9–21.
Locate the centroid
x
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded
x
Centroid. Perform the integration
1
A x
~ dA
L
4 ft
0
x(x–8x1
>
2+16)dx
y
x
16 ft
4 ft
4 ft
y (4 x )2
1
2
9–22.
Locate the centroid
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in
Centroid. Perform the integration,
y=
1
A y
~ dA
1
A dA
=
L
4 ft
0
1
2
(
4–x1
>
2
)
2
(
x–8x1
>
2+16
)
dx
L
4 ft
0
(x–8x1
>
2+16)dx
y
x
16 ft
4 ft
4 ft
y (4 x )2
1
2
9–23.
Locate the centroid
x
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in Fig.a
is dA =y dx =
a
–
h
a
2 x2+h
b
dx and its centroid is at
x
~=x
.
Centroid. Perform the integration,
h
ax
y
y  x2h
h
a2
902
*9–24.
Locate the centroid
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in Fig.a
is dA =y dx =
a
–
2 x2+h
b
dx and its centroid is at y
~=
=
1
a
–h
x2+h
b
.
Ans:
h
ax
y
y  x2h
h
a2
903
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.
SOLUTION
Area and Moment Arm:Here, The area of the differential
element is and its centroid is and
Evaluating the integrals, we have
Centroid:Applying Eq.9–6, we have
Equations of Equilibrium:The weight of the plate is
= 1920 lb.W=42.6710.25211802
y
‘=1
2 1x–8x1
2+162.
x
‘=xdA =ydx =1x–8x1
2+162dx
y=x–8x1
2+16.
g=180 lb>ft3.
16 ft
16 ft
A
x
B
z
C
y
y x 4
1
2
1
2
9–26.
Locate the centroid
x
of the shaded area.
SOLUTION
x
y
4 ft x2
y
1
4
905
9–27.
Locate the centroid
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in Fig.a
Ans:
y
x
4 ft x2
y
1
4
*9–28.
Locate the centroid
x
of the shaded area.
SOLUTION
Area And Moment Arm. Here,
y2=x
and y1=
1
100
x2. Thus the area of the
differential element shown shaded in Fig. a is dA =
(
y2–y1
)
dx =
(
x–
1
100
x2
)
dx
and its centroid is at
x
~=x
.
Centroid. Perform the integration
y
x
100 mm
100 mm
y
y
x
1
100 x2
Ans:
907
9–29.
Locate the centroid
of the shaded area.
y
x
100 mm
100 mm
y
y x
1
100 x2
908
9–30.
Locate the centroid
x
of the shaded area.
SOLUTION
y
x
h
a
y x
h
––
ay ( )(xb)
b
h
ab
909
9–31.
Locate the centroid
of the shaded area.
SOLUTION
y
y
x
h
a
y x
h
––
ay ( )(xb)
b
h
ab
Ans:
910
*9–32.
Locate the centroid of the shaded area.
SOLUTION
Area and Moment Arm:The area of the differential element is
and its centroid are
x=xdA =ydx =a sin x
x
p
y
x
a
a
y a sin
x
a
Ans:
9–33.
Locate the centroid of the shaded area.
SOLUTION
Area and Moment Arm: The area of the differential element is
y
y
a
y a sin
x
a
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
y2
2=2x2
y1=-x1
7850 kg>m3.
A
x
y
y2 2x
2 m
2 m
Ans:
SOLUTION
Area And Moment Arm. Here, y2=h–
h
a
n xn and y1=h–
h
a
x. Thus, the
area of the differential element shown shaded in Fig. a is dA
=
(
y
2–
y
1
)
dx
x
9–35.
Locate the centroid
x
of the shaded area.
y
x
a
y h xn
h
h
—
an
y h x
h
—
a
*9–36.
Locate the centroid
of the shaded area.
SOLUTION
Area And Moment Arm. Here, y2=h–
h
a
n xn and y1=h–
h
a
x. Thus, the
area of the differential element shown shaded in Fig. a is dA
=
(
y
y
dx
Centroid. Perform the integration
y=
1
A y
~ dA
1
A dA
=
La
0
1
2
a
2h–
h
an xn–
h
a x
bah
a x–
h
an xn
b
dx
L
a
0a
h
a x–h
an xn
b
dx
y
x
a
y h xn
h
h
—
an
y h x
h
—
a
9–37.
Locate the centroid
x
of the circular sector.
SOLUTION
Area And Moment Arm. The area of the differential element shown in Fig. a is
dA =
1
2
r2 du and its centroid is at x
~=
2
3
r cos u.
y
x
C
r
x
a
a
916
9–38.
SOLUTION
dA =1
2(r)rdu=1
2r2du
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:
9–39.
Locate the center of gravity of the volume.The material is
homogeneous.
z
y
y2=2z
2m
2m
SOLUTION
Volume and Moment Arm: The volume of the thin disk differential element is
and its centroid
z
‘
=z.dV =py2dz =p12z2dz =2pzdz
*9–40.
SOLUTION
Locate the centroid of the paraboloid.
y
y
z2=4y
4m
z
Ans: