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
ℎ
=
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
ℎ
=
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
∆
ℎ
Shear Modulus
𝐹
=
𝑆
∆
ℎ
•
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
ℎ
=
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
∆
ℎ
•
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
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
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