# Test Bank for Fundamentals of Hydraulic Engineering Systems, 5th Edition

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TEST QUESTIONS – CHAPTER #2
Short Answer Questions
1. State Pascalโs law.
Ans. A pressure applied at any point in a liquid at rest is transmitted equally and
undiminished in all directions to every other point in the liquid.
2. (T or F) The difference in pressure between any two points in still water is always equal to
the product of the density of water and the difference in elevation between the two points.
Ans. False โ specific weight, not density.
3. Gage pressure is defined as
a) the pressure measured above atmospheric pressure.
b) the pressure measured plus atmospheric pressure.
c) the difference in pressure between two points.
d) pressure expressed in terms of the height of a water column.
Ans. (a) is true
4. Some species of seals dive to depths of 400 m. Determine the pressure at that depth in N/m2
assuming sea water has a specific gravity of 1.03.
Ans. P = ฮณโh = (1.03)(9790 N/m3)(400 m) = 4.03โ106 N/m2
5. Pressure below the surface in still water (or hydrostatic pressure)
a) is linearly related to depth.
b) acts normal (perpendicular) to any solid surface.
c) is related to the temperature of the fluid.
d) at a given depth, will act equally in any direction.
e) all of the above.
f) (a) and (b) only.
Ans. (e)
6. (T or F) A single-reading manometer makes use of a reservoir of manometry fluid with a
large cross sectional area so that pressure calculations are only based on one reading.
Ans. True.
7. What is an open manometer?
Ans. A manometer is a pressure measurement device that utilizes fluids of known specific
gravity and differences in fluid elevations. An open manometer has one end open to the air.
8. (T or F) The total hydrostatic pressure force on any submerged plane surface is equal to the
product of the surface area and the pressure acting at the center of pressure of the surface.
Ans. False. The total hydrostatic pressure force on any submerged plane surface is equal to
the product of the surface area and the pressure acting at the centroid of the plane surface.
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9. A surface of equal pressure requires all of the following except:
a) points of equal pressure must be at the same elevation.
b) points of equal pressure must be in the same fluid.
c) points of equal pressure must be interconnected.
d) points of equal pressure must be at the interface of immiscible fluids.
Ans. (d); points of equal pressure do not need to be at and interface of fluids.
10. The center of pressure on inclined plane surfaces is:
a) at the centroid.
b) is always above the centroid.
c) is always below the centroid.
d) is not related to the centroid.
Ans. (c); points of equal pressure do not need to be at and interface of fluids.
11. (T or F) The location of the centroid of a submerged plane area and the location where the
resultant pressure force acts on that area are identical.
Ans. False. The resultant force acts at the center of pressure.
12. The equation for the determination of a hydrostatic force on a plane surface and its location
are derived using all of the following concepts except
a) integration of the pressure equation b) moment of inertia concept
c) principle of moments
d) Newtonโs 2nd Law
Ans. Since this deals with hydrostatics (i.e., no acceleration), (d) is the answer.
13. The equation for the righting moment on a submerged body is M = WฮGMฮsin ฮธ, where GM
= MB โ GB or MB + GB. Under what conditions is the sum used instead of the difference?
Ans. Use the sum when the center of gravity is below the center of buoyancy.
14. Given the submerged cube with area (A) on each face, derive the buoyant force on the cube if
the depth (below the surface of the water) to the top of the cube is x and the depth to the
bottom of the cube is y. Show all steps.
Ans. Fbottom= PavgโA =ฮณโyโA; Ftop= PavgโA =ฮณโxโA; Fbottom-Ftop= ฮณโ(y-x)โA=ฮณโVol
15. A 3 ft x 3 ft x 3 ft wooden cube (specific weight of 37 lb/ft 3) floats in a tank of water. How
much of the cube extends above the water surface? If the tank were pressurized to 2 atm
(29.4 psi), how much of the cube would extend above the water surface? Explain.
Ans. โFy=0; W = B; (37 lb/ft3)(3 ft)3 = (62.3 lb/ft3)(3 ft)2(y); y = 1.78 ft Note: The draft
does not change with pressure. That is, the added pressure on the top of the cube would
be compensated by the increased pressure in the water under the cube.
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16. The derivation of the flotation stability equation utilizes which principles? Note: More than
one answer is possible.
a) moment of a force couple
b) moment of inertia
nd
c) Newtonโs 2 Law
d) buoyancy
Ans. It utilizes (a), (b), and (d).
17. Rotational stability is a major concern in naval engineering. Draw the cross section of the
hull of a ship and label the three important points (i.e., centers) which affect rotational
stability.
Ans. See Figure 2.16.
18. A 4 m (length) by 3 m (width) by 2 m (height) homogeneous box floats with a draft 1.4 m.
What is the distance between the center of buoyancy and the center of gravity?
Ans. G is 1 m up from bottom and B is 0.7 m up from bottom. Thus, GB = 0.3 m.
19. Determine the waterline moment of inertia about the width of a barge (i.e., used to assess
stability from side to side about its width) if it is 30 m long, 12 m wide, and 8 m high?
Ans. Io =(30m)(12m)3/12 = 4320 m4
20. (T or F) Floatation stability is dependent on the relative positions of the center of gravity and
the center of buoyancy.
Ans. True.
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Problems
1. Given the submerged, inclined rod with a top and bottom area (dA), a length of L, and an
angle of incline of ฮธ, derive an expression that relates the pressure on the top of the rod to the
pressure on the bottom. Show all steps and define all variables.
Ans. Because the prism is at rest, all forces acting upon it must be in equilibrium in all
directions. For the force components in the inclined direction, we may write
๏ฅ F = P dA โ P dA + ๏งLdAsin ๏ฑ = 0
x
A
B
Note that Lโsinฮธ = h is the vertical elevation difference between the two points. The
above equation reduces to
PB โ PA = ๏งh
2. Collapse depth (or crush depth) is the submerged depth that a submarine canโt exceed
without collapsing due to the surrounding water pressure. The collapse depth of modern
submarines is not quite a kilometer (730 m). Assuming sea water to be incompressible (S.G.
= 1.03), what is the crush depth pressure in N/m2 and psi (lb/in2). Is the pressure you
computed absolute or gage pressure?
Ans. P = ฮณโh; where ฮณ = (1.03)(9810 N/m3) = 1.01×104 N/m3 (using the specific weight of
water at standard conditions since water gets very cold at great depths)
P = ฮณโh = (1.01×104 N/m3)(730 m) = 7.37×106 N/m2 = 1,070 psi (gage pressure)
To get absolute pressure, atmospheric pressure must be added.
3. Mercury (Hg) is a preferred measurement fluid in simple barometers since its vapor pressure
is low enough to be ignored. In addition, it is so dense (S.G. = 13.6), the tube height can be
shortened considerably. If a barometric pressure reading is 29.9 inches of Hg, determine the
pressure (in lb/in.2) and the comparable reading if the measurement fluid was water..
Ans. From Eqโn 2.4: P = (๏งHg)(h)
P = (13.6)(62.3 lb/ft3)(29.9 in.)(1 ft/12 in.)3 =14.7 lb/in.2
Also, h = P/๏ง = [(14.7 lb/in.2) / (62.3 lb/ft3)]ยท(12 in./1 ft)3 = 408 in. = 34.0 ft
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4. A storage tank (6 m x 6m x 6m) is filled with water. Determine the force on the bottom and
on each side.
Ans. The force exerted on the tank bottom is equal to the pressure on the bottom times the
area of the bottom. P = ฮณโh = (9.79 kN/m3)(6 m) = 58.7 kN/m2
F = PโA = (58.7 kN/m2)(36 m2) = 2,110 N
The force exerted on the sides of the tank may be found in like manner (pressure times
the area). However, the pressure is not uniform on the tank sides since P = ฮณโh.
Therefore, the average pressure is required. Since the pressure is a linear relationship,
the average pressure occurs at half the depth. Now,
Pavg = ฮณโhavg = (9.79 kN/m3)(3 m) = 29.4 kN/m2
F = Pavg โA = (29.4 kN/m2)(36 m2) = 1,060 N
5. A water container consists of a 10-ft-high, 1-ft-diameter pipe welded on top of a
cube (3ft x 3ft x 3ft). The container is filled with water (20ยฐC). Determine the weight
of the water and the pressure forces on the bottom and sides of the bottom cube.
Ans. Wtotal = ๏งยทVol = (๏ง)[Volcube + Volpipe]
Wtotal = (62.3 lb/ft3)[(3 ft)3 + (๏ฐ)(0.50 ft)2(10 ft)] = 2,170 lb
Pbottom = ๏งh = (62.3 lb/ft3)(13 ft) = 810 lb/ft2; Fbottom = (810 lb/ft2)(9 ft2) = 7,290 lb
Note: The weight of the water is not equal to the force on the bottom. Why? (Hint:
Draw a free body diagram of the 3 ft x 3 ft x 3 ft water body labeling all vertical forces
acting on it. Donโt forget the pressure from the container top. Now, the side force is:
Pavg = ฮณโhavg = (62.3 lb/ft3)(11.5 ft) = 716 lb/ft2
F = Pavg โA = (716 lb/ft2)(9 ft2) = 6,440 lb
6. A weight of 5,400 lbs is to be raised by a hydraulic jack. If the large piston has an area of 120
in.2 and the small piston has an area of 2 in.2, what force must be applied through a lever
having a mechanical advantage of 6 to 1?
Ans. From Pascalโs law, the pressure on the small piston is equal to the pressure on the large.
Fsmall/Asmall = Flarge/Alarge
Fsmall = [(Flarge)(Asmall)]/(Alarge) = [(5400 lb)(2 in2)]/(120 in2) = 90 lb
๏ The applied force = 90 lb/6 = 15 lb based on the mechanical advantage of the lever.
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7. Carbon tetrachloride (manometer liquid, M, in the figure below, S.G. = 1.6) is poured into a
U-tube with both ends open to the atmosphere. Then water is poured into one leg of the Utube until the water column (y) is 20 cm high. Determine the height of the carbon
tetrachloride (h) above the water-carbon tetrachloride interface (point 1).
y
Note: y = height of
water. Air is above
the Hg at point D.
Ans. A surface of equal pressure surface can be drawn at the interface (1-2).
Therefore, P1 = P2; (0.2 m)(๏ง) = (h)(๏งM)
h = [(0.2 m)(๏ง)]/(๏งM) = [(0.2 m)(ฮณ)]/[(ฮณ)(SGM)] = (0.2 m) / (1.6)
h = 0.125 m = 12.5 cm
8. The manometer depicted below is mounted on a city water supply pipe to monitor water
pressure. However, the field engineer suspects the manometer reading of h = 3 feet (Hg) may
be incorrect. If the pressure in the pipe is measured independently and found to be 16.8
lb/in.2 (psi), determine the correct value of the reading h.
Ans. A surface of equal pressure can be drawn at the mercury-water meniscus. Therefore,
Ppipe + (2 ft)(๏ง) = (h)(๏งHg)
(16.8 lb/in2)(144 in2/ft2) + (2 ft)(๏ง) = (h)(๏ง)(S.G.Hg)
(2.42 x 103 lb/ft2) + (2 ft)(62.3 lb/ft3) = (h)(62.3 lb/ft3)(13.6)
h = 3.00 ft (manometer is correct)
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9. Manometer computations for the figure above yield a pressure of 16.8 lb/in.2 (psi) if h = 3 ft.
If the fluid in the pipe was oil (S.G. = 0.80) under the same pressure, would the manometer
measurements (2 ft and 3 ft) still be the same? If not, what would the new measurements be?
Ans. The measurements will not be the same since oil is now in the manometer instead of
water. A surface of equal pressure can be drawn at the mercury-oil interface.
Ppipe + (2 ft + ฮh)(๏งoil) = (3 ft + 2ฮh)(๏งHg)
This is based on volume conservation. If the mercury-oil meniscus goes down ฮh on
the right, it must climb up ฮh on the left making the total difference 2ฮh. Now
(2.42 x 103 lb/ft2) + (2 ft + ฮh)(0.80)(62.3 lb/ft3) = (3 ft + 2ฮh)(13.6)(62.3 lb/ft3)
ฮh = -0.0135 ft New measurements:
2 feet becomes 2 โ 0.0135 = 1.99 ft and 3 feet becomes 3 – 2(0.0135) = 2.97 ft
10. Determine the air pressure (in kPa and cm of Hg) in the sealed left tank depicted in the figure
below if EA = 32.5 m.
Ans. Using the โswim throughโ technique, start at the sealed right tank where the pressure
is known.
Then โswim throughโ the tanks and pipes, adding pressure when
โswimmingโ down and subtracting when โswimmingโ up until you reach the left tank
where the pressure is not known. The computations are as follows:
20 kN/m2 + (4.5 m)(9.79 kN/m3) – (2.5 m)(1.6)(9.79 kN/m3) – (5 m)(0.8)(9.79 kN/m3) = Pleft
Pleft = -14.3 kN/m2 (or -14.3 kPa). Now to determine the pressure in cm of Hg, since
P = ฮณh or h = P/ฮณ; h = (-14.3 kN/m2) / [(SGHg)(ฮณ)]
h = (-14.3 kN/m2) / [(13.6)(9.79 kN/m3)] = -0.107 m = 10.7 cm (Hg)
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11. A vertical gate keeps water from flowing in a triangular irrigation channel. The channel has a
4-m top width and a 3-m depth. If the channel is full, what is the magnitude of the hydrostatic
force on the triangular gate and its location?
Ans. The hydrostatic force and its locations are (based on Table 2.1):
F = ๏ง ๏ h ๏ A = = (9790 N/m3)[(3m)/(3)]โ[6 m2] = 5.87 x 104 N = 58.7 kN
yP =
๏
๏
I0
(4m)(3m) 3 / 36
+y=
+ 1.00 m
๏(4m)(3m) / 2๏(1.00 m)
Ay
yp = 1.50 m (depth to center of pressure)
12. A 3-ft square (plane) gate is mounted into an inclined wall (45ยฐ). The center of the gate is
located 4 feet (vertically) below the water surface. Determine the magnitude of the
hydrostatic force (in lbs) and its location with respect to the water surface along the incline.
Ans. F = ๏ง ๏ h ๏ A = (62.3 lb/ft3)(4 ft)(3 ft)2
F = 2,240 lbs; Now noting that y = h / sin 45๏ฐ = 4ft/(sin 45ยฐ) = 5.66 ft
yP =
๏
๏
I0
(3 ft)4 / 12
+y=
+ 5.66 ft
Ay
(3 ft)2 (5.66 ft)
๏
๏
yp = 5.79 ft (distance from water surface to the center of pressure along incline)
13. A circular gate is installed on a vertical wall as shown in the figure below. Determine the
horizontal force, P, necessary to hold the gate closed if the gate diameter is 6 feet and h = 7
feet. Neglect friction at the pivot.
Ans. F = ๏ง ๏ h ๏ A = (62.3 lb/ft3)[(7 ft)]โ[ฯ(6 ft)2/4] = 1.23 x 104 lbs
yP =
๏
๏
I0
๏ฐ (6 ft ) 4 / 64
+y=
+ 7 ft ; yp = 7.32 ft (depth to the center of pressure)
Ay
๏ฐ (6 ft ) 2 / 4 (7 ft )
๏
๏
Thus, summing moments: โ Mhinge = 0
P(3 ft) โ (1.23 x 104 lbs)(7.32 ft โ 7 ft) = 0; P = 1.31 x 103 lbs
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14. A vertical plate, composed of a square
and a triangle, is submerged so that its
upper edge coincides with the water
surface (Figure P2.5.5). What is the
value of the ratio H/L such that the
pressure force on the square is equal to
the pressure force on the triangle?
Ans. Fsquare = ๏ง ๏ h ๏ A = ฮณ(L/2)(L2) = (ฮณ/2)โL3
Ftri = ๏ง ๏ h ๏ A = ฮณ(L+H/3)(LH/2) = (ฮณ/2)[L2H + LH2/3]; Setting the two forces equal:
Fsquare = Ytri ; substituting yields; (ฮณ/2)โL3 = (ฮณ/2)[L2H + LH2/3]
L2 – HL – H2/3 = 0; divide by H2 and solve quadratic
(L/H)2 โ (L/H) – 1/3 = 0; L/H = 1.26 or H/L = 0.791
15. The wicket dam, pictured on the right, is
5 m high and 3 m wide and is pivoted at
its center. Determine the reaction force
in the supporting member AB.
Ans. F = ๏ง ๏ h ๏ A = (9790 N/m3)(2.5m)[(5/cos 30ห)(3 m)] = 4.24 x 105 N = 424 kN
yp = I 0 + y =
Ay
๏(3m)(5.77m) / 12๏ + 2.89m = 3.85 m (inclined depth to pressure center)
3
๏(3m)(5.77m)๏(2.89m)
Summing moments about the base of the dam; โ M = 0
(424 kN)(5.77 m โ 3.85 m) โ (FAB)(5.77m/2) = 0; FAB = 282 kN
16. Calculate the magnitude and the location
of the resultant pressure force on the
annular gate shown in the figure to the
right if the cross-section of central hub is
a square that is 1 m by 1 m.
Ans. F = ๏ง ๏ h ๏ A = (9790 N/m3)(2.5 m)[(ฯ)(1.5 m)2โ(1.0 m)2] = 1.49 x 105 N = 149 kN
yp = I 0 + y = ๏{๏ฐ (3m) / 642} โ {(1m2)(1m) /12}๏ + 2.5m = 2.76 m (below the water surface)
4
Ay
3
๏๏ฐ (1.5m) โ (1m) ๏(2.5m)
10
17. Determine the relationship between ฮณ1
and ฮณ2 in the figure to the right if the
weightless triangular gate is in
equilibrium in the position shown.
(Hint: Use a unit length for the gate.)
Ans. Fincline = ๏ง ๏ h ๏ A = (ฮณ1)(0.5 m)[(1m/cos30ห)(1m)] = 0.577โฮณ1 N; Since (1m/cos 30ห)=1.15 m
yP =
๏
๏
I0
(1m)(1.15 m)3 / 12
+y=
+ (1.15 m / 2) = 0.767 m (inclined distance to center of pressure)
๏(1m)(1.15m)๏(1.15 m / 2)
Ay
Fright = ๏ง ๏ h ๏ A = (ฮณ2)(0.5 m)[(1m)(1m)] = 0.500โฮณ2 N; and y P = 0.667 m
โ Mhinge = 0; (0.577โฮณ1 N)(1.15m – 0.767 m) โ (0.500โฮณ2 N)(1 m – 0.667 m) = 0; ฮณ2 = 1.33โฮณ1 N
18. An inverted hemispherical shell of
diameter d = 3 feet as shown in the
figure to the right is used to cover a tank
filled with water at 20ยฐC. Determine the
minimum weight the shell needs to be to
hold itself in place (i.e., not be lifted up).
Ans. The vertical component of the total hydrostatic pressure force is equal to the weight of
the water column above it to the free water surface. In this case, it is the virtual or
displaced weight of water above the shell since the pressure is from below.
FV = ๏ง ๏ Vol = (62.3 lb/ft3)[(1/2)(4/3)(ฯ)(1.5 ft)3] = 440 lb
The weight of the cover must balance this upward pressure; thus W = 3,520 lb
19. The corner plate of a bargeโs hull is
curved with a radius of 1.75 m. The
depth of submergence (draft) is depicted
in the figure. However, the barge is
leaking and the water on the inside is up
to level A producing hydrostatic pressure
on the inside as well as the outside.
Determine the resultant horizontal
hydrostatic pressure force on plate AB
per unit length of the hull.
Ans. FH = (๏ง ๏ h ๏ A) right โ (๏ง ๏ h ๏ A)left = (9790 N/m3) [(1.75 m)(1 m)] (3.875 m โ 0.875 m)
FH = 5.14 x 104 N = 51.4 kN ๏ง (towards the barge)
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20. The corner plate of a bargeโs hull is
curved with a radius of 1.75 m. The
depth of submergence (draft) is depicted
in Figure P2.6.4. The barge is leaking
and the water on the inside is up to level
A producing hydrostatic pressure on the
inside as well as the outside. Determine
the resultant vertical hydrostatic force
on plate AB per unit length of hull.
Ans. The resultant vertical component of the total hydrostatic force is the weight of the water
column above the curved surface (water in the barge) subtracted from the virtual weight
of the water column above the curved surface up to the water level outside. Note that
this equals the water column above the curved surface between the two water levels.
FV = (๏ง ๏ Vol ) outside โ (๏ง ๏ Vol )inside = (9790 N/m3)[(1.75 m)(1 m)(3 m)]
FV = 5.14 x 104 N = 51.4 kN โ
21. Calculate the horizontal and vertical
hydrostatic forces on the curved surface
ABC in the figure to the right.
Ans. The horizontal component of the total hydrostatic pressure force equals the total
pressure on the vertical projection of the curved surface ABC:
FH = ๏ง ๏ h ๏ A = (ฮณ)(R)[(2R)(1)] = 2(ฮณ)(R)2
The vertical component of the total hydrostatic pressure force equals the weight of the
water above the curved surface ABC. The volume of water above this surface is:
Vol = (Aquaarter circle + Arectangle โ Aquarter circle )(unit length)
Vol = (Arectangle)(unit length) = (2R)(R)(1) = 2(R)2; FV = ๏ง ๏ Vol = ฮณ[2(R)2] = 2(ฮณ)(R)2
22. The tainter gate section shown in figure
to the right has a cylindrical surface with
a 12-m radius and is supported by a
structural frame hinged at O. The gate is
10 m long (in the direction perpendicular
to the page). Determine the magnitude,
direction, and location of the total
hydrostatic force on the gate
7
Ans. First find the height of the vertical projection of area: (R)(sin 45ห) = 8.49 m. Thus,
FH = ๏ง ๏ h ๏ A = (9790 N/m3)(4.25m)[(10 m)(8.49 m)]; FH = 3.53 x 106 N = 3,530 kN
The vertical component of the total pressure force is the weight of the water column
above the curved gate. The volume of water above the gate is:
Vol = (Arectangle – Atriangle – Aarc)(length)
Vol =[(12m)(8.49m) – (1/2)(8.49m)(8.49m) – (ฯ/8)(12m)2](10 m) = 92.9 m3
Now, FV = ๏ง ๏ Vol = (9790 N/m3)(92.9 m3) = 9.09 x 105 N = 909 kN; and the total force
is: F = [(3,530 kN)2 + (909 kN)2]1/2 = 3650 kN; ฮธ = tan-1 (FV/FH) = 14.4ห
.
Since all hydrostatic pressures pass through point O (i.e., they are all normal to the
surface upon which they act), then the resultant must also pass through point O.
23. A piece of irregularly shaped metal weighs 301 N. When the metal is completely submerged
in water, it weighs 253 N. Determine the specific weight and the specific gravity of the
metal.
Ans. The buoyant force equals the weight reduction. Thus, B = 301 N โ 253 N = 48.0 N
In addition, B = wt. of water displaced = ฮณโVol = (9790 N/m3)(Vol); Thus,
Vol = 4.90 x 10-3 m3 and ฮณmetal = W/Vol = 301N/(0.00490 m3) = 6.14 x 104 N/m3
S.G. = (6.14 x 104 N/m3)/(9,810 N/m3) = 6.26
24. A concrete block that has a total volume of 12 ft3 and a specific gravity of 2.67 is tied to one
end of a long cylindrical buoy as depicted in the figure below. The buoy is 10 ft long and is
2 ft in diameter. Unfortunately the water level has risen and the buoy is floating away with 1
ft sticking above the water surface. Determine the specific gravity of the buoy. The fluid is
brackish bay water (S.G. = 1.02).
Ans. W = 2.67ฮณ(12 ft3) + (SG)ฮณโฯ(1 ft)2(10 ft);
B = 1.02ฮณ(12 ft3) + 1.02ฮณโฯ(1 ft)2(9 ft)
โFy = 0; or W = B; 32.0 + 31.4(SG) = 12.2 + 28.8;
SG = 0.287
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25. A cube of ice measures 10 inches on each side. It has a density of 1.76 slugs/ft 3. Determine
the weight of the ice and the percentage of the cube below the waterline when it is floating.
Ans. Since ฯ = m/Vol; m = (Vol)(ฯ) = [(10/12)ft3](1.76 slugs/ft3) = 1.02 slugs
Also, W = mg = (1.02 slugs)(32.2 ft/sec2) = 32.8 lbs
ฮณice = W/Vol = 32.8 lbs/[(10/12)ft3] = 56.7 lbs/ft3
S.G. = ฮณice/ฮณ = 56.7 lbs/ft3/ 62.4 lbs/ft3 = 0.909
Therefore, 90.9% of the cube will be below the waterline.
26. A rectangular barge is 14 m long, 6 m wide, and 2 m deep. The center of gravity is 1.0 m
from the bottom and the barge drafts 1.5 m of seawater (S.G. = 1.03). Find the metacentric
height and the righting moments for a 8ยฐ angle of heel (or list).
Ans. The center of gravity (G) is given as 1 m up from the bottom of the barge.
The center of buoyancy (B) is 0.75 m up from the bottom since the draft is 1.5 m.
Therefore GB = 0.25 m, and GM is found using
GM = MB ๏ฑ GB =
I0
๏ฑ GB ; where Io is the waterline moment of inertia about the tilting
Vol
axis. Chopping off the barge at the waterline and looking down we have a rectangle
which is 14 m by 6 m. Thus,
GM = MB ๏ฑ GB =
๏
๏
(14m)(6m) / 12 โ 0.25m = 1.75 m;
I0
๏ฑ GB =
(14m)(6m)(1.5m)
Vol
3
Note: Vol is the submerged volume and a negative sign is used in the equation since G
is located above the center of buoyancy.
M = W ๏ GM ๏ sin ๏ฑ
M = [(1.03)(9790 N/m3)(14 m)(6m)(1.5m)](1.75m)(sin 8ห)
M = 309 kNโm (for a heel angle of 8ห)
8

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