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CHAPTER
5
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30
Dim
30 in
m
PROBLEM
5.1
Locate
the centroid
of the
plane
area shown.
1
300 mil)
-240
mm-
SOLUTION
Dimensions
in
mm
-27T
«E.
'
<
U
cfiy~\4
T
/TO
4-
J
L/or|,/ar
j^i
i4,
mm
2
x,
mm
y,
mm
x./4,
mm
3
y.A,
mm
1
6300
105
15
0.661
50x1
6
0.094500x1
6
2
9000
225
150
2.0250x1
6
1.35000x10*
£
15300
2.6865xl0
6
1.44450xl0
6
Then
X
£x/5l
2.6865x10*
ZA
ZA
15300
£JM.
1.44450x10*
15300
X
=
175.6
mm
<
Y
=
94.4
mm
8/17/2019 Cap 5, Novena Edc_text.pdf
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20
mm
30mm
-~r
36
mm
24
mm
PROBLEM 5.2
Locate the
centroid
of
the
plane area
shown.
SOLUTION
Dimensions
in mm
r
ZA
+
K
C,
-A
\0
A, mm x, mm
_y,
mm
xA, mm
3
yA,
mm
3
I 1200
10
30
12000
36000
2
540
30 36
16200
19440
1
1740
28200 55440
Then
X
ZxA
28200
XA
1740
gjvj
=
55440
XA
1740
X
=
16.21
mm
<
F
=
31.9mm
<
PROPRIETARY
MATERIAL
©
2010 The
McGraw-Hill
Companies, Inc.
All
rights reserved. No part
of
this Manual may be
displayed,
reproduced or
distributed
in
any
form
or by
any
means,
without the
prior written permission
of
the
publisher,
or
used
beyond
the
limited
distribution
to
teachers and
educatorspermittedby
McGraw-Hill
for
their
individual
course
preparation.
If
you
are
a student using
this
Manual,
you are using it
without permission.
542
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5/196
-12
in.-*+«
21
in.
-
15
in.
PROBLEM
5.3
Locate
the
centroid
of
the
plane area
shown.
SOLUTION
Dimensions
in
in.
zfirt
m
+
7*
4
in.
2
x,in.
y>
in.
x^, in;
M
in.
3
1
-xl2xl5
=
90
2
8
5
720
450
2
21x15
=
315
22.5
7.5
7087.5
2362.5
I
405.00
7807.5
2812.5
Then
-
=
£x,4
=
7807.5
ZA 405.00
Y^XyA^
2812.5
1^1
405.00
Z
=
19.28
in.
^
F
=
6.94
in.
^
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies,
Inc. All
rights
reserved.
Afo ywrc o///?/i-
Manual
may be
displayed,
reproduced or distributed
in
any
form
or
by
any
means,
without
the prior
written
permission
of
the
publisher,
or used beyond
the
limited
distribution
to teachers
and
educatorspermitted
by McGraw-Hill
for
their individual
course
preparation.
If
you
are a
student
using
this
Manual,
you
are
using
if without permission.
543
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3 in,
6 in.
6 in.
•
6 in.
6
in.
PROBLEM 5.4
Locate the
centroid of
the
plane
area
shown.
SOLUTION
4
WW
A, in.
2
x,m.
y,
in.
x^i,
in/
_y^,
in.
3
1
i(12)(6)
=
36
4
4
144
144
2
(6X3)
=
18
9
7.5 162
135
£
54
306
279
Then
XA
=
Y,xA
X
(54)
=
306
YA
-
XyA
F(54)
=
279
X
-
5.67 in.
-<
r
=
5.17in. <
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill Companies,
Inc.
All rights
reserved.
No
part
of
this
Manual
may be
displayed,
reproduced
or distributed in
any
form or
by
any means,
without the prior
written
permission
of
the
publisher,
or
used beyond
the limited
distribution to
teachers
and
educators
permitted by
McGraw-Hill
for
their
individual course
preparation.
If
you are
a
student
using
this
Manual,
you are using
it without
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544
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6 in. , 8 in.
PROBLEM
5.5
Locate
the
eentroid
of
the
plane area
shown.
S
in.
12 in.
SOLUTION
©
,.- .J
c,
|
1 \H*
*
ttiM.
A, in.
2
x,
in.
p,
in.
X/4, in.
3
jM,
in.
3
]
14x20
=
280
7
10
1960
2800
2
-^(4)
2
= 16^
6
12
-301.59
-603.19
I
229.73
1658.41
2196.8
Then
X
TsxA
1658.41
1A
229.73
£j^
=
2.196.8
XA
~
229.73
X
=
7.22
in.
^
F =
9.56
in.
<
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies,
Inc. All
rights
reserved.
No part
of
this
Manual
may
be
displayed,
reproduced or
distributed in
any
form
or by
any means,
without
the prior
written
permission
of
the
publisher, or
used beyond the
limited
distribution
to
teachers
andeducators
permitted
by McGraw-Hill
for
their
individual
course
preparation.
If
you
are a student
using
this
Manual,
you are using
it
without
permission.
545
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.120
mm
«Ar
=
75
/
mm
ilg
\
PROBLEM 5.6
Locate the
centroid
of
the
plane
area
shown.
SOLUTION
Z5mw
1
157.5
wwi
•i
|
0Omm
©
37.
5Vm
3-n
v\v»
HS>3».**»***3'K^*
W
*
J, ram
2
x,mm jvmm
xA,mm
3
yA,
mm
3
1
(120)(75)
=
4500 80
2.5
360
xlO
3
112.5
xlO
3
2
(75)(75)
=
5625
157.5
37.5
885.94
xlO
3
2
10.94
xlO
3
3
™^(75)
2
--4417.9
4
163.169
43.169
-720.86x1
3
-190.716xl0
3
X
5707.1
525.08
xlO
3
132.724
xlO
3
Then
XA
=
ZxA
^(5707.
1)
=
525.08
x
1
3
YA
-
ZyA
F(5707.1)
=
132.724xl0
3
X
=
92.0
mm <
Y
=
23.3
mm
^
PROPRIETARY
MATERIAL. ©
2010
The
McGraw-Hill
Companies, Inc.
All rights
reserved.
No
part
of
this
Manual may
be
displayed,
reproduced or
distributed
in
any
form
or by any
means, without the
prior
written
permission
of
the publisher,
or used beyond
the
limited
distribution to
teachers and
educators
permittedby McGraw-Hill
for
their
individual
coursepreparation.
If
you
are a
student
using
this
Manual,
you
are
using
it
without
permission.
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PROBLEM
5.7
Locate
the
centroid
of
the
plane area
shown.
16
in.
1
20 in.
SOLUTION
tM.
ft
II*
i
A
•*Z
tO
1M.
A,
in.
2
x,
in.
J
7
*
in.
xA,
in;
yA,
in.
3
1
—
(38)
2
=2268.2
.
16.1277
36581
2
-20x16-
-320 -10
8
3200
-2560
Z
1948.23
3200
3402.1
Then
X
*
xA
3200
Y
~ZA
1948.23
SJ^
3402.1
LA
1948.23
X
=
1.643
in.
^
7
=
17.46
in.
•
fee
displayed,
reproduced
or distributed
in any
form or
by
any
means, without
the prior
written
permission
of
the
publisher, or
used beyond
the
limited
distribution
to teachers and
educatorspermitted
by McGraw-Hill
for
their
individual
course
preparation.
If
you are
a
student
using
this
Manual,
you are
using
it
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PROBLEM
5.8
Locate the
centroid of
the
plane
area
shown.
SOLUTION
2.S1N.
©
4-
C,
\S
l»».
A,
in?
x,
in.
^>
'
n
*
X/4,
in?
yA,
in?
1
30x50
=
1500
15
25
22500
37500
2
-~(15)
2
=353.43
2
23.634
30
-8353.0
-10602.9
£
1
146.57
14147,0
26.897
Then
X
Y
=
ZxA_
14147.0
2^
1146.57
Xy^
26897
2,4
1146.57
X
=
12.34
in.
^
F
=
23.5
in.
<
PROPRIETARY
MATERIAL.
CO 2010 The
McGraw-Hill Companies,
Inc.
AH
rights
reserved. No
part
of
this
Manual
may
be
displayed,
reproduced
or
distributed
in
any
form
or by
any
means,
without
the
prior
written permission
of
the
publisher,
or used beyond
the
limited
distribution to
teachers
and
educators
permitted
by
McGraw-Hill
for
their
individual course
preparation.
If
you
are a student
using
this Manual,
you
are
using
if without
permission.
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() iniii
PROBLEM
5.9
Locate
the
centroid
of the
plane
area
shown.
60mm
SOLUTION
^l^-^&.^S
be
displayed,
reproduced
or
distributed
in
any
form
or
by
any means,
without the
prior
written
permission
of
the
publisher,
or used
beyond
the
limited
distribution
to teachers
and
educators
permitted
by McGraw-Hill
for
their individual
course
preparation.
If
you
are
a student using
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Manual,
you
are
using it
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549
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If
Semicllipse
.
70
nun
47 mm
47 mm
PROBLEM
5.10
Locate the
centroid ofthe
plane area
shown.
26
mm
SOLUTION
Dimensions
in mm
T%
kUi-t)
A,
mm
2
x,
mm
y,
mm
jtvf,
mm
3
jvi,
ram
3
1
~x47x26
=
1919.51
2
11.0347
21181
2
-x
94x70
=
3290
2
-15.6667
-23.333
-51543
-76766
Z
5209.5
-51543
-55584
Then
-
S*^
-51543
£/*
5209.5
£v^.
-55584
Z.A
5209.5
A'
«
-9.89
mm
^
F
=
-10.67 mm
<
PROPRIETARY
MATERIAL
C©
2010 The
McGraw-Hill
Companies,
Inc. AH
rights
reserved.
No
part
of
this Manual may
be
displayed,
reproduced or
distributed in any
form
or by
any
means,
without
the
prior
written
permission
of
the
publisher, or used
beyond
the
limited
distribution to
teachers
and
educators
permittedby
McGraw-Hill
for
their
individual course
preparation.
If
you
are
a
student using
this
Manual,
you
are
using it
withoutpermission.
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PROBLEM
5.11
Locate
the
centroid
of the
plane area shown.
/\
A
\
/'
\
.
Ju
iiui.
'i
•
Sin.
\
id
\/
SOLUTION
First
note that
symmetry
implies
X
=
M
,4,
in.
2
7,
in.
jv*, in.
3
1
-*
(8)
'
=-100.531
2
3.3953 -341.33
2
*
(12)2
=226.19
2
5.0930
1151.99
I 125.659
810.66
Then
r
2y^ 8 10.66
in.
3
ZA
125.66
in.
2
or
Y
=
6.45
in.
4
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc. All
rights
reserved.
M? part
of
this Manual
may be displayed,
reproduced
or
distributed
in any
form
or
by any means,
without
the
prior
written
permission
of
the
publisher, or used
beyond the limited
distribution
to
teachers
andeducatorspermitted
by
McGraw-Hill
for
their individual
coursepreparation.
If
you are
a student using
this
Manual,
you are
using it
without
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50 nun
15
mull
PROBLEM
5.12
Parabola J-
r
Vertex
\.-
*'
-
1
80
mm
*-
Locate
the centroid of the
plane area
shown.
X
SOLUTION
»>,
mm
X/4,
mm
3
yA,
mm
3
1.
(15)(80)
=
1200
40
.
7.5
48xl0
3
9xl0
3
2
i(50)(80)~
1333.33
60 30
80xI0
3
40xl0
3
I
2533.3
128xI0
3
49xl0
3
Then
X
(2533.3)
=
128
xlO
3
YA^ZyA
7(2533.3)
=
49
xlO
3
^
=
50.5
mm
^
F
=
19.34
mm
<
PROPRIETARY MATERIAL. &
2010 The
McGraw-Hill
Companies, Inc. All
rights
reserved. No part
of
this Manual may be
displayed,
reproduced
or distributed in anyform
or
by
any means,
without
the
prior
written
permission
of
the
publisher, or used
beyond
the limited
distribution to
teachers andeducatorspermitted by
McGraw-Hill
for
their
individual
course
preparation.
If
you are
a
student using
this
Manual,
you
are using it without
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20;
U
1
~
\
x
i
mm
1
f
x=ky*
20
nun
I
*—
30
mm-*
X
PROBLEM
5.13
Locate
the
centroid ofthe
plane
area
shown.
SOLUTION
i
.
4*y$
+***\
^fflffl
2
x, mm
^,
mm
xA, mm
3
yA,
mm
3
1
1x30x20 =
200
3
9
15
1800
3000
2
™(30)
2
=
706.86
4
12.7324
32.7324
9000.0
23137
X
906.86
10800 26137
Then
X
Y
ZxA
10800
ZA
906.86
2yA_ 26137
£>4
~
906.86
X
=
1.1.91
mm
4
F- 28.8
mm
^
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies,
inc.
All
rights
reserved. Ato
/w/
ofrftft
Manual
may be displayed,
reproduced
or
distributed
in any
form
or
by
any
means,
without the prior
written
permission
of
the
publisher, or
used beyond
the limited
distribution
to teachers
andeducators permitted
by McGraw-Hill
for
their individual
course preparation.
If
you are
a
student
using
this
Manual,
you are
using
it
without
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553
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f
/
^
\
*
20
in.
»
X
SOLUTION
Dimensions
in
in.
*1
©
•»+
Aio)
|M»>
tt
-no)
\o>
/*,
in.
2
x,in.
yM-
j/4,
in
3
yA,
in.
3
I
|x(20X20)
=
552
12
7.5
3200
2000
2
ZL
(20
)(20)^
15
6.0
-2000
-800
£
400
3
1200
1200
Then
X
ZxA
1200
Y.A
f
400^
I
3
J
XyA_ 1200
XA
(
400^
*
=
9.00
in.
^
F
=
9.00
in.
^
I
3
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill Companies,
Inc.
All
rights reserved.
No
pan
of
this Manual may
be
displayed,
reproduced or distributed in
any
form
or
by
any
means, without the
prior
written
permission
of
the
publisher, or used
beyond the limited
distribution to
teachers
and
educators
permitted by
McGraw-Hill
for
their individual
coursepreparation.
If
you
are
a
student
using
this
Manual,
you
are
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it
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y
PROBLEM
5.15
r
^Vertex
\ ^Parabola
Locate
the centroid
of the plane
area
shown.
60
nm
\
60 11
m
\4
fi§§
*—
—
75 mm
*\
SOLUTION
Then
and
iWvn
(fWs.fcA')
vwv*\
©
\
jb
)WM
/4,
mm
2
x,mm
j,
mm
xA, mm
3
—
—
^
jM, mm
1
-(75)(120)
=
6000
28.125
48
168750
288000
2
—
(75)(60)
=
-2250
25
20
-56250 -45000
I
3750
112500
243000
XT,A=I,xA
X(3750 mm
2
)
=
1
12500
mm
3
YXA
=
'LyA
7(3750
mm
2
)
=
243000
or
X-
30.0
mm
<
or
7
=
64.8
mm
4
PROPRIETARY
MATERIAL
©
2010 The
McGraw-Hill
Companies,
Inc.
All rights reserved. No part
of
this Manual may
be displayed,
reproduced
or distributed in
any
form
or
by any means, without
the prior
written permission
of
the
publisher, or
used beyond
the
limited
distribution to teachers
and
educators
permitted
by McGraw-Hill
for
their
individualcourse
preparation.
If
you
are a
student using
this Manual,
you
are
using it without permission.
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y
PROBLEM 5,16
Determine thej>
coordinate
of the centroid
of the shaded
area in
terms of
r\, r2,
and
a.
«r~
*\
z^xk
~T«
i
SOLUTION
First,
determine
the
location of the
centroid.
From Figure 5.8A:
_ 2
s*
n
(f
ff
)
A,
/r
ff
cos
a
3'(f-«)
Similarly
Then
cosa
3'(f-«)
4
#
ff r,
__
. 2
cos
a
Ly/4
=—
r
2
/r
3'(f-«)
-(r
2
3
-r,
3
)cosOf
a
r
2
cos«
.
3'(i-«)
ft
I
7
and
Now
Y
ft
a
YIA
=
ZyA
-{$
- Oleosa
K
=
2
cos
or
PROPRIETARY
MATERIAL ©
2010 The
McGraw-Hill
Companies, Inc.
Alt
rights reserved. No part
of
this
Manual
may
be
displayed
reproduced or distributed in
any
form
or by any means,
without
the prior
written permission
of
the
publisher, or used
beyond the
limited
distribution to teachers and
educators
permitted
by
McGraw-Hill
for
their
individual
coursepreparation.
If
you are
a
student
using
this
Manual,
you
are using it without
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«
\
1
PROBLEM 5.17
Show
that
as r
}
approaches r
2
,
the location
of the centroid
approaches
that
for
an
arc of
circle
of radius
(/]
+r,)/2.
SOLUTION
U®
First,
determine the location
of
the centroid.
2
sin
(f-
or)
From
Figure 5.8
A:
Similarly
* 3
(f-tf)
2
cos
3
2
(f-«)
_ 2
cos
#
^i=r'i
3
l
(f-
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PROBLEM
5.17 (Continued)
Using
Figure
5.8B,
Y ofan
arc of radius -~(t\ +r
2
)
is
y
=
1
sin(f-a)
=
— (r,
+ r
7
)
2
Vl
2;
(f-or)
1
.
. COS«f
= —(?•,
+7%)
2'
2
'(f-a)
(1)
Now
3
r
2 -n
a
fra-'i)('2
+'i'2
+
'i
2
)
(
f
2~
r
\)i
f
2
+f
\)
Let
>2
=
r,
=
=
r + A
~r-A
Then
/*
=
and
.3
>2
-I
3
.
_
(r + A)
2
+
(i-
+
A)(r
-
A)(r
-
A)
2
1
r
2 -n
2
(r
+ A)
+ (r~A)
3r
2
+A
2
2r
In the
limit
as
A
-
-
(i.e., t]
-
r
2
),
then
-I
2
3
2
=—
x—
(r,
+r
2
)
2 2
So that
F
=
2
3,
.cosa
=
—
x—
(k
+r
7 )
3 4
V)
2
~\~a
-
cosa
.
or Y={r
x
+r
2
)
-~-
A
K~2a
Which
agrees with Equation
CD-
PROPRIETARY MATERIAL, © 2010
The
McGraw-Hill Companies, Inc. Ail rights
reserved.
No
part
of
this
Manual may
be displayed,
reproduced or distributed in any
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or
by
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means,
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written
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/':
PROBLEM
5.18
For
the
area
shown,
determine
the
ratio
alb
for which
x
-
y.
SOLUTION
Then
or
or
Now
X
XZA^XxA
X\
~ab
6
£b
12
X
J
a
rLA^yA
Y
ah
ab
l
15
5
X
=
Y=>
1
2.
—a~~b
2 5
4 X
V
jM
y^
1
loft
3
3
8
1*
5
a
2
b
4
lab
2
5
2
2
1
—a
3
3
6
3
E
—a/;
6
12
£*1
15
a 4
or
—
-
—
-^
/?
5
PROPRIETARY
MATERIAL. ©
2010
The
McGraw-Hill Companies, inc. All rights reserved. A' /«»-/
of
(his
Manual may
be
displayed,
reproduced
or
distributed in any
form
or by
any
means, without the prior
written permission
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publisher, or used
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y
'*»
=
.12 in.
\
is
/
Bl _
PROBLEM
5.19
For
the
semiannular
area
of Problem 5.11,
determine
the
ratio r
2
/t'\
so
thaty
=
3/]/4.
X
SOLUTION
Then
or
Let
(Q
-vn
/T Y
^
]
2
l
3/r
3
2
71
o
2
2
4^
3^-
3
2
X
%4-t)
f('
2W)
YZA^ZyA.
3
7F
i^-^K^-n
3
)
9#
16
^
'i>
P
—[(/>
+
l)(p-l)]==(/>~l)(/>
2
+
/>
+
)
16
or
1
6/7
2
+
(1.6
-
9^)p
+
(1
6
-
9/r)
=
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc.
All
rights reserved. No
part
of
this Manual may be
displayed,
reproduced
or distributed
in
any
form
or by
any means,
without the
prior written
permission
of
the
publisher, or used beyond the
limited
distribution
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PROBLEM
5.19 (Continued)
(1
6
-
9/c)
±
V(l
6
-
9nf
-
4(1
6)(1
6
-
9n)
Then
p
2(16)
or
p
=
-0.5726
p
=
1 .3397
Taking
the positive
root
—
=
1
.340
^
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill Companies,
Inc.
AH
rights
reserved. No
pari
of
this
Manual
may
be
displayed,
reproduced or distributed in
any
form
or
by any
means, without the prior written permission
of
the
publisher, or used beyond
the
limited
distribution
to teachers
andeducators
permitted
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for
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coursepreparation.
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-300 mm
-
12
mm
PROBLEM
5.20
W0&
60 mm
I
'11
1.2 mm
—*\
12 mm
450
min
c
4
I Ij
X
1.2
mm
{«)
(/>)
A composite beam
is constructed by bolting
four
plates to four
60
x
60 x
12-mm angles as
shown.
The
bolts
are
equally
spaced
along
the beam, and
the
beam
supports a
vertical
load.
As
proved
in mechanics of materials,
the
shearing
forces
exerted
on
the
bolts
at
A
and
B
are
proportional
to
the
first moments with respect
to
the centroidal x
axis
of the
red shaded areas
shown,
respectively,
in Parts
a
and
b of
the figure. Knowing
that
the
force
exerted
on the
bolt at
A is 280
N, determine
the
force exerted
on the bolt
at
B.
SOLUTION
22C*w*
From
the
problem statement:
F is
proportional
to
Q
x
.
Therefore:
F„
(Q
x
)a
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1.50
iii.
2.00 in.
4.00 in.
.50
in.
t
,
/U
2.00 in.
0.75
in.
0.75
in.
.1.50
in.
2.00 in.
PROBLEM
5.22
The
horizontal
x
axis
is
drawn through
the
centroid
C
of the
area
shown,
and
it
divides the area into two
component areas A
{
and
A
2
. Determine
the
first moment of each
component area with respect
to the
x axis,
and
explain the results
obtained.
SOLUTION
First
determine
the location of the
centroid
C.
We
have
Then
or
Now
Then
and
A, in.
2
Y>
in.
_/
.
.
3
y
A,
in.
I
2
(ix2x..
5
)
=
3
0,5
1.5
II 1.5x5.5
=
8.25
2.75 22.6875
III
4.5x2
=
9 6.5
58.5
20.25
82.6875
Y LA
=
I,y'A
F'(20.25)
=
82.6875
F'
=
4.0833 in.
Q
x
=^y,A
(Or).
(Q
x
):
2
(5.5-4.0833)in.
[(1.5)(5.5-4.0833)]in.
2
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PROBLEM
5.22
(Continued)
Now
Or
=
(Q,\-
K&:-0)
PROPRIETARY
MATERIAL.
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lights reserved.
M> /wrt
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V
„-
-
(
1
11
}
C
X
c;
J
—
—
*~
PROBLEM
5.23
The
first moment of the shaded
area
with respect
to
the x
axis is denoted
by
Q
x
.
(a) Express
Q
x
in terms of
b, c,
and the
distance
y
from
the
base of
the
shaded
area to
the
x
axis,
(b)
For
what value of
y
is O
x
maximum,
and
what is
that
maximum value?
SOLUTION
Shaded area:
VA4///
97777Z
^(t+y)-y
y,
A--
=
%-^)
c
a
=
=
}H
CL
=
i(c
+
^)[6(c-^)]
^H
(fl) Q
x
=
=
i%
2
/) «
(b) For2
max
:
d
Q.
dy~
-0
or
-b(-2y)
=
Q
;•-
=
()
^
For y~0:
(fir)
=
2
(&)^fc
2
«
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill
Companies,
Inc.
AH
rights
reserved. Wo /wwf o/rfm
Mmim/
j»qv
'6c displayed,
reproduced
or
distributed
in
any
form
or
by
any means, without the prior
written
permission
of
the publisher,
or
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30 mm
30
mm
300mm
PROBLEM
5.24
A
thin,
homogeneous
wire is bent
to
form the
perimeter
of the figure
indicated. Locate
the
center
of gravity
of
the
wire
figure
thus
formed.
-240
mm-
SOLUTION
Dimensions
in
mm
Perimeter ofFigure 5.1
xw -J,
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20
mm
30
mm
36
nun
,t
24
nil )
PROBLEM
5.25
A thin,
homogeneous
wire
is
bent to
form the
perimeter of
the
figure
indicated. Locale
the
center
of
gravity of
the
wire
figure
thus
formed.
SOLUTION
First
note
that
because wire is
homogeneous,
its center of gravity will coincide with
the
centroid
ofthe
corresponding line.
nm
A
nm
A
1
|\$
/,,
mm
x,
mm
y,
mm
xL,
mm
2
yL,
mm
2
1
20 10
200
2
24
20
12 480
288
3 30 35 24 1050
720
®®
4
46.861
35
42
1640.14 1968.16
(l\ V
5
20
1.0 60 200 1200
6 60
30 1800
2
200.86
3570.1 5976.2
Then
=
£xL
1(200,86)
=
3570.1
-ZyL
F(200.86)
=
5976.2
l
=
17.77i
7
=
29.8 i
PROPRIETARY
MATERIAL
©
2010
The
McGraw-Hill
Companies, Inc. AH rights reserved.
No part
of
this
Manual
may
be
displayed,
reproduced
or distributed in
any
form
or by any means, without
the
prior
written permission
of
the publisher, or used beyond
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limited
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/
PROBLEM
5.26
—
l2in.-4*
21 in.
*
t
A
thin,
homogeneous
wire is bent
to
form the
perimeter
of
the
figure
indicated.
Locate
the
center
of gravity of the wire figure thus
formed.
/
1.5 in.
a
SOLUTION
First note that because
the
wire is homogeneous,
its center of gravity will coincide with
the
centroid
of
the
corresponding
line.
.&
©
L, in. x, in.
y,
in.
xL,m.
2
yL, in.
2
1 33
16.5
544.5
2 15
33
7.5
495
112.5
3 21 22.5 15 472.5
315
4
Vl2
2
+15
2
=19.2093
6
7.5
115,256
144.070
X 88.209
1627.26 571.57
Then
and
XZL
=
LxL
X(88.209)
=
1627.26
7(88.209)
=
571.57
or
X
=
18.45 in.
<
or
y=
6.48 in.
4
PROPRIETARY
MATERIAL.
©
2010 The McGraw-Hill Companies, Inc. AH
rights
reserved.
No part
of
this
Manual
may
be
displayed,
reproduced or distributed
in any
form
or
by
any
means,
without the prior
written
permission
of
the
publisher, or
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PROBLEM
5.27
A thin,
homogeneous wire
is
bent to
form the
perimeter
of the
figure
indicated. Locate
the
center
of
gravity of the wire
figure
thus formed.
20
in.
SOLUTION
First
note
that
because the
wire is homogeneous,
its
center of
gravity
will
coincide
with
the
centroid
of the
corresponding line.
K
6
=—
(38in.)
71
L, in.
x,
in.
y,
in. xL,
in.
2
yL,
in.
2
1
18
-29
-522
2
16
-20
8
-320
128
3 20
-10
16
-200
320
4
16 8
128
5
38 19
722
6
^(38)
=
119.381 24.192 2888.1
I
227.38
-320
3464.1
Then
X
liXL
-320
ZL
227.38
£7^,^
3464.1
£/.
~
227.38
X-
-1.407
in.
^
F
=
15.23 in.
^
PROPRIETARY MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc. All rights
reserved. tfo
port o/7/i/a- AAmua/
/«aj>
Ae displayed,
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or distributed
in any
form
or by
any
means, without the prior writ
ten
permission
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publisher, or used beyond the limited
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PROBLEM
5.28
A
uniform
circular
rod
of
weight
8
lb and radius
10
in.
is
attached
to a pin at Cand
to the
cable AB.
Determine
(a)
the
tension
in
the cable,
(b)
the
reaction
at C.
SOLUTION
For quarter
circle
(«)
+)XM
c
=
0:
W
~
-rr
=
2r
__2r
7t
T^W
(81b)
(b)
±^XF
x
=0:
T~C
x
=0
5.09lb-C
r
=0
+\XF
y
=0:
C
y
-W
=
Q
C
y
-&\b
=
d^Xof/j,
f
=
5.0.9 lb
^
C
v
=5.09
lb
—
C,
=
8
lb)
C
=
9.48
lb
^57.5°-*
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc.
All
rights reserved.
No part
of
this
Manual may be displayed,
reproduced or
distributed
in any
farm
or
by
any
means, without
theprior
written
permission
of
the
publisher; or used
beyond the limited
distribution
to teachers and
educatorspermitted
by
McGraw-Hill
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their
individualcoursepreparation.
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PROBLEM 5.29
Member
ABCDE
is
a
component of
a mobile and
is formed
from
a
single
piece
of
aluminum
tubing. Knowing
that
the
member
is supported at
Cand
that
/
=
2m,
determine the distance d so
that
portion
BCD of the member
is horizontal.
SOLUTION
First note
that
for
equilibrium, the
center
of
gravity
of the component must lie on
a
vertical
line through C.
Further,
because
the tubing is
uniform, the
center
of gravity
of
the component
will coincide with
the
centroid
of the corresponding
line. Thus, X
=
H
c
So that
Then
ZxL
=
d
0.75
2
cos55<
mx(0.75
m)
+
(0.75
-d)mx
(1.5
m)
+
1
>
*
(\.5-d)m~\
—
x2
mxcos55
or
(0.75
+
1.5
+
2)^/
(0.75)'
x(2m)
=
cos
55°
+
(0.75)(1
.5)
+
3 or
d
=
0.739
m <
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill Companies,
Inc.
All rights reserved. No part
of
this
Manual
may
be displayed
reproduced or
distributed
in
any
form
or
by
any
means, without
the prior
written
permission
of
the
publisher,
or
used
beyond
the
limited
distribution to
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educators
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PROBLEM
5.30
Member
ABCDE
is a component
of
a mobile
and is formed
from
a
single
piece
of
aluminum
tubing.
Knowing
that
the
member is
supported
at
C
and that
d
is
0.50 m, determine
the
length
/ of
aim
DE
so that this
portion
of
the member is
horizontal.
SOLUTION
First
note that
for equilibrium,
the
center
of
gravity
of
the
component
must
lie
on a vertical line
through C.
Further,
because
the
tubing
is
uniform,
the
center
of gravity
of the component
will coincide
with
the
centroid
of
the
corresponding
line.
Thus,
So that
or
HxL
=
—sin
20°
+
0.5
sin
35°
Imx(0.75
m)
or
+
(0.25
m
x
sin
35°)
x
(1
.5
m)
+
|l.0xsin35°-~]mx(/m)
=
-0.096193
+(sin35»-j)/
=
(xL),
xL)
AB
+(xL)
m
y^jDE
The equation
implies that the
center of gravity
ofDE
must
be
to
the right
of
C.
Then
/
2
-
1
.
1
47
1
5/ +
0.
192386
=
1.14715
±J(-U4715)
2
-4(0.192386)
or
/
or
/
=
0.204
m
Note
that
sin
35°
-
\l >
for
both values
of
/ so both values
are acceptable.
or
/
=
0.943
m
A
PROPRIETARY MATERIAL.
©
2010 The
McGraw-Hill
Companies,
Inc.
All rights
reserved.
No
part
of
this Manual
may
be
displayed
reproduced
or distributed
in any
form
or by
any means, without
the prior written
permission
of
the
publisher,
or used
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PROBLEM
5.31
B
W
W:»>^
Id
?
^
The
homogeneous
wire ^4^C is bent
into a
semicircular arc
and a
straight section
as
shown and
is attached
to a
hinge at A. Determine
the value
of 9
for which,
the
wire
is
in
equilibrium
for
the
indicated position.
\y
Cc
:C^
SOLUTION
First note
that
for equilibrium, the
center
of
gravity
of the
wire
must
lie on
a
vertical line through^.
Further,
because the
wire
is homogeneous, its
center
of
gravity
will
coincide
with the centroid
of
the
corresponding
line. Thus,
So
that
Then
or
X
=
ZxL
=
Q
'
'J
V
rto^B
rcosO
](/')+
rcos0
\(xr)-0
cos
6
4
\A-l7t
0.5492.1
or
#
=
56.7°
<
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill
Companies, Inc.
All rights
reserved.
A'f>
part
of
this Manual
may
be
displayed,
reproduced
or distributed in any
form
or by
any
means, without
the
prior written
permission
of
the
publisher,
or
used
beyond
the
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educators
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for
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individualcourse
preparation.
If
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:
—
kb-
/,-
li'
PROBLEM
5.32
Determine the
distance
h
for which
the centroid
of the
shaded
area
is
as far
above
line
BB'
as
possible
when
(a)
k
~
0.10,
(/?)*
=
0.80.
SOLUTION
kkbJ
A
y
yA
1
ha
2
i.
—a
3 6
2
~(kb)h
~h
3
-~kbh
2
6
£
~(a-kh)
t{a
2
~~kh
2
)
6
Then
or
and
or
YZA
=
LyA
~{a-kh)
2
r
-(a
2
-Aft
2
)
6
«
2
-
A/?
2
dY
1 -2*%
-
Mr)
-
(a
2
-
*ft
2
)(-*)
(1)
dh
3
2h(a-kh)-a
2
+kh
2
=0
Simplifying
Eq.
(2)
yields
k/r-2ah
+
a
2
^0
(a-khy
(2)
PROPRIETARY MATERIAL. ® 2010
The
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part
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may
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by
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PROBLEM 5.32
(Continued)
T
,
. 2a±J(-2a)
2
-4(k)(a
2
)
Then h
~
—
:
2k
=f[>^]
Note
that only
the
negative root is acceptable since
h
<
a. Then
(a)
k
=
0.10
O.IOL
J
(b)
k
=
0.80
I-Vl-0.8'
0.80
or
A
=
0.5
13a ^
or
//
=
0.691a
^
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill
Companies, Inc. All
rights reserved.
No pari
of
this Manual may
be
displayed,
reproduced
or distributed in
any
form
or by
any
means,
without
the
prior written
permission
of
the
publisher, or
used
beyond the limited
distribution to
teachers
and
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7
\
:':
:\
f
h
~
/,
B'
PROBLEM
5.33
Knowing
that
the
distance
h
has been selected to maximize the
distance
y
from line
BB'
to
the
centroid
of
the
shaded area,
show
that
y
~
2h/3.
SOLUTION
See solution
to Problem 5.32
for analysis
leading
to
the
following equations:
y_a
2
-kh
2
Xa-kh)
(1)
2h(a-kh)~a
2
+kh
2
=0
(2)
Rearranging Eq.
(2)
(which defines the
value
of/?
which
maximizes
/)
yields
a
2
-kh
2
=2h{a~kh)
Then
substituting
into
Eq.
(1)
(which
defines Y)
f
=
-—
x2h(a-kh)
3(a~kh)
or
F
= -/?
<
3
PROPRIETARY MATERIAL. ©
2010 The
McGraw-Hill
Companies, Inc. All
rights
reserved. No
part
of
this Manual may be
displayed
reproduced
or
distributed
in
any
form
or by any means,
without
the
prior
written
permission
of
the
publisher,
or
used
beyond
the
limited
distribution
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andeducators
permitted
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course
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PROBLEM
5.35
Determine
by
direct
integration
the
centroid
of the
area
shown. Express
your
answer
in terms
of
a and
h.
SOLUTION
At {a,
h)
or
or
Now
and
Then
and
)\
: h
=
-~ka
A
k
=
h
~~
2
jv
A
=
~
ma
m-
h
a
yEL
d.A
(^+^2)
(y
2 ~yO^
h
f
2
W
—(ax
—
x
)ax
h
h ,
—x
—
~x
a
a
clx
A=\dA^
VJL{ax-x
2
)dx
\*ElM
—(ax-x
)dx-~
J
I
iJliil
a*
a
2
1
3
~X
XT
2
3
_
—
—ah
6
a
3
1
4
—x
x
3
4
o
JjO'i
+
j
2
X0'
2
-*)*]
=
Ji(
2
2
-
yf)dx
=
~a
l
h
12
PROPRIETARY
MATERIAL
©
2010
The
McGraw-Hill
Companies,
Inc. All
rights
reserved.
A'o part
of
this Manual
may
be
displayed
reproduced
or distributed
in
any
form
or
by
any
means, without
the prior
written
permission
of
the publisher, or
used
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distribution
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andeducatorspermitted
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PROBLEM 5.35
(Continued)
xA~ \x
Fl
dA:
x -ah
.6
j
=
-a
2
h
1.2
x =—a
^1
2
yA^p
EL
d.A\
y
—ah
1
/2
—-aft
15 5
PROPRIETARY
MATERIAL. ©
2010
The
McGraw-Hill
Companies, Inc. AH
rights reserved. No
part
of
this
Manual
may
be
displayed,
reproduced
or distributed in any form
or by
any
means,
without the prior
written
permission
of
the
publisher, or
used
beyond
the
limited
distribution
to
teachers
andeducators
permittedby
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individual
course
preparation.
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are a student
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il
7
W
t
\
PROBLEM
5.36
Determine
by
direct
integration
the
centroid
ofthe
area
shown. Express
your
answer
in terms
of
a and h.
SOLUTION
For
the
element
(EL)
shown
At
Then
Now
Then
and
Hence
x-a,
y-h: h
=
ka
2
or k
1/3
x
=i*y
dA^xcfy
=
-^-y
m
dy
x
EL
=-x
2 2
h
m
y
1/3
\x
El
dA=l
4
h
m
m
-ah
-.Wf
-fL
V
W
di
,]
==
l^_f
3
^
A'
ft
2A»*1^
^J-I^-ll*
J^-f^y°*l=^
TO
X//
\x
IiL
dA\
h
m
\l
x\ —ah
\
=
—
2
/?
4
J
JO
-ah
2
7
y
A
~\yEi
dA
'-
y
'|
--«/?
\
=
:
-ah'
—
**.
u
r
NX\*\\\
i
m
i
*~*
-»
*
10
X
=
—
tf
7
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc.
All rights
reserved.
A'o/ra/-/
f
>/7//w M»»/a/
iimiv
/« displayed,
reproduced
or
distributed
in
any
form
or
by any
means, without
the prior
written permission
of
the
publisher,
or
used beyond
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a-
b-
PROBLEM 5.37
Determine
by
direct
integration
the
cenlroid
of
the
area
shown.
SOLUTION
For
the element
(EL) shown
and
Then
To
integrate, let
Then.
and
h
r~
i
a
dA^(b-y)dx
=
t
a
JCe-i
=X
\a-4a
2
-x
2
}
dx
-dx
Mel.
V
£L
1
yi:i^-(y
,+b
)
a
+ yj
la
2 ..2
a
~x
b
o
a
a~\a -x
\dx
x
=
a
sin
9:
*Ja
2
-
x
2
=
a
cos
6,
dx~a
cosOdd
pan
f}
A=
—
{a-
a
cos
9){acos$d$)
Jo
a
a
sin
-
a
—
+
sin
/r/2
£?6
1
tt
K^
=
r
a -4
2
-x
2
)dx
—
*-+-(«
-x
)
2\3/2
/r/2
*>»
PROPRIETARY
MATERIAL. © 2010
The
McGraw-Hill
Companies, Inc.
Ail
rights reserved. No
part
of
this
Maiiual
may be
displayed,
reproduced or
distributed in
any
form
or by
any
means, without the
prior
written
permission
of
the
publisher,
or
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PROBLEM
5,37
(Continued)
f^-ri(«
+
^[£(-^*
2
(
3^
a:
2a
6
2^
v
3
;
xA
=
\x
EL
dA:
x
yA=jy
EL
cM:
y
ah
ab\
1
7t
71
1
2
a'h
1
,2
ab'
—
2a
.
or
x
=
^
3(4-^)
2/?
or
y
=
^
3(4
-/r)
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc.
All
lights
reserved. JVe port
o/fAfo Manual
may be.
displayed,
reproduced
or distributed in
any
form
or by
any
means, without
the prior
written
permission
of
the publisher,
or used
beyond
the
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distribution
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andeducators
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,/\
PROBLEM 5.38
Determine
by
direct
integration the
centroid of
the
area shown.
:
ss
SOLUTION
First note that
symmetry implies
A
For
the
element (EL)
shown
2r
Then
and
y
E[
=
—
(Figure 5.8B)
n
dA
-
fcrdr
A-
\dA~
|
rcrdr-K
2
2/
( 2\
r
,
2
,
It
2
2
r
2
~
r
\
\n,,A-\l^r
dr)
=
2\f
--(
/
3
h ~n
So
yA~\y
E
iM'-
y -(^-'f)
=
-^-n»
or
^^
—
—
^
PROPRIETARY
MATERIAL.
© 2010 The
McGraw-Hill Companies,
Inc. All
rights reserved.
No part
of
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Manual may be
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distributed in
any
form
or by
any
means,
without
(he prior written
permission
of
the
publisher,
or
used
beyond
(he
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y
=
kix-(ir-
PROBLEM
5.40
Determine
by
direct
integration the
centroid
of the
area shown. Express
your answer
in terms of
a
and
b.
SOLUTION
At
Then
Now
0, y
=
b
k(0-a)
2
or
k.
x
b
v
=
\ix-a?
y
h
f
^2
-
=
—
(x-a)
-X-
M«L
and
cIA
=
ydx
=
—
—
(x
—
a)' dx
Then
and
A-fr-fa-faMx-af]
=
~ab
o
3
K^r
—{x-d)
2
dx
a
a
f
x
3
-
lax
2
+
a
2
x)dx
a
bix*
2
3
^
a
2
1
2,
21
4
3 2
12
K
clA
Jo
2a
2
1
,2
—
aft
-^(Jc-a)
2
^
2a'
7*-*
-0
Hence
xA
—
vA
—ah
3
J 12
'a
X/^dA:
x
\ymdA:
y[~abU-^ab
\ A
1
.2
4
'
10
PROPRIETARY
MATERIAL. © 20JO
The
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of
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Manual
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by
any
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Vi
=
h^
PROBLEM
5.41
Determine
by
direct
integration the
centroid
of the
area shown.
Express
your answer
in
terms ofa and
b.
SOLUTION
y
l
-k
]
x
2
but
b-k
x
a
2
y
l
=~x
2
y
2
-k
2
x
4
but
b~k
2
a
4
y
2
dA
=
(y
2
-y
x
)dx
b
.2
*
4\
dx
X
EL
~
X
yEL^-iyi+yi)
h^v
'
t
4\
X
—
dx
\-,M=[
4 6
X
X
4
6a
12
-a
l
b
PROPRIETARY MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc. All
rights reserved. No
part
of
this Manual may be displayed,
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or distributed in
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or by any means, without
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prior
written
permission
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or used beyond
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PROBLEM
5.41
(Continued)
jy^A=[
xA
=
x
F
,
dA
:
x
—
ha
,
J
LL
\
15
J
12
a
l
h x
=—a
yA=\y
£
i.dA:
y
r
2,\
2
,
2
15
ba
45
aff y^-b
<
3
PROPRIETARY
MATERIAL.
© 2010 The McGraw-Hill Companies,
Inc. All rights
reserved. No
part
of
this Manual
may be
displayed,
reproduced or
distributed
in
any
form
or
by
any
means, without
the
prior written permission
of
the
publisher,
or used beyond
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and
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A
^-i
+
i3
PROBLEM
5.42
Determine
by direct
integration
the
centroid
of
the
area
shown.
SOLUTION
We have
Then
and
/'*/
*
(
y
*-°V l.
dA
=
ydx
=
a
.2
\
X X
I
+
^r
f
2\
X
X
t/x
8
2L
f
X
X^
a\
1
+
—
dx
—
a
3
-ah
K^
=
r
f
2>
X X
l-
+ dx
~a
^
L
1/j
x
2
x
3
—
+
—
2/.
3.L
2
-.2/.
2 3.1
+
4/.
2
n2A
«/;
a*
^6L
Hence
x/1
ril-2::
+
3-
r
2\
.
X
X
c/x
3
4^
z?
/
4
,
cfir
(3
T
,.2 .3
4
5
X X X
X
x
4.
_
_
+
n2/.
I
7/ 2I
j
St
5
\x
EL
dA:
-aL
y
A
^\yr-iM-
y\-
3
11
x=~L
4
4
_ 33
.
y
=
—a
^
40
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc. AH
rights
reserved. Afo
/««•/
o/'rfe Manual
may be displayed,
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or distributed in any
form
or by any
means, without
the
prior
written
permission
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the
publisher, or used beyond
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x
=
kij*
%-
.'
b
,2
b
2
a
6.
X
PROBLEM
5.43
Determine
by
direct
integration the
centroid
of the
area,
shown.
Express
your
answer in terms of
a
and
b.
SOLUTION
For_y2 at
Then
Now
and for 0
8/17/2019 Cap 5, Novena Edc_text.pdf
53/196
PROBLEM
5,43
(Continued)
and
[x
Fl
dA~
\
l2
x\b~chU
f
.
J
J
Jn
Ja/2
yfa
a
2
j
dx
-ia/2
+
b
2
x x x
+
2 b
5*fa
+b\—
l
-
240
f ^
5/2
K^-J
5
V«
3a
4
/ \
5/2
+ (a)
5/2
00
3
«
V
1
+
—
)
4
all
(*-if
/>* -£
,/2/,
x
'/2
x
2
b—f=rdx
c
a
b\
x
1 x
+
+
-
•>«/2
2
a 2
^
1/2
\
Ja
a 2
b
2
\\
>'
on
/>2
f
—XT +
—
2a
2 2
V
x
2
1
(x
2a
3a
\
a
2
dx
,3\
all
b_
4a
'
a
)
>
a
6a{
2
2
Hence xA
-afr
48
^,,
:i
dA:
13
/>
7l
2,
—
a/>
=
a
b
24
J
240
*=—
a
=
0.546a
^
130
yA^
jy
EL
dA:
y\~ab
11
/2
—ao
48
y=—
6
=
0.423/;
^
26
PROPRIETARY MATERIAL.
©
2010 The
McGraw-Hill
Companies, Inc.
All rights
reserved. A'o
/>«/•/
o/rAra
A/
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54/196
PROBLEM
5.44
Determine
by
direct
integration the
centroid of
the
area
shown.
Express
your
answer in terms
of
a
and
b.
SOLUTION
For
V]
at
Then
By
observation
Now
and
for
0S.vEL
—
y
2'
a
—x
and dA
—
y
v
dx
—
—rx
dx
—y
2
=—\
2——
I and dA=
y
2
dx
—
b\
2
—
-
\dx
2 2
{
a
\
a
A
=\
dA
=&
d
**t
b
V
dx
26
a
+
b 2
—
-\2a
-0
ab
2b_
*>
a
I
2
M-«?^hD
4
+
6
/;|
2--|rfx
n2«
3a
-0
a
2
b + b\[(2a)
2
-(a)
2
]
+ ~-[(2a
2
)-(a)
7
2/
~a
b
6
'**».
PROPRIETARY
MATERIAL
©
2010 The
McGraw-Hill
Companies, Inc.
All rights reserved.
No
part
of
this Manual may be
displayed,
reproduced or distributed
in
any
form
or by any means, without
the
prior written permission
of
the
publisher,
or used
beyond
the
limited
distribution to teachers
and
educatorspermitted
by
McGraw-Hill
for
their
individual coursepreparation.
If
you
are a
student using this Manual,
you
are using it without permission.
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PROBLEM 5.44
(Continued)
jy£L
8/17/2019 Cap 5, Novena Edc_text.pdf
56/196
PROBLEM 5.45
A homogeneous
wire is bent into the
shape
shown. Determine by direct
integration
the* coordinate
of
its
centroid.
SOLUTION
First note that because
the
wire
is homogeneous, its
center
of
gravity coincides with
the centroid of the
corresponding
line
Now
Where
Then
x
EI
-
a cos
3
and
dL
—
^jdx
2
+
dy
2
x~a
cos
3
$: dx
=
~3a cos
2
sin
9d9
y-asur'
9:
dy
-
3a
sin
2
9 cos Od'6
dL
=
[(-3a
cos^
9
sin
0d6f
+
(3a
sin
z
9
cos
OdGf
]
=
3a cos
9
sin
0(cos
2
+ sin
2
0)
m
d9
2
-,1/2
3a cos
9 sin Odd
L
-
\d'L
-
3a cos
9 sin
9d9
~
3a
3
1
9
-sin
2
2
nil
-0
and
f- C*
n
3
x
EL
dL— \
a
cos~
(3acos
sin
0d0)
3a
2
-\7ll2
cos
5
-0
-la*
5
Hence
Alternative
Solution
xL
=
\x
E/
dL:
-I
3
x\
—a
2
x
=
acos~
9=>cos'
:
y
=
asm 9^>sm'
V
J
2/3
x =—a
^
5
x2/3
/
x2/3
^
+U
=1
or y~(a
m
-x
m
f
2
PROPRIETARY MATERIAL. © 2010
The
McGraw-Hill Companies,
inc.
AH rights reserved. No part
of
this Manual
may be
displayed,
reproduced
or
distributed in
anyform
or
by
any means,
without
the
prior
written
permission
of
the
publisher, or used
beyond
the limited
distribution to teachers andeducatorspermitted by McGraw-Hill
for
their
individual
coursepreparation.
If
you are
a
student
using
this Manual,
you are using
it
without permission.
594
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PROBLEM
5.45 (Continued)
Then
Now
dx
*s=*
and
*-Hi
dx
J\
+ [(a
m
-
x
m
f\~x-
m
)]
2
r
dx
Then
p-r
1/3
1/3
dx~a
If
2
and
Hence
5
PROPRIETARY
MATERIAL.
£5
2010
The
McGraw-Hill Companies,
Inc.
AH rights reserved. No
part
of
this Manual may
be
displayed,
reproduced or
distributed in
any
form
or
by
any
means, without
the
prior
written
permission
of
the publisher, or used beyond the
limited
distribution to teachers and
educatorspermittedby McGraw-Hill
for
their individualcourse
preparation.
If
you
are a
student using
this Manual,
you are
using
it
withoutpermission.
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PROBLEM
5.46
A
homogeneous
wire is
bent
into
the
shape
shown.
Determine
by
direct
integration
the x
coordinate of its centroid.
SOLUTION
First
note
that because
the
wire
is
homogeneous,
its center
of
gravity
coincides
with
the
centroid of
the
corresponding line
Now
Then
and
x
El
~rcos$
and di~rdd
r
el*'
4
-,
/, ^
Thus
•7/T/4
\x
EL
dL= r
cos
0(rdO)
rising
nIA
2
i_
=
-r
2
V2
xL- YxdL:
x
—
itr -~r
2
v2
_
2V2
3tf
PROPRIETARY
MATERIAL
©
2010 The
McGraw-Hill
Companies, Inc. All
rigiiis
reserved.
No
part
of
this Manual may be displayed,
reproduced
or distributed in any
form
or by
any means, without the prior
written
permission
of
the
publisher, or
used
beyond the limited
distribution
to
teachers
andeducatorspermitted by McGraw-Hill
for
their individual
coursepreparation.
If
you are a student using
this
Manual,
you
are
using it without permission.
596
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y
=
kx~
PROBLEM
5.47*
A homogeneous
wire is bent
into
the
shape
shown.
Determine
by
direct
integration
the a-
coordinate
of
its
centroid.
Express your
answer
in
terms of
a.
SOLUTION
First
note that because
the
wire
is
homogeneous,
its
center
of gravity
will coincide
with the centroid of
the
corresponding line.
We
have at
Then
and
Now
and
Then
and
x
=
a,
y
=
a
a
-
ka~
or k
1
3/2
-4
a
dy
=
3
dx
2^a
1/2
dL~J\
+
1
+
1
\dx
j
dx
f
3
V
->
.1/2
X
1/2
dx
24a
a + 9x
dx
L= |rfL
=
r'-^~^a
+
9xdx
J Jo
l4a
\
24a
a
27
1.4397
k
-x-(4a
+
9xf
2
3 9
V }
[(13)
3/2
-8]
M-f
yl4a +
9xdx
PROPRIETARY MATERIAL.
©
2010 The
McGraw-Hill
Companies,
Inc. All
rights
reserved. No part
of
this
Manual
may
be
displayed
reproduced or distributed in any
form
or
by
any
means, without
the
prior
written
permission
of
the
publisher,
or
used beyond the limited
distribution
to teachers and
educatorspermitted
by
McGraw-Hill
for
their
individual
coursepreparation.
If
you
are a student
using this
Manual,
you are
using it withoutpermission.
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PROBLEM
5.47*
(Continued)
Use integration
by
parts
with
u
—
x
dv~
\}4a
+
9x
dx
da
~dx v
-
—
(4a
+
9x)
27
3/2
Then
K
dL
xx—(4a
+
9xf
2
27
-0
{
l
'~(4a
+
9x)
i/2
dx
]
Jo
27
(4a
+
9x)
5/2
i3)
3/2
al
1_
27
°
27>/al45
27j
(t3)
3/2
~^[(13)
5/2
-32]
=
0.78566a
2
xL
=
\x
EL
dL
:
x(\
.4397
la)
=
0.78566a
2
or
x
=
0.546a
A
PROPRIETARY MATERIAL. ©
2010 The
McGraw-Hill
Companies,
Inc.
AH rights reserved. No part
of
this
Manual
may be
displayed,
reproduced or distributed in
any
form
or by
any
means, without the prior
written
permission
of
the publisher, or used beyond
the
limited
distribution
to
teachers
andeducators
permitted
by
McGraw-Hill
for
their
individual
course
preparation.
If
you
area
student using this Manual,
you
are
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598
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T
XX
l.--'.--.V:0
;
.
\
-«
H-<
»
L L
PROBLEM
5.48*
Determine
by
direct
integration
the
centroid
of the
area
shown.
SOLUTION
We have
and
Then
and
X
EL=
X
I a
kx
y
~
—
y-—
COS
LL
2
2
21,
KX
dA
=
ydx
=
a
cos
dx
21
./4
=
IcW
=
8/17/2019 Cap 5, Novena Edc_text.pdf
62/196
PROBLEM
5.48*
(Continued)
Also
\y
KI
M-[
xA
112
a
Ttx
—cos
—
2 2L{
21,
....
.
JIX
cos
— a cos
—
ax
. 2
4
In
0.20458^
2
I
K
(
dA
:
x
s
ah
v
*
j
yA~
jy
gL
dA:
4i
71
ah
-0.106374a/:
0.20458
A
or x=
0.2361
A
or
y
-
0.454a ^
PROPRIETARY
MATERIAL.
©
2010 The McGraw-Hill
Companies,
Inc.
All rights reserved.
AV;
yx/r/ o/ /Mv
M»wa/
»mj> Ae displayed,
reproduced
or distributed
in any
form
or
by
any
means,
without
the prior
written
permission
of
the publisher,
or
used beyond the limited
distribution to teachers
and
educators
permitted
by McGraw-Hill
far
their individual coursepreparation.
If
you are
a
student using
this Manual,
you
are using it
without
permission.
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PROBLEM
5.49*
Determine
by direct integration
the
centroid
of
the
area shown.
SOLUTION
We have
2 2
x
FL
-~rcosO
=—
ae°
cos
6
2 2
V
=
r sin $
—
~-ae°
sin
6
L
3
3
and
Then
dA
=
~{r){rd&)
=
-a
2
e
26
d0
A
M
?
n]
~a
2
e
ld
d0^a
2
2 2
—e
2
2ff
-0
a
2
{e
2
*-\)
and
=
133.623«
2
r*2
jx
EL
dA^
j*-ae
d
cos&\
~a
2
e
2O
d0
ro
ia
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