Gravity force W’ is vertical, and the lateral forces on all the
horizontal projection elements are normal to these elements
and hence normal to F’.
Thus the only horizontal forces on MNM’N” are F’ and Fx’,
and therefore,
F '  Fx'  0
and
Fx  Fx'  F '
a. Horizontal Force on a Curved Surface:



Hence the horizontal force in any given direction on any area
is equal to the force on the projection of that area onto a
vertical plane normal to the given direction.
The line of action of Fx must be the same as that of F’.
Above equation is applicable to liquid and gases.
b. Vertical Force on a Curved Surface:

We can find the vertical force Fz on a curved area, such as
MN in Fig a, by considering the volume of liquid enclosed by
the area and vertical elements extending to the free surface
(MNKJ) (Fig c). This volume of liquid is in static equilibrium.
K
J
b. Vertical Force on a Curved Surface:


Vertical forces on this volume of liquid are;
a. The force FG  pG A due to any gas (at pressure pG ) above
the liquid.
b. The gravity force W downward.
c. Fz’ the upward vertical force on the irregular area MN.
The force Fz’ (Fig-c) is equal and opposite to the force Fz
(Fig-a). Any other forces on the vertical elements are normal
to the elements, and so are horizontal. Therefore
Fz'  W  FG  0
Fz  Fz'  W  FG
b. Vertical Force on a Curved Surface:



Therefore the vertical force acting on any area is equal to the
weight of the volume of liquid above it, plus any
superimposed gas pressure force.
The line of action of Fz is the resultant of FG and W.
FG must pass through the centroid of the plan (surface
projection) area, and W must pass through the center of
gravity of the liquid volume.
c. Resultant Force on a Curved Surface:


In general, there is no single resultant force on an irregular
area, because the horizontal and vertical forces, as found in
the above discussion, may not be in the same plane.
But in certain cases these two forces will lie in the same plane
and then we can combine them into a single force.
Problem
Find the horizontal and vertical components of the force exerted by
the fluids on the horizontal cylinder in Fig. if :
a) The fluid to the left of cylinder is a gas confined in a closed tank at a
pressure of 35 kPa.
b) The fluid to the left of cylinder is water with a free surface at an
elevation coincident with the uppermost part of the cylinder.
Assume in both cases that the atmospheric pressure occurs to the right
of the cylinder.
Solution:
The net projectionon a verticalplaneof the portionof the cylindrical
surface underconsideration (see left diagram)is, from the right handed
diagram,ef  2  2cos30o  3.73m
(a) For thegas :
Fx  pA2  35kN / m 2 (3.73m)  130.5kN / m to the right
The verticalforce of the gas on thesurface ' ac' is equaland oppositeto that
on thesurface ' cd'.Hencethe net projectionon a horizontalplanefor the gas
is af  2sin30o  1m. Thus
Fz  pAx  35kN / m 2 (1m)  35kN / m upward
(b) For thefluid :
1

Fx  ghc A  9.81x x3.73 x3.73
2

 68.3kN/mto the right
Net Fz  upwardforce on surface cde - downwardforce on surface ca
 weightof volumeabcdefa - weightof volumeabca
 weightof cross- hatchedvolumeof liquid(1  2  3)
3  210


1

 9.81kN/m 
p 2  2 sin30o x 2 cos30o  (1x 2) m 2
2
 360

 100.0kN / m upward
2
3
1
2

210o  180o  30o