2ª MEDIDA
4.5. CRITERIOS DE INCLUSIÓN EN LOS DISTINTOS GRUPOS DEL ESTUDIO
Together the two weigh 13.35 N in water. What is the specific gravity of the wood?
SOLUTION:
BF = WWood and Metal in H20 s.g. = W_(in air)/(W_(in air)-W_(in water) )
BF = 13.35N = 17.80N/(17.80N-(-4.45)) = 0.8
BF = WWood + WMetal
13.35N = WWood + 17.80N WWood = -4.45N
s.g =
s.g =
s.g = 0.8
55. In Figure W is shown a thin-walled inverted box 1.60 m long and 30 cm square, which was full of air before being submerged. In position (a), it is being held by 0.0280 m3 of concrete anchor weighing 23540 N/m3. Determine the depth D1. At what other depth D2 (position B) is the system
in equilibrium?
SOLUTION:
Wair + Wrock = BF air + BFrock (0.3)(0.3)(1.6)(0.012) + (23.54)(0.028) = (0.3)(0.3) d (9.81) + (9.81)(0.028) 0.6608 – 0.2747 = 0.8829 d d = 0.44 m
56. In Figure X a gate 1.30 m square and 30 cm thick is hinged at A. It is subjected to water pressures on both sides. What force T is required to open the gate if it weighs 1780 N?
SOLUTION:
F = whA
F = (9.81)(1.8 - 0.3)(1.3)(0.3) F = 5.73885 KN
F2 = (9.81)(1.2 – 0.3)(1.3)(0.3) F2 = 3.44331 KN
W = 1.78 KN
BF = 5.73885 + 1.78 BF = 7.51385 KN
1.3 T = 5.74885(1.3) + 3.44331(1.3) + 1.78(1.3) – 7.51385(1.3)
T = 3.45 KN
57. The timber AB of Figure Y is 7.20 m long by 15 cm square. Find the specific weight of the timber and the total weight of the anchor (sp. gr. 2.40).
SOLUTION:
a.)
∑MA = 0
3.6 W – 2.6 BF = 0
(3.12)(ω)(0.15)(0.15)(7.2) – (2.6)(9.81)(6)(0.15)(0.15) = 0 ω = 6.44
sg = 6.44 / 9.81 sg = 0.7 b.)
Wt + Wa = BFt + BFa
(0.7)(0.15)(0.15)(7.2) + (2.4)(9.81)(VD) = (9.81)(6)(0.15)(0.15) + 9.81 (VD) VD = 0.15429
W = (2.4)(9.81)(0.15429) W = 363.25 N
58. A sphere 1 m in diameter floats half submerged in a tank of liquid (sp. gr. =0.80) (a) what is the weight of the sphere? (b) What is the minimum weight of the anchor (sp. gr. 2.40) that will require to submerge the sphere completely?
Solution:
In half submerged:
( )
( )
59. Figure Z shows a hemispherical shell covering a circular hole 1.30 m in diameter at the vertical side of a tank. If the shell weighs 12,450 N, what vertical force is necessary to lift the shell considering a friction factor of 0.30 between the wall and the shell?
60. An iceberg has a specific gravity of 0.92 and floats in salt water (sp. gr. 1.03). If the volume of ice above the water surface is 700 m3, what is the total volume of the iceberg?
SOLUTION:
s.g. (iceberg) = 0.92
s.g. (saltwater) = 1.03 VD = (s.g.body/s.g. fluid)(V) VD = VT -700 m3
∑
BF = W
WliquidVD = wbodyVT
(1.03)(9.81)(VT – 700) = 9.81(0.92) VT
VT = 6554.545455 m3
61. A concrete cube 60 cm on each edge (sp. gr. 2.40) rests on the bottom of a tank in which sea water stands to a depth of 5 m. The bottom edges of the block are sealed off so that no water is admitted under the block. Find the vertical pull required to lift the block.
SOLUTION:
P = ,ρghseawater –ρghconcrete} Area
= ,*(1000kg⁄m3 )(1.03)(9.81m/s2)(5m)] –
*(1000kg⁄m3)(2.4)(9.81m/s2)(0.6m)]}(0.6m)2 P =13,102.236N ≈13.1022kN
62. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/m3 is hinged at one end as shown in Figure AA. If the anchor weighs 23540 N/m3, determine the minimum total weight it must have.
SOLUTION:
[W1 + W2]V = Wtotal
[6280N/m3 + 23540 N/m3](0.15m × 0.15m × 7m) = Wtotal
[29820N/m3](0.1575m3) = Wtotal
4696.65N = Wtotal
Wtotal = 4696.65N ≈4700N
63. A cylinder weighing 445 N and having a diameter of 1 m floats in salt water (sp. gr.
1.03) with its axis vertical as in Figure BB. The anchor consists of 0.280 m3 of concrete weighing 23540N/m3. What rise in the tide r will be required to lift the anchor off the bottom?
Solution:
∑ BF = W
445 N + 23540(0.280) = 0 BF = 7036.2 N
W = (9.81)(1.03)(Vc + 0.280m2) W = 10.1043 (0.3156 + 0.785r)
W = 5.21 + 7.93r BF = W
7.0362 KN = 5.21 + 7.93r r = 0.23 m or 23cm
64. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One end is hinged to the wall and the other is left to float in water (Figure CC). For a = 60 cm, what is the length of the timber submerged in water?
Solution:
P = wh P = (9.81)(1.30) P = 13.2435 KN Wwood = wV
Wwood = (9.81)(0.5)(0.270) Wwood = 1.05948 P = F/A
P = 1.05948/(o.6)2 P = 2.94 KN
65. A metal block 30 cm square and 25 cm deep is allowed to float on a body of liquid which consists of 20 cm layer of water above a layer of mercury. The block weighs 18,850 N/m3. What is the position of the upper level of the block? If a downward vertical force of 1110 N is applied to the centroid of the block, what is the new position of the upper level of the block?
Solution:
66. Two spheres, each 1.2 m in diameter, weigh 4 and 12 kN, respectively. They are connected with a short rope and placed in water. What is the tension in the rope and what portion of the lighter sphere protrudes from the water? What should be the weight of the heavier sphere so that the lighter sphere will float halfway out of the water?
Solution:
Tension in the hoops T+BF_2=12 kN
T+(π(1.2)^3)/6(9.79)=12 kN T=3.14 kN
Portion of the lighter sphere protrudes from the water BF1=4kN+3.14 kN
BF1=7.14 kN
BF2=(π(1.2)^3)/6 (9.79)=8.56 kN By proportion:
(8.56-7.14)/8.56= V_1/V V1= (8.56-7.14)/8.56 V V1=0.1659
V1=16.59 % of V above the water surface
67. A flat-bottomed scow is built with vertical sides and sloping ends. Its length on deck is 24.40 m, on the bottom 19.80 m, its width 6.10 m and its vertical depth 3.70 m.
What is its depth of flotation in sea water (sp. gr. 1.03) if the scow weighs 2,220 kN?
Solution:
BF = W ωVD= 2220kN
(9.81)(1.03) [((24.40+19.80))/2] (6.10)(x)=2220kN X = 1.63m
68. A ship with cargo weighs 44,480 kN and draws 7.60 m of salt water. On crossing a bar at the entrance to a river her draft is decreased by 30 cm by the discharge of 2670 kN of water ballast. In going up the river to fresh water, 445 kN of coal burned. What will her draft be then and how much ballast must be required to increase it by 30 cm? Assume that the sides of the vessel near the water surface are vertical.
Solution:
69. A tank with vertical sides is 1.30 m square, 3 m deep and is filled to a depth of 2.70 m with water. By how much, if at all, will the pressure on one side of the tank be changed if a cube of wood (sp. gr. 0.50), measuring 60 cm on an edge, be placed in the water so as to float with one ace horizontal?
Solution:
P = 9.81(1.35) P = 13.2435 Wwood= ωV = 9.81(. ) = 2.12 kN
70. A cask which weighed 270 N was placed on platform scales and then nearly filled with water. A total load on the scales of 1425 N was read. Should the net weight of water as computed from these data be corrected by reason of the fact that a 7.5 cm diameter vertical shaft suspended from the ceiling above has its lower end
submerged in the water to a depth of 30 cm? If so, by what amount?
Solution:
71. . If the specific gravity of a body is 0.80, what proportional part of its total volume will be submerged below the surface of a liquid (sp. gr. 1.20) upon which it floats?
Solution:
BF = W ωVD = ωV
(9.81)(1.20)(V-x) = (9.81)(0.8)(V) 11.772V – 11.772x = 7.848V 3.924V/11.772 = 11.772x/11.772 X = 1/3 V
VSubmerged = V – x = V - 1/3 V VSubmerged= V
72. .A vertical cylindrical tank, open at the top, contains 45.50 m3 of water. It has a horizontal sectional area of 7.40 m2 and its sides are 12.20 m high. Into it is lowered another similar tank, having a sectional are of 5.60 m2 and a height of 12.20 m. The second tank is inverted so that its open end is down and it is allowed to rest on the bottom of the first. Find the maximum hoop tension in the outer tank. Neglect the thickness of the inner tank.
Solution:
P = ω h
P = (9.81) (6.15) P = 60.32 kPa FB = 60.32 (5.60) FB = 337.79 T = FB / 2 T = 337.79 / 2
T = 168. 896 kN/m
73. . A small metal pan of length 1 m, width 20 cm, and depth 4 cm floats in water.
When a uniform load of 15 N/m is applied as shown in Figure DD, the pan assumes the figure shown. Find the weight of the pan and the magnitude of the righting moment developed.
Solution:
74. .A ship of 39,140 kN displacement floats in sea water with the axis of symmetry vertical when a weight of 490 kN is midship. Moving the weight 3 m toward one side of the deck causes a plumb bob, suspended at the end of a string 4 m long, to move 24 cm. Find the metacentric height.
Solution:
Tanθ=
Θ= 3.43
RM= w(MGsinθ) MG=
MG=
MG= 0.627m
75. .A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the bottom of the scow.
(a) Determine the initial metacentric height. (b) If the scow tilts until one of the
longitudinal side is just at the point of submergence, determine the righting couple or the overturning couple.
Solution:
A) MG= MGo + GBo
MBo= (1+ ) ; θ=0 GBo= 2.75- (2.44-2) MBo= (1+ ) GBo= 1.53m MBo= 2.82m
MG= 2.84 – 1.53 MG= 1.33m
B)
OM= BF. X GBo= 2.75-1.22 X= MGsinθ = 1.53 MG= MBo-GBo
MBo=
(1+ ) MG=2.96-1.53 Θ= tan (
) = 1.43m Θ= 14.81° x= MGsinθ x= 1.43 sin (14.81) MBo=
(1+ ) x= 0.37m MBo= 2.96
OM= BF.x
= (9.81) (1.03)(9.15)(2.44)(15.25)(0.37) OM= 1256.9KN.m
76. A cylindrical caisson has an outside diameter of 6 m and floats in fresh water with its axis vertical. Its lower end is submerged to a depth of 6 m below the water surface. Find: (a) the initial metacentric height; (b) the righting couple when the caisson is tipped through an angle of 10°.
Solution:
77. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in fresh water. Its center of gravity is 4.60 m above the bottom. Determine the height of the scow if, with on side just at the point of submergence, the scow is in unstable position. 2.75 m. Its center of gravity, transversely and longitudinally, is at the center of the scow. If the sow is tipped transversely until one end is at the point of submergence, find the righting couple. solid timbers (sp. gr. 0.60). If a man weighing 890 N steps on the edge of the raft at the middle of one side, how much will the original water line on that side be depressed below the water surface?
Solution:
80. .The timber shown in Figure EE is 30 cm square and 6 m long, having a specific gravity of 0.50. A man standing at a point 60 cm from one end causes that end to be just submerged. Find the weight of the man.
Solution:
81. .A submarine of 10, 700 kN displacement has its center of gravity 30 cm from its center of volume. What is the righting couple when it is submerged in sea water and the angle of heel is 5˚.
Solution:
82. .A log 30 cm in diameter weighs 4900 N/m3. What is its shortest length so that it may float in water with the axis horizontal?
Solution:
Wlog = 4.9 (0.30)3 = 0.1323 kN BF = Wwood
9.81 (1) *π (0.15)2 h] = 4.9 h = 0.1908 m
h = 19.08 cm
83. .A block of wood is 15 cm square and 30 cm long. What is the specific gravity of the wood if the metacentre is at the same point as the center of gravity of the wood when the block is floating in water on its side? Would it be stable floating on its end? Explain.
Solution:
W = BF
9.81(0.15)(0.15)(.3)(S) = 9.81(1)(0.15)(0.3)(.10) S = 0.667
84. .Figure FF shows a scow equipped with a derrick with a boom 5.50 m long. What maximum weight could be picked by the boom at its end along the longitudinal side in order that water will not enter into the scow? Assume the weight of the boom to be negligible and consider its position to be always horizontal.
Solution:
MG = Mbo-Gbo
Mbo = I/V =0.1188 in.
I = bh3/12 =(10)3(16)/12 = 16000 m4
V = bhw = 10(16)(2.6) = 46 m3
Gbo = 3- = 1.7 m MG = Mbo + Gbo
MG = .12 + 1.7 MG = 1.82
W = ѡV
W = 9.81(1)(416) W = 4080.86 KN