Problem 2.98
The Altus dam in the figure below is made of
concrete with a density of 3
150 lbm/ft . The
coefficient of friction
µ
between the base of the
dam and the foundation is 0.65. Is the dam likely
to slide downstream? Consider a unit length of
the dam
()
b = 1 ft .
Solution 2.98
The total vertical force acting downward is
Using the results of Problem 2.97
The dam weights are
Headwater
10′
87′
4′
Tailwater
Slope =
0.1:1
EI. 1553.00′
EI. 1555.00′
EI. 1564.00′
Slope = 0.6:1
EI. 1475.00
FHV
w1w3
W2
Problem 2.99
Find the magnitude and location of
the net horizontal force on the gate
shown in the figure below. The gate
width is 5.0 m.
Solution 2.99
The hydrostatic horizontal force ‘
H
F is
The location ‘h of ‘
H
F is
The hydrostatic horizontal force ”
H
F is
3.0 m
2.0 m
Water Water
2.0 m
A
B
1.0 m
4.0 m
45°
The location of h of H
F is
The magnitude of the net horizontal force H
F
The location of the net horizontal force F above the base is denoted by h and is found by
noting the moment of the resultant is equal to the moment of the components or
DISCUSSION Note that the resultant of the two opposing forces is not located between
Problem 2.100
Find the magnitude and location of the net vertical force on the gate shown in the figure
below. The gate width is 5.0 m.
Solution 2.100
The hydrostatic vertical force ‘
V
F is the weight of the
3.0 m
2.0 m
Water Water
2.0 m
A
B
1.0 m
4.0 m
45°
ℓ
W
BB
The hydrostatic vertical force ”
V
F is the weight of the water above the gate to the level B-B.
The magnitude of the vertical force V
F
The location of F is found by first finding the locations of ”‘
and .
VV
FF First
Recognizing that =
1gives
2
The location of the resultant force from the left side of the gate is denoted by V and is
found from
DISCUSSION Noting that the resultant vertical force, =98100 N
V
F is the weight of a
volume of water measuring 2.0 m=, width 5.0 m , and height 1.0 m=, is there a quick way
to find ?
V
F
Problem 2.101
Find the total vertical force on the cylinder shown in the figure below.
Solution 2.101
The net force F on the cylinder is due to the water and is
Then
Water
18 cm
5 cm
D
= 6 cm
p
atm
d
=
3 cm
3 cm
H
x
H
y
ℓ
Problem 2.102
A 3-m -wide, 8-m -high rectangular gate is located at the end of a rectangular passage that
is connected to a large open tank filled with water as shown in the figure below. The gate is
hinged at its bottom and held closed by a horizontal force, H
F
, located at the center of the
gate. The maximum value for H
F is
3
500 kN. (a) Determine the maximum water depth,
h, above the center of the gate that can exist without the gate opening. (b) Is the answer the
same if the gate is hinged at the top? Explain your answer.
Solution 2.102
For gate hinged at bottom
()
HH
and
so that
Hinge
F
H
h
4 m
4 m
For gate hinged at top
and
Problem 2.103
A gate having the cross section shown in the figure below is 4 ft wide and is hinged at C.
The gate weighs
1
8,000 lb, and its mass center is 1.67 ft to the right of the plane BC.
Determine the vertical reaction at A on the gate when the water level is 3 ft above the base.
All contact surfaces are smooth.
Solution 2.103
γ
=
1c
FhA where =1.5ft
c
h
and acts at the center of the bottom gate surface.
For equilibrium,
3 ft
9 ft
5 ft
Hinge
AB
C
Water surface
Cy
Problem 2.104
The massless, 4-ft -wide gate shown in the figure below pivots about the frictionless hinge O.
It is held in place by the 2000 lb counterweight,
W
. Determine the water depth,
h
.
Solution 2.104
Thus,
To locate R
F,
Pivot O
Gate
2 ft
h
3 ft
Width = 4 ft
Water
𝒲
FR
Problem 2.105
A −200 lb homogeneous gate 10 ft wide and 5 ft long is hinged at point A and held in place
by a 12-ft -long brace as shown in the figure below. As the bottom of the brace is moved to
the right, the water level remains at the top of the gate. The line of action of the force that
the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted
on the gate by the brace as a function of the angle of the gate,
θ
, for
θ
≤≤ °0 90. (b) Repeat
the calculations for the case in which the weight of the gate is negligible. Comment on the
result as
θ
→0.
Solution 2.105
(a) For the free-body-diagram of the gate (see figure),
L
and
12 ft
Brace
Gate
5 ft
Water
A
Moveable
stop
θ
12 ft = L
Gate
= 5 ft
Water
Moveable
stop
ℓ
3
ϕ
FR
FB
ℓ
where w is the gate width. Thus, Eq. (1) can be written as
so that
φ
Since
φ
θ
=sin sin
L and =5ft, =12 ftL
φ
(b) For =0W, Eq. (3) reduces to
φ
θ
(degree) F(B) (lb) (W=200 lb) F(B) (lb) (W=0 lb)
90.0 2843 2843
85.0 2745 2736
80.0 2651 2633
45.0 2165 2085
40.0 2125 2032
35.0 2094 1985
30.0 2075 1945
2.0 3858 1836
As
θ
→0the value of B
F can be determined from Eq. (4),
3000
3500
4000
4500
Theta, deg
it follows that
φφ
θ
=− =−
2
22
5
cos 1 sin 1 sin
12
and therefore
Problem 2.106
An open tank has a vertical partition and on one side contains gasoline with a density
ρ
=3
700 kg / m at a depth of 4 m , as shown in the figure below. A rectangular gate that is
4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly
added to the empty side of the tank. At what depth,
h
, will the gate start to open?
Solution 2.106
γ
=
Rg g cg g
FhA; where g refers to gasoline.
()( )
=×
=×
32
3
kg m
700 9.81 2 m 4 m 2 m
ms
110 10 N=110 kN
Rg
F
Stop
Partition
Hinge
Water Gasoline
4 m
h
F
Rw
F
Rg
γ
Problem 2.107
A horizontal 2-m-diameter conduit is half filled with a liquid ( =1. 6SG ) and is capped at
both ends with plane vertical surfaces. The air pressure in the conduit above the liquid
surface is
2
00 kPa. Determine the resultant force of the fluid acting on one of the end caps,
and locate this force relative to the bottom of the conduit.
Solution 2.107
Thus,
A
1
~ area of end
Thus,
Problem 2.108
A 4-ft by 3-ft massless rectangular gate is used to close the end of the water tank shown in
the figure below. A 200-lb weight attached to the arm of the gate at a distance ℓ from the
frictionless hinge is just sufficient to keep the gate closed when the water depth is
2
ft, that
is, when the water fills the semicircular lower portion of the tank. If the water were deeper,
the gate would open. Determine the distance .
Solution 2.108
Gate
200 lb
Hinge Hinge
ℓ
Water
2 ft
radius
3 ft
4 ft
1 ft
ℓ
Hy
Hx