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

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

( 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

(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

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

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

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

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

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

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

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

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

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

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

Vertical scale

Vertical slidingweight

Mast

Jockey weight

METACENTRIC HEIGHT APPARATUS

Balancing weight

Pontoon

Jockey weight

Tilt anglescale

Plumb bob

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

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

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.

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.

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.

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.

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

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

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

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

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

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

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

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)

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)

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

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

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

'

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

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

(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