9–1.
SOLUTION
Ans.
Ans.y = 0 (By symmetry)
= 124 mm
=(300)2 Csin u D 2p
3-2p
3
300a43pb
x = Lx'dL
LdL=L
2p3
-2p3
300 cos u (300du)
L2p3
-2p3
300d u
y' = 300 sin u
x' = 300 cos u
dL = 300 d u
Locate the center of mass of the homogeneous rod bentinto the shape of a circular arc.
y
x
30�
300 mm
30�
.
.
.
.
Determine the mass and the location of the center ofmass of the rod if its mass per unit length is
.m = m0(1 + x>L)x
y
x
L
9–3.
.
.
9–5.
Locate the centroid of the uniform rod. Evaluate theintegrals using a numerical method.
SOLUTION
Differential Element. The length of the element shown shaded in Fig. a is
Here,
The centroid of the element is located at and . Applying
Eq. 9–5 we have
Ans.
Ans.y = LLy~ dL
LLdL
= La>2
0a cos
p
a xB1 + p
2 sin 2 p
a x dx
La>2
0 B1 + p 2 sin
2 p
a x dx
=0.6191a2
1.1524a= 0.537a
x = LL
~x dL
LLdL
= La>2
0xB1 + p
2 sin 2 p
a x dx
La>2
0 B1 + p 2 sin
2 p
a x dx
=0.3444a2
1.1524a= 0.299a
yc = y = a cos p
a xxc = x
dL = C1 + a -p sin p
a xb2
dx = C1 + p2 sin2 p
a x dx
dy
dx= aa -sin
p
a xb ap
ab = -p sin
p
a x
dL = 2dx2 + dy2 = C1 + ady
dxb2
dx
( x, y ) y
x
a
y � a cos x
a––2
––ap
9–6.
Locate the centroid of the area.y
SOLUTIONArea and Moment Arm: The area of the differential element is
and its centroid is .
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
=25
m
y = LAy'
dA
LAdA
=L
2m
-2m
12¢1 -
14
x2≤ ¢1 -14
x2≤dx
L2m
-2m¢1 -
14
x2≤dx
x = 0
y' =
y
2=
12a1 -
14
x2bdA = ydx = a1 -14
x2bdx
y
x2 m
1 m
y 1 – x21–4
9–7.
SOLUTIONDifferential Element:The area element parallel to the x axis shown shaded in Fig. awill be considered. The area of the element is
Centroid: The centroid of the element is located at and .
Area: Integrating,
y' = yx
' =x
2=
a
2h1>2 y1>2
dA = xdy =a
h1>2 y1>2 dy
Determine the area and the centroid of the parabolic area.x
x
h
a
y x2h––a2
y
Ans.
Ans.x = LAx'
dA
LAdA
=L
h
0¢ a
2h1>2y1>2≤ ¢ a
h1>2 y1>2 dy≤23
ah
= Lh
0
a2
2hy dy
23
ah
=
a2
2h¢y2
2≤ ` h
0
23
ah
=38
a
A = LAdA = L
h
0
a
h1>2 y1>2 dy =2a
3h1>2 Ay3>2 B 2 h0
=23
ah
x
h
a
y � x2 h––a2
y
9–8.
Locate the centroid of the parabolic area.
SOLUTION
Differential Element: The area element parallel to the x axis shown shaded in Fig. awill be considered. The area of the element is
Centroid: The centroid of the element is located at .
Area: Integrating,
Ans.
Ans.y = LAy '
dA
LAdA
=L
h
0ya a
h1>2 y
1>2
dyb23
ah
=L
h
0
a
h1>2 y
3>2
dy
23
ah
=
2a
5h1>2 y
5>2 20
h
23
ah
=35
h
A = LA dA = L
h
0
a
h1>2 y1>2 dy =2a
3h1>2 Ay3>2 B 2 h0
=23
ah
x'
=x
2=
a
2h1>2 y
1>2 and y
'= y
dA = x dy =a
h1>2 y1>2 dy
y
9–9.
Locate the centroid of the area.
SOLUTION
Differential Element: The differential area element parallel to the y axis shownshaded in Fig. a will be considered. The area of the element is
Area: Integrating the area of the differential elements gives
Ans.
Centroid: The centroid of the element is located at Applying Eq. 9–4, we have
Ans.x = LAx'
dA
LAdA
= L8
0x Ax2>3dx B19.2
= L8
0x5>3dx
19.2=c38
x 8>3 d 20
8
19.2= 5 m
x' = x.
A = LA dA = L
8
0x2>3 dx = c3
5 x5>3 d 2 8
0= 19.2 m2
dA = y dx = x2>3 dx
x y
y � x
x
8
4
2––3
m m
m m m
m
m
m
y
y � x
x
8
4
2––3
9–10.
Locate the centroid of the area.
SOLUTION
Area: Integrating the area of the differential element gives
Ans.
Centroid: The centroid of the element is located at . Applying Eq. 9–4, we have
Ans.=
c 314
x7>3 d 20
8
19.2= 1.43 m
y = LAy'
dA
LAdA
=L
8
0
12
x2>3 Ax2>3 Bdx
19.2=L
8
0
12
x4>3dx
19.2
y'
= y>2 =12
x2>3
A = LA dA = L
8
0x2>3 dx = c3
5 x5>3 d 2 8
0= 19.2 m2
y
m m
m m
m
m
m
m
9–11.
Locate the centroid of the area.
SOLUTION
Ans. x = LA x'
dA
LA dA
= Lb
0 h
b2 x3 dx
Lb
0 h
b2 x2 dx
=B h
4b2 x4R0
b
B h
3b2 x3R0
b=
34
b
x' = x
dA = y dx
x
bx
y
h
y � x2h—b2
bx
y
h
y � x2h—b2
9–12.
Locate the centroid of the shaded area.
SOLUTION
Ans. y = LA y'
dA
LA dA
= Lb
0
h2
2b4 x4 dx
Lb
0 h
b2 x2 dx
=B h2
10b4 x5R0
b
B h
3b2 x3R0
b=
310
h
y' =
y
2
dA = y dx
y
9–13.
Locate the centroid of the shaded area.
SOLUTION
Ans. x = 6.00 m
x = LA x'
dA
LA dA
= L8
0 x a x2
16 b dx
L8
0 a 1
16 x2b dx
x' = x
dA = 14 - y2dx = a 116
x2b dx
x y
x8 m
4 my � 4 � x2 1––
16
9–14.
Locate the centroid of the shaded area.
SOLUTION
Ans. y = 2.80 m
y = LA y'
dA
LA dA
=
12
L8
0 ¢8 -
x2
16 b a x2
16b dx
L8
0 a 1
16 x2b dx
y =4 + y
2
dA = 14 - y2dx = a 116
x2b dx
y y
x8 m
4 my � 4 � x2 1––
16
9–15.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at
Area: Integrating,
We have
Ans. x = LA x'
dA
LA dA
= Lb
ax a c2
x dxb
c2 ln ba
= Lb
ac2 dx
c2 ln ba
=c2x 2 b
a
c2 ln ba
=
b - a
ln ba
A = LA dA = L
b
a
c2
x dx = c2 ln x 2 b
a= c2 ln
ba
x'
= x
dA = y dx =c2
x dx
x y
x
a
b
xy � c2
9–16.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at
Area: Integrating,
Ans.
We have
Ans.=
-c4
2x3b
a
c2 ln ba
=
c2(b - a)
2ab ln ba
y = LA yc dA
LA dA
= Lb
aa c
2
2xb a c
2
x dxb
c2 ln ba
= Lb
a
c4
2x2 dx
c2 ln ba
A = LA dA = L
b
a
c2
x dx = c2 ln x 2 b
a= c2 ln
ba
y'
=y
2=
c2
2x.
dA = y dx =c2
x dx
y y
x
a
b
xy � c2
9–17.
SOLUTION
and its centroid is
Centroid: Applying Eq. 9–4 and performing the integration, we have
Ans.=n + 1
2 n + 2a
=
h¢x2
2-
xn+2
1n + 22an ≤ `0
a
h x -xn+1
1n + 12an `0
a
x = LAx'
dA
LAdA
=L
a
0xBh¢1 -
xn
an ≤dxR
La
0h¢1 -
xn
an ≤dx
x' = x.dA = 1h - y2dx = h¢1 -
xn
an ≤dx
Locate the centroid of the shaded area.x y
x
a
h
y = h xnan
Area and Moment Arm: T he area of the differential element is
9–18.
Locate the centroid of the shaded area.
SOLUTION
Ans. x =ah
2-
h
n + 2ba2
ah -h
n + 1ba
=n + 1
21n + 22 a
=Bh
2 x2 -
h1xn+22an1n + 22R0
a
Bhx -h1xn+12
an1n + 12R0
a
x = LA x'
dA
LA dA
=L
a
0 ¢hx -
h
an xn+1≤ dx
La
0 ¢h -
h
an xn≤ dx
x' = x
dA = y dx
x y
xa
y � h � xn
h
h—an
y
xa
y � h � xn
h
h—an
9–19.
Locate the centroid of the shaded area.
SOLUTION
Ans. y =
2n2
21n + 1212n + 12 h
n
n + 1
=n
2n + 1 h
=
12
Bh2x -2h21xn+12an1n + 12 +
h21x2n+12a2n12n + 12R0
a
Bhx -h1xn+12
an1n + 12R0
a
y = LA y'
dA
LA dA
=
12
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
9–20.
SOLUTION
Ans.y = LAy'
dA
LAdA
=
12 L
a
0
h2
a2n x2n dx
La
0
han xn dx
=h2(a2n + 1)
2a2n(2n + 1)
h(an + 1)an(n + 1)
=n + 1
2(2n + 1)h
y =y
2
dA = y dx
Locate the centroid of the area.y y
x
h
a
y xnh––an
9–21.
Locate the centroid of the area.
SOLUTION
Ans.x = LA x'
dA
LA dA
=
14
a3
23
a2=
38
a
LA x'
dA = La
0 x2
2 dy = L
a
0 a
2 y dy =
14
a3
LA dA = L
a
0 x dy = L
a
0 1ay1>2dy = 1a a2
3 a3>2b =
23
a2
x' =
x
2
dA = x dy
x
a
a
x
y
y � x21–a
9–22.
Locate the centroid of the area.
SOLUTION
Ans.y = LA y'
dA
LA dA
=
25
a3
23
a2=
35
a
LA y'
dA = La
0 xy dy = L
a
0 1ay3>2dy = 1a a2
5 a5>2b =
25
a3
y' = y
dA = x dy
y
a
a
x
y
y � x21–a
9–23.
Locate the centroid of the quarter elliptical area.
SOLUTION
Ans.x = LA x'
dA
LA dA
=13 a2bp4 ab
=4a
3 p
LA x'
dA =12
Lb
0 ¢a2 -
a2
b2 y2≤ dy =12
Ba2y -a2
3b2 y3R0
b
=13
a2b
LA dA = L
a
0 Bb2 -
b2
a2 x2 dx =b
2a cx2a2 - x2 + a2 sin-1
xad
0
a
=p
4 ab
y' = y
x' =
x
2
dA = y dx
x y
x
a
� � 1
b
y2—b2
x2—a2
9–24.
Locate the centroid of the quarter elliptical area.
SOLUTION
Ans.y = LA y'
dA
LA dA
=13 ab2
p4 ab
=4b
3 p
LA y'
dA =12
La
0 ¢b2 -
b2
a2 x2≤ dx =12
Bb2x -b2
3a2 x3R0
a
=13
ab2
LA dA = L
a
0 Bb2 -
b2
a2 x2 dx =b
2a cx2a2 - x2 + a2 sin-1
xad
0
a
=p
4 ab
y' =
y
2
x' = x
dA = y dx
y y
x
a
� � 1
b
y2—b2
x2—a2
m2
m3
m
2 m
2 m
.
y
x
a
a––2
––a
y � a cos xp
9–26.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at .
Area: Integrating,
Ans.
We have
Ans.=
a 2
p x sin
p
a x +
a3
p 2 cos
p
a x 2
0
a>2
a2
p
= ap - 22p
ba
x = La>2
0x'
dA
LA dA
= La>2
0xaa cos
p
a x dxb
a2
p
= La>2
0ax cos
p
a x dx
a2
p
A = LA dA = L
a>2
0a cos
p
a x dx =
a2
p sin
p
a x 2
0
a>2=
a2
p
x' = x
dA = y dx = a cos p
a x dx
x
9–27.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at
Area: Integrating,
Ans.
We have
Ans. =
a2
4 La>2
0acos
2pa
x + 1b dx
a2
p
=
a 2
4 a a
2p sin
2pa
x + xb 20
a>2
a2
p
=p
8 a
y = LAy'
dA
LA dA
= La>2
0aa
2 cos p
a xb aa cos
p
a x dxb
a2
p
= La>2
0
a2
2 cos2
p
ax dx
a2
p
A = LA dA = L
a>2
0a cos
p
a x dx =
a2
p sin p
a x 2
0
a>2=
a2
p
y' =
y
2=
a
2 cos p
a x.
dA = y dx = a cos p
a x dx
y y
x
a
a––2
––a
y � a cos xp
1 m
1 m
y
x
y � x2––3
y � x3––2
9–28.
Locate the centroid of the area.
SOLUTION
Differential Element: The differential element parallel to the y axis shown shadedin Fig. a will be considered. The area of the element is
Centroid: The centroid of the element is located at .
Ans.
We have
Ans.=¢3
8 x8>3 -
27
x7>2≤ `0
1m
15
=2556
m
x = LAx' dA
LAdA
= L1 m
0x(x2>3 - x3>2) dx
15
= L1 m
0(x5>3 - x5>2) dx
15
A = LAdA = L
1 m
0(x2>3 - x3>2) dx = ¢3
5 x5>3 -
25
x5>2≤ ` 10=
15
m2
x' = x
dA = (y1 - y2) dx = (x2>3 - x3>2) dx
x
9–29.
Locate the centroid of the area.
SOLUTION
Differential Element: The differential element parallel to the y axis shown shadedin Fig. a will be considered. The area of the element is
Centroid: The centroid of the element is located at .
Ans.
We have
Ans.=2556
m
=
12
B 37
x7>3 -x4
4R ` 1 m
0
15
y- = LAy' dA
LAdA
= L1 m
0 12
(x2>3 + x3>2)(x2>3 - x3>2) dx
15
= L1 m
0 12
(x4>3 - x3) dx
15
A = LAdA = L
1 m
0(x2>3 - x3>2) dx = ¢3
5 x5>3 -
25
x5>2≤ ` 10=
15
m2
y' =
12
(y1 + y2) =12
(x2>3 + x3>2)
dA = (y1 - y2) dx = (x2>3 - x3>2) dx
y
1 m
1 m
y
x
y � x2––3
y � x3––2
9–30.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at .
Area: Integrating,
Ans.
We have
Ans. =
16
x3 -1
16 x4 2
0
2 m
0.3333= 1 m
x- = LAxc dA
LAdA
=L
2 m
0x¢1
2 x -
14
x2≤
dx
0.3333=L
2 m
0¢1
2 x2 -
14
x3≤
dx
0.3333
= 0.333 m2
A = LAdA = L
2 m
0¢1
2 x -
14
x2≤ dx =14
x2 -1
12 x3 2
0
2 m
= 0.3333m2
x' = x
dA = (y1 - y2) dx = ¢12
x -14
x2≤
dx
x y
x
y � x2
y � x1 m
2 m
1––2
1––4
9–31.
Locate the centroid of the area.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The area of the element is
Centroid: The centroid of the element is located at
Area: Integrating,
Ans.
We have
Ans. =
12¢ 1
12x3 -
180
x5≤ 20
2 m
0.3333= 0.4 m
y = LAy' dA
LAdA
=L
2 m
0
12¢1
2x +
14
x2≤ ¢12
x -14
x2≤
dx
0.3333=L
2 m
0
12
¢14
x2 -1
16x4≤
dx
0.3333
= 0.3333 m2 = 0.333 m2
A = LAdA = L
2 m
0¢ 1
2 x -
14
x2≤ dx =14
x2 -1
12 x3 2
0
2 m
= 12¢1
2 x +
14
x2≤y' =
12
(y1 - y2)
dA = (y1 - y2) dx = ¢12
x -14
x2≤
dx
y y
x
y � x2
y � x1 m
2 m
1––2
1––4
9–32.
Locate the centroid of the shaded area.
SOLUTION
Area and Moment Arm: The area of the differential element isand its centroid are
Ans. =p
2 a
=ca3 sin
xa
- xaa2 cos xab d `
0
pa
a -a2 cos xab `
0
pa
x = LA x'
dA
LA dA
= Lpa
0xaa sin
xa
dxb
Lpa
0a sin
xa
dx
x = xdA = ydx = a sin xa
dx
x
p
y
xa
a
y � a sin xa
9–33.
Locate the centroid of the shaded area.
SOLUTION
Area and Moment Arm: The area of the differential element isand its centroid are
Ans. y = LA ydA
LA dA
= Lpa
0
a2
sin xa
aa sin xa
dxb
Lpa
0a sin
xa
dx
=c14
a2ax -12
a sin 2xab d `
0
pa
a -a2 cos xab `
0
pa=p
8 a
y =y
2=
a
2 sin
xa
.dA = ydx = a sin xa
dx
y
p
y
xa
a
y � a sin xa
9–34.
The steel plate is 0.3 m thick and has a density ofDetermine 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 ©Fy = 0;
Ax = 33.9 kN
-Ax + 47.92 sin 45° = 0©Fx = 0;:+NB = 47.92 = 47.9 kN
-1.25711107.812 + NB A222 B = 0+ ©MA = 0;
W = 785019.81214.667210.32 = 107.81 kN
A = 4.667 m2
y = LA y'
dA
LA dA
= L2
0
22x - x
2 A22x + x B dx
L2
0 A22x + x B dx
=cx2
2-
16
x3 d0
2
c2223
x3>2 +12
x2 d0
2= 0.143 m
x = LA x'
dA
LA dA
= L2
0 x A22x + x B dx
L2
0 A22x + x B dx
=c222
5 x5>2 +
13
x3 d0
2
c2223
x3>2 +12
x2 d0
2= 1.2571 = 1.26 m
y' =
y2 + y1
2=22x - x
2
x' = x
dA = 1y2 - y12 dx = A22x + x B dx
y22 = 2x2
y1 = -x1
7850 kg>m3.
A
B
x
y
y � �x
y2 � 2x
2 m
2 m
2 m
9 5.
Locate the centroid xc of the shaded area.
Given:
a 4 m�
b 4 m�
Solution:
A
0
a
xbxa
bxa������
2
��
��
����
d�
xc1A
0
a
xx bxa
bxa������
2
��
��
����
d�
xc 1.80 m�
3
Ans.
–
9 6.
Locate the centroid yc of the shaded area.
Given:
a 4 m�
b 4 m�
Solution:
A
0
a
xbxa
bxa������
2
��
��
����
d�
yc1A
0
a
x12
bxa
bxa������
2�
��
��
bxa
bxa������
2
��
��
����
d� yc 1.80 m�
3
Ans.
–
9–37.
If the density at any point in the quarter circular plate isdefined by = 0xy, where 0 is a constant, determine themass and locate the center of mass ( , ) of the plate. Theplate has a thickness t.
SOLUTION
Differential Element: The element parallel to the y axis shown shaded in Fig. a willbe considered. The mass of this element is
Mass: Integrating,
Ans.
Center of Mass: The center of mass of the element is located at . ApplyingEq. 9–2 we have
Ans.
By considering the element parallel to the x axis shown shaded in Fig. b,. In this case, .
Applying Eq. 9–2, we have
Ans.y = Lm y~ dm
Lm dm
= Lr
0y cr0 t(r
2y - y
3)dy d14
r0 r
4 t= L
r
0r0 t(r
2y
2 - y
4) dy
14
r0 r4t
=815
r
yc = ydm = r dV = r0xy(tx dy) = r0 tx2y dy = r0 t(r 2y - y
3) dy
=
r0 ta r
2x
3
3-
x5
5b 2 r
0
14
r0 r4t
=815
r
x = Lm x~ dm
Lm dm
= Lr
0x cr0 t(r
2x - x
3)dx d14
r0 r
4 t
= Lr
0r0 tar2x2 - x4bdx
14r0 r4t
xc = x
m = Lm dm = L
r
0r0 t(r
2x - x
3) dx = r0 ta r
2x
2
2-
x
4
4b 2
0
r
=14
r0 r4 t
= r0 t(r 2x - x
3) dx
= r0 txy2dx
dm = r dV = r0 xy(ty dx)
yxrrr
y
x
r
x2 � y2 � r2
9–38.
Determine the location of the centroid C of the cardioid,r = a11 - cos u2.
r
SOLUTION
Ans.= LArx'
dA
LAdA
=3.927 a 3
32p a 2
= 0.833 a
=23
a 3
Lp
0(1 -
-
cos u)3 cos u du = -3.927 a 3
LArx dA = 2L
p
0a2
3r cos ub a 1
2b(a 2)(1 - cos u)2 du
A = 2Lp
0
12
(a 2)(1 - cos u)2 du =32p a 2
dA =12
r2 du
r
Cu
r
~
r~
9–39.
SOLUTION
Volume and Moment Arm: The volume of the thin disk differential element is
Centroid: Applying Eq. 9–3 and performing the integration, we have
Ans.=
py3
3 0
py2
2 0
m=
y = LVy'dV
LV dV= L0
y3p1 y2dy4
L4 m
0p1 y2dy
Locate the centroid of the paraboloid.y
y
z =2 y
m
m
z
and its centroid y' = y.dV = pz dy2 = p1 y2dy6
6 m
6
6
66 m
66
4 m
6
6
6
(6y)½
6 m
9–40.
Locate the center of gravity of the volume. The material ishomogeneous.
z
y
y 2 = 2z2 m
2 m
SOLUTION
Volume and Moment Arm: The volume of the thin disk differential element isand its centroid
Centroid: Due to symmetry about z axis
Ans.
Applying Eq. 9–3 and performing the integration, we have
Ans.=
2pz3
3 0
2 m
2pz2
2 0
2 m=
43
m
z' = Lv
z'dV
Lv dV= L
2 m
0z12pzdz2
L2 m
02pzdz
x' = y' = 0
z' = z.dV = py2dz = p12z2dz = 2pzdz
9–41.
Locate the centroid of the hemisphere.z
SOLUTIONVolume and Moment Arm: The volume of the thin disk differential element is
and its centroid
Centroid: Applying Eq. 9–3 and performing the integration, we have
Ans.=
pa2z2
2-z4
4 0
a
p a2z -z3
3 0
a=
38a
z' = LV
z'dV
LV dV= L
a
0z3p1a2 - z22dz4
La
0p1a2 - z22dz
z' = z.dV = py2dz = p1a2 - z22dz
y
z
x
a
y 2 + z 2 = a 2
.
.
9–44.
SOLUTION
Ans.
(By symmetry) Ans.x = z = 0
y = LVy'dV
LVdV=
pa2b2
42pa2b
3
=38b
Ly'dV = L
b
0p a2ya1 -
y2
b2 b dy = p a2 cy2
2-y4
4b2 db
0=p a2b2
4
LdV = Lb
0p a2 a1 -
y2
b2 b dy = p a2 cy -y3
3b2 db
0=
2pa2b
3
dV = p z2 dy
Locate the centroid of the ellipsoid of revolution. z
x
b
y
a
� 1�y2
b2
z2
a2
9
Locate the center of gravity zc of thefrustum of the paraboloid.The materialis homogeneous.
Given:
a 1 m�
b 0.5 m�
c 0.3 m�
Solution
V
0
a
z� b2 za
b2 c2� ���
��
����
d�
zc1V
0
a
zz� b2 za
b2 c2� ���
��
����
d�
zc 0.422 m�
45.
Ans.
–
9–46.
The hemisphere of radius r is made from a stack of very thinplates such that the density varies with height where k is a constant. Determine its mass and the distanceto the center of mass G.
SOLUTION
Mass and Moment Arm: The density of the material is The mass of the thindisk differential element is and itscentroid Evaluating the integrals, we have
Ans.
Centroid: Applying Eq. 9–3, we have
Ans.z = Lm z'
dm
Lm dm
=2pkr5>15
pkr4>4 =8
15 r
= pk¢ r2z3
3-
z5
5≤ `
0
r
=2pkr5
15
Lm z'
dm = Lr
0z5kz3p(r2 - z2) dz46
= pk¢ r2z2
2-
z4
4≤ `
0
r
=pkr4
4
m = Lm dm = L
r
0kz3p(r2 - z2) dz4
z' = z.
dm = rdV = rpy2dz = kz3p(r2 - z2) dz4r = kz.
r = kz,
z
y
z
G
x
_r
9–47.
SOLUTION
Ans.
Ans.x = y = Lx'
dV
LdV
=
a3 h
12p a2h
12
=ap
=p a2
4 h2 aa h3
3pb =
a3 h
12
=p a2
4 h2
4a
3phah4 -
3h4
2+ h4 -
h4
4b
Lx'
dV =pa 2
4 h2Lh
0
4 r
3p(h - z)2 dz =
pa2
4 h2Lh
0
4a
3 p h(h3 - 3h2 z + 3hz2 - z3) dz
z = Lz'
dV
LdV
=
p a2 h2
48p a2h
12
=h
4
=p a2
4 h2 a h4
12b =
pa2h2
48
Lz'
dV =p a2
4 h2 Lh
0(h2 - 2hz + z2) z dz =
p a2
4 h2 ch2 z2
2- 2h
z3
3+
z4
4dh
0
=p a2
4 h2 ah3
3b =
pa2 h
12
LdV =p a2
4 h2 Lh
0(h2 - 2hz + z2) dz =
p a2
4 h2 ch2z - hz2 +z3
3dh
0
dV =p
4r2 dz =
p a2
4 h2 (h - z)2 dz
r =a
h(h - z)
z' = z
Locate the centroid of the quarter-cone.
y
z
x
h
a
9–48.
SOLUTION
Volume and Moment Arm: From the geometry,
The volume of the thin disk differential element is
and its centroid
Centroid: Applying Eq. 9–3 and performing the integration, we have
Ans.=R2 + 3r2 + 2rR
4 R2 + r2 + rRh
=
p
h2 B1r - R22¢ z4
4≤ + 2Rh1r - R2¢ z3
3≤ + R2h2¢ z2
2≤ R `
0
h
p
h2 1r - R22 z3
3+ 2Rh1r - R2 z2
2+ R2h21z2 `
0
h
z' = LV
z'
dV
LVdV
=L
h
0zb p
h2 31r - R22z2 + 2Rh1r - R2z + R2h24dz r
Lh
0
p
h2 3 r - R21 2z2 + 2Rh1r - R2z + R2h24dz
z = z.
=p
h2 c1r - R22z2 + 2Rh1r - R2z + R2h2 ddz
dV = py2dz = p c a 1r - R2z + Rh
hb2 ddz
y =1r - R2z + Rh
h.
y - r
R - r=
h - z
h,
Locate the centroid of the frustum of the right-circularcone.
z z
x y
h
r
R
9–49.
The king’s chamber of the Great Pyramid of Giza is locatedat its centroid. Assuming the pyramid to be a solid, provethat this point is at Suggestion: Use a rectangulardifferential plate element having a thickness dz and area(2x)(2y).
z = 14 h,
z
h
a
a
a
a
x y
SOLUTION
(QED)z' = L
z'dV
L dV=a2 h2
3
4 a2 h3
=h
4
L z'dV = L
h
0
4 a2
h2 1h - z22z dz =4 a2
h2 Bh2 z2
2- 2h
z3
3+z4
4R
0
h
=a2 h2
3
LdV = Lh
0
4 a2
h2 1h - z22 dz =4 a2
h2 Bh2z - hz2 +z3
3R
0
h
=4 a2 h
3
x = y =a
h1h - z2
dV = 12x212y2 dz = 4xy dz
9–50.
SOLUTION
Ans.z = Lz'dV
LdV=
abc2
24abc
6
=c
4
Lz'dV =
12L
c
0z aa1 -
zcbba1 -
zcbdz =
abc2
24
LdV = Lc
0
12
(x)(y)dz =12L
c
0aa1 -
zcbba1 -
zcb dz =
abc
6
z = ca1 -1byb = ca1 -
1axb
Determine the location of the centroid for thetetrahedron. Hint: Use a triangular “plate” element parallelto the x–y plane and of thickness dz.
z
y
z
x
ab
c
9–51.
The truss is made from five members, each having a lengthof 4 m and a mass of If the mass of the gusset platesat the joints and the thickness of the members can beneglected, determine the distance d to where the hoistingcable must be attached, so that the truss does not tip(rotate) when it is lifted.
7 kg>m.
SOLUTION
Ans.d = x =©x'M©M
=420140
= 3 m
©M = 4(7)(5) = 140 kg
©x'M = 4(7)(1+ 4 + 2 + 3 + 5) = 420 kg # m x
y
4 m
4 m4 m
4 m
4 m
CB
A D
60�
d
.
.
9–53.
Locate the centroid ( , ) of the cross section. All thedimensions are measured to the centerline thickness of eachthin segment.
SOLUTION
Centroid: The centroid of each composite segment is shown in Fig. a.
Ans.
Ans. =106.57(103)
882.84= 121 mm
y =©y'
L
©L=
0(200) + 50(100) + 200a22002 + 2002b + 150(300)
200 + 100 + 22002 + 2002 + 300
=68.28(103)
882.84= 77.3 mm
x =©x
'L
©L=
100(200) + 200(100) + 100a22002 + 2002b + 0(300)
200 + 100 + 22002 + 2002 + 300
yx y
x
300 mm
100 mm
200 mm
9–54.
Locate the centroid ( ) of the metal cross section. Neglectthe thickness of the material and slight bends at the corners.
yx,
Due to symmetry about y axis, Ans.
Ans.y =©y'
L
©L=
53457.56917.63
= 58.26 mm = 58.3 mm
x- = 0
50 mm
x
150 mm
100 mm 100 mm50 mm 50 mm
y
Segment L (mm) y (mm)'
y'
L (mm2)
1 50p 168.17 26415.93
2 180.28 75 13520.82
3 400 0 0
4 180.28 75 13520.82
© 917.63 53457.56
SOLUTIONCentroid: The length of each segment and its respective centroid are tabulated below.
.
.
.
.
.
9–56.
SOLUTION
Ans.x =©x'm©m
=3964.222.092
= 179 mm
= 3964.2 kg.mm
©x'm = 150{2 C7.85(10)3(0.3)(0.2)(0.02) D}+350 C2.71(10)3(0.3)(0.2)(0.02) D
©m = 2 C7.85(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 togetherand fastened to the wall. Each plate has a constant width inthe z direction of 200 mm and thickness of 20 mm. If thedensity of A and B is and for C,
determine the location of the center ofmass. Neglect the size of the bolts.
xral = 2.71 Mg>m3,rs = 7.85 Mg>m3,
300 mm
200 mm100 mm
A
BC
x
y
9–57.
To determine the location of the center of gravity of theautomobile it is first placed in a level position, with the twowheels on one side resting on the scale platform P. In thisposition the scale records a reading of . Then, one side iselevated to a convenient height c as shown. The newreading on the scale is . If the automobile has a totalweight of W, determine the location of its center of gravityG .(x, y)
W2
W1
SOLUTIONEquation of Equilibrium: First,we will consider the case in which the automobile isin a level position. Referring to the free-body diagram in Fig. a and writing themoment equation of equilibrium about point A,
a Ans.
From the geometry in Fig. c, and . Using the result of
and referring to the free-body diagram in Fig. b, we can write the moment equation
of equilibrium about point .A¿
xcos u =2b2 - c2
bsin u =
c
b
+©MA = 0; W1(b) - W(x) = 0 x =W1
Wb
b
Pc
G–y
–x
W2
a
Ans.y =b(W2 - W1)2b2 - c2
cW
+©MA¿ = 0; W2Bb¢2b2 - c2
b≤ R - W¢2b2 - c2
b≤ aW1
Wbb - Wa c
bby = 0
9–58.
SOLUTION
Ans.y =©y'A©A
=1 907 981.05
12 353.98= 154 mm
= 12 353.98 mm2
©A = 15(150) + 150(15) + p(50)2
= 1 907 981.05 mm2
©y'A = 7.5(15) (150) + 90(150) (15) + 215(p) (50)2
Determine the location of the centroidal axis of thebeam’s cross-sectional area. Neglect the size of the cornerwelds at A and B for the calculation.
x xy
50 mm
A
C
B
15 mm
15 mm
150 mm
150 mm
y
xx
-
Locate the centroid of the cross-sectional area ofthe built-up beam.
y y
x
450 mm
150 mm150 mm
200 mm
20 mm
20 mm
9–59.
Locate the centroid of the cross-sectional area ofthe built-up beam.
y
200 mm
20 mm50 mm
150 mm
y
x
200 mm
300 mm
10 mm
20 mm 20 mm
10 mm
9–60.
9–61.
Locate the centroid ( ) of the member’s cross-sectionalarea.
yx,
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
Thus,
Ans.
Ans.y =©y'
A
©A=
287 0009 050
= 31.71 mm = 31.7 mm
x =©x'
A
©A=
698 5009 050
= 77.18 mm = 77.2 mm
y
x
100 mm60 mm
30 mm
50 mm
40 mm
Segment A (mm2) x'
(mm) y'
(mm) x'
A (mm3) y'
A (mm3)
11213021902 20 30 27 000 40 500
2 30(90) 45 45 121 500 121 500
3 100(50) 110 25 550 000 125 000
© 9 050 698 500 287 000
9–62.
Locate the centroid of the bulb-tee cross section.y
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
Thus,
Ans.y =©y'A©A
=28 078 12596 562.5
= 290.78 mm = 291 mm
_y
C
x
400 mm
300 mm
75 mm
100 mm
75 mm
50 mm
225 mm 225 mm
Segment A 1mm22 y'1mm2 y
'A 1mm32
1 450(50) 600 13 500 000
2 475(75) 337.5 12 023 437.5
312122521752 125 1 054 687.5
4 300(100) 50 1 500 000
© 96 562.5 28 078 125
.
.
9–64.
SOLUTION
Ans.y =© y'A©A
=19.44(106)
0.144(106)= 135 mm
©A = 900(80) + 100(360) + 2 c12
(100) (360) d = 0.144(106) mm2
© y' A = 900(80) (40) + 100(360) (260) + 2 c12
(100) (360) (200) d = 19.44(106) mm3
Locate the centroid of the concrete beam having thetapered cross section shown.
y 300 mm300 mm
300 mm
80 mm
360 mmx
100 mm
C
y
9
Locate the center of gravity G(xc, yc) of the streetlight.Neglect the thickness of each segment. The mass perunit length of each segment is given.
Given:
a 1 m� �AB 12kgm
�
b 3 m� �BC 8kgm
�
c 4 m� �CD 5kgm
�
d 1 m� �DE 2kgm
�
e 1 m� f 1.5 m�
Solution:
M �AB c �BC b� �CD a e��d2
����
���
� �DE f��
xc1M
�CD�d2
d2d�
���
���
�CD e de2
����
���
� �DE f d e�f2
����
���
���
��
�
yc1M
�CD a c b�a2
����
���
�d2
c b� a�2d�
����
���
� e c b� a� d�( )���
��
�DE f c b� a� d�( ) �BC b cb2
����
���
� �AB cc2
��
�������
����
�
xc
yc
���
���
0.200
4.365���
���
m�
65.
Ans.
–
9–66.
Thus,
Ans.y' =
©y'A©A
=4 000 500
14 700= 272.14 mm = 272 mm
Locate the centroid of the cross-sectional area of thebeam constructed from a channel and a plate. Assume allcorners are square and neglect the size of the weld at A.
y
y
70 mm
20 mm
10 mm
350 mm
325 mm
C
A
325 mm
Segment A (mm2) y'
(mm) y'A (mm3)
1 350(20) 175 1 225 000
2 630(10) 355 2 236 500
3 70(20) 385 539 000
© 14 700 4 000 500
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
9–67.
An aluminum strut has a cross section referred to as a deephat. Locate the centroid of its area. Each segment has athickness of 10 mm.
y
SOLUTION
Centroid: The area of each segment and its respective centroid are tabulated below.
Thus,
Ans.y =©y'A©A
=3600
=
–y
120 mm
30 mm
x'
30 mm
C
Segment A (mm2) y'
(mm) y'A (mm3)
1 40(10) 5 2000
2 120(20) 144 000
3
© 3600 000
80(10)
60
115 92 000
238
100 mm
238 00066.1 mm
10 mm 20 mm
60 mm
110 mm
80 mm
110 mm 115 mm
9–68.
Locate the centroid of the beam’s cross-sectional area.
SOLUTION
Centroid: The centroid of each composite segment is shown in Fig. a. We have
Ans.y =© y
'A
©A=
5(180)(10) + 2[37.5(75)(10)] + 2[70(40)(10)]
180(10) + 2(75)(10) + 2(40)(10)= 29.6 mm
y
x
y
50 mm
100 mm
10 mm10 mm10 mm
10 mm
10 mm
100 mm
75 mm
50 mm
9–69.
Determine the location of the centroid C of the shadedarea which is part of a circle having a radius
Using symmetry, to simplify, consider just the top half:
r.x
SOLUTION
Ans.x =©x'
A
©A=
r3
3 sin3 a
12 r2 Aa - sin 2 a
2 B =23 r sin3a
a - sin 2a2
=12
r2aa -sin2a
2b
©A =12
r2 a -12
(r sin a) (r cos a)
=r3
3sin3 a
=r3
3sin a -
r3
3sin a cos2a
©x'
A =12
r2 aa 2r
3asin ab -
12
(r sin a) (r cos a) a23
r cos ab
y
xC
r
x
a
a
9–70.
SOLUTIONCentroid : The area and the centroid for segments 1 and 2 are
Listed in a tabular form, we have
y'
2 = aa
2-t
2bcos 45° +
t
2cos 45°=2241a + t2
A2 = at
y'
1 = aa - t
2+t
2bcos 45° +
t
2cos 45°=2241a + 2t2
A1 = t1a - t2
Locate the centroid for the cross-sectional area of theangle.
y
Thus,
Ans.=22 a2 + at - t2
2 2a - t
y =©y'A©A
=
22t21a2 + at - t22
t12a - t2
aa–y
t t
C
Segment A y'
y'A
1 t1a - t2 2241a + 2t2
22t41a2 + at - 2t22
2 at 2241a + t2
22t41a2 + at2
© t12a - t2 22t21a2 + at - t22
9–71.
SOLUTION
Ans.y =©y'
A
©A=
393 1124575.6
= 85.9 mm
©A = p(25)2 + 15(110) + pa352b2
= 4575.6 mm2
©y'
A = p(25)2(25) + 15(110)(50 + 55) + pa352b2a50 + 110 +
352b = 393 112 mm3
Determine the location of the centroid of the beam’s cross-sectional area. Neglect the size of the corner welds at A and Bfor the calculation.
y
35 mm
50 mm
110 mm
15 mmC
A
By
Uniform blocks having a length L and mass m arestacked one on top of the other, with each block overhangingthe other by a distance d, as shown. If the blocks are gluedtogether, so that they will not topple over, determine thelocation of the center of mass of a pile of n blocks.x
L
d
2d
y
x
9–72.
.
Uniform blocks having a length L and mass m arestacked one on top of the other, with each blockoverhanging the other by a distance d, as shown. Show thatthe maximum number of blocks which can be stacked inthis manner is .n 6 L>d
L
d
2d
y
x
9–73.
.
The assembly is made from a steel hemisphere,, and an aluminum cylinder,. Determine the mass center of the
assembly if the height of the cylinder is .h = 200 mmral = 2.70 Mg>m3rst = 7.80 Mg>m3
160 mm
h
z
y
x
80 mm
z
G
_
9–74.
.
The assembly is made from a steel hemisphere,, and an aluminum cylinder,. Determine the height h of the cylinder
so that the mass center of the assembly is located at.z = 160 mm
ral = 2.70 Mg>m3rst = 7.80 Mg>m3
160 mm
h
z
x
80 mm
z
G
_
9–75.
.
9
Locate the centroid yc for the beam’s cross-sectional area.
Given:
a 120 mm�
b 240 mm�
c 120 mm�
Solution:
A a b�( )5c 3b c�
yc1A
a b�( )2
25c 2b c
b2������
b cb3������
��
��
� yc 229 mm�
76.
Ans.
–
9 77.
Locate the centroid yc for the strut’scross-sectional area.
Given:
a 40 mm�
b 120 mm�
c 60 mm�
Solution:
A�b2
22a c�
yc1A
�b2
24b3����
���
2a cc2������
��
��
� yc 56.6 mm� Ans.
–
.
9–79.
Locate the center of mass of the block. Materials 1, 2, and 3have densities of 2.70 Mg m3, 5.70 Mg m3, and 7.80 Mg m3,respectively.
SOLUTION
Centroid: We have
Ans.
Ans.
Ans. =19.962(10-3)
0.8838= 0.02259 m = 22.6 mm
z =g z
' m
gm=
0.07 c2700a12
(0.03)(0.06)(0.02)b d + 0.04 C5700 A0.04(0.06)(0.02) B D + 0.01 C7800 A0.06(0.06)(0.02) B D2700a1
2 (0.03)(0.06)(0.02)b + 5700 A0.04(0.06)(0.02) B + 7800 A0.06(0.06)(0.02) B
=26.028(10-3)
0.8838= 0.02945 m = 29.5 mm
y =gy
' m
gm=
0.02 c2700a12
(0.03)(0.06)(0.02)b d + 0.03 C5700 A0.04(0.06)(0.02) B D + 0.03 C7800 A0.06(0.06)(0.02) B D2700a1
2 (0.03)(0.06)(0.02)b + 5700 A0.04(0.06)(0.02) B + 7800 A0.06(0.06)(0.02) B
=20.07(10-3)
0.8838= 0.02271 m = 22.7 mm
x =gx
' m
gm=
0.01 c2700a12
(0.03)(0.06)(0.02)b d + 0.01 C5700 A0.04(0.06)(0.02) B D + 0.03 C7800 A0.06(0.06)(0.02) B D2700a1
2 (0.03)(0.06)(0.02)b + 5700 A0.04(0.06)(0.02) B + 7800 A0.06(0.06)(0.02) B
>>>
x
y
z
60 mm20 mm
60 mm
40 mm
30 mm
20 mm
1
2
3
9–80.
Locate the centroid of the homogenous solid formed byboring a hemispherical hole into the cylinder that is cappedwith a cone.
SOLUTIONCentroid: Since the solid is made of a homogeneous material, its center of masscoincides with the centroid of its volume.The centroid of each composite segment isshown in Figs. a and b. Since segment (3) is a hole, its volume should be considerednegative. We have
Ans. =2.7422(10-3)p
9(10-3)p= 0.3047 m = 305 mm
z =©z
'V
©V=p(0.2)(0.152)(0.4) +
13
p(0.4 + 0.075)(0.152)(0.3) + a38
(0.15)b a -
23
p(0.153)bp(0.152)(0.4) +
13
p(0.152)(0.3) + a -
23
p(0.153)b
z
y
z
x
150 mm
150 mm
400 mm
300 mm
9–81.
Locate the center of mass of the solid formed by boring ahemispherical hole into a cylinder that is capped with a cone.The cone and cylinder are made of materials having densitiesof and , respectively.
SOLUTION
Centroid: The center of mass of each composite segment is shown in Figs. a and b.Since segment (3) is a hole, its mass should be considered negative. We have
Ans. =12.8545p35.775p
= 0.3593 m = 359 mm
z =©z' m©m
=0.2 c2700 cp(0.152)(0.4) d
ad + (0.4 + 0.075)7800 c1
3 p(0.152)(0.3) d + a3
8 (0.15)b c -2700a2
3 p(0.153)b d
2700 cp(0.152)(0.4) d + 7800 c13
p(0.152)(0.3) d + c -2700a23
p(0.153)b d
2.70 Mg>m37.80 Mg>m3
z
y
z
x
150 mm
150 mm
400 mm
300 mm
9–82.
Determine the distance h to which a 100-mm diameter holemust be bored into the base of the cone so that the center ofmass of the resulting shape is located at Thematerial has a density of
Choosing the positive root,
Ans.h = 323 mm
h2 - 0.230 h - 0.0300 = 0
0.4313 - 0.2875 h = 0.4688 - 1.25 h2
13 p10.152210.52A0.5
4 B - p10.05221h2Ah2 B13 p10.152210.52 - p10.05221h2 = 0.115
8 Mg>m3.z = 115 mm.
z
y
x
C
150 mm50 mm
h
500 mm
_z
9–83.
SOLUTION
Ans.z =©z' V©V
=1.463 (10-3)
0.01139= 0.12845 m = 128 mm
= 0.01139 m3
©V =13p (0.15)2 (0.5) - p (0.05)2 (0.05)
= 1.463(10-3) m4
©z'V =13p (0.15)2 A0.5 B a0.5
4b - p (0.05)2 (0.05) a0.05
2b
Determine the distance to the centroid of the shape whichconsists of a cone with a hole of height bored intoits base.
h = 50 mmz z
y
x
C
150 mm50 mm
h
500 mm
z
Determine the distance to the centroid of thesolid which consists of a cylinder with a hole of length
bored into its base.h = 50 mm
x y
h
120 mm
40 mm
20 mm
9–84.
.
Determine the distance h to which a hole must bebored into the cylinder so that the center of mass of theassembly is located at . The material has adensity of .8 Mg>m3
x = 64 mm
y
x
h
120 mm
40 mm
20 mm
9–85.
..
9–86.
Locate the center of mass of the assembly. The assemblyconsists of a cylindrical center core, A, having a density of7.90 Mg/m3, and a cylindrical outer part, B, and a cone cap,C, each having a density of 2.70 Mg m3.
SOLUTION
Center of mass: The assembly is broken into four composite segments, as shown inFigs. a, b, and c. Since segment (3) is a hole to segments (1) and (2), its mass shouldbe considred negative. We have
Ans.=121.58p262.6p
= 0.4630 m = 463 mm
z =gzcm
gm=
2700(0.3) cp(0.32)(0.6) d + 2700(0.6 + 0.2) c13p(0.32)(0.8) d + (0.5) c -2700 cp(0.12)(1) d d + 7900(0.5) cp(0.12)(1) d
2700 cp(0.32)(0.6) d + 2700 c13p(0.32)(0.8) d + c -2700 cp(0.12)(1) d d + 7900 cp(0.12)(1) d
>
z z
x y
600 mm
400 mm
400 mm
100 mm
300 mm
AC
B
.
9–88.
A hole having a radius r is to be drilled in the center of thehomogeneous block. Determine the depth h of the hole sothat the center of gravity G is as low as possible.
h
a
a2
a2
a2
a2
G
r
SOLUTION
Solving for the smaller root,
Ans.=a3 - a22a2 - pr2
pr2
h =2a3 - 2(2a3)2 - 4(pr2)a4
2(pr2)
pr2h2 - 2a3 h + a4 = 0
dz
dh=
12B (a3 - pr2h)(-2pr2h) - (a4 - pr2h2)(-pr2)
(a3 - pr2h)2 R = 0
z =©zV
©V=
12 (a4 - pr2 h2)
a3 - pr2h=
a4 - pr2h2
2(a3 - pr2h)
©zV = Aa2 Ba3 - Ah2 B Cpr2 h D = 12(a4 - pr2 h2)
©V = a3 - pr2h
9–89.
Locate the center of mass of the assembly. The cylinderand the cone are made from materials having densities of
and , respectively.9 Mg>m35 Mg>m3
z
SOLUTIONCenter of mass: The assembly is broken into two composite segments, as shown inFigs. a and b.
Ans.=1060.601407.4
= 0.754 m = 754 mm
z =©z'
m
©m=
5000(0.4) Cp(0.22)(0.8) D + 9000(0.8 + 0.15) c13p(0.42)(0.6) d
5000 Cp(0.22)(0.8) D + 9000 c13p(0.42)(0.6) d
z
x
0.8 m
0.6 m0.4 m
0.2 m
y
The water tank AB has a hemispherical top and isfabricated from thin steel plate. Determine the volumewithin the tank.
1.5 m
1.6 m
0.2 m
B
A
1.6 m
9–90.
The water tank AB has a hemispherical roof and isfabricated from thin steel plate. If a liter of paint can cover
of the tank’s surface, determine how many liters arerequired to coat the surface of the tank from A to B.3 m2
1.5 m
1.6 m
0.2 m
B
A
1.6 m
9–91.
9–92.
Determine the outside surface area of the hopper.
SOLUTION
Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9, with , , , and
as indicated in Fig. a,
Ans.= 47.15 m2 = 47.1 m2
A = u©NL = 2p [1.5(4) + 0.8523.13]
N2 =0.2 + 1.5
2= 0.85 m
N1 = 1.5 mL2 = 21.22 + 1.32 = 23.13 mL1 = 4mu = 2p rad
0.2 m
4 m
z
1.2 m
1.5 m
9–93.
The hopper is filled to its top with coal. Determine thevolume of coal if the voids (air space) are 30 percent of thevolume of the hopper.
SOLUTION
Ans.V = 22.1 m3
V = © u r~ A = 2p [(0.75)(1.5)(4) + (0.1)(.2)(1.2) + (0.6333)a12b(1.3)(1.2)](0.70)
0.2 m
4 m
z
1.2 m
1.5 m
9–94.
SOLUTION
Ans.V = 4.25(106) mm3
V = ©ur'A = 2p(350)(60)(20) + 2p(320)(40)(20)
The rim of a flywheel has the cross section A–A shown.Determine the volume of material needed for its construction.
A
A
60 mm
20 mm
40 mm
20 mm
300 mm
Section A–A
9 95.
Determine the surface area of the tank, which consists of a cylinder and hemispherical cap.
Given:
a 4 m�
b 8 m�
Solution:
A 2� a b2a�
�a2
����
���
�
A 302 m2�
–
9 96.
Determine the volume of the tank, which consists of a cylinder and hemispherical cap.
Given:
a 4 m�
b 8 m�
Solution:
V 2�4a3�
�a2
4
���
���
a2
b a( )���
��
�
V 536 m3� Ans.
–
9–97.
The process tank is used to store liquids duringmanufacturing. Estimate the outside surface area of thetank. The tank has a flat top and the plates from which thetank is made have negligible thickness.
SOLUTION
Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9 with ,
, , , , and as indicated in Fig. a,
Ans.= 188.49 m2 = 188 m2
A = u©NL = 2p [1.5(3) + 3(6) + 1.5(5)]
N3 = 1.5 mN2 = 3 mN1 = 1.5 mL3 = 232 + 42 = 5 mL2 = 6 mL1 = 3 m
u = 2p
6 m
4 m
3 m 3 m
9–98.
The process tank is used to store liquids duringmanufacturinghas a flat top and the plates from which the tank is made have negligible thickness.
. Estimate the volume of the tank. The tank
SOLUTION
Ans.V = 207.3 m3 = 207 m3
V = ©urA = 2pB1 a12b (3) (4) + 1.5(3)(6)
6 m
4 m
3 m 3 m
~ R
The hopper is filled to its top with coal. Estimatethe volume of coal if the voids (air space) are 35 percent ofthe volume of the hopper.
0.2 m
4 m
z
1.2 m
1.5 m
9–99.
.
.
9–101.
SOLUTION
Ans.
Ans.V = ©urA = 2p(b)(a)2 = 2pba2
A = ©urL = 2p(b)(4a) = 8pba
= 8pba
= 4p cba -a2
2sin 45° + ba +
a2
2sin 45° d
A = ©urL = 2 c2pab -a
2sin 45°b(a) d + 2 c2pab +
a
2sin 45°b(a) d
Determine the surface area and the volume of the ringformed by rotating the square about the vertical axis
Also
.b
a
a
45
~
~
� �.
Determine the volume of material needed to make the casting.
������
�� ���
�� ��
�� �� ����
���������
� 2 �� 2�4
���
���
� ��2
�4 ���
3 ��
���
���
� 2 ��� 2 ���� ����2
���
���
�� 2�2
���
���
� ��2
� ��4 ���
3 ��
���
���
���
��
���
� 1.40 �3�
40 �
60��
�3(10 )6
–
Determine the surface area from A to B of the tank.
A
1 m
z
B
1.5 m
3 m
9–103.
.
Determine the volume within the thin-walled tankfrom A to B.
A
1 m
z
B
1.5 m
3 m
9–104.
.
9–106.
Determine the volume of an ellipsoid formed by revolvingthe shaded area about the x axis using the second theoremof Pappus–Guldinus. The area and centroid y of the shadedarea should first be obtained by using integration.
SOLUTION
Area and Centroid: The differential element parallel to the x axis is shown shadedin Fig. a. The area of this element is given by
Integrating,
The centroid can be obtained by applying Eq. 9–4 with
Volume: Applying the second theorem of Pappus–Guldinus and using the resultsobtained above, we have
Ans.V = 2pyA = 2pa 4b
3pb apab
4b =
23pab2
y = LAy'
dA
LAdA
= Lb
0y c a
b2b2 - y2 dy dpab
4
=4b
3p
y' = y.y
A = LAdA = L
b
0
a
b2b2 - y2 dy =
pab
4
dA = xdy =a
b2b2 - y2 dy
x
y
b
a
� � 1x2––a2
y2––b2
9–107.
Using integration, determine both the area and thecentroidal distance of the shaded area. Then, using thesecond theorem of Pappus–Guldinus, determine thevolume of the solid generated by revolving the area aboutthe y axis.
x
SOLUTION
Ans.
Ans.
Thus,
Ans.V = ur'
A = 2p 0.6 1.333 = 5.03 m3
x' = L
x'
dA
L dA
=0.8
1.333= 0.6 m
L x'
dA = L2
0
y4
8dy = c y5
40d
0
2
= 0.8 m3
A = LdA = L2
0
y2
2dy = cy3
6d
0
2
= 1.333 = 1.33 m2
dA = x dy
y' = y
x' =
x
2
C
y
x
2 m
2 m
_x
y2 = 2x
9–108.
SOLUTION
Surface Area: This problem requires that . Applying the theorem of
Pappus and Guldinus, Eq. 9–7, with , ,
and , we have
Ans.h = 106 mm
12
[2p(25)(158.11)] = 2pah
6b ¢210
3h≤
12
(ur1 L1) = ur2L2
r2 =h
6r1 = 25 mmL2 =
Ch2 + a
h
3b
2
=210
3h,
L1 = 2502 + 1502 = 158.11 mmu = 2p
12A1 = A2
Determine the height h to which liquid should be pouredinto the conical paper cup so that it contacts half the surfacearea on the inside of the cup.
100 mm
h
150 mm
.
9
Determine the surface area and the volume of the conical solid.
Solution:
A 2a3
2a2
2��
A 3�a2�
V 212
a2
���
���
32
a���
���
36
a2����
���
�
V�4
a3�
110.
Ans.
Ans.
–
9–111.
SOLUTION
Ans.h = 29.9 mm
10.77h + 0.2154h2 = 513.5
2pb 5(10) + a10 +h
5bBa
2h5b
2
+ h2 r =12
(2p)(1127.03)
x =20h50
=2h5
= 2p(1127.03) mm2
A = uz rL = 2p{202(20)2 + (50)2 + 5(10)}
Determine the height h to which liquid should be poured intothe cup so that it contacts half the surface area on the insideof the cup. Neglect the cup’s thickness for the calculation.
50 mm
10 mm
h
30 mm
~
9–112.
The water tank has a paraboloid-shaped roof. If one liter ofpaint can cover 3 m2 of the tank, determine the number ofliters required to coat the roof.
SOLUTION
Length and Centroid: The length of the differential element shown shaded in Fig. a is
where
Integrating,
The centroid of the line can be obtained by applying Eq. 9–5 with
Surface Area: Applying the first theorem of Pappus and Guldinus and using theresults obtained above with
Thus, the amount of paint required is
Ans.# of liters =459.39
3= 153 liters
A = 2prL = 2p(6.031)(12.124) = 459.39 m2
r = x = 6.031 m, we have
x = LLx~ dL
LLdL
= L12 m
0x c 1
482482 + x2 dx d
12.124=
73.11412.124
= 6.031 m
xc = x.x
L = LLdL = L
12 m
0
1482482 + x2 dx = 12.124 m
dL = B1 + a - 148
xb2
dx = B1 +x2
482 dx =
1482482 + x2 dx
dy
dx= -
148
x. Thus,
dL = 2dx2 + dy2 = A1 + adydxb2
dx
x
y
2.5 m
12 m
y � (144 � x2)1––96
9–113.
A steel wheel has a diameter of 840 mm and a cross sectionas shown in the figure. Determine the total mass of thewheel if r = 5 Mg>m3.
30 mm
30 mm
80 mm
Section A–A
100 mm
250 mm420 mm
840 mm
60 mm
A
A
SOLUTION
Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–12, with
evah ew dna
The mass of the wheel is
Ans.= 138 kg
m = rV = 51103238.775110-32p4
= 8.775p110-32m3
V = u©rA = 2p30.09510.0032 + 0.23510.00752 + 0.3910.00624A3 = 10.1210.062 = 0.006 m2,A2 = 0.2510.032 = 0.0075 m2
A1 = 0.110.032 = 0.003 m2,r3 = 0.39 m,r2 = 0.235 m,r1 = 0.095 m,u = 2p,
9–114.
Determine the surface area of the roof of the structure if itis formed by rotating the parabola about the y axis.
SOLUTION
Centroid: The length of the differential element is
and its centroid is . Here, . Evaluating the
integrals, we nave
Applying Eq. 9–5, we have
Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–7, with ,, , we have
Ans.A = urL = 2p(9.178) (23.663) = 1365 m2
r = x = 9.178L = 23.663 mu = 2p
x = LLx'
dL
LLdL
=217.18123.663
= 9.178 m
LLx' '
dL = L16 m
0x¢C1 +
x2
64≤dx = 217.181 m2
L = L dL = L16 m
0¢C1 +
x2
64≤dx = 23.663 m
dy
dx= -
x
8x = x= ¢C1 + ady
dxb2≤dx
dL = 2dx2 + dy2
16 m
y
x
16 m
y 16 (x2/16)
.
9–116.
The loading acting on a square plate is represented by aparabolic pressure distribution. Determine the magnitudeof the resultant force and the coordinates of the pointwhere the line of action of the force intersects the plate.Also, what are the reactions at the rollers B and C and theball-and-socket joint A? Neglect the weight of the plate.
1x, y2
SOLUTION
Ans.
Ans.
Ans.
Ans.
Ans.Ay = 17.1 kN
+ c ©F = 0; Ay - 42.67 + 12.8 + 12.8 = 0
By = Cy = 12.8 kN
©Mx = 0; 42.67(2.40) - 2By (4) = 0
©My = 0; By = Cy
FR = 10.67(4) = 42.7 kN
y = Ly dA
L dA=
25.610.67
= 2.40 m
L y'
'
dA = L4
02y3/2 dy = B 4
5y5/2R4
0= 25.6 kN
L dA = L4
02y1/2 dy = B 4
3y3/2R4
0= 10.67 kN/m
x = 0 (Due to symmetry)
dA = p dy
y = y2 m4 m
4 m
x
y
p
p 2y1/2
A
BC
4 kPa
9–117.
SOLUTIONResultant Force and its Location: The volume of the differential element is
and its centroid is at and .
Ans.
Ans.
Ans.y = LFRy'
dFR
LFRdFR
=32.024.0
= 1.33 m
x = LFRx'
dFR
LFRdFR
=48.024.0
= 2.00 m
=23B ax2
2b 2 3m
0a2y2 -
y3
3b 2 4 m
0R = 32.0 kN # m
LFR
y'
dFR = L3 m
0
23
(xdx)L4 m
0y(4 - y)dy
=23B ax3
3b 2 3 m
0a4y -
y2
2b 2 4m
0R = 48.0 kN # m
LFR
xdFR = L3 m
0
23
(x2dx)L4 m
0(4 - y)dy
=23B ax2
2b 2 3
0
m a4y -y2
2b 2 4
0
mR = 24.0 kN
FR = LFk
dFR = L3 m
0
23
(xdx)L4 m
0(4 - y)dy
y' = yx
' = xdV = d FR = pdxdy =23
(xdx)[(4 - y)dy]
The load over the plate varies linearly along the sides of theplate such that Determine the resultantforce and its position on the plate.1x, y2
p = 23[x14 - y2] kPa.
p
3 m
4 m
y
x
8 kPa
9–118.
The rectangular plate is subjected to a distributed load overits entire . The load is defined by the expression
where represents the pressure acting at the center of the plate. Determine the magnitude and location of the resultant force acting on the plate.
p0
p = p0
sin (px>a)
xb
a
p0
y
p
SOLUTIONResultant Force and its Location: The volume of the differential element is
.
Ans.
Since the loading is symmetric, the location of the resultant force is at the center ofthe plate. Hence,
Ans.x =a
2y =
b
2
=4ab
p2 p0
= p0B a - ap
cos pxab 2 a
0a - bp
cos px
bb 2 b
0R
FR = LFR
dFR = p0La
0asin
pxa
dxbLb
0asin
py
bdyb
dV = dFR = pdxdy = p0 asin pxa
dxb asin
py
bdyb
sin (py>b),surface
.
.
.
MN
MN
9–120.
The tank is filled with water to a depth of Determine the resultant force the water exerts on side A andside B of the tank. If oil instead of water is placed in the tank,to what depth d should it reach so that it creates the sameresultant forces? and rw = 1000 kg>m3.ro = 900 kg>m3
d = 4 m.
SOLUTION
For water
At side A:
Ans.
At side B:
Ans.
For oil
At side A:
Ans.d = 4.22 m
FRA=
12
(17 658 d)(d) = 156 960 N
= 17 658 d
= 2(900)(9.81)d
wA = b ro g d
FRB=
12
(117 720)(4) = 235 440 N = 235 kN
= 117 720 N>m= 3(1000)(9.81)(4)
wB = b rw g d
FRA=
12
(78 480)(4) = 156 960 N = 157 kN
= 78 480 N/m
= 2(1000)(9.81) (4)
wA = b rw g d
dBA
3 m 2 m
9–121.
When the tide water A subsides, the tide gate automaticallyswings open to drain the marsh B. For the condition of hightide shown, determine the horizontal reactions developedat the hinge C and stop block D. The length of the gate is6 m and its height is 4 m. rw = 1.0 Mg/m3.
SOLUTION
Fluid Pressure: The fluid pressure at points D and E can be determined using Eq. 9–13,
Thus,
Resultant Forces:
Equations of Equilibrium:
Ans.
Ans.Cx = 46.6 kN
264.87 - 117.72 - 100.55 - Cx = 0:+ ©Fx = 0;
Dx = 100.55 kN = 101 kN
264.87132 - 117.7213.3332 - Dx 142 = 0a+©MC = 0;
FR2=
121117.722122 = 117.72 kN
FR1=
121176.582132 = 264.87 kN
wE = 29.43162 = 176.58 kN>m
wD = 19.62162 = 117.72 kN>m
pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2
pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2
p = rgz.
AB
C
D
3 m2 m
4 m
9 1
The symmetric concrete “gravity” dam is held in place by its own weight. If the density of concrete is�c and water has a density �w, determine the smallest distance d at its base that will prevent the damfrom overturning about its end A.The dam has a width w.
Units Used:
Mg 103 kg� MN 106 N�
Given:
a 1.5 m� �c 2.5Mg
m3�
b 9 m�
�w 1.0Mg
m3�w 8 m�
Solution:
Guesses
d 3 m� Fh 1 MN�
Fv 1 MN� W 1 MN�
Given
Fvd a
4b w�w g�
Fh12�w g b wb�
W �c g w a bd a
2���
���
b���
��
�
Wd2
Fv dd a
6��
����
� Fhb3
0�
Fv
Fh
W
d
������
������
Find Fv Fh� W� d�� ��
Fv
Fh
W
�����
�����
0.379
3.178
4.545
����
����
MN� d 3.65 m�
22.
Ans.
–
9–123.
SOLUTION
Ans.
Ans.h =23
(6) = 4 m
F =12
(6)(8)(58 860) = 1.41(106) N = 1.41 MN p = 6(1)(103)(9.81) = 58 860 N/m2
Determine the magnitude of the resultant hydrostaticforce acting on the dam and its location, measured fromthe top surface of the water. The width of the dam is 8 m;rw = 1.0 Mg>m3.
6 m
9–124.
The concrete dam in the shape of a quarter circle.Determine the magnitude of the resultant hydrostatic forcethat acts on the dam per meter of length. The density ofwater is
SOLUTION
Loading: The hydrostatic force acting on the circular surface of the dam consists ofthe vertical component and the horizontal component as shown in Fig. a.
Resultant Force Component: The vertical component consists of the weightof water contained in the shaded area shown in Fig. a. For a 1-m length of dam,we have
The horizontal component consists of the horizontal hydrostatic pressure.Since the width of the dam is constant (1 m), this loading can be representedby a triangular distributed loading with an intensity of
at point C, Fig. a.
Thus, the magnitude of the resultant hydrostatic force acting on the dam is
Ans.FR = 3Fh2 + Fv
2 = 344.145
2 + 18.95
2 = 48.0 kN
Fh =12
(29.43)(3) = 44.145 kN
1000(9.81)(3)(1) = 29.43 kN>mwC = rghCb =
Fh
Fv = rgAABCb = (1000)(9.81)B(3)(3) -p
4 (32)R(1) = 18947.20 N = 18.95 kN
Fv
FhFv
rw = 1 Mg>m3.
3 m
���.
The storage tank contains oil having a specific weight �. If the tank has width w, calculate theresultant force acting on the inclined side BC of the tank, caused by the oil, and specify itslocation along BC, measured from B. Also compute the total resultant force acting on the bottomof the tank.����������
�� 103 ���
������
� 9 �
�3��
� 8���
! 6��� � 4���
" 10��� � 3���
� 2��� # 4���
���������
!$ ! �� ���� !$ 72573�
����#�
� !% ! �� � ��( )��� !% 540 ��
�
&'� !$ ���� &'�12
!% !$� �� ���� &�� � !� �� ���� &��12�� !� �� ����
The resultant force
&() &'� &'���� &(* &�� &����� &( &()2 &(*
2��� &( 2.77���
The location h measured from point B ����� ' 1#��� �����
&���2� &��
2 ��3
�� &'��2�� &'�
2 ��3
�� &()� '�
�2 �2�
� &(*� '�
�2 �2�
��=
' &��� '( )�� ' 5.22��On the bottom of the tank
&��� � !� #� � �� ��( )��� &��� 3.02���
Ans.
Ans.
Ans.
–
9–126.
The tank is filled to the top with water havinga density of Determine the resultant forceof the water pressure acting on the flat end plate C of thetank, and its location, measured from the top of the tank.
rw = 1.0 Mg/m3.1y = 0.5 m2
0.5 m
x2 + y2 = (0.5)2
x
y
C
SOLUTION
Ans.
Hence, measured from the top of the tank,
Ans.d¿ = 0.5 + 0.125 = 0.625 m
d =-0.481
3.85= -0.125 m
F1-d2 = Ly dF
= -0.481 kN m
y
4C2510.522 - y263 D-0.5
0.5 -10.522
8cy210.522 - y2 + 10.522 sin-1 y
0.5d-0.5
0.5 rLA
y dF = 219.812L0.5
-0.510.5y - y221210.522 - y2dy = 19.62b c - 0.5
32510.522 - y263 d
-0.5
0.5
+
F = 3.85 kN
+219.812
3C2510.522 - y263 D-0.5
0.5
=9.81
2cy210.522 - y2 + 0.52 sin-1 a y
0.5b d
-0.5
0.5
F = 219.812L0.5
-0.510.5 - y21210.522 - y2 dy
dF = p dA = 11219.81210.5 - y2 2x dy
9–127.
SOLUTIONFluid Pressure: The fluid pressure at an arbitrary point along y axis can bedetermined using Eq. 9–13, .
Resultant Force and its Location: Here, . The volume of thedifferential element is . the
Evaluating integrals using Simpson’s rule, we have
Ans.
Ans.y =LFRy
'dFR
LFRdFR=
-866.76934.2
= -0.125 m
= -866.7 N # mLFRyd FR = 17658 L
0.5 m
-0.5 my(0.5 - y)(21 - 4y2)dy
= 6934.2 N = 6.93 kN
FR = LFRd FR = 17658 L0.5 m
-0.5 m(0.5 - y)(21 - 4y2)dy
dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y2] dyx = 21 - 4y2
(0.5 - y) = 8829(0.5 - y)p = g(0.5 - y) = 900(9.81)
The tank is filled with a liquid which has a density ofDetermine the resultant force that it exerts on
the elliptical end plate, and the location of the center ofpressure, measured from the x axis.
900 kg>m3.
0.5 m
1 m
0.5 m
1 m
x
y
4 y2 x2 1
The underwater tunnel in the aquatic center isfabricated from a transparent polycarbonate materialformed in the shape of a parabola. Determine the magnitudeof the hydrostatic force that acts per meter length along thesurface AB of the tunnel. The density of the water is
.rw = 1000 kg/m3
y
x
2 m 2 m
2 m
4 m
y � 4 � x2 A
B
9–128.
9–129.
SOLUTION
Ans.F = 391 kN>mF = 9.81112162162 + 211219.812c3132 -
p
41322 d
The semicircular tunnel passes under a river which is 9 m deep.Determine the vertical resultant hydrostatic force acting permeter of length along the length of the tunnel. The tunnel is 6m wide; rw = 1.0 Mg/m3.
6 m
9 m
9–130.
The arched surface AB is shaped in the form of a quartercircle. If it is 8 m long, determine the horizontal and verticalcomponents of the resultant force caused by the wateracting on the surface. rw = 1.0 Mg>m3.
SOLUTION
Ans.
Ans.Fy = 470.88 + 67.37 = 538 kN
Fx = 156.96 + 470.88 = 628 kN
w = c(2)2 -14p(2)2 d (8)(1000)(9.81) = 67.37 kN
F1 = 1000(9.81)(2)a12b(2)(8) = 156.96 kN
F2 = 1000(9.81)(3)(2)(8) = 470.88 kN
F3 = 1000(9.81)(3)(2)(8) = 470.88 kN
3 m
2 m
B
A
9–131.
A circular V-belt has an inner radius of 600 mm and a cross-sectional area as shown. Determine the volume of materialrequired to make the belt.
SOLUTION
Ans.= 22.7(10)-3 m3
= 2p C0.65(2)a12b (0.025)(0.075) + 0.6375(0.05)(0.075) D
V = ©urA
50 mm25 mm 25 mm
75 mm
600 mm
~
9–132.
SOLUTION
Ans.= 1.25 m2
= 2 p[0.6(0.05) + 2(0.6375)(2(0.025)2 + (0.075)2) + 0.675(0.1)]
A = ©urL
A circular V-belt has an inner radius of 600 mm and across-sectional area as shown. Determine the surface areaof the belt.
50 mm25 mm 25 mm
75 mm
600 mm
~
9–133.
100 mm
25 mm
25 mm
x
25 mm
y
50 mm 50 mm
y
75 mm75 mm
C
Locate the centroid of the beam’s cross-sectional area.y
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
Segment A (mm2)1 300(25) 112.5 843 7502 100(50) 50 250 000
12 500 1 093 750
Thus,
Ans.y =©yA
©A=
1 093 75012 500
= 87.5 mm
©
y'
A (mm3)y'
'
(mm)
9–134.
SOLUTIONVolume and Moment Arm: The volume of the thin disk differential element is
and its centroid is at .
Centroid: Due to symmetry about the z axis
Ans.
Applying Eq. 9–5 and performing the integration, we have
Ans.=paaaz2
2-z3
6b ` 2a
0
paaaz- z2
4b ` 2a
0
=23a
z' = LV
z'dV
LVdV= L
2a
0z cpaaa- z
2bdz d
L2a
0paaa- z
2bdz
x' = y' = 0
z' = zdV = py2dz = x caaa- z
2b ddz = paaa- z
2bdz
Locate the centroid of the solid. z
xy
a
y2 � a (a �z2)
9 1
Locate the centroid xc of thetriangular area.
Solution:
Ab h2
�
xc2
b h0
a
xxha
x����
d
a
b
xxh
b ab x( )
����
d����
���
� xca b�
3�
35.
Ans.
–
.
9
Determine the location (xc, yc) of the center of mass of the turbine and compressor assembly. Themass and the center of mass of each of the various components are indicated below.
Given:
a 0.75 m� M1 25 kg�
b 1.25 m� M2 80 kg�
c 0.5 m� M3 30 kg�
d 0.75 m� M4 105 kg�
e 0.85 m�
f 1.30 m�
g 0.95 m�
Solution:
M M1 M2� M3� M4��
xc1M
M2 a M3 a b�( )� M4 a b� c�( )�� �� xc 1.594 m�
yc1M
M1 d M2 e� M3 f� M4 g�� �� yc 0.940 m�
137.
Ans.
Ans.
–
9
Each of the three homogeneous plates welded to the rod has a density � and a thickness a. Determinethe length l of plate C and the angle of placement, �, so that the center of mass of the assembly lies onthe y axis. Plates A and B lie in the x–y and z–y planes, respectively.
Units Used:
Mg 1000 kg�
Given:
a 10 mm� f 100 mm�
b 200 mm� g 150 mm�
c 250 mm� e 150 mm�
� 6Mg
m3�
Solution: The thickness and density are uniform
Guesses � 10 deg� l 10 mm�
Given
b ff2������
g lg2������
cos �� � 0� c ee2������
g lg2������
sin �� �� 0�
l
����
���
Find l ��� �� l 265 mm� � 70.4 deg�
138.
Ans.
–
9–139.
The gate AB is 5 m wide. Determine the horizontal andvertical components of force acting on the pin at B and thevertical reaction at the smooth support A. rw = 1.0 Mg>m3.
SOLUTIONFluid Pressure: The fluid pressure at points A and B can be determined using Eq. 9–15, .
Thus,
Resultant Forces:
Equations of Equilibrium:
Ans.
Ans.
Ans.By = kN
+ c©Fy = 0; 1 - a35b - 1471.5a3
5b - By = 0
Bx = kN =
:+ ©Fx = 0; 490.5a45b + 1 a 4
5b - Bx = 0
Ay = kN = . MN
+©MB = 0; 1471.5(2.5) + - Ay(3) = 0
FR2=
12
(490.5 - = kN
FR1= = 1471.5 kN
wB = = kN/m
wA = 8. ( ) = . kN/m
PB = 1.0(10 )(9.81)(6)3 = 2 = kN/m2
PA = 1.0(10 )(9.81)(10)3 = 2 = kN/m2
p = rgz4 m
3 m
A
B
98 100 N/m 98.10
58 860 N/m 58.86
9 1 5 490 5
58.86(5) 294.3
294.3(5)
294.3)(5) 490.5
490.5(3.333)
17 2.71 1 77
471.5
1569.6 1.57 MN
771.2 490.5
594
WB = 294.3 kN/m
1471.5 kN
490.5 kN
490.5 kN/m
6 m
9–140.
The pressure loading on the plate is described by the functionDetermine the magnitude
of the resultant force and coordinates of the point where theline of action of the force intersects the plate.
p = 5-240>1x + 12 + 3406 Pa.
SOLUTION
Resultant Force and its Location: The volume of the differential element is
and its centroid in .
Ans.
Ans.
Due to symmetry,
Ans.y = 3.00 m
x =LFRx
'dFR
LFRdFR=
20 880.137619.87
= 2.74 m
= 20880.13 N #m= [-1440[x - In(x + 1)] + 1020x2]|50
m
LFRxd FR = L5 m
06xa -
240x + 1
+ 340b= 7619.87 N = 7.62 kN
= 6[-240 In(x + 1) + 340x2]|50m
FR = LFRdFR = L5 m
06a -
240x + 1
+ 340bdx
x' = xdV = dFR = 6pdx = 6 a -
240x+1
+ 340bdx
p
x
y
6 m
5 m
100 Pa
300 Pa
