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 buoyantforce has a magnitude equal to the weight of thefluid displaced by the body and is directedvertically upward.

• Buoyant force is a force that results from afloating or submerged body in a fluid.

• The force results from different pressures on thetop 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, orvertically upward and is known as the BuoyantForce.

The force is equal to the weight of the fluid itdisplaces.

The buoyant forces acts through the centroid ofthe 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” 4

STABILITY: SUBMERGED OBJECT

If the Centre of Gravity is below the centre ofbuoyancy this will be a righting moment and thebody will tend to return to its equilibriumposition (Stable).

If the Centre of Gravity is above the centre ofbuoyancy ,an overturning moment is producedand the body is (unstable).

Note that, As the body is totally submerged, theshape of displaced fluid is not altered when thebody is tilted and so the centre of buoyancyunchanged 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 aspositive.

If M lies below G an overturning moment isproduced, equilibrium is unstable and GM isregarded as negative.

If M coincides with G, the body is in neutralequilibrium.

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DETERMINATION OF METACENTRIC HEIGHT1- Practically :

2- Theoretically: MG = BM + OB – OG……..........(2)

In Water

OB = 0.5 dbV.

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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 shipmodel is floated, the weight of the model can bechanged by adding or removing weights.Adjustable mass is used for tilting the ship,plump line is attached to the mast to measurethe 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 ensurethat the plumb line on the zero mark.

4. Apply a weight of 50 g on the bridge pieceloading pin then measure and record the angle oftilting 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

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

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. 19

θ.3500)123(PGM =

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

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W

LWmWbWbWvmOG

∑

++++=

)2

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

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3500

)2

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

++++=

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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 = 100.020 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