Engineering Mechanics

Chapter 36 : Equilibrium of Floating Bodies „„„„„ 751

EXERCISE 36.1

1. A wooden block of volume 3 cubic metres floats in water. The specific gravity of the block is 0.6.
   What load should be placed on it so that it may be completely immersed in water.
   [Ans. 11.76 kN]


2. A block of wood floats in water with 6 cm projecting above the surface of water. If the same
   block is made to float in glycerine of specific gravity 1.4, it projects 18 cm above the surface of
   glycerine.
   Find the specific gravity of the wood. [Ans. 0.7]


3. A cylindrical buoy of 3 metres diameter, 4 metres long is weighing 4 tonnes. Show that it
   cannot float vertically in water.


4. A cylindrical buoy of 2 metres diameter is 3 metres long. Determine the state of its equilibrium,
   if the buoy weighs 2 tonnes. [Ans. Unstable equilibrium]

36.12.MAXIMUM LENGTH OF A BODY FLOATING VERTICALLY IN WATER

We see that a cube of wood (having specific gravity less than unity can float in water, in any
position. If we maintain any two sides (say breadth and thickness), of the cube, constant and go on
gradually increasing the third side (say length) and try to float the block vertically in water, we see
that the block can float vertically in water up to some length. If we increase the length of the block,
beyond this length, we find that it cannot float vertically in water; through it can float longitudinally.

This maximum permissible length of the block, floating vertically in water, may be found out
by keeping the body in stable equilibrium. Or in other words, this can also be found out by avoiding
the unstable equilibrium of the floating body. For doing so, the metacentre (M) should be above
centre of gravity (G) of the floating body (a condition of stable equilibrium) or the metacentre (M)
may coincide with the centre of gravity (G) of the floating body (a condition of neutral equilibrium
i.e., by avoiding the unstable equilibrium).

Example 36.8. A wooden cylinder of a circular section and uniform density with specific
gravity 0·6, is required to float in an oil of specific gravity 0·9. If the cylinder has a diameter
(d) and length (l), show that (l) cannot exceed (0.75 d) for the cylinder to float with its longi-
tudinal axis vertical.

Solution. Given: Sp. gr. of cylinder = 0·6 and sp. gr. of oil = 0.9.

Let l= Length of cylinder, and

d= Dia of cylinder.

We know that depth of immersion of the cylinder

=

12

0.6

0.9 3

l

××=l

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

OB=

12

233

ll

×=

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

OG=

2

l

∴ BG=OG – OB = –

236

lll

= ...(i)

Fig. 36.8.