11–33
11–43
Solution An elastic air balloon submerged in water is attached to the base of the tank. The change in the tension force
of the cable is to be determined when the tank pressure is increased and the balloon diameter is decreased in accordance
with the relation P = CD-2.
Assumptions 1 Atmospheric pressure acts on all surfaces, and thus it can be
ignored in calculations for convenience. 2 Water is an incompressible fluid. 3
The weight of the balloon and the air in it is negligible.
Analysis The tension force on the cable holding the balloon is determined
from a force balance on the balloon to be
B
balloon
B
cable FWFF
P1=100 kPa
D1=30 cm
11–34
11–44
P1=0.1 “MPa”
11–35
11–45
Solution A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point
B. The force exerted to the plate by the ridge is to be determined.
Assumptions Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience.
11–46
Solution A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point
B. The force exerted to the plate by the ridge is to be determined.
Assumptions Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience.
m 1.33 3
)m 2(2
3
2h
yP
11–47E
Solution A semicircular tunnel is to be built under a lake. The total hydrostatic force acting on the roof of the tunnel
is to be determined.
Assumptions Atmospheric pressure acts on both sides of the
tunnel, and thus it can be ignored in calculations for
tunnel) theof sideeach (on lbf 10398.1
ft/slbm 32.2
lbf 1
ft) 800ft ft)(20 2/20130)(ft/s 2.32)(lbm/ft 4.62(
)2/(
8
2
23
ARsgAghAPFF CavexH
lbf 10596.2
ft/slbm 32.2
lbf 1
ft) 800ft ft)(40 130)(ft/s 2.32)(lbm/ft 4.62(
8
2
23
top
AghAghAPF Cavey
Weight of fluid block on each side within the control volume (downward):
lbf 1
ft) 2000)(4/(
223
22
RRggmgW
V
= 2.596 108 lbf. A more conservative approximation would be to estimate the force on the bottom of the lake if the tunnel
were not there. This yields FV = 2.995 108 lbf. The actual force is between these two estimates, as expected.
Fy
11–48
Solution A hemispherical dome on a level surface filled with water is to be lifted by attaching a long tube to the top
and filling it with water. The required height of water in the tube to lift the dome is to be determined.
Assumptions 1 Atmospheric pressure acts on both sides of the dome, and thus it can be ignored in calculations for
convenience. 2 The weight of the tube and the water in it is negligible.
R
R
Rm
R
R
mm
hdomewaterdome
2
3
2
]6/4[
R = 2 m
11–38
11–49
Solution The water in a reservoir is restrained by a triangular wall. The total force (hydrostatic + atmospheric) acting
on the inner surface of the wall and the horizontal component of this force are to be determined.
Assumptions 1 Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience. 2 Friction at the hinge is negligible.
2
8
9.64 10 N
Noting that
m 77.11
N 1
m/skg 1
60sin)m/s 81.9)(kg/m 1000(
N/m 000,100
60sin
2
23
2
0
g
P
m 17.1
m 77.11
2
m 87.28
012
m) 87.28(
2
m 87.28
0
sin2
12
2
2
0
2
g
P
b
s
bb
sy p
11–50
Solution An iceberg floating in seawater is considered. The volume fraction of the iceberg submerged in seawater is
to be determined, and the reason for their turnover is to be explained.
Assumptions 1 The buoyancy force in air is negligible. 2 The density of iceberg and seawater are uniform.
Analysis (a) 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,
W = FB
submergedfluidtotalbody
VV
gg
88%or 880.0
1042
917
seawater
iceberg
fluid
body
total
submerged
V
V
Therefore, 88% of the volume of the iceberg is submerged in this case.
(b) Heat transfer to the iceberg due to the temperature difference between the
seawater and an iceberg causes uneven melting of the irregularly shaped iceberg.
The resulting shift in the center of mass causes the iceberg to turn over.
Discussion The submerged fraction depends on the density of
seawater, and this fraction can differ in different seas.
Sea
FB
W
Iceberg
11–40
11–51
Solution The density of a wood log is to be measured by tying lead weights to it until both the log and the weights
are completely submerged, and then weighing them separately in air. The average density of a given log is to be determined
by this approach.
Analysis The weight of a body is equal to the buoyant force when the body is floating in a fluid while being
completely submerged in it (a consequence of vertical force balance from static equilibrium). In this case the average
Therefore,
lead log lead log
total
water log lead
total lead log water
ave
m m m m
m
VV
V V V
where
33
lead
lead 3
lead
34 kg 3.0089 10 m
11,300 kg/m
m
V
Lead, 34 kg
Log, 1540 N
FB
Water
11–52
Solution A rectangular gate that leans against the floor with an angle of 45 with the horizontal is to be opened from
its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be
determined.
m 2m 5.0
2
m 3
5.0
2
m 7071.0
45sin
m 5.0
and m 243.4
45sin
m 3
h
h
sb
C
The average pressure on a surface is the pressure at the centroid (midpoint) of
the surface, and multiplying it by the plate area gives the resultant hydrostatic on
3 m
A
kN 5.499
m/skg 1000
The distance of the pressure center from the free surface of water along the plane of
the gate is
m 359.3
)2/243.47071.0(12
243.4
2
243.4
7071.0
)2/(122
22
bs
bb
syP
The distance of the pressure center from the hinge at point B is
m 652.27071.0359.3 syL PP
Taking the moment about point B and setting it equal to zero gives
2/ 0 FbLFM PRB
Solving for F and substituting, the required force to overcome the pressure is
kN 4.624
m 4.243
m) kN)(2.652 5.499(2
2 b
LF
FPR
In addition to this, there is the weight of the gate itself, which must be added. In the 45o direction,
kN 942.1)45cos(
m/skg 1000
kN 1
)m/s 81.9(kg) 280()45cos()45cos( 2
2
mgWFgate
Thus, the total force required in the 45o direction is the sum of these two values,
direction 45 in the kN 3.626942.14.624 kN 626
total
F
Discussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the
required force can be reduced by applying the force at a lower point on the gate. The weight of the gate is nearly negligible
compared to the pressure force in this example; in reality, a heavier gate would probably be required.
FR
B
0.5 m
11–42
11–53
Solution A rectangular gate that leans against the floor with an angle of 45 with the horizontal is to be opened from
its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be
determined.
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
11-54
11-55
Solution We are to discuss how to measure the volume of a rock without using any volume measurement devices.
Analysis The volume of a rock can be determined without using any volume measurement devices as follows: We
weigh the rock in the air and then in the water. The difference between the two weights is due to the buoyancy force, which
techniques.