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HYDROSTATIC FORCE ON SUBMERGEDHYDROSTATIC FORCE ON SUBMERGED
SURFACESSURFACES
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HYDROSTATIC FORCE ON SUBMERGEDHYDROSTATIC FORCE ON SUBMERGED
SURFACESSURFACES
Need to determine: (1) Force magnitude
(2) Force direction(3) The location of the acting force
rr
'rr
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
ApdFdrr
=
Static Fluids NO Shear stress
Against the surface
= AR ApdF
rr
g
dh
dp= p = p0 +g h
Hence: p and A must be expressed in terms of (x,y)
FORCE MAGNITUDE (FORCE MAGNITUDE (FFRR))
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MOMENTS ABOUT O (0,0)MOMENTS ABOUT O (0,0)
== AAR ApdrFdrF'rr
r
r
r
r
r
rr
'rr
= AR ypdAF'y
=
AR xpdAF'x
Maka didapat:
THE LOCATION OF THE ACTING FORCE
dimana: ;'zk'yj'xi'r ++=r
;dAkAd =r
;FkFRR
=r
( ) ( ) ( ) +=+=+AA
R kpdAyjxiFdyjxiFk'yj'xi
r
maka:
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= AR
ypdAF
'y1
= AR
xpdAF
'x1
Ap
hdAApF
atm
A
atmR
+=
THE LOCATION OF THE ACTING FORCE
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EXAMPLE PROBLEM 3.5EXAMPLE PROBLEM 3.5
Resultant force,
GIVEN: Rectangular gate, hinged alongA, w = 5m
FIND:
SOLUTION:
, of the water on the gateRFr
Does the resultant force pass through the center
of gravity?
NO!
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= AR ApdF
rr
g
dh
dp=
Basic Eqs.:
== AAR pwdApdF rr
k wdAd =r
k
( ) +== L
AR wdsinDgApdF
0
030 rr
k
588=RFr
k kN {Force acts in negative z direction}
h
Fdr
From the diagram:
h = D + sin 300
p = g(D + sin 300) (p = gh)
Atmospheric
pressure is neglected
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Finding the line of action of the resultant force, .FRr
= AR pdAF'
( ) +=== L L
RRA
R
dsinDFgwpwd
FpdA
F'
0 0
03011
+=
+= 0
32
0
032
3032
3032
sinLDL
F
gwsin
D
F
gw'
R
L
R
= 2.22 m; and y = m.m.sin
m
sin
D ' 22622230
2
30 00 =+=+
m.FF
wpdA
F
wpdA
w
FxpdA
F'x R
RA
RA
RA
R
52222
11=====
j.i.'r 22652 +=r
m
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
==sss AAA
ydAdAyhdA sinsin
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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From solid mechanics, the location of the center of
gravity (centroid of the area) measured from the free
surface is:
=A
c ydAA
y1
AhAysinhdA ccA
==
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
ApF cR =
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
Center of pressure on an inclined plane surface:
AyI
Ay
dAy
'yc
xx
c
A==
2
Ay
Iy'y
c
xx
c+=
Ay
Ix'x
c
yx
c
+=
where:
y = y-location of acting force
x = x-location of acting force
yc = y-location of surface center of
gravity
xc = x-location of surface center of
gravity
= surface moment of inertiaaround an axis parallel to x-axis
passing through CG
= cross moment of inertiaaround axes passing through CG
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES
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Centroidal moments of inertia for various cross sections: (a) rectangle,
(b) circle, (c) triangle, and (d) semicircle.
BUOYANCY AND STABILITYBUOYANCY AND STABILITY
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Buoyancy is the net pressure force acting on a submerged
body.
The magnitude of the buoyant force is equal to the weight of
the fluid displaced by the body.
bbb VgVF ==
Buoyant force passes through the center of buoyancy, which
is at the centroid of the displaced fluid.
BUOYANCY AND STABILITYBUOYANCY AND STABILITY
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Stability of floating bodies
EXAMPLE PROBLEM 3.8EXAMPLE PROBLEM 3.8
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GIVEN: Crown volume, V = 18.9 in.3
FIND: The average material density of the crown
SOLUTION:
Net force on the string, Fnet = 4.7 lbf
netF
r
buoyancyF
r
+ gravityFr
=