MSU Physics 231 Fall 2015 1 Physics 231 Topic 10: Solids & Fluids Alex Brown Nov 4-9 2015.

MSU Physics 231 Fall 2015 1

Physics 231Topic 10: Solids & Fluids

Alex Brown Nov 4-9 2015


Key Concepts: Solids & FluidsStates of Matter

Density and matter states

Solids and ElasticityYoung’s & bulk moduli

Fluid Pressure & Motion

Continuity equation

Bernoulli’s equation

Buoyancy & Archimedes Principle

Surface Tension and ViscosityPoiseuille’s law

Covers chapter 10 in Rex & Wolfson


States of matterSolid 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


Phase Transformations


Solids

amorphousordered

crystalline


The Deformation of SolidsStress: Related to the force causing the deformation:

Force per unit area causing the deformationStrain: Measure of the degree of deformation Measure of the amount of deformationFor small stress, strain and stress are linearly correlated.

Strain = ConstantStress

Constant: elastic modulus

The elastic modulus depends on:• Material that is deformed• Type of deformation (a different modulus is defined

fordifferent types of deformations)

strainst

ress

elastic regime

elastic limit

inelastic regime

breaking point


The Young’s modulus (tensile)

positive) is ( /

/

] Unitless![ L/L :strain tensile

](Pa) Pascal[N/mF/A :stress tensile

strain tensile

stress tensile

0

0

0

2

LLA

LF

LL

AFY

Y

Beyond the elastic limit an object is permanently deformed (it does not return to its original shape if the stress is removed).

L

0L


The Shear Modulus

xA

Fh

hx

AFS

S

/

/

] Unitless![x/h :strainshear

](Pa) Pascal[N/mF/A :stressshear

strainshear

stressshear

2

x


Bulk Modulus

Compressibility: 1/(Bulk modulus)

pressure/

negative) is ( //

/

V/V :strain volume

](Pa) Pascal[N/mF/A :stress volume

strain volume

stress volume

00

0

2

AFP

VVV

P

VV

AFB

B

V0 V

0V


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

Water 0.21x1010


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?

5m

20

0

2

0 /

)/(

/

/

pillar

statue

pillarstatue

RY

gLML

LL

RgM

LL

AFY

Rpillar = 0.5m L0 = 5m Y = 1.0x1010 Pa Mstatue = 1.0x104 kg

L = 6.2x10-5 m


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 blockis: F = n (mblockg)

n = total number of blocksmblock = mass of one block = 5x103 kgg = 9.8 m/s2

Ultimate strength: 2x107= F/A= n (mblockg)/(1 m2)

n = 408 blocks.


Moving Earth’s Crust (shear)

30 m

100 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 layerof the earth does not move? (shear modulus S = 1.5x1010 Pa)

hx

AFS

/

/

strainshear

stressshear

F/A = S(x/h) = 4.5x109 Pa


An Example (bulk)

What force per unit area needs to be applied to compress1 m3 water by 1%? (B=0.21x1010 Pa)

0/

/

VV

AFB

V/V0 = -0.01 so, F/A = 2.1x107 Pa !!!


Density

)/( 3mkgV

M

)4(/

Cwatermaterialspecific o

density

specific density

material density (kg/m3) Specific density (kg/m3)

water 1.00x103 1.00

oxygen 1.43 0.00143

lead 11.3x103 11.3


Pressure

Pressure = P = F/A (N/m2=Pa)

Same Force, different pressure


An ExampleA 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 pressureexerted 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)


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 surfaceof only 0.25A is also placed on the floor. What is the ratioof the pressure exerted on the floor by block 1 to thepressure exerted on the floor by block 2 (I.e. P1/P2)?

a) 1/8b) ¼c) ½d) 1e) 2


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 surfaceof only 0.25A is also placed on the floor. What is the ratioof the pressure exerted on the floor by block 1 to thepressure exerted on the floor by block 2 (I.e. P1/P2)?

a) 1/8b) ¼c) ½d) 1e) 2

P1= F1/A1 = Mg/A

P2= F2/A2 = 2Mg/(0.25A) = 8 Mg/A

P1/P2 = 1/8


Force and Pressure

A P=0 vacuumF

Air (Pa=1.0x105 Pa)

Fp

Force due to pressure difference: F = P AIf A=0.010 m2 then

a force Fp = (1.0x105) x 0.010 = 1000 N (225 pounds)is needed to pull the lid.

What is the force needed to move the lid?


Magdeburg’s Hemispheres

Otto von Guericke (Mayor of Magdeburg, 17th century)


Pascal’s principleA 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.


Pascal’s principle

P = F1/A1 = F2/A2

If A2>>A1 then F2>>F1.

Hydraulic Lift

If we apply a smallforce F1, we can exerta very large Force F2.


Pressure & Depth (column of water in water)

Horizontal direction:P1=F1/A P2=F2/A F1=F2 (no net force)So, P1=P2

Forces in the up directionFt = - PtA (top)Fb = -Mg + PbA = - g A h + PbA (bottom)

Since the column of water is not moving: Ft + Fb = 0 - PtA - g Ah + PbA = 0Pb = Pt + g h

water

F1 F2

top PtA

PbAW=Mg

Pt=Patm atmospheric, A: surface area, M: mass

bottom


Pressure and Depth:

Ph = P0 + g h

Where:Ph = the pressure at depth hP0 = the pressure at depth 0 = density of the liquidg = 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.


A SubmarineA submarine is built in such a way that it can stand pressuresof 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


Does the shape of the container matter?

NO!!


Pressures at same heights are the same

P0

P=P0+gh

h

P0

P=P0+gh

h

P=P0+gh

h


Pressure Measurement:The mercury barometer

P0 = m g h

m = mercury = 13.6x103 kg/m3


Buoyant force: B

ht

hb

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)

P0

Archimedes (287 BC) principle: the magnitude of the buoyantforce is equal to the weight of the fluid displaced by the object

B

W


Comparing densities

B = g V buoyant force

W = Mo g = o g V weight of object

o = W = B stationary

o > W > B object goes down

o < W < B object goes up

B

W


A floating object (B=W)

h

A

B

W

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)


P = P0+ g hh = distance between liquid surface and the point where you measure P

P0

P

h

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 =

Pressure at depth h

Buoyant force for submerged object

Buoyant force for floating objecthBThe 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)

W=mg

W=mg


quizA block of weight Mg is placed in waterand found to stay submerged as shownin the picture. The water is thenreplaced by another liquid of lowerdensity. What will happen if the block is placed in the liquid of lower density?

a) the block will float on the surface of the liquidb) the block will be partially submerged and partially above the liquidc) the block will again be submerged as shown in the pictured) the block will sink to the bottom

initially B = Vo g W = M g B = Wlower density liquid: W remains the same, B becomes smallerthe block will sink to the bottom B < W

W=Mg


ProblemAn air mattress 2m long 0.5m wide and 0.08m thick and hasa 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


T1 T2

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

o = density of object = density of fluid V = volume of object


Density of a liquidAn 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


Tricky…

h

l


Without the balloon: B = W = Vd g Vd = volume of water displaced

With the balloon B + Fb= WThe 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.

W

a)B Fb=Fballoon


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 bythe 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


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.


Fluid flow - equation of continuity

A1, A2,

v1v21

2

the mass flowing into area 1 (M1) must be the same as themass 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 Av = constantA1v1 = A2v2

x1

x2


Bernoulli’s equation

P: pressure½v2: kinetic Energy per unit volume gy: potential energy per unit volume

P1,A1,y1

P2,A2,y2

v1

v2

P1 + ½ v12 + g y1= P2 + ½ v2

2 + g y2

P + ½ v2 + g y = constant

P: pressure y: height: density g: gravitational v: velocity acceleration


(1) P + ½ v2 + g y = constant

(2) A v = constant

equal

A. Incompressible fluid, so density is constantB. (2) AA>AB so vA<vB

(1) vA < vB so PA > PB

greater

C. Must be the same, else the liquid would get ‘stuck’ between A and B, o

equalless than

See B.

(1) Since y = constant P + ½ v2 = constant


Moving cansP0

P0

Top view Before air is blown in betweenthe cans, the cans remainat rest and the air in between the cans is at rest (0 velocity) P = PoP

Bernoulli’s law: P + ½v2 = P0 so P = P0 - ½v2

So P < P0 Because of the pressure differenceleft and right of each can, they move inward

When air is blown in between thecans, the velocity is not equal to 0.(ignore the y part)

case:1: no blowing2: blowing


Ph = P0 + gh

h

x

y

If h=1m and y=3m what is x?Assume that the holes are smalland the water level doesn’t dropnoticeably.

P0

A Hole in a Tank

Ph + g y = P0+ g (y+h)


h

x1

y

If h=1 m and y=3 m what is x?

Use Bernoulli’s law

PA + ½vA2 + gyA= PB + ½vB

2 + gyB

at A: PA = P0 vA=? yA = y

at B: PB = P0 vB=0 yB = y + h

vA2 = 2gh vA = 4.43 m/s

A

P0 B


x1

3m

vA

In the horizontal direction:x(t) = x0 + v0xt + ½at2 = 0 + 4.43t + 0

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

0

MSU Physics 231 Fall 2015 49PHY 231

49

ViscosityViscosity: stickiness of a fluidOne layer of fluid feels a largeresistive force when slidingalong another one or along asurface of for example a tube.


ViscosityContact surface A

fixed

movingF=Av/d

=coefficient of viscosityunit: Ns/m2

or poise=0.1 Ns/m2

Fv

d

Fluid T (oC) Viscosity (Ns/m2)

water 20 1.0x10-3

blood 37 2.7x10-3

oil 30 250x10-3


Poiseuille’s Law (not covered in homework or exams)

How fast does a fluid flowthrough a tube?

Rate of flow Q= v/t=R4(P1-P2)

8L(unit: m3/s)

RvP1

P2

L


Example

P=106 Pa P=105 Pa

Flow rate QTube length: L=3 m=1.50 Ns/m2

What should the radius of the tube be to allow for Q = 0.5 m3/s ?

Rate of flow Q=R4(P1-P2)

8L

R=[ (8Q L) / (P1-P2) ]1/4 = 0.05 m


QuizA 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) (W+V)/Sb) W/Sc) (W+V)/(KS)d) W/(KS)e) (W+V)g/S with g=9.81 m/s2


Clicker QuestionA 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) (W+V)/Sb) W/Sc) (W+V)/(KS)d) W/(KS)e) (W+V)g/S with g=9.81 m/s2

P=F/A F=V+W A=KS


Measurement of Pressure

The open-tube manometer.The pressure at A and B is the same:PA = PB

P = P0 + ghso 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

MSU Physics 231 Fall 2015 1 Physics 231 Topic 10: Solids & Fluids Alex Brown Nov 4-9 2015.

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Transcript of MSU Physics 231 Fall 2015 1 Physics 231 Topic 10: Solids & Fluids Alex Brown Nov 4-9 2015.