SOLUTION Continued
Then
2
22
1ft
PROBLEM 5.81
Determine the reactions at the beam supports for the given
loading when wO = 400 lb/ft.
SOLUTION
I
II
11
(12 ft) (400 lb/ft)(12 ft) 2400 lb
22
1(300 lb/ft)(12 ft) 1800 lb
2
O
Rw
R
= = =
= =
0: (2400 lb)(1ft) (1800 lb)(3 ft) (7 ft) 0
B
MCΣ= − + =
428.57 lbC=
0: 428.57 lb 2400 lb 1800 lb 0
y
FBΣ= + − − =
3771lbB=
3770 lb=B
consent of McGraw–Hill Education.
PROBLEM 5.82
Determine (a) the distributed load wO at the end A of the beam
ABC for which the reaction at C is zero, (b) the corresponding
reaction at B.
SOLUTION
For
,
O
w
I
II
1(12 ft) 6
2
1(300 lb/ft)(12 ft) 1800 lb
2
OO
Rw w
R
= =
= =
(a) For
0,C=
0: (6 )(1ft) (1800 lb)(3 ft) 0
BO
Mw
Σ= − =
900 lb/ft
O
w=
(b) Corresponding value of
I
:R
I
6(900) 5400 lbR= =
0: 5400 lb 1800 lb 0
y
FBΣ= − − =
7200 lb=B
consent of McGraw–Hill Education.
PROBLEM 5.83
For the machine element shown, locate the z coordinate of
the center of gravity.
SOLUTION
3
, mmV
, mmx
, mmz
4
, mmxV
4
, mmzV
I Rectangular plate
3
(10)(90)(38) 34.2 10= ×
19 45 649.8 × 103 1539 × 103
II Half cylinder
23
(20) (10) 6.2832 10
2
π
= ×
46.5
20 292.17 × 103 125.664 × 103
III –(Cylinder)
23
(12) (10) 4.5239 10
π
− =−×
38 20 −171.908 × 103 −90.478 × 103
IV Rectangular prism
3
(30)(10)(24) 7.2 10= ×
5 78 36 × 103 561.6 × 103
V Triangular prism
3
1(30)(9)(24) 3.24 10
2= ×
13 78 42.12 × 103 252.72 × 103
Σ 46.399 × 10
3
848.18 × 10
3
2388.5 × 10
3
Dimensions in mm
SOLUTION Continued
34
33
2388.5 10 mm
46.399 10 mm
Z V zV
zV
Z
V
Σ=Σ
Σ×
= =
Σ
×
51.5 mmZ=
consent of McGraw–Hill Education.
PROBLEM 5.84
A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is
1030 kg/m3 and of steel is
3
7860 kg/m ,
locate the center of gravity of the awl.
SOLUTION
First, note that symmetry implies
0YZ= =
I
33
I
3
II
32
II
3
III
32
III
5(12.5 mm) 7.8125 mm
82
(1030 kg/m ) (0.0125 m)
3
4.2133 10 kg
52.5 mm
(1030 kg/m ) (0.025 m) (0.08 m)
4
40.448 10 kg
92.5 mm 25 mm 67.5 mm
(1030 kg/m ) (0.0035 m) (0.
4
x
W
x
W
x
W
π
π
π
−
−
= =

=

= ×
=

=

= ×
= −=

= − 

3
05 m)
0.49549 10 kg
−
=−×
IV
3 22 3
IV
V
32 3
V
182.5 mm 70 mm 112.5 mm
(7860 kg/m ) (0.0035 m) (0.14 m) 10.5871 10 kg
4
1
182.5 mm (10 mm) 185 mm
4
(7860 kg/m ) (0.00175 m) (0.01 m) 0.25207 10 kg
3
x
W
x
W
π
π
−
−
= −=

= = ×


=+=

= = ×


consent of McGraw–Hill Education.
SOLUTION Continued
, kgW
, mmx
, kg mmxW ⋅
I
3
4.123 10
−
×
7.8125
3
32.916 10
−
×
II
3
40.948 10−
×
52.5
3
2123.5 10−
×
III
3
0.49549 10
−
−×
67.5
3
33.447 10
−
−×
IV
3
10.5871 10−
×
112.5
3
1191.05 10
−
×
V
3
0.25207 10
−
×
185
3
46.633 10−
×
Σ
3
55.005 10
−
×
3
3360.7 10
−
×
We have
33
: (55.005 10 kg) 3360.7 10 kg mmX W xW X
−−
Σ=Σ × = × ⋅
or
61.1mmX=
(from the end of the handle)
consent of McGraw–Hill Education.
PROBLEM 5.81
Determine the reactions at the beam supports for the given
loading when wO = 400 lb/ft.
SOLUTION
I
II
11
(12 ft) (400 lb/ft)(12 ft) 2400 lb
22
1(300 lb/ft)(12 ft) 1800 lb
2
O
Rw
R
= = =
= =
0: (2400 lb)(1ft) (1800 lb)(3 ft) (7 ft) 0
B
MCΣ= − + =
428.57 lbC=
0: 428.57 lb 2400 lb 1800 lb 0
y
FBΣ= + − − =
3771lbB=
3770 lb=B
consent of McGraw–Hill Education.
PROBLEM 5.82
Determine (a) the distributed load wO at the end A of the beam
ABC for which the reaction at C is zero, (b) the corresponding
reaction at B.
SOLUTION
For
,
O
w
I
II
1(12 ft) 6
2
1(300 lb/ft)(12 ft) 1800 lb
2
OO
Rw w
R
= =
= =
(a) For
0,C=
0: (6 )(1ft) (1800 lb)(3 ft) 0
BO
Mw
Σ= − =
900 lb/ft
O
w=
(b) Corresponding value of
I
:R
I
6(900) 5400 lbR= =
0: 5400 lb 1800 lb 0
y
FBΣ= − − =
7200 lb=B
consent of McGraw–Hill Education.
PROBLEM 5.83
For the machine element shown, locate the z coordinate of
the center of gravity.
SOLUTION
3
, mmV
, mmx
, mmz
4
, mmxV
4
, mmzV
I Rectangular plate
3
(10)(90)(38) 34.2 10= ×
19 45 649.8 × 103 1539 × 103
II Half cylinder
23
(20) (10) 6.2832 10
2
π
= ×
46.5
20 292.17 × 103 125.664 × 103
III –(Cylinder)
23
(12) (10) 4.5239 10
π
− =−×
38 20 −171.908 × 103 −90.478 × 103
IV Rectangular prism
3
(30)(10)(24) 7.2 10= ×
5 78 36 × 103 561.6 × 103
V Triangular prism
3
1(30)(9)(24) 3.24 10
2= ×
13 78 42.12 × 103 252.72 × 103
Σ 46.399 × 10
3
848.18 × 10
3
2388.5 × 10
3
Dimensions in mm
SOLUTION Continued
34
33
2388.5 10 mm
46.399 10 mm
Z V zV
zV
Z
V
Σ=Σ
Σ×
= =
Σ
×
51.5 mmZ=
consent of McGraw–Hill Education.
PROBLEM 5.84
A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is
1030 kg/m3 and of steel is
3
7860 kg/m ,
locate the center of gravity of the awl.
SOLUTION
First, note that symmetry implies
0YZ= =
I
33
I
3
II
32
II
3
III
32
III
5(12.5 mm) 7.8125 mm
82
(1030 kg/m ) (0.0125 m)
3
4.2133 10 kg
52.5 mm
(1030 kg/m ) (0.025 m) (0.08 m)
4
40.448 10 kg
92.5 mm 25 mm 67.5 mm
(1030 kg/m ) (0.0035 m) (0.
4
x
W
x
W
x
W
π
π
π
−
−
= =

=

= ×
=

=

= ×
= −=

= − 

3
05 m)
0.49549 10 kg
−
=−×
IV
3 22 3
IV
V
32 3
V
182.5 mm 70 mm 112.5 mm
(7860 kg/m ) (0.0035 m) (0.14 m) 10.5871 10 kg
4
1
182.5 mm (10 mm) 185 mm
4
(7860 kg/m ) (0.00175 m) (0.01 m) 0.25207 10 kg
3
x
W
x
W
π
π
−
−
= −=

= = ×


=+=

= = ×


consent of McGraw–Hill Education.
SOLUTION Continued
, kgW
, mmx
, kg mmxW ⋅
I
3
4.123 10
−
×
7.8125
3
32.916 10
−
×
II
3
40.948 10−
×
52.5
3
2123.5 10−
×
III
3
0.49549 10
−
−×
67.5
3
33.447 10
−
−×
IV
3
10.5871 10−
×
112.5
3
1191.05 10
−
×
V
3
0.25207 10
−
×
185
3
46.633 10−
×
Σ
3
55.005 10
−
×
3
3360.7 10
−
×
We have
33
: (55.005 10 kg) 3360.7 10 kg mmX W xW X
−−
Σ=Σ × = × ⋅
or
61.1mmX=
(from the end of the handle)
consent of McGraw–Hill Education.