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9–1.
SOLUTION
Ans.
Ans.y=0 (By symmetry)
=124 mm
=
(300)2
C
sin u
D
2p
3
-2p
3
300a4
3pb
x=
Lx
'dL
LdL
=
L2p
3
-2p
3
300 cos u(300du)
L2p
3
-2p
3
300du
y
'
=300 sin u
x
'
=300 cos u
dL =300 du
Locate t
h
e center of mass of t
h
e
h
omogeneous ro
d
b
ent
into the shape of a circular arc.
y
x
30
300 mm
30
9–2.
Determine the location (x, y) of the centroid of the wire.
SOLUTION
y
2
2 ft
4 ft
9–3.
Locate the center of gravity
x
of the homogeneous rod. If
the rod has a weight per unit length of 100 N
>
m, determine
the vertical reaction at A and the x and y components of
reaction at the pin B.
A
B
x
1 m
1 m
y x2
y
*9–4.
Locate the center of gravity
y
of the homogeneous rod.
A
B
x
1 m
1 m
y x2
y
9–5.
Determine the distance
y
to the center of gravity of the
homogeneous rod.
SOLUTION
1
L dL
y
y 2x3
x
2 m
1 m
884
9–6.
Locate the centroid of the area.y
SOLUTION
Centroid: Due to symmetry
Ans.
Applying Eq. 9–4 and performing the integration, we have
Ans.=
¢
x
2-x3
12 +x5
160
≤
`
2m
-2m
¢
x-x3
12
≤
`
2m
-
2m
=2
5m
y=LA
y
'dA
L
A
dA
=
L2m
-2m
1
2
¢
1-1
4x2
≤¢
1-1
4x2
≤
dx
L2m
-2m
¢
1-1
4x2
≤
dx
x=0
y
x
2m
1m
y1– x
2
1
–
4
5
885
9–7.
SOLUTION
Centroid: The centroid of the element is located at and .
Area: Integrating,
y
'
=yx
'
=
x
2=
a
2h1>2y1>2
Determine the area and the centroid of the parabolic area.
x
h
y x2
h
––
a2
y
Ans.x=
L
A
x
'dA
LA
dA
=
Lh
0
¢
a
2h1>2y1>2
≤¢
a
h1>2y1>2dy
≤
2
3ah
=
Lh
0
a2
2hydy
2
3ah
=
a2
2h
¢
y2
2
≤
`
h
0
2
3ah
=
3
8a
a
2a
2
Ans:
x=
3
8
a
886
*9–8.
Locate the centroid of the shaded area.
SOLUTION
Ans:
y=
p
8
a
x=0
y
x
L
a
y a cos L
px
2
L
2
887
9–9.
Locate the centroid
x
of the shaded area.
SOLUTION
2
Ans:
x=
3
2
m
y
x
4 m
4 m
x2
y
1
4
9–10.
Locate the centroid
y
of the shaded area.
SOLUTION
5
y
x
4 m
4 m
x2
y
1
4
889
9–11.
Locate the centroid of the area.
SOLUTION
Ans. x=
L
A
x
' dA
LA
dA
=
Lb
0
h
b2 x3 dx
Lb
0
h
b2 x2 dx
=
B
h
4b2 x4
R
0
b
B
h
3b2 x3
R
0
b
=
3
4 b
x
'
=x
dA =y dx
x
b
x
y
y
x2
h
—
b2
4
890
*9–12.
b
x
y
h
y
x2
h
—
b2
Locate the centroid of the shaded area.
SOLUTION
Ans. y=
L
A
y
' dA
LA
dA
=
Lb
0
h2
2b4 x4 dx
Lb
0
h
b2 x2 dx
=
B
h2
10b4 x5
R
0
b
B
h
3b2 x3
R
0
b
=
3
10 h
y
'
=
y
2
dA =y dx
y
10
891
9–13.
Locate the centroid of the shaded area.
SOLUTION
Ans. x=6 m
x=
L
A
x
'dA
LA
dA
=
L8
0
x ax2
16 b dx
L8
0
a1
16 x2b dx
x
'
=x
dA =14-y2dx =a1
16 x2b dx
x
y
4 m
1
Ans:
x=6 m
892
9–14.
Locate the centroid of the shaded area.
SOLUTION
Ans. y=2.8 m
y=L
A
y
'dA
LA
dA
=
1
2 L8
0
¢
8-x2
16 bax2
16 b dx
L8
0
a1
16 x2b dx
y=
4+y
2
dA =14-y2dx =a1
16 x2b dx
y
y
4 m
1
Ans:
y=2.8 m
893
9–15.
SOLUTION
Ans.x=
L
A
xdA
LA
dA
=
0.25
0.6278
=0.398 m
=0.25 m3
LA
xdA =L1
0
xa1.359 -0.5 ex2bdx
x=x
y=0.5e12
=1.359 m
Locate the centroid of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.
x
y= 0.5e
x2
y
Ans:
x=0.398
m
894
*9–16.
Locate the centroid of the shaded area.Solve the problem by
evaluating the integrals using Simpson’s rule.
y
SOLUTION
Ans.y=L
A
ydA
LA
dA
=0.6278
0.6278 =1.00 m
=1
2L1
0a1.847 -0.25 e2x2bdx =0.6278 m3
LA
ydA =L1
0a1.359 +0.5 ex2
2b
A
1.359 -0.5 ex2
B
dx
L
A
dA =L1
0
(1.359 -y)dx =L1
0a1.359 -0.5ex2bdx =0.6278 m2
y= 0.5e
x2
y
Ans:
y=1.00 m
9–17.
y
y x
8 in.
2
––
3
Locate the centroid of the area.
SOLUTION
Centroid:The centroid of the element is located at .Applying
Eq. 9–4, we have
Ans.
=
c3
14 x7>3d20
8 in.
19.2
=1.43 in.
y=
L
A
y
'
dA
L
A
dA
=
L8 in.
0
1
2 x2>3
A
x2>3
B
dx
19.2 =
L8 in.
0
1
2 x4>3dx
19.2
y
'
=y>2=
1
2 x2>3
A
0
5 x5>3d28
0
y
896
9–18.
Locate the centroid of the area.
SOLUTION
ah-h
n+1ba
=
B
h
2 x2-
h1xn+22
an1n+22
R
0
a
B
hx -
h1xn+12
an1n+12
R
0
a
x=L
A
x
' dA
L
A
dA
=
La
0
¢
hx -h
an xn+1
≤
dx
La
0
¢
h-h
an xn
≤
dx
x
dA =y dx
x
y
y h xn
h
h
—
an
2(2+n)
Ans:
x=
a(1 +n)
2(2
+
n)
897
9–19.
y
y h xn
h
h
—
an
Locate the centroid of the area.
SOLUTION
2n2
n+1
=
1
2
B
h2x-
2h21xn+12
an1n+12+
h21x2n+12
a2n12n+12
R
R
0
a
B
hx -
h1xn+12
an1n+120
a
y=L
A
y
' dA
LA
dA
=
1
2 La
0
¢
h2-2
h2
an xn+h2
a2n x2n
≤
dx
La
0
¢
h-h
an xn
≤
dx
y
'=y
2
dA =y dx
y
Ans:
y=
hn
2n+1
898
*9–20.
SOLUTION
y
'dA
LA
dA
1
2La
h2
a2nx2ndx
La
0
h
anxndx
h2(a2n+1)
an(n+1)
y=y
2
dA =ydx
L
ocate the centroid of the shaded area.
y
y
h
yx
n
h
––
a
n
Ans:
y=
hn +1
2(2n
+
1)
