1. Engineering Mechanics: Statics
Chapter 9:
Center of Gravity and Centroid
Engineering Mechanics: Statics

2. Chapter Objectives
To discuss the concept of the center of gravity, center of mass, and the centroid.
To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape.
To use the theorems of Pappus and Guldinus for finding the area and volume for a surface of revolution.
To present a method for finding the resultant of a general distributed loading and show how it applies to finding the resultant of a fluid.

3. Chapter Outline
Center of Gravity and Center of Mass for a System of Particles
Center of Gravity and Center of Mass and Centroid for a Body
Composite Bodies
Theorems of Pappus and Guldinus
Resultant of a General Distributed Loading
Fluid Pressure

4. 9.1 Center of Gravity and Center of Mass for a System of Particles
Locates the resultant weight of a system of particles
Consider system of n particles fixed within a region of space
The weights of the particles
comprise a system of parallel
forces which can be replaced
by a single (equivalent) resultant
weight having defined point G
of application

5. 9.1 Center of Gravity and Center of Mass for a System of Particles
Resultant weight = total weight of n particles
Sum of moments of weights of all the particles about x, y, z axes = moment of resultant weight about these axes
Summing moments about the x axis,
Summing moments about y axis,

6. 9.1 Center of Gravity and Center of Mass for a System of Particles
Although the weights do not produce a moment about z axis, by rotating the coordinate system 90° about x or y axis with the particles fixed in it and summing moments about the x axis,
Generally,

7. 9.1 Center of Gravity and Center of Mass for a System of Particles
Where represent the coordinates of the center of gravity G of the system of particles,
represent the coordinates of each particle in the system and represent the resultant sum of the weights of all the particles in the system.
These equations represent a balance between the sum of the moments of the weights of each particle and the moment of resultant weight for the system.

8. 9.1 Center of Gravity and Center of Mass for a System of Particles
Center Mass
Provided acceleration due to gravity g for every particle is constant, then W = mg
By comparison, the location of the center of gravity coincides with that of center of mass
Particles have weight only when under the influence of gravitational attraction, whereas center of mass is independent of gravity

9. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Center Mass
A rigid body is composed of an infinite number of particles
Consider arbitrary particle
having a weight of dW

10. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Center Mass
γ represents the specific weight and dW = γdV

11. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Density ρ, or mass per unit volume, is related to γ by γ = ρg, where g = acceleration due to gravity
Substitute this relationship into this equation to determine the body’s center of mass

12. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Defines the geometric center of object
Its location can be determined from equations used to determine the body’s center of gravity or center of mass
If the material composing a body is uniform or homogenous, the density or specific weight will be constant throughout the body
The following formulas are independent of the body’s weight and depend on the body’s geometry

13. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Volume
Consider an object subdivided into volume elements dV, for location of the centroid,

14. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Area
For centroid for surface area of an object, such as plate and shell, subdivide the area into differential elements dA

15. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Line
If the geometry of the object such as a thin rod or wire, takes the form of a line, the balance of moments of differential elements dL about each of the coordinate system yields

16. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Line
Choose a coordinate system that simplifies as much as possible the equation used to describe the object’s boundary
Example
Polar coordinates are appropriate for area with circular boundaries

17. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Symmetry
The centroids of some shapes may be partially or completely specified by using conditions of symmetry
In cases where the shape has an axis of symmetry, the centroid of the shape must lie along the line
Example
Centroid C must lie along the
y axis since for every element
length dL, it lies in the middle

18. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Symmetry
For total moment of all the elements about the axis of symmetry will therefore be cancelled
In cases where a shape has 2 or 3 axes of symmetry, the centroid lies at the intersection of these axes

19. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Procedure for Analysis
Differential Element
Select an appropriate coordinate system, specify the coordinate axes, and choose an differential element for integration
For lines, the element dL is represented as a differential line segment
For areas, the element dA is generally a rectangular having a finite length and differential width

20. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Procedure for Analysis
Differential Element
For volumes, the element dV is either a circular disk having a finite radius and differential thickness, or a shell having a finite length and radius and a differential thickness
Locate the element at an arbitrary point (x, y, z) on the curve that defines the shape

21. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Procedure for Analysis
Size and Moment Arms
Express the length dL, area dA or volume dV of the element in terms of the curve used to define the geometric shape
Determine the coordinates or moment arms for the centroid of the center of gravity of the element

22. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Procedure for Analysis
Integrations
Substitute the formations and dL, dA and dV into the appropriate equations and perform integrations
Express the function in the integrand and in terms of the same variable as the differential thickness of the element
The limits of integrals are defined from the two extreme locations of the element’s differential thickness so that entire area is covered during integration

23. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.1
Locate the centroid of
the rod bent into the
shape of a parabolic arc.

24. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Differential element
Located on the curve at the arbitrary point (x, y)
Area and Moment Arms
For differential length of the element dL
Since x = y2 and then dx/dy = 2y
The centroid is located at

25. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

26. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.2
Locate the centroid of
the circular wire
segment.

27. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Differential element
A differential circular arc is selected
This element intersects the curve at (R, θ)
Length and Moment Arms
For differential length of the element
For centroid,

28. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

29. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.3
Determine the distance
from the x axis to the
centroid of the area
of the triangle

30. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Differential element
Consider a rectangular element having thickness dy which intersects the boundary at (x, y)
Length and Moment Arms
For area of the element
Centroid is located y distance from the x axis

31. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

32. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.4
Locate the centroid for
the area of a quarter
circle.

33. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 1
Differential element
Use polar coordinates for circular boundary
Triangular element intersects at point (R,θ)
Length and Moment Arms
For area of the element
Centroid is located y distance from the x axis

34. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

35. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 2
Differential element
Circular arc element having thickness of dr
Element intersects the
axes at point (r,0) and
(r, π/2)

36. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
Centroid is located y distance from the x axis

37. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

38. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.5
Locate the centroid of
the area.

39. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 1
Differential element
Differential element of thickness dx
Element intersects curve at point (x, y), height y
Area and Moment Arms
For area of the element
For centroid

40. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

41. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 2
Differential element
Differential element of thickness dy
Element intersects curve at point (x, y)
Length = (1 – x)

42. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
Centroid is located y distance from the x axis

43. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

44. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.6
Locate the centroid of
the shaded are bounded
by the two curves
y = x
and y = x2.

45. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 1
Differential element
Differential element of thickness dx
Intersects curve at point (x1, y1) and (x2, y2), height y
Area and Moment Arms
For area of the element
For centroid

46. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

47. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 2
Differential element
Differential element of thickness dy
Element intersects curve at point (x1, y1) and (x2, y2)
Length = (x1 – x2)

48. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
Centroid is located y distance from the x axis

49. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

50. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.7
Locate the centroid for the
paraboloid of revolution,
which is generated by
revolving the shaded area
about the y axis.

51. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 1
Differential element
Element in the shape of a thin disk, thickness dy, radius z
dA is always perpendicular to the axis of revolution
Intersects at point (0, y, z)
Area and Moment Arms
For volume of the element

52. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
For centroid
Integrations

53. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Method 2
Differential element
Volume element in the form of thin cylindrical shell, thickness of dz
dA is taken parallel to the axis of revolution
Element intersects the
axes at point (0, y, z) and
radius r = z

54. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
Centroid is located y distance from the x axis

55. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Integrations

56. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Example 9.8
Determine the location of
the center of mass of the
cylinder if its density
varies directly with its
distance from the base
ρ = 200z kg/m3.

57. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
View Free Body Diagram
Solution
For reasons of material symmetry
Differential element
Disk element of radius 0.5m and thickness dz since density is constant for given value of z
Located along z axis at point (0, 0, z)
Area and Moment Arms
For volume of the element

58. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
For centroid
Integrations

59. 9.2 Center of Gravity and Center of Mass and Centroid for a Body
Solution
Not possible to use a shell element for integration since the density of the material composing the shell would vary along the shell’s height and hence the location of the element cannot be specified

60. 9.3 Composite Bodies
Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular
A body can be sectioned or divided into its composite parts
Provided the weight and location of the center of gravity of each of these parts are known, the need for integration to determine the center of gravity for the entire body can be neglected

61. 9.3 Composite Bodies Accounting for finite number of weights Where
represent the coordinates of the center of gravity G of the composite body
represent the coordinates of the center of gravity at each composite part of the body
represent the sum of the weights of all the composite parts of the body or total weight

62. 9.3 Composite Bodies
When the body has a constant density or specified weight, the center of gravity coincides with the centroid of the body
The centroid for composite lines, areas, and volumes can be found using the equation
However, the W’s are replaced by L’s, A’s and V’s respectively

63. 9.3 Composite Bodies Procedure for Analysis Composite Parts
Using a sketch, divide the body or object into a finite number of composite parts that have simpler shapes
If a composite part has a hole, or a geometric region having no material, consider it without the hole and treat the hole as an additional composite part having negative weight or size

64. 9.3 Composite Bodies Procedure for Analysis Moment Arms
Establish the coordinate axes on the sketch and determine the coordinates of the center of gravity or centroid of each part

65. 9.3 Composite Bodies Procedure for Analysis Summations
Determine the coordinates of the center of gravity by applying the center of gravity equations
If an object is symmetrical about an axis, the centroid of the objects lies on the axis

66. 9.3 Composite Bodies
Example 9.9
Locate the centroid of the wire.

67. 9.3 Composite Bodies Solution Composite Parts Moment Arms
Location of the centroid for each piece is determined and indicated in the diagram

68. 9.3 Composite Bodies Solution Summations Segment L (mm) x (mm) y (mm)
z (mm)
xL (mm2)
yL (mm2)
zL (mm2) 1
188.5 60
-38.2
11 310
-7200 2 40 20
800 3
-10
-200
Sum
248.5
-5600

69. 9.3 Composite Bodies
Solution
Summations

70. 9.3 Composite Bodies Example 9.10
Locate the centroid of the plate area.

71. 9.3 Composite Bodies Solution Composite Parts
Plate divided into 3 segments
Area of small rectangle considered “negative”

72. 9.3 Composite Bodies Solution Moment Arm
Location of the centroid for each piece is determined and indicated in the diagram

73. 9.3 Composite Bodies Solution Summations Segment A (mm2) x (mm) y (mm)
xA (mm3)
yA (mm3) 1
4.5 2 9
-1.5
1.5
-13.5
13.5 3 -2
-2.5 5 -4
Sum
11.5 14

74. 9.3 Composite Bodies
Solution
Summations

75. 9.3 Composite Bodies Example 9.11 Locate the center of mass of the
composite assembly. The conical
frustum has a density of
ρc = 8Mg/m3 and the hemisphere
has a density of ρh = 4Mg/m3.
There is a 25mm radius
cylindrical hole in the center.

76. 9.3 Composite Bodies Solution Composite Parts
View Free Body Diagram
Solution
Composite Parts
Assembly divided into 4 segments
Area of 3 and 4 considered “negative”

77. 9.3 Composite Bodies Solution Moment Arm
Location of the centroid for each piece is determined and indicated in the diagram
Summations
Because of symmetry,

78. 9.3 Composite Bodies Solution Summations Segment m (kg) z (mm)
zm (kg.mm) 1
4.189 50 2
1.047
-18.75 3
-0.524
125 4
-1.571
Sum
3.141
45.815

79. 9.3 Composite Bodies
Solution
Summations

80. 9.4 Theorems of Pappus and Guldinus
A surface area of revolution is generated by revolving a plane curve about a non-intersecting fixed axis in the plane of the curve
A volume of revolution is generated by revolving a plane area bout a nonintersecting fixed axis in the plane of area
Example
Line AB is rotated about
fixed axis, it generates
the surface area of a
cone (less area of base)

81. 9.4 Theorems of Pappus and Guldinus
Example
Triangular area ABC rotated
about the axis would
generate the volume of
the cone
The theorems of Pappus and Guldinus are used to find the surfaces area and volume of any object of revolution provided the generating curves and areas do not cross the axis they are rotated

82. 9.4 Theorems of Pappus and Guldinus
Surface Area
Area of a surface of revolution = product of length of the curve and distance traveled by the centroid in generating the surface area

83. 9.4 Theorems of Pappus and Guldinus
Volume
Volume of a body of revolution = product of generating area and distance traveled by the centroid in generating the volume

84. 9.4 Theorems of Pappus and Guldinus
Composite Shapes
The above two mentioned theorems can be applied to lines or areas that may be composed of a series of composite parts
Total surface area or volume generated is the addition of the surface areas or volumes generated by each of the composite parts

85. 9.4 Theorems of Pappus and Guldinus
Example 9.12
Show that the surface area of a sphere is
A = 4πR2
and its volume
V = 4/3 πR3

86. 9.4 Theorems of Pappus and Guldinus
View Free Body Diagram
Solution
Surface Area
Generated by rotating semi-arc about the x axis
For centroid,
For surface area,

87. 9.4 Theorems of Pappus and Guldinus
Solution
Volume
Generated by rotating semicircular area about the x axis
For centroid,
For volume,

88. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Consider the flat plate subjected to the loading function ρ = ρ(x, y) Pa
Determine the force dF acting on the differential area dA m2 of the plate, located at the differential point (x, y)
dF = [ρ(x, y) N/m2](d A m2)
= [ρ(x, y) d A]N
Entire loading represented as
infinite parallel forces acting on
separate differential area dA

89. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
This system will be simplified to a single resultant force FR acting through a unique point on the plate

90. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Magnitude of Resultant Force
To determine magnitude of FR, sum the differential forces dF acting over the plate’s entire surface area dA
Magnitude of resultant
force = total volume under
the distributed loading
diagram

91. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Location of Resultant Force
Line of action of action of the resultant force passes through the geometric center or centroid of the volume under the distributed loading diagram

92. 9.6 Fluid Pressure
According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions
Magnitude of ρ measured as a force per unit area, depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface
ρ = γz = ρgz
Valid for incompressible fluids
Gas are compressible fluids and thus the above equation cannot be used

93. 9.6 Fluid Pressure Consider the submerged plate
3 points have been specified

94. 9.6 Fluid Pressure
Since point B is at depth z1 from the liquid surface, the pressure at this point has a magnitude of ρ1= γz1
Likewise, points C and D are both at depth z2 and hence ρ2 = γz2
In all cases, pressure acts normal to the surface area dA located at specified point
Possible to determine the resultant force caused by a fluid distribution and specify its location on the surface of a submerged plate

95. 9.6 Fluid Pressure Flat Plate of Constant Width
Consider flat rectangular plate of constant width submerged in a liquid having a specific weight γ
Plane of the plate makes an angle with the horizontal as shown

96. 9.6 Fluid Pressure Flat Plate of Constant Width
Since pressure varies linearly with depth,
the distribution of pressure over the plate’s
surface is represented by a trapezoidal
volume having an
intensity of ρ1= γz1
at depth z1 and
ρ2 = γz2 at depth z2

97. 9.6 Fluid Pressure
Magnitude of the resultant force FR = volume of this loading diagram and FR has a line of action that passes through the volume’s centroid, C
FR does not act at the centroid of the plate but at
point P called the center of
pressure
Since plate has a constant
width, the loading diagram
can be viewed in 2D

98. 9.6 Fluid Pressure Flat Plate of Constant Width
Loading intensity is measured as force/length and varies linearly from
w1 = bρ1= bγz1 to w 2 = bρ2 = bγz2
Magnitude of FR = trapezoidal area
FR has a line of action that passes through the area’s centroid C

99. 9.6 Fluid Pressure Curved Plate of Constant Width
When the submerged plate is curved, the pressure acting normal to the plate continuously changes direction
For 2D and 3D view of the loading distribution,
Integration can be used to determine FR and location of center of centroid C or pressure P

100. 9.6 Fluid Pressure Curved Plate of Constant Width Example
Consider distributed loading acting on the curved plate DB

101. 9.6 Fluid Pressure Curved Plate of Constant Width Example
For equivalent loading

102. 9.6 Fluid Pressure Curved Plate of Constant Width
The plate supports the weight of the liquid Wf contained within the block BDA
This force has a magnitude of
Wf = (γb)(areaBDA)
and acts through the centroid of BDA
Pressure distributions caused by the liquid acting along the vertical and horizontal sides of the block
Along vertical side AD, force FAD’s magnitude = area under trapezoid and acts through centroid CAD of this area

103. 9.6 Fluid Pressure Curved Plate of Constant Width
The distributed loading along horizontal side AB is constant since all points lying on this plane are at the same depth from the surface of the liquid
Magnitude of FAB is simply the area of the rectangle
This force acts through the area centroid CAB or the midpoint of AB
Summing three forces,
FR = ∑F = FAB + FAD + Wf

104. 9.6 Fluid Pressure Curved Plate of Constant Width
Location of the center of pressure on the plate is determined by applying
MRo = ∑MO
which states that the moment of the resultant force about a convenient reference point O, such as D or B = sum of the moments of the 3 forces about the same point

105. 9.6 Fluid Pressure Flat Plate of Variable Width
Consider the pressure distribution acting on the surface of a submerged plate having a variable width

106. 9.6 Fluid Pressure Flat Plate of Variable Width
Resultant force of this loading = volume described by the plate area as its base and linearly varying pressure distribution as its altitude
The shaded element may be used if integration is chosen to determine the volume
Element consists of a rectangular strip of area dA = x dy’ located at depth z below the liquid surface
Since uniform pressure ρ = γz (force/area) acts on dA, the magnitude of the differential force dF
dF = dV = ρ dA = γz(xdy’)

107. 9.6 Fluid Pressure Flat Plate of Variable Width
Centroid V defines the point which FR acts
The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations
This point should not be mistaken for centroid of the plate’s area

108. 9.6 Fluid Pressure Example 9.13
Determine the magnitude and location of the
resultant hydrostatic force acting on the submerged
rectangular plate AB. The
plate has a width of 1.5m;
ρw = 1000kg/m3.

109. 9.6 Fluid Pressure Solution The water pressures at depth A and B are
Since the plate has constant
width, distributed loading
can be viewed in 2D
For intensities of the load at
A and B,

110. 9.6 Fluid Pressure Solution
For magnitude of the resultant force FR created by the distributed load
This force acts through the
centroid of the area
measured upwards from B

111. 9.6 Fluid Pressure Solution
Same results can be obtained by considering two components of FR defined by the triangle and rectangle
Each force acts through its associated centroid and has a magnitude of
Hence

112. 9.6 Fluid Pressure Solution
Location of FR is determined by summing moments about B

113. 9.6 Fluid Pressure Example 9.14
Determine the magnitude of the resultant
hydrostatic force acting on the surface of a seawall
shaped in the form of a parabola. The wall is 5m
long and ρw = 1020kg/m2.

114. 9.6 Fluid Pressure Solution
The horizontal and vertical components of the resultant force will be calculated since
Then
Thus

115. 9.6 Fluid Pressure Solution
Area of the parabolic sector ABC can be determined
For weight of the wafer within this region
For resultant force

116. 9.6 Fluid Pressure Example 9.15
Determine the magnitude and location of
the resultant force acting on the triangular
end plates of the wafer of the water trough.
ρw = 1000 kg/m3

117. 9.6 Fluid Pressure Solution
View Free Body Diagram
Solution
Magnitude of the resultant force F = volume of the loading distribution
Choosing the differential volume element,
For equation of line AB
Integrating

118. 9.6 Fluid Pressure Solution
Resultant passes through the centroid of the volume
Because of symmetry
For volume element

119. Chapter Summary Center of Gravity and Centroid
Center of gravity represents a point where the weight of the body can be considered concentrated
The distance to this point can be determined by a balance of moments
Moment of weight of all the particles of the body about some point = moment of the entire body about the point
Centroid is the location of the geometric center of the body

120. Chapter Summary Center of Gravity and Centroid
Centroid is determined by the moment balance of geometric elements such as line, area and volume segments
For body having a continuous shape, moments are summed using differential elements
For composite of several shapes, each having a known location for centroid, the location is determined from discrete summation using its composite parts

121. Chapter Summary Theorems of Pappus and Guldinus
Used to determine surface area and volume of a body of revolution
Surface area = product of length of the generating curve and distance traveled by the centroid of the curve to generate the area
Volume = product of the generating area and the distance traveled by the centroid to generate the volume

122. Chapter Summary Fluid Pressure
Pressure developed by a fluid at a point on a submerged surface depends on the depth of the point and the density of the liquid according to Pascal’s law
Pressure will create a linear distribution of loading on a flat vertical or inclined surface
For horizontal surface, loading is uniform
Resultants determined by volume or area under the loading curve

123. Chapter Summary Fluid Pressure
Line of action of the resultant force passes through the centroid of the loading diagram

124. Chapter Review

125. Chapter Review

126. Chapter Review

127. Chapter Review

128. Chapter Review

129. Chapter Review