Equilibrium Stability

Buoyancy:

When a body is immersed in fluid an upward force is exerted by the fluid on the body. This upward force is equal to the weight of the fluid displaced by the body is called the force of buoyancy or buoyancy.

Center of Buoyancy:

It is defined as the point through which the force of buoyancy is supposed to act. As the force of buoyancy is a vertical force and is equal to the weight of the fluid displaced by the body the center of buoyancy will be the center of gravity of the fluid displaced.

Floating concept:

When any boat displaces a weight of water equal to its own weight, it floats. This is often called the “principle of floatation”, a floating object displaces a weight of fluid equal to its own weight. Every ship, submarine must be designed to displaced weight of fluid at least equal to its own weight.

Experimental Method for determination of metacentric height:

Let,

W= Weight of vessel including w₁

G = Center of gravity of the vessel

B = Center of buoyancy of the vessel

The weight w₁   is moved across the vessel towards right through a distance x as shown in figure B. The vessel will be tilted. The angle of heel θ is measured by means of a plumbline and a protractor attached on the vessel.The new center of gravity of the vessel will be G₁ as the weight w₁ has been moved toward the right.

Above the center of buoyancy will change to B₁ as the vessel has tilted.

Under equilibrium the moment caused by the movement of load w₁ through a distance x must be equal to the movement caused by the shift of center of gravity from G to G₁ .

Thus,

The moment due to change of G

= GG₁ * W

= W * GMtanθ

 The moment due to movement of w₁

= w_{1 *} x

From above equation,

w_{1 *} x = W * GMtanθ

or, GM = (w_{1 *} x)/ Wtanθ

Analytical Method for Metacentric height:

In figure A, it shows the position of a floating body in equilibrium. The location of center of gravity and center buoyancy in this position is at G and B. The floating body is given a small angular displacement in the clockwise direction. This is shown in figure B. The new center of buoyancy is B_1. The vertical line through B₁ cuts the normal axis at M. Hence, M is the metacenter and GM is metacentric height.

Couple due to wedges

Consider towards the right of the axis a small strip of thickness dx at a distance x from O as shown in figure B.

The height of strip x ∠ BOB^’ = x* θ [∴∠BOB^’ = ∠AOA^’ = ∠BMB₁^’=θ]

∴Area of strip = Height * Thickness

                            = x* θ*dx

If L is the length of the flaoting body,then

Volume of strip = Area * L

                          = x* θ*L*dx

∴Weight of strip = ρg*Volume

                            = ρg *x* θ*L*dx

Moment of the couple = Weight of each strip * Distance between these weigth

                                    = ρg x θLdx*[x+x]

                                    = 2ρg x² θLdx

∴Moment of couple for the whole wedge=∫2ρg x² θLdx ——(i)

Moment of couple due to shifting of center of buoyancy from B to B₁

= F_B * BB₁

= F_B * BM* θ [∴ BB₁ = BM* θ , if θ is very small]

= W * BM* θ [∴ F_B = W] ————(ii)

Hence,equating equation (i) and (ii)

W * BM* θ =∫2ρg x² θLdx

Or, W * BM* θ = 2ρgθ∫ x² Ldx

Or, W * BM = 2ρg∫ x² Ldx

But,

Ldx = dA [∴ Elemental area on the water line shown in figure C]

∴W*BM=2ρg∫ x² dA

But, figure C it is clear that 2∫ x² dA is the second moment of area of the plane of the body at water surface about the axis Y-Y.

Therefore,

W*BM = ρgI [∴ where I = 2∫ x² dA]

∴ BM = (ρgI)/ W

But,

W= Weight of the body

    = ρg * Volume of the body submerged in water

    = ρg * ∀

∴ BM = (ρgI)/ ρg * ∀

Or, BM = I/∀

Now,

GM = BM – BG

        = (I/∀)  – BG

∴Metacentric height = GM = (I/∀)  – BG