1. §2.9 Pressure Prism Area of the plane is rectangular
Since p0=0 at surface p = p0+ρgh at bottom
FR= pCGA = (p0+ρghCG)A =ρg (h/2) A where p0=0
=ρg (h/2) b·h = 1/2 (ρgh)·b·h = (Volume of pressure prism)
The resultant force must pass through the centroid of the pressure prism (1/3 h above the base)
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2. = F1+F2= (ρgh1A) + [(1/2)ρg(h2-h1)A]
FR= (Volume of trapezoidal
pressure prism )
= (volume of pressure prism ABDE)
+(volume of pressure prism BCD)
= F1+F2= (ρgh1A) + [(1/2)ρg(h2-h1)A]
FRyR = F1y1+F2y (F1+F2) yR = F1[(h2-h1)/2]+F2 [(h2-h1)/3]
Only if the area of plane
is rectangular
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3. B A D C Example 2.8 Given：as figure SG=0.9(oil) square 0.6m * 0.6m
Pgage = 50Kpa = PS
Find: the magnitude and location of the
resultant force on the attached plate.
Solution：
Trapezoidal pressure prism, ABCDO
FR = (vol. of trapezoidal pressure
prism)
= (vol. of ABDO)+(Vol. of BCD)
= F1+F2 B A D C
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4. F1=(PS+ρgh1)A = {50. 103+[0. 9. (9. 81. 103N/m3). 2m]}. 6. 6 = 24. 4
F1=(PS+ρgh1)A = {50*103+[0.9*(9.81*103N/m3)*2m]}* * 0.6 = 24.4 * 103 N
F2=(Vol. of BCD)=[ ] A
= *(0.6m*0.6m)
=0.954*103N
FR=F1+F2=25.4*103N
By summing moments around an axis through point O,so that FRy0=F1*0.3m+F2*0.2m
(25.4*103N)y0=24.4*103* *103N*0.2m  y0 = 0.296m
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5. §2.10 Hydrostatic Force on a Curved Surface
FH = F2 = ρghCACAAC = PCG(AV)proj
FV = F1 + w = ρg (aboveAB) +ρg ( ABC)
FR =
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6. Example2.6 Given : Determine : The magnitude and line of FR Solution:
Length(b)
=1m
Determine : The magnitude and line of FR
Solution:
FH = F2 = ρghCACAAC = 62.4
FV=W=ρg(ABC)=
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7. §2.11 Buoyancy, Flotation, and Stability
  § Archimedes’ Principle
  ─Buoyant force : when a body is completely submerged in a
fluid, or floating so that it is only partially submerged, the
resultant fluid force acting on the body is called the
buoyant force.
─Buoyant force direction: upward vertical force
because pressure increases with depth and pressure forces
acting from below are larger than the pressure force acting
from above.
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8. For unform density of body immersed  point B = Center of mass
─The first law (immersed body)
 In equilibrium
 dFB= dFup –dFdown
=ρgy2dA–ρgy1dA
=ρg(y2-y1)dA = ρghdA = γhdA
= weight of fluid equilibrium to body volume
 For unform density of body immersed  point B = Center of mass
For variable density of body immersed  point B≠ Center of mass
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9. ─The second law (floating body) dFB=dFup-dFdown
=[γ1h+ γ2(y2-h)]dA- γ 1y1dA
= γ1(h-y1)dA+ γ2(y2-h)dA
= γ1dV1+ γ 2dV2
For a layered fluid
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10. Example 2.10 As the figures on right (ρg)seawater =10.1*103N/m3
 Given:
As the figures on right
(ρg)seawater =10.1*103N/m3
Note:(ρg)fresh water
= 9.80*103N/m3
Weight of buoy = 8.50*103N
Dbuoy = 1.5m
Determine:
What is the tension of the cable?
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11. Where FB = (ρg)water Buoy = (10.1*103N/m3)(4/3)π*(1.5/2)3(m3)
Solution:
For equilibrium ΣF=0
 T+W – FB = 0
T = FB – W
Where FB = (ρg)water Buoy
= (10.1*103N/m3)(4/3)π*(1.5/2)3(m3)
= 1.785*104N
W = 8.50*103N
 T = 1.785*104 – 8.50*103
= 9.35*103N
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12. §2.11.2 Stability “Distrub” the floating body slightly
(a)develops a restoring moment which will return it to its
origin positionstable
(b)otherwise  unstable
The basic principle of the static-stability calculations
(1)The basic floating position
(2)The body is titled a small angle Δθ
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14. M MG>0(M above CG) =>Stable MG<0(M below CG) =>Unstable M
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15. §2.12 pressure variation in a Fluid with Rigid-Body Motion
 The general equation of motion,
No shearing stress
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16. §2.12.1 Linear Motion Free surface slope = dz/dy acceleration az ay
Fluid
at rest
P1 constant
P3 P2 pressure lines
Fluid with
acceleration az ay
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17. No shearing stress
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18. Along a line of constant pressure , dp=0
All lines of constant pressure will be parallel to the free surface
If from Eq(2.26)
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23. Example 2.11 Given:Right figure ay only Questions:
(1) P = (ay) at point 1
(2) ay = ? if fuel level on point 1
Solution:
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24. since
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25. § Rigid-Body Rotation
since
……(2.30 )
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26. at (r,z) = (0,0), P=P0
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27. For a constant pressure surface at air liquid surface, P=P1b d c
volume(ace)
Volume(abcde)
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