978-0133915426 Chapter 9 Part 3

subject Type Homework Help
subject Pages 14
subject Words 2873
subject Authors Russell C. Hibbeler

Unlock document.

This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
page-pf1
919
9–41.
SOLUTION
The volume of the thin disk differential element is
and its centroid
Centroid: Applying Eq. 9–5 and performing the integration, we have
Ans.=
R2+3r2+2rR
4R2+r2+rR h
=
p
h2
B
1r-R22
¢
z4
4
+2Rh1r-R2
¢
z3
3
+R2h2
¢
z2
2
≤R
`
0
h
p
h21r-R22z3
3+2Rh1r-R2z2
2+R2h21z2
`
0
h
z=L
V
z
'dV
LV
dV
=
Lh
0
z
b
p
h231r-R22z2+2Rh1r-R2z+R2h24dz
r
Lh
0
p
h23r-R22z2+2Rh1r-R2z+R2h24dz
z=z.
=
p
h2c1r-R22z2+2Rh1r-R2z+R2h2ddz
dV =py2dz =pca1r-R2z+Rh
hb2ddz
y=1r-R2z+Rh
h.
Locate the centroid of the frustum of the right-circular
cone.
zz
h
r
R
4(R2
+
r2
+
rR)
page-pf2
9–42.
z
z (y 1)
y
––
6
1 ft
Determine the centroid of the solid.
SOLUTION
Centroid:The centroid of the element is located at .We have
Ans.=2.61 ft
p
36cy6
6
-
2
5y5+y4
4d23 ft
0
p
36cy5
5
-y4
2
+y3
3d23 ft
0
y=L
V
y
~dV
LV
dV
=L3 ft
0
ycp
36
A
y
4-2y
3+y2
B
dy d
L3 ft
0
p
36
A
y4-2y3+y2
B
dy
=L3 ft
0
p
36
A
y
5-2y4
+y3
B
dy
L3 ft
0
p
36
A
y4-2y3+y2
B
dy
=
yc=y
y
page-pf3
921
9–43.
SOLUTION
Ans.
Ans.x=y=Lx
'dV
L
dV
=
a3h
12
pa2h
12
=a
p
=
pa2
4h2aah
3
3pb=a3h
12
=
pa2
4h2
4a
3phah4-
3h4
2+h4-h4
4b
Lx
'dV =
pa2
4h2Lh
0
4r
3p
(h-z)2dz =
pa2
4h2Lh
0
4a
3ph(h3-3h2z+3hz2-z3)dz
z=Lz
'dV
LdV
=
pa2h2
48
pa2h
12
=h
4
=
pa2
4h2ah4
12 b=
pa2h2
48
Lz
'dV =
pa2
4h2Lh
0
(h2-2hz +z2)zdz=
pa2
4h2ch2z2
2-2hz3
3+
z4
4dh
0
=
pa2
4h2ah3
3b=
pa2h
12
LdV =
pa2
4h2Lh
0
(h2-2hz +z2)dz =
pa2
4h2ch2z-hz2+
z3
3dh
0
dV =
p
4r2dz =
pa2
4h2(h-z)2dz
r=
h(h - z)
'=z
Locate the centroid of the quarter-cone.
z
h
Ans:
z=
h
4
x=y=
a
p
page-pf4
922
*9–44.
The hemisphere of radius ris made from a stack of very thin
plates such that the density varies with height
where kis a constant. Determine its mass and the distance
to the center of mass G.
SOLUTION
disk differential element is and its
centroid Evaluating the integrals, we have
Ans.
Centroid: Applying Eq. 9–3, we have
Ans.z=
L
m
z
'dm
Lm
dm
=
2pkr5>15
pkr4>4
=
8
15 r
=pk
¢
r2z3
3-
z5
5
`
0
r
=
2pkr5
15
L
m
z
'dm =Lr
0
z5kz3p(r2-z2) dz46
=pk
¢
r2z2
2-
z4
4
`
0
r
=
pkr4
4
m=L
m
dm =Lr
0
kz3p(r2-z2) dz4
z
'
=z.
dm =rdV =rpy2dz =kz3p(r2-z2) dz4
r=kz,
z
z
G
_
r
Ans:
m=
pkr
4
4
z=
r
page-pf5
923
9–45.
Locate the centroid
z
of the volume.
SOLUTION
z
y2 0.5z
1 m
2 m
3
page-pf6
9–46.
SOLUTION
Ans.
(By symmetry) Ans.x
=
z
=
0
y=
L
V
y
'dV
LV
dV
=
pa2b2
4
2pa2b
3
=
3
8b
Ly
'dV =Lb
0
pa2ya1-
y2
b2bdy =pa2cy2
2-
y4
4b2db
0
=
pa2b2
4
LdV =Lb
0
pa2a1-
y2
b2bdy =pa2cy-
y3
3b2db
0
=
2pa2b
3
dV =pz2dy
Locate t
h
e centro
id
of t
h
e e
lli
pso
id
of revo
l
ut
i
on.
z
x
b
y
a
1
y
2
b
2
z
2
a
2
page-pf7
9–47.
Locate the center of gravity of the solid.
SOLUTION
Centroid:The centroid of the element is located at .We have
Ans.
p
64 az5
5b216 in.
0
p
64 az4
4b216 in.
0
=12.8 in.=
L16 in.
0
p
64 z4 dz
L16 in.
0
p
64 z3 dz
z=
L
V
z
~dV
LV
dV
=
L16 in.
0
zcp
64 z
3 dz d
L16 in.
0
p
64 z3 dz
=
zc=z
z
16 in.
x
z
z 4y
2
––
3
page-pf8
*9–48.
Locate the center of gravity
y
of the volume. The material is
homogeneous.
y
4 in.
10 in.
z
10 in.
1 in.
z y2
1
100
page-pf9
9–49.
Locate the centroid of the spherical segment.
z
SOLUTION
Ans.z=0.675 a
=
p
B
a2az2
2b-az4
4b
R
a
a
2
p
B
a2(z)-az3
3b
R
a
a
2
=
p
B
a4
2-
a4
4-
a4
8+
a4
64
R
p
B
a3-a3
3-a3
2+a3
24
R
=
p
B
9a4
64
R
p
B
5a3
24
R
z=L
V
z
'dV
LV
dV
=
pLa
a
2
z(a2-z2)dz
pLa
a
2
(a2-z2)dz
z=z
dV =py2dz =p(a2-z2)dz
z
x
y
a
1
2
C
a
z
2
a
2
y
2
page-pfa
928
9–50.
SOLUTION
Ans.z=
Lz
'dV
LdV
=
abc2
24
abc
6
=
c
4
Lz
'dV =
1
2Lc
0
zaa1-
z
cbba1-
z
cbdz =
abc
2
24
LdV =Lc
0
2(x)(y)dz =
2Lc
0
aa1-
cbba1-
cbdz =
6
z=ca1-
byb=ca1-
axb
Determine the location of the centroid for the
tetrahedron. Hint: Use a triangular “plate” element parallel
to the xyplane and of thickness dz.
z
z
a
b
c
Ans:
z=
c
4
page-pfb
929
9–51.
The truss is made from five members, each having a length
of 4mand a mass of If the mass of the gusset plates
at the joints and the thickness of the members can be
neglected, determine the distance dto where the hoisting
cable must be attached, so that the truss does not tip
(rotate) when it is lifted.
7kg>m.
Ans.d=x=©x
'M
©M=420
140=3m
©M=4(7)(5) =140 kg
y
4m
4m
4m
4m
C
B
d
Ans:
d=3 m
page-pfc
*9–52.
Determine the location
(x, y, z)
of the centroid of the
homogeneous rod.
SOLUTION
ΣL
900 mm
x
y
z
600 mm
200 mm
100 mm
30
page-pfd
931
9–53.
A rac
k
i
s ma
d
e from ro
ll
-forme
d
s
h
eet stee
l
an
d
h
as t
h
e
cross section shown. Determine the location of the
centroid of the cross section. The dimensions are indicated
at the center thickness of each segment.
1x,y2
SOLUTION
Ans.
Ans.y=
©y
'L
©L
=9550
235
=40.6 mm
x=©x
'L
©L
=5737.50
235 =24.4 mm
©y
'L=0(15) +25(50) +50(15) +65(30) +80(30) +40(80) +0(15) =9550 mm2
©x
'L=7.5(15) +0(50) +7.5(15) +15(30) +30(30) +45(80) +37.5(15) =5737.50 mm2
©L=15 +50 +15 +30 +30 +80 +15 =235 mm
y
x
30 mm
15 mm 15 mm
80 mm
50 mm
x=24.4 mm
y=40.6 mm
page-pfe
932
9–54.
Locate the centroid () of the metal cross section. Neglect
the thickness of the material and slight bends at the corners.
y
x
,
Due to symmetry about yaxis, Ans.
Ans.y=
©y
'L
©L
=53457.56
917.63 =58.26 mm =58.3 mm
x
-=0
50 mm
150 mm
y
'y
'L(mm2)
1 50p168.17 26415.93
2 180.28 75 13520.82
3 400 0 0
4 180.28 75 13520.82
©917.63 53457.56
SOLUTION
Centroid: The length of each segment and its respective centroid are tabulated below.
Ans:
x=0
y=58.3
mm
page-pff
933
9–55.
SOLUTION
Ans.
Due to symmetry,
Ans.
Ans.z=©z
'L
©L
=200 000
1471.24
=136 mm
©z
'L=200(500) +200(500) +0ap
2b(300) =200 000 mm2
y=112 mm
x=©x
'L
©L
=165 000
1471.24 =112 mm
©L=500 +500 +ap
2b(300) =1471.24 mm
©x
'L=150(500) +0(500) +2 (300)
pap
2b(300) =165 000 mm2
Locate the center of gravity of the homogeneous
wire.
1x,y,z2
z
y
x
400mm
300mm
Ans:
x=112 mm
y=112 mm
z=136 mm
page-pf10
934
*9–56.
SOLUTION
Ans.x=©x
'm
©m=3964.2
22.092 =179 mm
=3964.2 kg.mm
©x
'm=150{2
C
7.85(10)3(0.3)(0.2)(0.02)
D
}+350
C
2.71(10)3(0.3)(0.2)(0.02)
D
©m=2
C
7.85(10)3(0.3)(0.2)(0.02)
D
+2.71(10)3(0.3)(0.2)(0.02) =22.092 kg
The steel and aluminum plate assembly is bolted together
and fastened to the wall. Each plate has a constant width in
the zdirection of 200 mm and thickness of 20 mm. If the
density of Aand Bis and for C,
determine the location of the center of
mass. Neglect the size of the bolts.
xral =2.71 Mg>m3,
rs=7.85 Mg>m3,
200 mm
100 mm
A
BC
x
y
Ans:
x=179 mm
page-pf11
935
9–57.
Locate the center of gravity of the streetlight. Neglect
the thickness of each segment. The mass per unit length of
each segment is as follows:
and rDE =2kg>m.rCD =5kg>m,
rBC =8kg>m,rAB =12 kg>m,
G1x,y2
SOLUTION
Ans.
Ans.y=
©y
'm
©m
=405.332
92.854
=4.37 m
+9 (1) (5) +9(1.5) (2) =405.332 kg #m
©y
'm=2 (4) (12) +5.5 (3)(8) +7.5(1) (5) +a8+2(1)
pbap
2b(5)
x=©x
'm
©m=18.604
92.854 =0.200 m
©m=4 (12)+3 (8) +1(5) +p
2(5)+1(5) +1.5 (2) =92.854 kg
+1.5 (1) (5) +2.75 (1.5) (2) =18.604 kg #m
©x
'm=0(4)(12) +0(3)(8) +0(1)(5) +a1-2(1)
pbap
2b(5)
1m
1m
1m
90
1m
C
B
A
DE
y
x
1.5 m
3m
4m
G(x,y)
Ans:
x=0.200
m
y=4.37
m
page-pf12
936
9–58.
SOLUTION
Ans.y=
©y
'A
©A=1 907 981.05
12 353.98 =154 mm
=12 353.98 mm2
©A=15(150) +150(15) +p(50)2
=1 907 981.05 mm2
©y
'A=7.5(15) (150) +90(150) (15) +215(p) (50)2
Determ
i
ne t
h
e
l
ocat
i
on of t
h
e centro
id
a
l
ax
i
s of t
h
e
beam’s cross-sectional area. Neglect the size of the corner
welds at Aand Bfor the calculation.
xxy
50 mm
A
C
B
15 mm
15 mm
150 mm
150 mm
y
xx
-
Ans:
y=154
mm
page-pf13
9–59.
Locate the centroid (
x,
y
) of the shaded area.
SOLUTION
y
x
6 in.
6 in.
6 in.
6 in.
page-pf14
*9–60.
Locate the centroid
y
for the beam’s cross-sectional area.
120 mm
240 mm
x
y

Trusted by Thousands of
Students

Here are what students say about us.

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.