EXPERIMENT (2)
BUOYANCY & FLOTATION
(METACENTRIC HEIGHT)
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By:
Eng. Motasem M. Abushaban.
Eng. Fedaa M. Fayyad.
ARCHIMEDES’ PRINCIPLE

•
•
Archimedes’ Principle states that the buoyant
force has a magnitude equal to the weight of the
fluid displaced by the body and is directed
vertically upward.
Buoyant force is a force that results from a
floating or submerged body in a fluid.
The force results from different pressures on the
top and bottom of the object.
W is the weight of the shaded area
F1 and F2 are the forces on the plane surfaces
FB is the buoyant force the body exerts on the fluid
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ARCHIMEDES’ PRINCIPLE
The force of the fluid on the body is opposite, or
vertically upward and is known as the Buoyant
Force.
 The force is equal to the weight of the fluid it
displaces.
 The buoyant forces acts through the centroid of
the displaced volume

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The location is known as the center of buoyancy.
STABILITY: SUBMERGED OBJECT
Stable Equilibrium: if when displaced returns to equilibrium position.
Unstable Equilibrium: if when displaced it returns to a new equilibrium
position.
Stable Equilibrium:
C > CG, “Higher”
Unstable Equilibrium:
C < CG, “Lower”
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STABILITY: SUBMERGED OBJECT
If the Centre of Gravity is below the centre of
buoyancy this will be a righting moment and the
body will tend to return to its equilibrium
position (Stable).
 If the Centre of Gravity is above the centre of
buoyancy ,an overturning moment is produced
and the body is (unstable).
 Note that, As the body is totally submerged, the
shape of displaced fluid is not altered when the
body is tilted and so the centre of buoyancy
unchanged relative to the body.

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BUOYANCY AND STABILITY: FLOATING OBJECT
Slightly more complicated as the location of the center buoyancy can change:
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METACENTRE AND METACENTRIC HEIGHT
Metacentre point (M): This point, about which
the body starts oscillating.
 Metacentric Height: Is the distance between the
centre of gravity of floating body and the
metacentre.

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STABILITY OF FLOATING OBJECT
If M lies above G a righting moment is produced,
equilibrium is stable and GM is regarded as
positive.
 If M lies below G an overturning moment is
produced, equilibrium is unstable and GM is
regarded as negative.
 If M coincides with G, the body is in neutral
equilibrium.

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DETERMINATION OF METACENTRIC HEIGHT
1- Practically :
2- Theoretically:
MG = BM + OB – OG……..........(2)
In Water
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V
OB = 0.5
b.d
h
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PURPOSE:

To determine the metacentric height of a flat
bottomed vessel in two parts:
PART (1) : for unloaded and for loaded pontoon.
PART (2) : when changing the center of gravity of
the pontoon.
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EXPERIMENTAL SET-UP:

The set up consists of a small water tank having
transparent side walls in which a small ship
model is floated, the weight of the model can be
changed by adding or removing weights.
Adjustable mass is used for tilting the ship,
plump line is attached to the mast to measure
the tilting angle.
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PART (1)
Determination of floatation
characteristic for unloaded and for
loaded pontoon.
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PROCEDURE
1. Assemble the pontoon by positioning the bridge
piece and mast.
2. Weigh the pontoon and determine the height of
its center of gravity up the line of the mast.
3. Fill the hydraulic bench measuring tank with
water and float the pontoon in it, then ensure
that the plumb line on the zero mark.
4. Apply a weight of 50 g on the bridge piece
loading pin then measure and record the angle of
tilting and the value of applied weight
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PROCEDURE
5. Repeat step 4 for different weights; 100, 150, &
200 g, and take the corresponding angle of tilting.
6. Repeat the above procedure with increasing the
bottom loading by 2000 gm and 4000 gm.
7. Record the results in the table.
8. Calculate GM practically where , W has three
cases.
9. Draw a relationship between θ (x-axis) and GM
(y-axis), then obtain GM when θ equals zero.
10. Calculate GM theoretically.
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Pontoon measurement:
- Pontoon dimension : Depth (D) = 170 mm
Length (L) = 380 mm, Width (W) = 250 mm.
-The height of the center of gravity of the pontoon is OGvm = 125 mm from
outer surface of vessel base.
- The balance weight is placed at x = 123 mm from pontoon center line.
- The weight of the pontoon and the mast Wvm = 3000 gm
Bilge
Weight
Off balance
wt.
Wb (gm)
P (gm)
0.00
50
Mean
Def.
θ
(degree)
Exp.
GM
(mm)
GM at
θ =0
from
graph
BM
OB
Theo.
GM
(mm)
(mm)
(mm)
100
150
200
2000.00
50
x1 = 30
100
150
200
4000.00
100
x1 = 37.5
150
200
250
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PURPOSE:

To determine the metacentric height of a flat
bottomed vessel in two parts:
PART (1) : for unloaded and for loaded pontoon.
PART (2) : when changing the center of gravity of
the pontoon.
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Remember:
- Pontoon dimension : Depth (D) = 170 mm
Length (L) = 380 mm,
Width (W) = 250 mm.
- The height of the center of gravity of the pontoon is
OGvm = 125 mm from outer surface of vessel base.
- The balance weight is placed at x = 123 mm from
pontoon center line.
- The weight of the pontoon and the mast Wvm = 3000 gm
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PROCEDURE

PART (2) : when changing the center of gravity of the pontoon.
1. Replace the bilge weights by 4x 50 gm weights.
2. Apply a weight of 300gm on a height of 190 mm from
the pontoon surface.
3. Apply weights of 40, 80 &120 gms on the bridge piece
loading pin, then record the corresponding tilting angle.
4. Calculate GM practically where
GM 
P (123)
3500.
5. Draw a relationship between θ in degrees (x-axis) and
GM Practical (y-axis), then obtain GM when θ equals
zero.
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PROCEDURE
6. Move 50 gm bilge weight to the mast ahead, then
repeat steps 3,4&5.
7. Repeat step 6 moving 100, 150 & 200 gm bilge
weight to the mast.
8. Determine the height of the center of gravity for
each loading condition according to equation
L
Wvm(125)  Wb(35)  Wb1(190)  Wm(790  )
2
OG 
W
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L
3000(125)  300(190)  Wb(35)  Wm(790  )
2
OG 
3500
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8. Calculate GM theoretically according to equation
GM (Th.) = BM + OB – OG
Notice: BM & OB are constants for all loading conditions, since
the dimensions & the weight of pontoon do not alter.
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Table (2) \ Part (2)
Off balance wt.
Mean Def.
Exp. GM
BM
OG
Theo. GM
P (gm)
θ (degree)
(mm)
(mm)
(mm)
(mm)
Mast Weight = 0.0
40
2.40
80
4.88
120
7.50
Mast Weight = 50.0
40
3.45
80
7.23
120
10.50
Mast weight = 100.0
20
3.28
40
6.35
80
12.00
Mast Weight = 150.0
10
3.70
20
10.23
40
14.78
Mast weight = 200.0
Unstable
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QUESTIONS
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