phys 1441 – section 002 lecture # 23

Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

1

PHYS 1441 – Section 002Lecture #23

Monday, Dec. 6, 2010Dr. Jaehoon Yu

• Similarities Between Linear and Rotational Quantities

• Conditions for Equilibrium• How to Solve Equilibrium Problems?• A Few Examples of Mechanical Equilibrium• Elastic Properties of Solids• Density and Specific Gravity

Announcements• The Final Exam

– Date and time: 11am – 1:30pm, Monday Dec. 13– Place: SH103– Comprehensive exam

• Covers from CH1.1 – what we finish Wednesday, Dec. 8• Plus appendices A.1 – A.8• Combination of multiple choice and free response

problems • Bring your Planetarium extra credit sheet to the class next

Wednesday, Dec. 8, with your name clearly marked on the sheet!


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3

Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.

Quantities Linear RotationalMass Mass Moment of Inertia

Length of motion Distance Angle (Radian)Speed

AccelerationForce Force TorqueWork Work WorkPower

MomentumKinetic Energy Kinetic Rotational

I

rvt

t

vat

t

Furm

r τrI

ur

W F d rr

P Fur⋅vr τP

2

21 mvK 2

21 IKR

LM

W τ

purmv

r LurI

ur


4

Conditions for EquilibriumWhat do you think the term “An object is at its equilibrium” means?

Fur

∑

The object is either at rest (Static Equilibrium) or its center of mass is moving at a constant velocity (Dynamic Equilibrium).

Is this it?

When do you think an object is at its equilibrium?

Translational Equilibrium: Equilibrium in linear motion

The above condition is sufficient for a point-like object to be at its translational equilibrium. However for an object with size this is not sufficient. One more condition is needed. What is it?

Let’s consider two forces equal in magnitude but in opposite direction acting on a rigid object as shown in the figure. What do you think will happen?

CMd

d

F

-F

The object will rotate about the CM. Since the net torque acting on the object about a rotational axis is not0.

For an object to be at its static equilibrium, the object should not have linear or angular speed.

τ r

0CMv 0

0

0


5

More on Conditions for EquilibriumTo simplify the problem, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case?

The six possible equations from the two vector equations turns to three equations.

What happens if there are many forces exerting on an object?

0F ur

0τ r0xF 0zτ

O

F1

F4

F3

F 2

F5

r5 O’r’

If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis.

0yF

Why is this true?Because the object is not moving, no matter what the rotational axis is, there should not be any motion. It is simply a matter of mathematical manipulation.

AND


6

How do we solve static equilibrium problems?

1. Select the object to which the equations for equilibrium are to be applied.

2. Identify all the forces and draw a free-body diagram with them indicated on it with their directions and locations properly indicated

3. Choose a convenient set of x and y axes and write down force equation for each x and y component with correct signs.

4. Apply the equations that specify the balance of forces at equilibrium. Set the net force in the x and y directions equal to 0.

5. Select the most optimal rotational axis for torque calculations Selecting the axis such that the torque of one or more of the unknown forces become 0 makes the problem much easier to solve.

6. Write down the torque equation with proper signs.7. Solve the force and torque equations for the desired unknown

quantities.

7Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Example for Mechanical EquilibriumA uniform 40.0 N board supports the father and the daughter each weighing 800 N and 350 N, respectively, and is not moving. If the support (or fulcrum) is under the center of gravity of the board, and the father is 1.00 m from CoG, what is the magnitude of the normal force n exerted on the board by the support?

Since there is no linear motion, this system is in its translational equilibriumF D

n1m x

Therefore the magnitude of the normal force nDetermine where the child should sit to balance the system.The net torque about the fulcrum by the three forces are

τ

Therefore to balance the system the daughter must sit x

Fx∑ 0

yF BM g 0FM g DM gn

mgMgM

D

F 00.1 mm 29.200.1350800

00.1 gM F xgM D 0

N11903508000.40

MBgMDgMFg

0 gM B 0n


Example for Mech. Equilibrium Cont’d Determine the position of the child to balance the system for different position of axis of rotation.

Since the normal force is

The net torque about the axis of rotation by all the forces are

τ

Therefore x

nThe net torque can be rewritten

τ

What do we learn?

No matter where the rotation axis is, net effect of the torque is identical.

F Dn

MBgMFg MFg

1m x

x/2

Rotational axis

2/xgM B 0gMgMgM DFB

2/00.1 xgM F 2/xn 2/xgM D

2/xgM B 2/00.1 xgM F

2/xgMgMgM DFB 2/xgM D

xgMgM DF 00.1 0

mgMgM

D

F 00.1 mm 29.200.1350800


Ex. 9.8 for Mechanical EquilibriumA person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.

xF

Since the system is in equilibrium, from the translational equilibrium condition

From the rotational equilibrium condition

O

FB

FUmg

d

lτ

Thus, the force exerted by the biceps muscle is

dFB

Force exerted by the upper arm is UF

0 yF mgFF UB 0

lmgdFF BU 0 0lmg

BFd

lmg N583

00.30.350.50

mgFB N5330.50583


Elastic Properties of Solids

Elastic Modulus

We have been assuming that the objects do not change their shapes when external forces are exerting on it. It this realistic?

No. In reality, the objects get deformed as external forces act on it, though the internal forces resist the deformation as it takes place.

Deformation of solids can be understood in terms of Stress and Strain

Stress: A quantity proportional to the force causing the deformation.Strain: Measure of the degree of deformation

It is empirically known that for small stresses, strain is proportional to stress

The constants of proportionality are called Elastic Modulus

Three types of Elastic Modulus

1. Young’s modulus: Measure of the elasticity in a length2. Shear modulus: Measure of the elasticity in an area3. Bulk modulus: Measure of the elasticity in a volume

stressstrain

Monday, Nov. 30, 2009 PHYS 1441-002, Fall 2009 Dr. Jaehoon Yu

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Example 9 – 7 A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not), determine the forces exerted on the ladder by the ground and the wall.

xFFBD

First the translational equilibrium, using components

Thus, the y component of the force by the ground is

mg

FW

FGx

FGyO

GyF

Gx WF F 0

yF Gymg F 0

mg 12.0 9.8 118N N The length x0 is, from Pythagorian theorem

2 20 5.0 4.0 3.0x m


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Example 9 – 7 cont’d

OτFrom the rotational equilibrium 0 2 4.0Wmg x F 0Thus the force exerted on the ladder by the wall is

WF

Thus the force exerted on the ladder by the ground is

The x component of the force by the ground is

44Gx WF F N

GF

0 24.0

mg x 118 1.5 44

4.0N

0x Gx WF F F Solve for FGx

2 2Gx GyF F 2 244 118 130N

The angle between the ground force to the floor

1tan Gy

Gx

FF

1 118tan 70

44

o


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A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board.

Ex. A Diving Board

τ

2

530 N 3.90 m1480 N

1.40 mF

2F

yF 1 1480 N 530 N 0F

1 950 NF

First the torque eq. 2 2F WW 0So the force by the fulcrum is

How large is the torque by the bolt?None Why?

2

WW

Now the force eq. 1F 2F W 0

So the force by the bolt is

Because the lever arm is 0.


Finger Holds Water in StrawYou insert a straw of length L into a tall glass of your favorite beverage. You place your finger over the top of the straw so that no air can get in or out, and then lift the straw from the liquid. You find that the straw strains the liquid such that the distance from the bottom of your finger to the top of the liquid is h. Does the air in the space between your finger and the top of the liquid have a pressure P that is (a) greater than, (b) equal to, or (c) less than, the atmospheric pressure PA outside the straw?

What are the forces in this problem?

Gravitational force on the mass of the liquid

mg

Force exerted on the top surface of the liquid by inside air pressure

gF mg A L h gr

inF inp A

pinA

Force exerted on the bottom surface of the liquid by the outside air outF Ap A

Since it is at equilibrium

pAA

out g inF F F 0 0A inp A g L h A p Ar

in Ap p g L hr Cancel A and solve for pin

Less

So pin is less than PA by rg(L-h).


15

Young’s Modulus

AFexStress Tensile

Let’s consider a long bar with cross sectional area A and initial length Li.

Fex=Fin

Young’s Modulus is defined as

What is the unit of Young’s Modulus?

Experimental Observations

1. For a fixed external force, the change in length is proportional to the original length

2. The necessary force to produce the given strain is proportional to the cross sectional area

Li

A:cross sectional areaTensile stress

Lf=Li+LFex After the stretch FexFin

Tensile strainiLL

Strain Tensile

YForce per unit area

Used to characterize a rod or wire stressed under tension or compression

Elastic limit: Maximum stress that can be applied to the substance before it becomes permanently deformed

Strain TensileStress Tensile

i

ex

LL

AF


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Strain VolumeStress VolumeB

iVV

AF

iVV

P

Bulk Modulus

AF

applies force theArea Surface

Force NormalPressure

Bulk Modulus characterizes the response of a substance to uniform squeezing or reduction of pressure.

Bulk Modulus is defined as

Volume stress =pressure

After the pressure change

If the pressure on an object changes by P=F/A, the object will undergo a volume change V.

V V’F

FF

FCompressibility is the reciprocal of Bulk Modulus

Because the change of volume is reverse to change of pressure.


17

Example for Solid’s Elastic PropertyA solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is submerged?

The pressure change P is

Since bulk modulus is

iVV

P

B

The amount of volume change isB

iPVV

From table 12.1, bulk modulus of brass is 6.1x1010 N/m2

757 100.2100.1100.2 if PPP

Therefore the resulting volume change V is

3410

7

106.1106.1

5.0100.2 mVVV if

The volume has decreased.


18

Density and Specific GravityDensity, r (rho), of an object is defined as mass per unit volume

r Unit? Dimension?

3/ mkg][ 3ML

Specific Gravity of a substance is defined as the ratio of the density of the substance to that of water at 4.0 oC (rH2O=1.00g/cm3).

SG Unit? Dimension?

None None

What do you think would happen of a substance in the water dependent on SG?

1SG1SG

Sink in the waterFloat on the surface

2

substance

H O

rr

M V


19

Fluid and PressureWhat are the three states of matter? Solid, Liquid and Gas

Fluid cannot exert shearing or tensile stress. Thus, the only force the fluid exerts on an object immersed in it is the force perpendicular to the surface of the object.

AFP

How do you distinguish them? Using the time it takes for a particular substance to change its shape in reaction to external forces.

What is a fluid? A collection of molecules that are randomly arranged and loosely bound by forces between them or by an external container.

We will first learn about mechanics of fluid at rest, fluid statics. In what ways do you think fluid exerts stress on the object submerged in it?

This force by the fluid on an object usually is expressed in the form of the force per unit area at the given depth, the pressure, defined as

Note that pressure is a scalar quantity because it’s the magnitude of the force on a surface area A.

What is the unit and the dimension of pressure?

Expression of pressure for an infinitesimal area dA by the force dF is

dFPdA

Unit:N/m2

Dim.: [M][L-1][T-2]Special SI unit for pressure is Pascal

2/11 mNPa


20

Example for PressureThe mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress.

The volume density of water at the normal condition (0oC and 1 atm) is 1000kg/m3. So the total mass of the water in the mattress is

Since the surface area of the mattress is 4.00 m2, the pressure exerted on the floor is

m

P

Therefore the weight of the water in the mattress is W

b) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor.

MWVr kg31020.1300.000.200.21000

mg N43 1018.18.91020.1

AF

A

mg 3

4

1095.200.4

1018.1


Buoyant Forces and Archimedes’ PrincipleWhy is it so hard to put an inflated beach ball under water while a small piece of steel sinks in the water easily?

The water exerts force on an object immersed in the water. This force is called the buoyant force.

How large is the buoyant force?

Let‘s consider a cube whose height is h and is filled with fluid and in its equilibrium so that its weight Mg is balanced by the buoyant force B.

This is called Archimedes’ principle. What does this mean?

The magnitude of the buoyant force always equals the weight of the fluid in the volume displaced by the submerged object.

B

BMgh

The pressure at the bottom of the cube is larger than the top by rgh.

PTherefore,

Where Mg is the weight of the fluid in the cube.

gF Mg

AB / ghr

B PA ghAr Vgr

B gFVgr Mg


More Archimedes’ PrincipleLet’s consider the buoyant force in two special cases.

Let’s consider an object of mass M, with density r0, is fully immersed in the fluid with density rf .

Case 1: Totally submerged object

The total force applies to different directions depending on the difference of the density between the object and the fluid.

1. If the density of the object is smaller than the density of the fluid, the buoyant force will push the object up to the surface.

2. If the density of the object is larger than the fluid’s, the object will sink to the bottom of the fluid.

What does this tell you?

The magnitude of the buoyant force is

BMgh

B

The weight of the object is gF

Therefore total force in the system is F

Vgfr

Mg Vg0r

gFB Vgf 0rr


More Archimedes’ PrincipleLet’s consider an object of mass M, with density r0, is in static equilibrium floating on the surface of the fluid with density rf , and the volume submerged in the fluid is Vf.

Case 2: Floating object

Since the object is floating, its density is smaller than that of the fluid.

The ratio of the densities between the fluid and the object determines the submerged volume under the surface.

What does this tell you?

The magnitude of the buoyant force isBMgh

B

The weight of the object is gF

Therefore total force of the system is F

Since the system is in static equilibrium gV ffr

gV ffr

Mg gV00r

gFB gVgV ff 00rr

gV00r

frr0

0VV f

0


Example for Archimedes’ PrincipleArchimedes was asked to determine the purity of the gold used in the crown. The legend says that he solved this problem by weighing the crown in air and in water. Suppose the scale read 7.84N in air and 6.86N in water. What should he have to tell the king about the purity of the gold in the crown?

In the air the tension exerted by the scale on the object is the weight of the crown airTIn the water the tension exerted by the scale on the object is waterT

Therefore the buoyant force B is B

Since the buoyant force B is BThe volume of the displaced water by the crown is cV

Therefore the density of the crown is cr

Since the density of pure gold is 19.3x103kg/m3, this crown is not made of pure gold.

mg N84.7

Bmg N86.6

waterair TT N98.0

gVwwr gVcwr N98.0

wVgN

wr98.0

34100.18.91000

98.0 m

c

c

Vm

gVgm

c

cgVc

84.7 33

4 /103.88.9100.1

84.7 mkg


Example for Buoyant ForceWhat fraction of an iceberg is submerged in the sea water?

Let’s assume that the total volume of the iceberg is Vi. Then the weight of the iceberg Fgi is

Since the whole system is at its static equilibrium, we obtain

giF

Let’s then assume that the volume of the iceberg submerged in the sea water is Vw. The buoyant force B caused by the displaced water becomes

B

gViirTherefore the fraction of the volume of the iceberg submerged under the surface of the sea water is i

w

VV

About 90% of the entire iceberg is submerged in the water!!!

gViir

gVwwr

gVwwr

w

i

rr

890.0/1030/917

3

3

mkgmkg


Flow Rate and the Equation of ContinuityStudy of fluid in motion: Fluid Dynamics

If the fluid is water: • Streamline or Laminar flow: Each particle of the fluid

follows a smooth path, a streamline• Turbulent flow: Erratic, small, whirlpool-like circles called

eddy current or eddies which absorbs a lot of energy

Two main types of flow

Water dynamics?? Hydro-dynamics

Flow rate: the mass of fluid that passes the given point per unit time /m t

since the total flow must be conserved

1mt

1 1V

tr

1 1 1A lt

r

1 1 1A vr

1 1 1A vr 1mt

Equation of Continuity

2mt

2 2 2A vr


Example for Equation of ContinuityHow large must a heating duct be if air moving at 3.0m/s through it can replenish the air in a room of 300m3 volume every 15 minutes? Assume the air’s density remains constant.

Using equation of continuity

1 1 1A vr

Since the air density is constant

1 1A v Now let’s imagine the room as the large section of the duct

1A 2 2

1

/A l tv

2

1

Vv t

2300 0.113.0 900

m

2 2 2A vr

2 2A v

2 2

1

A vv


Bernoulli’s PrincipleBernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high.

Amount of the work done by the force, F1, that exerts pressure, P1, at point 1

1W

Work done by the gravitational force to move the fluid mass, m, from y1 to y2 is

1 1F l 1 1 1P A lAmount of the work done by the force in the other section of the fluid is

2W

3W 2 1mg y y

2 2 2P A l


Bernoulli’s Equation cont’dThe total amount of the work done on the fluid is

W 1 1 1P A l From the work-energy principle

22

12

mv

Since the mass m is contained in the volume that flowed in the motion

1 1A l and m

2 2221

2A l vr

1 1 1P A l

1 11 lP A

Thus,

2 2 2P A l 2 1mgy mgy 1W 2W 3W

21

12

mv 2 2 2P A l 2 1mgy mgy

2 2A l 1 1A lr 2 2A lr

1211

12

vA lr

2 22 lP A 2 2 1 12 1gyA l A l gyr r


Bernoulli’s Equation cont’d

We obtain

2 2 1 1 1 1 2 2 2 2 12 22 1 1 2 2 11

1 12 2

A l A l A l A l A lv v P P g A ly gyr r r r

Re-organize

21 1 1

12

P v gyr r Bernoulli’s Equation

21 1 1

12

P v gyr r

22 2 2

12

P v gyr r

2 22 1 1 2 2 1

1 12 2

v v P P gy gyr r r r

Since

Thus, for any two points in the flow

For static fluid 2P

For the same heights 2 22 1 1 2

12

P P v vr

The pressure at the faster section of the fluid is smaller than slower section.

Pascal’s Law

.const

1 1 2P g y yr 1P ghr

Result of Energy conservation!


Example for Bernoulli’s EquationWater circulates throughout a house in a hot-water heating system. If the water is pumped at the speed of 0.5m/s through a 4.0cm diameter pipe in the basement under a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter pipe on the second 5.0m above? Assume the pipes do not divide into branches.

Using the equation of continuity, flow speed on the second floor is

2v

Using Bernoulli’s equation, the pressure in the pipe on the second floor is

2P

5 3 2 2 313.0 10 1 10 0.5 1.2 1 10 9.8 52

5 22.5 10 /N m

1 1

2

A vA

2

1 12

2

r vr

20.0200.5 1.2 /

0.013m s

1P 2 21 2

12

v vr 1 2g y yr

phys 1441 – section 002 lecture # 23

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