Physics 231 Topic 9: Solids & Fluids Wade Fisher

Physics 231
Topic 9: Solids & Fluids
Wade Fisher
Oct
Nov
22011
2011
MSU28Physics
231 Fall
1
Key Concepts: Solids & Fluids
States of Matter
Density and matter states
Solids and Elasticity
Young’s & bulk moduli
Fluid Pressure & Motion
Continuity equation
Bernoulli’s equation
Buoyancy & Archimedes Principle
Surface Tension and Viscosity
Poiseuille’s law
Covers chapter 10 in Rex &
Wolfson
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States of matter
Solid
Liquid
Gas
Plasma
difficult to
deform
easy to
deform
easy to
deform
easy to
deform
difficult to
compress
difficult to
compress
easy to
compress
easy to
compress
difficult to
flow
easy to flow easy to flow easy to flow
not charged not charged not charged charged
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Phase Transformations
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Solids
amorphous
ordered
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crystalline
5
The Deformation of Solids
Strain = ConstantStress
Constant: elastic modulus
stress
Stress: Related to the force causing the deformation:
Force per unit area causing the deformation
Strain: Measure of the degree of deformation
Measure of the amount of deformation
For small stress, strain and stress are linearly correlated.
inelastic regime
breaking point
elastic limit
elastic regime
The elastic modulus depends on:
strain
• Material that is deformed
• Type of deformation (a different modulus is defined for
different types of deformations)
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The Young’s modulus
tensile stress
Y
tensile strain
2
tensile stress : F/A [N/m  Pascal (Pa)]
tensile strain : L/L0 [ Unitless! ]
FL0
F/A
Y

L / L0 AL
Beyond the elastic limit an object is
permanently deformed (it does not
return to its original shape if the
stress is removed).
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An Example
An architect wants to design a 5m high circular pillar with a
radius of 0.5 m that holds a bronze statue that weighs 1.0E+04
kg. He chooses concrete for the material of the pillar
(Y=1.0E+10 Pa). How much does the pillar compress?
F/A
Y

L / L0
5m
2
M statue g /(R pillar
)
L / L0
M statue gL0
L 
2
YR pillar
R = 0.5m L0 = 5m
Y = 1.0E+10 Pa M = 1.0E+04 kg
L = 6.2E-05 m
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Another Example
A builder is stacking 1 m3 cubic concrete blocks. Each block
weighs 5E+03 kg. The ultimate strength of concrete for
compression is 2E+07 Pa. How many blocks can he stack before
the lowest block is crushed?
The force on the low end of the lowest block
is: F = N(mblockg)
N = total number of blocks
mblock = mass of one block
g = 9.81 m/s2
Ultimate strength: 2E+07= F/A
= N(mblockg)/(1 m2)
N=408 blocks.
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Ultimate strength
Ultimate strength: maximum force per unit area a material
can withstand before it breaks or fractures.
Different for
compression and tension.
Material
Tensile Strength
(N/m2)
Compressive
Strength (N/m2)
Steel
5.0x108
5.5x108
Bone
1.2x108
1.5x108
Concrete
2x106
2x107
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The Shear Modulus
x
shear stress
S
shear strain
shear stress : F/A [N/m 2  Pascal (Pa)]
shear strain : x/h [ Unitless! ]
F/A
Fh
S

x / h Ax
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Moving Earth’s Crust
100 m
30 m
A tectonic plate in the lower crust (100 m deep) of
the earth is shifted during an earthquake by 30m.
What is the shear stress involved, if the upper layer
of the earth does not move? (S=1.5E+10 Pa)
shear stress F / A
S

shear strain x / h
F/A = S(x/h) = 4.5E+09 Pa
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Bulk Modulus
volume stress
B
volume strain
volume stress : F/A [N/m 2  Pascal (Pa)]
volume strain : V/V 0
F / A
P
B

V / V0
V / V0
P  pressure
Compressibility: 1/(Bulk modulus)
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An Example
What force per unit area needs to be applied to compress
1 m3 water by 1%? (B=0.21E+10 Pa)
F / A
B
V / V0
V/V0=0.01 so, F/A=2.1E+07 Pa !!!
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Some Elastic Moduli
Material
Young’s
Modulus
(N/m2)
Shear
Modulus
(N/m2)
Bulk
Modulus
(N/m2)
Steel
20x1010
8.4x1010
16x1010
Bone
1.8x1010
8.0x1010
-
Aluminum
7x1010
2.5x1010
7x1010
0.21x1010
Water
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Density
M

V
density
specific density
3
(kg / m )
 specific   material /  water( 4
o
C)
density (kg/m3)
material
water
1.00x103
oxygen
1.43
lead
11.3x103
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Pressure
Pressure = F/A (N/m2=Pa)
Same Force, different pressure
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An Example
A nail is driven into a piece of wood with a force of 700N.
What is the pressure on the wood if Anail = 1 mm2?
A person (weighing 700 N) is lying on a bed of such
nails (his body covers 1000 nails). What is the pressure
exerted by each of the nails?
Pnail = F/Anail = 700N/1E-06m2=7E+08 Pa
Pperson = F/(1000Anail)
= 700/(1E3 x 1E-6) = 7E+05 Pa
(about 7 times the atmospheric
pressure)
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Clicker Question
A object with weight of W (in N) is resting on a table with
K legs each having a contact surface S (m2) with the floor.
The weight of the table is V (in N). The pressure P exerted
by each of the legs on the floor is:
a)
b)
c)
d)
e)
(W+V)/S
W/S
(W+V)/(KS)
W/(KS)
(W+V)g/S with g=9.81 m/s2
P=F/A
F=V+W
A=KS
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Force and Pressure
Air (P=1.0E+5 Pa)
A
P=0
Vacuum
F
What is the force needed to move the lid?
Force due to pressure difference: Fpressure=PA
If A=0.01 m2 (about 10 by 10 cm) then
a force F = (1.0E+5)x0.01 = 1000N is needed to pull
the lid.
Fpressure-difference= PA
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Magdeburg’s Hemispheres
Otto von Guericke (Mayor of Magdeburg, 17th century)
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Previously
Solids:
General:
FL0
F/A
Y

L / L0 AL
Young’s modulus
F/A
Fh
Shear modulus
S

x / h Ax
F / A
P
Bulk modulus
B

V / V0
V / V0 Also fluids
P  pressure
P=F/A (N/m2=Pa)
=M/V (kg/m3)
Fpressure-difference=PA
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the
whole fluid and all the walls of the container that
hold the fluid.
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Clicker Quiz!
A block of mass M is lying on the floor. The contact
surface between the block and the floor is A.
A second block, of mass 2M but with a contact surface
of only 0.25A is also placed on the floor. What is the ratio
of the pressure exerted on the floor by block 1 to the
pressure exerted on the floor by block 2 (I.e. P1/P2)?
a)
b)
c)
d)
e)
1/8
¼
½
1
2
P1=F1/A1=Mg/A
P2=F2/A2=2Mg/(0.25A)=8Mg/A
P1/P2=(Mg/A)/(8Mg/A)=1/8
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Pascal’s principle
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the
whole fluid and all the walls of the container that
hold the fluid.
HOLDS FOR A FLUID FULLY ENCLOSED ONLY
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Pascal’s principle
In other words then before: a change in pressure applied
to a fluid that is enclosed in transmitted to the whole
fluid and all the walls of the container that hold the fluid.
So, if we apply a small
force F1, we can exert
a very large Force F2.
P = F1/A1 = F2/A2
If A2>>A1
F2>>F1.
then
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Pressure & Depth
water
F1
surface
Mg
F2
PatmA
Horizontal direction:
P1=F1/A P2=F2/A
F1=F2 (no net force)
So, P1=P2
atm: atmospheric, A: surface area, M: mass
Vertical direction:
Ftop=PatmA
Fbottom=PbottomA-Mg=PbottomA-gAh
Since the column of water is not moving:
Ftop-Fbottom=0
PatmA=PbottomA-gAh
PbottomA Pbottom=Patm+ gh
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Pressure and Depth:
Pdepth=h = Pdepth=0 + gh
Where:
Pdepth=h: the pressure at depth h
Pdepth=0: the pressure at depth 0
=density of the liquid
g=9.81 m/s2
h=depth
Pdepth=0 = Patmospheric = 1.013x105 Pa = 1 atm =760 Torr
From Pascal’s principle: If P0 changes then the pressures
at all depths changes with the same value.
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A Submarine
A submarine is built in such a way that it can stand pressures
of up to 3x106 Pa (approx 30 times the atmospheric
pressure). How deep can it go?
Pdepth=h = Pdepth=0+ gh
3E+06 = (1.0E+05) +(1.0E+03)(9.81)h
h=296 m
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Does the shape of the container
matter?
NO!!
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Measurement of Pressure
The open-tube manometer.
The pressure at A and B is
the same:
P = P0 + gh
so h = (P-P0)/(g)
If the pressure P = 1.01 atm, what
is h? (the liquid is water)
h = (1.01-1)x(1.0E+05)/(1.0E+03x9.81)=
= 0.1 m
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Pressure Measurement:
The mercury barometer
P0= mercurygh
mercury=13.6E+03 kg/m3
mercury,specific=13.6
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Pressures at same heights are
the same
P0
P0
h
P=P0+gh
h
h
P=P0+gh
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P=P0+gh
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P0
htop
B
hbottom
w
Buoyant force: B
Ptop =P0+ wghtop
Pbottom =P0+ wghbottom
p = wg(htop-hbottom)
-
F/A = wgh
F = wghA=wgV
B =wgV=Mwaterg
Fg=w=Mobjg
If the object is not moving:
B=Fg so: wgV=Mobjg
Archimedes (287 BC) principle: the magnitude of the buoyant
force is equal to the weight of the fluid displaced by the object
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Comparing densities
B =fluidgV Buoyant force
B
w =Mobjectg=objectgV
Stationary: B=w
object= fluid
w
If object> fluid the object goes down!
If object< fluid the object goes up!
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A floating object
A
B
w
h
w=Mobjectg=objectVobjectg
B=weight of the fluid displaced by
the object
=Mwater,displacedg
= waterVdisplacedg
= waterhAg
h: height of the object under water!
The object is floating, so there is no net force (B=w):
objectVobject= waterVdisplaced
h= objectVobject/(waterA) only useable if part of the object
is above the water!!
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A)
?? N
An example
B)
?? N
7 kg iron sphere of
the same dimension
as in A)
1 kg of water inside
thin hollow sphere
Two weights of equal size and shape, but different mass are
submerged in water. What are the weights read out?
B=waterVdisplacedg
w=sphereVsphereg
A) B= waterVsphereg w=waterVsphereg so B=w and 0 N is read out!
B) B= waterVsphereg=Mwater sphereg
w= ironVsphereg=Miron sphereg=7Mwater sphereg
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T=w-B=6*1*9.8
=58.8 N
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Another one
An air mattress 2m long 0.5m wide and 0.08m thick and has
a mass of 2.0 kg. A) How deep will it sink in water? B) How
much weight can you put on top of the mattress before it
sinks? water=1.0E+03 kg/m3
A) h= objectVobject/(waterA)
h=Mobject/(1.0E+03*2*0.5)=2.0/1.0E+03=2.0E-03m=2mm
B) if the objects sinks the mattress is just completely
submerged: h=thickness of mattress.
0.08=(Mweight+2.0)/(1.0E+03*2*0.5)
So Mweight=78 kg
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Density of a liquid
An object with a density of 2395 kg/m3 and mass of
0.0194 kg is hung from a scale and submerged in a liquid.
The weight read from the scale is 0.128 N.
What is the density of the liquid?
F=ma=0 (while hanging in the liquid, a=0)
T+B-Mobjectg=0 (the tension in the wire of the scale and the
buoyant force both push the object up)
0.128+B-0.0194*9.81=0 T: tension=weight read from scale
B=0.0623
B=liquidVobjectg Vobject=mobject/object=0.0194/2395=8.1x10-6 m3
So: B=liquid x 8.1x10-6 x 9.81
Combine: 0.0623= liquid x 8.1x10-6 x 9.81
So: liquid =784 kg/m3
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P0
Pressure at depth h
P = P0+ fluidgh
h: distance between liquid surface
and the point where you measure P
h
P
Buoyant force for submerged object
B = fluidVobjectg = Mfluidg = wfluid
The buoyant force equals the weight of the
amount of water that can be put in the
volume taken by the object.
If object is not moving: B=wobject object= fluid
Buoyant force for floating object
The buoyant force equals the weight of the amount of water
that can be put in the part of the volume of the object that is
under water.
objectVobject= waterVdisplaced h= objectVobject/(waterA)
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w
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quiz (extra credit)
a)
b)
c)
d)
A block of weight w is placed in water
and found to stay submerged as shown
in the picture. The liquid is then
replaced by another liquid of lower
density. What will happen if the
block is placed in the liquid of lower density?
the block will float on the surface of the liquid
the block will be partially submerged and partially above
the liquid
the block will again be submerged as shown in the picture
the block will sink to the bottom
initially B = fluidVobjectg w=Mblockg B=w
lower density liquid: w remains the same, B becomes smaller
the block will sink to the bottom
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Tricky…
h
l
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a)
Fballoon
Fbuoyancy
Without the balloon:
Fbuoyancy = Fgravity
waterVdisplacedg = Fgravity
Fg
With the balloon
Fbuoyancy + Fballoon= Fgravity
The balloon is trying to pull the boat out of the water.
waterVdisplacedg = Fgravity-Fballoon
Vdisplaced= (Fgravity-Fballoon)/(waterg)
If Fballoon =0 then Vdisplaced increases (the boat sinks deeper)
As a result, the water level rises.
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b)General equation to use:
Fbuoyancy = Fgravity
waterVdisplacedg = Fgravity Vdisplaced=Fgravity/(waterg)
Initially there is water in the boat:
Vdisplaced=Fgravity/(waterg)=(mwater+mrest of boat)g/(waterg)
=mwater/water + mrest of boat/water
=Vwater thrown out + mrest of boat/water
When the water is thrown out of the boat:
Vdisplaced= mrest of boat/water (same Eq. As above but with mwater=0)
So by throwing the water, the displaced volume reduces by
the volume of the water thrown out of the boat. The boat
should rise and the water level go down BUT the volume of
water is thrown back into the lake so it is filled up again!
Answer: water level is unchanged
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c) waterVdisplacedg = Fgravity Vdisplaced=Fgravity/(waterg)
So Vdisplaced=(mboat+manchor)g/(waterg)=(mboat+manchor)/water
If the anchor is thrown overboard the gravitational force
is still acting on it and nothing chances.
Note: if the anchor hits the ground, what happens?
Consider the normal force acting on the anchor…
Does it help to lift the anchor?
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equation of continuity
x2
x1
v1
1
A1,1
2
v2
A2,2
the mass flowing into area 1 (M1) must be the same as the
mass flowing into area 2 (M2), else mass would accumulate
in the pipe).
M1= M2
1A1x1= 2A2x2 (M=V=Ax)
1A1v1t =2A2v2t (x=vt)
1A1v1 =2A2v2
if  is constant (liquid is incompressible) A1v1 =A2v2
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Bernoulli’s equation
P2,A2,y2
v2
P1,A1,y1
v1
P1+½v12+gy1= P2+½v22+gy2
P+½v2+gy=constant
P: pressure
: height
v: velocity
y: height
g: gravitational
acceleration
P: pressure
½v2:kinetic Energy per unit volume
gy: potential energy per unit volume
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1) P+½v2+gy=constant
2) A1v1 =A2v2
A. Incompressible fluid,
so density is constant
equal
B. AA>AB so vA<vB use 2)
vA<vB so PA>PB use 1)
C. Must be the same, else
the liquid would get
‘stuck’ between A and B,
or a ‘hole’ would open
between A and B
greater
equal
less than
See B.
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Moving cans
P0
Top view
case:
1: no blowing
2: blowing
P0
P1
Before air is blown in between
the cans, P0=P1; the cans remain
at rest and the air in between
the cans is at rest (0 velocity)
P1+½v12+gy1= Po
When air is blown in between the
cans, the velocity is not equal to 0.
P2+½v22 (ignore y)
Bernoulli’s law:
P1+½v12+gy1= P2+½v22+gy2
P0=P2+½v22 so P2=P0-½v22
So P2<P0
Because of the pressure difference
left and right of each can, they move inward
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A Hole in a Tank
P0
Pdepth=h =Pdepth=0+ gh
y
h
If h=1m & y=3m what is x?
Assume that the holes are small
and the water level doesn’t drop
noticeably.
x
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If h=1 m and y=3 m what is X?
P0 B
y
h
Use Bernoulli’s law
A
PA+½vA2+gyA= PB+½vB2+gyB
At A: PA=P0 vA=? yA=y=3
At B: Pb=P0 vB=0 yB=y+h=4
P0+½vA2+g3=P0+g4
vA=(g/2)=2.2 m/s
x1
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vA
3m
0
Each water element of mass m has the same
velocity vA. Let’s look at one element m.
vA=(g/2)=2.2 m/s
In the horizontal direction:
x(t)=x0+v0xt+½at2=2.2t
x1
In the vertical direction:
y(t)=y0+v0yt+½at2=3-0.5gt2
= 0 when the water hits the ground, so
t=0.78 s
so x(0.78)=2.2*0.78=1.72 m
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Viscosity
Viscosity: stickiness of a fluid
One layer of fluid feels a large
resistive force when sliding
along another one or along a
surface of for example a tube.
PHY 231
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52
Viscosity
Contact surface A
F
v
moving
F=Av/d
d
=coefficient of viscosity
unit: Ns/m2
or poise=0.1 Ns/m2
fixed
T (oC)
Fluid
Viscosity 
(Ns/m2)
water
20
1.0x10-3
blood
oil
37
30
2.7x10-3
250x10-3
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Poiseuille’s Law
P1
R
v
P2
How fast does a fluid flow
through a tube?
L
Rate of flow Q= v/t=
R4(P1-P2)
8L
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(unit: m3/s)
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Example
Flow rate Q=0.5 m3/s
Tube length: 3 m
=1500E-03 Ns/m2
P=105 Pa
P=106 Pa
What should the radius of the tube be?
Rate of flow Q=
R4(P1-P2)
8L
R=[8QL/((P1-P2))]1/4=0.05 m
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For Next Week
Chapter 11:
Waves & Sound
No Homework due
Study for the exam!
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