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Problem 2.84
Panel AB in Fig. P2.84 is a parabola with its maximum at point A . It is 150 cm wide into the
paper. Neglect atmospheric pressure. Find (a) the vertical and (b) the horizontal water forces on
the panel.
Solution 2.84
(b) The horizontal force is calculated from the vertical projection of the panel (from point A
down to the bottom). This is a rectangle, 75 cm by 150 cm, and its centroid is 37.5 cm below A,
or (25 + 37.5) = 62.5 cm below the surface. Thus
Problem 2.85
Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle
panel at the bottom of the water tank in Fig. P2.85.
Solution 2.85
The horizontal component is
Problem 2.86
The quarter circle gate BC in Fig. P2.86 is hinged at C. Find the horizontal force P required to
hold the gate stationary. Neglect the weight of the gate.
Solution 2.86
The horizontal component of water force is
3
H CG
F h A (9790 N/m )(1 m)[(2 m)(3 m)] 58,740 N
= = =
Problem 2.87
The bottle of champagne (SG = 0.96) in Fig. P2.87 is under pressure as shown by the mercury-
manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the
bottom of the bottle.
Solution 2.87
First, from the manometer, compute the gage pressure at section AA in the champagne 6 inches
above the bottom:
Problem 2.88*
Gate ABC is a circular arc, sometimes called a Tainter gate , which can be raised and lowered by
pivoting about point O . See Fig. P2.88. For the position shown, determine (a) the hydrostatic
force of the water on the gate and (b) its line of action. Does the force pass through point O ?
Solution 2.88
The horizontal hydrostatic force is based on vertical projection:
Problem 2.89
The tank in Fig. P2.89 contains benzene and is pressurized to 200 kPa (gage) in the air gap.
Determine the vertical hydrostatic force on circular-arc section AB and its line of action.
Solution 2.89
Assume unit depth into the paper. The vertical force is the weight of benzene plus the force due to
the air pressure:
Problem 2.90
The tank in Fig. P2.90 is 120 cm long into the paper. Determine the horizontal and vertical
hydrostatic forces on the quarter-circle panel AB. The fluid is water at 20C. Neglect atmospheric
pressure.
Solution 2.90
For water at 20C, take
= 9790 N/m3.
Problem 2.91
The hemispherical dome in Fig. P2.91 weighs 30 kN and is filled with water and attached to the
floor by six equally-spaced bolts. What is the force in each bolt required to hold down the
dome?
Solution 2.91
Assuming no leakage, the hydrostatic force required equals the weight of missing water, that is,
the water in a 4-m-diameter cylinder, 6 m high, minus the hemisphere and the small pipe:
Problem 2.92
A 4-m-diameter water tank consists of two half-cylinders, each weighing 4.5 kN/m, bolted
together as in Fig. P2.92. If the support of the end caps is neglected, determine the force induced
in each bolt.
Solution 2.92
Consider a 25-cm width of upper cylinder, as seen below. The water pressure in the bolt plane is
1
p h (9790)(4) 39160 Pa
= = =
Problem 2.93*
In Fig. P2.93 a one-quadrant spherical shell of radius R is submerged in liquid of specific weight
and depth h R. Find an analytic expression for the resultant hydrostatic force, and its line of
action, on the shell surface.
Solution 2.93
The two horizontal components are identical in magnitude and equal to the force on the quarter-
circle side panels, whose centroids are (4R/3
) above the bottom:
Problem 2.94
Find an analytic formula for the vertical and horizontal forces on each of the semicircular panels
AB in Fig. P2.94. The width into the paper is b. Which force is larger? Why?
Solution 2.94
It looks deceiving, since the bulging panel on the right has more water nearby, but these two
forces are the same, except for their direction. The left-side figure is the same as Example 2.9,
22
Problem 2.95*
The uniform body A in Fig. P2.95 has width b into the paper and is in static equilibrium when
pivoted about hinge O . What is the specific gravity of this body if (a) h = 0 and (b) h = R ?
Solution 2.95
The water causes a horizontal and a vertical force on the body, as shown:
,
23
H
RR
F Rb at above O
=
Problem 2.96
In Fig. P2.96 the curved section AB is 5 m wide into the paper and is a 60º circular arc of radius
2 m. Neglecting atmospheric pressure, calculate the vertical and horizontal hydrostatic forces on
arc AB.
Solution 2.96
For water take γ = 9790 N/m3. Find the distances AC and BC:
Problem 2.97
The contractor ran out of gunite mixture and finished the deep corner, of a 5-m-wide swimming
pool, with a quarter-circle piece of PVC pipe, labeled AB in Fig. P2.97. Compute the horizontal
and vertical water forces on the curved panel AB.
Solution 2.97
For water take
= 9790 N/m3. (a) The horizontal force relates to the vertical projection of the
curved panel AB:
m
Problem 2.98
The curved surface in Fig. P2.98 consists of two quarter-spheres and a half cylinder. A side view
and front view are shown. Calculate the horizontal and vertical forces on the surface.
Solution 2.98
For water take γ = 9790 N/m3. The horizontal force involves the projected area, in the front view
– two half-circles and a square:
Problem 2.99
The mega-magnum cylinder in Fig. P2.99 has a hemispherical bottom and is pressurized with air
to 75 kPa (gage) at the top. Determine (a) the horizontal and (b) the vertical hydrostatic forces on
the hemisphere, in lbf.
Solution 2.99
Since the problem asks for BG units, convert the air pressure to BG:
75,000 Pa 47.88 = 1566lbf/ft2.
Problem 2.100
Pressurized water fills the tank in Fig. P2.100. Compute the net hydrostatic force on the conical
surface ABC.
Solution 2.100
The gage pressure is equivalent to a fictitious water level
h = p/
= 150000/9790 = 15.32 m above the gage or 8.32 m above AC. Then the vertical force on
the cone equals the weight of fictitious water above ABC:
