Lecture 9: Solids and Fluids

baconossifiedMechanics

Oct 29, 2013 (4 years and 2 months ago)

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Lecture 9: Solids and Fluid
Mechanics

Serway

and
Vuille

Ch.
9.1
-
9.6

Outline


States of matter


Density and pressure


Deformation of solids


Buoyant force


Archimedes’ principle

States of Matter


Matter

can

be

classified

in

three

basic

states
:

solid,

liquid,

and

gas
.


S
olids

have

a

definite

volume

and

shape

and

tend

to

maintain

its

familiar

shape
.


Liquids

have

a

definite

volume

but

no

definite

shape



i
.
e
.

liquids

fill

the

shape

of

their

container
.


Gases

neither

have

a

definite

volume

nor

a

definite

shape,

but

they

share

many

properties

with

liquids
.

Density


The

density

ρ

of

an

object

having

uniform

composition

is

given

by




𝑉





SI

Unit
:

kg/m
3


The

densities

of

most

liquids

and

solids

vary

slightly

with

changes

in

temperature

and

pressure
.


The

densities

of

gases

vary

greatly

with

temperature

and

pressure
.

Pressure


If

F

is

the

magnitude

of

a

force

exerted

perpendicular

to

a

given

surface

of

area

A
,

then

the

average

pressure

P

is

given

by

𝑃

𝐹





SI

Unit
:

N/m
2

or

Pa


The

force

exerted

by

a

fluid

on

an

object

is

always

perpendicular

to

the

surfaces

of

the

object
.

Example 1


Calculate

the

weight

of

a

cylindrical

column

of

water

with

height


=
40
.
0



and

radius


=
1
.
00


.


Calculate

the

force

exerted

by

air

on

a

disk

of

radius

1
.
00

m

at

the

water’s

surface
.


What

pressure

at

a

depth

of

40
.
0

m

supports

the

water

column?

Example 1 Solution


To

calculate

the

weight

of

the

cylindrical

column

of

water,

we

first

determine

its

mass



=
𝑉
=



2

=

1
.
00



10
3

/

3
1
.
00


2
40
.
0


=
1
.
26



10
5





The

weight

of

the

water

is

given

by



=

=
1
.
26



10
5



9
.
80


/

2
=
1
.
23



10
6



Example 1 Solution


To

determine

the

force

at

the

water’s

surface,

we

note

that

the

downward

force

is

caused

by

air

pressure
.


Using

the

definition

of

pressure,

we

have

𝑃
=
𝐹



𝐹
=
𝑃
=
𝑃


2
=

1
.
01



10
5

𝑃𝑎
1
.
00


2
=
3
.
17



10
5




Example 1 Solution


To

determine

the

pressure

at

a

depth

of

40
.
0

m,

we

must

first

determine

the

upward

force

at

the

bottom

of

the

water

column


𝐹
=

𝐹
𝑤


+
𝐹

=
0

𝐹

=
𝐹
𝑤
+

=
3
.
17



10
5


+
1
.
23



10
6


=
1
.
55



10
6





The

pressure

can

be

determined

by

𝑃
=
𝐹


=
1
.
55



10
6


1
.
00


2
=
4
.
93



10
5

𝑃𝑎

Stress and Strain


The

elastic

properties

of

solids

are

discussed

in

terms

of

stress

and

strain


Stress

is

the

force

per

unit

area

causing

the

deformation

of

a

solid
.


Strain

is

a

measure

of

the

amount

of

deformation
.


For

sufficiently

small

stresses,

we

have

 
=
𝑎𝑖





 𝑎𝑖


The

elastic

modulus

is

a

measure

of

the

stiffness

of

a

solid,

analogous

to

the

spring

constant

of

a

spring
.


There

are

three

types

of

solid

deformations
:



Tensile

deformation


Shear

deformation


Bulk

deformation

Tensile Deformation


Consider

a

long

bar

of

cross
-
sectional

area

A

and

length


0


When

an

external

force

is

applied

along

the

bar,

the

bar

attains

an

equilibrium

in

which

its

length

increases

and

the

external

force

is

balanced

by

internal

forces
.


Under

these

circumstances,

the

bar

is

under

tensile

stress

and

the

tensile

strain

is

related

to

the

change

in

length

of

the

bar
.


The

tensile

stress

is

related

to

the

tensile

strain

as

follows

𝐹

=
𝑌



0

Young’s Modulus

𝐹

=
𝑌



0


Y

is

the

constant

of

proportionality

called

Young’s

modulus
.


A

material

having

a

large

Young’s

modulus

is

difficult

to

stretch

or

compress
.


Experiments

show

that



the

change

in

length

for

a

fixed

external

force

is

proportional

to

the

original

length



the

force

necessary

to

produce

a

given

strain

is

proportional

to

the

cross
-
sectional

area
.


Shear Deformation


Another

type

of

deformation

occurs

when

an

object

is

subjected

to

a

force

parallel

to

the

cross
-
sectional

area
.


The

applied

force

is

called

a

shear

stress

and

the

shear

strain

is

related

to

horizontal

distance

that

the

sheared

face

moves
.


The

shear

stress

is

related

to

the

shear

strain

according

to

𝐹

=
𝑆




Shear Modulus

𝐹

=
𝑆





S

is

the

constant

of

proportionality

called

the

shear

modulus
.



A

material

having

a

large

shear

modulus

is

difficult

to

bend
.



Bulk Deformation


Suppose

the

external

forces

acting

on

an

object

are

all

perpendicular

to

the

surface

on

which

the

force

acts

and

are

distributed

uniformly

over

the

surface

of

the

object
.


Under

this

circumstance,

the

object

is

under

a

volume

stress

and

undergoes

a

change

in

volume

(
volume

strain
)

without

a

change

in

shape
.


The

volume

stress

is

related

to

the

volume

strain

as

follows



𝐹

=



𝑉
𝑉

Bulk Modulus


𝐹

=



𝑉
𝑉


B

is

the

constant

of

proportionality

known

as

the

bulk

modulus
.


A

material

having

a

large

bulk

modulus

doesn’t

compress

easily
.


An

increase

(decrease)

in

pressure

causes

a

decrease

(increase)

in

volume


Table of Elastic Modulus


Example 2


A

plank

2
.
00

cm

thick

and

15
.
0

cm

wide

is

firmly

attached

to

the

railing

of

a

ship

by

clamps

so

that

the

rest

of

the

board

extends

2
.
00

m

horizontally

over

the

sea

below
.


A

man

of

mass

80
.
0

kg

is

forced

to

stand

on

the

very

end
.


If

the

end

of

the

board

drops

by

5
.
00

cm

because

of

the

man’s

weight,

find

the

shear

modulus

of

the

wood
.

Example 2 Solution


For

this

plank,

the

cross

sectional

area

is

given

by


=
0
.
02


0
.
15


=
0
.
003


2


The

original

height

of

the

board

over

the

sea

is

given

by


=
2
.
00




The

force

applied

parallel

to

the

cross
-
sectional

area

is

the

weight

of

the

man

given

by

𝐹

=

=
80
.
0


9
.
8


/

2
=
784




Because

of

the

weight

of

the

man,

the

board

bends

a

distance

given

by



=
0
.
05



Example 2 Solution


The

shear

stress

is

related

to

the

shear

strain

according

to

𝐹

=
𝑆





Solving

for

the

shear

modulus

S

gives

𝑆
=
𝐹ℎ



=
784


2
.
00


0
.
003


2
0
.
05


=
1
.
05



10
7


/

2

Example 3


Determine

the

elongation

of

the

rod

if

it

is

under

a

tension

of

5
.
8

x

10
3

N
.

Example 3 Solution


When

the

rod

comes

to

equilibrium,

the

tension

will

be

uniform

throughout

its

length

with

both

types

of

metal

subject

to

the

stretching

force
.


The

total

elongation

of

the

rod

can

be

written

as




=



+




Example 3 Solution


Using

the

expression

for

Young’s

modulus,

we

have




=



+








=
𝐹

0

𝑌


+
𝐹

0

𝑌


=
𝐹


0

𝑌

+

0

𝑌






=
𝐹


0

𝑌

+

0

𝑌

=
5
.
8


10
3


2
.
0



10

3


2
1
.
3


7
.
0



10
10

𝑃𝑎
+
2
.
6


11



10
10

𝑃𝑎
=
1
.
9




Variation of Pressure with Depth


Consider

the

fluid

contained

within

the

volume

indicated

to

the

right
.


Since

the

fluid

is

at

rest

in

a

container,

the

fluid

is

in

static

equilibrium

and

thus,

all

points

at

the

same

depth

must

have

the

same

pressure
.


It

can

be

shown

that

the

pressure

at

y
2

is

related

to

the

pressure

at

y
1

𝑃
2
=
𝑃
1
+


1


2

Variation of Pressure with Depth


Atmospheric

pressure

is

caused

by

the

weight

of

air

above

a

particular

location
.


The

pressure

P

at

any

depth

h

below

the

surface

of

the

water

is

given

by

𝑃
=
𝑃
0
+
ℎ


The

pressure

P

at

depth

h

below

the

surface

of

a

liquid

open

to

the

atmosphere

is

greater

than

atmospheric

pressure

by

the

amount

ℎ
.

Pascal’s Principle


Since

the

pressure

in

a

fluid

depends

on

depth,

any

increase

in

pressure

at

the

surface

must

be

transmitted

to

every

point

in

the

fluid
.


This

is

formerly

stated

by

Pascal’s

principle


A

change

in

pressure

applied

to

an

enclosed

fluid

is

transmitted

undiminished

to

every

point

of

the

fluid

and

to

the

walls

of

the

container
.

Example 4


In

a

car

lift

used

in

a

service

station,

compressed

air

exerts

a

force

on

a

small

piston

of

circular

cross

section

having

a

radius

of


1
=
5
.
00


.


This

pressure

is

transmitted

by

an

incompressible

liquid

to

a

second

piston

of


2
=
15
.
0


.


(a)

What

force

must

the

compressed

air

exert

on

the

small

piston

in

order

to

lift

a

car

weighing

1
.
33



10
4


?

Neglect

the

weights

of

the

pistons
.


(b)

What

air

pressure

will

produce

a

force

of

that

magnitude?

Example 4 Solution


Pascal’s

principle

states

that

the

change

in

pressure

is

transmitted

undiminished

to

every

point

of

the

fluid

and

to

the

walls

of

the

container
.

Therefore,

𝑃
1
=
𝑃
2


𝐹
1

1
=
𝐹
2

2



The

necessary

force

on

the

small

piston

is

given

by

𝐹
1
=

1

2
𝐹
2
=

5
.
00



10

2


2

1
5
.
00



10

2


2
1
.
33



10
4


=
1
.
48



10
3




The

air

pressure

needed

to

produce

F
1

is

given

by

𝑃
=
𝐹
1

1
=
1
.
48



10
3




5
.
00



10

2


2
=
1
.
88



10
5

𝑃𝑎

Example 5


The

figure

below

shows

the

essential

parts

of

a

hydraulic

brake

system
.


The

area

of

the

piston

in

the

master

cylinder

is

1
.
8

cm
2

and

that

of

the

piston

in

the

brake

cylinder

is

64

cm
2
.


The

coefficient

of

friction

between

the

shoe

and

the

wheel

drum

is

0
.
50
.


If

the

wheel

has

a

radius

of

34

cm,

determine

the

frictional

torque

about

the

axle

when

a

force

of

44

N

is

exerted

on

the

brake

pedal

Example 5 Solution


The

brake

pedal

exerts

a

force

of

44

N

on

the

piston,

which

creates

an

increase

in

pressure

of


𝑃
=
𝐹
 

 
=
44


1
.
8



10

4


2
=
2
.
44



10
5

𝑃𝑎


According

to

Pascal’s

principle,

the

increase

of

pressure

on

the

brake

pedal

will

be

transmitted

undiminished

to

the

brake

cylinder
.


Thus,

there

will

be

an

increased

force

on

the

shoe

of

𝐹
ℎ
=

𝑃

 
=
2
.
44



10
5

𝑃𝑎
6
.
4



10

4


2
=
1
.
56



10
2



Example 5 Solution


The

force

exerted

by

the

fluid

serves

as

the

normal

force

to

the

brake

shoe
.


The

frictional

force

between

the

shoe

and

drum

will

produce

a

frictional

torque

about

the

axle

of

magnitude

𝜏
=




𝜏
=
𝜇

𝐹
ℎ

=
0
.
5
1
.
56



10
2


0
.
34


=
27





Archimedes’ Principle


Archimedes’

principle

can

be

stated

as

follows


Any

object

completely

or

partially

submerged

in

a

fluid

is

buoyed

up

by

a

force

with

magnitude

equal

to

the

weight

of

the

fluid

displaced

by

the

object
.


Archimedes’

principle

is

based

on

the

concept

of

the

buoyant

force
.


The

buoyant

force

is

the

upward

force

that

arises

in

fluids

due

to

pressure

differences

between

the

upper

and

lower

sides

of

an

object
.

The Buoyant Force


Consider

the

sphere

submerged

in

the

fluid
.


Since

all

points

at

the

same

depth

have

the

same

pressure,

then

the

horizontal

forces

cancel

with

each

other
.


Because

pressure

increases

with

depth,

the

forces

on

the

bottom

of

the

sphere

are

larger

than

those

on

top,

causing

a

net

upward

force
.

The Buoyant Force


Since

the

sphere

is

not

accelerating,

the

upward

buoyant

force

is

balanced

by

the

weight

of

the

sphere
.


If

the

mass

of

the

cannon

ball

is

larger

(smaller)

than

the

mass

of

the

sphere,

the

cannon

ball

will

sink

(rise)


Since

the

buoyant

force

is

due

to

the

surrounding

fluid,

the

magnitude

of

the

buoyant

force

is

given

by


=

 
𝑉
 


Submerged Objects


When

an

object

is

fully

submerged

in

a

fluid,

the

upward

buoyant

force

acting

on

the

object

has

a

magnitude

of


=

 
𝑉




The

downward

gravitational

force

acting

on

the

object

has

a

magnitude

of


=

=


𝑉




Therefore,

the

net

force

on

the

object

is

𝐹

=



=

 



𝑉




Therefore,

if

the

density

of

the

object

is

less

(greater)

than

the

density

of

the

fluid,

then

the

net

force

exerted

on

the

object

is

positive

(negative)

and

the

object

accelerates

upward

(downward)
.

Floating Objects


When

an

object

is

floating

in

a

fluid,

the

upward

buoyant

force

is

balanced

by

the

downward

gravitational

force

acting

on

the

object
.


The

magnitude

of

the

buoyant

force

is

given

by


=

 
𝑉
 



Since

the

buoyant

force

is

balanced

by

the

gravitational

force,

this

gives


 
𝑉
 

=


𝑉






 
=
𝑉
 
𝑉



Conceptual Question 1


Imagine

holding

two

bricks

under

water
.

Brick

A

is

just

beneath

the

surface

of

the

water,

while

brick

B

is

at

a

greater

depth
.

The

force

needed

to

hold

brick

B

in

place

is



A
:

Larger

than

the

force

required

to

hold

brick

A

in

place


B
:

The

same

as

the

force

required

to

hold

brick

A

in

place


C
:

Smaller

than

the

force

required

to

hold

brick

A

in

place



Solution
:

(B)

The

buoyant

force

on

each

brick

is

equal

to

the

weight

of

the

water

it

displaces

and

does

not

depend

on

depth

Conceptual Question 2


Two

identical

cups

are

filled

to

the

same

level

with

water
.

One

of

the

two

cups

has

ice

cubes

floating

in

it
.

Which

weighs

more?



A
:

The

cup

without

ice

cubes


B
:

The

cup

with

ice

cubes


C
:

The

two

weigh

the

same



Answer
:

(C)

According

to

Archimedes

principle,

each

ice

cube

displaces

an

amount

of

water

that

is

exactly

equal

to

its

own

weight

and

thus,

the

cups

weigh

the

same
.

Conceptual Question 3


Two

identical

cups

are

filled

to

the

same

level

with

water
.

Solid

steel

balls

are

at

the

bottom

in

one

of

the

glasses
.

Which

of

the

two

glasses

weighs

more?




A
:

The

cup

without

steel

balls


B
:

The

cup

with

steel

balls


C
:

The

two

weigh

the

same



Answer
:

(B)

Since

steel

balls

sink

to

the

bottom

of

the

cup,

the

buoyant

force

(which

is

equal

to

the

weight

of

the

displaced

water)

is

not

sufficient

to

counter

the

weight

of

the

steel

balls
.

Therefore,

the

cup

with

steel

balls

weight

more
.

Conceptual Question 4


Consider

an

object

that

floats

in

water

but

sinks

in

oil
.

When

the

object

floats

in

water,

half

of

it

is

submerged
.

If

we

slowly

pour

oil

on

top

of

the

water

so

that

it

completely

covers

the

object,

the

object

will



A
:

Move

up


B
:

Stay

in

the

same

place


C
:

Move

down



Solution
:

(A)

When

the

oil

is

poured

over

the

object,

it

displaces

some

oil
.

Therefore

the

object

feels

a

buoyant

force

from

the

oil

in

addition

to

the

buoyant

force

from

the

water

and

rises

higher
.

Example 6


A

raft

is

constructed

of

wood

having

a

density

of

6
.
00

x

10
2

kg/m
3
.


Its

surface

area

is

5
.
70

m
2
,

and

its

volume

is

0
.
60

m
3
.


When

the

raft

is

placed

in

fresh

water,

to

what

depth

is

the

bottom

of

the

raft

submerged
.

Example 6 Solution


Since

the

raft

is

in

equilibrium,

Newton’s

2
nd

law

gives


𝐹
=





=
0


=



=


𝑉





By

Archimedes

principle,

the

magnitude

of

the

buoyant

force

is

also

equal

to

the

weight

of

the

displaced

volume

of

water


=

𝑤 
𝑉
𝑤 

=

𝑤 
ℎ

Example 6 Solution


Setting

both

expressions

equal

to

each

other

gives



𝑉


=

𝑤 
ℎ



=


𝑉


𝑤 

=
6
.
00



10
2


/

3
0
.
600


3
1
.
00



10
3


/

3
5
.
70


3
=
0
.
0632




Example 7


A

light

spring

of

force

constant


=
160


/


rests

vertically

on

the

bottom

of

a

large

beaker

of

water
.


A

5
.
00
-
kg

block

of

wood

(density

=

650

kg/m
3
)

is

connected

to

the

spring,

and

the

block
-
spring

system

is

allowed

to

come

to

static

equilibrium
.


What

is

the

elongation

Δ
L?

Example 7 Solution


When

the

block

is

held

in

equilibrium,

there

are

three

forces

acting

on

the

block
:

the

gravitational

force,

the

spring

force,

and

the

buoyancy

force
.


Newton’s

2
nd

law

gives


𝐹
=


𝐹


𝐹

=
0


𝑤 
𝑉


𝑤
𝑉




=
0




=



𝑤 


𝑤

𝑉
=




𝑤 

𝑤

1




=
5
.
00


9
.
80


/

2
160


/

1
.
00



10
3

/

3
650


/

3

1
=
0
.
165