BUOYANCY; PRINCIPLE OF ARCHIMEDES
W
L.S. (γ,
N
m3
)
Body of volume, V
Fb
Principle of Archimedes: “ a body submerged in a liquid of specific weight γ is buoyed
up by a force equal to the weight of the displaced liquid,”
Fb  V  
where: V = volume of the submerged body
γ = specific weight of the liquid
Note: Fb is called the “ buoyant force” and its direction is vertically upward.
Actual Weight of a Body ( Weight in Air )
W  VB   b  VB sB   w 
where:
VB 
B 
sB 
w 
volume of the body
specific weight of the body
specific gravity of the body
specific weight of water
Apparent Weight of a Body (Weight in Liquid)
W’ = W - Fb
Flotation; Stability of Floating Bodies
W
G
O
Bo 
Fb
(a) The body is in Upright Position
W  Fb  V  
G = Center of gravity of the body
Bo = Center of buoyancy ( centroid of the submerged
portion )
( b) The body is in Tilted
Position
M (Metacenter)
W
x
B
A
G 
O
A’
B’
B1 Bo
Fb

r
B1 = new center of Buoyancy
r = horizontal shifting of Bo
x = moment arm of W or Fb
θ = angle of tilt
(b) Tilted Position
M (Metacenter)

W
x
B
A
G
O
B1
The Righting or Overturning
Couple, C
 _____

C  W  x  W  MG sin  


A’
B’
Bo
 

r
NOTE: C is a righting moment if M falls above G, an overturning moment if M falls
below G. MG is known as the “metacentric height”.
M (Metacenter)

W
Fb
x
G O
B
A
B1
Fb
A’
B’
Bo
 
Fb

r
S
The shifting of the original upward buoyant force Fb in the wedge A’OB’ to
Fb in the wedge AOB causes a shift in Fb from Bo to B1, a horizontal distance r
Hence,
Fb  r  Fb  S  V  r    S

r

V
S
_____
Also,
r  MBo sin 
_____
Then,
MBo sin  

Note:
_____
S
V
_____
 S
MBo 
V sin 
For small angle θ,
_____
and Fb    
Where ν is the additional volume AOB.
S is the distance between the centroids of AOB
and A’OB’.
 S
(approximately )
V 
ν = volume of the wedge A’OB’
V = volume of the submerged body
S = horizontal distance between the centroid
of A’OB’ and AOB
θ = angle of tilt
MBo 
where:
r  MBo sin 
______
Metacentric Height, MG
_____
_____
_____
MG  MBo  GBo

+ if Bo is above G
- if Bo is below G

_____
NOTE: GBo is usuallya knownvalue
_____
If θ is negligible,
MBo is given as
_____
Io
MBo 
V
Where: I o is the moment of inertia of the waterline section relative
to a line through O.
_____
Io
Derivation of MBo 
V
M (Metacenter)
W
Consider now a small prism of the wedge AOB,
dA
at a distance x from O, having a horizontal area dA.
For small angles the length of this prism = xθ
x
(approximately). The buoyant force produced
G O
B
By this immersed prism is xdA, and
A
2
The moment of this force about O is x dA.
B1 Bo
The sum of all these moments for both wedges
 
Fb
Must be equal to γνS or
  x 2 dA  S  Vr
Fb
_____
But for small angles r  MBo  (approximately )
Hence
_____


 x dA  V  MBo 
A’
B’

r
S
2
Fb
But  x 2 dAis the moment of inertia,I o , of the water-line section about the longitudinal axis through O (approximately constant for small angles of heel). Therefore
_____
MBo 
Io
,
V
_____
The metacentric height
_____
_____
MG  MBo  GBo
SAMPLE CALCULATIONS

M

Position of weight on the mast
G (center of gravity)

_____
_____
MB GB
__
Y1
Yh
O

__
Y
h
_____
MB
h
2
h
2
B
Sketch showing various distances on the pontoon
DETERMINATION OF THEORETICAL METACENTRIC HEIGHT FROM THE GEOMETRY OF
THE PONTOON:

Lb3 400mm200mm
0.40m 0.20m   2.67 x104 m4
Io 


12
12
12
3
Displaced Volume
9.81N
kg
W

V
  g 1,000 kg  9.81 m
m3
s2
2.52kg 
 2.52 x103 m3
3
2.67 x10 4 m 4
I
 0.1059m  106mm

MB 
3
3
2.52 x10 m
V
_____
Depth of displaced water
V 2.52 x103 m 4
 0.0315m
h 
Lb 0.4m  0.20m
_____
The center of buoyancy force below the water surface and the distance OB will be
_____
OB 
h 0.0315m  0.01575m  15.75mm

2
2
_____
The Metacenter is above the water surface and distance MOis
_____
_____
____
MO  MB  OB  106  15.75
 90.25mm
In the case when the height of the mast, Y1  100mm and the
__
height of the center of gravity ( by experiment) , Y  69mm
_____
Thus, the theoretical metacentric height MGth
_____
_____
_____
MGth  MB  GB
 __ h 
MGth  MB   Y  
2

31.5  Y1 _ _

 106   69 
Y

2 

h
 37.02mm
_____
_____
_____
 Position of weight on the mast
M (Metacenter)

_____ _____
G (center of gravity)

MB GB
__
Yh
O

B
h
2
h
2
BecauseMGth is positive, this shows that the pontoon is stable.
_____
MB
Determination of Metacentric Height by Experiment
M
x
G
w
G
d
O


w

B1 
B
θ
Fb
_____
Fb  W
The metacentric height
is determined experimentally as shown in the figure
above. When shifting the jockey weight w to the left side of the pontoon at a distance
x, the pontoon tilts to a small angle θ causing the metacentric height to rotate slightly
around the longitudinal axis of the pontoon . Likewise, the buoyancy force Fb shifted a
horizontal distance d from G. Hence, the moment produced by w must be equal to
_____
moment of Fb ,
w x
MG 
W tan 
w cos  x  d W
w x (For small angle of tilt)
 _____

  MG sin  W




W 
MG
METACENTRIC HEIGHT APPARATUS
Vertical scale
Mast
Vertical sliding
weight
Jockey weight
Balancing weight
Pontoon
Tilt angle
scale
Plumb bob
_____
DETERMINATION OF METACENTRIC HEIGHT, MGBY EXPERIMENT
Typical Data:
In the case of vertical sliding weight on the mast is at the height, Y1  100mm.
Distance of jockey weight w from center of pontoon , x = 80 mm
Angle of tilt, θ = 6.80˚
Convert angle of tilt into radian

6.80 
 6.80 0.11868radian
180
Then,
From equation (2), the experimental metacentric height is,
x
80mm
 674.06mm

 0.11868radian
_____
MG exp 
w x
0.20kg
 674.06mm  53.49mm
 
2.52kg
W 
_____
MG exp is positve, this shows that the pontoon at that tilt angle is stable.
TEST PROCEDURES:
Data recording:
- Pontoon weight,
- Jockey weight,
- Adjustable vertical weight
- Pontoon width,
- Pontoon length,
W = 2.50 kg
w = 0.20 kg
= 0.40 kg
D = 200 mm
L = 400 mm
Determining the Center of Gravity of the Pontoon
Center of gravity ( CG)
Scale

Adjustable vertical
weight
Mast
Support
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 distance
from the scale on the mast.
4. Place knife edge support under the mast and
move it to a position of equilibrium and record
the height ( center of gravity) where the knife
edge is position on the scale.
Taking Readings with the Pontoon in a Water Tank
1. Initial Set Up
When placing the pontoon in the water ensure that the position of
the jockey weight horizontal adjustments is in the middle of the
pontoon and the pontoon is sitting level in the water. The pontoon
should be in a vertical position and have no angle of tilt ( zero
degrees in the tilt angle scale). If not, adjust the balancing weight
until the angle of tilt is “0”.
2. The jockey weight can change the position of the pontoon in the
water and in order to take some experimental readings we move
the jockey weight in steps from its central position horizontally and
record the tilt angle of the pontoon from the scale on the pontoon
in degrees.
3. Each time we move the jockey weight from its central position we
must record on the data sheets supplied the distance measured from
its central position and the angle of tilt.
4. We also change the adjustable vertical weight height on the mast
and record its measurement along with the jockey weight distance
from its central position, the angle of tilt at different values and
record all the data on the sheets provided.
5. Step (3) and (4) can be repeated many times to obtain a satisfactory
conclusion.
SAMPLE DATA SHEET
METACENTRIC HEIGHT APPARATUS
Position of jockey weight in a horizontal position (cm.)
2
4
6
8
10
12
14
16
18
Distance x of the jockey weight measured from the center of the pontoon (mm)
80
Height of
weight
on the
Tilt Angle
mast,
θ
____mm. (degrees)
x/θ
Height of (mm/rad.)
center of Metacentr
gravity, ic Height
(mm)
____mm.
60
40
20
0
20
40
60
80
Example 1. An iceberg weighing 8.95 kN/m3 floats in sea water,
γ = 10.045 kN/m3, with a volume of 595 m3 above the surface. What is
The total volume of the iceberg?
Solution:
kN 
kN 


Vs 10.045 3   V  8.95 3 
m 
m 


W
but
W.S.
Vs  V  595m3
kN

V  595m 10.045 kN
xV
  8.95
m
m
3

1.095V
Let Vs = volume submerged Fb
V = total volume
(a) ∑Fv = 0,
Fb = W
Vsx γw = V x γi
where: γw = specific weight of sea water
γi = specific weight of iceberg
3

3
kN
 5,976.775kN
3
m
V  5,458.242m3
Example 2. A sphere 0.90 m in diameter floats half submerged in a tank
of 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 that
will be required to submerge the sphere completely?
Solution:
O.S.
W
W
0.45 m

0.45 m
O.S.
Fb
Wa
Figure (a)
Figure (b)
Fb
(a) Consider Figure (a)
Fv  0,
Fv W  0
Fv  W  V
Fba
kN 
2
3 
W    0.45m   0.80 x9.81 3 
m 
3

W  1.498kN
O.S.
W
Va  0.093m3
0.45 m
therefore
Wa  Va a
Fb
Wa
Figure (b)
Fba


 kN 
Wa  0.093m3  24 3 
 m 
Wa  2.232kN
(b) Consider Figure (b),
Fv  0,
Fb  Fba  W  Wa  0,
V  Va  1.498kN  Va a  0
kN 
kN 
kN 
4


3 
  0.45m   0.80 x9.81 3   Va  0.80 x9.81 3   1.498kN  Va  24 3   0
m 
m 
m 
3



kN
16.152 3 Va  1.498kN
m
Example 3. A cylinder weighing 500 N and having a diameter of 0.90 m
floats 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. What
rise in tide r, will be required to lift the anchor off the bottom?
W
Solution:
new W.S.
r
Fb
Wa
Fba
0.30 m
Fv  0,
Fb  Fba  W  Wa  0
V  Va  500 N  Va a  0
W
new W.S.
r
0.30 m
Fb
Wa
Fba
V  Va  500 N  Va a  0

4
0.90m2 0.30m  r  9810


4




N 
N 
N 
3 
3 

0
.
3
m
9810

500
N

0
.
3
m
24
,
000
0




3
3
3 
m 
m 
m 


0.90m2  9810

N 
r  2,884.744 N
3 
m 
r  0.462m
Example 4. Timber AC hinged at A having a length of 10 ft., cross sectional area of
3 in.2 and weighing 3 lbs. Block attached to the end C having a volume of 1 ft.3, and
weighing 67 lbs.
Required: Angle θ for equilibrium.
(2) Buoyant force on the block,
csc θ
FbB  V w  162.4  62.4lb
A
1’
10 – csc θ

10’
WT
5cosθ
θ
F
bT
1


10  10  csc  cos 
2


 10  csc  

 cos 
2


C
WB=67 lbs
FbB
10cosθ
Solution:
(1) Buoyant force on the timber,
3
FbT  V w 
10  csc 62.4
144
FbT  1.310  csc 
csc θ
A
1’
csc2   6.15
10 – csc θ

WT
5cosθ
θ
FbT
1


10  10  csc  cos 
2


 10  csc  

 cos 
2


csc  2.48
sin   0.40
C
WB=67 lbs
  23.8
FbB
10cosθ
(3) ∑MA = 0,
 10  csc 
WT 5 cos    WB 10 cos    FbT 
 cos   FbB 10 cos    0
2


 10  csc 
35  6710  1.310  csc 
  62.410  0
2


15  670  0.65 100  csc2   624  0


Example 5. A vessel going from salt into fresh water sinks two inches, then after
burning 112,500 lb of coal, rises one inch. What is the original weight of the vessel?
W
W
W – 112,500 lb
d + 2/12
d + 1/12
d
Fb
(a) Salt water ( γ = 64 lb/ft3)
Fb
Fb
(b) Fresh water (γ = 62.4 lb/ft3)
(c ) Fresh water after
losing 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)
W
W
W – 112,500 lb
d + 2/12
d + 1/12
d
Fb
(a) Salt water ( γ = 64 lb/ft3)
Fb
Fb
(b) Fresh water (γ = 62.4 lb/ft3)
(c ) Fresh water after
losing 112,500 lb
4. In salt water, W = Fb
W = Va ( 64 )
( 1)
5. In fresh water, W = Fb
W = [Va + (2/12)(A)](62.4)
(2)
6. In fresh water after losing 112,500 lb,
W – 112,500 = [Va +(1/12)(A)](62.4)
(3)
W 1 
W

  A 62.4  0.975W  10.4 A
7. Substitute eq. 1 to eq. 2 and eq. 3,
 64 6 
(4)
0.025W  10.4 A
W 1 
W  112,500   
A 62.4  0.975W  5.2 A
 64 12 
0.025W  5.2 A  112,500
8. Solve eqs. (4) and (5) simultaneously, we obtain
W  9 x106 lb
(5)
Example 6. A ship of 4,000 tons displacement floats in sea water with
its 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 plumb
bob, suspended at the end of a string 12 feet long, to move 9 inches.
Find the metacentric height.
Solution:

12’
θ
9”
9
1. Solve the angle of tilt,   Arc tan 12
12
 3.58
2. Righting Moment = W (MG x sinθ)
50 x10  4050MGx sin 3.58
MG  1.977 ft
Example 7. A ship with a horizontal sectional area at the waterline of 76,000 ft2 has a
draft of 40.5 ft. in sea water (γs =64 lb/ft3). In fresh water it drops 41.4 ft. Find the
weight 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 the
sills with a clearance of 0.60 ft.?
Solution:
W
W’
W
original fresh W.S
sea W.S
fresh W.S
1 ft
ΔV2
ΔV1
0.90 ft
40.5 ft
41.4 ft
41 ft
41.4 ft
sills
Fbs
(a) SEA WATER
Fb f
(b) FRESH WATER
Fb ' f
0.60 ft
(c) SHIP IN THE LOCK
W – original weight of the ship (including cargo)
W’ – new weight of the ship when part of the cargo has been disposed
Fbs - buoyant force in sea water, Fb f - buoyant force in fresh water
W
W’
W
sea W.S
original fresh W.S
ΔV2
1 ft
fresh W.S
ΔV1
0.90 ft
40.5 ft
41 ft
41.4 ft
41.4 ft
sills
Fbs
Fb ' f  W '
(a) SEA WATER
Fb f
Fb ' f
0.60 ft
(b) FRESH WATER
(c) SHIP IN THE LOCK
ΔV1 = additional submerged volume = 0.90(76,000) = 68,400 ft3
ΔV2 = 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);
(2) Using position (b)
Fbs  W
Fb f  W
V  s   W
V  V1  f  W
V  68,40062.4  W
V 64  W
(1)
(2)
(3) Solve equations (1) and (2) simultaneously,
64V  62.4V  68,400
and
V  2,667,600 ft 3
The ship’s displacement in sea water
V  V1  2,736,000 ft 3
The ship’s displacement in fresh water
Therefore,
or
W’
W  64V  1.70726 x10 lb
8
original fresh W.S
1 ft
W  62.4V  V1   1.70726 x10 lb
8
(4) Using position (c);
41 ft
W ' Fb ' f  V  V1  V2 62.4
ΔV2
41.4 ft
sills
 2,736,000  76,00062.4  1.65984 x10 lb
8
0.60 ft
(5) Weight of disposed cargo = W – W’
W  W '  0.04742 x108 lb
 4,742,000lb
1LT
 4,742,000lbx
2,200lb
Fb ' f
(c) SHIP IN THE LOCK
 2,155.5 LONG TONS
Example 8.
W.S
W
G
A
4’
A
Bo
9’
8’
4’
15’
Fb
Given: Rectangular scow 50’ x 30’ x 12’
as shown with the given draft and center
of gravity.
Required: Righting or overturning moment
When one side, as shown, is at the point
Of submergence
15’
Solution:
A
W.S
A’
C  W MG sin  
M
G
O
Bo
where: W  V
A’
θ
A
4”
W  5030862.4
W  748,800lb
4
  14.93
15
 
  Arc tan
MG  MBo  GBo
MG  MBo  5
MBo 
vS
V sin 
1
15450 2 30
2
3

MBo 
50308sin 14.93
MBo  9.7 ft
Therefore,
MG  9.7 ft  5 ft  4.7 ft
Then,
C  748,800(4.7 sin 14.93)
C  906,600 ft  lb