Fluid Mechanics
Topic 4: Buoyancy and Floatation
INTRODUCTION
Buoyancy - when a body is immersed in a fluid, an
upward force is exerted by the fluid on the body. This
upward force is equal to the weight of the fluid displaced by
the body and is called the force of buoyancy or simply
buoyancy.
Solution. Given:
Width
= 2.5m
Depth
= 1.5m
Length = 6.0m
3
Volume of the block = 2.5 x 1.5 x 6.0 = 22.500m
Density = 650kg/m3
Weight of the block = ρ x g x Volume
3
2
3
= 650kg/m x 9.81m/s x 22.5m
= 143,471.250N
Fb  WObject
1000
And according to Archimedes Principles any body
completely or partially submerged in a fluid (gas or liquid) at
rest is acted upon by an upward, or buoyant, force, the
magnitude of which is equal to the weight of the fluid
displaced by the body. The volume of displaced fluid is
equivalent to the volume of an object fully immersed in a
fluid or to that fraction of the volume below the surface for
an object partially submerged in a liquid. The weight of the
displaced portion of the fluid is equivalent to the magnitude
of the buoyant force. The buoyant force on a body floating in
a liquid or gas is also equivalent in magnitude to the weight
of the floating object and is opposite in direction; the object
neither rises nor sinks.
Centre of Buoyancy - it is defined as the point, through
which the force of buoyancy is supposed to act. As the force
of buoyancy is a vertical force and is equal to the weight of
the fluid displaced by the body, the centre of buoyancy will
be the centre of gravity of the fluid displaced.
NUMERICAL PROBLEMS
1. Find the volume of the water displaced and position of
centre of buoyancy for a wooden block of width 2.5m and
depth of 1.5m, when it floats horizontally in water. The
3
density of wooden block is 650kg/m and its length 6.0m.
x
kg 
m
9.81 2 2.5m  6m  x   143471.250 N
3 
m 
s 
x  0.975m
Volume of displaced water  w  d  x
 2.5m  6m  0.975m  14.625m 3
Computing for the centre of buoyancy
Centre of Buoyancy 
Note:
x 0.975m

 0.488m
2
2
The distance is from the base of the
object
2. A body of dimensions 1.5m x 1.0m x 2m weighs 1962N in
water. Find its weight in air. What will be it specific gravity?
Solution. Given:
3
Volume of body
= 1.5m x 1.0m x 2m = 3.0m
Weight of body in water = 1962N
Volume of water displaced = volume of body = 3.0m3
According to Archimedes Principle
Weight of the body in air
= Weight of the body in water
+ Weight of water displaced
3
2
Weight of water displaced (Fb) = 1000kg/m x 9.81m/s x
3
3.0m
= 29430N
Wair  Wliquid  Fb
Wair  1962 N  29430 N
Wair  31392 N
w
31392 N
mbody  air 
 3200kg
m
g
9.81 2
s
m 3200kg
kg
 body 

 1066.6667 3 \
3
v
3.0m
m
kg
 body 1066.667 m 3
s.g body 

 1.067
kg
 water
1000 3
m
GM 
Meta-Centre - it is define as the point about which body
starts oscillating when the body is tilted by a small angle. The
meta-centre may also be defined as the point at which the
line of action of the force of buoyancy will meet the normal
axis of the body when the body is given a small angular
displacement.
Meta-Centric Height - the distance between the metacentre
of a floating body and the centre of gravity of the body.
Where:
IG
Vsubmerged
 BG
BG = AG – AB
BG = 0.4m – 0.28m = 0.12m
IG = Moment of inertia about an axis passing
through C.G and parallel to the base
bd 3
GM  12  BG
bd  x
2m1m 
12
GM 
 0.12m
2m  1m  0.56m
GM  0.029m
3
Note:
NUMERICAL EXAMPLES
1. A block of wood of specific gravity 0.7 floats in water.
Determine the meta-centric height of the block if its size is
2m x 1m x 0.8m.
+GM indicates that the meta centre
is at the top of CG
-GM indicates that the meta centre
is at the bottom of CG
2. A solid cylinder of diameter 4.0m has a height of 3meters.
Find the metacentric height of the cylinder when it’s floating
in water with its axis vertical. The s.g. of cylinder is 0.6.
x
Solution. Given:
Dimension of block
= 2.0m x 0.8m x 1.0m
Specific gravity of wood = 0.7
Computing for height of object submerged in the liquid (x)
Fb  WObject
. kg 
1000
m
kg 
m

 9.81 2 2m  1m  x   0.71000 3  9.81 2 2m  1m  0.8m 
m 
s 
m 
s 

x  0.560m
3
Computing for metacentric height (GM)
AB 
Where:
Where:
h 0.8m

 0.4m
2
2
AG is the distance from the bottom of the
object to the centre of gravity.
= 4.0m
= 3.0m
= 0.6
Computing for metacentric height (GM)
x 0.56m

 0.28m
2
2
AB is the distance from the bottom of the
object to the centre of buoyancy
AG 
Solution. Given:
Diameter of cylinder
Height of cylinder
Specific gravity
Fb  WObject
1000
2
2


kg 
m   4m
kg 
m   4m




9
.
81

x

0
.
6
1000
9
.
81
 3.0m 






3
2 
3
2 

m 
s  4
m 
s  4



x  1.8m
AB 
x 1.8m

 0.9m
2
2
AG 
h 3m

 1.5m
2
2
GM 
Where:
IG
Vsubmerged
 BG
(c) Neutral Equilibrium – If
at the same point.
FB = W, and CB and CG are
BG = AG – AB
BG = 1.5m – 0.9m = 0.6m
d 4
GM 
64  BG
d 2
x 
4
 4.0m 4
64
 0.6m
 4.0m 2
1.8m 
4
GM  0.044m
GM 
CONDITIONS OF EQUILIBRIUM OF FLOATING AND
SUBMERGED BODIES
A sub-merged or a floating body is said to be stable if it
comes back to its original position after a slight disturbance.
The relative position of the centre of gravity (CG) and the
centre of buoyancy (CB) of a body determines the stability of
a submerged body.
*STABILITY OF FLOATING BODY
(a) Stable Equilibrium - If the point M is above G, the
floating body will be in stable equilibrium. If slight
angular displacement is given to the floating body in the
clockwise direction, the centre of buoyancy shifts from
B to B1 such that the vertical line through B1 cuts at M.
Then the buoyant force FB through B1 and weight W
through G constitute a couple acting in the
anti-clockwise direction and thus bringing the floating
body in the original position.
*STABILITY OF SUBMERGED BODY
(a) Stable Equilibrium - When W = FB and point B is
above G, the body is said to be in stable equilibrium.
(b) Unstable Equilibrium – If the point M is below G,
the floating body will be in unstable equilibrium. The
disturbing couple is acting in the clockwise direction.
The couple due to buoyant force F B and W is also acting
in the clockwise direction and thus overturning the
floating body.
(b) Unstable Equilibrium – If W = FB, but the centre of
buoyancy (CB) is below centre of gravity (CG). A slight
displacement to the body, in the clockwise direction,
gives the couple due to W and FB also in the clockwise
direction. Thus the body does not return to its original
position and hence the body is in unstable equilibrium.
(c) Neutral Equilibrium – If the point M is at the centre
of gravity, the floating body will be in neutral
equilibrium.
NUMERICAL EXAMPLE
1. A solid cylinder of diameter 4.0m has a height of
4.0m. Find the meta-centric height of the cylinder if the
specific gravity of the material of cylinder is 0.6 and it is
floating in water with its axis vertical. State whether the
equilibrium is stable or unstable.
Solution. Given:
Diameter
Height
Specific Gravity
= 4m
= 4m
= 0.6
Computing for metacentric height (GM)
Fb  WObject
1000
kg
m3
2
2


kg 
m   4m 
m   4m 


 x   0.61000 3  9.81 2 
 4.0m 
 9.81 2 
4
s  4
m
s








x  2.4m
AB 
x 2.4m

 1.2m
2
2
AG 
GM 
Where:
h
4m

 2m
2
2
IG
Vsubmerged
 BG
BG = AG – AB
BG = 2m – 1.2m = 0.8m
d 4
GM 
64  BG
d 2
x 
4
 44
GM 
64
 42
 0.8m
2.4
4
GM  0.383m
-sign means that the meta-centre (M) is below the centre of
gravity (CG). Thus the cylinder is in unstable equilibrium.