Mechanical Engineering Principles

242 MECHANICAL ENGINEERING PRINCIPLES

21.14 The stability of floating bodies

For most ships and boats the centre of buoyancy (B)
of the vessel is usually below the vessels’ centre of
gravity (G), as shown in Figure 21.13(a). When this
vessel is subjected to a small angle of keel (θ), as
shown in Figure 21.13(b), the centre of buoyancy
moves to the positionB′,where

W

W

G

B

W

W

M

G

B B′

(a) (b)

q

Figure 21.13

BM=the centre of curvature of the centre of

buoyancy=

I

V

, (given without proof)

GM=the metacentric height,
M=the position of the metacentre,
I=the second moment of area of the water
plane about its centreline, and
V=displaced volume of the vessel.
The metacentric height GM can be found by
a simple inclining experiment, where a weightP
is moved transversely a distancex, as shown in
Figure 21.14.

G

P

B

GG′

BB′ P

M

x

(a) (b)

Figure 21.14

From rotational equilibrium considerations,

W(GM)tanθ=Px

Therefore, GM=

Px

W

cotθ( 21. 1 )

whereW=the weight of the vessel, and

cotθ=

1

tanθ

Problem 13. A naval architect has carried

out hydrostatic calculations on a yacht,

where he has found the following:

M =mass of yacht=100 tonnes,

KB=vertical distance of the centre of

buoyancy(B)above the keel(K)

= 1 .2 m (see Figure 21.15),

BM=distance of the metacentre(M)above

the centre of buoyancy= 2 .4m.

M

G

B

K

Figure 21.15

He then carries out an inclining experiment,

where he moves a mass of 50 kg through a

transverse distance of 10 m across the

yacht’s deck. In doing this, he finds that the

resulting angle of keelθ= 1 °. What is the

metacentric height (GM) and the position of

the centre of gravity of the yacht above the

keel, namelyKG? Assumeg= 9 .81 m/s^2.

P=50 kg× 9. 81 = 490 .5N,

W=100 tonnes× 1000

kg

tonne

× 9. 81

m

s^2

=981 kN,

x=10 m,

θ= 1 °from which, tanθ= 0 .017455 and

cotθ=

1

tanθ

= 57. 29