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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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