Assignment 1
Answer’s Scheme
2.1C
Intensive properties do not depend on the size (extent) of the system but extensive properties do
depend on the size (extent) of the system. An example of an intensive property is temperature. An
example of an extensive property is mass.
2.6C
Ru is the universal gas constant that is the same for all gases, whereas R is the specific gas constant
that is different for different gases. These two are related to each other by R  Ru / M , where M is
the molar mass (also called the molecular weight) of the gas. Since molar mass has dimensions of
mass per mole, R and Ru do not have the same dimensions or units.
2.9
At specified conditions, oxygen behaves as an ideal gas. The gas constant of oxygen is obtained from
Table A-1, R = 0.2598 kPam3/kgK. According to the ideal gas equation of state,
v
RT (0.2598 kPa  m 3 /kg  K)(27  273 K)

 0.260 m3 /kg
P
300 kPa
In ideal gas calculations, it saves time to write the gas constant in appropriate units.
2.10
The gas constant of air is R  0.287
kJ  kPa  m3 
kPa  m3
.

  0.287
kg  K  kJ 
kg  K
The initial and final absolute pressures in the tire are
P1 = Pg1 + Patm = 140 + 100 = 240 kPa
Tire
0.0740 m3
30C
140 kPa
P2 = Pg2 + Patm = 210 + 100 = 310 kPa
Treating air as an ideal gas, the initial mass in the tire is
m1 
P1V
(240 kPa)(0.074 0 m 3 )

 0.2042 kg
RT1 (0.287 kPa  m 3 /kg  K)(303 K)
Noting that the temperature and the volume of the tire remain constant, the final mass in the tire
becomes
m2 
P2V
(310 kPa)(0.074 0 m 3 )

 0.2638 kg
RT2 (0.287 kPa  m 3 /kg  K)(303 K)
Thus the amount of air that needs to be added is
m  m2  m1  0.2638  0.2042  0.0596 kg
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
2.11
The gas constant of air is R  0.287
kJ  kPa  m3 
kPa  m3
.

  0.287
kg  K  kJ 
kg  K
Initially, the absolute pressure in the tire is
P1  Pg  Patm  210  100  310 kPa
Treating air as an ideal gas and assuming the volume of the
tire to remain constant, the final pressure in the tire is
determined from
Tire
25C
210 kPa
323 K
P1V1 P2V 2
T


 P2  2 P1 
(310 kPa)  336 kPa
T1
T2
T1
298 K
Thus the pressure rise is
P  P2  P1  336  310  26.0 kPa
The amount of air that needs to be bled off to restore pressure to its original value is
m1 
m2 
P1V
(310 kPa)(0.025 m 3 )

 0.0906 kg
RT1 (0.287 kPa  m 3 /kg  K)(298 K)
P2V
(310 kPa)(0.025 m 3 )

 0.0836 kg
RT2 (0.287 kPa  m 3 /kg  K)(323 K)
m  m1  m 2  0.0906  0.0836  0.0070 kg
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
2.2C
The specific gravity, or relative density, is defined as the ratio of the density of a substance to the
density of some standard substance at a specified temperature (the standard is water at 4°C, for
which H2O = 1000 kg/m3). That is, SG   /  H2O . When specific gravity is known, density is
determined from   SG   H2O .
Specific gravity is dimensionless and unitless [it is just a number without dimensions or units
2.3C
The original specific weight is
1 
W
V
If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V /2.
The specific weight of one of these halves is
W /2
 1
V /2

which is the same as the original specific weight. Hence, specific weight is an intensive property.
If specific weight were an extensive property, its value for half of the system would be halved.
2.7
Knowing the weight, the mass and the density of the fluid are determined to be
m
W
225 N

g 9.80 m/s 2

m
V

 1 kg  m/s 3 

  23.0 kg
 1N



23.0 kg
 0.957 kg/L
24 L
Note that mass is an intrinsic property, but weight is not.
2.8
At specified conditions, air behaves as an ideal gas.
The gas constant of air is R  0.287
kJ  kPa  m3 
kPa  m3
(see also Table A-1).

  0.287
kg  K  kJ 
kg  K
The definition of the specific volume gives
v
V
m

0.100 m 3
 0.100 m 3 /kg
1 kg
Using the ideal gas equation of state, the pressure is
Pv  RT

P
RT
v

(0.287 kPa  m3 /kg  K)(27  273.15 K)
 861kPa
0.100 m3 /kg
In ideal gas calculations, it saves time to convert the gas constant to appropriate units.
2.11
An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and
the amount of air that must be bled off to reduce the temperature to the original value are to be
determined.
Assumptions: 1) At specified conditions, air behaves as an ideal gas. 2) The volume of the tire
remains constant.
The gas constant of air is R  0.287
kJ  kPa  m3 
kPa  m3
.

  0.287
kg  K  kJ 
kg  K
Initially, the absolute pressure in the tire is
P1  Pg  Patm  210  100  310 kPa
Treating air as an ideal gas and assuming the volume of the
tire to remain constant, the final pressure in the tire is
determined from
Tire
25C
210 kPa
323 K
P1V1 P2V 2
T


 P2  2 P1 
(310 kPa)  336 kPa
T1
T2
T1
298 K
Thus the pressure rise is
P  P2  P1  336  310  26.0 kPa
The amount of air that needs to be bled off to restore pressure to its original value is
m1 
m2 
P1V
(310 kPa)(0.025 m 3 )

 0.0906 kg
RT1 (0.287 kPa  m 3 /kg  K)(298 K)
P2V
(310 kPa)(0.025 m 3 )

 0.0836 kg
RT2 (0.287 kPa  m 3 /kg  K)(323 K)
m  m1  m 2  0.0906  0.0836  0.0070 kg
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
2.14
The density of water is 1000 kg/m3.
The methanol’s weight, density, and specific gravity are
𝑊 = 𝑚𝑔 = 40 × 9.81 = 392.40 N
𝜌=
𝑚
=
𝑉
SG =
40 kg
= 784 kg/m3
1 m3
51 L ×
1000 L
𝜌
𝜌H2 O
=
784 kg/m3
= 0.784
1000 kg/m3
The force needed to accelerate the tank at the given rate is
𝐹 = 𝑚𝑎 = (392.40 N) × (0.25
m
) = 98.1 N
s2
2.17C
In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure
drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low
pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon is
a common cause for drop in performance and even the erosion of impeller blades.
The word “cavitation” comes from the fact that a vapor bubble or “cavity” appears in the liquid. Not
all cavitation is undesirable. It turns out that some underwater vehicles employ “super cavitation” on
purpose to reduce drag.
2.18C
Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with
pressure. The higher the pressure, the higher the saturation or boiling temperature.
This fact is easily seen by looking at the saturated water property tables. Note that boiling
temperature and saturation pressure at a given pressure are equivalent.
2.19C
If the pressure of a substance increases during a boiling process, the temperature also increases
since the boiling (or saturation) temperature of a pure substance depends on pressure and increases
with it.
We are assuming that the liquid will continue to boil. If the pressure is increased fast enough, boiling
may stop until the temperature has time to reach its new (higher) boiling temperature. A pressure
cooker uses this principle.
2.22
The vapor pressure of water at 20C is 2.339 kPa.
To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the
vapor (or saturation) pressure at the given temperature. That is,
Pmin  Psat@20C  2.339kPa
Therefore, the lowest pressure that can exist in the pump is 2.339 kPa.
Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is
greater at higher fluid temperatures.
2.23
The vapor pressure of water at 30C is 4.246 kPa.
To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the
vapor (or saturation) pressure at the given temperature. That is,
Pmin  Psat@30C  4.246kPa
Therefore, the pressure should be maintained above 4.246 kPa everywhere in flow.
Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is
greater at higher fluid temperatures.
2.24
The vapor pressure of water at 20C is 2.339 kPa.
To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or
saturation) pressure at the given temperature, which is
Pv  Psat @ 20C  2.339 kPa
The minimum pressure in the pump is 2 kPa, which is less than the vapor pressure. Therefore, a there is danger
of cavitation in the pump.
Note that the vapor pressure increases with increasing temperature, and thus there is a greater
danger of cavitation at higher fluid temperatures.
2.25C
Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids
at rest do not possess any flow energy.
Flow energy is not a fundamental quantity, like kinetic or potential energy. However, it is a useful
concept in fluid mechanics since fluids are often forced into and out of control volumes in practice.
2.27C
The macroscopic forms of energy are those a system possesses as a whole with respect to some
outside reference frame. The microscopic forms of energy, on the other hand, are those related to
the molecular structure of a system and the degree of the molecular activity, and are independent
of outside reference frames.
We mostly deal with macroscopic forms of energy in fluid mechanics.
2.29C
The internal energy of a system is made up of sensible, latent, chemical, and nuclear energies. The
sensible internal energy is due to translational, rotational, and vibrational effects.
We deal with the flow of a single phase fluid in most problems in this textbook; therefore, latent,
chemical, and nuclear energies do not need to be considered.
2.30C
Thermal energy is the sensible and latent forms of internal energy. It does not include chemical or
nuclear forms of energy. In common terminology, thermal energy is referred to as heat. However,
like work, heat is not a property, whereas thermal energy is a property.
Technically speaking, “heat” is defined only when there is heat transfer, whereas the energy state of
a substance can always be defined, even if no heat transfer is taking place.
2.33
The total energy of a flowing fluid is given by (Eq. 28)
𝑉2
𝑒 =ℎ+
+ 𝑔𝑧
2
The enthalpy of the vapor at the specified temperature can be found in any thermo text to be 2745.9 kJ/kg.
Then the total energy is determined as
m 2
(50 s )
J
m
J
𝑒 = 2745.9 × 103 +
+ (9.81 2 ) × (10 m) ≅ 2.7472 × 106 = 2747.2 kJ/kg
kg
2
s
kg
Note that only 0.047% of the total energy comes from the combination of kinetic and potential energies, which
explains why we usually neglect kinetic and potential energies in most flow systems.
2.68C
Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the
internal frictional force that develops between different layers of fluids as they are forced to move
relative to each other. Viscosity is caused by the cohesive forces between the molecules in liquids,
and by the molecular collisions in gases. In general, liquids have higher dynamic viscosities than
gases.
The ratio of viscosity  to density  often appears in the equations of fluid mechanics, and is defined
as the kinematic viscosity,  =  /.
2.69C
Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called
Newtonian fluids. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids.
In the differential analysis of fluid flow, only Newtonian fluids are considered in this textbook.
2.71C
(a) The dynamic viscosity of liquids decreases with temperature. (b) The dynamic viscosity of gases
increases with temperature.
A good way to remember this is that a car engine is much harder to start in the winter because the
oil in the engine has a higher viscosity at low temperatures.
2.75
The velocity profile of a fluid flowing though a circular pipe is given. The friction drag force exerted
on the pipe by the fluid in the flow direction per unit length of the pipe is to be determined.
The viscosity of the fluid is constant.
The wall shear stress is determined from its definition to be
 w  
du
dr
rR
  u max
d 
rn 
 nr n 1
1 




u
max
dr  R n  r  R
Rn

r R
nu max
R
Note that the quantity du /dr is negative in pipe flow, and
the negative sign is added to the w relation for pipes to
make shear stress in the positive (flow) direction a positive
quantity. (Or, du /dr = - du /dy since y = R – r). Then the
friction drag force exerted by the fluid on the inner
surface of the pipe becomes
F   w Aw 
nu max
(2R) L  2n u max L
R
u(r) = umax(1-rn/Rn)
R
r
0
umax
Therefore, the drag force per unit length of the pipe is
F / L  2n u max .
Note that the drag force acting on the pipe in this case is independent of the pipe diameter.
2.91C
The magnitude of the pulling force at the surface of a liquid per unit length is called surface tension
s. It is caused by the attractive forces between the molecules. The surface tension is also surface
energy (per unit area) since it represents the stretching work that needs to be done to increase the
surface area of the liquid by a unit amount.
Surface tension is the cause of some very interesting phenomena such as capillary rise and insects
that can walk on water.
2.94C
The pressure inside a soap bubble is greater than the pressure outside, as evidenced by the stretch
of the soap film.
You can make an analogy between the soap film and the skin of a balloon.
2.96
The surface tension s is given for two cases to be 0.08 and 0.12 N/m.
Considering that an air bubble in a liquid has only one interface, he pressure difference between the
inside and the outside of the bubble is determined from
Pbubble  Pi  P0 
2 s
R
liquid
Substituting, the pressure difference is determined to be:
(a) s = 0.08 N/m:
2(0.08 N/m)
Pbubble 
 2133 N/m 2  2.13 kPa
0.00015/2 m
(b) s = 0.12 N/m:
Pbubble 
Air
bubble
P
2(0.12 N/m)
 3200 N/m 2  3.20 kPa
0.00015/2 m
Note that a small gas bubble in a liquid is highly pressurized.
The smaller the bubble diameter, the larger the pressure
inside the bubble.
2.97
The surface tension of solution is given to be s = 0.039 N/m.
The work associated with the stretching of a film is the surface tension work, and is expressed in
differential form as Ws   s dAs . Noting that surface tension is constant, the surface tension work is
simply surface tension multiplied by the change in surface area,
Ws   s ( A2  A1 )  2
2
s ( D2
The factor 2 is due to having two surfaces in contact with air. Substituting, the
required work input is determined to be

Air
 D12 )
Soap
bubble
P

 1 kJ

-7
Ws  2 (0.039 N/m) (0.069 m) 2  (0.060 m) 2 
  2.84  10 kJ
 1000 N  m 
Note that when a bubble explodes, an equivalent amount of energy is released to the environment.
2.101
Assumptions: 1) There are no impurities in the liquid, and no contamination on the surfaces of the
wire frame. 2) The liquid is open to the atmospheric air.
Substituting the numerical values, the surface tension is determined from the surface tension force
relation to be
s 
F
0.024 N

 0.15 N/m
2b 2(0.08 m)
b
The surface tension depends on temperature. Therefore, the value
determined is valid at the temperature of the liquid.
Liquid
film
3-2C
Pascal’s law states that the pressure applied to a confined fluid increases the pressure throughout
by the same amount. This is a consequence of the pressure in a fluid remaining constant in the
horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack.
3.4C
The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure
relative to an absolute vacuum is called absolute pressure.
Most pressure gages (like your bicycle tire gage) read relative to atmospheric pressure, and therefore
read the gage pressure.
3-6
Friction between the piston and the cylinder is negligible.
(a) The gas pressure in the piston–cylinder device depends on the atmospheric pressure and the
weight of the piston. Drawing the free-body diagram of the piston as shown in Fig. 3–6 and balancing
the vertical forces yield
PA  Patm A  W
Solving for P and substituting,
P  Patm 
mg
(40 kg)(9.81 m/s 2 ) 
1 kN
 95 kPa 
2
2

A
0.012 m
 1000 kg  m/s
 1 kPa 

 128 kPa
 1 kN/m 2 

(b) The volume change will have no effect on the free-body diagram drawn in part (a), and therefore
we do not expect the pressure inside the cylinder to change – it will remain the same.
F
If the gas behaves as an ideal gas, the absolute temperature doubles when the volume is doubled at
constant pressure.
3-7
The absolute pressure in the chamber is determined from
Pabs  Patm  Pvac  92  36  56 kPa
We must remember that “vacuum pressure” is the negative of gage pressure – hence the negative
sign.
3-10
The density of mercury is given to be 13,600 kg/m3.
The atmospheric pressure is determined directly from
Patm  gh

1N
 (13,600 kg/m 3 )(9.81 m/s 2 )( 0.735 m) 
2
 1 kg  m/s
 98.1 kPa
 1 kPa


 1000 N/m 2 

We round off the final answer to three significant digits. 100 kPa is a fairly typical value of
atmospheric pressure on land slightly above sea level.
3-11
The variation of the density of the liquid with depth is negligible.
The gage pressure at two different depths of a liquid can be expressed as P1  gh1 and P2  gh2 .
Taking their ratio,
P2 gh2 h2


P1
gh1 h1
h1
h2
Solving for P2 and substituting gives
h
12 m
P2  2 P1 
(28 kPa)  112 kPa
h1
3m
Note that the gage pressure in a given fluid is proportional to depth.
1
2
3-12
The liquid and water are incompressible.
The specific gravity of the fluid is given to be SG = 0.78. We take the density of water to be 1000
kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of
water,
  SG   H 2O  (0.78)(100 0 kg/m 3 )  780 kg/m 3
Patm
(a) Knowing the absolute pressure, the atmospheric
pressure can be determined from
h
Patm  P  gh
 1 kPa

 (175 kPa) - (1000 kg/m 3 )(9.81 m/s 2 )(8 m) 

2
 1000 N/m 
 96 .52 kPa  96.5 kPa
P
(b) The absolute pressure at a depth of 8 m in the other liquid is
P  Patm  gh
 1 kPa

 (96.52 kPa)  (780 kg/m 3 )(9.81 m/s 2 )(8 m) 

2
 1000 N/m 
 157 .7 kPa  158 kPa
Note that at a given depth, the pressure in the lighter fluid is lower, as expected.
3-16
The density of mercury is given to be  = 13,590 kg/m3.
The atmospheric (or barometric) pressure can be expressed as
Patm   g h

1N
 (13,590 kg/m 3 )(9.807 m/s 2 )(0.740 m) 
 1 kg  m/s 2

 98 .6 kPa
 1 kPa

 1000 N/m 2





Then the absolute pressure in the tank is
Pabs  Pgage  Patm  350  98 .6  449 kPa
This pressure is more than four times as much as standard atmospheric pressure.
3-21
The variation of air density and the gravitational acceleration with
altitude is negligible.
The density of air is given to be  = 1.20 kg/m3.
Taking an air column between the top and the bottom of the
mountain and writing a force balance per unit base area, we obtain
Wair / A  Pbottom  Ptop 
( gh) air  Pbottom  Ptop 
 h 
h
 100,000 N/m 2

1 bar
(1.20 kg/m 3 )(9.81 m/s 2 ) 
(0.980  0.790 ) bar
 1 kg  m/s 2


1N

Pbottom  Ptop
g

  1614 m


which is also the distance climbed.
A similar principle is used in some aircraft instruments to measure elevation.
3-22
The variation of air density with altitude is negligible.
The density of air is given to be  = 1.18 kg/m3. The density of mercury is 13,600 kg/m3.
Atmospheric pressures at the top and at the bottom of the building are
730 mmHg
Ptop  (ρ g h) top
Pbottom
 1N
 1 kPa 
 (13,600 kg/m3 )(9.807 m/s 2 )(0.730 m) 
2 
2 
 1 kg  m/s  1000 N/m 
 97.36 kPa
 (  g h )bottom
 1N
 1 kPa 
 (13,600 kg/m3 )(9.807 m/s 2 )(0.755 m) 
2 
2 
 1 kg  m/s  1000 N/m 
 100.70 kPa
h
755 mmHg
Taking an air column between the top and the bottom of the building, we write a force balance per
unit base area,
Wair / A  Pbottom  Ptop
and
(  gh )air  Pbottom  Ptop
 1N
 1 kPa 
(1.18 kg/m3 )(9.807 m/s 2 )(h) 
 (100.70  97.36) kPa
2 
2 
 1 kg  m/s  1000 N/m 
which yields h = 288.6 m  289 m, which is also the height of the building.
There are more accurate ways to measure the height of a building, but this method is quite simple
3-25
The variation of the density of water with depth is negligible.
The specific gravity of sea water is given to be SG = 1.03. The
density of water at 0C is 1000 kg/m3.
The density of the seawater is obtained by multiplying its
specific gravity by the density of water,
  SG   H 2O  (1.03)(100 0 kg/m 3 )  1030 kg/m 3
The pressure exerted on the surface of the submarine cruising 70 m below the free surface of the sea
is the absolute pressure at that location:
P  Patm  gh
 1 kPa
 101 kPa  (1030 kg/m 3 )(9.81 m/s 2 )(70 m) 
 1000 N/m 2

 808 .3 kPa  800 kPa




where we have rounded the final answer to three significant digits.
This is more than 10 times the value of atmospheric pressure at sea level.
3-26
Drawing the free body diagram of the piston and balancing the vertical forces yields
PA  Patm A  W  Fspring
Thus,
P  Patm 
mg  Fspring
A
(4 kg)(9.807 m/s 2 )  60 N  1 kPa 
 (95 kPa) 
 123.4 kPa  123 kPa

2 
35 104 m 2
 1000 N/m 
Discussion
This setup represents a crude but functional way to control the pressure in a tank.
3-39
Both water and oil are incompressible substances.
The density of oil is given to be oil = 790 kg/m3. We take the density of water to be w =1000 kg/m3.
The height of water column in the left arm of the manometer is given to be hw1 = 0.70 m. We let the
height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 6hw2. Noting that
both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed
as
Pbottom  Patm   w ghw1
and
Pbottom  Patm   w ghw2   a gha
Setting them equal to each other and simplifying,
 w ghw1   w ghw2   a gha

 w hw1   w hw2   a ha

Noting that ha = 6hw2 and we take a =oil, the water and oil
column heights in the second arm are determined to be
0.7 m  hw2  (790/1000) 6hw2 
hw1  hw2  (  a /  w )ha
Water
oil
ha
hw2  0.122 m
0.7 m  0.122 m  (790/1000) ha 
ha  0.732 m
hw1
hw2
Note that the fluid height in the arm that contains oil is higher. This
is expected since oil is lighter than water.
3.37
Assumptions: 1)The density of blood is constant. 2) The gage pressure of blood is 120 mmHg.
The density of blood is given to be  = 1040 kg/m3.
For a given gage pressure, the relation P  gh can be expressed for
Blood
mercury and blood as P   blood ghblood and P   mercury ghmercury . Setting
these two relations equal to each other we get
P   blood ghblood   mercury ghmercury
Solving for blood height and substituting gives
hblood 
 mercury
 blood
hmercury 
13,600 kg/m 3
1040 kg/m 3
(0.12 m)  1.57 m
3-57C
Dams are built much thicker at the bottom because the pressure force increases with depth, and the
bottom part of dams are subjected to largest forces.
Dam construction requires an enormous amount of concrete, so tapering the dam in this way saves a
lot of concrete, and therefore a lot of money.
h
3-63
We take the density of water to be 1000 kg/m3 throughout.
The average pressure on a surface is the pressure at the centroid
(midpoint) of the surface, and is determined to be
Pavg  ghC  gh / 2

1N
 (1000 kg/m 3 )(9.81 m/s 2 )( 60 / 2 m) 
 1 kg  m/s 2


  294 ,300 N/m 2


2h/3
FR
Then the resultant hydrostatic force acting on the dam becomes
h=60 m
h/3
FR  Pavg A  (294 ,300 N/m 2 )(60 m  360 m)  6.36  10 9 N
Resultant force per unit area is pressure, and its value at the top and the bottom of the dam
becomes
Ptop  ghtop  0 N/m 2

1N
Pbottom  ghbottom  (1000 kg/m 3 )(9.81 m/s 2 )(60 m) 
 1 kg  m/s 2


  588,600 N/m 2


The values above are gage pressures, of course. The gage pressure at the bottom of the dam is about
590 kPa, which is almost six times greater than standard atmospheric pressure.
3-65
Assumptions: Atmospheric pressure acts on both sides of the dam, and thus it can be ignored in
calculations for convenience.
We take the density of water to be 1000 kg/m3 throughout.
We consider the free body diagram of the liquid block enclosed by the circular surface of the dam
and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal
plane surfaces as well as the weight of the liquid block are:
Horizontal force on vertical surface:
Fy = 0
FH  Fx  Pavg A  ghC A  g ( R / 2) A

1N
 (1000 kg/m 3 )(9.81 m/s 2 )( 7 / 2 m)(7 m  70 m) 
 1 kg  m/s 2

R=7m




FH
 1.682  10 7 N
W
Vertical force on horizontal surface is zero since it coincides with the
free surface of water. The weight of fluid block per m length is
FV  W  gV  g[ w  R 2 / 4]

1N
 (1000 kg/m 3 )(9.81 m/s 2 )[( 70 m)  (7 m) 2 /4]
 1 kg  m/s 2





 2.643  10 7 N
Then the magnitude and direction of the hydrostatic force acting on the surface of the dam become
FR  FH2  FV2  (1.682  10 7 N) 2  (2.643  10 7 N) 2  3.13  10 7 N
tan 
FV 2.643  10 7 N

 1.571 
  57.5 
FH 1.682  10 7 N
Therefore, the line of action of the hydrostatic force passes through the center of the curvature of the dam,
making 57.5 downwards from the horizontal.
3-67
We take the density of water to be 1000 kg/m3 throughout.
0.3 m
0.7 m
water
0.9 m
0.7 m

We first determine the angle;
Sin( ) 
0.9
 0.9,   64 .16 o
1

2
1

FR  hcg A  9810   0.3  0.7   Sin(64 .16 )  0.7  0.7  1658 N
3
2


In order to locate FR on the gate xcp, and ycp must be found.
xcp  xcg 
I xyc
ycg A
For simplicity, we can consider x axis to be passing through center of gravity of the gate, so that
xcg=0.
I xyc 
0.7  0.7  2  0 0.7 2
 3.334 10 3 m 4
72
2
ycg  0.3  0.7  0.766 m
3
A  0.5  0.7 2  0.245 m2
xcp  0 
I xc 
3.334 10 3
 1.776 10 2 m  1.77 cm
0.766  0.245
0.7  0.7 3
 6.67 10 3 m 4
36
y cp  0.766 
6.67  10 3
 0.801 m
0.766  0.245
3-68
We take the density of water to be 1000 kg/m3 throughout.
The average pressure on a surface is the pressure at the centroid
(midpoint) of the surface, and is determined to be
Pave  PC  ghC  g (h / 2)

1 kN
 (1000 kg/m 3 )(9.81 m/s 2 )(5 / 2 m) 
 1000 kg  m/s 2


  24 .53 kN/m 2


Then the resultant hydrostatic force on each wall becomes
FR  Pave A  (24.53 kN/m2 )(6 m  5 m)  735.9 m
The line of action of the force passes through the pressure center,
which is 2h/3 from the free surface,
yP 
2h 2  (5 m)

 3.333 m
3
3
Taking the moment about point A and setting it equal to zero gives
M
A
0

FR ( s  y P )  Fridge AB
Solving for Fridge and substituting, the reaction force is determined to be
Fridge 
s  yP
AB
FR 
(1  3.333 ) m
(735 .9 kN)  638 kN
5m
3-58C
The horizontal component of the hydrostatic force acting on a curved surface is equal (in both
magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the
curved surface.
We could also integrate pressure along the surface, but the method discussed here is much simpler
and yields the same answer.
3-73
Assumptions : 1) Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in
calculations for convenience. 2) The weight of the gate is negligible.
We take the density of water to be 1000 kg/m3 throughout.
The average pressure on a surface is the pressure at the centroid
(midpoint) of the surface, and is determined to be
2.4 m
Pavg  ghC  g (h / 2)
W

1N
 (1000 kg/m )(9.81 m/s )( 2.4 / 2 m) 
 1 kg  m/s 2

3
2




s =2.1 m
A
 11,770 N/m 2
B
Then the resultant hydrostatic force acting on the dam becomes
FR  Pavg A  (11,770 N/m 2 )( 2.4 m  1.5 m)  42,380 N
FR
h=2.4 m
The line of action of the force passes through the
pressure center, which is 2h/3 from the free surface,
2h 2  (2.4 m)

 1.6 m
3
3
yP 
Taking the moment about point A and setting it equal to zero gives
M
A
0

FR ( s  y P )  W AB
Solving for W and substituting, the required weight is determined to be
W
s  yP
AB
FR 
(2.1  1.6) m
(42,380 N)  65,340 N
2.4 m
The corresponding mass is
m
F 65,340 N

g 9.81 m/s 2
 1 kg  m/s 2


1N


  6660 kg


3-87C
The upward force a fluid exerts on an immersed body is called the buoyant force. The buoyant force
is caused by the increase of pressure in a fluid with depth. The magnitude of the buoyant force
acting on a submerged body whose volume is V is expressed as FB   f gV . The direction of the
buoyant force is upwards, and its line of action passes through the centroid of the displaced
volume.
If the buoyant force is greater than the body’s weight, it floats
3-90C
The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as
FB   f gV , which is independent of the shape of the body. Therefore, the buoyant forces acting on
the cube and sphere made of copper submerged in water are the same since they have the same
volume.
The two objects have the same volume because they have the same mass and density.
3-94
Assumptions: 1 )The buoyancy force in air is negligible. 2) The top surface of the ice block is parallel
to the surface of the sea.
The specific gravities of ice and seawater are given to be 0.92 and 1.025, respectively, and thus the
corresponding densities are 920 kg/m3 and 1025 kg/m3.
The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of
vertical force balance from static equilibrium). Therefore, in this case the average density of the body
must be equal to the density of the fluid since

W = FB
V submerged
V total
 body gV total   fluid gV submerged

Ice block
 body
Sea
 fluid
W
h
The cross-sectional of a cube is constant, and thus the “volume
ratio” can be replaced by “height ratio”. Then,
hsubmerged
htotal

 body
 fluid


h
 ice
h  0.25  water

h
0.92

h  0.25 1.025
FB
where h is the height of the ice block below the surface. Solving for h gives
h
25 cm
(0.92 )(0.25)
 2.19 m
1.025  0.92
Note that 0.92/1.025 = 0.89756, so approximately 90% of the volume of an ice block remains under
water. For symmetrical ice blocks this also represents the fraction of height that remains under
water.
3-95
The density of shell is given to be 1600 kg/m3 and that for water is 1000 kg/m3.
R2
R1
The weight of the shell:
WS  mg  




4 3
4
R2  R13 g  1600 
0.06 3  0.05 3  9.81
3
3
WS  5.98 N
The buoyancy force:
Fb   w submerged  9810   submerged
Since WS  Fb ,
5.98  9810 submerged ,  submerged  5.096  10 4 m 3
 submerged


5.096  10 4
 100  67.4 %
4
0.06 3
3
3,91C
A submerged body whose center of gravity G is above the center of buoyancy B, which is the
centroid of the displaced volume, is unstable. But a floating body may still be stable when G is
above B since the centroid of the displaced volume shifts to the side to a point B’ during a
rotational disturbance while the center of gravity G of the body remains unchanged. If the point B’
is sufficiently far, these two forces create a restoring moment, and return the body to the original
position.
Stability analysis like this is critical in the design of ship hulls, so that they are least likely to capsize.
3-96
The density of water is 1000 kg/m3.
Analysis
40 cm
10 cm
20 cm
water
cord
From the figure below,
R
r
2R 40
and r 


 13.33 cm
3
3
30 20
R=20 cm
The displaced volume of water is
1
1
  r 2 h    0.1333 2  0.2  3.72 10 3 m3
3
3
Therefore the buoyancy force acting on the cone is
Fb    9810 3.72 103  36.5 N
10 cm
r
20 cm
For the static equilibrium, we write
F  Wc  Fb
F  16.5  36.5
F  36.5 16.5  20 N
3-98
Assumptions: 1) The rock is c completely submerged in water. 2) The buoyancy force in air is
negligible.
The density of granite rock is given to be 2700 kg/m3. We take the density of water to be 1000 kg/m3.
The weight and volume of the rock are

1N
W  mg  (170 kg)(9.81 m/s 2 )
2
 1 kg  m/s
170 kg
m
V  
 0.06296 m 3
 2700 kg/m 3
The buoyancy force acting on the rock is

  1668 N


Water
W
FB
Fnet =W - FB
 1N

FB   water gV  1000 kg/m 3  9.81 m/s 2  0.06296 m 3  
 618 N
2 
 1 kg  m/s 
The weight of a body submerged in water is equal to the weigh of
the body in air minus the buoyancy force,
Win water  Win air  FB  1668  618  1050 N
This force corresponds to a mass of m 

Win water
1050 N  1 N

 107 kg . Therefore, a
2 
2 
g
9.81 m/s  1 kg  m/s 
person who can lift 107 kg on earth can lift this rock in water.
3-100
Assumptions 1 The dynamic effects of the waves are disregarded. 2 The buoyancy force in air is
negligible.
The density of sea water is given to be 1.031000 = 1030 kg/m3. We take the density of water to be
1000 kg/m3.
Analysis
The weight of the unloaded boat is

1 kN
Wboat  mg  (8560 kg)(9.81 m/s 2 )
 1000 kg  m/s 2


  84.0 kN


The buoyancy force becomes a maximum when the entire hull of the
boat is submerged in water, and is determined to be

1 kN
FB,lake   lake gV  (1000 kg/m 3 )(9.81 m/s 2 )(180 m 3 )
 1000 kg  m/s 2

FB,sea

1 kN
  sea gV  (1030 kg/m 3 )(9.81 m/s 2 )(180 m 3 )
 1000 kg  m/s 2


  1766 kN



  1819 kN


The total weight of a floating boat (load + boat itself) is equal to
the buoyancy force. Therefore, the weight of the maximum load
is
Wload, lake  FB , lake  Wboat  1766  84  1682 kN
Wload,sea  FB,sea  Wboat  1819  84  1735 kN
The corresponding masses of load are
mload,lake 
Wload,lake
g

1682 kN  1000 kg  m/s 2
1 kN
9.81 m/s 2 

  171,500 kg


Wload
FB
Wboat
mload,sea 
Wload,lsea
g

1735 kN  1000 kg  m/s 2
1 kN
9.81 m/s 2 

  176,900 kg


Note that this boat can carry nearly 5400 kg more load in the sea than it can in fresh water. Fullyloaded boats in sea water should expect to sink into water deeper when they enter fresh water, such
as a river where the port may be.
3-101C
A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no
motion between fluid layers relative to each other) in the fluid body.
When there is no relative motion between fluid particles, there are no viscous stresses, and pressure
(normal stress) is the only stress.
3-105
Assumptions: 1) The road is horizontal so that acceleration has no vertical
component (az = 0). 2) Effects of splashing, breaking, driving over bumps,
and climbing hills are assumed to be secondary, and are not considered. 3)
The acceleration remains constant.
We take the x-axis to be the direction of motion, the z-axis to be the upward
vertical direction. The tangent of the angle the free surface makes with the
horizontal is
tan  
ax
g  az
Solving for ax and substituting,
a x  ( g  a z ) tan  (9.81 m/s2  0) tan12  2.09 m/s2
Discussion Note that the analysis is valid for any fluid with constant density since we used no
information that pertains to fluid properties in the solution.
3-106
Assumptions 1 The acceleration
remains constant. 2 Water is an
incompressible substance.
Properties We take the density
of water to be 1000 kg/m3.
Analysis
The pressure difference between two points 1 and 2 in an incompressible fluid is given by
or P1  P2   ( g  a z )(z 2  z1 )
P2  P1  a x ( x2  x1 )   ( g  a z )(z 2  z1 )
since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have P2  Patm and
z 2  z1  h and thus
P1, gage  Pbottom   ( g  a z )h
Tank A: We have az = 0, and thus the pressure at the bottom is

1 kN
PA, bottom  gh A  (1000 kg/m 3 )(9.81 m/s 2 )(8 m) 
2
 1000 kg  m/s

  78 .5 kN/m 2


Tank B: We have az = +5 m/s2, and thus the pressure at the bottom is

1 kN
PB, bottom   ( g  a z )h B  (1000 kg/m 3 )( 9.81  5 m/s 2 )( 2 m) 
1000
kg  m/s 2


  29 .6 kN/m 2


Therefore, tank A has a higher pressure at the bottom.
We can also solve this problem quickly by examining the relation Pbottom   ( g  a z )h . Acceleration
for tank B is about 1.5 times that of Tank A (14.81 vs 9.81 m/s2), but the fluid depth for tank A is 4
times that of tank B (8 m vs 2 m). Therefore, the tank with the larger acceleration-fluid height
product (tank A in this case) will have a higher pressure at the bottom.
3-111
Solution
Two water tanks
filled with water, one stationary
and the other moving upwards at
constant acceleration. The tank
with the higher pressure at the
bottom is to be determined.
Assumptions 1 The acceleration
remains constant. 2 Water is an
incompressible substance.
Tank
A
8
m
2
Tank
B

z
Wat 1
er
2
m
0
az = 5
m/s2
2

Wat  1
er
g
Properties We take the density
of water to be 1000 kg/m3.
Analysis
The pressure difference between two points 1 and 2 in an incompressible fluid is given by
or P1  P2   ( g  a z )(z 2  z1 )
P2  P1  a x ( x2  x1 )   ( g  a z )(z 2  z1 )
since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have P2  Patm and
z 2  z1  h and thus
P1, gage  Pbottom   ( g  a z )h
Tank A: We have az = 0, and thus the pressure at the bottom is

1 kN
PA, bottom  gh A  (1000 kg/m 3 )(9.81 m/s 2 )(8 m) 
2
 1000 kg  m/s

  78 .5 kN/m 2


Tank B: We have az = +5 m/s2, and thus the pressure at the bottom is

1 kN
PB, bottom   ( g  a z )h B  (1000 kg/m 3 )( 9.81  5 m/s 2 )( 2 m) 
2
 1000 kg  m/s
Therefore, tank A has a higher pressure at the bottom.

  29 .6 kN/m 2

