FLUID STATICS - gtu.darshan.ac.in

Department of Mechanical Engineering Prepared By: Sagar S. Kanjiya Darshan Institute of Engineering & Technology, Rajkot. Page 2.1

2 FLUID STATICS

Course Contents

2.1 Introduction

2.2 Types of Pressures

2.3 Pascal’s Law

2.4 Pressure Variation in Fluid at Rest

2.5 Measurement of Pressure

2.6 Simple Manometers

2.7 Differential Manometers

2.8 Hydrostatic force Introduction

2.9 Total Pressure and Centre of Pressure

2.10 Introduction of Buoyancy

2.11 Buoyancy and Centre of Buoyancy

2.12 Meta-Centre and Meta Centric

Height

2.13 Analytical Method for Meta Centric

Height

2.14 Condition of Equilibrium of a

Submerged Body

2.15 Condition of Equilibrium of a

Floating Body

2.16 Experimental Method for Meta Centric

Height

2.17 Oscillation of a Floating Body

2.18 References

Fluid Mechanics & Hydraulics (3140611) 2. Fluids Statics

Prepared By: Sagar S. Kanjiya Department of Mechanical Engineering .2 Darshan Institute of Engineering & Technology, Rajkot.

MEASUREMENT OF PRESSURE

2.1 Pressure

− When a certain mass of fluids is held in static equilibrium by confining it within solid

boundaries, it exerts force along direction perpendicular to the boundary in contact.

This force is called fluid pressure.

− “It is defined as a normal force exerted by a fluid per unit area.”

− SI Unit N/m2 and is known as Pascal represented by Pa.

Figure 1 Normal force exerted by a fluid to the boundary of surface

2.2 Types of Pressures 1) Atmospheric Pressure

− Air above the surface of liquids exerts pressure on the exposed surface of the liquid

and normal to the surface. This pressure exerted by the atmosphere is called

atmospheric pressure.

− Atmospheric pressure at a place depends on the elevation of the place and the

temperature.

− Atmospheric pressure is measured using an instrument called ‘Barometer’.

Figure 2 Relationship between pressures

2) Absolute Pressure

− It is defined as the pressure which is measured with reference to absolute zero

pressure. It is given by abs gauge atmp p p= +

2. Fluids Statics Fluid Mechanics & Hydraulics (3140611)


3) Gauge Pressure

− It is defined as the pressure which is measured above the atmospheric pressure.

− It can be measured by pressure measuring instrument in which atmospheric pressure

is taken as datum and it marked as zero.

4) Vacuum Pressure

− It is defined as the pressure below the atmospheric pressure.

2.3 Pascal’s Law

− Pascal’s law: “Pressure or Intensity of pressure at a point in a static mass of fluid is

equal in all directions.”

Figure 3 Forces on a fluid element

− Let us consider an arbitrary fluid element of wedge shape ( )ABC = in fluid mass at

rest with three planes around a point shown in Fig. 2.3. Let xp , yp and zp are the

intensity of pressure acting on the face in different directions with width of element

perpendicular to the plane of paper is unity ( )1dz = . The system of forces should be in

equilibrium.

− The forces acting on the elements are:

a) Pressure forces normal to the surfaces and,

b) Weight of the fluid element in the vertical direction.

− Pressure force on the face AB Area of face ABx xp p dy dz= =

− Pressure force on the face BC Area of face BCz zp p ds dz= =

− Pressure force on the face AC Area of face ACy yp p dx dz= =

− Weight of element ( ) ( )Mass of element Volume 2

dx dyg g dz g

= = =

− Resolving the forces in x direction, 0xF = thus we have,

( ) ( )sin 90 0x zp dy dz p ds dz − − =

( )1 1 cos 0x zp dy p ds − = ( )cosds dy =

x zp p= ……………… (1)


Prepared By: Sagar S. Kanjiya Department of Mechanical Engineering Page 2.4 Darshan Institute of Engineering & Technology, Rajkot.

− Resolving the forces in y direction, 0yF = thus we have,

( ) ( )cos 90 02

y z

dx dyp dx dz p ds dz dz g

− − − =

( )1 1 sin 1 02

y z

dx dyp dx p ds g

− − = ( )sinds dx =

− Neglect the weight of very small element,

y zp p= ...................... (2)

− From equation (1) and (2) we have,

x y zp p p= =

− The above equation shows that intensity of pressure at a point in a static mass of fluid

is equal in all directions.

2.4 Pressure Variation in Fluid at Rest (Hydrostatic Law)

− Hydrostatic law: “Rate of increase of pressure in vertically downward direction must

be equal to the weight density of the fluid at that point”.

− Consider the small cylindrical fluid element as shown in Fig.

− Let, A =Cross sectional area of element

Z = Height of fluid element

p = Pressure on face AB

Z = Distance of fluid element from free surface

Figure 4 Forces on a fluid element

− The forces acting on the elements are:

a) Pressure force on face AB p A= and acting perpendicular to face AB in the

downward direction.

b) Pressure force on face CDp

p Z Az

= +

and acting perpendicular to face

CD in the upward direction.



c) Weight of fluid element ( )Density g Volume g A Z= =

d) Pressure forces on surface BC and AD are equal and opposite.

− For equilibrium of fluid element, we have

( ) 0

0

0

pp A p Z A g A Z

z

pp A p A Z A g A Z

z

pZ A g A Z

z

− + + =

− − + =

− + =

Weight density of fluidp

gz

= =

……………… (1)

− Above equation states that rate of increase of pressure in vertically downward

direction must be equal to the weight density of the fluid at that point. This is known

as hydrostatic law.

− By integrating above equation (1),

dp gdz =

p gz= And p

zg

= where z is called pressure head

2.5 Measurement of Pressure

− Various devices used to measure fluid pressure can be classified into,

1. Manometers 2. Mechanical gauges

2.5.1 Manometers

− Manometers are the pressure measuring devices which are based on the principal of

balancing the column of the liquids whose pressure is to be measured by the same

liquid or another liquid.

Classification of Manometers

− Manometers are broadly classified into:

a) Simple Manometers

b) Differential Manometers

2.5.2 Mechanical Gauges

− Mechanical gauges consist of an elastic element which deflects under the action of

applied pressure and this movement will operate a pointer on a graduated scale.

− The mechanical pressure gauges are:

a) Diaphragm pressure gauge

b) Bourdon tube pressure gauge

c) Dead weight pressure gauge

d) Below pressure gauge



Burdon tube pressure gauge

− This gauge is consist of elastic bent in circular arc, fixed at one end and free at other

end as shown in Fig. The fixed end is attached at the side of application of pressure

and free end attached with the sector through adjustable link. The sector is in mesh

with the pinion which is fixed with the pointer on the calibrated scale.

− When the pressure inside the tube is increase, tube will uncoil. So that the pointer

gives a reading on the scale due to movement of the pinion through sector and free

end of tube.

− Tube material: Brass, copper, stainless steel etc.

− Advantages: Simple construction, Low cost, high pressure range (up to 700 kPa),

Accuracy is good

− Disadvantage: Susceptibility to shock and vibration, spring constant effect is major

consideration.

Figure 5 Bourdon tube gauge

2.6 Simple Manometers

− Simple monometers consists of glass tube having one of its end connected to a point

where pressure is to be measured and other end is open to atmosphere.

− Types of Simple manometers are:

a) Piezometer

b) U-tube manometer

c) Single column manometer

d) Inclined column manometer



2.6.1 Piezometer

− It consists of a glass tube inserted in the wall of the vessel or pipe at the level of point

at which the intensity of pressure is to be measured as shown in Fig. The other end of

the piezometer is exposed to air. The height of the liquid in the piezometer gives the

pressure head from which the intensity of pressure can be calculated.

Figure 6 Piezometer

− If at a point A, the height of liquid say water h in piezometer tube, then pressure at

point A is given by gh according to the Hydrostatic law.

− So, In equilibrium condition, pA = gh

2.6.2 U-tube Manometer

− A U-tube manometer consists of a glass tube bent in U-Shape, one end of which is

connected to the point at which pressure is to be measured and the other end is

exposed to atmosphere. U-tube consists of a liquid of specific of gravity is greater than

the specific gravity of the liquid whose pressure intensity is to be measured.

a) For Gauge Pressure

− Let B is the point at which pressure is to be measured, whose value is p . The datum

line is A A− as shown in Fig. (a).

− Let 1h = Height of the light liquid above the datum line

2h = Height of the heavy liquid above the datum line

1S = Specific gravity of light liquid and 1 is density of light liquid

2S = Specific gravity of heavy liquid and 2 is density of heavy liquid

− As the pressure is the same for the horizontal surface. Hence pressure above the

horizontal datum line A A− in the left column and in the right column of U-tube

manometer should be same.



Figure 7 U-tube manometer

− Pressure above datum line above A A− in the left column 1 1p gh= +

− Pressure above datum line above A A− in the right column 2 2gh=

− Comparing above equations, 1 1 2 2p gh gh + =

2 2 1 1p gh gh = −

b) For Vacuum Pressure

− For measuring vacuum pressure, the level of the heavy liquid in the manometer will be

as shown in Fig.(b).

− Pressure above datum line above A A− in the left column 2 2 1 1gh gh p = + +

− Pressure above datum line above A A− in the right column 0=

− Comparing above equations, 2 2 1 1 0gh gh p + + =

( )2 2 1 1p gh gh = − +

2.6.3 Micro manometer

a) Single Column Manometer

− A single column manometer is a modified form of U-tube manometer in which

reservoir having large cross sectional area (100 times) as compared to cross sectional

area of U – tube connected to it as shown in Fig.

− For any change in pressure, change in the level of manometeric liquid in the reservoir

is small and change in level of manometric liquid in the U- tube is large. The other limb

may be vertical or inclined. Thus there are two type of single column manometer as:

a) Vertical single column manometer

b) Inclined single column manometer

a) Vertical Single Column Manometer

− Fig. 8 shows the vertical single column manometer. Let X X− be the datum line in

the reservoir and in the right limb of the manometer, when it is not connected to the

pipe. When the manometer is connected to the pipe, due to high pressure at A, the

heavy liquid in the reservoir will be pushed downwards and will rise in the right limb.



Figure 8 Vertical single column manometer

− Let h = Fall of the heavy liquid in reservoir

2h = Rise of heavy liquid in right limb

1h =Height of centre of pipe above X X−

Ap = Pressure at A, which is to be measured

A= Cross section area of the reservoir

a = Cross section area of the right limb

1S = Specific gravity of liquid in pipe and 1 is density of liquid in pipe

2S = Specific gravity of heavy liquid in reservoir and right limb and 2 is density of

liquid in reservoir

− Fall of heavy liquid in reservoir will cause a rise of heavy liquid level in the right limb.

2A h a h =

2a hh

A

= ………… (1)

− Now consider the datum lineY Y− . Then pressure in the right limb above Y Y−

( )2 2g h h= + …………. (2)

− Then pressure in the left limb above Y Y−

( )1 1 Ag h h p= + + …….. (3)

− Comparing equation (2) and (3),

( ) ( )

2 2 1 1

2 1 2 2 1 1

A

A

g h h g h h p

p h g g gh gh

+ = + +

= − + −

− Put value of 2a hh

A

= in above equation, we get

22 1 2 2 1 1A

a hp g g gh gh

A

= − + −



− As the area A is very large compared to a , hence a

A becomes very small and can be

neglected.

− Then 2 2 1 1Ap gh gh = − …… (4)

b) Inclined Single Column Manometer

− Fig. 9 shows the inclined single column manometer. This manometer is more sensitive.

Due to inclination the distance moved by the heavy liquid in the right limb will be

more.

− Let L= length of heavy liquid moved in right limb form X X−

= Inclination of right limb with horizontal

2h = Vertical rise of heavy liquid in right limb from sinX X L − =

Figure 9 Inclined single column manometer

− From equation (4), the pressure at A is,

2 2 1 1

2 1 1sin

A

A

p gh gh

p g gh

= −

= −

2.7 Differential Manometers

− Differential manometers are used to measure the pressure difference between two

points. It is consists of a U-tube, containing heavy liquid, whose two ends are

connected to the two points, whose difference of pressure is to be measured.

− Types of differential manometers are:

a) U-tube Differential manometers

b) Inverted U-tube differential manometers

2.7.1 U-Tube Differential Manometers

− In Fig. 10 (a), the two points A and B are at different level and also contained liquids

of different specific gravity. These points are connected to the U-tube manometer.

Let the pressure at A and B are AP and Bp .



Figure 10 U-tube differential manometer

− Let h = Difference of mercury level

y = Distance of the centre of B, from the mercury level in the right limb

x = Distance of the centre of A, from the mercury level in the right limb

1 = Density of liquid at A

2 = Density of liquid at B

g =Density of heavy liquid

− Taking at datum line at X X− .

− Pressure above X X− in the left limb ( )1 Ag h x p= + +

− Pressure above X X− in the right limb 2g Bgh gy p = + +

− Comparing above equation, we have

( )

( )

1 2

2 1

A g B

A B g

g h x p gh gy p

p p gh gy g h x

+ + = + +

− = + − +

− In Fig. 10 (b), the two points A and B are at same level and contained liquids of

density.

− Pressure above X X− in the right limb 1g Bgh gx p = + +

− Pressure above X X− in the left limb ( )1 Ag h x p= + +


( )

( )

1 1

1 1

g B A

A B g

gh gx p g h x p

p p gh gx g h x

+ + = + +

− = + − +

2.7.2 Inverted U-Tube Differential Manometers

− It consists of an inverted U-tube having two ends are connected to the pipes at points

A and B whose difference of pressure is to be measured as shown in Fig. 2.10. It is

used for measuring the difference of low pressures. Let the pressure at A is more than

the pressure at B.



Figure 11 Inverted U-tube differential manometer

− Let 1h = Height of liquid in left limb below the datum line X X−

2h = Height of liquid in right limb

h = Difference of light liquid

1 = Density of liquid at A

2 = Density of liquid at B

s =Density of light liquid

Ap = Pressure at A

Bp = Pressure at B

− Pressure below datum line X X− in the left limb is,

1 1Ap g h= −

− Pressure below datum line X X− in the right limb is,

2 2B sp g h g h = − −


1 1 2 2

1 1 2 2

A B s

A B s

p gh p gh gh

p p gh gh gh

− = − −

− = − −



HYDROSTATIC FORCE ON PLANE AND CURVED SURFACE

2.8 Introduction

− This chapter deal with the fluids at rest conditions with no relative motion between

adjacent or neighbouring fluid layers. The velocity gradient and shear stress are zero

du

dy

=

. Then the forces acting on the fluid partials will be:

1. Due to pressure of fluid normal to the surface.

2. Due to gravity (or self-weight of fluid particles).

2.9 Total Pressure and Centre of Pressure

− Total pressure is defined as the force exerted by a static fluid on a surface either

plane or curved when the fluid comes in contact with the surface and this force

always normal to the surface.

− Centre of pressure is defined as the point of application of the total pressure on the

surface. There are four cases of submerged surfaces on which the total pressure

force and centre of pressure is to be determined.

− The submerged surfaces may be:

1. Vertical plane surface

2. Horizontal plane surface

3. Inclined plane surface

4. Curved surface

2.9.1 Vertical Plane Surface Submerged in Liquid

− Consider a plane surface of arbitrary shape immerged in a liquid as shown in Fig. 12.

− Let, A=Total area of the surface

h−

=Distance of C.G of the area from free surface of liquid

G =Centre of gravity of plane surface

P =Centre of pressure

*h =Distance of centre of pressure from free surface of liquid

Figure 12 Vertical plane surface submerged in liquid



a) Total pressure force (F)

− The total pressure on the surface may be determined by dividing the entire surface

into a number of small parallel strips.

− Consider a strip of thickness dh and width b at depth of h from free surface of

liquid as shown in Fig. 12.

− Pressure intensity on the strip, p gh=

− Area of strip, dA b dh=

− Total pressure force on strip, dF p Area gh b dh= =

− Total pressure force on whole surface,

F dF gh b dh g b h dh = = =

But, b h dh h dA = = Moment of surface area about the free surface of liquid

=Area of surface Distance of C.G from free surface A h−

=

F g h A−

= ............. (1)

b) Centre of Pressure (h*)

− Centre of pressure is calculated by using the principle of moments which states that

the moment of the resultant force about an axis is equal to the sum of moments of

the components about the same axis.

− The resultant force F is acting at P , at distance h from free surface of the liquid

as shown in Fig.12. Hence moment of the force F about free surface of

liquid F h= ........... (2)

− Hence moment of the force dF about free surface of the liquid,

dF h= gh b dh h= ( )dF gh b dh=

− Sum of moment of all such forces about free surface of liquid is,

gh b dh h=

2 2g bh dh g h dA = = ( )bdh dA=

− But 2 2h dA bh dh= = second moment of inertia of the surface about free surface of

liquid = 0I

− Sum of moment about free surface, 2

0g h dA gI = = ....................... (3)

− Equating (2) and (3), we get

0F h gI =

0g h A h gI −

= F g h A−

=

0Ih

Ah−

= .................. (4)



− But from theorem of parallel axis, 2

0 = GI I Ah−

+

− Substituting 0I in equation (4), we get

2

GI Ahh

Ah

−

−

+=

GIh h

Ah

−

− = + ............... (5)

2.9.2 Horizontal Plane Surface Submerged in Liquid

− Consider a plane horizontal surface immersed in a static fluid is shown in Fig. 13. As

every point of the surface at the same depth from the free surface of the liquid, the

pressure intensity will be equal on the entire surface is given by p gh= .

− Let, A=Total area of the surface



h−

=Distance of C.G of the area from free surface of liquid = h is the depth of

the surface

*h =Distance of centre of pressure from free surface of liquid = h

Figure 13 Horizontal plane surface submerged in liquid

− Total pressure force, F on the surface,

p Area g h A= =



2.9.3 Inclined Plane Surface Submerged in Liquid

− Consider an inclined plane surface of arbitrary shape immerged in a liquid such a way

that the surface makes an angle with the free surface as shown in Fig. 14.

Figure 14 Inclined plane surface submerged in liquid

− Let, A=Total area of an inclined surface

=An angle at which the immersed surface is inclined with the liquid surface



y−

=Distance of C.G of the surface from axis O-O

y=Distance of centre of pressure from axis O-O

h =Depth of small strip from free surface of liquid

h−

=Depth of C.G of the area from free surface of liquid

h=Depth of C.G of the area from free surface of liquid

a) Total pressure force (F)

− Consider the small strip of area dA at a depth h from the free surface at a distance

y from the axis 0 0− .

− The pressure acting on strip is,

sinp gh gy = = ( )siny h =

− The pressure force acting on strip is,

sindF P dA gy dA = =

− Total pressure force on whole surface is,

sin sinF dF gy dA g ydA = = =

− Where ydA y A−

= = moment of area about axis 0 0−

sinF g y A −

= siny h− −

=



F g h A−

=

b) Centre of Pressure (h*)

− The pressure force acting on strip is,

sindF P dA gy dA = =

− The moment of pressure force dF about axis O-O

2sin sindF y gy dA y g y dA = = =

− Sum of moment of total pressure force about axis O-O

2sing y dA =

− But 2

0y dA I= = moment of inertia of surface about axis O-O

0sing I = ....................... (1)

− Moment of total pressure force F about axis 0 0− is also given by,

F y= ............................. (2)

− Equating two values in equation (1) and (2)

0sinF y g I =

0sing Iy

F

=

0sin

sin

g Ih

g h A

−

= F g h A

− =

and

sin

hy

=

h2

0

sinI

h A

−

= ...................... (3)

− But by the theorem of parallel axis,

2

0 = GI I A y−

+ sin

hy

−−

=

Thus 2

0 2=

sinG

AhI I

−

+

− Substitute the value of 0I in equation (3)

2 2

2

sin

sinG

Ahh I

h A

−

−

= +

2sinGIh h

Ah

−

− = + ...................... (4)

2.9.4 Curved Surface Submerged in Liquid

− Consider a curved surface AB wholly submerged in static fluid as shown in Fig. 15.

− Let dA is the area of small strip at a vertical depth h form the free surface of the

liquid.



− Pressure intensity on the elemental area dA is P gh= .

− The pressure force which is acting normal to the curved surface is,

dF P dA ghdA= = ......................... (1)

Figure 15 Curved surface submerged in liquid

− Total pressure force on the curved surface is,

F dF ghdA= =

− But here as the direction of the forces on the small area are not in the same

direction but varies from point to point. Thus integration of equation for curved

surface is impossible.

− Thus resolving the force dF in to two components xdF and ydF in the x and y

direction respectively. The total force in x and y direction, xF and yF are obtaining

by integrating xdF and ydF .

− Then total pressure forces on the curved surface is,

2 2

x yF F F= +

− The direction of resultant force F with horizontal is,

1tany

x

F

F − =

− Now resolving dF given by equation (3.10) in x and y direction

sin sinxdF dF ghdA = = ( )dF ghdA=

cos cosydF dF ghdA = =

− Total pressure force acting in ' 'x and ' 'y direction are

sin sin sinx xF dF dF ghdA g hdA = = = = ............. (2)

cos cos cosy yF dF dF ghdA g hdA = = = = ........... (3)

− Fig. 15 (b) shows the enlarge area dA . From this figure EFG

− EF dA= sinFG dA = cosEG dA =



− Thus in equation (2), sindA FG = = vertical projection of the area dA and hence

the expression sinxF g hdA = represents the total pressure force on the

projected area of curved surface on the vertical plane.

− Also cosdA EG = = horizontal projection of the area dA and hence the expression

coshdA is the volume of the liquid contained in the elementary area dA upto free

surface of the liquid. Thus coshdA is the total volume contained between the

curved surface extended up to free surface.

− Hence cosxF g hdA = is the total weight of liquid supported by curved surface

up to free surface of liquid or liquid lying in portion ABOCDA).

− In Fig. 16 curved surface not supporting any fluid. In such cases, yF is equal to the

weight of the imaginary liquid supported by AB up to free surface of the liquid. The

direction of yF will be taken in upward direction.

Figure 16 Curved surface not supporting any fluid



BUOYANCY AND FLOTATION

2.10 Introduction of Buoyancy

− Objects feels lighter and weighs less in a liquid that it does in air. This observation

suggest that a fluid exerts an upward force on a body immerged in it. This force that

tends to lift the body is called buoyancy force.

2.11 Buoyancy and Centre of Buoyancy 2.11.1 Buoyancy

− When a body is immersed in a fluid an upward force 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.

2.11.2 Centre of Buoyancy

− It is defined as the point through which the force of buoyancy is supposed to be act.

As the force of buoyancy is a vertical force is equal to the weight of the fluid

displaced by the body, the center of buoyancy will be the center of gravity of the

fluid displaced.

2.12 Meta-Centre and Meta Centric Height 2.12.1 Meta-center

− It is defined as the point about which a body starts oscillating when the body is tilted

by a small angle. The Meta center 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.

− Consider a body floating in a liquid as shown in Fig. 17 (a). Let the body is in

equilibrium and G is the center of gravity and B the center of buoyancy. For

equilibrium, both the points lie on the normal axis, which is vertical.

Figure 17 Meta-center

− Let the body is given a small angular displacement in the clockwise direction as

shown in Fig. 17 (b). The center of buoyancy, which is the center of gravity of the

displaced liquid or center of gravity of the portion of the body submerged in liquid,

will now be shifted towards right from the normal axis. Let it is at 1B . The line of



action of the force of buoyancy in new position, will intersect the normal axis of the

body at some point say M . This point is called meta center.

2.12.2 Meta Centric height

− It is the distance MG between the meta center of the floating body and the center

of gravity of the body.

2.13 Analytical Method for Meta Centric Height

− Consider a body floating in a liquid as shown in Fig. 18 (a). Let the body is in

equilibrium and G is the center of gravity and B the center of buoyancy. For

equilibrium, both the points lie on the normal axis, which is vertical.

− Let the body is given a small angular displacement in the clockwise direction. The

new center of buoyancy is at1B . The vertical line through

1B cuts the normal axis

at M . This M is called meta center and GM is meta-centric height.

Figure 18 Meta-centric height of floating body

− The angular displacement of the body in the clockwise direction causes the wedge-

shaped prism 'BOB on the right of the axis to go inside the water while the identical

wedge shaped prism represented by 'AOA emerges out of the water on the left of

the axis. This wedge represents a gain in buoyant force on the right side and

corresponding loss of buoyant force on the left side. The gain is represented by a

vertical force BdF acting through the C.G of the prism 'BOB while the loss is

represented by an equal and opposite force BdF acting vertically downward through

the centroid of 'AOA .



− The couple due to these buoyant forces BdF tends to rotate the ship in the

counterclockwise direction. Also the moment caused by the displacement of the

center of buoyancy from B to 1B is also in the counterclockwise direction. Thus

these two couple must be equal.

− Consider towards the right of the axis a small a small strip of thickness dx at a

distance x from the O as shown in Fig. 18 (b).

− Area of strip = Height Thickness x dx=

− Volume of strip = Area L x L dx =

− Weight of strip g= Volume gx Ldx =

− Similarly, if a small a small strip of thickness dx at a distance x from the O towards

the left of the axis is considered, the weight of the strip will be gx Ldx . Two

weights are acting in the opposite direction and hence constitute a couple.

− Moment of the couple= Weight of the strip distance between these two weights

22

gx Ldx x x

gx Ldx

= +

=

− Moment of the couple for the whole wedge

22 gx Ldx = …………… (1)

− Moment of couple due to shifting of center of buoyancy from B to 1B

1B

B

F BB

F BM

=

=

W BM = …………………… (2)

− But these two couple are the same. Hence equating equations (1) and (2),We get

2

2

2

2

W BM gx Ldx

W BM g x Ldx

=

=

22W BM g x Ldx = but Ldx dA=

22W BM g x dA =

− From Fig. 18 (c), 22 x dA I= = second moment of area of the plan of the body at

water surface about the axisY Y− . Thus

W BM gI

gIBM

W

=

=

− But W = Weight of the body= Weight of the fluid displaced by the body

g= Volume of the fluid displaced by the body

g= Volume of the body submerged in water

g=



gI I

BMg

= =

I

GM BM BG BG= − = −

− Thus meta-centric height is given by,

I

GM BG= −

………………… (3)

2.14 Condition of Equilibrium of a Submerged Body

− The position of center of gravity and center of buoyancy in case of a completely sub-

merged body are fixed.

a) Stable Equilibrium

− Consider a balloon which is completely submerged in air. Let lower portion of the

balloon contained heavy material, so that its center of gravity is lower than center of

buoyancy.

− Let the weight of balloon is W acting through G vertically downwards direction

while the buoyant force BF is acting vertically up through B . For the equilibrium of

the balloon BW F= . Now if the balloon is given an angular displacement in the

clockwise direction as shown in Fig. 19 (a), then W and BF constitute a couple acting

in the anti-clockwise direction and brings the balloon in the original position. Thus

the balloon in this position is in stable equilibrium.

− Thus if BW F= and point B is aboveG , the body is said to be in stable equilibrium.

Figure 19 Stabilities of sub-merged bodies

b) Unstable Equilibrium

− If BW F= and point B is belowG , the body is said to be in unstable equilibrium as

shown in Fig. 19 (b). If the slight angular displacement given to the balloon in the

clockwise direction, then W and BF constitute a couple also acting in the clockwise

direction. Thus the balloon does not return to its original position and it is in

unstable equilibrium.



c) Neutral Equilibrium

− If BW F= and point B and G are at the same point as shown in Fig. 19 (c), then the

body is said to be in neutral equilibrium.

2.15 Condition of Equilibrium of a Floating Body

− The stability of a floating body is determined from the position of Meta-center. In

case of floating body, the weight of the body is equal to weight of the liquid

displaced.

a) Stable Equilibrium

− If the point M is aboveG , the floating body will be in stable equilibrium as shown in

Fig. 20 (a). If the slight angular displacement given to the balloon in the clockwise

direction, the center of buoyancy sifted from B to 1B such that the vertical line

through b 1B cuts at M . Then the buoyant force BF through 1B and weight W

through G constitute a couple acting in the anticlockwise direction. Thus bringing

the floating body in original position.

Figure 20 Stabilities of floating bodies

b) Unstable Equilibrium

− If the point M is belowG , the floating body will be in unstable equilibrium as

shown in Fig. 20 (b). If the slight angular displacement given to the balloon in the

clockwise direction. Then the couple due to buoyant force BF and weight W

constitute a couple also acting in the clockwise direction and thus overturning the

floating body.

c) Neutral Equilibrium

− If the point M is center of gravity of the body, the floating body will be in neutral

equilibrium as shown in Fig. 20 (c).



2.16 Experimental Method for Meta Centric Height

− The meta-centric height of the floating vessel can be determined, provided that we

know center of gravity of the floating vessel. Let 1w is known weight placed over the

center of the vessel as shown in Fig. 21 (a).

Figure 21 Meta centric height of the floating body

− Let W = Weight of the vessel including 1w

G = Centre of gravity of the vessel

B = Centre of buoyancy of the vessel

− The weight 1w is moved across the vessel towards right through a distance x as

shown in Fig 21. The vessel will be tilted. The angle is measured by means of a

plumbline and a protractor attached on the vessel. The new center of gravity will

shifted to 1G as the weight has been moved towards the right. Also the center of

buoyancy will change to 1B as the vessel has tilted. Under equilibrium, the moment

cause by movement of the load 1w through a distance x must be equal to the

moment caused by the shift of the center of gravity from G to 1G . Thus

− The moment due to change of 1 tanG GG W W GM = =

− The moment due to movement of

1 1

1

1

tan

tan

w w x

w x WGM

w xGM

W

=

=

=

2.17 Oscillation of a Floating Body

− Consider a floating body, which is tilted through an angle by an overturning couple

as shown in Fig 22. Let the overturning couple is suddenly removed. The body will

start oscillating. Thus body will be in a state of oscillation as if suspended at the

meta-center M . The only force acting on the body is due to the restoring couple due

to the weight W of the body force of buoyancy BF .



Figure 22 Oscillation of the floating body

− Restoring couple Distant GAW=

sinW GM = …………… (1)

− This couple tries to decrease the angle

− Angular acceleration of the body, 2

2

d

dt

= −

− -ve sign indicates that restoring couple tries to decrease the angle .

− Torque due to inertia = moment of inertia about Y Y− angular

acceleration

2

2Y Y

dI

dt

−

= −

− But 2

Y Y

WI K

g− =

− where, W = Weight of the body, K = Radius of gyration

− Inertia torque 2 2

2 2

2 2

W d W dK K

g dt g dt

= − = −

…………… (2)

− Equating (1) and (2), we get

22

2

2 2

2

sin

sin

W dW GM K

g dt

K dGM

g dt

= −

= −

− For small angle ,

2 2

2

K dGM

g dt

= −

2 2

20

K dGM

g dt

+ =

− Divided by2K

g, we get

2

2 20

d GM g

dt K

+ =



− The above equation differential equation of second degree. The solution is

1 22 2

. .sin cos

GM g GM g tC t C

K K

= + ………… (3)

− where 1C and

2C are constant of integration

− The values of 1C and 2C are obtained from boundary conditions which are

I. At 0, 0t = =

II. At , 02

Tt = =

− Where T is the time period of one complete oscillation.

− Substituting the 1st boundary condition in (3), we get

1 20 0 1C C= +

2 0C =

− Substituting the 2nd boundary condition in (3), we get

1 2

.0 0 sin

2

GM g TC

K= = =

− But 1C can not be zero and the other alternative is

2

.

2

GM g T

K = then

2

2.

KT

GM g= …………… (4)

2.18 References Yunus A. Cengel & John M. Simbala, “Fluid Mechanics: Fundamentals & Applications”,

4th Edition, 2017, McGraw-Hill Education.

R. K. Bansal, “Fluid Mechanics & Hydraulic Machines”, 3rd Edition, 2007, Laxmi

Publication.

************

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− Swami Vivekananda

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