Solution manual vector mechanics engineers dynamics 8th beer chapter 05
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Chapter 5, Solution 1.
\
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
200 × 150 = 30000
−100
250
− 30 000000
6 750 000
2
400 × 300 = 120000
200
150
24 000 000
18000000
Σ
150 000
21000 000
24 750000
X =
ΣxA 21 000000
=
mm
ΣA
150000
or X = 140.0 mm
Y =
ΣyA 24 750000
=
mm
ΣA
150 000
or Y = 165.0 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 2.
A,in 2
x ,in.
y ,in.
xA,in 3
yA,in 3
1
10 × 8 = 80
5
4
400
320
2
1
× 9 × 12 = 54
2
13
4
702
216
Σ
134
1102
536
Then
X =
ΣxA 1102
=
ΣA
134
and
Y =
ΣyA 1102
=
ΣA
134
or
X = 8.22 in.
or Y = 4.00 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 3.
Then
A, mm 2
x , mm
xA, mm3
1
1
× 90 × 270 = 12 150
2
2
( 90 ) = 60
3
729 000
2
1
× 135 × 270 = 18 225
2
Σ
30375
X =
90 +
ΣxA 3189375
mm
=
ΣA
30375
1
(135) = 135
3
2 460 375
3 189 375
or X = 105.0 mm
For the whole triangular area by observation:
Y =
1
( 270 mm )
3
or Y = 90.0 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 4.
A,in 2
x ,in.
1
1
( 21)( 24 ) = 252
2
2
(13)( 40 ) = 520
Σ
2
( 21) = 14
3
21 +
1
(13) = 27.5
2
xA,in 3
y ,in.
40 −
1
( 24 ) = 32
3
20
772
Then
yA,in 3
3528
8064
14 300
10 400
17 828
18 464
X =
ΣxA 17828
=
in.
ΣA
772
or
Y =
ΣyA 18464
=
in.
ΣA
772
or Y = 23.9 in.
X = 23.1 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 5.
A, mm 2
x , mm
2
1
π ( 225 )
2
1
( 375)( 225) = 42 188
2
4
−
= 39 761
4 ( 225 )
3π
= − 95.493
125
y , mm
xA, mm3
yA, mm3
95.493
− 3 796 900
3 796 900
5 273 500
3 164 100
1 476 600
6 961 000
75
81 949
Σ
Then
X =
ΣxA 1476600
mm
=
ΣA
81 949
or X = 18.02 mm
Y =
ΣyA 6961 000
mm
=
ΣA
81 949
or
Y = 84.9 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 6.
1
2
3
−
π
4
−
A,in 2
x ,in.
y ,in.
xA,in 3
yA,in 3
17 × 9 = 153
8.5
4.5
1300.5
688.5
2
× ( 4.5 ) = −15.9043 8 −
π
4
( 6 )2 = − 28.274
Σ
4 × 4.5
4 × 4.5
= 6.0901 9 −
= 7.0901 − 96.857
3π
3π
−112.761
− 298.19
−182.466
905.45
393.27
10.5465
6.4535
108.822
Then
X =
ΣxA
905.45
=
ΣA 108.822
and
Y =
ΣyA 393.27
=
ΣA 108.22
or
X = 8.32 in.
or Y = 3.61 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 7.
A,in 2
1
π (16 )
4
2
= 201.06
2
− ( 8 )( 8 ) = − 64
Σ
137.06
ΣxA 1109.32
=
in.
ΣA
137.06
Then
X =
and
Y = X by symmetry
x ,in.
4 (16 )
3π
= 6.7906
4
xA,in 3
1365.32
− 256
1109.32
or
X = 8.09 in.
or Y = 8.09 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 8.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
35 343
63.662
0
2 250 006
0
2
− 4417.9
31.831
− 31.831
−140 626
140 626.2
Σ
30925.1
2 109 380
140 626.2
Then
X =
ΣxA 2109 380
=
ΣA
30 925.1
and
Y =
ΣyA 140 625
=
ΣA
30 925.1
or
X = 68.2 mm
or Y = 4.55 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 9.
A
−
1
Therefore, for X =
4
π
2
Σ
π2
2
π
( 2r
4
2
2
r12
r22
− r12
4r1
3π
r13
π 2 4r1
− r1
=−
3
4 3π
4r2
3π
π 2 4r2 2r23
r2
=
3
2 3π
1
2r23 − r13
3
(
)
ΣxA 4r1
=
:
ΣΑ
3π
(
(
or
xA
)
)
4 2r23 − r13
4r1
=
3π
3π 2r22 − r12
or
x
π =
)
r 3
r13 2 2 − 1
r1
4
=
2
r
3π
r12 2 2 − 1
r1
r
2ρ 3 − 1
, where ρ = 2
2
r1
2ρ − 1
2 ρ 3 − 2πρ 2 + (π − 1) = 0.
Solving numerically for ρ and noting that ρ > 1:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
r2
= 3.02
r1
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Chapter 5, Solution 10.
First, determine the location of the centroid.
y2 =
From Fig. 5.8A:
=
y1 =
Similarly
Then
Σ yA =
(π
2 sin 2 − α
r2 π
3
−α
2
(
2
cos α
r2 π
3
−α
2
)
(
)
(
)
2
cos α
r1 π
3
−α
2
2
cosα
r2 π
3
−α
2
(
)
(
)
A2 =
A1 =
( π2 − α ) r12
( π2 − α ) r22 − 23 r1
(
cosα
−α
2
π
)
2 3
r2 − r13 cosα
3
π
π
Σ A = − α r22 − − α r12
2
2
=
and
π
= − α r22 − r12
2
Y Σ A = Σ yA
(
Now
( π2 − α ) r22
)
π
2 3
Y − α r22 − r12 =
r2 − r13 cos α
2
3
(
)
Y =
(
)
2 r23 − r13 cos α
3 r22 − r12 π2 − α
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
)
( π2 − α ) r12
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Using Figure 5.8B, Y of an arc of radius
1
( r1 + r2 ) is
2
Y =
=
(π
sin − α
1
( r1 + r2 ) π 2
2
−α
2
(
)
)
1
cos α
(r1 + r2 ) π
2
−α
2
(
(
( r2 − r1 ) r22 + r1 r2 + r12
r23 − r13
=
r22 − r12
( r2 − r1 )( r2 + r1 )
Now
=
(1)
)
)
r22 + r1 r2 + r12
r2 + r1
r2 = r + ∆
Let
r1 = r − ∆
r =
Then
1
( r1 + r2 )
2
2
and
( r + ∆ ) + ( r + ∆ )( r − ∆ ) + ( r − ∆ )
r23 − r13
=
2
2
r2 − r1
(r + ∆) + (r − ∆)
=
2
3r 2 + ∆ 2
2r
In the limit as ∆ → 0 (i.e., r1 = r2 ), then
r23 − r13
3
= r
2
2
2
r2 − r1
=
so that
Y =
3 1
× (r1 + r2 )
2 2
2 3
cos α
× ( r1 + r2 ) π
3 4
−α
2
Which agrees with Eq. (1).
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
or Y =
1
cos α
!
( r1 + r2 ) π
2
−α
2
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Chapter 5, Solution 11.
Then
X =
A,in 2
x ,in.
xA,in 3
1
27
8.1962
221.30
2
15.5885
3.4641
54.000
3
−18.8495
3.8197
−71.999
Σ
23.739
ΣxA 203.30
=
ΣA
23.739
203.30
or
X = 8.56 in.
and by symmetry
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Y =0
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Chapter 5, Solution 12.
1
2
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
( 240 )(150 ) = 18 000
2
160
50
2 880 000
900 000
3
( 240 ) = 180
4
3
(150 ) = 45
10
−2160000
−540 000
720 000
360 000
−
1
( 240 )(150 ) = 12 000
3
6000
Σ
Then
X =
ΣxA 720000
=
mm
ΣA
6000
Y =
ΣyA 360000
=
mm
ΣA
6000
or X = 120.0 mm
or
Y = 60.0 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 13.
A,in 2
x ,in.
y ,in.
1
(18)(8) = 144
−3
4
− 432
576
2
1
( 6 )( 9 ) = 27
2
2
−3
54
−81
−5.0930
−3.8197
− 432.00
− 324.00
−810.00
171.00
3
Σ
Then
π
4
(12 )( 9 ) = 84.823
255.82
X =
ΣxA −810.00
=
in.
255.82
ΣA
Y =
ΣyA 171.00
=
in.
ΣA
255.82
xA,in 3
or
yA,in 3
X = − 3.17 in.
or Y = 0.668 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 14.
X = 90 mm
First, by symmetry
1
2
−
3
−
A, mm 2
y , mm
yA, mm3
(180 )(120 ) = 21 600
60
1 296 000
π
4
π
4
( 90 )(120 ) = − 8482.3
120 −
4 × 120
= 69.070
3π
−585 870
( 90 )(120 ) = − 8482.3
120 −
4 × 120
= 69.070
3π
−585 870
4635.4
Σ
Y =
ΣyA 124 260
=
4635.4
ΣA
124 260
or Y = 26.8 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 15.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
18 240
−4
12
72 960
218 880
2
−1920
− 56
54
107520
−103 680
3
− 4071.5
− 41.441
− 41.441
168 731
186 731
Σ
12 248.5
−134171
−53 531.1
Then
and
X =
ΣxA −134171
=
ΣA
12 248.5
Y =
ΣyA −53 531
=
ΣA 12 248.5
or
X = −10.95 mm
or Y = − 43.7 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 16.
\
A, mm 2
2
( 200 )( 200 ) = 26 667
3
1
−
2
2
(100 )( 50 ) = − 3333.3
3
23 334
Σ
xA, mm3
yA, mm3
x , mm
y , mm
75
70
2 000 000
1866 690
37.5
− 20
−125 000
66 666
1875 000
1 933 360
Then X =
ΣxA 1875 000
=
mm
ΣA
23 334
or X = 80.4 mm
Y =
ΣyA 1 933 360
=
mm
ΣA
23 334
or Y = 82.9 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 17.
Locate first Y :
Note that the origin of the X axis is at the bottom of the whole area.
A, in 2
Y =
yA, in 3
1
8 × 15 = 120
7.5
900
2
− 4 × 10 = − 40
8
− 320
Σ
Then
y , in.
80
580
ΣyA 580
=
= 7.2500 in.
ΣA
80
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Now, to find the first moment of each area about the x-axis:
Area I:
QI = ΣyA =
7.75
5.75
− ( 4 × 5.75 ) ,
(8 × 7.75) +
2
2
or QI = 174.125 in 3 !
Area II:
QII = ΣyA = −
7.75
4.25
− ( 4 × 4.25 ) ,
(8 × 7.25) −
2
2
or QII = −174.125 in 3 !
Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 18.
A, mm 2
Y =
yA, mm3
1
(80 )( 20 ) = 1600
90
144 000
2
( 20 )(80 ) = 1600
40
64 000
Σ
Then
y , mm
3200
208 000
ΣyA 208 000
=
= 65.000 mm
ΣA
3200
Now, for the first moments about the x-axis:
Area I
QI = ΣyA = 25 ( 80 × 20 ) + 7.5 ( 20 × 15 ) = 42 250 mm3 ,
or QI = 42.3 × 103 mm3 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Area II
QII = ΣyA = − 32.5 ( 20 × 65 ) = 42 250 mm3 ,
or QII = 42.3 × 103 mm3 !
Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 19.
(a) With Qx = Σ yA and using Fig. 5.8 A,
(
)
2 r sin π − θ
r 2 π2 − θ −
Qx = 3 π 2
−
θ
2
2
= r 3 cos θ − cos θ sin 2 θ
3
(
) ( 32 r sin θ ) 12 × 2r cos θ × r sin θ
(
)
or Qx =
(b) By observation, Qx is maximum when
and then
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
2 3
r cos3 θ
3
θ =0
Qx =
2 3
r
3
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Chapter 5, Solution 20.
From the problem statement: F is proportional to Qx . Therefore:
FA
FB
=
, or
( Qx ) A ( Qx )B
FB =
( Qx )B
F
( Qx ) A A
For the first moments:
Then
( Qx ) A
12
= 225 + ( 300 × 12 ) = 831 600 mm3
2
( Qx )B
12
= ( Qx ) A + 2 225 − ( 48 × 12 ) + 2 ( 225 − 30 )(12 × 60 ) = 1 364 688 mm 3
2
FB =
1364688
( 280 N ) ,
831600
or FB = 459 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 5, Solution 21.
Because the wire is homogeneous, its center of gravity will coincide with the centroid for the
corresponding line.
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
400
200
0
80 000
0
2
300
400
150
120 000
45 000
3
600
100
300
60 000
180 000
4
150
− 200
225
− 30 000
33 750
5
200
−100
150
− 20 000
30 000
6
150
0
75
0
11 250
Σ
1800
210 000
300 000
Then
X =
ΣxL
210 000
=
= 116.667 mm
ΣL
1800
or X = 116.7 mm
and
Y =
ΣyL 300 000
=
= 166.667 mm
ΣL
1800
or Y = 166.7 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 22.
L, in.
x , in.
y , in.
xL, in 2
y , in 2
1
19
9.5
0
180.5
0
2
15
14.5
6
217.5
90
3
4
10
10
40
40
4
10
5
8
50
80
5
8
0
4
0
32
Σ
56
488
242
Then
X =
ΣxL
488
=
ΣL
56
or X = 8.71 in.
and
Y =
ΣyA 242
=
56
ΣA
or Y = 4.32 in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
