Equilibrium of Floating Bodies

Equilibrium of Floating Bodies Points : chapter equilibrium of floating bodies, centre of buoyancy, example, metacentre, metacentric height, analytical method for metacentric height, kinds of equilibrium of a floating body, stable equilibrium, unstable equilibrium Buoyancy: The tendency of a fluid to uplift a submerged body, because of the upward thrust of the fluid, is known as the force of buoyancy or simply buoyancy. Ills always equal to the weight of the fluid displaced by the body. It will be interesting -to know that if the force of buoyancy is greater than the weight of the body, it will be pushed up tilt the weight of the fluid displaced is equal to the weight of the body, and the body will float. But if the force of buoyancy is less than the weight of the body, it will sink down. Centre of Buoyancy The centre of buoyancy is the point, through which the force of buoyancy supposed to act. It is always the centre of gravity of the volume of the liquid displaced other words, the centre of buoyancy is the centre of area of the immersed section. Example 1 A uniform body 3 m long, 2 m wide and I m deep floats in water. If the depth of immersion is 0.6 m, what is the weight of the body?

Solution:
Given
Length 3 m; Width 2 m; Depth 1 m and depth of immersion 0.6 m.
We know that volume of water displaced
=3 x 2 x 0.6 =3.6 m³
and weight of the body = Weight of water displaced
= 9.81 x3.6 = 35.3 kN Ans. Example 2 A floating buoy in harbour is to be assisted in floating upright by a submerged weight of concrete attached to the bottom of the buoy. How many cubic metres of concrete weighing 23 kNJm3 must be provided to get a net downward pull of 325 kN from the weight? Take specific weight of sea water as 10 KN/m³

Solution:
Given
Specific weight of concrete =23 kN/m³

Downward pull a 3.25 kN and specific weight of water. *w = 10 kN/m³
We know that the weight of I m3 of concrete in sea water
=23.10— 13kN/m³ and
the weight of concrete required.

= Downward pull 2.25/Weight of concrete in water 2.24/13 = 0.25m³ Example 3 A block of wood 4 m long, 2 m wide and I m deep is floating horizontally in water. If the density of the wood be 6-87 kN/m³, find, the volume of the water displaced and position of the centre of buoyancy.

Solution:
Given
Size of wooden block 4 m x 2 m x 1 m and density of wood — 6.87 kN/m³.
Volume of water displaced
We know that the volume of the wooden block
= 4 x 2 x 1 = 8m³
and it is weight = 8 x 687 = 55kN

Volume of water displaced
= Weight of Block/Density of water = 55/9.81 = 5.6 m²

Position of centre of buoyancy
We also know that the depth of immersion
= volume/Sectional Area = 5.6/4x2=0.7 m²
and centre of buoyancy = 0.7/2 = 0.35 m from the base Ans Example 4 A wooden block of 4 mx 1 mx 0.5 m in size and of specific gravity 0.75 is floating In water. Find the weight of concrete of specific weight 24 kN/m³ that may be placed on the block, which will Immerse the wooden block completely.

Solution:
Given
Size of wooden block = 4 m x 1 m x 0,5 m; Specific gravity of block
= 0-75 and specific weight of concrete — 24 kN/m³,
Let W = Weight of the concrete required to be placed on the wooden block.
We know that the volume of the wooden block

= 4 x 1 x 0.5 = 2 m³
And it is weight = 981 x (0.75 x 2) = 14.72kN
Total weight of the block and concrete
=14.72 +W kN

We also know that when the block is completely immersed in water, volume of water displaced =2m²

Upward thrust when the block is completely immersed in water
=9.81x2 = 19.62kN

Now equating the total weight of block and concrete with the upward thrust,
14.72 + W = 19.62 or W= 19.62 - 14.72 – 4.9 kN Ans 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 centre of buoyancy (B) and e.g., (0) of the body, and the vertical line through the new centre of buoyancy (B) Metacentric Height The distance between the centre of gravity of a floating body and the metacentre (I.e., distance GM as shown) is called metacentric height.

As a matter of fact, the metacentric height of a floating body is a direct measure of its stability. Or in other words, more the metacentric height of a floating body, more it will be stable. In the modem design offices, the metacentric height of a boat or ship is accurately calculated to check its stability. Some values of metacentric height are given below:

Merchant ships= up to 10m
Sailing ships = upto1-5m
Battle ships = upto20 m
River craft = up to 3-5 m Analytical Method for Metacentric Height Consider a ship floating freely in water. Let the ship be given a clockwise rotation through a small angle 8 (in radians) as shown in The immersed section has now changed from acde to acd iei.

The original centre of buoyancy, B has now changed to a new position /?,.It may be noted that the triangular wedge aom has come out of water, whereas the triangular wedge ocn has gone under water. Since the volume of water displaced remains the same, therefore the two triangular wedges must have equal areas. A little consideration will show that as the triangular wedge aom has come out of water, thus decreasing the force of buoyancy on the left, therefore it tends to rotate the vessel in an anti-clockwise direction about 0. Similarly, as the triangular wedge ocn has gone under water, thus increasing the force of buoyancy on the right, therefore it again tends to rotate the vessel in an anti-clockwise direction. It is thus obvious, that these forces of buoyancy will form a couple, which will tend t rotate the vessel in anti-clockwise direction about 0. If the angle (8 ), through which the body is given rotation, is extremely small, then the ship may be assumed to rotate about M { i . e., metacentre).

Let I = length of the ship
B= Breadth of the ship
θ = Very small angle ( in radian) through which the ship is rotated and
V = Volume of water displaced by the ship

From the geometry o the figure, we find that

Am = cn= bθ/2

Volume of wedge of water aom

=1/2(b/2 x am) 1= ½ (b/2 x bθ/2) i (am = bθ/2)
Weight of this wedge of water
= wb²θ1/8

And arm of the couple
= Wb² θ1/8 x2/3 b= Wb²θ1/8

And moment of the disturbing couple
= w.VxBB₁

Substituting the values of 1b³/12 = (i.e moment of inertia of the plant of the ship and
AB₁ = BM x θ in the above equation.
w.I.θ = w x V (BM x θ)
BM = 1/V = Moment of inertia to the plan/ Volume of water displaced

Now metacentric height
GM = BM ± BC Example 5 A rectangular pontoon of 5 m long, 3 m wide and 1-2 m deep is immersed 0-8 m sea water. If the density of sea water is 10 kN/m², find the metacentric height of the pontoon.

Solution:
Given
/=5m, b =3 m; d = 1-2 m ; Depth of immersion = 0-8 m and **w = 10
We know that distance of centre of buoyancy from the bottom of the block.
OB = 0 8/2=04 m the distance of e.g. from the bottom of the block
OG= 1-2/2=0-6m
BG = OG – OB = 06 - 0-4 = 02m

We also know that moment of inertia of the rectangular plan about central axis and parallel to the long side.

If the moment of inertia of the section parallel to the short side is taken, then the metacentric height will be than this. Since the metacentric height plays an important role in finding out the stability of a floating body Eh will be discussed in succeeding pages), it is, therefore general practice to find out the smaller metacentric it of the two. For doing so the moment of inertia of a rectangular section is always taken about the central and parallel to the long side. Such a moment of inertia is obtained by taking the cube of the breadth. BM = 1/V = 11.25/12 = 0.94 m
and metacenti height,
GM = BM-BG =0 94-0-2 = 0-74 m Ms. Example 6 A solid cylinder of 2 m diameter and 1 m height is made up of a material of specific gravity 0 7 and floats in water. Find its metacentric height.

Solution:
Given
Dia. of cylinder (d) = 2 m; Height of cylinder (h) = 1 m and sp. gr. of. cylinder = 0-7
We know that the depth of immersion of the block
07 x 1 = 07

Distance of centre of buoyancy from the bottom of the block,
CB = 07/2 = 035 m
and distance of e.g. from the bottom of the block,
OG= 1/2=0-5 m
BG=OG – OB – 05 – 035 = 015m

We also know that the moment of inertia of the circular plan about the central axis, Kinds of equilibrium of a floating body A body is said to be in equilibrium, when it remains in a steady state, while floating in a liquid. Following are the three conditions of equilibrium of a floating body:
1. Stable equilibrium,
2. Unstable equilibrium and
3. Netural Stable equilibrium A body is said to be in a stable equilibrium, if it returns back to its original position, when given i a small angular displacement. This happens when the metacentre (A/) is higher than the centre of gravity (G) of the floating body. Unstable equilibrium: A body is said to be in an unstable equilibrium, if it does not return back to its original position and heels farther away, when given a small angular displacement. This happens when the metacentre (Af) is lower than the centre of gravity (G) of the floating body.