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Fig. P2.39
Problem 2.40
In Fig. P2.40, if pressure gage A reads 20 lbf/in2 absolute, find the pressure in the closed air
space B. The manometer fluid is Meriam red oil, SG = 0.827
Solution 2.40
For water take γ= 62.4 lbf/ft2. Neglect hydrostatic changes in the air. Proceed from A to B:
Problem 2.41
The system in Fig. P2.41 is at 20C. Determine the pressure at point A in lbf/ft2.
Solution 2.41
Take the specific weights of water and mercury from Table 2.1. Write the hydrostatic formula
Problem 2.42
Very small pressure differences pA − pB can be measured accurately by the two-fluid differential
manometer in Fig. P2.42. Density ρ2 is only slightly larger than that of the upper fluid ρ1 . Derive
an expression for the proportionality between h and pA − pB if the reservoirs are very large.
Solution 2.42
Apply the hydrostatic formula from A to B:
Problem 2.43
The traditional method of measuring blood pressure uses a sphygmomanometer, first recording
the highest (systolic) and then the lowest (diastolic) pressure from which flowing “Korotkoff”
sounds can be heard. Patients with dangerous hypertension can exhibit systolic pressures as high
as 5 lbf/in2. Normal levels, however, are 2.7 and 1.7 lbf/in2, respectively, for systolic and diastolic
pressures. The manometer uses mercury and air as fluids.
(a) How high in cm should the manometer tube be?
(b) Express normal systolic and diastolic blood pressure in millimeters of mercury.
Solution 2.43
(a) The manometer height must be at least large enough to accommodate the largest systolic
pressure expected. Thus apply the hydrostatic relation using 5 lbf/in2 as the pressure,
Problem 2.44
Water flows downward in a pipe at 45°, as shown in Fig. P2.44. The pressure drop p1 − p2 is
partly due to gravity and partly due to friction. The mercury manometer reads a 6-in height
difference. What is the total pressure drop p1 − p2 in lbf/in2 ? What is the pressure drop due to
friction only between 1 and 2 in lbf/in2 ? Does the manometer reading correspond only to friction
drop? Why?
Solution 2.44
Let “h” be the distance down from point 2 to the mercury-water interface in the right leg. Write
the hydrostatic formula from 1 to 2:
Problem 2.45
In Fig. P2.45, determine the gage pressure at point A in Pa. Is it higher or lower than
atmospheric?
Solution 2.45
Take
= 9790 Nm3 for water and 133100 Nm3 for mercury. Write the hydrostatic formula
between the atmosphere and point A:
A atm
Problem 2.46
In Fig. P2.46 both ends of the manometer are open to the atmosphere. Estimate the specific
gravity of fluid X.
Solution 2.46
The pressure at the bottom of the manometer must be the same regardless of which leg we approach
through, left or right:
Problem 2.47
The cylindrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an
exit pressure of 175 kPa. At the instant shown, the air pressure is 110 kPa and H = 35 cm. The
pump stops when it can no longer raise the water pressure. For isothermal air compression,
estimate H at that time.
Solution 2.47
At the end of pumping, the bottom water pressure must be 175 kPa:
Problem 2.48
The system in Fig. P2.48 is open to 1 atm on the right side. (a) If L = 120 cm, what is the air
pressure in container A? (b) Conversely, if pA = 135 kPa, what is the length L?
Solution 2.48
(a) The vertical elevation of the water surface in the slanted tube is (1.2m)(sin55) = 0.983 m.
Then the pressure at the 18-cm level of the water, point D, is
Problem 2.49
3
101350 (9790 )(0.983 0.18 ) 109200
D atm water
N
p p z Pa m Pa
m
= + = + − =
Conduct the following experiment to illustrate air pressure. Find a thin wooden ruler
(approximately 1 ft in length) or a thin wooden paint stirrer. Place it on the edge of a desk or
table with a little less than half of it hanging over the edge lengthwise. Get two full-size sheets of
newspaper; open them up and place them on top of the ruler, covering only the portion of the
ruler resting on the desk as illustrated in Fig. P2.49.
( a ) Estimate the total force on top of the newspaper due to air pressure in the room.
( b ) Careful! To avoid potential injury, make sure nobody is standing directly in front of the
desk. Perform a karate chop on the portion of the ruler sticking out over the edge of the desk.
Record your results.
( c ) Explain your results.
Solution 2.49
(a) Newsprint is about 27 in (0.686 m) by 22.5 in (0.572 m). Thus the force is:
Fig. P2.48
Problem 2.50
A small submarine, with a hatch door 30 inches in diameter, is submerged in seawater. (a) If the
water hydrostatic force on the hatch is 69,000 lbf, how deep is the sub? (b) If the sub is 350 ft
deep, what is the hydrostatic force on the hatch?
Solution 2.50
In either case, the force is pCGAhatch. Stay with BG units. Convert 30 inches = 2.5 ft. For
Problem 2.51
Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure,
compute the force F on the gate and its center-of-pressure position X.
Solution 2.51
The centroidal depth of the gate is
Problem 2.52
Example 2.5 calculated the force on plate AB and its line of action, using the moment-of-inertia
approach. Some teachers say it is more instructive to calculate these by direct integration of the
pressure forces. Using Figs. 2.52 and E2.5a, (a) find an expression for the pressure variation
p(
) along the plate; (b) integrate this pressure to find the total force F; (c) integrate the
moments about point A to find the position of the center of pressure.
Solution 2.52
(a) Point A is 9 ft deep, and point B is 15 ft deep, and
= 64 lbf/ft3.
Thus pA = (64lbf/ft3)(9ft) = 576 lbf/ft2 and pB = (64lbf/ft3)(15ft) = 960 lbf/ft2. Along the 10-ft
Problem 2.53
The Hoover Dam, in Arizona, encloses Lake Mead, which contains 10 trillion gallons of water.
The dam is 1200 ft wide and the lake is 500 ft deep. (a) Estimate the hydrostatic force on the
dam, in MN. (b) Explain how you might analyze the stress in the dam due to this hydrostatic
force.
Solution 2.53
Convert to SI. The depth down to the centroid is 250 ft = 76.2 m. A crude estimate of the dam’s
Problem 2.54
In Fig. P2.54, the hydrostatic force F is the same on the bottom of all three containers, even
though the weights of liquid above are quite different. The three bottom shapes and the fluids are
the same. This is called the hydrostatic paradox. Explain why it is true and sketch a free-body of
each of the liquid columns.
Solution 2.54
The three freebodies are shown below. Pressure on the side-walls balances the forces. In (a),
Problem 2.55
Gate AB in Fig. P2.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. The
water is at 20°C. Compute (a) the force on stop B; and (b) the reactions at A if h = 9.5 ft.
Solution 2.55
The centroid of AB is 2.0 ft below A, hence the centroidal depth is
Problem 2.56
In Fig. P2.55, gate AB is 5 ft wide into the paper, and stop B will break if the water force on it
equals 9200 lbf. For what water depth h is this condition reached?
Solution 2.56
The formulas must be written in terms of the unknown centroidal depth hCG:
Problem 2.57
The square vertical panel ABCD in Fig. 2.57 is submerged in water at 20ºC. Side AB is at least
1.7 m below the surface. Determine the difference between the hydrostatic forces on subpanels
ABD and BCD.
Solution 2.57
Let H be the distance down from the surface to line AB. Take γwater = 9790 N/m3. The
subpanel areas are each 0.18 m2. Then the difference between these two subpanel forces is
Problem 2.58
In Fig. P2.58, the cover gate AB closes a circular opening 80 cm in diameter. The gate is held
closed by a 200 - kg mass as shown. Assume standard gravity at 20°C. At what water level h will
the gate be dislodged? Neglect the weight of the gate.
Solution 2.58
The centroidal depth is exactly equal to h and force F will be upward on the gate. Dislodging
occurs when F equals the weight:
Problem 2.59*
Gate AB has length L, width b into the paper, is hinged at B, and has negligible weight. The
liquid level h remains at the top of the gate for any angle
. Find an analytic expression for the
force P, perpendicular to AB, required to keep the gate in equilibrium in Fig. P2.59.
Solution 2.59
The centroid of the gate remains at distance L2 from A and depth h2 below
Problem 2.60
In Fig. P2.60, vertical, unsymmetrical trapezoidal panel ABCD is submerged in fresh water with
side AB 12 ft below the surface. Since trapezoid formulas are complicated, (a) estimate,
reasonably, the water force on the panel, in lbf, neglecting atmospheric pressure. For extra
credit, (b) look up the formula and compute the exact force on the panel.
Solution 2.60
For water, take γ = 62.4 lbf/ft3. The area of the panel is ½ (6+9)(8) = 60 ft2. (a) The panel
centroid should be slightly below the mid-panel, say, about 4.5 ft below AB. Then we estimate
Problem 2.61*
Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg, 1.2 m wide into the paper, hinged at A
and resting on smooth bottom B. All fluids are at 20C. For what water depth h will the force at
point B be zero?
Solution 2.61
Let
= 12360 Nm3 for glycerin and 9790 Nm3 for water. The centroid of
AB is 0.433 m vertically below A, so hCG = P2.0 − 0.433 = 1.567 m, and we may compute the
glycerin force and its line of action:
Problem 2.62
Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper, hinged at B with a stop at A.
The gate is 1-in-thick steel, SG = 7.85. Compute the 20°C water level h for which the gate
will start to fall.
Solution 2.62
Only the length (h csc 60) of the gate lies below the water. Only this part contributes to the
hydrostatic force shown in the freebody below.
