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9–61.
Determine the location
y
of the centroid C of the beam
having the cross-sectional area shown.
A
C
B
15 mm
15 mm
15 mm
150 mm
150 mm
100 mm
y
x
x
940
9–62.
Locate the centroid (
x,
y
) of the shaded area.
y
6 in.
3 in.
6 in.
6 in.
x
SOLUTION
9–63.
Determine the location of the centroid of the beam’s cross-
sectional area. Neglect the size of the corner welds at Aand B
for the calculation.
y
35 mm
A
942
*9–64.
Locate the centroid (
x,
y
) of the shaded area.
x
y
3 in.
1 in.
3 in.
3 in.
943
9–65.
Determine the location (
x,
y
) of the centroid C of the area.
SOLUTION
x
y
1.5 in.
1.5 in.
1.5 in.
1.5 in.
1.5 in.
944
9–66.
SOLUTION
Ans.y=
©y
'A
©A
=40
20
=2 in.
©A=6(4) -1(1) -3(1) =20 in2
©y
'A=6(4)(2) -1(1)(0.5) -3(1)(2.5) =40 in3
Determine the location of the centroid Cfor a beam having
the cross-sectional area shown. The beam is symmetric with
respect to the yaxis.
y
2in. 2 in.1in. 1 in.
1 in.
3in.
y
x
C
y
Ans:
y=2 in.
9–67.
Locate the centroid of the cross-sectional area of the
beam constructed from a channel and a plate. Assume all
corners are square and neglect the size of the weld at A.
y
y
70 mm
20 mm
10 mm
350 mm
C
A
946
*9–68.
SOLUTION
Ans.y=
©y
'A
©A=36
48 =0.75 in.
=36 in3
©y
'A=2(1) a1
2b(1)(3) +1.5(6)(3) +2(2) a1
2b(1)(3)
©A=1
2(8) (12)=48 in2
A tr
i
angu
l
ar p
l
ate ma
d
e of
h
omogeneous mater
i
a
l
h
as a
constant thickness which is very small. If it is folded over as
shown,determine the location of the plate’s center of
gravity G.
y
G
1in.
1in.
3in.
3in.
z
1in.
1in.
y
x
6in.
3in.
3in.
y
z
9–69.
Atriangular plate made of homogeneous material has a
constant thickness which is very small. If it is folded over as
shown, determine the location of the plate’s center of
gravity G.
z
SOLUTION
G
z
1in.
1in.
3in.
y
9–70.
Locate t
h
e center of mass of t
h
e for
k
e
d
l
ever,w
hi
c
h
is made from a homogeneous material and has the
dimensions shown.
z
SOLUTION
'A
-a5-4(2)
3pba1
2p(2)2b+6.5(2)(3)(0.5) =33.651 in3
©z
'A=2.5
2(2.5) (0.5) +a5-4(2.5)
3pba1
2p(2.5)2b
©A=2.5(0.5) +c1
2p(2.5)2-1
2p(2)2d+2 [(3)(0.5)] =7.7843 in2
3 in.
2 in.
G
0.5 in.
2.5 in.
0.5 in.
z
xy
z
949
9–71.
Determine the location of the centroid Cof the shaded
area which is part of a circle having a radius
r.
x
Ans.x=©x
'A
©A=
r3
3sin3a
1
2
r2
A
a-sin 2 a
2
B
=
2
3rsin3a
a-sin 2a
2
=1
2r2aa-sin2a
2b
©A=1
2r2a-1
2(rsin a)(rcos a)
=r3
3sin3a
=r3
3sin a-r3
3sin acos2a
©x
'A=1
2r2aa2r
3a
sin ab-1
2(rsin a)(rcos a)a2
3rcos ab
y
C
r
a
Ans:
x =
2
3
r
sin
3
a
a
-
sin 2a
2
950
*9–72.
A toy skyrocket consists of a solid conical top,
a hollow cylinder, and a
stick having a circular cross section,
Determine the length of the stick, x, so that the center of
gravity Gof the skyrocket is located along line aa.
rs=300 kg>m3.
rc=400 kg>m3,rt=600 kg>m3,
Solving for the positive root gives
Ans.x=490 mm
1060.29x2-212.058
A
103
B
x-150.80
A
106
B
=0
-116.24
A
106
B
-x2(1060.29) =-267.0
A
106
B
-212.058
A
103
B
x
x=©x
'
m
©m
=
-116.24(106)-x2(1060.29)
2.670(106)+2120.58x
=-100
=2.670
A
106
B
+2120.58xkg #mm3/m3
©m=c1
3p(5)2(20) d(600) +p
A
52-2.52
B
(100)(400) +
C
xp(1.5)2
D
(300)
=-116.24
A
106
B
-x2(1060.29) kg #mm4>m3
4bca1
3bp(5)2(20) d(600) -50
2
x
a
a
G
100mm
5mm
3mm
10 mm
20 mm
Ans:
x=490 mm
951
9–73.
Listed in a tabular form, we have
y
'
2=aa
2-t
2bcos 45° +t
2cos 45° =22
41a+t2
A2=at
y
'
1=aa-t
2+t
2bcos 45° +t
2cos 45° =22
41a+2t2
Locate the centroid for the cross-sectional area of the
angle.
y
aa –
y
C
'y
'A
952
9–74.
Determine the location () of the center of gravity of the
three-wheeler.The location of the center of gravity of each
component and its weight are tabulated in the figure. If the
three-wheeler is symmetrical with respect to the x–y plane,
determine the normal reaction each of its wheels exerts on
the ground.
y
x
,
1.
2.
3.
4.
Rear wheels
Mechanical components
Frame
Front wheel
18 lb
85 lb
120 lb
8lb
y
BA x
2ft
1.50 ft 1.30 ft
1ft
2.30 ft 1.40 ft
0.80 ft
1
4
3
2
SOLUTION
Ans.
Ans.
a
Ans.
Ans.N
A=86.9 lb
N
A+2172.12-231 =0+c©F
y=0;
N
B=72.1 lb
21N
B214.52-23112.812=0+©MA=0;
y=©y
'W
©W=398.9
231 =1.73 ft
=398.9 lb #ft
©y
'W=1.301182+1.51852+211202+1182
x=©x
'W
©W=648.5
231 =2.81 ft
©W=18 +85 +120 +8=231 lb
=648.5 lb #ft
©x
'W=4.51182+2.31852+3.111202
Ans:
x=2.81
ft
y=1.73
ft
NB=72.1
lb
NA=86.9
lb
9–75.
Locate the center of mass of the homogeneous
block assembly.
(x,y,z)
y
z
x150 mm
250 mm
200 mm
150 mm
150 mm
100 mm
SOLUTION
*9–76.
T
h
e s
h
eet meta
l
part
h
as t
h
e
di
mens
i
ons s
h
own. Determ
i
ne
the location of its centroid.1x,y,z2
SOLUTION
©A=4(3) +1
2(3)(6) =21 in2
y
z
A
D
C
3in.
4in.
955
9–77.
The sheet metal part has a weight per unit area of
and is supported by the smooth rod and at C. If the cord is
cut, the part will rotate about the yaxis until it reaches
equilibrium. Determine the equilibrium angle of tilt,
measured downward from the negative xaxis, that AD
makes with the axis.-x
2lb
>
ft
2
z
D
C
3in.
Ans:
u=53.1°
956
9–78.
The wooden table is made from a square board having a
weight of 15 lb.Each of the legs weighs 2 lb and is 3 ft long.
Determine how high its center of gravity is from the floor.
Also,what is the angle,measured from the horizontal,
through which its top surface can be tilted on two of its
legs before it begins to overturn? Neglect the thickness of
each leg.
SOLUTION
Ans.u=tan-1a2
2.48 b=38.9°
'W
4ft
3ft
4ft
Ans:
z=2.48
ft
u=38.9°
957
9–79.
The buoy is made from two homogeneous cones each
having a radius of 1.5 ft. If find the distance to
the buoy’s center of gravity G.
zh =1.2 ft,
SOLUTION
Ans.z
-=©z
'V
©V=8.577
12.25 =0.70 ft
=12.25 ft3
©V=1
3p(1.5)2(1.2) +1
3p(1.5)2(4)
=8.577 ft4
©z
'V=1
3p(1.5)2(1.2) a-1.2
4b+1
3p(1.5)2(4) a4
4b
h
G
4ft
1.5 ft
z
Ans:
z=0.70
ft
958
*9–80.
SOLUTION
Ans.h
=
2.00 ft
-0.5890 h2+9.4248 =1.1781 h+4.7124
z
'=©z
'V
©V=-0.5890 h2+9.4248
2.3562 h+9.4248 =0.5
=2.3562 h+9.4248
©V=1
3p(1.5)2(h)+1
3p(1.5)2(4)
=-0.5890 h2+9.4248
©z
'V=1
3p(1.5)2(h)a-h
4b+1
3p(1.5)2(4) a4
4b
The buoy is made from two homogeneous cones each
having a radius of 1.5 ft. If it is required that the buoy’s
center of gravity Gbe located at determine the
height hof the top cone.
z=0.5 ft,
h
G
4ft
1.5 ft
z
Ans:
h=2.00
ft
