BUOYANT FORCE

When a body is either wholly or partially immersed in a fluid, a

lift is generated due to the net vertical component of hydrostatic
pressure forces experienced by the body. This lift is called the
buoyant force and the phenomenon is called buoyancy. The
buoyant force FB acts upward through the centroid C of the
displaced volume and is equal in magnitude to the weight W of
the displaced fluid, but is opposite in direction

Prof. Sreeja Sadasivan

Archimedes Principle

Prof. Sreeja Sadasivan

Prof. Sreeja Sadasivan
 WHEN WILL A BODY FLOAT ?

For a body to float on a fluid the weight of the fluid

displaced by the body should be atleast equal to its
own weight

Free surface CG

CB

Fb
Prof. Sreeja Sadasivan
 < 1

= 1

> 1

For a body to float, the buoyant force (weight of the

fluid displaced), should be greater than the weight of
the solid body Prof. Sreeja Sadasivan
Prof. Sreeja Sadasivan
Prof. Sreeja Sadasivan
 Stability of submerged and floating bodies

Depending upon the relative locations of G and B, a floating or

submerged body attains three different states of equilibrium

Let us suppose that a body is given a small angular displacement

and then released. Then it will be said to be in

Stable Equilibrium: If the body returns to its original position by

retaining the originally vertical axis as vertical.

Unstable Equilibrium: If the body does not return to its original

position but moves further from it.

Neutral Equilibrium: If the body neither returns to its original

position nor increases its displacement further, it will simply
adopt its new position.
Prof. Sreeja Sadasivan
 Stability of Submerged Object

When a body is submerged in a liquid (or a fluid), the

equilibrium requires that the weight of the body acting
through its Center of Gravity should be co-linear with
the Buoyancy Force acting through the Center of
Buoyancy. If the Body is Not Homogeneous in its
distribution of mass over the entire volume, the location
of Center of Gravity (G) does not coincide with the
Center of Volume (B). Depending upon the relative
locations of (G) and (B), the submerged body attains
different states of equilibrium: Stable, Unstable and
Neutral.

Prof. Sreeja Sadasivan

 Stability of Submerged Object
The rotational stability of an immersed body depends on the relative locations
of the centre of gravity G of the body and the centre of buoyancy B. An
immersed body is stable if the body is bottom-heavy and thus point G is
directly below point B. If the body is tilted slightly in any direction, the buoyant
force and the weight always produce a restoring couple trying to return the
body to its original position
A stable design for a submarine calls for the engines and the cabins for the
crew to be located at the lower half in order to shift the weight to the bottom as
much as possible

Prof. Sreeja Sadasivan

An immersed body whose centre of gravity G is directly above
point B is unstable, and any disturbance will cause this body to
turn upside down. if point G is above point B, any disturbance
from the equilibrium position will create a destroying couple
which will turn the body away from its original position.

A body for which G and B coincide is neutrally stable. When the

centre of gravity G and centre of buoyancy B coincides, the body
will always assume the same position in which it is placed and
hence it is in neutral equilibrium

it can be concluded that a submerged body will be in

stable, unstable or neutral equilibrium if its centre of
gravity is below, above or coincident with the centre of
buoyancy respectively.

Prof. Sreeja Sadasivan

Prof. Sreeja Sadasivan
 Stability of Floating Object

Stable conditions of the floating body can be achieved, under certain

conditions even though (G) is above (B). When a floating body undergoes
angular displacement about the horizontal position, the shape of the
immersed volume changes and so, the Center of Buoyancy moves relative to
the body.

Prof. Sreeja Sadasivan

 Stability of Floating Object
A measure of stability for floating bodies is the metacentric height,
which is the distance between the centre of gravity and the
metacentre

Stable equilibrium : If the point M is above G ( GM > 0)

Neutral equilibrium : If the point M is at G (GM = 0)
Unstable equilibrium : If the point M is below G (GM < 0)
Prof. Sreeja Sadasivan
 METACENTRE
METACENTRE is defined as the point about which a body
starts oscillating when a body is tilted by a small angle. It is
also defined as the point at which the line of action of the
force of buoyancy will meet the normal axis of the body
when body is given a small angular displacement.
Metacentric height (GM) is given by the formula as follows:

Prof. Sreeja Sadasivan

 DETERMINATION OF META-CENTRIC HEIGHT
Consider a floating object as shown. It is given a small tilt angle(θ) from the initial
state. Increase in the volume of displacement on the right hand side displaces the
Center of Buoyancy from (B) to (B1)

Prof. Sreeja Sadasivan

The shift in the center of Buoyancy results in the Restoring Couple = W (BM tan θ);

Since FB= W;

FB This is the moment caused by the movement of Center of Buoyancy from (B) to (B1)

Volume of the liquid displaced by the object remains same.

Area AOA1=Area DOD1

Weight of the wedge AOA1(which emerges out)=Weight of the wedge DOD1(that was
submerged)

Let (l) and (b) be the length and breadth of the object. .

Weight of each wedge shaped portion of the liquid = dFB

= (w/2)(b/2)(b/2)(tanθ)(l)

=(w(b^2)l tan θ)/8

There exists a buoyant force dFB upwards on the wedge (ODD1) and dFB downwards on
the wedge (OAA1) each at a distance of (2/3)(b/2)=(b/3) from the center.
Prof. Sreeja Sadasivan
Prof. Sreeja adasivan
A block of wood of specific gravity=0.7 floats in water.
Determine the meta-centric height if it’s size is 1m×1m × 0.8m.

Let the depth of immersion=h

Weight of the wooden block = 5494 N

Weight of water displaced

= 1000 × 9.81× 1×1 × h
= 5494N

Therefore, h=0.56 m; AB =0.28 m;

AG=0.4 m; BG=0.12 m;

GM=(Iyy /V) - BG

Iyy =(1 / 12) m4,

V=0.56m3
GM = 1 / (12 × 0.56) - 0.12 = 0.0288m

(The body is Stable)

Prof. Sreeja Sadasivan
A rectangular barge of width ‘b’ and a submerged depth ‘H’ has its center of gravity at
the water line. Find the meta-centric height in terms of (b/H) and show that for
stable equilibrium of the barge, (b/ H) >√6

Prof. Sreeja Sadasivan

Prof. Sreeja Sadasivan
Prof. Sreeja Sadasivan
1)Find the volume of the water displaced and position of centre of buoyancy
for a wooden block of width 2.5 m and of depth 1.5 m, when it floats
horizontally in water. The density of wooden block is 650 kg/m3 and its length
6 m.

2) A solid cylinder of diameter 4 m has a height of 3 m. Find the meta-centric

height of the cylinder when it is floating in water with its axis vertical. The
specific gravity of the cylinder is 0.6.

Prof. Sreeja Sadasivan

A cone of base diameter d and height H floats in water with
the axis vertical and vertex downwards. If the sp gr of the
cone material is S, show that for stable equilibrium

If S = 0.7, what would be the minimum value of R/H for

stable equilibrium?

Prof. Sreeja Sadasivan

Hydrometer

Prof. Sreeja Sadasivan

Prof. Sreeja Sadasivan
Prof. Sreeja Sadasivan