Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
9–41.
SOLUTION
Volume and Moment Arm: From the geometry,
The volume of the thin disk differential element is
y=1r-R2z+Rh
h.
y-r
R-r=h-z
h,
Locate the centroid of the frustum of the right-circular
cone.
zz
xy
h
r
R
920
9–42.
y
x
z
z (y 1)
y
––
6
3 ft
1 ft
Determine the centroid of the solid.
SOLUTION
Differential Element:The thin disk element shown shaded in Fig. awill be considered. The
volume of the element is
y
Ans:
9–43.
SOLUTION
z
'=z
Locate the centroid of the quarter-cone.
y
z
h
a
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
Mass and Moment Arm: The density of the material is The mass of the thin
disk differential element is and its
centroid Evaluating the integrals, we have
z
'
=z.
dm =rdV =rpy2dz =kz3p(r2-z2) dz4
r=kz.
r=kz,
z
y
z
G
x
_
r
Ans:
923
9–45.
Locate the centroid
z
of the volume.
SOLUTION
Volume And Moment Arm. The volume of the thin disk differential element shown
shaded in Fig. a is
dV =
p
y2 dz =
p
(0.5z)dz
and its centroid is at
z
~=z
.
Centroid. Perform the integration
z
y
y2 0.5z
1 m
2 m
x
9–46.
SOLUTION
Locate t
h
e centro
id
of t
h
e e
lli
pso
id
of revo
l
ut
i
on.
z
x
9–47.
Locate the center of gravity of the solid.
SOLUTION
Differential Element:The thin disk element shown shaded in Fig. awill be
considered. The volume of the element is
z
16 in.
z
z 4y
2
––
3
Ans:
926
*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
SOLUTION
9–49.
Locate the centroid of the spherical segment.
z
SOLUTION
z=z
dV =py2dz =p(a2-z2)dz
z
y
a
z
1
—
2
C
a
z
2
a
2
y
2
Ans:
928
9–50.
SOLUTION
z=ca1-
1
byb=ca1-
1
axb
Determine the location of the centroid for the
tetrahedron. Hint: Use a triangular “plate” element parallel
to the x–yplane and of thickness dz.
z
y
z
a
b
c
Ans:
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.
SOLUTION
y
4m
4m
4m
4m
C
B
d
Ans:
930
*9–52.
Determine the location
(x, y, z)
of the centroid of the
homogeneous rod.
SOLUTION
Centroid. Referring to Fig. a, the length of the segments and the locations of their
respective centroids are tabulated below
Segment L(mm)
x
~(mm)
y
~(mm)
z
~(mm)
x
~
L(mm
2
)y
~
L(mm
2
)z
~
L(mm
2
)
1 200 0 0 100 0 0 20.0
(
10
3
)
x
y
z
600 mm
200 mm
100 mm
30
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
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
,
50 mm
x
150 mm
100 mm 100 mm50 mm 50 mm
y
Segment L (mm) y (mm)
'y
'L(mm2)
SOLUTION
Centroid: The length of each segment and its respective centroid are tabulated below.
Ans:
9–55.
SOLUTION
©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.
934
*9–56.
SOLUTION
©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,
300 mm
200 mm
100 mm
A
BC
x
y
Ans:
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
©x
'm=0(4)(12) +0(3)(8) +0(1)(5) +a1-2(1)
pbap
2b(5)
Ans:
x=0.200
m
936
9–58.
SOLUTION
©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
A
C
B
15 mm
15 mm
150 mm
150 mm
y
xx
-
Ans:
937
9–59.
Locate the centroid (
x,
y
) of the shaded area.
SOLUTION
Centroid. Referring to Fiq. a, the areas of the segments and the locations of their
respective centroids are tabulated below
Segment
A(in.2)
x
∼(in.)
y
∼(in.)
x
∼A(in.3)
y
∼A(in.3)
Thus,
Ans:
y
x
6 in.
6 in.
6 in.
6 in.
*9–60.
Locate the centroid
y
for the beam’s cross-sectional area.
120 mm
120 mm
240 mm
240 mm
240 mm
x
y
SOLUTION
Centroid. The locations of the centroids measuring from the x axis for segments
1 and 2 are indicated in Fig. a. Thus
