fluid mechanics - buoyancy
DESCRIPTION
Fluid mechanics for studentsTRANSCRIPT
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BUOYANCY; PRINCIPLE OF ARCHIMEDESL.S. (, )3mNW
Body of volume, V
bFPrinciple of Archimedes: a body submerged in a liquid of specific weight is buoyedup by a force equal to the weight of the displaced liquid,
where: V = volume of the submerged body or volume displaced liquid = specific weight of the liquid
Note: is called the buoyant force and its direction is vertically upward.
VFb
bF
-
Actual Weight of a Body ( Weight in Air )
where: volume of the bodyspecific weight of the bodyspecific gravity of the bodyspecific weight of water
Apparent Weight of a Body (Weight in Liquid)W = W - Fb
wBBbB sVVW BV B Bs w
Actual Weight of a Body ( Weight in Air )
where: volume of the bodyspecific weight of the bodyspecific gravity of the bodyspecific weight of water
Apparent Weight of a Body (Weight in Liquid)W = W - Fb
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Flotation; Stability of Floating Bodies
W
bF
oB
GO
(a) The body is in Upright Position
G = Center of gravity of the body= Center of buoyancy ( centroid of the submergedportion )
VFW b
oB
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( b) The body is in TiltedPosition
= new center of Buoyancyr = horizontal shifting ofx = moment arm of W or
M
O
1B oB
W
G
x
AAB B
(Metacenter)( b) The body is in Tilted
Position
= new center of Buoyancyr = horizontal shifting ofx = moment arm of W or
r
bF
1B
oB
bF
= angle of tilt
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(b) Tilted Position
G
W
1B oB
x
AB
AB
(Metacenter)
O
M
NOTE: C is a righting moment if M falls above G, an overturning moment if M fallsbelow G. MG is known as the metacentric height.
rThe Righting or OverturningCouple, C
sin_____MGWxWC
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O
1B oB
G
x
AB
W
bF
bF
AB
M(Metacenter)
rS
bF
The shifting of the original upward buoyant force in the wedge AOB toin the wedge AOB causes a shift in from , a horizontal distance r
Hence,bF
bF bF oB 1BtoSFrF bb SrV SVr
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Also,
Then,
For small angle ,
where: = volume of the wedge AOBV = volume of the submerged bodyS = horizontal distance between the centroid
of AOB and AOB = angle of tilt
sin_____
oMBr
SV
MBo sin_____
sin
_____
VSMBo
VSMBo
_____ )( elyapproximat
Note:
Where is the additional volume AOB.S is the distance between the centroids of AOBand AOB.
sin_____
oMBr and bFAlso,
Then,
For small angle ,
where: = volume of the wedge AOBV = volume of the submerged bodyS = horizontal distance between the centroid
of AOB and AOB = angle of tilt
VSMBo
_____ )( elyapproximat
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Metacentric Height,
NOTE:
If is negligible, is given as
Where: is the moment of inertia of the waterline section relativeto a line through O.
______
MG_______________
oo GBMBMG + if Bo is above G- if Bo is below G _____
oGB is usually a known value_____
oMB
Metacentric Height,
NOTE:
If is negligible, is given as
Where: is the moment of inertia of the waterline section relativeto a line through O.
_____
oMB
VIMB oo
_____
oI
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Derivation ofVIMB oo
_____
O
1B oB
G
bF
AB
W
bF
M(Metacenter)
AB
dAx
Consider now a small prism of the wedge AOB,at a distance x from O, having a horizontal area dA.For small angles the length of this prism = x(approximately). The buoyant force producedBy this immersed prism isThe moment of this force about O isThe sum of all these moments for both wedgesMust be equal to S or
But for small anglesHence
But is the moment of inertia, of the water-line section about the longitu-dinal axis through O (approximately constant for small angles of heel). Therefore
The metacentric height
,dAx and.
2 dAx
rS
bF
bF
Consider now a small prism of the wedge AOB,at a distance x from O, having a horizontal area dA.For small angles the length of this prism = x(approximately). The buoyant force producedBy this immersed prism isThe moment of this force about O isThe sum of all these moments for both wedgesMust be equal to S or
But for small anglesHence
But is the moment of inertia, of the water-line section about the longitu-dinal axis through O (approximately constant for small angles of heel). Therefore
The metacentric height
VrSdAx 2)(
_____
elyapproximatMBr o
_____
2oMBVdAx
dAx 2 ,oI,
_____
VIMB oo _______________
oo GBMBMG
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VESSEL WITH RECTANGULAR SECTION
1B oB
x
AB
AB
(Metacenter)
O
MbF
GB/2
(B/2)(sec)
A
B O
2
__ Sx
(B/2) (cos)
1B oB
r
bF
B
S
2
__ Sx
sin
_____
VSMBo
BDLV where:
Recall:
LBBv
tan222
1
tan81 2LBv
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Considering triangle AOB
B/2A
B O
(B/2)(sec)
2
__ Sx
(B/2) (cos)Multiplying both sides by 2 we obtain,
seccos3
Bs
cos
1cos
3B
s
cos
1cos3
2Bs
Then from the formula,2__ Sx
From geometry, the centroid of the triangle isdefined by the coordinates of the vertices:
3321
__ xxxx
3
sec2
cos2
0
2
BB
s
sin
_____
VSMBo
Then from the formula,
sincos
1cos3
tan81 22
_____
BDL
BLBMBo
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sincos
1cos3
tan81 22
_____
BDL
BLBMBo
sincos
1coscos
sin24
23
_____
BDL
LB
MBo
222
_____
cos
1cos24
D
BMBo
1tan124
22
_____ D
BMBo
22_____ tan224
D
BMBo
22
_____
tan2212
D
BMBo
222
_____
cos
1cos24
D
BMBo
2
2_____
cos
1124DBMBo
22_____ sec124
D
BMBo
22
_____
tan2212
D
BMBo
2tan1
12
22_____
DBMBo
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SAMPLE CALCULATIONS
MG (center of gravity)
O
B
1Y__
Y
h 2h
2h
_____
MB
__________
GBMB hY __
Position of weight on the mast
Sketch showing various distances on the pontoonDETERMINATION OF THEORETICAL METACENTRIC HEIGHT FROM THE GEOMETRY OFTHE PONTOON:DETERMINATION OF THEORETICAL METACENTRIC HEIGHT FROM THE GEOMETRY OFTHE PONTOON:
12200400
12
33 mmmmLbIo 441067.2 mx 1220.040.0 3mm
Displaced Volume
gWV
23 81.9000,1
81.952.2
s
m
m
kgkg
Nkg
331052.2 mx
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VIMB _____ 33
44
1052.21067.2
mx
mx
mmm 1061059.0 Depth of displaced water
LbVh
mm
mx
20.04.01052.2 43
m0315.0The center of buoyancy force below the water surface and the distance will be
_____
OB
2
_____ hOB 2
0315.0 m mm75.15m01575.02
_____ hOB 2
0315.0 mThe Metacenter is above the water surface and distance
_____
MO is______________
OBMBMO 75.15106 mm25.90
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In the case when the height of the mast,height of the center of gravity ( by experiment) ,
Thus, the theoretical metacentric height
is positive, this shows that the pontoon is stable.
mmY 1001 andmmY 69
__ the
_____
thMG_______________
GBMBMGth
2
____________ hYMBMGth
2
5.3169106
M (Metacenter)G (center of gravity)
O1Y_____
MB
__________
GBMB hY __
Position of weight on the mast
In the case when the height of the mast,height of the center of gravity ( by experiment) ,
Thus, the theoretical metacentric height
is positive, this shows that the pontoon is stable.
2
5.3169106
mm02.37
Because_____
thMG
O
B
1Y__
Y
h 2h
2h
-
O
G
Determination of Metacentric Height by Experiment
w
B
WFb
G
M
w
x
1B
d
bF WFb The metacentric height is determined experimentally as shown in the figureabove. When shifting the jockey weight w to the left side of the pontoon at a distancex, the pontoon tilts to a small angle causing the metacentric height to rotate slightlyaround the longitudinal axis of the pontoon . Likewise, the buoyancy force shifted ahorizontal distance d from G. Hence, the moment produced by wmust be equal tomoment of ,bF
bF
Wdxw cosWMG
sin_____tan
_____
WxwMG
x
Ww (For small angle of tilt)
_____
MG
bF
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Vertical scale
Vertical slidingweight
Mast
Jockey weight
METACENTRIC HEIGHT APPARATUS
Balancing weight
Pontoon
Jockey weight
Tilt anglescale
Plumb bob
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DETERMINATION OF METACENTRIC HEIGHT, BY EXPERIMENTTypical Data:In the case of vertical sliding weight on the mast is at the height,Distance of jockey weight w from center of pontoon , x = 80 mmAngle of tilt, = 6.80
Convert angle of tilt into radian
Then,From equation (2), the experimental metacentric height is,
is positve, this shows that the pontoon at that tilt angle is stable.
_____
MG
.1001 mmY
80.618080.6
radian11868.0
DETERMINATION OF METACENTRIC HEIGHT, BY EXPERIMENTTypical Data:In the case of vertical sliding weight on the mast is at the height,Distance of jockey weight w from center of pontoon , x = 80 mmAngle of tilt, = 6.80
Convert angle of tilt into radian
Then,From equation (2), the experimental metacentric height is,
is positve, this shows that the pontoon at that tilt angle is stable.
radianmmx
11868.080
mm06.674
x
WwMG exp
_____
mmkgkg 06.674
52.220.0 mm49.53
exp
_____
MG
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TEST PROCEDURES:Data recording:
- Pontoon weight, W = 2.50 kg- Jockey weight, w = 0.20 kg- Adjustable vertical weight = 0.40 kg- Pontoon width, D = 200 mm- Pontoon length, L = 400 mm
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Determining the Center of Gravity of the Pontoon
Center of gravity ( CG)
ScaleAdjustable vertical
weight
MastSupport
Procedures:1. Tilt the pontoon as shown in figure.2. Attach the plum bob on the angle scale.3. Move or adjust the vertical weight to a
required distance and record that distancefrom the scale on the mast.
4. Place knife edge support under the mast andmove it to a position of equilibrium and recordthe height ( center of gravity) where the knifeedge is position on the scale.
SupportProcedures:1. Tilt the pontoon as shown in figure.2. Attach the plum bob on the angle scale.3. Move or adjust the vertical weight to a
required distance and record that distancefrom the scale on the mast.
4. Place knife edge support under the mast andmove it to a position of equilibrium and recordthe height ( center of gravity) where the knifeedge is position on the scale.
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Taking Readings with the Pontoon in a Water Tank1. Initial Set Up
When placing the pontoon in the water ensure that the position ofthe jockey weight horizontal adjustments is in the middle of thepontoon and the pontoon is sitting level in the water. The pontoonshould be in a vertical position and have no angle of tilt ( zerodegrees in the tilt angle scale). If not, adjust the balancing weightuntil the angle of tilt is 0.
2. The jockey weight can change the position of the pontoon in thewater and in order to take some experimental readings we movethe jockey weight in steps from its central position horizontally andrecord the tilt angle of the pontoon from the scale on the pontoonin degrees.
1. Initial Set UpWhen placing the pontoon in the water ensure that the position ofthe jockey weight horizontal adjustments is in the middle of thepontoon and the pontoon is sitting level in the water. The pontoonshould be in a vertical position and have no angle of tilt ( zerodegrees in the tilt angle scale). If not, adjust the balancing weightuntil the angle of tilt is 0.
2. The jockey weight can change the position of the pontoon in thewater and in order to take some experimental readings we movethe jockey weight in steps from its central position horizontally andrecord the tilt angle of the pontoon from the scale on the pontoonin degrees.
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3. Each time we move the jockey weight from its central position wemust record on the data sheets supplied the distance measured fromits central position and the angle of tilt.
4. We also change the adjustable vertical weight height on the mastand record its measurement along with the jockey weight distancefrom its central position, the angle of tilt at different values andrecord all the data on the sheets provided.
5. Step (3) and (4) can be repeated many times to obtain a satisfactoryconclusion.
3. Each time we move the jockey weight from its central position wemust record on the data sheets supplied the distance measured fromits central position and the angle of tilt.
4. We also change the adjustable vertical weight height on the mastand record its measurement along with the jockey weight distancefrom its central position, the angle of tilt at different values andrecord all the data on the sheets provided.
5. Step (3) and (4) can be repeated many times to obtain a satisfactoryconclusion.
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SAMPLE DATA SHEETMETACENTRIC HEIGHT APPARATUS
2 4 6 8 10 12 14 16 18
80 60 40 20 0 20 40 60 80Height ofweighton themast,
____mm.Tilt Angle
(degrees)
x/(mm/rad.)Metacentric Height(mm)
Position of jockey weight in a horizontal position (cm.)
Distance x of the jockey weight measured from the center of the pontoon (mm)
Height ofcenter ofgravity,____mm.
2 4 6 8 10 12 14 16 18
80 60 40 20 0 20 40 60 80Height ofweighton themast,
____mm.Tilt Angle
(degrees)
x/(mm/rad.)Metacentric Height(mm)
Position of jockey weight in a horizontal position (cm.)
Distance x of the jockey weight measured from the center of the pontoon (mm)
Height ofcenter ofgravity,____mm.
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Example 1. An iceberg weighing 8.95 kN/m3 floats in sea water, = 10.045 kN/m3, with a volume of 595 m3 above the surface. What isThe total volume of the iceberg?Solution:
W.S.
W
33 95.8045.10 m
kNVm
kNV s
3595 mVV s but
xVm
kNm
kNmV 33
3 95.8045.10595
FbLet Vs = volume submergedV = total volume(a) Fv = 0,Fb = WVsx w = V x iwhere: w = specific weight of sea wateri = specific weight of iceberg
xVm
kNm
kNmV 33
3 95.8045.10595
3242.458,5 mV
kNm
kNV 775.976,5095.1 3
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Example 2. A sphere 0.90 m in diameter floats half submerged in a tankof oil ( s=0.80). (a) What is the total vertical pressure on the sphere?(b) What is the minimum weight of an anchor weighing 24 kN/m3 thatwill be required to submerge the sphere completely?Solution:
0.45 m
W
O.S.
O.S. W
Fb
0.45 m
Fb
FbWa
abF
Figure (a) Figure (b)(a) Consider Figure (a)
,0 vF0 WF v
VWF v
33 81.980.045.032
m
kNxmW
kNW 498.1
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O.S. W
FbWa
abF
0.45 m
Figure (b)
(b) Consider Figure (b),
3093.0 mV a
thereforeaaa VW
33 24093.0 mkNmWakNWa 232.2
(b) Consider Figure (b),,0 vF
,0 abb WWFF a0498.1 aaa VkNVV
024498.181.980.081.980.045.034
3333
m
kNVkNm
kNxV
m
kNxm aa
kNVm
kNa 498.1152.16 3
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Example 3. A cylinder weighing 500 N and having a diameter of 0.90 mfloats in salt water ( s=1.03) with its axis vertical as shown in the figure.The anchor consists of 0.30 m3 of concrete weighing 24 kN/m3. Whatrise in tide r, will be required to lift the anchor off the bottom?
0.30 mrnew W.S.
W
Fb
Solution:,0 vF
Wa
abF
0.30 mFb 0 abb WWFF a0500 aaa VNVV
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Wa
abF
0.30 mrnew W.S.
W
Fb
0000,243.0500
981003.13.0981003.130.090.04
33
33
32
m
NmN
m
Nxm
m
Nxrmm
0500 aaa VNVV abF
mr 426.0
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Example 4. Timber AC hinged at A having a length of 10 ft., cross sectional area of3 in.2 and weighing 3 lbs. Block attached to the end C having a volume of 1 ft.3, andweighing 67 lbs.Required: Angle for equilibrium.
1A
CWTTbF
csc 10 csc
WB=67 lbs5cos
(2) Buoyant force on the block,wb VF B 4.621 lb4.62
10
BbF
TbF WB=67 lbs5cos coscsc10
2110
Solution:(1) Buoyant force on the timber,
wb VF T 4.62csc101443
csc103.1
TbF
cos2csc10
10cos
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1A
C
BbF
WT
TbF
csc 10 csc
WB=67 lbs5cos
coscsc102110
cos2csc10
15.6csc2 48.2csc
40.0sin 8.23
BbF10cos
cos2csc10
(3) MA = 0, 0cos10cos
2csc10
cos10cos5
BT bbBT FFWW
0104.622csc10
csc103.1106753
0624csc10065.067015 2
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Example 5. A vessel going from salt into fresh water sinks two inches, then afterburning 112,500 lb of coal, rises one inch. What is the original weight of the vessel?
dd + 2/12 d + 1/12
(a) Salt water ( = 64 lb/ft3) (b) Fresh water ( = 62.4 lb/ft3) (c ) Fresh water afterlosing 112,500 lb
W
Fb
W
Fb Fb
W 112,500 lb
(c ) Fresh water afterlosing 112,500 lb
Solution:1. In figure (a), submerged volume is, Va = Axd ft3
where: A = cross-sectional area of the vessel ( ft2 )2. In figure (b), submerged volume is Vb = Va + (2/12)(A)3. In figure (c), submerged volume is Vc = Va + (1/12)(A)
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dd + 2/12 d + 1/12
(a) Salt water ( = 64 lb/ft3) (b) Fresh water ( = 62.4 lb/ft3) (c ) Fresh water afterlosing 112,500 lb
W
Fb
W
Fb Fb
W 112,500 lb
4. In salt water, W = FbW = Va ( 64 )5. In fresh water, W = FbW = [Va + (2/12)(A)](62.4)6. In fresh water after losing 112,500 lb,
W 112,500 = [Va +(1/12)(A)](62.4)7. Substitute eq. 1 to eq. 2 and eq. 3,
( 1)4. In salt water, W = FbW = Va ( 64 )5. In fresh water, W = FbW = [Va + (2/12)(A)](62.4)6. In fresh water after losing 112,500 lb,
W 112,500 = [Va +(1/12)(A)](62.4)7. Substitute eq. 1 to eq. 2 and eq. 3,
( 1)
(2)
(3)4.62
61
64
AWW AW 4.10975.0 AW 4.10025.0 (4)
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4.62121
64500,112
AWW AW 2.5975.0
500,1122.5025.0 AW (5)
8. Solve eqs. (4) and (5) simultaneously, we obtainlbxW 6109
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Example 6. A ship of 4,000 tons displacement floats in sea water withits axis of symmetry vertical when a weight of 50 tons is midship.Moving the weight 10 feet towards one side of the deck causes a plumbbob, suspended at the end of a string 12 feet long, to move 9 inches.Find the metacentric height.Solution:
1. Solve the angle of tilt,1212
9tanArc 58.3
12
9
2. Righting Moment = W (MG x sin) 58.3sin40501050 MGxx
ftMG 977.1
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Example 7. A ship with a horizontal sectional area at the waterline of 76,000 ft2 has adraft of 40.5 ft. in sea water (s =64 lb/ft3). In fresh water it drops 41.4 ft. Find theweight of the ship. With an available depth of 41 ft. in a river above the sills of a lock,how many long tons of the cargo must the ship be relieved off so that it will pass thesills with a clearance of 0.60 ft.?Solution:
V1fresh W.S
W
V2sea W.S
W
1 ftoriginal fresh W.S
W
40.5 ft
V1fresh W.S
41.4 ft0.90 ft
fbF
'
V2sea W.S
sbF
(a) SEA WATER (b) FRESH WATER
1 ft
0.60 ft
41 ft
sills
41.4 ft
(c) SHIP IN THE LOCKW original weight of the ship (including cargo)W new weight of the ship when part of the cargo has been disposed
- buoyant force in sea water, - buoyant force in fresh watersbF fbF
fbF
-
40.5 ft
sea W.S
W
sbF
W W
V1fresh W.S
41.4 ft0.90 ft
fbF fbF
'0.60 ft
41 ft
original fresh W.S1 ft
sills
V2
41.4 ft
''
WFfb (a) SEA WATER (b) FRESH WATER (c) SHIP IN THE LOCKV1 = additional submerged volume = 0.90(76,000) = 68,400 ft3V2 = volume of the ship at the waterline which rose up when it was relieved off the cargo= 1 (76,000) = 76,000 ft3
V = original volume submerged ( in sea water )(1) Using position (a);
WFsb WV s WV 64 (1)
(2) Using position (b)WF
fb WVV f 1 WV 4.62400,68 (2)
-
(3) Solve equations (1) and (2) simultaneously, 400,684.6264 VV
3600,667,2 ftV The ships displacement in sea water3
1 000,736,2 ftVV and The ships displacement in fresh water
Therefore, VW 64 lbx 81070726.1or 14.62 VVW lbx 81070726.1 V2
original fresh W.S1 ft
W
0.60 ft
(c) SHIP IN THE LOCK
sills41 ft
41.4 ft(4) Using position (c);fbFW '' 4.6221 VVV
4.62000,76000,736,2 lbx 81065984.1(5) Weight of disposed cargo = W W
lbxWW 81004742.0'lb000,742,4
lbLTlbx
200,21000,742,4 5.155,2 LONG TONS
fbF
'
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Example 8.
49
W
Fb
4
8
A AG
Bo
W.S
15 15
Given: Rectangular scow 50 x 30 x 12as shown with the given draft and centerof gravity.
Required: Righting or overturning momentwhen one side, as shown, is at the pointof submergenceSolution:
4
A
A
A AOG
Bo
W.S
Solution: sinMGWC
where: VW 4.6283050W
lbW 800,748
154
tanArc 93.14
oo GBMBMG 5 oMBMG
M
-
sinVvSMBo
93.14sin83050
303250415
21
oMB
ftMBo 7.9
4A AO
G
Bo
W.SA
A
30
8
30'32S
= 14.93
Therefore,ftftMG 57.9 ft7.4
Then,)93.14sin7.4(800,748 C
lbftC 600,906
30
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Example 9. A rectangular scow 10 m wide, 16 m long, and 4.0 m high has a draft insea water (s = 1.03) of 2.5 m. Its center of gravity is 2.80 m above the bottom of thescow. Determine the following:(a) The initial metacentric height,(b) The righting or overturning moment when the scow tilts until one side is just at
the point of submergence.
G
Bo
S.W.4.0 m
Solution:mMBo 333.3
mmGBo 25.180.2 2.8 m
1.25 mBo
10 m
2.5 m4.0 m
(a) Initial Metacentric Height
2tan1
12
22 D
BMBo where = 0
20tan1
5.21210 22
oMB
mGBo 55.1The initial metacentric height , MG
oo GBMBMG 55.1333.3 MG
mMG 783.1
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(b)
1.5 mO
G
Bo
M
4.0 m1.25 m
2.8 m
2.5 m
55.1
tan
5.0 m5.0 m
The metacentric height MGoo GBMBMG
55.1483.3 MGmMG 933.1
Since MG > MBo, the moment is righting moment. TheRighting moment is,55.1
tan 699.16
2tan1
12
22 D
BMBo
255.11
5.21210 22
oMB
mMBo 483.3
Since MG > MBo, the moment is righting moment. TheRighting moment is, sinMGWRM
where: VFW b 5.2161081.903.1 xxxW
kNW 72.041,4 699.16sin933.172.041,4RM
mkNRM 92.244,2