PROBLEM 5.73
Locate the centroid of the plane area shown.
SOLUTION
2
, mmA
, mmx
, mmy
3
, mmxA
3
, mmyA
1
126 54 6804×=
9
27
61,236
183,708
2
1126 30 1890
2× ×=
30
64
56,700 120,960
3
172 48 1728
2××=
48
16
−
82,944
27,648
−
Σ
10,422
200,880
277,020
Then
X A xAΣ=Σ
22
(10,422 m ) 200,880 mmX=
or
19.27 mmX=
and
Y A yAΣ=Σ
23
(10,422 m ) 270,020 mmY=
or
26.6 mmY=
consent of McGraw–Hill Education.
PROBLEM 5.74
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Rectangle 16 in. by 8 in.
Area 2: Semicircle radius of 5 in.
2
, in
A
, inx
, iny
3
,inxA
3
,inyA
1 128 0 4 0 512
2 -39.27 0 5.878 0 -230.83
Σ
88.73 0 281.17
X A xAΣ=
0 in.X=
Y A yAΣ=Σ
23
(88.73 in ) 281.17 inY=
3.17 in.Y=
consent of McGraw–Hill Education.
PROBLEM 5.75
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, mm
, mmx
, mmy
2
, mmxL
2
, mmyL
1
22
72 48 86.533+=
36 −24 3115.2 −2076.8
2 132 72 18 9504.0 2376.0
3
22
126 30 129.522+=
9 69 1165.70 8937.0
4 54 −54 27 −2916.0 1458.0
5 54 −27 0 −1458.0 0
Σ
456.06 9410.9 10,694.2
Then
X L xLΣ=Σ
(456.06) 9410.9
X=
or
20.6 mmX=
Y L yLΣ=Σ
(456.06) 10,694.2Y=
or
23.4 mmY=
consent of McGraw–Hill Education.
PROBLEM 5.76
Member ABCDE is a component of a mobile and is formed from a
single piece of aluminum tubing. Knowing that the member is
supported at C and that d is 0.50 m, determine the length l of arm DE
so that this portion of the member is horizontal.
SOLUTION
2
consent of McGraw–Hill Education.
PROBLEM 5.77
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and h.
SOLUTION
For the element (el) shown at
2 , x ay h= =
or
( )
2
2h ka=
2
= 4
h
ka
∴
el
xx=
1
2
el
yy=
dA ydx=
2
3
22
22
0
0
Then 3
44
7
12
a
a
h hx
A dA x dx
aa
ah
= = =
=
∫∫
2
4
22
22
2
and 4
44
15
16
a
a
el a
a
h hx
x dA x x dx
aa
ha
= =
=
∫∫
( )
22
2
22
24
24
2
25 2
4
1
2
1
24 32
31
5 160
32
a
el a
a
a
a
a
y dA y dx
hx h
dx x dx
aa
hx ah
a
=
= =
= =
∫∫
∫∫
Then
2
7 15
: 12 16
el
xA x dA x ah a h
= =
∫
or
1.607xa=
2
7 31
:12 160
EL
yA y dA y ah ah
= =
∫
or
0.332yh=
consent of McGraw–Hill Education.
PROBLEM 5.78
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
2 22
11 1 1 2
3 33
22 2 2 3
3
2
21 2
12
3
2
2
but
but
()
1()
2
2
EL
EL
b
y kx b ka y x
a
b
y kx b ka y x
a
bx
dA y y dx x dx
a
a
xx
y yy
bx
xa
a
= = =
= = =
=−= −
=
= +
= +
3
2
20
34
20
3
2
2
0
4
3
20
45
20
2
34
1
12
45
1
20
a
a
a
EL
a
a
bx
A dA x dx
a
a
bx x
a
a
ba
bx
x dA x x dx
a
a
bx
x dx
a
a
bx x
a
a
ab
= = −
= −
=
= −
= −
= −
=
∫∫
∫∫
∫
consent of McGraw–Hill Education.
SOLUTION Continued
33
22
22
0
26
4
42
0
25 7 2
42
0
2
2
1
5 35
27
a
EL
a
a
b xb x
y dA x x dx
aa
aa
bx
x dx
aa
bx x ab
aa
=+−
= −
= −=
∫∫
∫
2
11
:12 20
EL
xA x dA x ba a b
= =
∫
3
5
xa=
2
11
:12 35
EL
yA y dA y ba ab
= =
∫
12
35
yb=
PROBLEM 5.79
A
3
4–
in.–diameter hole is drilled in a piece of 1–in.–thick steel; the hole
is then countersunk as shown. Determine the volume of steel removed
during the countersinking process.
SOLUTION
The required volume can be generated by rotating the area shown about the y-axis. Applying the second
theorem of Pappus-Guldinus, we have
2
3 11 1 1 1
2 in. in. in.
8 34 2 4 4
V xA
π
π
=
= + ×× ×
3
0.0900 inV=
consent of McGraw–Hill Education.
PROBLEM 5.80
A manufacturer is planning to produce 20,000 wooden pegs
having the shape shown. Determine how many gallons of
paint should be ordered, knowing that each peg will be given
two coats of paint and that one gallon of paint covers 100 ft2.
SOLUTION
The number of gallons of paint needed is given by
2
1 gallon
Number of gallons (Number of pegs)(Surface area of 1 peg) (2 coats)
100 ft
=
or
2
Number of gallons 400 ( ft )
ss
AA=�
We have
0.875 in.
0.5
sin 2 0.875
R
α
=
=
or
2 34.850° 17.425°
22
s
A YL yL
αα
ππ
= =
= = Σ
, in.L
, in.y
2
, in.yL
1
0.25
0.125
0.03125
2
0.5
0.25
0.125
3
0.0625
0.25 0.3125 0.28125
2
+=
0.0175781
4
3 0.875(1 cos34.850) 0.1875 2.6556−− −=
0.3125
0.82988
5
0.1875 0.29452
2
π
×=
2 0.1875
0.5 0.38063
π
×
−=
0.112103
6
2 (0.875)
α
0.875sin17.425 sin17.425
α
°×°
0.137314
2
1.25312 in.yLΣ=
consent of McGraw–Hill Education.
PROBLEM 5.73
Locate the centroid of the plane area shown.
SOLUTION
2
, mmA
, mmx
, mmy
3
, mmxA
3
, mmyA
1
126 54 6804×=
9
27
61,236
183,708
2
1126 30 1890
2× ×=
30
64
56,700 120,960
3
172 48 1728
2××=
48
16
−
82,944
27,648
−
Σ
10,422
200,880
277,020
Then
X A xAΣ=Σ
22
(10,422 m ) 200,880 mmX=
or
19.27 mmX=
and
Y A yAΣ=Σ
23
(10,422 m ) 270,020 mmY=
or
26.6 mmY=
consent of McGraw–Hill Education.
PROBLEM 5.74
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Rectangle 16 in. by 8 in.
Area 2: Semicircle radius of 5 in.
2
, in
A
, inx
, iny
3
,inxA
3
,inyA
1 128 0 4 0 512
2 -39.27 0 5.878 0 -230.83
Σ
88.73 0 281.17
X A xAΣ=
0 in.X=
Y A yAΣ=Σ
23
(88.73 in ) 281.17 inY=
3.17 in.Y=
consent of McGraw–Hill Education.
PROBLEM 5.75
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, mm
, mmx
, mmy
2
, mmxL
2
, mmyL
1
22
72 48 86.533+=
36 −24 3115.2 −2076.8
2 132 72 18 9504.0 2376.0
3
22
126 30 129.522+=
9 69 1165.70 8937.0
4 54 −54 27 −2916.0 1458.0
5 54 −27 0 −1458.0 0
Σ
456.06 9410.9 10,694.2
Then
X L xLΣ=Σ
(456.06) 9410.9
X=
or
20.6 mmX=
Y L yLΣ=Σ
(456.06) 10,694.2Y=
or
23.4 mmY=
consent of McGraw–Hill Education.
PROBLEM 5.76
Member ABCDE is a component of a mobile and is formed from a
single piece of aluminum tubing. Knowing that the member is
supported at C and that d is 0.50 m, determine the length l of arm DE
so that this portion of the member is horizontal.
SOLUTION
2
consent of McGraw–Hill Education.
PROBLEM 5.77
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and h.
SOLUTION
For the element (el) shown at
2 , x ay h= =
or
( )
2
2h ka=
2
= 4
h
ka
∴
el
xx=
1
2
el
yy=
dA ydx=
2
3
22
22
0
0
Then 3
44
7
12
a
a
h hx
A dA x dx
aa
ah
= = =
=
∫∫
2
4
22
22
2
and 4
44
15
16
a
a
el a
a
h hx
x dA x x dx
aa
ha
= =
=
∫∫
( )
22
2
22
24
24
2
25 2
4
1
2
1
24 32
31
5 160
32
a
el a
a
a
a
a
y dA y dx
hx h
dx x dx
aa
hx ah
a
=
= =
= =
∫∫
∫∫
Then
2
7 15
: 12 16
el
xA x dA x ah a h
= =
∫
or
1.607xa=
2
7 31
:12 160
EL
yA y dA y ah ah
= =
∫
or
0.332yh=
consent of McGraw–Hill Education.
PROBLEM 5.78
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
2 22
11 1 1 2
3 33
22 2 2 3
3
2
21 2
12
3
2
2
but
but
()
1()
2
2
EL
EL
b
y kx b ka y x
a
b
y kx b ka y x
a
bx
dA y y dx x dx
a
a
xx
y yy
bx
xa
a
= = =
= = =
=−= −
=
= +
= +
3
2
20
34
20
3
2
2
0
4
3
20
45
20
2
34
1
12
45
1
20
a
a
a
EL
a
a
bx
A dA x dx
a
a
bx x
a
a
ba
bx
x dA x x dx
a
a
bx
x dx
a
a
bx x
a
a
ab
= = −
= −
=
= −
= −
= −
=
∫∫
∫∫
∫
consent of McGraw–Hill Education.
SOLUTION Continued
33
22
22
0
26
4
42
0
25 7 2
42
0
2
2
1
5 35
27
a
EL
a
a
b xb x
y dA x x dx
aa
aa
bx
x dx
aa
bx x ab
aa
=+−
= −
= −=
∫∫
∫
2
11
:12 20
EL
xA x dA x ba a b
= =
∫
3
5
xa=
2
11
:12 35
EL
yA y dA y ba ab
= =
∫
12
35
yb=
PROBLEM 5.79
A
3
4–
in.–diameter hole is drilled in a piece of 1–in.–thick steel; the hole
is then countersunk as shown. Determine the volume of steel removed
during the countersinking process.
SOLUTION
The required volume can be generated by rotating the area shown about the y-axis. Applying the second
theorem of Pappus-Guldinus, we have
2
3 11 1 1 1
2 in. in. in.
8 34 2 4 4
V xA
π
π
=
= + ×× ×
3
0.0900 inV=
consent of McGraw–Hill Education.
PROBLEM 5.80
A manufacturer is planning to produce 20,000 wooden pegs
having the shape shown. Determine how many gallons of
paint should be ordered, knowing that each peg will be given
two coats of paint and that one gallon of paint covers 100 ft2.
SOLUTION
The number of gallons of paint needed is given by
2
1 gallon
Number of gallons (Number of pegs)(Surface area of 1 peg) (2 coats)
100 ft
=
or
2
Number of gallons 400 ( ft )
ss
AA=�
We have
0.875 in.
0.5
sin 2 0.875
R
α
=
=
or
2 34.850° 17.425°
22
s
A YL yL
αα
ππ
= =
= = Σ
, in.L
, in.y
2
, in.yL
1
0.25
0.125
0.03125
2
0.5
0.25
0.125
3
0.0625
0.25 0.3125 0.28125
2
+=
0.0175781
4
3 0.875(1 cos34.850) 0.1875 2.6556−− −=
0.3125
0.82988
5
0.1875 0.29452
2
π
×=
2 0.1875
0.5 0.38063
π
×
−=
0.112103
6
2 (0.875)
α
0.875sin17.425 sin17.425
α
°×°
0.137314
2
1.25312 in.yLΣ=
consent of McGraw–Hill Education.