ME 221 Statics
Lecture #13
Sections 4.1 – 4.2
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Lecture 13
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Homework #5
• Chapter 3 problems:
– 48, 51, 55, 57, 61, 62, 65, 71, 72 & 75
– May use MathCAD, etc. to solve
– Due Friday, October 3
ME221
Lecture 13
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Distributed Forces (Loads);
Centroids & Center of Gravity
• The concept of distributed loads will be
introduced
• Center of mass will be discussed as an
important application of distributed loading
– mass, (hence, weight), is distributed throughout
a body; we want to find the “balance” point
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Distributed Loads
Two types of distributed loads exist:
– forces that exist throughout the body
• e. g., gravity acting on mass
• these are called “body forces”
– forces arising from contact between two bodies
• these are called “contact forces”
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Contact Distributed Load
• Snow on roof, tire on road, bearing on race,
liquid on container wall, ...
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Center of Gravityz
z
w1(x
˜1,y˜ 1,z˜1)
w3(x
˜3,y˜ 3,z˜3)
w2(x
˜2,y˜ 2,z˜2)
w5(x
˜5,y˜ 5,z˜5)
w4(x
˜4,y˜ 4,z˜4)
z
y
x
y
x
y
x
The weights of the n particles comprise a system of parallel forces. We can
replace them with an equivalent force w located at G(x,y,z), such that:
~
~
~
x w=x~1w1+x
w
+x
w
+x
2 2
3 3
4w4
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Or
x
n
 ~
xi wi
i 1
n
 wi
i 1
,
y
n
 ~
yi wi
i 1
n
 wi
i 1
,
z
n
 ~
zi wi
i 1
n
 wi
i 1
Where ~
x, ~
y, ~
z are the coordinates of each
point. Point G is called the center of gravity
which is defined as the point in the space where
all the weight is concentrated.
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CG in Discrete Sense
20
10
??
??
??
Where do we hold the bar to balance it?
Find the point where the system’s weight may
be balanced without the use of a moment.
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Discrete Equations
y
r
dw
Define a reference frame
z
x
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~
xi wi

x
~
x dw
x
wi
 dw

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Center of Mass
The total mass is given by M
M 
m
i
i
Mass center is defined by
m x
i i
xc.m. 
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i
M
m y
i i
; yc.m. 
i
M
m z
i i
; zc.m. 
Lecture 13
i
M
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Continuous Equations
Take our volume, dV, to be infinitesimal.
Summing over all volumes becomes an integral.
1
M
 dV
VV

1
1
1
xc.m. 
xdV ; yc.m. 
ydV ; zc.m. 
zdV
VV
VV
VV



Note that dm = dV . Center of gravity deals with forces and
gdm is used in the integrals.
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If  is constant
x
x dv
~
 dv
,
y
y dv
~
 dv
,
z
z dv
~
 dv
•These coordinates define the geometric center
of an object (the centroid)
•In case of 2-D, the geometric center can be
defined using a differential element dA
x
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x dA
~
 dA
,
y
y dA
~
 dA
Lecture 13
,
z
z dA
~
 dA
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If the geometry of an object takes the form
of a line (thin rod or wire), then the
centroid may be defined as:
x
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x dL
~
 dL
,
y
y dL
~
 dL
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z
z dL
~
 dL
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Procedure for Analysis
1-Differential element
Specify the coordinate axes and choose an appropriate
differential element of integration.
•For a line, the differential element is dl
•For an area, the differential element dA is generally a
rectangle having a finite height and differential width.
•For a volume, 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 differential thickness.
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2- Size
Express the length dl, dA, or dv of the element in terms of
the coordinate used to define the object.
3-Moment Arm
Determine the perpendicular distance from the coordinate
axes to the centroid of the differential element.
4- Equation
Substitute the data computed above in the appropriate
equation.
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Symmetry conditions
•The centroid of some objects may be partially or
completely specified by using the symmetry conditions
•In the case where the shape of the object has an axis of
symmetry, then the centroid will be located along that line of
symmetry.
y
x
In this case, the centoid is located along the y-axis
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In cases of more than one axis of symmetry, the
centroid will be located at the intersection of these axes.
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