Chapter 2

Pressure Distribution
in a Fluid

Motivation. Many fluid problems do not involve motion. They concern the pressure
distribution in a static fluid and its effect on solid surfaces and on floating and sub-
merged bodies.
When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure
variation is due only to the weight of the fluid. Assuming a known fluid in a given
gravity field, the pressure may easily be calculated by integration. Important applica-
tions in this chapter are (1) pressure distribution in the atmosphere and the oceans,
(2) the design of manometer, mechanical, and electronic pressure instruments, (3) forces
on submerged flat and curved surfaces, (4) buoyancy on a submerged body, and (5) the
behavior of floating bodies. The last two result in Archimedes’ principles.
If the fluid is moving in rigid-body motion, such as a tank of liquid that has been
spinning for a long time, the pressure also can be easily calculated because the fluid
is free of shear stress. We apply this idea here to simple rigid-body accelerations in
Sec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter of
fact, pressure also can be analyzed in arbitrary (nonrigid-body) motions V(x, y, z, t),
but we defer that subject to Chap. 4.

2.1 Pressure and Pressure In Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohr’s
Gradient circle reduces to a point. In other words, the normal stress on any plane through a
fluid element at rest is a point property called the fluid pressure p, taken positive for
compression by common convention. This is such an important concept that we shall
review it with another approach.
Figure 2.1 shows a small wedge of fluid at rest of size x by z by s and depth
b into the paper. There is no shear by definition, but we postulate that the pressures
px, pz, and pn may be different on each face. The weight of the element also may
be important. The element is assumed small, so the pressure is constant on each

65
66 Chapter 2 Pressure Distribution in a Fluid

z (up)

pn
θ
Δs
Element weight:
Δz dW = ρ g( 12 b Δx Δz)
px
Δx θ
x
O
Fig. 2.1 Equilibrium of a small Width b into paper
wedge of fluid at rest. pz

face. Summation of forces must equal zero (no acceleration) in both the x and z
directions.
g Fx  0  px b z  pnb s sin ␪
(2.1)
g Fz  0  pzb x  pnb s cos ␪  12␳gb x z
But the geometry of the wedge is such that
s sin ␪  z s cos ␪  x (2.2)
Substitution into Eq. (2.1) and rearrangement give
px  pn pz  pn  12␳g z (2.3)
These relations illustrate two important principles of the hydrostatic, or shear-free,
condition: (1) There is no pressure change in the horizontal direction, and (2) there
is a vertical change in pressure proportional to the density, gravity, and depth change.
We shall exploit these results to the fullest, starting in Sec. 2.3.
In the limit as the fluid wedge shrinks to a “point,’’ z n 0 and Eqs. (2.3)
become
px  pz  pn  p (2.4)
Since ␪ is arbitrary, we conclude that the pressure p in a static fluid is a point prop-
erty, independent of orientation.

Pressure Force on a Fluid Pressure (or any other stress, for that matter) causes a net force on a fluid element
Element when it varies spatially.1 To see this, consider the pressure acting on the two x faces
in Fig. 2.2. Let the pressure vary arbitrarily
p  p(x, y, z, t)
The net force in the x direction on the element in Fig. 2.2 is given by
p p
dFx  p dy dz  ap  dxb dy dz   dx dy dz
x x

1
An interesting application for a large element is seen in Fig. 3.7.
 2.2 Equilibrium of a Fluid Element 67

dz

∂p
p dy dz (p+ dx) dy dz
∂x
dy
x

Fig. 2.2 Net x force on an element dx

due to pressure variation. z

In like manner the net force dFy involves p/y, and the net force dFz concerns
p/z. The total net-force vector on the element due to pressure is
p p p
d Fpress  ai  j  k b dx dy dz (2.5)
x y z
We recognize the term in parentheses as the negative vector gradient of p. Denoting f
as the net force per unit element volume, we rewrite Eq. (2.5) as
fpress  p (2.6)
  
where   gradient operator  i j k
x y z
Thus it is not the pressure but the pressure gradient causing a net force that must be
balanced by gravity or acceleration or some other effect in the fluid.

2.2 Equilibrium of a Fluid The pressure gradient is a surface force that acts on the sides of the element. There
Element may also be a body force, due to electromagnetic or gravitational potentials, acting
on the entire mass of the element. Here we consider only the gravity force, or weight
of the element:
dFgrav  ␳g dx dy dz
(2.7)
or fgrav  ␳g
In addition to gravity, a fluid in motion will have surface forces due to viscous
stresses. By Newton’s law, Eq. (1.2), the sum of these per-unit-volume forces equals
the mass per unit volume (density) times the acceleration a of the fluid element:

a f  fpress  fgrav  fvisc  p  ␳g  fvisc  ␳a (2.8)

This general equation will be studied in detail in Chap. 4. Note that Eq. (2.8) is a vector
relation, and the acceleration may not be in the same vector direction as the velocity.
For our present topic, hydrostatics, the viscous stresses and the acceleration are zero.
68 Chapter 2 Pressure Distribution in a Fluid

p (Pascals)
High pressure:
120,000
p = 120,000 Pa abs = 30,000 Pa gage
30,000
Local atmosphere:
90,000 p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum

40,000

Vacuum pressure:
50,000
p = 50,000 Pa abs = 40,000 Pa vacuum

50,000

Fig. 2.3 Illustration of absolute, Absolute zero reference:

0
gage, and vacuum pressure p = 0 Pa abs = 90,000 Pa vacuum
(Tension)
readings.

Gage Pressure and Vacuum Before embarking on examples, we should note that engineers are apt to specify pres-
Pressure: Relative Terms sures as (1) the absolute or total magnitude or (2) the value relative to the local ambi-
ent atmosphere. The second case occurs because many pressure instruments are of
differential type and record, not an absolute magnitude, but the difference between
the fluid pressure and the atmosphere. The measured pressure may be either higher
or lower than the local atmosphere, and each case is given a name:
1. p  pa Gage pressure: p(gage)  p  pa
2. p pa Vacuum pressure: p(vacuum)  pa  p
This is a convenient shorthand, and one later adds (or subtracts) atmospheric pressure
to determine the absolute fluid pressure.
A typical situation is shown in Fig. 2.3. The local atmosphere is at, say, 90,000 Pa,
which might reflect a storm condition in a sea-level location or normal conditions at
an altitude of 1000 m. Thus, on this day, pa  90,000 Pa absolute  0 Pa gage  0 Pa
vacuum. Suppose gage 1 in a laboratory reads p1  120,000 Pa absolute. This value
may be reported as a gage pressure, p1  120,000  90,000  30,000 Pa gage. (One
must also record the atmospheric pressure in the laboratory, since pa changes grad-
ually.) Suppose gage 2 reads p2  50,000 Pa absolute. Locally, this is a vacuum
pressure and might be reported as p2  90,000  50,000  40,000 Pa vacuum.
Occasionally, in the problems section, we will specify gage or vacuum pressure to
keep you alert to this common engineering practice. If a pressure is listed without
the modifier gage or vacuum, we assume it is absolute pressure.

2.3 Hydrostatic Pressure If the fluid is at rest or at constant velocity, a  0 and fvisc  0. Equation (2.8) for
Distributions the pressure distribution reduces to
p  ␳g (2.9)
This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their
viscosity, because the viscous term vanishes identically.
 2.3 Hydrostatic Pressure Distributions 69

Recall from vector analysis that the vector p expresses the magnitude and direc-
tion of the maximum spatial rate of increase of the scalar property p. As a result, p
is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.9) states that a
fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere
normal to the local-gravity vector. The maximum pressure increase will be in the
direction of gravity—that is, “down.” If the fluid is a liquid, its free surface, being at
atmospheric pressure, will be normal to local gravity, or “horizontal.” You probably
knew all this before, but Eq. (2.9) is the proof of it.
In our customary coordinate system z is “up.” Thus the local-gravity vector for
small-scale problems is
g  gk (2.10)
2
where g is the magnitude of local gravity, for example, 9.807 m/s . For these coor-
dinates Eq. (2.9) has the components
p p p
0 0  ␳g  ␥ (2.11)
x y z
the first two of which tell us that p is independent of x and y. Hence p/z can be
replaced by the total derivative dp/dz, and the hydrostatic condition reduces to
dp
 ␥
dz
2
or p2  p1   冮 ␥ dz
1
(2.12)

Equation (2.12) is the solution to the hydrostatic problem. The integration requires an
assumption about the density and gravity distribution. Gases and liquids are usually
treated differently.
We state the following conclusions about a hydrostatic condition:
Pressure in a continuously distributed uniform static fluid varies only with vertical
distance and is independent of the shape of the container. The pressure is the same
at all points on a given horizontal plane in the fluid. The pressure increases with
depth in the fluid.
An illustration of this is shown in Fig. 2.4. The free surface of the container is atmo-
spheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a hori-
zontal plane and are interconnected by the same fluid, water; therefore all points have
the same pressure. The same is true of points A, B, and C on the bottom, which all
have the same higher pressure than at a, b, c, and d. However, point D, although at
the same depth as A, B, and C, has a different pressure because it lies beneath a dif-
ferent fluid, mercury.

Effect of Variable Gravity For a spherical planet of uniform density, the acceleration of gravity varies inversely
as the square of the radius from its center

g  g0 a 0 b
r 2
(2.13)
r
70 Chapter 2 Pressure Distribution in a Fluid

Atmospheric pressure:

Free surface
Fig. 2.4 Hydrostatic-pressure distri-
bution. Points a, b, c, and d are at Water
equal depths in water and therefore
have identical pressures. Points A,
a b c d
B, and C are also at equal depths Depth 1
in water and have identical pres-
sures higher than a, b, c, and d. Mercury
Point D has a different pressure
from A, B, and C because it is not A B C D
connected to them by a water path. Depth 2

where r0 is the planet radius and g0 is the surface value of g. For earth, r0 ⬇ 3960
statute mi ⬇ 6400 km. In typical engineering problems the deviation from r0 extends
from the deepest ocean, about 11 km, to the atmospheric height of supersonic trans-
port operation, about 20 km. This gives a maximum variation in g of (6400/6420)2,
or 0.6 percent. We therefore neglect the variation of g in most problems.

Hydrostatic Pressure in Liquids Liquids are so nearly incompressible that we can neglect their density variation in
hydrostatics. In Example 1.6 we saw that water density increases only 4.6 percent at
the deepest part of the ocean. Its effect on hydrostatics would be about half of this,
or 2.3 percent. Thus we assume constant density in liquid hydrostatic calculations, for
which Eq. (2.12) integrates to

Liquids: p2  p1  ␥ (z2  z1) (2.14)

p2 p1
or z1  z2  
␥ ␥

We use the first form in most problems. The quantity ␥ is called the specifi c weight
of the fluid, with dimensions of weight per unit volume; some values are tabulated in
Table 2.1. The quantity p/␥ is a length called the pressure head of the fluid.

Table 2.1 Specific Weight of Some

c weight ␥
Specifi
Common Fluids at 68°F ⴝ 20°C
Fluid lbf/ft3 N/m3
Air (at 1 atm) 0.0752 11.8
Ethyl alcohol 49.2 7,733
SAE 30 oil 55.5 8,720
Water 62.4 9,790
Seawater 64.0 10,050
Glycerin 78.7 12,360
Carbon tetrachloride 99.1 15,570
Mercury 846 133,100
 2.3 Hydrostatic Pressure Distributions 71

+b p ≈ pa –b γair
Air

Free surface: Z = 0, p = pa
0

Water
g
Fig. 2.5 Hydrostatic-pressure distri- –h p ≈ pa + hγwater
bution in oceans and atmospheres.

For lakes and oceans, the coordinate system is usually chosen as in Fig. 2.5, with
z  0 at the free surface, where p equals the surface atmospheric pressure pa. When
we introduce the reference value (p1, z1)  ( pa, 0), Eq. (2.14) becomes, for p at any
(negative) depth z,
Lakes and oceans: p  pa  ␥z (2.15)
where ␥ is the average specific weight of the lake or ocean. As we shall see, Eq. (2.15)
holds in the atmosphere also with an accuracy of 2 percent for heights z up to 1000 m.

EXAMPLE 2.1
Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of
60 m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa
at this maximum depth.

Solution
• System sketch: Imagine that Fig. 2.5 is Newfound Lake, with h  60 m and z  0 at
the surface.
• Property values: From Table 2.1, ␥water  9790 N/m3. We are given that patmos  91 kPa.
• Solution steps: Apply Eq. (2.15) to the deepest point. Use SI units, pascals, not kilopascals:

N
pmax  pa  ␥ z  91,000 Pa  (9790 )(60 m)  678,400 Pa ⬇ 678 kPa Ans.
m3
• Comments: Kilopascals are awkward. Use pascals in the formula, then convert the answer.

The Mercury Barometer The simplest practical application of the hydrostatic formula (2.14) is the barometer
(Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and
inverted while submerged in a reservoir. This causes a near vacuum in the closed upper
end because mercury has an extremely small vapor pressure at room temperatures
72 Chapter 2 Pressure Distribution in a Fluid

p1 ≈ 0
(Mercury has a very
low vapor pressure.)

z1 = h

p2 ≈ pa
( The mercury is in
contact with the
atmosphere.) p
h= γa
M
z

pa
z2 = 0
pM

Mercury

(a) (b)
Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury
column is proportional to patm; (b) a modern portable barometer, with digital readout, uses the
resonating silicon element of Fig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.)

(0.16 Pa at 20
C). Since atmospheric pressure forces a mercury column to rise a dis-
tance h into the tube, the upper mercury surface is at zero pressure.
From Fig. 2.6, Eq. (2.20) applies with p1  0 at z1  h and p2  pa at z2  0:
pa  0  ␥M (0  h)
pa
or h (2.16)
␥M

At sea-level standard, with pa  101,350 Pa and ␥M  133,100 N/m3 from Table 2.1,
the barometric height is h  101,350/133,100  0.761 m or 761 mm. In the United
States the weather service reports this as an atmospheric “pressure” of 29.96 inHg
(inches of mercury). Mercury is used because it is the heaviest common liquid. A
water barometer would be 34 ft high.

Hydrostatic Pressure in Gases Gases are compressible, with density nearly proportional to pressure. Thus density must
be considered as a variable in Eq. (2.12) if the integration carries over large pressure
changes. It is sufficiently accurate to introduce the perfect-gas law p  ␳RT in Eq. (2.12):
dp p
 ␳g   g
dz RT
 2.3 Hydrostatic Pressure Distributions 73

Separate the variables and integrate between points 1 and 2:

2 2

冮 冮
dp p g dz
 ln 2   (2.17)
1
p p1 R 1
T
The integral over z requires an assumption about the temperature variation T(z). One
common approximation is the isothermal atmosphere, where T  T0:
g(z2  z1)
p2  p1 exp c  d (2.18)
RT0
The quantity in brackets is dimensionless. (Think that over; it must be dimensionless,
right?) Equation (2.18) is a fair approximation for earth, but actually the earth’s mean
atmospheric temperature drops off nearly linearly with z up to an altitude of about
36,000 ft (11,000 m):
T ⬇ T0  Bz (2.19)
Here T0 is sea-level temperature (absolute) and B is the lapse rate, both of which vary
somewhat from day to day.

The Standard Atmosphere By international agreement [1] the following standard values are assumed to apply
from 0 to 36,000 ft:
T0  518.69
R  288.16 K  15
C
B  0.003566
R/ft  0.00650 K/m
This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.19)
into (2.17) and integrating, we obtain the more accurate relation

Bz g/(RB)
p  pa a1  b
g
where  5.26 (air)
T0 RB
(2.20)
g
1
␳  ␳o a1  b RB
Bz kg
where ␳o  1.2255 3 , po  101,350 Pa
To m

in the troposphere, with z  0 at sea level. The exponent g/(RB) is dimensionless

(again it must be) and has the standard value of 5.26 for air, with R  287 m2/(s2  K).
The U.S. standard atmosphere [1] is sketched in Fig. 2.7. The pressure is seen to
be nearly zero at z  30 km. For tabulated properties see Table A.6.

EXAMPLE 2.2
If sea-level pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m,
using (a) the exact formula and (b) an isothermal assumption at a standard sea-level tem-
perature of 15
C. Is the isothermal approximation adequate?
74 Chapter 2 Pressure Distribution in a Fluid

Solution
Part (a) Use absolute temperature in the exact formula, Eq. (2.20):
(0.00650 K/m)(5000 m) 5.26
p  pa c 1  d  (101,350 Pa)(0.8872)5.26
288.16 K
 101,350(0.5328)  54,000 Pa Ans. (a)
This is the standard-pressure result given at z  5000 m in Table A.6.

Part (b) If the atmosphere were isothermal at 288.16 K, Eq. (2.18) would apply:

p ⬇ pa exp a b  (101,350 Pa) exp e  f

gz (9.807 m/s2)(5000 m)
RT 3 287 m2/(s2 # K)4 (288.16 K)
 (101,350 Pa) exp (0.5929) ⬇ 56,000 Pa Ans. (b)

This is 4 percent higher than the exact result. The isothermal formula is inaccurate in the
troposphere.

Is the Linear Formula Adequate The linear approximation from Eq. (2.14), ␦p ⬇ ␳g ␦z, is satisfactory for liquids,
for Gases? which are nearly incompressible. For gases, it is inaccurate unless ␦z is rather small.
Problem P2.26 asks you to show, by binomial expansion of Eq. (2.20), that the error
in using constant gas density to estimate ␦p from Eq. (2.14) is small if
2T0
␦z  (2.21)
(n  1)B

60 60

50 50

40 40
1.20 kPa
Altitude z, km

Altitude z, km

30 30

20.1 km Eq. (2.24)

20 20
–56.5°C

Eq. (2.27)
10 11.0 km
Eq. (2.26) 10

Troposphere
15°C 101.33 kPa
Fig. 2.7 Temperature and pressure
distribution in the U.S. standard 0 – 60 – 40 – 20 0 +20 0 40 80 120
atmosphere. (From Ref. 1.) Temperature, °C Pressure, kPa
 2.4 Application to Manometry 75

where To is the local absolute temperature, B is the lapse rate from Eq. (2.19), and
n  g/(RB) is the exponent in Eq. (2.20). The error is less than 1 percent if ␦z 200 m.

2.4 Application to Manometry From the hydrostatic formula (2.14), a change in elevation z2  z1 of a liquid is equiv-
alent to a change in pressure ( p2  p1)/␥. Thus a static column of one or more liquids
or gases can be used to measure pressure differences between two points. Such a device
is called a manometer. If multiple fluids are used, we must change the density in the
formula as we move from one fluid to another. Figure 2.8 illustrates the use of the for-
mula with a column of multiple fluids. The pressure change through each fluid is cal-
culated separately. If we wish to know the total change p5  p1, we add the succes-
sive changes p2  p1, p3  p2, p4  p3, and p5  p4. The intermediate values of p
cancel, and we have, for the example of Fig. 2.8,
p5  p1  ␥0 (z2  z1)  ␥w (z3  z2)  ␥G (z4  z3)  ␥M (z5  z4) (2.22)
No additional simplification is possible on the right-hand side because of the different
densities. Notice that we have placed the fluids in order from the lightest
on top to the heaviest at bottom. This is the only stable configuration. If we attempt to
layer them in any other manner, the fluids will overturn and seek the stable arrangement.

Pressure Increases Downward The basic hydrostatic relation, Eq. (2.14), is mathematically correct but vexing to engi-
neers because it combines two negative signs to have the pressure increase downward.
When calculating hydrostatic pressure changes, engineers work instinctively by simply
having the pressure increase downward and decrease upward. If point 2 is a distance h
below point 1 in a uniform liquid, then p2  p1  ␳gh. In the meantime, Eq. (2.14)
remains accurate and safe if used properly. For example, Eq. (2.22) is correct as shown,
or it could be rewritten in the following “multiple downward increase” mode:
p5  p1  ␥0 0 z1  z2 0  ␥w 0 z2  z3 0  ␥G 0 z3  z4 0  ␥M 0 z4  z5 0
That is, keep adding on pressure increments as you move down through the layered
fluid. A different application is a manometer, which involves both “up” and “down”
calculations.

Known pressure p1
z = z1

Oil, ρo
z2 p2 – p1 = – ρog(z 2 – z1)

Water, ρw
z z3 p3 – p2 = – ρw g(z 3 – z 2)

Glycerin, ρG
z4 p4 – p3 = – ρGg(z 4 – z 3)

Mercury, ρM
Fig. 2.8 Evaluating pressure z5 p5 – p4 = – ρMg(z 5 – z 4)
changes through a column of Sum = p5 – p1
multiple fluids.
76 Chapter 2 Pressure Distribution in a Fluid

Open, pa

z 2 , p2 ≈ pa
ρ1
zA, pA A

Jump across
z1, p1 p = p1 at z = z1 in fluid 2
Fig. 2.9 Simple open manometer
for measuring pA relative to atmo-
spheric pressure. ρ2

Application: A Simple Manometer Figure 2.9 shows a simple U-tube open manometer that measures the gage pressure
pA relative to the atmosphere, pa. The chamber fluid ␳1 is separated from the atmo-
sphere by a second, heavier fluid ␳2, perhaps because fluid A is corrosive, or more
likely because a heavier fluid ␳2 will keep z2 small and the open tube can be shorter.
We first apply the hydrostatic formula (2.14) from A down to z1. Note that we can
then go down to the bottom of the U-tube and back up on the right side to z1, and the
pressure will be the same, p  p1. Thus we can “jump across” and then up to level z2:

pA  ␥1 0 zA  z1 0  ␥2 0 z1  z2 0  p2 ⬇ patm (2.23)
Another physical reason that we can “jump across” at section 1 is that a continuous
length of the same fluid connects these two equal elevations. The hydrostatic relation
(2.14) requires this equality as a form of Pascal’s law:
Any two points at the same elevation in a continuous mass of the same static fluid
will be at the same pressure.
This idea of jumping across to equal pressures facilitates multiple-fluid problems. It
will be inaccurate however if there are bubbles in the fluid.

EXAMPLE 2.3
The classic use of a manometer is when two U-tube legs are of equal length, as in Fig. E2.3,
and the measurement involves a pressure difference across two horizontal points. The typical
application is to measure pressure change across a flow device, as shown. Derive a formula
for the pressure difference pa  pb in terms of the system parameters in Fig. E2.3.

Flow device

(a) (b)

␳1 h

␳2
E2.3
 2.4 Application to Manometry 77

Solution
Using Eq. (2.14), start at (a), evaluate pressure changes around the U-tube, and end up
at (b):

pa  ␳1gL  ␳1gh  ␳2gh  ␳1gL  pb

or pa  pb  (␳2  ␳1)gh Ans.

The measurement only includes h, the manometer reading. Terms involving L drop out.
Note the appearance of the difference in densities between manometer fluid and working
fluid. It is a common student error to fail to subtract out the working fluid density ␳1—a
serious error if both fluids are liquids and less disastrous numerically if fluid 1 is a gas.
Academically, of course, such an error is always considered serious by fluid mechanics
instructors.

Although Example 2.3, because of its popularity in engineering experiments, is

sometimes considered to be the “manometer formula,” it is best not to memorize it
but rather to adapt Eq. (2.14) to each new multiple-fluid hydrostatics problem. For
example, Fig. 2.10 illustrates a multiple-fluid manometer problem for finding the dif-
ference in pressure between two chambers A and B. We repeatedly apply Eq. (2.14),
jumping across at equal pressures when we come to a continuous mass of the same
fluid. Thus, in Fig. 2.10, we compute four pressure differences while making three
jumps:

pA  pB  ( pA  p1)  ( p1  p2)  ( p2  p3)  ( p3  pB)

(2.24)
 ␥1(zA  z1)  ␥2(z1  z2)  ␥3(z2  z3)  ␥4(z3  zB)

The intermediate pressures p1,2,3 cancel. It looks complicated, but really it is merely
sequential. One starts at A, goes down to 1, jumps across, goes up to 2, jumps across,
goes down to 3, jumps across, and finally goes up to B.

ρ3

Jump across
z 2, p2 z 2, p2

ρ1
zA, pA A

B zB, pB
Jump across
z1, p1 z1, p1
Fig. 2.10 A complicated multiple- Jump across
fluid manometer to relate pA to pB. z 3, p3 z 3, p3
This system is not especially prac- ρ2
tical but makes a good homework
ρ4
or examination problem.
78 Chapter 2 Pressure Distribution in a Fluid

EXAMPLE 2.4
Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B
is 87 kPa, estimate the pressure at A in kPa. Assume all fluids are at 20
C. See Fig. E2.4.

SAE 30 oil Gage B

Mercury 6 cm

A
5 cm
Water
flow 11 cm
4 cm

E2.4

Solution
• System sketch: The system is shown in Fig. E2.4.
• Assumptions: Hydrostatic fluids, no mixing, vertical “up” in Fig. E2.4.
• Approach: Sequential use of Eq. (2.14) to go from A to B.
• Property values: From Table 2.1 or Table A.3:

␥water  9790 N/m3; ␥mercury  133,100 N/m3; ␥oil  8720 N/m3

• Solution steps: Proceed from A to B, “down” then “up,” jumping across at the left mer-
cury meniscus:

pA  ␥w 0 z 0 w  ␥m 0 zm 0  ␥o 0 z 0 o  pB
or pA  (9790 N/m )(0.05 m)  (133,100 N/m3)(0.07 m)  (8720 N/m3)(0.06 m)  87,000
3

or pA  490  9317  523  87,000 Solve for pA  96,350 N/m2 ⬇ 96.4 kPa Ans.

• Comments: Note that we abbreviated the units N/m2 to pascals, or Pa. The intermediate
five-figure result, pA  96,350 Pa, is unrealistic, since the data are known to only about
three significant figures.

In making these manometer calculations we have neglected the capillary height

changes due to surface tension, which were discussed in Example 1.8. These effects
cancel if there is a fluid interface, or meniscus, between similar fluids on both sides of
the U-tube. Otherwise, as in the right-hand U-tube of Fig. 2.10, a capillary correction
can be made or the effect can be made negligible by using large-bore ( 1 cm) tubes.

2.5 Hydrostatic Forces on The design of containment structures requires computation of the hydrostatic forces
Plane Surfaces on various solid surfaces adjacent to the fluid. These forces relate to the weight of
fluid bearing on the surface. For example, a container with a flat, horizontal bottom
 2.5 Hydrostatic Forces on Plane Surfaces 79

Free surface p = pa

h (x, y)
hCG
Resultant
force:
F = pCG A

ξ= h
sin θ

Side view y
CG
x dA = dx dy
Fig. 2.11 Hydrostatic force and
center of pressure on an arbitrary CP
plane surface of area A inclined at
Plan view of arbitrary plane surface
an angle ␪ below the free surface.

of area Ab and water depth H will experience a downward bottom force Fb  ␥HAb.
If the surface is not horizontal, additional computations are needed to find the hori-
zontal components of the hydrostatic force.
If we neglect density changes in the fluid, Eq. (2.14) applies and the pressure on
any submerged surface varies linearly with depth. For a plane surface, the linear
stress distribution is exactly analogous to combined bending and compression of a
beam in strength-of-materials theory. The hydrostatic problem thus reduces to sim-
ple formulas involving the centroid and moments of inertia of the plate cross-
sectional area.
Figure 2.11 shows a plane panel of arbitrary shape completely submerged in a liq-
uid. The panel plane makes an arbitrary angle ␪ with the horizontal free surface, so
that the depth varies over the panel surface. If h is the depth to any element area dA
of the plate, from Eq. (2.14) the pressure there is p  pa  ␥h.
To derive formulas involving the plate shape, establish an xy coordinate system in
the plane of the plate with the origin at its centroid, plus a dummy coordinate ␰ down
from the surface in the plane of the plate. Then the total hydrostatic force on one side
of the plate is given by

F 冮 p dA  冮 ( p a  ␥h) dA  pa A  ␥ h dA 冮 (2.25)

The remaining integral is evaluated by noticing from Fig. 2.11 that h  ␰ sin ␪
and, by definition, the centroidal slant distance from the surface to the plate is

冮 ␰ dA
1
␰CG 
A
80 Chapter 2 Pressure Distribution in a Fluid

Therefore, since ␪ is constant along the plate, Eq. (2.25) becomes

冮
F  pa A  ␥ sin ␪ ␰ dA  pa A  ␥ sin ␪ ␰CG A

Finally, unravel this by noticing that ␰CG sin ␪  hCG, the depth straight down from
the surface to the plate centroid. Thus

F  paA  ␥hCGA  ( pa  ␥hCG)A  pCGA (2.26)

The force on one side of any plane submerged surface in a uniform fluid equals the
pressure at the plate centroid times the plate area, independent of the shape of the
plate or the angle ␪ at which it is slanted.
Equation (2.26) can be visualized physically in Fig. 2.12 as the resultant of a lin-
ear stress distribution over the plate area. This simulates combined compression and
bending of a beam of the same cross section. It follows that the “bending” portion of
the stress causes no force if its “neutral axis” passes through the plate centroid of
area. Thus the remaining “compression” part must equal the centroid stress times the
plate area. This is the result of Eq. (2.26).
However, to balance the bending-moment portion of the stress, the resultant force
F acts not through the centroid but below it toward the high-pressure side. Its line
of action passes through the center of pressure CP of the plate, as sketched in
Fig. 2.11. To find the coordinates (xCP, yCP), we sum moments of the elemental force
p dA about the centroid and equate to the moment of the resultant F. To compute
yCP, we equate

FyCP  冮 yp dA  冮 y( p a 冮
 ␥␰ sin ␪) dA  ␥ sin ␪ y␰ dA

Pressure distribution

pav = p
CG

p (x, y)
Fig. 2.12 The hydrostatic pressure
force on a plane surface is equal,
regardless of its shape, to the
resultant of the three-dimensional Arbitrary
linear pressure distribution on that plane surface
of area A
surface F  pCG A. Centroid of the plane surface
 2.5 Hydrostatic Forces on Plane Surfaces 81

The term 兰 pa y dA vanishes by definition of centroidal axes. Introducing ␰  ␰CG  y,

we obtain

冮
FyCP  ␥ sin ␪ a␰CG y dA  冮 y dAb  ␥ sin ␪ I
2
xx

where again 兰 y dA  0 and Ixx is the area moment of inertia of the plate area about
its centroidal x axis, computed in the plane of the plate. Substituting for F gives the
result

Ixx
yCP  ␥ sin ␪ (2.27)
pCGA

The negative sign in Eq. (2.27) shows that yCP is below the centroid at a deeper level
and, unlike F, depends on angle ␪. If we move the plate deeper, yCP approaches the
centroid because every term in Eq. (2.27) remains constant except pCG, which increases.
The determination of xCP is exactly similar:

FxCP  冮 xp dA  冮 x 3p a  ␥(␰CG  y) sin ␪4 dA

冮
 ␥ sin ␪ xy dA  ␥ sin ␪ Ixy

where Ixy is the product of inertia of the plate, again computed in the plane of the
plate. Substituting for F gives

Ixy
xCP  ␥ sin ␪ (2.28)
pCGA

For positive Ixy, xCP is negative because the dominant pressure force acts in the third,
or lower left, quadrant of the panel. If Ixy  0, usually implying symmetry, xCP  0
and the center of pressure lies directly below the centroid on the y axis.

Gage Pressure Formulas In most cases the ambient pressure pa is neglected because it acts on both sides of
the plate; for example, the other side of the plate is inside a ship or on the dry side
of a gate or dam. In this case pCG  ␥hCG, and the center of pressure becomes inde-
pendent of specific weight:

Ixx sin ␪ Ixy sin ␪

F  ␥hCGA yCP   xCP   (2.29)
hCGA hCGA

Figure 2.13 gives the area and moments of inertia of several common cross sections for
use with these formulas. Note that ␪ is the angle between the plate and the horizon.
82 Chapter 2 Pressure Distribution in a Fluid

L
y A = bL y A = π R2
2
x Ixx =
bL3 x Ixx =
π R4
12 R 4
L R
Ixy = 0 Ixy = 0
2

b b
2 2
(a) (b)

A = bL A = πR
2
2L 2 2
y 3
bL3 Ixx = 0.10976R 4
Ixx =
x 36
y Ixy = 0
L b(b – 2s)L 2
Ix y = x
3 72
Fig. 2.13 Centroidal moments of 4R
inertia for various cross sections: b b R R 3π
2 2
(a) rectangle, (b) circle, (c) trian-
gle, and (d) semicircle. (c) (d)

EXAMPLE 2.5
The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at
point A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force
P exerted by the wall at point A, and (c) the reactions at the hinge B.

Wall
pa

Seawater:
64 lbf/ft 3

15 ft A

pa
Gate
6 ft
B θ

E2.5a Hinge 8 ft
 2.5 Hydrostatic Forces on Plane Surfaces 83

Solution
Part (a) By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or
at elevation 3 ft above point B. The depth hCG is thus 15  3  12 ft. The gate area is
5(10)  50 ft2. Neglect pa as acting on both sides of the gate. From Eq. (2.26) the hydro-
static force on the gate is

F  pCG A  ␥hCG A  (64 lbf/ft3)(12 ft)(50 ft2)  38,400 lbf Ans. (a)

Part (b) First we must find the center of pressure of F. A free-body diagram of the gate is shown in
Fig. E2.5b. The gate is a rectangle, hence

bL3 (5 ft)(10 ft)3

Ixy  0 and Ixx    417 ft4
12 12

The distance l from the CG to the CP is given by Eq. (2.29) since pa is neglected.

Ixx sin ␪ (417 ft4)(106 )

l  yCP     0.417 ft
hCG A (12 ft)(50 ft2)

A
P

F 5 ft

l CG
B θ CP L = 10 ft
Bx

Bz
E2.5b

The distance from point B to force F is thus 10  l  5  4.583 ft. Summing the moments
counterclockwise about B gives
PL sin ␪  F(5  l )  P(6 ft)  (38,400 lbf )(4.583 ft)  0

or P  29,300 lbf Ans. (b)

Part (c) With F and P known, the reactions Bx and Bz are found by summing forces on the gate:

g Fx  0  Bx  F sin ␪  P  Bx  38,400 lbf (0.6)  29,300 lbf

or Bx  6300 lbf
g Fz  0  Bz  F cos ␪  Bz  38,400 lbf (0.8)

or Bz  30,700 lbf Ans. (c)

This example should have reviewed your knowledge of statics.
84 Chapter 2 Pressure Distribution in a Fluid

The solution of Example 2.5 was achieved with the moment of inertia formu-
las, Eqs. (2.29). They simplify the calculations, but one loses a physical feeling
for the forces. Let us repeat Parts (a) and (b) of Example 2.5 using a more visual
approach.

EXAMPLE 2.6
Repeat Example 2.5 to sketch the pressure distribution on plate AB, and break this distri-
bution into rectangular and triangular parts to solve for (a) the force on the plate and (b) the
center of pressure.

Solution
Part (a) Point A is 9 ft deep, hence pA  ␥hA  (64 lbf/ft3)(9 ft)  576 lbf/ft2. Similarly, Point B
is 15 ft deep, hence pB  ␥hB  (64 lbf/ft3)(15 ft)  960 lbf/ft2. This defines the linear
pressure distribution in Fig. E2.6. The rectangle is 576 by 10 ft by 5 ft into the paper. The
triangle is (960  576)  384 lbf/ft2  10 ft by 5 ft. The centroid of the rectangle is 5 ft
down the plate from A. The centroid of the triangle is 6.67 ft down from A. The total force
is the rectangle force plus the triangle force:

F  a576 2 b (10 ft)(5 ft)  a b(10 ft)(5 ft)

lbf 384 lbf
ft 2 ft2
 28,800 lbf  9600 lbf  38,400 lbf Ans. (a)

576 lbf/ft2

F
A
t
5f
960 lbf/ft2
l 6 ft
l
5-
8 ft
E2.6 B

Part (b) The moments of these forces about point A are

MA  (28,800 lbf)(5 ft)  (9600 lbf)(6.67 ft)  144,000  64,000  208,000 ft # lbf

MA 208,000 ft # lbf
Then 5 ft  l    5.417 ft hence l  0.417 ft Ans. (b)
F 38,400 lbf
Comment: We obtain the same force and center of pressure as in Example 2.5 but with
more understanding. However, this approach is awkward and laborious if the plate is
not a rectangle. It would be difficult to solve Example 2.7 with the pressure distribu-
tion alone because the plate is a triangle. Thus moments of inertia can be a useful
simplification.
 2.5 Hydrostatic Forces on Plane Surfaces 85

EXAMPLE 2.7
A tank of oil has a right-triangular panel near the bottom, as in Fig. E2.7. Omitting pa, find
the (a) hydrostatic force and (b) CP on the panel.

pa

5m Oil: ρ = 800 kg/m 3

30° 11 m
4m
6m
pa

CG CP

8m 4m

2m
E2.7 4m

Solution
Part (a) The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and one-
third over (2 m) from the lower left corner, as shown. The area is
1
2 (6 m)(12 m)  36 m2
The moments of inertia are
bL3 (6 m)(12 m)3
Ixx    288 m4
36 36
b(b  2s)L2 (6 m)3 6 m  2(6 m)4 (12 m)2
and Ixy    72 m4
72 72
The depth to the centroid is hCG  5  4  9 m; thus the hydrostatic force from Eq. (2.26) is
F  ␳ghCG A  (800 kg /m3)(9.807 m /s2)(9 m)(36 m2)
 2.54  106 (kg # m)/s2  2.54  106 N  2.54 MN Ans. (a)
Part (b) The CP position is given by Eqs. (2.29):
Ixx sin ␪ (288 m4)(sin 30
)
yCP     0.444 m
hCG A (9 m)(36 m2)
Ixy sin ␪ (72 m4)(sin 30
)
xCP     0.111 m Ans. (b)
hCG A (9 m)(36 m2)
The resultant force F  2.54 MN acts through this point, which is down and to the right
of the centroid, as shown in Fig. E2.7.
86 Chapter 2 Pressure Distribution in a Fluid

2.6 Hydrostatic Forces on The resultant pressure force on a curved surface is most easily computed by separating
Curved Surfaces it into horizontal and vertical components. Consider the arbitrary curved surface
sketched in Fig. 2.14a. The incremental pressure forces, being normal to the local area
element, vary in direction along the surface and thus cannot be added numerically. We
could sum the separate three components of these elemental pressure forces, but it
turns out that we need not perform a laborious three-way integration.
Figure 2.14b shows a free-body diagram of the column of fluid contained in the
vertical projection above the curved surface. The desired forces FH and FV are
exerted by the surface on the fluid column. Other forces are shown due to fluid
weight and horizontal pressure on the vertical sides of this column. The column of
fluid must be in static equilibrium. On the upper part of the column bcde, the hor-
izontal components F1 exactly balance and are not relevant to the discussion. On
the lower, irregular portion of fluid abc adjoining the surface, summation of hori-
zontal forces shows that the desired force FH due to the curved surface is exactly
equal to the force FH on the vertical left side of the fluid column. This left-side
force can be computed by the plane surface formula, Eq. (2.26), based on a verti-
cal projection of the area of the curved surface. This is a general rule and simpli-
fies the analysis:
The horizontal component of force on a curved surface equals the force on the
plane area formed by the projection of the curved surface onto a vertical plane nor-
mal to the component.
If there are two horizontal components, both can be computed by this scheme. Sum-
mation of vertical forces on the fluid free body then shows that
FV  W1  W2  Wair (2.30)
We can state this in words as our second general rule:
The vertical component of pressure force on a curved surface equals in magnitude
and direction the weight of the entire column of fluid, both liquid and atmosphere,
above the curved surface.

Wair
d e

Curved surface
F1 W1 F1
projection onto
FV vertical plane
c b

FH W2
FH FH FH
Fig. 2.14 Computation of
hydrostatic force on a curved a
surface: (a) submerged curved
surface; (b) free-body diagram of FV
fluid above the curved surface. (a) (b)
 2.6 Hydrostatic Forces on Curved Surfaces 87

Thus the calculation of FV involves little more than finding centers of mass of a
column of fluid—perhaps a little integration if the lower portion abc in Fig. 2.14b has
a particularly vexing shape.

EXAMPLE 2.8
A dam has a parabolic shape z/z0  (x/x0)2 as shown in Fig. E2.8a, with x0  10 ft and
z0  24 ft. The fluid is water, ␥  62.4 lbf/ft3, and atmospheric pressure may be omitted.
Compute the forces FH and FV on the dam and their line of action. The width of the dam
is 50 ft.

pa = 0 lbf/ft2 gage

FV
z0

z FH

x
x0 2

E2.8a
( (
x
z = z0 x
0

Solution
• System sketch: Figure E2.8b shows the various dimensions. The dam width is b  50 ft.
• Approach: Calculate FH and its line of action from Eqs. (2.26) and (2.29). Calculate FV
and its line of action by finding the weight of fluid above the parabola and the centroid
of this weight.
• Solution steps for the horizontal component: The vertical projection of the parabola lies
along the z axis in Fig. E2.8b and is a rectangle 24 ft high and 50 ft wide. Its centroid is
halfway down, or hCG  24/2  12 ft. Its area is Aproj  (24 ft)(50 ft)  1200 ft2. Then,
from Eq. (2.26),

FH  ␥ hCG Aproj  a62.4 b (12 ft)(1200 ft2)  898,560 lbf ⬇ 899  103 lbf
lbf
ft3
The line of action of FH is below the centroid of Aproj, as given by Eq. (2.29):

Ixx sin ␪ (1/12)(50 ft)(24 ft)3 sin 90

yCP, proj     4 ft
hCG Aproj (12 ft)(1200 ft2)
Thus FH is 12  4  16 ft, or two-thirds of the way down from the surface (8 ft up from
the bottom).
• Comments: Note that you calculate FH and its line of action from the vertical projec-
tion of the parabola, not from the parabola itself. Since this projection is vertical, its angle
␪  90
.
• Solution steps for the vertical component: The vertical force FV equals the weight of
water above the parabola. Alas, a parabolic section is not in Fig. 2.13, so we had to look
88 Chapter 2 Pressure Distribution in a Fluid

it up in another book. The area and centroid are shown in Fig. E2.8b. The weight of this
parabolic amount of water is

FV  ␥ Asectionb  a62.4 b c (24 ft)(10 ft) d (50 ft)  499,200 lbf ⬇ 499  103 lbf
lbf 2
ft3 3

z0 = 24 ft
2 x0z 0
Area =
3
3z0
5
FV

Parabola

0 3x 0 x0 = 10 ft
E2.8b 8

This force acts downward, through the centroid of the parabolic section, or at a distance
3x0 /8  3.75 ft over from the origin, as shown in Figs. E2.8b,c. The resultant hydrostatic
force on the dam is

F  (FH2  FV2)1/2  3 (899E3 lbf )2  (499E3 lbf )2 4 1/2  1028  103 lbf at 29
Ans.

This resultant is shown in Fig. E2.8c and passes through a point 8 ft up and 3.75 ft over
from the origin. It strikes the dam at a point 5.43 ft over and 7.07 ft up, as shown.
• Comments: Note that entirely different formulas are used to calculate FH and FV. The
concept of center of pressure CP is, in the writer’s opinion, stretched too far when applied
to curved surfaces.

z
Resultant = 1028 × 103 lbf acts along z = 10.083–0.5555x

3.75 ft
FV = 499 × 103 lbf

Parabola z = 0.24x2
FH = 899 × 103 lbf 29°

7.07 ft
8 ft

x
E2.8c 0 5.43 ft
 2.7 Hydrostatic Forces in Layered Fluids 89

B
A EXAMPLE 2.9
Find an algebraic formula for the net vertical force F on the submerged semicircular pro-
FU
jecting structure CDE in Fig. E2.9. The structure has uniform width b into the paper. The
liquid has specific weight ␥.
C

R D Solution
The net force is the difference between the upward force FL on the lower surface DE and
the downward force FU on the upper surface CD, as shown in Fig. E2.9. The force FU
E equals ␥ times the volume ABDC above surface CD. The force FL equals ␥ times the vol-
FL ume ABDEC above surface DE. The latter is clearly larger. The difference is ␥ times the
volume of the structure itself. Thus the net upward fluid force on the semicylinder is
E2.9
␲ 2
F  ␥fluid (volume CDE)  ␥fluid Rb Ans.
2

This is the principle upon which the laws of buoyancy, Sec. 2.8, are founded. Note that the
result is independent of the depth of the structure and depends upon the specific weight of
the fl
uid, not the material within the structure.

2.7 Hydrostatic Forces in The formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for a
Layered Fluids fluid of uniform density. If the fluid is layered with different densities, as in Fig. 2.15,
a single formula cannot solve the problem because the slope of the linear pressure

Plane z=0
F1= p A1 surface
CG1 pa
ρ1 < ρ2
Fluid 1
p = pa – ρ1gz

z 1, p1

p1 = pa – ρ1gz1
F2= p A
CG 2 2
ρ2
Fluid 2

z 2 , p2

p = p1 – ρ2 g(z –z 1)
Fig. 2.15 Hydrostatic forces on a
surface immersed in a layered fluid
must be summed in separate pieces. p2 = p1 – ρ 2 g(z 2 –z 1)
90 Chapter 2 Pressure Distribution in a Fluid

distribution changes between layers. However, the formulas apply separately to each
layer, and thus the appropriate remedy is to compute and sum the separate layer forces
and moments.
Consider the slanted plane surface immersed in a two-layer fluid in Fig. 2.15. The
slope of the pressure distribution becomes steeper as we move down into the denser
second layer. The total force on the plate does not equal the pressure at the centroid
times the plate area, but the plate portion in each layer does satisfy the formula, so
that we can sum forces to find the total:

F  g Fi  g pCGiAi (2.31)

Similarly, the centroid of the plate portion in each layer can be used to locate the cen-
ter of pressure on that portion:

␳ig sin ␪i Ixxi ␳ig sin ␪i Ixyi

yCPi   xCPi   (2.32)
pCGi Ai pCGi Ai

These formulas locate the center of pressure of that particular Fi with respect to the
centroid of that particular portion of plate in the layer, not with respect to the cen-
troid of the entire plate. The center of pressure of the total force F   Fi can then
be found by summing moments about some convenient point such as the surface. The
following example will illustrate this.

EXAMPLE 2.10
A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mer-
cury. Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the
fluid on the right-hand side of the tank.

Solution
Part (a) Divide the end panel into three parts as sketched in Fig. E2.10, and find the hydro-
static pressure at the centroid of each part, using the relation (2.26) in steps as in
Fig. E2.10:

pCG1  (55.0 lbf/ft3)(4 ft)  220 lbf/ft2

pCG2  (55.0)(8)  62.4(3)  627 lbf/ft2
pCG3  (55.0)(8)  62.4(6)  846(2)  2506 lbf/ft2

These pressures are then multiplied by the respective panel areas to find the force on each
portion:

F1  pCG1A1  (220 lbf/ft2)(8 ft)(7 ft)  12,300 lbf

F2  pCG2A2  627(6)(7)  26,300 lbf
F3  pCG3A3  2506(4)(7)  70,200 lbf
F  g Fi  108,800 lbf Ans. (a)
 2.8 Buoyancy and Stability 91

pa = 0
z=0

Oi 7 ft 4 ft
l: 5
5.0
lbf
/ft 3
8 ft
Wa 11 ft
ter (1)
(62
.4)
Me
rcu 6 ft 16 ft
ry (2)
(84
6)
4 ft (3)
E2.10

Part (b) Equations (2.32) can be used to locate the CP of each force Fi, noting that ␪  90
and
sin ␪  1 for all parts. The moments of inertia are Ixx1  (7 ft)(8 ft)3/12  298.7 ft4, Ixx2 
7(6)3/12  126.0 ft4, and Ixx3  7(4)3/12  37.3 ft4. The centers of pressure are thus at

␳1gIxx1 (55.0 lbf/ft3)(298.7 ft4)

yCP1     1.33 ft
F1 12,300 lbf
62.4(126.0) 846(37.3)
yCP2    0.30 ft yCP3    0.45 ft
26,300 70,200

This locates zCP1  4  1.33  5.33 ft, zCP2  11  0.30  11.30 ft, and zCP3 
16  0.45  16.45 ft. Summing moments about the surface then gives

g FizCPi  FzCP

or 12,300(5.33)  26,300(11.30)  70,200(16.45)  108,800zCP

1,518,000
or zCP    13.95 ft Ans. (b)
108,800

The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft
below the surface.

2.8 Buoyancy and Stability The same principles used to compute hydrostatic forces on surfaces can be applied to
the net pressure force on a completely submerged or floating body. The results are
the two laws of buoyancy discovered by Archimedes in the third century B.C.:

1. A body immersed in a fluid experiences a vertical buoyant force equal to the

weight of the fluid it displaces.
2. A floating body displaces its own weight in the fluid in which it floats.