Transcript ch10

Physics 231
Topic 10: Solids & Fluids
Alex Brown
Nov
4-9
2015
MSU
Physics
231 Fall
2015
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
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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)
L
L0
tensile stress
Y
tensile strain
2
tensile stress : F/A [N/m  Pascal (Pa)]
tensile strain : L/L0 [ Unitless! ]
F L0
F/A
Y

L / L0 A L
(L is positive)
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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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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Bulk Modulus
V0
V0  V
volume stress
B
volume strain
volume stress : F/A [N/m 2  Pascal (Pa)]
volume strain : V/V0
F / A
P
B

V / V0
V / V0
(V is negative)
P  F / A  pressure
Compressibility: 1/(Bulk modulus)
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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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An Example (tensile)
An architect wants to design a 5m high circular pillar with a
radius of 0.5 m that holds a bronze statue that weighs 1.0x104
kg. He chooses concrete for the material of the pillar
(Y=1.0x1010 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
Rpillar = 0.5m
Y = 1.0x1010 Pa
L0 = 5m
Mstatue = 1.0x104 kg
L = 6.2x10-5 m
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Another Example (tensile)
A builder is stacking 1 m3 cubic concrete blocks. Each block
weighs 5x103 kg. The ultimate strength of concrete for
compression is 2x107 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 = 5x103 kg
g = 9.8 m/s2
Ultimate strength: 2x107= F/A
= n (mblockg)/(1 m2)
n = 408 blocks.
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Moving Earth’s Crust (shear)
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? (shear modulus S = 1.5x1010 Pa)
shear stress F / A
S

shear strain x / h
F/A = S(x/h) = 4.5x109 Pa
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An Example (bulk)
What force per unit area needs to be applied to compress
1 m3 water by 1%? (B=0.21x1010 Pa)
F / A
B
V / V0
V/V0 = -0.01
so,
F/A = 2.1x107 Pa !!!
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Density
M

V
density
specific density
3
(kg / m )
 specific   material /  water ( 4
o
C)
density (kg/m3)
Specific
density (kg/m3)
water
1.00x103
1.00
oxygen
1.43
0.00143
lead
11.3x103
11.3
material
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Pressure
Pressure = P = 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/1x10-6m2 = 7x108 Pa
Pperson = F/(1000Anail)
= 700/(1000x10-6 1E-6) = 7x105 Pa
(about 7 times the atmospheric
pressure)
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Clicker Quiz!
A block (1) of mass M is lying on the floor. The contact
surface between the block and the floor is A.
A second block (2), 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
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Clicker Quiz!
A block (1) of mass M is lying on the floor. The contact
surface between the block and the floor is A.
A second block (2), 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) = 8 Mg/A
P1/P2 = 1/8
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Force and Pressure
Air (Pa=1.0x105 Pa)
A
P=0 vacuum
F
Fp
What is the force needed to move the lid?
Force due to pressure difference: F = P A
If A=0.010 m2 then
a force Fp = (1.0x105) x 0.010 = 1000 N (225 pounds)
is needed to pull the lid.
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Magdeburg’s Hemispheres
Otto von Guericke (Mayor of Magdeburg, 17th century)
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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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Hydraulic Lift
Pascal’s principle
If we apply a small
force F1, we can exert
a very large Force F2.
P = F1/A1 = F2/A2
If A2>>A1 then
F2>>F1.
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Pressure & Depth (column of water in water)
water
F1
top
F2
PtA
Horizontal direction:
P1=F1/A P2=F2/A
F1=F2 (no net force)
So, P1=P2
Pt=Patm atmospheric, A: surface area, M: mass
Forces in the up direction
Ft = - PtA (top)
Fb = -Mg + PbA = -  g A h + PbA (bottom)
bottom
W=Mg
PbA
Since the column of water is not moving:
Ft + Fb = 0
- PtA -  g Ah + PbA = 0
Pb = Pt +  g h
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Pressure and Depth:
Ph = P0 +  g h
Where:
Ph = the pressure at depth h
P0 = the pressure at depth 0
 = density of the liquid
g = 9.8 m/s2
h = depth
P0 = Patm = 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?
Ph = P0+  g h
30x105 = (1.0x105) + (1.0x103)(9.81)h
h=296 m
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Does the shape of the container
matter?
NO!!
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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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Pressure Measurement:
The mercury barometer
P0 = m g h
m = mercury = 13.6x103 kg/m3
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Buoyant force: B
P0
ht
B
hb
W
Pt = P0 +  g ht (top)
Pb = P0 +  g hb (bottom)
Pb - Pt =  g (hb-ht)
=  g h
B = Fb – Ft = (Pb – Pt) A
=  g h A =  g V
= Mf g (weight of
displaced fluid)
W = Mo g (weight of object)
If the object is not moving:
B = W so: Mo= Mf
(for a submerged object)
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 = g V
B
W
W = M o g = o g V
buoyant force
weight of object
o = 
W=B
stationary
o > 
W>B
object goes down
o < 
W<B
object goes up
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A floating object (B=W)
A
B
W
h
W = Mo g
(weight of object)
B = weight of the fluid with density 
displaced by the object
= Md g =  Vd g
Thus
Vd = Mo / volume displaced
For a block with top area A and
h = height of the object under water, then Vd = A h
h = Mo / ( A) = (oVo ) / ( A)
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P0
Pressure at depth h
P = P0+  g h
h = distance between liquid surface
and the point where you measure P
h
P
Buoyant force for submerged object
B =  Vo g = Mf g
The buoyant force equals the weight of the
amount of fluid that can be put in the volume
taken by the object.
If object is not moving: B = W o = 
Buoyant force for floating object
The buoyant force equals the weight of the amount of fluid
that can be put in the part of the volume of the object that is
under water. Vd = Mo/ (any object)
h = oVo/( A) = Mo/( A) (for a block area A)
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W=mg
B h
W=mg
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quiz
W=Mg
A block of weight Mg is placed in water
and found to stay submerged as shown
in the picture. The water is then
replaced by another liquid of lower
density. What will happen if the
block is placed in the liquid of lower density?
a) the block will float on the surface of the liquid
b) the block will be partially submerged and partially above
the liquid
c) the block will again be submerged as shown in the picture
d) the block will sink to the bottom
initially B =  Vo g
W= Mg
B=W
lower density liquid: W remains the same, B becomes smaller
the block will sink to the bottom B < W
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Problem
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?
A) h = Mo/( A) = 2.0/[1.0x103 (2)(0.5)] = 0.002 m = 2 mm
B) if the objects sinks the mattress is just completely
submerged: h = thickness of mattress.
h = Mo/( A)
0.08 = (Mweight+2.0)/[1.0x103(2)(0.5)]
so Mweight=78 kg
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T1
T2
o = density of object
 = density of fluid
V = volume of object
T1 = Mog = oVg
T2 = Mog – Mfg = oVg - Vg
(T2 /T1 ) = R = 1 - ( /o )
 = o (1-R)
(o / ) = 1/(1-R)
= specific density when the fluid is water  = w
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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?
T1 = Mog = (0.0194)(9.81) = 0.190
T2 = 0.128
R = T2 /T1 = 0.674
 = o (1-R) = (2395) (1-0.674) = 781 kg/m3
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Tricky…
h
l
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a)
Fb=Fballoon
Without the balloon:
B = W =  Vd g
Vd = volume of water displaced
B
W
With the balloon B + Fb= W
The balloon is trying to pull the boat out of the water.
 Vd g = W - Fb
Vd = (W - Fb)/( g)
If Fb = 0 then Vd increases (the boat sinks deeper)
As a result, the water level rises.
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b) water thrown out (B = W)
Vd = Mo / volume displaced
Initially there is water in the boat:
Vd = (mw + mboat) / = (mw/ ) + (mboat/)
= Vwater thrown out + (mboat/ )
When the water is thrown out of the boat:
Vd = mboat/ (same Eq. As above but with mw = 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) Anchor thrown out (W = B)
Vd = Mo / volume displaced
So Vd = (mboat + manchor) /
If the anchor is thrown overboard the gravitational force
is still acting on it and nothing chances.
If the anchor hits the ground, what happens?
Vd decreases – the water level falls.
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Fluid flow - equation of continuity
x2
x1
v1
1
A1,
2
v2
A2,
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. (liquid is incompressible: 1 = 2 = )
M1= M2
(M = V = Ax)
 A1 x1 =  A2 x2 (x = v t)
A1v1 t = A2v2 t
A1v1 = A2v2
Av = constant
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Bernoulli’s equation
P2,A2,y2
v2
P1 + ½  v12 +  g y1= P2 + ½  v22 +  g y2
P1,A1,y1
P + ½  v2 +  g y = constant
v1
P: pressure
: density
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 +  g y = constant
(2) A v = constant
(1) Since y = constant
P + ½  v2 = constant
A. Incompressible fluid,
so density is constant
equal
B. (2) AA>AB so vA<vB
greater
(1) vA < vB so PA > PB
equal
less than
C. Must be the same, else
the liquid would get
‘stuck’ between A and B, o
See B.
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Moving cans
P0
Top view
P
case:
1: no blowing
2: blowing
P0
Before air is blown in between
the cans, the cans remain
at rest and the air in between
the cans is at rest (0 velocity)
P = Po
When air is blown in between the
cans, the velocity is not equal to 0.
(ignore the y part)
Bernoulli’s law:
P + ½v2 = P0 so P = P0 - ½v2
So P < 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
y
Ph = P0 + gh
h
Ph +  g y = P0+  g (y+h)
If h=1m and 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
y
B
h
Use Bernoulli’s law
A
PA + ½vA2 + gyA= PB + ½vB2 + gyB
at A: PA = P0 vA=?
yA = y
at B: PB = P0 vB=0
yB = y + h
vA2 = 2gh
vA = 4.43 m/s
x1
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vA
In the horizontal direction:
x(t) = x0 + v0xt + ½at2 = 0 + 4.43t + 0
3m
0
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(t=0.78) = (4.43)(0.78) = 3.45 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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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
water
blood
oil
20
37
30
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Viscosity 
(Ns/m2)
1.0x10-3
2.7x10-3
250x10-3
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Poiseuille’s Law
(not covered in homework or exams)
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
Tube length: L=3 m
=1.50 Ns/m2
P=105 Pa
P=106 Pa
What should the radius of the tube be to allow for
Q = 0.5 m3/s ?
Rate of flow Q=
R4(P1-P2)
8L
R=[ (8Q  L) / (P1-P2) ]1/4 = 0.05 m
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Quiz
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
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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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Measurement of Pressure
The open-tube manometer.
The pressure at A and B is
the same:
PA = PB
P = P0 + gh
so h = (P-P0)/(g)
If the pressure P = 1.1 atm, what
is h? (the liquid is water)
h = (1.10-1.00)x105/(1.0x103x9.81) = 1.0 m
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