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EXPERIMENT (2)

BUOYANCY & FLOTATION (METACENTRIC HEIGHT)

By:

Eng. Motasem M. Abushaban.

Eng. Fedaa M. Fayyad.

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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 areaF1 and F2 are the forces on the plane surfacesFB 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

The location is known as the center of buoyancy.

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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: Unstable Equilibrium:

C > CG, “Higher” 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

OB = 0.5 db

V

.

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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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PROCEDURE1. 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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Bilge Weight

Off balance wt.

Mean Def.

Exp. GM

GM at θ =0

BM OBTheo. GM

Wb (gm) P (gm)θ

(degree)(mm)

from graph

(mm) (mm) (mm)

0.00 50          

  100          

  150          

  200          

2000.00 50          

x1 = 30 100          

  150          

  200          

4000.00 100          

x1 = 37.5 150          

  200          

  250          

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 fromouter 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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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

5. Draw a relationship between θ in degrees (x-axis) and

GM Practical (y-axis), then obtain GM when θ equals

zero.

.3500

)123(PGM

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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

W

LWmWbWbWvm

OG

)2

790()190(1)35()125(

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3500

)2

790()35()190(300)125(3000L

WmWbOG

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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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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          

Table (2) \ Part (2)

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QUESTIONS