Module 4. Floating bodies

Lesson 10

EQUILIBRIUM OF FLOATING BODIES, METACENTRIC HEIGHT

10.1  Introduction

In the previous lesson, the position (i.e. center of gravity) of a solid cone affecting its stability was discussed. This lesson will cover the equilibrium states of the floating bodies and the factors affecting it.

10.2  Equilibrium of Floating Body

1. Stable equilibrium

2. Unstable equilibrium

3. Neutral equilibrium

10.2.1  Stable equilibrium

When a body is given a small angular displacement, i.e. it is tilted slightly by some external force and then it returns back to original position due to internal forces. Such equilibrium is called stable equilibrium.

10.2.2  Unstable equilibrium

If a body does not return to its original position from the slightly displaced angular position and moves father away when give a small angular displacement such equilibrium is called an unstable equilibrium.

10.2.3  Neutral equilibrium

The body remains at rest in any position to which it may be displaced, no net force tends to return the body to its original state or to drive it further away from the original position, is called neutral equilibrium. Fig. 10.1 shows the case of neutral equilibrium

10.3  Metacentre

When a small angular displacement is given to a body floating in a liquid, it starts oscillation about some point M. This point about which the body starts oscillating is called the metacentre (Fig. 10.2).

Metacentre (M) may be defined as the point of intersection of the axis of body passing through centre of gravity (G) and original centre of buoyancy (B) and a vertical line passing through the new centre of buoyancy (B’) of the titled position of the body.

10.4  Metacentric height

Metacentric height: The distance between the centre of gravity of a floating body and the metacentre, i.e. distance GM is called meta-centric height (Fig. 10.2). Relation between centre of gravity and metacentre in different three types of equilibrium:

(a) Stable equilibrium

 In this, position of metacentre (M) remains higher than centre of gravity of body.

(b) Unstable equilibrium

 In this position of metacentre (M) remains lower than centre of gravity of body.

(c) Neutral equilibrium

The position of metacentre (M) coincides with centre of gravity of body.

10.5  Conditions of Stability in Air Balloon

Case 1: When G is lower than B

The balloon having G (centre of gravity) below B (centre of buoyancy) is a condition of stable equilibrium as shown in Fig. 10.3. Under equilibrium condition the balloon will return back to its original position if it is tilted a bit due to wind etc.

Case 2: When G is above B

Fig. 10.4 shows a air balloon that will be highly unstable when centre of gravity (G) is above centre of buoyancy (B).

10.6 Stability of ship

Fig. 10.5 Condition of stability of ship

The above cases were of submerged body. For floating body particularly ship the construction of ship body plays an important role in its stability. It is such so as to withstand high wave and tide (Fig. 10.5).

10.7 Calculation of Metacentric Height

In the Fig. 10.6:

G = Centre of gravity

O = The point at which line BM and top liquid surface intersect

M = Metacentre

B = Centre of buoyancy 

BM = Distance between centre of buoyancy and metacentre

Fig. 10.6 Metacentric height

Where I = area moment of inertia of the cross sectional area at the surface of fluid (Fig. 10.7).

Fig. 10.7 a×b is the cross sectional area at the surface of liquid

Area moment of Inertia

V= volume of the displaced fluid.

 Metacentric Height GM = BM - BG

If GM is +ve then G is above point B.

If GM is –ve then G is below point B.

10.8  Numericals

Q1. A wooden block of specific gravity = 0.75 floats in water if the size of the block is 1.2 m × 0.6 m × 0.5 m. Find its metacentric height i.e. GM.

Solution:

          BM = I/V

V × 1000 × 9.81 = (1.2 × 0.6 × 0.5) × 750 × 9.81

          V = 0.27 m³

          OG = 0.5/2 = 0.25 m

          h × 0.6 × 1.2 = 0.27

          h =

          OB = h/2 = 0.1875 m

          BG = OG - OB = 0.25 - 0.1875 = 0.0625 m

          GM = BM - BG

          BM =

          GM = 0.08 - 0.0625 = 0.0175 m = 1.75 cm.    

Q2. A solid cylinder 3 m in diameter and 3 m height is floating in water with its axis vertical. If the specific gravity of cylinder is 0.7 then find its meta centric height, i.e. GM.

Solution:

Q3. A wooden cube side 1 m is floating in water. Specific gravity of cube is 0.7; Find its meta centric height.

Solution:

          1 × 1 × 1 × 700 × 9.81 = V × 1000 × 9.81

          V = 0.7 m³

          h × 1 × 1 = 0.7

          h = 0.7 m

OB = h/2 = 0.35 m

          OG = 1/2 = 0.5 m

          BG = OG - OB = 0.5 - 0.35 = 0.15 m

BM = I/V