Engineering Mechanics

Chapter 36 : Equilibrium of Floating Bodies „„„„„ 749

36.8.TYPES OF EQUILIBRIUM OF A FLOATING BODY

We have already discussed in articles 5·10 and 5·11 the conditions and types of equilibrium.
The same conditions of equilibrium are also applicable for the floating bodies. Thus like the general
types of equilibrium, a floating body may also be in any one of the following types of equilibrium :

1.Stable equilibrium, 2.Unstable equilibrium and 3. Neutral equilibrium
In this chapter we shall discuss the above mentioned types of equilibrium with respect to the
metacentre of the floating body.

36.9.STABLE EQUILIBRIUM

A body is said to be in a stable equilibrium, if it returns back to its original position, when
given a small angular displacement. This happens when the metacentre (M) is higher than the centre
of gravity (G) of the floating body.

36.10.UNSTABLE EQUILIBRIUM

A body is said to be in an unstable equilibrium, if it does not return back to its original position
and heels farther away, when given a small angular displacement. This happens when the metacentre
(M) is lower than the centre of gravity (G) of the floating body.

36.11.NEUTRAL EQUILIBRIUM

A body is said to be in neutral equilibrium, if it occupies a new position and remains at rest in
this new position, when given a small angular displacement. This happens when the metacentre (M)
coincides with the centre of gravity (G) of the floating body.

Example 36.6. A solid cylinder of 3 metres diameter has a height of 3 metres. It is made up
of a material whose specific gravity is 0·8 and is floating in water with its axis vertical.
Find its metacentric height and state whether its equilibrium is stable or unstable.
Solution. Given: Diameter of cylinder = 3 m; Height of cylinder = 3 m
and specific gravity = 0·8
We know that depth of immersion of the cylinder
= 0·8 × 3 = 2·4 m

and distance of centre of buoyancy, from the bottom of the cylinder,

∴ OB=

2·4

1· 2 m

2

=

Distance of c.g. from the bottom of the cylinder,

OG=

3

1·5 m

2

=

∴ BG=OG – OB = 1·5 – 1·2 m

= 0·3 m

Moment of inertia of the circular section,

I=

(3)^44 1·27 m

64

π

=π

and volume of water displaced,

V=

(3)^23 2·4 5·4 m

4

π

×=π

∴ BM=

1· 2 7

0·235 m

5·4

I

V

π

==

π

and metacentric height, GM=BM – BG = 0·235 – 0·3 = – 0·065 m.

Fig. 36.6.