Static Surface ForcesStatic Surface Forces · 2012-02-06 · Static Surface ForcesStatic Surface Forces hinge 8m8 m water? 4 m Static Surface ForcesStatic Surface Forces Forces on

Static Surface ForcesStatic Surface ForcesStatic Surface ForcesStatic Surface Forces

hingehinge

8 m8 m waterwater8 m8 m

?? 4 m4 m

Static Surface ForcesStatic Surface ForcesStatic Surface ForcesStatic Surface Forces

� Forces on plane areas� Forces on plane areas� Forces on curved surfaces� Buoyant force� Forces on curved surfaces� Buoyant force� Buoyant force� Stability of floating and submerged bodies� Buoyant force� Stability of floating and submerged bodies

Forces on Plane AreasForces on Plane AreasForces on Plane AreasForces on Plane Areas

� Two types of problems� Horizontal surfaces (pressure is )

� Two types of problems� Horizontal surfaces (pressure is )� Horizontal surfaces (pressure is _______)� Inclined surfaces

� T k

� Horizontal surfaces (pressure is _______)� Inclined surfaces

� T k

constantconstant

dzdp

� Two unknowns� ____________

� Two unknowns� ____________Total forceTotal force

dz

� ____________� Two techniques to find the line of action of

� ____________� Two techniques to find the line of action of

Line of actionLine of actionq

the resultant force� Moments

qthe resultant force� Moments� Moments� Pressure prism� Moments� Pressure prism

Forces on Plane Areas: Forces on Plane Areas: Horizontal surfacesHorizontal surfaces

Sid ih

What is the force on the bottom of this tank of water?

Side viewpAdAppdAFR p = h

h

hhAFR

weight of overlying fluid!weight of overlying fluid!FR =

h = __________________________Vertical distance to free surface

AF is normal to the surface and towards th f if i iti

g y gg y gR

the surface if p is positive.

F passes through the of the area.centroidTop view

p sses oug e ________ o e e .centroid

Forces on Plane Areas: Inclined Forces on Plane Areas: Inclined SurfacesSurfaces

� Direction of force� Direction of force Normal to the plane

� Magnitude of force� integrate the pressure over the area

� Magnitude of force� integrate the pressure over the area� integrate the pressure over the area� pressure is no longer constant!� integrate the pressure over the area� pressure is no longer constant!

� Line of action� Moment of the resultant force must equal the

� Line of action� Moment of the resultant force must equal the o e o e esu o ce us equ e

moment of the distributed pressure forceo e o e esu o ce us equ e

moment of the distributed pressure force

HW#2HW#2HW#2HW#2

� 1.89, 2.2, 2.6, 2.27, 2.29, 2.30, 2.38, 2.41, � 1.89, 2.2, 2.6, 2.27, 2.29, 2.30, 2.38, 2.41, 2.44, 2.472.44, 2.47

Forces on Plane Areas: Inclined Forces on Plane Areas: Inclined SurfacesSurfaces

OO

xxFree surfaceFree surface

A’cx

RxAhF cR ch

B’B’ OO

R

centroidBB OO

The origin of the y

yycy

Rycenter of pressurecenter of pressure axis is on the free surface

Magnitude of Force on Inclined Magnitude of Force on Inclined gPlane Area

gPlane Area

pdAFR sinyhp

ydAFR sin Ac ydA

Ay 1

yy AA

sincR AyF

yy

AhFR hc is the vertical distance between free AhF cR

ApFR p is the pressure at the centroid of the area

csurface and centroid

ApF cR pc is the pressure at the __________________centroid of the area

First MomentsFirst MomentsFirst MomentsFirst Moments

xdA Moment of an area A about the y axis

1

A

Location of centroidal axisAc xdA

Ax 1

1

Location of centroidal axis

1c A

y ydAA

h31

For a plate of uniform thickness the intersection of the centroidal axes is also the center of gravity

Second MomentsSecond MomentsSecond MomentsSecond Moments

Al ll d f thmoment of inertiaAlso called _______________ of the area

dAyI 2

moment of inertia

Ax dAyI

2Ixc is the 2nd moment with respect to an

2cxcx AyII axis passing through its centroid and

parallel to the x axis.P ll l i thParallel axis theorem

Product of InertiaProduct of InertiaProduct of InertiaProduct of Inertia

� A measure of the asymmetry of the area� A measure of the asymmetry of the area

IAI

Axy xydAI Product of inertia

xycccxy IAyxI

yIxyc = 0

Ixyc = 0

y y

If x = xc or y = yc is an axis of symmetry then the product of i ti I i

x x

inertia Ixyc is zero.

Properties of AreasProperties of AreasProperties of AreasProperties of Areas

ba A ab

ay 3baI 0I

ycaIxc

A ab2cy

12xcI 0xycI

a

yaIxc 2

abA b d

3

36xcbaI

2

272xycbaI b d 3c

ay

yc

b

xc 23c

b dx 36 72

d2A R

4

4xcRI

Ryc

Ixc0xycI

d

cy R4yc

Properties of AreasProperties of AreasProperties of AreasProperties of Areas

2RA

4Ry 4RI

yc

RIxc

0I

b

2A

3cy

8xcI R 0xycI

3

4xcbaI

A abayc

b

Ixc0xycI cy a

4yc

43c

Ry

4

16xcRI

2

4RA

Ryc

3 164

Forces on Plane Areas: Forces on Plane Areas: Center of Pressure: xRCenter of Pressure: xR

� The center of pressure is not at the centroid (because pressure is increasing with depth)

� The center of pressure is not at the centroid (because pressure is increasing with depth)(because pressure is increasing with depth)� x coordinate of center of pressure: xR

(because pressure is increasing with depth)� x coordinate of center of pressure: xR

ARR xpdAFx1

Moment of resultant force = sum of Moment of resultant force = sum of moment of distributed forcesmoment of distributed forces

A

RR xpdA

Fx 1

1

sinAyF cR sinyp

A

cR dAxy

Ayx

sin

sin1

1

Ac

R xydAAy

x 1

Center of Pressure: xCenter of Pressure: xCenter of Pressure: xRCenter of Pressure: xR

R xydAA

x 1xy xydAI Product of inertia

I xy IAyxI

Ac

R yAy Axy y

Parallel axis theoremAy

xc

xyR xycccxy IAyxI

IA

Parallel axis theorem

AyIAyx

xc

xycccR

y

cc

xycR x

AyI

x

y

xcy x

Center of Pressure: yCenter of Pressure: yCenter of Pressure: yRCenter of Pressure: yR

ARR ypdAFy

1

Sum of the moments

A

RR ypdA

Fy 1

sinAyF cR sinyp

1

Ac

R dAyAy

y

sinsin

1 2

1

Ac

R dAyAy

y 21 Ax dAyI 2

AyIyc

xR AyII cxcx

2 Parallel axis theorem2

xc cR

c

I y Ayy A

xcR c

c

Iy yy A

Inclined Surface FindingsInclined Surface FindingsInclined Surface FindingsInclined Surface Findings0

� The horizontal center of pressure and the h i l id h h f

� The horizontal center of pressure and the h i l id h h f c

xycR x

AI

x i id

0

horizontal centroid ________ when the surface has either a horizontal or vertical axis of horizontal centroid ________ when the surface has either a horizontal or vertical axis of

cc

R Aycoincide

symmetry� The center of pressure is always _______ the

symmetry� The center of pressure is always _______ the xc

R yIy below>0

centroid� The vertical distance between the centroid and

centroid� The vertical distance between the centroid and

cc

R yAy

y

the center of pressure _________ as the surface is lowered deeper into the liquidthe center of pressure _________ as the surface is lowered deeper into the liquid

decreases(yc increases)p q

� What do you do if there isn’t a free surface?p q

� What do you do if there isn’t a free surface?(yc )

Example using MomentsExample using MomentsExample using MomentsExample using Moments

An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force F applied at the bottom of the gate is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the t h th th id ? N l t th i ht f th tatmosphere on the other side? Neglect the weight of the gate.

Solution Schemehinge

water8 mMagnitude of the force applied by the waterMagnitude of the force applied by the water

��

F 4 m

pp ypp y

�� Location of the resultant forceLocation of the resultant force

�� Find F using moments about hingeFind F using moments about hinge

Magnitude of the ForceMagnitude of the ForceMagnitude of the ForceMagnitude of the Force

ApF cR hingehinge

waterwater8 m

Depth to the centroid

abA

h = FF 4 mFRFR

10 m Depth to the centroid

pc = ___

hc = _____

ch

10 m

abhF cR N

a = 2.5 m

c

m 2m 2.5πm 10mN 9800 3

RF

F 1 54 MNb = 2 m

FR= ________1.54 MN

Location of Resultant ForceLocation of Resultant ForceLocation of Resultant ForceLocation of Resultant Force

cxc

R yAy

Iy hingehinge

waterwater8 m

cc hy c Ay

FF 4 mFrFr

Slant distance Slant distance

3baI ba 3

bA

________cy 12.5 m12.5 m to surfaceto surface

4xcI aby

yyc

cR 4

a2 m2 5 2

abA a = 2.5 m

ccR y

ayy4

m 12.54

m 2.5 cR yy cp

_______ cR yy 0.125 m0.125 m __Rx cxb = 2 m

Force Required to Open GateForce Required to Open GateForce Required to Open GateForce Required to Open Gate

How do we find the How do we find the hingehingewaterwater8 m

required force?required force?FF 4 m

FrFr

Moments about the hinge 0hingeM

g=Fltot - FRlcp=Fltot - FRlcp

2.5 mlcp=2.625 mtot

cpR

llF

F l

m5

m 2.625N 10 x 1.54 6

F

ltotcpcp

F = ______ b = 2 m

m 5

809 kN809 kN

Forces on Plane Surfaces ReviewForces on Plane Surfaces ReviewForces on Plane Surfaces ReviewForces on Plane Surfaces Review

� The average magnitude of the pressure force � The average magnitude of the pressure force is the pressure at the centroid

� The horizontal location of the pressure forceis the pressure at the centroid

� The horizontal location of the pressure force� The horizontal location of the pressure force was at xc (WHY?) ____________________

� The horizontal location of the pressure force was at xc (WHY?) ____________________ The gate was symmetrical about at least one of the centroidal axes

The gate was symmetrical about at least one of the centroidal axes___________________________________

� The vertical location of the pressure force is ___________________________________

� The vertical location of the pressure force is about at least one of the centroidal axes.about at least one of the centroidal axes.

below the centroid. (WHY?) ___________ below the centroid. (WHY?) ___________ Pressure increases with depth.______________________________________increases with depth.

Forces on Plane Areas: Pressure Forces on Plane Areas: Pressure PrismPrism

� A simpler approach that works well for � A simpler approach that works well for areas of constant width (_________)

� If the location of the resultant force isareas of constant width (_________)

� If the location of the resultant force isrectangles

� If the location of the resultant force is required and the area doesn’t intersect the free s rface then the moment of inertia

� If the location of the resultant force is required and the area doesn’t intersect the free s rface then the moment of inertiafree surface, then the moment of inertia method is about as easyfree surface, then the moment of inertia method is about as easy

Forces on Plane Areas: Pressure Forces on Plane Areas: Pressure PrismPrism

h O

h

Free surfaceh1

1h2h AhF

AF

Force = VolumeForce Volume of pressure prism

1Center of pressure hp

A

RR xpdA

Fx 1

xdxR1

ydyR1 is at centroid of

pressure prism

Example 1: Pressure PrismExample 1: Pressure PrismExample 1: Pressure PrismExample 1: Pressure Prism

m

h/cos

Damh=

10 m

FR

24º

h

F VolumeF VolumeFR = VolumeFR = VolumeFR = (h/cos)(h)(w)/2FR = (h/cos)(h)(w)/2(10 /0 913 )(9800 / 3*10 )( 0 )/2(10 /0 913 )(9800 / 3*10 )( 0 )/2

hFR = (10 m/0.9135)(9800 N/m3*10 m)(50 m)/2FR = (10 m/0.9135)(9800 N/m3*10 m)(50 m)/2FR = 26 MNFR = 26 MN

Example 2: Pressure PrismExample 2: Pressure PrismExample 2: Pressure PrismExample 2: Pressure Prism

OO

water8 m

hinge

4 m x 5 m (rectangular conduit)2

m

m

4 m( g )

5 m

12

8

m

5 m yy

Solution 2: Pressure PrismSolution 2: Pressure PrismSolution 2: Pressure PrismSolution 2: Pressure Prism

h1Magnitude of force

221 hh

5 m

h2m) m)(4 m)(5 )(10N/m 0098( 3RF

FR = 1.96 MN1.96 MN 5 mLocation of resultant force a

R ________measured from hingemeasured from hinge

2 hawahawa

21

2 hhway

232

2321 hawahawaFy RR

36m 10aw

yR

m12m8m5FRFR

3m 12

6m 8

m 10m 5

Ry yR = _______2.667 m2.667 m

Forces on Curved SurfacesForces on Curved SurfacesForces on Curved SurfacesForces on Curved Surfaces

� Horizontal component� Horizontal component� Vertical component� Vertical component

Forces on Curved Surfaces: Forces on Curved Surfaces: Horizontal ComponentHorizontal Component

� What is the horizontal component of � What is the horizontal component of pressure force on a curved surface equal to?pressure force on a curved surface equal to?

� The center of pressure is located using the moment of inertia or press re prism

� The center of pressure is located using the moment of inertia or press re prismthe moment of inertia or pressure prism technique.the moment of inertia or pressure prism technique.

� The horizontal component of pressure force on a closed body is .

� The horizontal component of pressure force on a closed body is .zerozeroforce on a closed body is _____.force on a closed body is _____.zerozero

Forces on Curved Surfaces: Forces on Curved Surfaces: Vertical ComponentVertical Component

� What is the magnitude of the vertical component of force on

� What is the magnitude of the vertical component of force onvertical component of force on the cup?vertical component of force on the cup?

hF = pA

r

p = hp = h

F = hr2 =W! rF = hr2 =W!

What if the cup had sloping sides?What if the cup had sloping sides?

Forces on Curved Surfaces: Forces on Curved Surfaces: Vertical ComponentVertical Component

The vertical component of pressure force on a The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and curved surface is equal to the weight of liquid vertically above the curved surface and y fextending up to the (virtual or real) free surface

y fextending up to the (virtual or real) free surfacesurface.

Streeter, et. alsurface.

Streeter, et. al

Example: Forces on Curved Example: Forces on Curved pSurfaces

pSurfaces

Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.on a 1 m wide section of the circular arc.

FV = 3 m W1W1 + W2W1 + W2

water= (3 m)(2 m)(1 m) +1/4(2 m)2(1 m)= (3 m)(2 m)(1 m) +1/4(2 m)2(1 m)

V

2 m

3 m 11 2

= 58.9 kN + 30.8 kN= 58.9 kN + 30.8 kN2 m W2

58.9 kN + 30.8 kN58.9 kN + 30.8 kN= 89.7 kN= 89.7 kN

FH = cp A

= (4 m)(2 m)(1 m)= (4 m)(2 m)(1 m)

xcc hp

= (4 m)(2 m)(1 m)= (4 m)(2 m)(1 m)= 78.5 kN= 78.5 kN y

Example: Forces on Curved Example: Forces on Curved pSurfaces

pSurfaces

The vertical component line of action goes through the centroid of the volume of water above the surface. A

Take moments about a vertical axis through ATake moments about a vertical axis through A 3 m W14Raxis through A.axis through A.

water 2 m

3 m 143

R

21 3)m 2(4)m 1( WWFx VR

2 m W2

3

)kN 8.30(3

)m 2(4)kN 9.58)(m 1(

0 948 ( d f A) ith it d f 89 7 kN0 948 ( d f A) ith it d f 89 7 kN

)kN 7.89(3Rx

= 0.948 m (measured from A) with magnitude of 89.7 kN= 0.948 m (measured from A) with magnitude of 89.7 kN

Example: Forces on Curved Example: Forces on Curved pSurfaces

pSurfaces

AThe location of the line of action of the horizontal component is given by

3 m W1

p g y

xcR c

Iy yy A

b

water 2 m

3 m 1cy A3

12xcbaI

b

a

2 m W212xc

xcI (1 m)(2 m)3/12 = 0.667 m4(1 m)(2 m)3/12 = 0.667 m4

cy 40 667 m

x4 m4 m

0.667 m 4 m 4.083 m4 m 2 m 1 mRy

y

Example: Forces on Curved Example: Forces on Curved pSurfaces

pSurfaces

m 78.5 kN78.5 kN4.083 m

0.94

8 m horizontalhorizontal

89.7 kN89.7 kN

0

verticalvertical

119.2 kN119.2 kN resultantresultant

Cylindrical Surface Force CheckCylindrical Surface Force CheckCylindrical Surface Force CheckCylindrical Surface Force Check

C0.948 m 89.7kN

� All pressure forces pass h h i C

� All pressure forces pass h h i CC

1 083

through point C. � The pressure force

li b

through point C. � The pressure force

li b1.083 m applies no moment about point C.

� Th lt t t

applies no moment about point C.

� Th lt t t78.5kN

� The resultant must pass through point C.

� The resultant must pass through point C.

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 00

Curved Surface TrickCurved Surface TrickCurved Surface TrickCurved Surface Trick

� Find force F required to open � Find force F required to open Athe gate.

� All the horizontal force isthe gate.

� All the horizontal force is3 m W1

� All the horizontal force is carried by the hinge

h f d f

� All the horizontal force is carried by the hinge

h f d f water 2 m

W2FFOO� The pressure forces and force F

pass through O. Thus the hinge � The pressure forces and force F

pass through O. Thus the hinge 2

force must pass through O!� Hinge carries only horizontal

force must pass through O!� Hinge carries only horizontal

W1 + W2W1 + W2

� Hinge carries only horizontal forces! (F = ________)

� Hinge carries only horizontal forces! (F = ________)

Curved Surface TrickCurved Surface TrickCurved Surface TrickCurved Surface Trick

Traditional way: 0.948 m 89.7kN

)083.1)(5.78()948.02)(7.89(

C:0 hingeM

1.083 m)2(

)083.1)(5.78()948.02)(7.89(F

(k )

78 5kN

(kN) 7.89F

Same as F = W1+W2 78.5kNF

Same as Fv W1+W2

Solution SchemeSolution SchemeSolution SchemeSolution Scheme

� Determine pressure datum and location in fl id h i ( i i )

� Determine pressure datum and location in fl id h i ( i i )fluid where pressure is zero (y origin)

� Determine total acceleration vector (a) fluid where pressure is zero (y origin)

� Determine total acceleration vector (a) ( )including acceleration of gravity

� Define h tangent to acceleration vector (call

( )including acceleration of gravity

� Define h tangent to acceleration vector (call� Define h tangent to acceleration vector (call this vertical!)

� Determine if surface is normal to a

� Define h tangent to acceleration vector (call this vertical!)

� Determine if surface is normal to a� Determine if surface is normal to a, inclined, or curved

� Determine if surface is normal to a, inclined, or curved

Static Surface Forces SummaryStatic Surface Forces SummaryStatic Surface Forces SummaryStatic Surface Forces Summary

� Forces caused by gravity (or ) b d f

� Forces caused by gravity (or ) b d fl l i_______________) on submerged surfaces

� horizontal surfaces (normal to total _______________) on submerged surfaces� horizontal surfaces (normal to total

total acceleration

acceleration)� inclined surfaces (y coordinate has origin at

f f )

acceleration)� inclined surfaces (y coordinate has origin at

f f )

RF hA Location where p = pref

free surface)� curved surfaces

free surface)� curved surfaces c

c

xcR y

AyIy AhF cR

� Horizontal component� Vertical component (________________________)� Horizontal component� Vertical component (________________________)

AhF cR weight of fluid above surface

� Virtual surfaces…� Virtual surfaces…

Buoyant ForceBuoyant ForceBuoyant ForceBuoyant Force

� The resultant force exerted on a body by a � The resultant force exerted on a body by a static fluid in which it is fully or partially submergedstatic fluid in which it is fully or partially submergedg� The projection of the body on a vertical plane is

always

g� The projection of the body on a vertical plane is

always zero

different

always ____.� The vertical components of pressure on the top

and bottom surfaces are

always ____.� The vertical components of pressure on the top

and bottom surfaces are

zero

differentand bottom surfaces are _________and bottom surfaces are _________

Buoyant Force: Thought Buoyant Force: Thought y gExperiment

y gExperiment

Place a thin wall balloon filled with water in a tank of water.Place a thin wall balloon filled with water in a tank of water.

FB

zeroWhat is the net force on the balloon?What is the net force on the balloon? zero

no

balloon? _______Does the shape of the balloon matter?

balloon? _______Does the shape of the balloon matter? no

W i h f

matter? ________What is the buoyant force on th b ll ?

matter? ________What is the buoyant force on th b ll ? Weight of water displaced

F V

the balloon? _____________ _________the balloon? _____________ _________

FB=VWhere is the line of action of the buoyant force? __________Where is the line of action of the buoyant force? __________Thru centroid of balloon

Buoyant Force: Line of ActionBuoyant Force: Line of ActionBuoyant Force: Line of ActionBuoyant Force: Line of Action

� The buoyant force acts through the centroid f th di l d l f fl id ( t f

� The buoyant force acts through the centroid f th di l d l f fl id ( t fof the displaced volume of fluid (center of

buoyancy)of the displaced volume of fluid (center of buoyancy) xdV Vx 1x xdV c

V

xdV Vx cV

x xdVV

= volume= volumed = distributed forced = distributed force

xc = centroid of volumexc = centroid of volume

Buoyant Force: ApplicationsBuoyant Force: ApplicationsBuoyant Force: ApplicationsBuoyant Force: ApplicationsF1F1 F2F2

� Using buoyancy it is � Using buoyancy it is 1 > 21 > 2

WW11

WW22

possible to determine:possible to determine: WW WW

WeightWeightVolumeVolume

� _______ of an object� of an object� _______ of an object� of an objectVolumeVolume

Specific gravitySpecific gravity Force balance� _______ of an object� _______________ of

bj

� _______ of an object� _______________ of

bj1 1F V W 2 2F V W an objectan object

(With F F and given)(With F1, F2 , 1, and 2 given)

Buoyant Force: ApplicationsBuoyant Force: ApplicationsBuoyant Force: ApplicationsBuoyant Force: Applications

1 1F V W 2 2F V W (force balance)Equate weightsEquate weights Equate volumes

1 1 2 2F V F V 1 2W F W FV

Equate weightsEquate weights Equate volumes

1 2 2 1

2 1

V F FF FV

1 2

2 1 2 1 2 1W F W F

2 1

1 2

V

1 2 2 1

2 1

F FW

1 2F FV

Suppose the specific weight of the first fluid is zero1 2

2

V

1FW

Buoyant Force (Just for fun)Buoyant Force (Just for fun)Buoyant Force (Just for fun)Buoyant Force (Just for fun)

A sailboat is sailing on Lake Bryan. The A sailboat is sailing on Lake Bryan. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Lake Bryan increase or decrease?overboard to lighten the boat. Does the water level of Lake Bryan increase or decrease?level of Lake Bryan increase or decrease?Why?_______________________________ level of Lake Bryan increase or decrease?Why?_______________________________

----------- ________----------- ________The anchor displaces less water when The anchor displaces less water when

____________________________________

____________________________________

it is lying on the bottom of the lake than it did when in the boat.it is lying on the bottom of the lake than it did when in the boat.________________________________________

Rotational Stability of Rotational Stability of ySubmerged Bodies

ySubmerged Bodies

� A completely � A completely submerged body is stable when its submerged body is stable when its

BB

G

center of gravity is the center

center of gravity is the centerbelowbelow B

G

G_____ the center of buoyancy_____ the center of buoyancybelowbelow

End of Chapter QuestionEnd of Chapter QuestionEnd of Chapter QuestionEnd of Chapter Question

� Write an equation for � Write an equation for the pressure acting on the bottom of a the pressure acting on the bottom of a

dconical tank of water.� Write an eq ation for

conical tank of water.� Write an eq ation for

d1

� Write an equation for the total force acting

� Write an equation for the total force acting Lon the bottom of the tank.on the bottom of the tank.

L

d2

End of ChapterEnd of ChapterEnd of ChapterEnd of Chapter

� What didn’t you understand so far about � What didn’t you understand so far about statics?

� Ask the person next to youstatics?

� Ask the person next to you� Ask the person next to you� Circle any questions that still need answers� Ask the person next to you� Circle any questions that still need answers

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