(^744) „„„„„ A Textbook of Engineering Mechanics
and volume of the water displaced =
Weight of block (^56) 5·71 m 3
Density of water 9·8

Position of the centre of buoyancy
We know that the depth of immersion

Volume of water displaced 5·71
0·71 m
Sectional area 4 2

×
and centre of buoyancy =
0·71
0·355 m
2
= Ans.
Example 36.3. A piece of steel of specific gravity 7·8 floats in mercury of specific gravity
13·6. If sufficient water is added just to cover the steel, what fraction of the steel will be below the
surface of mercury?
Solution. Given: Specific gravity of steel = 7·8 and specific gravity of mercury = 13·6.
Let x= Part of the steel piece inside the mercury.
∴ (1 – x) = Part of the steel piece outside the mercury, i.e., inside water.
Consider one cubic metre of the steel piece. We know that weight of the body
(9·8 × 7·8) × 1 = Weight of the fluid displaced
= Weight of mercury displaced + Weight of water displaced
= (9·8 × 13·6) × x + 9·8 (1 – x)
76·44 = 133·28 x + 9·8 – 9·8 x
123·48 x= 66·64
or x=
66·64
0·54
123·48

∴ Fraction of steel inside the mercury

0·54
0·54
1
= Ans.
36.5.METACENTRE
Whenever a body, floating in a liquid, is given a small
angular displacement, it starts oscillating about some point.
This point, about which the body starts oscillating, is called
metacentre.
In other words, the metacentre may also be defined as
the intersection of the line passing through the original cen-
tre of buoyancy and centre of gravity of the body, and the
vertical line through the new centre of buoyancy as shown in
Fig. 36.1.
36.6.METACENTRIC HEIGHT
The distance between the centre of gravity (G) of a floating body, and the metacentre (M)
i.e., distance GM as shown in Fig. 36.1 is called metacentric height.
Fig. 36.1. Metacentre.