Engineering Mechanics

Chapter 36 : Equilibrium of Floating Bodies „„„„„ 747

Example 36.4. A block of wood of specific gravity 0·8 and size 1·2 m × 0·4 m × 0·3 m floats
in water. Determine its metacentric height, for tilt about its longitudinal axis.

Solution. Given: Sp. gr. of wood = 0·8; Length of wooden block (l) = 1·2 m; Breadth of the
block (b) = 0·4 m and height or depth of the block (d) = 0·3 m.

We know that depth of immersion of the block

= 0·8 × 0·3 = 0·24 m

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

OB=

0·24

0·12 m

2

=

∴ Distance of c.g. from the bottom of the block,

OG=

0·3

0·15 m

2

=

∴ BG= OG – OB = 0·15 – 1·2 m
= 0·03 m ...(i)
We also know that moment of inertia of rectangular section
about the central axis and parallel to the long side,

I*=

33

1· 2 ( 0 · 4 ) 0·0064 m 4

12 12

lb

==

and volume of water displaced, V= 1·2 × 0·4 × 0·24 = 0·1152 m^3

∴ BM=

0·0064

0·056 m

0·1152

I

V

==

and metacentric height, GM = BM – BG = 0·056 – 0·03 = 0·026 m = 26 mm Ans.

Example 36.5. A buoy has the cylindrical upper portion of 2 metres diameter and 1·2 metre
deep. The lower portion, which is curved, displaces a volume of 400 litres of water and its centre of
buoyancy is situated 1·3 metre below the top of the cylinder. The centre of gravity of the whole buoy
is 0·8 m below the top of the cylinder and the total displacement of water is 2·6 m^3.

Find the metacentric height of the buoy.
Solution. Given: Dia. of buoy = 2 m; Depth of buoy = 1·2 m; Volume of curved portion =
400 litres = 0·4 m^3 ; Centre of buoyancy of the curved portion below the top of the cylinder (OB 1 ) =
1·3 m; Centre of gravity of the whole buoy below the top of the cylinder (OG) = 0·8 m and total
volume of water displaced = 2·6 m^3.

Fig. 36.3.

• If moment of inertia of a section parallel to the short side is taken, then the metacentric height will be
  more than this. Since metacentric height plays an important role in finding out the stability of a floating
  body (which will be discussed in succeeding pages), it is, therefore, general practice to find out the
  smaller metacentric height of the two.
  For doing so, the moment of inertia of a rectangular section is always taken about the central axis and
  parallel to the long side. Such a moment of inertia is obtained by taking the cube of the breadth.