Chapter 36 : Equilibrium of Floating Bodies „„„„„ 755

From the figure, we find that distance of centre of buoyancy from the apex O,

OB=

3

0.75

4

l

= l

and distance of centre of gravity from the apex O,

OG=

3

0.75

4

L

= L

∴ Volume of liquid displaced,

V=^32

1

tan

3

παl

and moment of inertia of the circular section about the liquid level,

I=^44 (2 tan )

64 64

dl

ππ

×= α = (tan )^44

4

l

π

α

Now the value of BM and metacentric height is found out as usual.

We know that BM=

44

2

32

(tan )

(^4) 0.75 tan
1
tan
3
I l
l
V l
π
α
==α
πα
.
Example 36.10. A wooden cone of specific gravity 0.8 is required to float vertically in
water. Determine the least apex angle, which shall enable the cone to float in stable equilibrium.
Solution. Given: Sp. gr. of cone = 0.8.
Let L= Length of the cone,
l= Length of the cone immersed
in water, and
2 α= Apex angle of the cone.
We know that weight of the cone
= Volume of cone × specific weight
of cone

(^132) tan (0.8 9·8)
3
πα××L
and weight of water displaced = Volume of water displaced × specific weight of water

(^132) tan (1·0 9·8)
3
πα××l
Since the cone is floating in water, therefore the weight of the cone is equal to the weight
of the water displaced. Therefore
(^132) tan (0·8 9·8)
3
πα××L =^32
1
tan (1·0 9·8)
3
πα××l
∴ l=L (0.8)1/3
Distance of the centre of buoyancy from the apex,
OB= 0.75 l = 0.75 L (0.8)1/3
Fig. 36.11.