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PROBLEM 9.66
Show that the resultant of the hydrostatic forces acting on a submerged
plane area
A
is a force
P
perpendicular to the area and of magnitude
sin
P
Ay
pA
γ
θ
==
, where
is the specific weight of the liquid and
p
is the pressure at the centroid
C
of the area. Show that
P
is applied at a
point
,
p
C
called the center of pressure, whose coordinates are
/
px
y
x
IA
y
=
and
/
yI
A
y
=
, where
xy
I
xy dA
∫
=
(see Sec. 9.8). Show also that the
difference of ordinates
p
yy
−
is equal to
2
/
x
ky
′
and thus depends upon
the depth at which the area is submerged.
SOLUTION
The pressure
p
at an arbitrary depth
( )
sin
y
is
( )
sin
py
=
so that the hydrostatic force
dP
exerted on an infinitesimal area
dA
is
( )
sin
dP
y
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
 Spring '08
 Jenkins
 Statics

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