Problem 2.63
The tank in Fig. P2.63 has a 4-cm-diameter plug at the bottom on the right. All fluids are at
20°C. The plug will pop out if the hydrostatic force on it is 25 N. For this condition, what will be
the reading h on the mercury manometer on the left side?
Solution 2.63
The water depth when the plug pops out is
2
CG CG
(0.04)
F 25 N h A (9790)h 4
= = =
Problem 2.64*
Gate ABC in Fig. P2.64 has a fixed hinge line at B and is 2 m wide into the paper. The gate will
open at A to release water if the water depth is high enough. Compute the depth h for which the
gate will begin to open.
Solution 2.64
Let H = (h − 1 meter) be the depth down to the level AB. The forces on AB and BC are shown in
Problem 2.65*
Gate AB in Fig. P2.65 is semicircular, hinged at B, and held by a horizontal force P at point A.
What force P is required for equilibrium?
Solution 2.65
The centroid of a semi-circle is at 4R/3
1.273 m off the bottom, as shown in the sketch at
right. Thus it is 3.0 − 1.273 = 1.727 m down from the force P. The water force F is
Problem 2.66
Dam ABC in Fig. P2.66 is 30 m wide into the paper and is concrete (SG P2.40). Find the
hydrostatic force on surface AB and its moment about C. Assuming no seepage of water under the
dam, could this force tip the dam over? How does your argument change if there is seepage
under the dam?
Solution 2.66
The centroid of surface AB is 40 m deep, and the total force on AB is
CG
F h A (9790)(40)(100 30)
= =
Problem 2.67*
Generalize Prob. P2.66 as follows. Denote length AB as H , length BC as L , and angle ABC as θ.
Let the dam material have specific gravity SG. The width of the dam is b . Assume no seepage of
water under the dam. Find an analytic relation between SG and the critical angle θc for which the
dam will just tip over to the right. Use your relation to compute θ c for the special case SG = 2.4
(concrete).
Solution 2.67
By geometry, L = Hcos
and the vertical height of the dam is Hsin
. The force F on surface AB
is
(H/2)(sin
)Hb, and its position is at 2H/3 down from point A, as shown in the figure. Its
Problem 2.68
Isosceles triangle gate AB in Fig. P2.68 is hinged at A and weighs 1500 N. What horizontal force P
is required at point B for equilibrium?
Solution 2.68
The gate is 2.0/sin50 = 2.611 m long from A to B and its area is 1.3054 m2. Its centroid is 1/3 of the
way down from A, so the centroidal depth is 3.0 + 0.667 m. The force on the gate is
Problem 2.69
Consider the slanted plate AB of length L in Fig. P2.69. (a) Is the hydrostatic force F on the
plate equal to the weight of the missing water above the plate? If not, correct this hypothesis.
Neglect the atmosphere. (b) Can a “missing water” theory be generalized to curved surfaces of
this type?
Solution 2.69
(a) The actual force F equals the pressure at the centroid times the plate area: But the weight of
the “missing water” is
2
sin sin
22
CG plate CG
L
F p A h L b L b L b
= = = =
Problem 2.70
The swing-check valve in Fig. P2.70 covers a 22.86-cm diameter opening in the slanted wall.
The hinge is 15 cm from the centerline, as shown. The valve will open when the hinge moment is
50 N-m. Find the value of h for the water to cause this condition.
Solution 2.70
For water, take
= 9790 N/m3. The hydrostatic force on the valve is
Problem 2.71*
In Fig. P2.71 gate AB is 3 m wide into the paper and is connected by a rod and pulley to a
concrete sphere (SG = 2.40). What diameter of the sphere is just sufficient to keep the gate
closed?
Solution 2.71
The centroid of AB is 10 m down from the surface, hence the hydrostatic
force is
Problem 2.72
In Fig. P2.72 gate AB is circular. Find the moment of the hydrostatic force on this gate about
axis A.
Solution 2.72
The gate centroid is 3+1 = 4 m down from the surface. The hydrostatic force is thus
Problem 2.73
Weightless gate AB is 5 ft wide into the paper and opens to let fresh water out when the ocean
tide is falling. The hinge at A is 2 ft above the freshwater level. At what ocean level h will the
gate first open? Neglect the gate weight.
Solution 2.73
There are two different hydro-static forces and two different lines of action. On the water side,
w CG
F h A (62.4)(5)(10 5) 15600 lbf
= = =
Problem 2.74
Find the height H in Fig. P2.74 for which the hydrostatic force on the rectangular panel is the
same as the force on the semicircular panel below.
Solution 2.74
Find the force on each panel and set them equal:
Problem 2.75
The cap at point B on the 5-cm-diameter tube in Fig. P2.75 will be dislodged when the
hydrostatic force on its base reaches 22 lbf. For what water depth h does this occur?
Solution 2.75
Convert the cap force to SI units: 22 lbf x 4.4482 = 97.9 N. Then the “dislodging: pressure just
under cap B will be
Problem 2.76
Panel BC in Fig. P2.76 is circular. Compute (a) the hydrostatic force of the water on the panel,
(b) its center of pressure, and (c) the moment of this force about point B .
Solution 2.76
(a) The hydrostatic force on the gate is:
Problem 2.77
The circular gate ABC in Fig. P2.77 has a 1-m radius and is hinged at B . Compute the force P
just sufficient to keep the gate from opening when h = 8 m. Neglect atmospheric pressure.
Solution 2.77
The hydrostatic force on the gate is
2
CG
F h A (9790)(8 m)( m )
==
Problem 2.78
Panels AB and CD in Fig. P2.78 are each 120 cm wide into the paper. (a) Can you deduce, by
inspection, which panel has the larger water force? (b) Even if your deduction is brilliant,
calculate the panel forces anyway.
Solution 2.78
(a) The writer is unable to deduce by inspection which panel force is larger. CD is longer than
AB, but its centroid is not as deep. If you have a great insight, let me know.
Problem 2.79
Gate ABC in Fig. P2.79 is 1-m-square and hinged at B. It opens automatically when the water
level is high enough. Determine the lowest level h which the gate will open. Neglect atmospheric
pressure. Is this result independent of the liquid density?
Solution 2.79
The gate will open when the hydrostatic force F on the gate is above B, that is, when
Problem 2.80*
A concrete dam (SG = 2.5) is made in the shape of an isosceles triangle, as in Fig. P2.80.
Analyze this geometry to find the range of angles
for which the hydrostatic force will tend to
tip the dam over at point B. The width into the paper is b.
Solution 2.80*
The critical angle is when the hydrostatic force F causes a clockwise moment equal to the
counterclockwise moment of the dam weight W. The length L of the slanted side of the dam is
L = h/sin
. The force F is two-thirds of the way down this face. The moment arm of the
Problem 2.81
For the semicircular cylinder CDE in Example. 2.9, find the vertical hydrostatic force by integrating
the vertical component of pressure around the surface from
= 0 to
=
.
Solution 2.81
A sketch is repeated here. At any position
, as in Fig. P2.81, the vertical component of pressure
is p cos
. The depth down to this point is h+R(1– cos
), and the local pressure is
times this
depth.
Problem 2.82*
The dam in Fig. P2.82 is a quarter-circle 50 m wide into the paper. Determine the horizontal and
vertical components of hydrostatic force against the dam and the point CP where the resultant
strikes the dam.
Solution 2.82
The horizontal force acts as if the dam were vertical and 20 m high:
H CG vert
F h A
=
Problem 2.83*
Gate AB in Fig. P2.83 is a quarter-circle 10 ft wide into the paper and hinged at B. Find the force
F just sufficient to keep the gate from opening. The gate is uniform and weighs 3000 lbf.
Solution 2.83
The horizontal force is computed as if AB were vertical:
2
H CG vert
F h A (62.4)(4 ft)(8 10 ft )
= =