899
9–21.
Locate the centroid
x
of the shaded area.
SOLUTION
x
Centroid. Perform the integration
x=
1
A x
~ dA
1
A dA
=
L
4 ft
0
x(x–8x1
>
2+16)dx
L
4 ft
0
(
x–8x1
>
2+ 16) dx
=
a
x
3
3–16
5 x5
>
2+8x2
b`
0
4 ft
a
x2
2–16
3 x3
>
2+16x
b`0
4 ft
=1
3
5
ft Ans.
Ans:
x=1
3
5
ft
y
x
16 ft
4 ft
4 ft
y (4 x )2
1
2
900
9–22.
Locate the centroid
of the shaded area.
SOLUTION
55
y
x
16 ft
4 ft
4 ft
y (4 x )2
1
2
901
9–23.
Locate the centroid
x
of the shaded area.
x
a
2 x2+h
Centroid. Perform the integration,
x=
1
A x
~ dA
1
A dA
=
La
0
x
a
–h
a2 x2+h
b
dx
L
a
0a
–h
a
2
x2+h
b
dx
=
a
–h
4a2 x4+h
2 x2
b`
0
a
a
–h
3a
2 x3+hx
b`0
a
=
3
8
a Ans.
Ans:
x=
3
8
a
h
ax
y
y  x2h
h
a2
902
*9–24.
Locate the centroid
of the shaded area.
5
h
ax
y
y  x2h
h
a2
9–25.
The plate has a thickness of 0.25 ft and a specific weight of
Determine the location of its center of
gravity. Also, find the tension in each of the cords used to
support it.
g=180 lb>ft3.
16 ft
B
z
904
9–26.
Locate the centroid
x
of the shaded area.
x
4
Centroid. Perform the integration
x=
1
A x
~ dA
1
A dA
=
L
4 ft
0
x
a
1
4 x2 dx
b
L
4 ft
0
1
4
x2dx
=
a
1
16 x4
b`4 ft
0
a
1
12 x3
b`
4 ft
0
=3 ft
Ans.
Ans:
x
–=3 ft
y
4 ft x2
y
1
4
905
9–27.
Locate the centroid
of the shaded area.
SOLUTION
5
y
4 ft x2
y
1
4
906
*9–28.
Locate the centroid
x
of the shaded area.
SOLUTION
x
1
Centroid. Perform the integration
x=
1
A x
~ dA
1
A dA
=
L100 mm
0
x
a
x–1
100 x2
b
dx
L
100 mm
0a
x–1
100
x2
b
dx
=
a
x
3
3–1
400 x4
b`100 mm
0
a
x2
2–1
300 x3
b`
100 mm
0
=50.0 mm
Ans.
y
x
100 mm
100 mm
y
y
x
1
100 x2
Ans:
x=50.0 mm
9–29.
Locate the centroid
of the shaded area.
SOLUTION
1>2
y
x
100 mm
100 mm
y
y x
1
100 x2
908
9–30.
Locate the centroid
x
of the shaded area.
3
y
x
h
a
y x
h
––
ay ( )(xb)
b
h
ab
x=
1
3
(a+b)
909
9–31.
Locate the centroid
of the shaded area.
SOLUTION
y
a
h
Centroid. Perform the integration,
y=
1
A y
~ dA
1
A dA
=
Lh
0
y
a
b–b
h y
b
dy
L
h
0a
b–b
h y
b
dy
=
a
b
2 y2–b
3h y3
b`h
0
a
by –b
2h y2
b`
h
0
=
1
6 bh2
1
2
bh
=
h
3 Ans.
y
x
h
a
y x
h
––
ay ( )(xb)
b
h
ab
y=
h
3
910
*9–32.
Locate the centroid of the shaded area.
SOLUTION
Ans. =
p
2
a
=
ca3 sin x
a–xaa2 cos x
abd
`
0
pa
a–a2 cos x
ab
`
0
pa
x=
L
A
x
‘dA
LA
dA
=
Lpa
0
xaa sin x
a dxb
Lpa
0
a sin x
a dx
x
y
a
y a sin
x
a
2
911
9–33.
Locate the centroid of the shaded area.
SOLUTION
ydA
L
A
dA
a
L
0
a sin x
a dx
c1
1
`
pa
a–a2 cos x
ab
`
0
y
y
a
y a sin
x
a
Ans:
y=
pa
8
912
9–34.
The steel plate is 0.3 m thick and has a density of
Determine the location of its center of mass.
Also compute the reactions at the pin and roller support.
SOLUTION
Ans.
Ans.
a
Ans.
Ans.
Ans.Ay=73.9 kN
Ay+47.92 cos 45° –107.81=0 + c ©F
y=0;
Ax=33.9 kN
–Ax+47.92 sin 45° =0©F
x=0;
:
+
N
B=47.92 =47.9 kN
–1.25711107.812+N
B
A
222
B
=0+ ©M
A=0;
W=785019.81214.667210.32=107.81 kN
A=4.667 m2
y=L
A
y
‘ dA
LA
dA
=L2
0
22x–x
2
A
22x+x
B
dx
L2
0
A
22x+x
B
dx
=cx2
2–1
6 x3d0
2
c222
3x3>2+1
2 x2d0
2
=0.143 m
x=L
A
x
‘ dA
LA
dA
=L2
0
x
A
22x+x
B
dx
L2
0
A
22x+x
B
dx
=c222
5 x5>2+1
3 x3d0
2
c222
3 x3>2+1
2 x2d0
2
=1.2571=1.26 m
y
‘=y2+y1
2=22x–x
2
x
‘=x
7850 kg>m3.
x
y
y2 2x
2 m
Ans:
x=1.26
m
y=0.143
m
NB=47.9
kN
Ax=33.9
kN
A
y
=73.9
kN
913
SOLUTION
x
h
h
a
n xn–
a
a
a
n xn
Centroid. Perform the integration
x=
1
A x
~ dA
1
A dA
=
La
0
x
a
h
a x–h
an xn
b
dx
L
a
0a
h
a x–h
an xn
b
dx
=
c
h
3a x3–h
an(n+2) xn+2
d `
0
a
c
h
2a x2–h
an(n
+
1) xn+1
b`
a
0
=
ha
2
(n–1)
3(n+2)
ha(n–1)
2(n
+
1)
=
c2(n+1)
3(n
+
2)
d
a Ans.
9–35.
Locate the centroid
x
of the shaded area.
y
x
a
y h xn
h
h
—
an
y h x
h
—
a
Ans:
x=
c2(n+1)
3(n
+
2)
d
a
914
*9–36.
Locate the centroid
of the shaded area.
SOLUTION
h
h
3(2n
+
1)
y
x
a
y h xn
h
h
—
an
y h x
h
—
a
3(2n
+
1)
915
9–37.
Locate the centroid
x
of the circular sector.
3
a
y
x
C
r
x
a
a
Ans:
x=
2
3
ar sin
a
a
b
916
9–38.
Ans.x=
0.7854 a3
a2
=0.785 a
L
A
xdA =
2
3L45°
0
r3cos udu=
2
3L45°
0
A
2a2
B
3/2 cos u(cos 2u)3/2du=0.7854 a3
x=
L
A
xdA
LA
dA
=
2L45°
0
A
2
3rcos u
BA
1
2r2du
B
a2=
2
3L45°
0
r3cos udu
a2
A=2L45°
0
1
2(2a2cos 2u)du=a2
C
sin 2u
D
45°
0=a2
Determine the location of the centroid Cfor the loop of the
lemniscate, ,.(–45° …u…45°)r2=2a2cos 2u
r
O
C
_
r
r
r
2
=2a
2
cos 2
θ
θ
Ans:
x=0.785 a
917
9–39.
Locate the center of gravity of the volume.The material is
homogeneous.
z
y2=2z
2m
2m
SOLUTION
Ans.=
2p
z3
30
2 m
2p
z2
20
2 m
=
4
3m
z=
Lv
dV
=
L2 m
0
2pzdz
Ans:
x=y=0
z=
4
3
m
918
*9–40.
SOLUTION
Ans.=
4p
y3
30
4 m
4p
y2
20
4 m
=2.67 m
LV
L4 m
0
Locate the centroid of the paraboloid.
y
y
z2=4y
4m
z
Ans:
y=2.67 m