Chapter 3

Fluid Statics: Pressure intensity and pressure head:

pressure and specific weight relationship, absolute

and gauge pressure, Forces on submerged planes &

curved surfaces and their applications, drag and lift

forces, Buoyancy and Floatation

Dr. Muhammad Ashraf Javid

Assistant Professor

Department of Civil and Environmental Engineering

1

Fluid Engineering Mechanics

Fluid Statics

2

Fluid Statics: It is the branch of fluid mechanics that deals

with the behavior/response of fluid when they are at rest.

Pressure, (average pressure intensity): It is the normal force

exerted per unit area. It is denoted by P and is given by;

Units

SI: N/m2 (called Pascal)

BG: lb/ft2 or lb/in2 (called psi)

CGS: dyne/cm2

1 bar=105N/m2=105Pascal

A

F

area

forceP

Pressure vs Water depth/height

3

Consider a strip or column of a cylindrical fluid,

h= height or depth of strip of fluid

γ= specific weight of fluid

dA=cross-sectional area of strip

dV=volume of strip

dW=weight of strip

Pressure at base of strip=dF/dA=dW/dA

P= γdV/dA

P= γdA.h/dA

P=γh

h

Pressure vs Water depth/height

4

P=γh

P α h

For h=0, P=0

For h=h, P=γh

h

Pressure distribution

diagram/pressure profile

As you know atmospheric pressure reduces, as we move to higher elevations. Is it because of h, as h reduces, P also reduces.

PASCAL’S LAW

5

“Pressure at any point in fluid is same in all directions when

the fluid is at rest”

Consider a wedge shape element of fluid

having dimension dx, dy and dz along x, y

and z axis.

dI= dimension of inclined plane making

an angle α with the vertical

Px, Py, Pz and P are pressure acting in x, y,

z and perpendicular to inclined surface

dW=weight of the element

z y

x

Py(dxdz)

Px(dydz)

P(dldz)

dx

dy

dz

α

α α P(dldz)

cosα

P(dldz)

sinα

dW

z y

x

PASCAL’S LAW

6

Py(dxdz)

Px(dydz)

P(dldz)

α α P(dldz)

cosα

P(dldz)

sinα

PP

odydzPdydzP

dldyodldydldzPdydzP

odldzPdydzP

F

x

x

x

x

x

)(

/cos/)(

cos)(

0

dW

PP

odxdzPdxdzP

dwdldxodldxdldzPdxdzP

odldzPdWdxdzP

F

y

y

y

y

y

)(

0&/sin/)(

sin)(

0

PASCAL’S LAW

7

Similarly by applying the conditions in z direction, it can be proved

that

Hence,

The above states that the pressure acting on fluid particle is same in

all directions when the fluid is at rest.

PPz

PPPP zyx

Absolute and Gauge Pressure

8

Atmospheric pressure

Gauge pressure

Vacuum/negative pressure

Absolute pressure

Atmospheric pressure: Pressure exerted by atmosphere

Gauge pressure: Pressure more than atmospheric pressure

Vacuum/negative pressure: Pressure less than atmospheric pressure

Absolute pressure: Pressure measure relative to absolute zero

vacatmabs

gatmabs

PPP

PPP

Atmospheric Pressure

9

It is defined as weight of air per unit

surface area of earth.

It decreases with increase in

elevation w.r.t. surface of earth.

Standard atmospheric pressure at

mean-sea-level (MSL) is

=101.3KN/m2

=1.013bar

=14.7psi

=760mm of Hg

=33.9ft of water

=10.3m of water

Measurement of Atmospheric Pressure

10

Barometer: It is device used to

measure the atmospheric pressure at

any point on the earth.

There are two types of barometer

(i) Liquid barometer

It measures the pressure with help of

column of liquid

(ii) Aneroid barometer

It measures atmospheric pressure by

its action on an elastic lid of evacuated

box.

Liquid Barometer

11

It consists of a transparent tube which

is open from one end only. The tube is

filled with liquid and is inserted in a jar

also containing same liquid. The liquid

initially drops in tube due to gravity but

stabilizes at certain level under the

action of atmospheric forces. The

atmospheric pressure is then measured

as height of liquid at which it stabilizes.

Three forces acting on fluid are

Patm(A)=Force of atmospheric Pressure

W=Weight of liquid

Pvap(A)= Force of vapour pressure

A=Cross-sectional area of tube

Weight

of liquid

Force of

Patm

Vapor

pressure

Liquid/

Liquid Barometer

12

Three forces acting on fluid are

Patm(A)=Force of atmospheric Pressure

W=Weight of liquid

Pvap(A)= Force of vapour pressure

A=Cross-sectional area of tube

For Equilibrium

W

Patm.A

Pvap.A

Liquid/

h

0; APWAPoFy vapatm

0 APAhAP vapatm AhW

vapatm PhP

Liquid Barometer

13

Generally, mercury is preferred liquid because its vapour pressure is

minimum. Moreover, its specific gravity is very high so that size

(height) of barometer required is small.

However, for other liquid vapour pressure must be considered in

estimation.

The barometer using mercury is called mercury barometer and

while using water is called water barometer.

Size of barometer tube should be more than ½ inches (or 13mm)

to avoid capillarity.

Absolute Pressure

14

Gauge Pressure (Pg): It is the pressure measured relative to

atmospheric pressure (Patm) and is always above the atmospheric

pressure

It may be defined as normal compressive force per unit area

Vacuum Pressure (Pvac): It is the pressure measured relative to

atmospheric pressure and is less than the atmospheric pressure

It may be defined as normal tensile force per unit area

Absolute Pressure(Pabs): It is the pressure measured from absolute

zero

vacatmabs

gatmabs

PPP

PPP

Problem

15

Q.3.2.2

4600m

Surface

γ=10.05kN/m3

2/46700

460005.10

mkN

P

hP

Problem

16

Q.3.2.2

Problem

17

Q.3.2.2

Problem

18

Q.3.2.4

psiP

P

hSPhPP woil

9.57

)12/(684.6288.032

2

2

2

112

P2=?

P1=32psi Stream

68ft

S=0.88

γ=Sγw

Problem

19

Q.3.3.1

5m

Surface

γ=10 kN/m3

2/1628 mkNhPi γ=8 kN/m3

2m

interface Pi

Pb 2

2211

/1.655)81.9(28 mkNP

hhP

b

b

Problem

20

Q.3.3.3

h=?

Surface

γ=12N/m3

mh

h

hP

8442

1210003.101

P=101.3kPa

Determine the depth of gaseous atmosphere to cause 101.3kN/m2 over surface of

earth?

Problem

21

psiaP

psiaOHftP

vap

atm

09.2

47.144.33 2

Weight

of liquid

Force of

Patm

Vapor

Pressure

Alcohol

w

vapatm

vapatmvapatm

vapatm

vapatm

S

PPh

PP

A

APAPh

APAhAP

APWAP

oFy

0

0

;

h=?

Measurement of Pressure

22

The following devices are used for pressure measurement

1. Piezo-meter

2. Manometer

a) Simple manometer

B) Differential manometer

3. Mechanical Pressure Transducer (Bourden gauge)

4. Electrical Pressure Transducer

1. Piezometer

23

It is used to measure pressure in

pipes or vessels.

In it simplest form, it consists of a

transparent tube open from other

ends

The diameter of tube should > ½” to

avoid capillarity action

Piezometers may be connected to

sides or bottom of pipe to avoid

eddies that are produced in the top

region of pipe

Limitations:

It must only be used for liquids

It should not be used for high pressure

It cannot measure vacuum (-ve) pressure

When connected to pipes, the

water level rises in it which

gives a measure of pressure.

2. Manometer

24

a). Simple Manometer

Figure shows a set up of simple

manometer.

It consists of a U-shaped tube, part

of which is filled with manometric

fluid.

One end of tube is connected with

the pipe whose pressure is required

to be determined.

Due to pressure, level of

manometric fluid rises on one side

while it falls on other side.

The difference in levels is measured

to estimate the pressure.

A

Fluid, γf

Manometric

fluid, γm

Y

z

Y=Manometric reading

γf =Specific weight of fluid in pipe

γm =Specific weight of

manometric fluid

2. Manometer

25

Manometric Fluids

1. Mercury

2. Oils

3. Salt solution etc

Properties of manometric

Fluid

1. Manometric fluid should not be

soluble/intermixalbe with fluid

flowing in pipe whose pressure is

required to be determined.

2. Lighter fluid should be used if

more precision is required.

A

Fluid,

Manometric

fluid,

Y

z

2. Manometer

26

Pressure measurement

It is also equation of

gauge pressure

A

Fluid, γf

Manometric

fluid, γm

Y

z

Sign Convention

-ve: upward direction

+ve: downward direction

AatmAabs

atmmfAabs

PPP

PYZP

Patm

ZYP

YZP

fmA

mfA

0

PA

2. Manometer

27

b). Differential manometer

It is used to measure difference of pressure.

Case 1: when two vessels/pipes are at same level

ZA

ZB

Y

A B PA

PB

Fluid A, γA Fluid B, γB

Manometric

Fluid , γm

AAmBBBA

BBBmAAA

BA

ZYZPP

PZYZP

PP

?

2. Manometer

28

YPP

YYPP

ZZYPP

ZYZPP

if

fmBA

fmBA

BAfmBA

AfmBfBA

fAB

2. Manometer

29

b). Differential Manometer

Case 1I: when two vessels/pipes are at different level

ZA ZB

Y

A

B

PA

PB

Fluid A, γA Fluid B, γB

Manometric

Fluid , γm

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

2. Manometer

30

BAfmBA

AfmBfBA

fAB

ZZYPP

ZYZPP

if

2. Manometer

31

b). Differential manometer

Case 1II: when manometer is inverted

ZA

ZB

Y

A

B

PA

PB

Fluid A, γA

Fluid B, γB

Manometric

Fluid , γm

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

In this case, lighter fluid should

be used as manometeric fluid.

2. Manometer

32

b). Differential manometer

Case 1II: when manometer is inverted

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

BAfmBA

AfmBfBA

fAB

BfmAABA

BBfmAAA

ZZYPP

ZYZPP

if

ZYZPP

PZYZP

In this case, lighter fluid should

be used as manometeric fluid.

Advantages and Limitation of Manometers

33

Advantages

Easy to fabricate

Less expansive

Good accuracy

High sensitivity

Require little maintenance

Not affected by vibration

Specially suitable for low

pressure and low differential

pressure

Easy to change sensitivity by

changing manometric fluid

Limitations

Usually bulky and large in size

Being fragile, get broken easily

Reading of manometer is get

affected by temperature,

altitude and gravity

Capillary action is created due

to surface action

Meniscus has to be measured

accurately for better accuracy.

3. Mechanical Pressure Transducer

34

Transducer is a device which is used to transfer energy from one

system to other

Mechanical pressure transducer converts pressure system to

displacement in mechanical measuring system

Bourden Gauge is used to measure high pressure either positive

or negative. It gives pressure directly in psi or Pascal units

Bourden Gauge

35

The essential mechanical

element in this gage is the

hollow, elastic curved tube

which is connected to the

pressure source as shown

in Fig.

As the pressure within the

tube increases the tube tends

to straighten, and although

the deformation is small, it

can be translated into the

motion of a pointer on a

dial as illustrated.

Fig. Bourden gauge

Bourden Gauge

36

Since it is the difference in pressure between the outside

of the tube and the inside of the tube that causes the

movement of the tube, the indicated pressure is gage

pressure.

The Bourdon gage must be calibrated so that the dial

reading can directly indicate the pressure in suitable units

such as psi, psf, or pascals.

A zero reading on the gage indicates that the measured

pressure is equal to the local atmospheric pressure.

This type of gage can be used to measure a negative gage

pressure (vacuum) as well as positive pressures.

3. Mechanical Pressure Transducer

37

Elevation Correction

Bourden gauge gives pressure at the

center of dial. So to calculate pressure

at point A,

Where

(γ)z=elevation correction

zPP gA

z

A

Pg

4. Electrical Pressure Transducer

38

It converts displacement of mechanical measuring system to an

electrical signal.

Its gives continuous record of pressure when converted to a strip

chart recorder.

Data can be displayed using computer data acquisition system.

Problem

39

Problem

40

Problem

41

Problem

42

Problem

43

Problem

44

45

Forces on Immersed Bodies

Forces on Immersed Bodies

46

Hydrostatic Force: It is the resultant force of pressure exerted by

liquid at rest on any side of submerged body.

It is the summation of product of uniform pressures and elementary

areas of submerged body

It is equal to the product of submerged area and pressure at the

centroid of the submerged area (to be discussed later)

pAdAppdAF

Forces on Plane Area

47

Center of pressure

The point of application of resultant

force of pressure on a submerged area

is called center of pressure.

Forces on Plane Area

48

A = total submerged area

F = hydrostatic force

θ =angle of submerged plane with free surface

hc=depth of center of area

hp=depth of center of pressure

yc=inclined depth to center of area

yp=inclined depth to center of

pressure

dA=elementary area

dF=force of pressure (hydrostatic

force) on elementary area

dAhdApdF

Projection of area of vertical plane

Forces on Plane Area

49

Lets’ choose an elementary area so that pressure over it is uniform. Such

an element is horizontal strip, of width, x so . The pressure force,

dF on the horizontal strip is

Integrating

Where, yc is the distance along the sloping plane to the centroid, C, of the

area A. If hc is the vertical depth to the centroid, then we have;

dAhdApdF

xdydA

ydAdAyhdApdAdF sinsin

sinyh

hp

A

ydAyc

AyF c sin

AhF c

Forces on Plane Area

50

Thus, we find the total force on any plane area submerged in a liquid

by multiplying the specific weight of the liquid by the product of the

area and the depth of its centroid.

The value of F is independent of the angle of inclination of the plane

as long as the depth of its centroid is unchanged.

Since is pressure at the centroid, we can also say that total

pressure force on any plane area submerged in a liquid is the

product of the area and the pressure at the centroid.

AhF c

ch

Center of Pressure

51

In order to determine location of center of pressure, yp, from OX,

let’s take the moment of elementary area around OX

Integrating

Where, ‘I’ is the 2nd moment of submerged area about OX

Where ‘ycA’ is called static moment of area

dAyyhdAypdAyydFdM sin

IyF

dAydAyyydF

p

sin*

sinsin 2

Ay

I

Ay

I

F

Iy

cc

p

sin

sinsin AyF c sin

ydFyF p*

Center of Pressure

52

Now, according to parallel axis theorem,

Where, Ic is 2nd moment of area about centroidal axis.

From this equation we again see that the location of center of pressure, P,

is independent of the angle θ.

When the plane is truly vertical, i.e., θ=90o

It is concluded that center of pressure is always below the center of area

except when the plane is horizontal. When the plane is horizontal center of

pressure and center of area lies at the same time.

Ay

AyI

Ay

Iy

c

cc

c

p

2 cc AyII 2

Ay

Iyy

c

ccp

Ah

Ihh

c

ccp

Lateral Position of Center of Pressure

53

When at least one centroidal axis is the axis of symmetry then Ixy

becomes equal to zero. i.e. xp = 0

It means center of pressure is lying on the symmetrical axis, just

below the center of area.

c

xy

pAy

Ix

Hydrostatic force formulas

54

Ay

Iyy

c

ccp

AhF c AyF c sin

Ah

Ihh

c

ccp

55

Problem

56

A plane surface is circular with a diameter of 2m. If it is vertical and the top

edge is 0.5m below the water surface, find the magnitude of the force on

one side and the depth of center of pressure.

Solution:

Ah

Ihh

c

ccp

AyF c sin

AhF cAy

Iyy

c

ccp

Free surface

0.5m

D=2m

mD

hc 5.12

5.0

kNF

F

AhF c

2.46

24

5.1810.9 2

mh

DDh

Ah

Ihh

p

p

c

ccp

667.1

4/5.1/64/5.1 24

Problem

57

A rectangular plate 1.5 m by 1.8 m is at an angle of 30o with the horizontal,

and the 1.5 m side is horizontal. Find the magnitude of force on one side of

the plate and the depth of its center of pressure when the top edge is (a)

at the water surface (b) 0.3m below water surface

(a)

1.8 ft

1.5 m

30o

hp

yp

hc yc

4ft

sinyh

ftyh o

cc 130sin2sin lbAhF c 12484514.62

ftbdbdAy

Iyy

c

ccp 67.22/12/2 3 fthp 33.1

Problem

58

(b)

hp

yp

hc yc

sinyh

fth o

c 230sin21sin21 lbAhF c 25004524.62

ftbdbdAy

Iy

c

cp 33.44/12/44 3

ftyh o

pp 167.230sin33.4sin

0.3 m

sin/cc hy 1.8 m

1.5 m

Forces on Curved Surface

59

Horizontal force on curved surface

Vertical force on curved surface

area horizontal equivalenton force chydrostati'' FFxFx

surface above liquid of volumeofWeight ' WFzFz

Forces on Curved Surface

60

Resultant Force

22

zx FFF

x

z

F

F1tan

Drag and Lift Force

61

Lift is the component of aerodynamic

force perpendicular to the relative wind.

Drag is the component of aerodynamic

force parallel to the relative wind.

Weight is the force directed downward

from the center of mass of the airplane

towards the center of the earth.

Thrust is the force produced by the

engine. It is directed forward along the

axis of the engine.

Drag and Lift Force

62

The drag force acts in a direction that is

opposite of the relative flow velocity.

Affected by cross-section area (form

drag)

Affected by surface smoothness

(surface drag)

The lift force acts in a direction that is

perpendicular to the relative flow.

CD= Coefficient of drag

CL= Coefficient of lift

A=projected area of body

normal to flow

V= relative wind velocity

Problem

63

=1.8m

a = 0.45 m, d = 1.8 m, and b = 1.2 m

=1.2m =.45m

Problem

64

1.

2

1.

2

1.2

65

Buoyancy and Floatation

Buoyancy and Floatation

66

When a body is immersed fully or partially in a fluid, it is subjected to an

upward force which tends to lift (buoy) it up.

The tendency of immersed body to be lifted up in the fluid due to an

upward force opposite to action of gravity is known as buoyancy.

The force tending to lift up the body under such conditions is known as

buoyant force or force of buoyancy or up-thrust.

The magnitude of the buoyant force can be determined by Archimedes’

principle which states

“ When a body is immersed in a fluid either fully or partially, it is

buoyed or lifted up by a force which is equal to the weight of fluid

displaced by the body”

Buoyancy and Floatation

67

Lets consider a body

submerged in water as shown

in figure.

The force of buoyancy

“resultant upward force or

thrust exerted by fluid on

submerged body” is given

Water surface

11 hP

212 hhP

2h

1h 1F

2F

dA=Area of cross-section of

element

γ= Specific weight of liquid

volumeF

dAhF

dAhdAhhF

FFF

B

B

B

B

2

121

12

Buoyancy and Floatation

68

=Weight of volume of liquid displaced by

the body (Archimedes's Principle)

Force of buoyancy can also be determined as difference

of weight of a body in air and in liquid.

Let

Wa= weight of body in air

Wl=weight of body in liquid

FB=Wa-Wl

volumeFB

Buoyancy and Floatation

69

Center of Buoyancy (B): The point of application

of the force of buoyancy on the body is known as the

center of buoyancy.

It is always the center of gravity of the volume of fluid

displaced.

Water surface

CG or G C or B

CG or G= Center of gravity

of body C or B= Centroid of

volume of liquid displaced

by body

Types of equilibrium of Floating Bodies

70

Stable Equilibrium:

If a body returns back to its original position due to internal

forces from small angular displacement, by some external force,

then it is said to be in stable equilibrium.

Note: Center of gravity of the volume (centroid) of fluid displaced is also

the center of buoyancy.

Types of equilibrium of Floating Bodies

71

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

position from the slightly displaced angular displacement and heels farther

away, then it is said to be in unstable equilibrium

Note: Center of gravity of the volume (centroid) of fluid displaced is also

center of buoyancy.

Types of equilibrium of Floating Bodies

72

Neutral Equilibrium: If a body, when given a small angular

displacement, occupies new position and remains at rest in this new

position, it is said to be in neutral equilibrium.

Note: Center of gravity of the volume (centroid) of fluid displaced is also

center of buoyancy.

W

FB

CG

C

Metacenter and Metacentric Height

73

Center of Buoyancy (B) The point of application of the force of buoyancy on the body is known as the center of buoyancy.

Metacenter (M): The point about which a body in stable equilibrium start to oscillate when given a small angular displacement is called metacenter.

It may also be defined as point of intersection of the axis of body passing

through center of gravity (CG or G) and original center of buoyancy (B) and

a vertical line passing through the center of buoyancy (B’) of tilted position

of body.

FB FB

B

‘

Metacenter and Metacentric Height

74

Metacentric height (GM): The

distance between the center of

gravity (G) of floating body and the

metacenter (M) is called

metacentric height. (i.e., distance

GM shown in fig)

GM=BM-BG

‘ B

FB

Condition of Stability

75

For Stable Equilibrium

Position of metacenter (M) is above of center of gravity (G)

For Unstable Equilibrium

Position of metacenter (M) is below of center of gravity (G)

For Neutral Equilibrium

Position of metacenter (M) coincides center of gravity (G)

Restoring moment Overturning

moment

Determination of Metacentric height

76

The metacentric height may be determined by the following two

methods

1. Analytical method

2. Experimental method

Determination of Metacentric height

77

In Figure shown AC is the

original waterline plane and B

the center of buoyancy in the

equilibrium position.

When the vessel is tilted

through small angle θ, the

center of buoyancy will move to

B’ as a result of the alteration in

the shape of displaced fluid.

A’C’ is the waterline plane in

the displaced position.

FB

Determination of Metacentric height

78

To find the metacentric height

GM, consider a small area in plan dA

at a distance x from O. The height of

elementary area is given by xθ.

Therefore, volume of the elementary

area becomes

The upward force of buoyancy on

this elementary area is then

Moment of dFB generated by

element about an axis passing

through O is given by;

dAxdV

dAxdFB

IdFx

dAxdAxxdFx

B

B

.

. 2

x

dA

FB

Prism OC’C has gone

inside the liquid and

OAA’ has come out

from liquid. There is an

increase in force of

buoyancy on right side

and decrease in left side.

dFB = gain or loss in FB

= Couple formed by triangular prism

Determination of Metacentric height

79

Moment developed by FB due to

shift from B to B’ is

For equilibrium,

Moment developed by FB due to shift

from B to B’ = Couple formed by

triangular prism

BMVBBF

mequilibriustableandanglesmallfor

SinBMVBBF

B

B

'

sin,

'

V

IBM

BMVI

BGBMGM

FB

NUMERICALS

80

Determine the volume of an object that weights 22 N in water

and 30 N in oil (s=0.82). What is the specific weight of the

object?

Solution:

For water

FB = W in air – W water

γw(Vol.)displaced = W in air – W water

γw(Vol.)object = W in air – 22 (1)

For Oil

FB = W in air – W oil

γoil(Vol.)displaced = W in air – W oil

γoil (Vol.)object = W in air – 30 (II)

By subtracting (II) from 1

(Vol.)object = 4.53 x 10-3 m3

W in air = 66.44 N

γobject = 14.67 N/m3

NUMERICALS

81

Q. 1 A wooden block of specific gravity 0.75 floats in water. If the size of block is 1mx0.5mx0.4m, find its meta centric height

0.5m

0.4m

1m

h

Solution: Given Data:

Size of wooden block= 1mx0.5mx0.4m,

Specific gravity of wood=0.75

Specific weight of wood=0.75(9.81)=7.36kN/m3

Weight of wooden block=(specific

weight)x(volume)

Weight of wooden block=7.36(1x0.5x0.4)=1.47kN

Let h is depth of immersion=?

For equilibrium

Weight of water displaced = weight of wooden

block

9.81(1x0.5xh)=1.47 >> h=0.3m

NUMERICALS

82

0.5m

0.4m

1m

h

Distance of center of buoyancy=OB=0.3/2=0.15m

Distance of center of gravity=OG=0.4/2=0.2m

Now; BG=OG-OM=0.2-0.15=0.05m

Also; BM=I/V

I=moment of inertia of rectangular section

I=(1)x(0.5)3/12=0.0104 m4

V=volume of water displaced by wooden block

V=(1)x(0.5)x(0.3)=0.15m3

BM=I/V=0.0104/0.15=0.069m

Therefore, meta centric height=GM=BM-BG

GM=0.069-0.05=0.019m

G

B

O

0.5m

NUMERICALS

83

Q 2. A solid cylinder 2m in diameter and 2m high is floating in water with its

axis vertical. If the specific gravity of the material of cylinder is 0.65, find its

meta-centric height. State also whether the equilibrium is stable or unstable.

2m

2m 1.3m G

B

O

Solution: Given Data:

Size of solid cylinder= 2m dia, & 2m height

Specific gravity solid cylinder=0.65

Let h is depth of immersion=?

For equilibrium

Weight of water displaced = weight of wooden

block

9.81(π/4(2)2(h))=9.81(0.65).(π/4(2)2(2))

h=0.65(2)=1.3m

NUMERICALS

84

2m

2m 1.3m G

B

O

Center of buoyancy from O=OB=1.3/2=0.65m

Center of gravity from O=OG=2/2=1m

BG=1-0.65=0.35m

Also; BM=I/V

Moment of inertia=I=(π/64)(2)4=0.785m4

Volume displaced=V=(π/4)(2)2(1.3)=4.084m3

BM=I/V=0.192m

GM=BM-BG=0.192-0.35=-0.158m

-ve sign indicate that the metacenter (M) is below the center of gravity (G), therefore, the cylinder is in unstable equilibrium