THE STABILITY OF A FLOATING BODY

Objective:
The objectives of the experiment was to determine the metacentric height of a floating body and the position of stability of a pontoon.

Introduction:
The question of the stability of a body such as a ship, which floats in the surface of a liquid is one of the obvious importance. Whether the equilibrium is stable, neutral or unstable is determined by the height of its center of gravity, and in this experiment, the stability of a pontoon may be determined with its centre of gravity at various heights.

Apparatus:
Pontoon, water tank , an adjustable weight, small magnetic weights and a plumbline.

a)

jockey weight

Plumbline

Method:
Theory: - metacentric height, GM = [pic] where W = total weight of the floating assembly (i.e., weight of pontoon and the applied weights). Wi = adjustable weight Sx = distance moved of adjustable weight from central position St = angle of tilt.

Let the angle of tilt, corresponding to a small displacement Sx, from the central position (i.e., from the plumbline) be st. the whole system tilts from the original position by this angle.

For one chosen height of the adjustable weight, the height of the centre of gravity of the whole floating assembly from its base OG, was measured: The pontoon was suspended from a hole at one side of the soil. The jockey weight was positioned on the line of symmetry such that the pontoon hung with its base almost vertical.

A plumbline was hung from the suspension point. The height of the centre of gravity, G, of the whole suspended assembly was located at the point of intersection between the plumb-line and the line of symmetry of the pontoon. Thus, the position, G, of the jockey height was determined.

The dimensions of the pontoon and the various components were taken, after which the pontoon was floated in water. Small magnetic weights were used to trim the assembly to even keel, whiles the jockey weight still remained on the line of symmetry. The jockey was then displaced in steps across the width of the pontoon, and the corresponding angle of tilt recorded at each step. This procedure was then repeated with the jockey traveled at a number of different heights.

a) Floating pontoon tilted by movement of jockey weight b) Determination of position of centre of gravity.
Overturning moment due to moment of load W = Wsx
Righting moment = WGm x st
For equilibrium in the tilted position, the righting moment must be equal to the overburning moment.
Thus W x Gm x St = wsx :. GM = [pic]
It can also be shown from analytical determination that the distance from the centre of buoyancy, B, to the metacentre point M, i.e., BM = [pic]
Centre of buoyancy B to the metacentre M, i.e. BM [pic].
Where I = second moment, about the axis of symmetry of the water plane area of the body, and V = volume of liquid displaced.
This value, computed from theory is seen to be reasonably verified by experimental results.

Results:
Total mass of pontoon without weight, MT = 2390g
Centre of gravity of pontoon = 92mm
Mass of adjustable weight = Ma = 392.0g
Mass of small magnetic weight = Mm = 44.0g

|Dimensions of Pontoon |
|Length: 360mm |Distance between rows: 60mm |
|Width: 202mm |Distance between slots: 75mm |
|Vertical height: 406mm | |

|Displacements |-37.5 |
|(sx) | |
|2.31 |5 2.8 |
|3.33 |44.6 |
|4.10 |36.4 |
|5.00 |28.2 |
|5.82 |20.6 |

DETERMINATION OF OG

M

G

C

B

Deductions from graph

Slope, m = [pic]
From the graph, the intercept is 8.065mm/deg
Hence equation of the line is [pic] = (-0.11Ogwj + 8.065) x w-3m/deg ………… ( [pic] = [pic]OGwj + [pic](BM – OB) ………… ( Comparing ( and ( [pic](BM – OB) = 0.06X10-3m/deg
But OB = depth of immersion (OC) [pic] = [pic] = 19.6 mm = 19.6x10-3m But [pic] = 0.11
[pic] 0.11 (BM – 19.6 X 10-3) = 8.065 X 10-3 [pic] BM = [pic]+ 19.6 x 10-3 = 0.093m or BM = 93.0mm

From BM = [pic], the theoretical value can be computed where, I = second moment about axis of symmetry of water plane area of the body V = volume of liquid displayed I = [pic] LD3 where L = length and D = width of pontoon.
[pic] I = [pic] x 360 x 10-3x (293x10-3)3 = [pic]x 10-3m4 V = Base area x Depth of immersion = 0.36 x 0.203 x 0.0392 = 2.865 x 10-3m3
:. BM = [pic]= [pic]10-3 x [pic] = 0.0876m = 87.6 x 10-3m GM = (BM – OB) – OGwj
But practically, BM = 0.093m Also, OB = [pic]= 0.0196m
Therefore GM = (0.093 – 0.0196) – OGwj = 0.0734 – OGwj

From the equation [pic] = (-0.11OGwj + 8.065) x 10-3 When [pic] = O [pic] -0.11OGwj + 8.065 = 0 [pic] OGwj = [pic]mm
This 73.32mm is the maximum value of OGwj

Sources of Error

1. The plumbline was shaking 2. The place of experiment was sloppy.

Precautions

1. It was ensured that the plumbline was stable before readings were taken. 2. It was ensured that the working surface was stable.

Conclusion

Extrapolating the graph 2 helped to give the values for points which are outside the range of value used and obtained in the experiment.
It can also be concluded that as the stability increases the volume of fluid displaced becomes very small.

100% accuracy was considered in carrying out the experiment

For a liquid with a greater density than that of water, the volume displayed by the pontoon will be greater that of water.

This would cause the metacentric height to increase. Hence stability of pontoon is increased

Reference:

-----------------------
O

[pic]