Pharos Univ.
ME 259 Fluid Mechanics
Static Forces on Inclined and
Curved Surfaces
Main Topics
The Basic Equations of Fluid
Statics
Pressure Variation in a Static
Fluid
Hydrostatic Force on Submerged
Surfaces
Buoyancy
The Basic Equations
of Fluid Statics
Body Force
The Basic Equations
of Fluid Statics
Surface Force
The Basic Equations
of Fluid Statics
Surface Force
The Basic Equations
of Fluid Statics
Surface Force
The Basic Equations
of Fluid Statics
Total Force
The Basic Equations
of Fluid Statics
Newton’s Second Law
The Basic Equations
of Fluid Statics
Pressure-Height Relation
2.3.1 Pressure and head
Free surface
P2 = Patm
h
ya
P1
y
 In a liquid with a free surface the pressure at any depth h
measured from the free surface can be found by applying
equation (2.3) to the figure.
 From equation (2.3): P1 – P2= g (ya-y)
But
Thus,
ya-y = h , and
P2 = Patm (atmospheric pressure since it is at free surface).
P1 – Patm= gh
or
P1 = Patm + gh (abs)
or in terms of gauge pressure (Patm= 0),:
P1 = gh = h
(2.4)
(2.5)
Pressure Variation in a
Static Fluid
 Incompressible Fluid: Manometers
Pressure Variation in a
Static Fluid
Compressible Fluid: Ideal Gas
Need additional information,
e.g., T(z) for atmosphere
Differential Manometer
 The liquids in manometer will rise
or fall as the pressure at either
end changes.
P1 = PA + 1ga
P2 = PB + 1g(b-h) + mangh
P1 = P2 (same level)
PA + 1ga = PB + 1g(b-h) + mangh
or PA - PB = 1g(b-h) + mangh - 1ga
PA- PB = 1g(b-a) + gh(man - 1)
Figure 2.13:
Differential manometer
Hydrostatic Force on
Submerged Surfaces
Plane Submerged Surface
Center of Pressure
 Line of action of resultant
force FR=PCA lies
underneath where the
pressure is higher.
 Location of Center of
Pressure is determined by
the moment.
I
y p  yC 
xx ,C
yc A
 Ixx,C is tabulated for simple
geometries.
Hydrostatic Force on
Submerged Surfaces
Plane Submerged Surface
We can find FR, and y´ and x´,
by integrating, or …
Hydrostatic Force on
Submerged Surfaces
 Plane Submerged Surface
• Algebraic Equations – Total Pressure Force
Hydrostatic Force on
Submerged Surfaces
 Plane Submerged Surface
• Algebraic Equations – Net Pressure Force
Table 2.1 Second Moments of Area
Shape
Rectangle
b
Area
Ig
bh
bh3/12
bh/2
bh3/36
d2/4
d4/64
hh
G
G
G
G
Triangle
Gh
hG
h/3
h/3
GG
b
b
Circle
dd
GG
GG
 A 6-m deep tank contains 4 m of water and 2-m of oil as
shown in the diagram below. Determine the pressure at
point A and at the tank bottom. Draw the pressure diag.
Pressure at oil water interface (PA)
water = 1000 kg/m3
PA = Patm + Poil (due to 2 m of oil)
= 0 + oilghoil = 0 + 0.98 x 1000 x 9.81 x 2
= 15696 Pa
PA = 15.7 kPa (gauge)
SG of oil = 0.98
Pressure at the bottom of the tank;
PB = PA + waterghwater
PB = 15.7x1000 + 1000 x 9.81 x 4
= 54940 Pa
PB = 54.9 kPa (gauge)
Pressure Diagram
Patm = 0
2m
oil
PA=15.7 kPa
A
water
PA
4m
B
PB = 54.9 kPA
Hydrostatic Force on
Submerged Surfaces
Curved Submerged Surface
Hydrostatic Force on
Submerged Surfaces
 Curved Submerged Surface
• Horizontal Force = Equivalent Vertical Plane Force
• Vertical Force = Weight of Fluid Directly Above
(+ Free Surface Pressure Force)
Hydrostatic Forces on Curved
Surfaces
 FR on a curved surface is more involved
since it requires integration of the pressure
forces that change direction along the
surface.
 Easiest approach: determine horizontal and
vertical components FH and FV separately.
Forces on Curved Surfaces
h1
h2
Submerged Curved Surface
Resultant force:Horizontal and vertical components
Horizontal component:
• FH = *s*w*(h + s/2),Where,
h = Depth to the top of rectangle (beginning of curve surface)
s = projected rectangle height
w = projected rectangle length or width
• Center of pressure
hp = hC + IC/(hCA)
hC = h + s/2
Vertical Component
FV = *Volume  *A*w Where,
FR 
FH2  FV2
 FV 

  tan 
 FH 
A = entire area of fluid
w = projected rectangle length or width
1
Hydrostatic Buoyant Force
Archimedes’ principle
 When a body is submerged or floating, the resultant force by the fluid is called
the buoyancy force. This buoyancy force is acting vertically upward
 The buoyancy force is equal to the weight of the fluid displaced by body.
 The buoyancy force acts at the centroid of the displaced volume of fluid.
 A floating body displaces a volume of fluid whose weight - body weight
For equilibrium: + ΣFy = 0
Fb – W = 0
or
Fb = W
Therefore we can write ;
W = mg
W = mg
Fb = weight of fluid displaced by the body
Or Fb = W = mg = g
Where Fb = buoyant force
 = displaced volume of fluid
G
B
Volume of
displaced
fluid
W = weight of fluid
Fb= W
G
B
Fb = W
Buoyancy
Buoyancy
For example, for a hot air
balloon
Buoyancy and Stability
Buoyancy is due to the fluid
displaced by a body. FB=fgV.
Archimedes principal : The buoyant
force = Weight of the fluid displaced
by the body, and it acts through the
centroid of the displaced volume.
Buoyancy and Stability
 Buoyancy force FB is
equal only to the
displaced volume
fgVdisplaced.
 Three scenarios
possible
1. body<fluid: Floating body
2. body=fluid: Neutrally
buoyant
3. body>fluid: Sinking body
Stability of Immersed Bodies
 Rotational stability of immersed bodies depends
upon relative location of center of gravity G and
center of buoyancy B.
• G below B: stable
• G above B: unstable
• G coincides with B: neutrally stable.
Stability of Floating Bodies
 If body is bottom heavy
(G lower than B), it is
always stable.
 Floating bodies can be
stable when G is higher
than B due to shift in
location of center
buoyancy and creation
of restoring moment.
 Measure of stability is
the metacentric height
GM. If GM>1, ship is
stable.