PROBLEM 2.1
Two forces are applied to an eye bolt fastened to a beam. Determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
(a)
(b)
R = 8.4 kN
We measure:
α = 19°
R = 8.4 kN
1
19°
PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing that the
tension is 500 N in AB and 160 N in AD, determine graphically the
magnitude and direction of the resultant of the forces exerted by the stays
at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
We measure:
α = 51.3°, β = 59°
(a)
(b)
We measure:
R = 575 N, α = 67°
R = 575 N
2
67°
PROBLEM 2.3
Two forces P and Q are applied as shown at point A of a hook support.
Knowing that P = 15 lb and Q = 25 lb, determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
(a)
(b)
R = 37 lb, α = 76°
We measure:
R = 37 lb
3
76°
PROBLEM 2.4
Two forces P and Q are applied as shown at point A of a hook support.
Knowing that P = 45 lb and Q = 15 lb, determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
(a)
(b)
We measure:
R = 61.5 lb, α = 86.5°
R = 61.5 lb
4
86.5°
PROBLEM 2.5
Two control rods are attached at A to lever AB. Using trigonometry and
knowing that the force in the left-hand rod is F1 = 120 N, determine
(a) the required force F2 in the right-hand rod if the resultant R of the
forces exerted by the rods on the lever is to be vertical, (b) the
corresponding magnitude of R.
SOLUTION
Graphically, by the triangle law
F2 ≅ 108 N
We measure:
R ≅ 77 N
By trigonometry: Law of Sines
F2
R
120
=
=
sin α
sin 38° sin β
α = 90° − 28° = 62°, β = 180° − 62° − 38° = 80°
Then:
F2
R
120 N
=
=
sin 62° sin 38° sin 80°
or (a) F2 = 107.6 N
(b)
5
R = 75.0 N
PROBLEM 2.6
Two control rods are attached at A to lever AB. Using trigonometry and
knowing that the force in the right-hand rod is F2 = 80 N, determine
(a) the required force F1 in the left-hand rod if the resultant R of the
forces exerted by the rods on the lever is to be vertical, (b) the
corresponding magnitude of R.
SOLUTION
Using the Law of Sines
F1
R
80
=
=
sin α
sin 38° sin β
α = 90° − 10° = 80°, β = 180° − 80° − 38° = 62°
Then:
F1
R
80 N
=
=
sin 80° sin 38° sin 62°
or (a) F1 = 89.2 N
(b) R = 55.8 N
6
PROBLEM 2.7
The 50-lb force is to be resolved into components along lines a-a′ and
b-b′. (a) Using trigonometry, determine the angle α knowing that the
component along a-a′ is 35 lb. (b) What is the corresponding value of
the component along b-b′ ?
SOLUTION
Using the triangle rule and the Law of Sines
sin β
sin 40°
=
35 lb
50 lb
(a)
sin β = 0.44995
β = 26.74°
α + β + 40° = 180°
Then:
α = 113.3°
(b) Using the Law of Sines:
Fbb′
50 lb
=
sin α
sin 40°
Fbb′ = 71.5 lb
7
PROBLEM 2.8
The 50-lb force is to be resolved into components along lines a-a′ and
b-b′. (a) Using trigonometry, determine the angle α knowing that the
component along b-b′ is 30 lb. (b) What is the corresponding value of
the component along a-a′ ?
SOLUTION
Using the triangle rule and the Law of Sines
(a)
sin α
sin 40°
=
30 lb
50 lb
sin α = 0.3857
α = 22.7°
(b)
α + β + 40° = 180°
β = 117.31°
Faa′
50 lb
=
sin β
sin 40°
 sin β 
Faa′ = 50 lb 

 sin 40° 
Faa′ = 69.1 lb
8
PROBLEM 2.9
To steady a sign as it is being lowered, two cables are attached to the sign
at A. Using trigonometry and knowing that α = 25°, determine (a) the
required magnitude of the force P if the resultant R of the two forces
applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
Have:
α = 180° − ( 35° + 25° )
= 120°
Then:
P
R
360 N
=
=
sin 35° sin120° sin 25°
or (a) P = 489 N
(b) R = 738 N
9
PROBLEM 2.10
To steady a sign as it is being lowered, two cables are attached to the sign
at A. Using trigonometry and knowing that the magnitude of P is 300 N,
determine (a) the required angle α if the resultant R of the two forces
applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have:
360 N
300 N
=
sin α
sin 35°
sin α = 0.68829
α = 43.5°
β = 180 − ( 35° + 43.5° )
(b)
= 101.5°
Then:
R
300 N
=
sin101.5° sin 35°
or R = 513 N
10
PROBLEM 2.11
Two forces are applied as shown to a hook support. Using trigonometry
and knowing that the magnitude of P is 14 lb, determine (a) the required
angle α if the resultant R of the two forces applied to the support is to be
horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have:
20 lb
14 lb
=
sin α
sin 30°
sin α = 0.71428
α = 45.6°
β = 180° − ( 30° + 45.6° )
(b)
= 104.4°
Then:
R
14 lb
=
sin104.4° sin 30°
R = 27.1 lb
11
PROBLEM 2.12
For the hook support of Problem 2.3, using trigonometry and knowing
that the magnitude of P is 25 lb, determine (a) the required magnitude of
the force Q if the resultant R of the two forces applied at A is to be
vertical, (b) the corresponding magnitude of R.
Problem 2.3: Two forces P and Q are applied as shown at point A of a
hook support. Knowing that P = 15 lb and Q = 25 lb, determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have:
Q
25 lb
=
sin15° sin 30°
Q = 12.94 lb
β = 180° − (15° + 30° )
(b)
= 135°
Thus:
R
25 lb
=
sin135° sin 30°
 sin135° 
R = 25 lb 
 = 35.36 lb
 sin 30° 
R = 35.4 lb
12
PROBLEM 2.13
For the hook support of Problem 2.11, determine, using trigonometry,
(a) the magnitude and direction of the smallest force P for which the
resultant R of the two forces applied to the support is horizontal,
(b) the corresponding magnitude of R.
Problem 2.11: Two forces are applied as shown to a hook support. Using
trigonometry and knowing that the magnitude of P is 14 lb, determine
(a) the required angle α if the resultant R of the two forces applied to the
support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
(a) The smallest force P will be perpendicular to R, that is, vertical
P = ( 20 lb ) sin 30°
= 10 lb
(b)
P = 10 lb
R = ( 20 lb ) cos 30°
= 17.32 lb
13
R = 17.32 lb
PROBLEM 2.14
As shown in Figure P2.9, two cables are attached to a sign at A to steady
the sign as it is being lowered. Using trigonometry, determine (a) the
magnitude and direction of the smallest force P for which the resultant R
of the two forces applied at A is vertical, (b) the corresponding magnitude
of R.
SOLUTION
We observe that force P is minimum when α is 90°, that is, P is horizontal
Then:
(a) P = ( 360 N ) sin 35°
or P = 206 N
And:
(b) R = ( 360 N ) cos 35°
or R = 295 N
14
PROBLEM 2.15
For the hook support of Problem 2.11, determine, using trigonometry, the
magnitude and direction of the resultant of the two forces applied to the
support knowing that P = 10 lb and α = 40°.
Problem 2.11: Two forces are applied as shown to a hook support. Using
trigonometry and knowing that the magnitude of P is 14 lb, determine
(a) the required angle α if the resultant R of the two forces applied to the
support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the force triangle and the Law of Cosines
R 2 = (10 lb ) + ( 20 lb ) − 2 (10 lb )( 20 lb ) cos110°
2
2
= 100 + 400 − 400 ( −0.342 )  lb 2
= 636.8 lb 2
R = 25.23 lb
Using now the Law of Sines
10 lb
25.23 lb
=
sin β
sin110°
 10 lb 
sin β = 
 sin110°
 25.23 lb 
= 0.3724
So:
β = 21.87°
Angle of inclination of R, φ is then such that:
φ + β = 30°
φ = 8.13°
R = 25.2 lb
Hence:
15
8.13°
PROBLEM 2.16
Solve Problem 2.1 using trigonometry
Problem 2.1: Two forces are applied to an eye bolt fastened to a beam.
Determine graphically the magnitude and direction of their resultant
using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle, the Law of Cosines and the Law of Sines
We have:
α = 180° − ( 50° + 25° )
= 105°
Then:
R 2 = ( 4.5 kN ) + ( 6 kN ) − 2 ( 4.5 kN )( 6 kN ) cos105°
2
2
= 70.226 kN 2
or
Now:
R = 8.3801 kN
8.3801 kN
6 kN
=
sin105°
sin β
 6 kN 
sin β = 
 sin105°
 8.3801 kN 
= 0.6916
β = 43.756°
R = 8.38 kN
16
18.76°
PROBLEM 2.17
Solve Problem 2.2 using trigonometry
Problem 2.2: The cable stays AB and AD help support pole AC. Knowing
that the tension is 500 N in AB and 160 N in AD, determine graphically
the magnitude and direction of the resultant of the forces exerted by the
stays at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
From the geometry of the problem:
α = tan −1
2
= 38.66°
2.5
β = tan −1
1.5
= 30.96°
2.5
θ = 180° − ( 38.66 + 30.96° ) = 110.38
Now:
And, using the Law of Cosines:
R 2 = ( 500 N ) + (160 N ) − 2 ( 500 N )(160 N ) cos110.38°
2
2
= 331319 N 2
R = 575.6 N
Using the Law of Sines:
160 N
575.6 N
=
sin γ
sin110.38°
 160 N 
sin γ = 
 sin110.38°
 575.6 N 
= 0.2606
γ = 15.1°
φ = ( 90° − α ) + γ = 66.44°
R = 576 N
17
66.4°
PROBLEM 2.18
Solve Problem 2.3 using trigonometry
Problem 2.3: Two forces P and Q are applied as shown at point A of a
hook support. Knowing that P = 15 lb and Q = 25 lb, determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines
We have:
γ = 180° − (15° + 30° )
= 135°
Then:
R 2 = (15 lb ) + ( 25 lb ) − 2 (15 lb )( 25 lb ) cos135°
2
2
= 1380.3 lb 2
R = 37.15 lb
or
and
25 lb 37.15 lb
=
sin β
sin135°
 25 lb 
sin β = 
 sin135°
 37.15 lb 
= 0.4758
β = 28.41°
Then:
α + β + 75° = 180°
α = 76.59°
R = 37.2 lb
18
76.6°
PROBLEM 2.19
Two structural members A and B are bolted to a bracket as shown.
Knowing that both members are in compression and that the force is
30 kN in member A and 20 kN in member B, determine, using
trigonometry, the magnitude and direction of the resultant of the forces
applied to the bracket by members A and B.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines
γ = 180° − ( 45° + 25° ) = 110°
We have:
Then:
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2
2
= 1710.4 kN 2
R = 41.357 kN
and
20 kN
41.357 kN
=
sin α
sin110°
 20 kN 
sin α = 
 sin110°
 41.357 kN 
= 0.4544
α = 27.028°
φ = α + 45° = 72.028°
Hence:
R = 41.4 kN
19
72.0°
PROBLEM 2.20
Two structural members A and B are bolted to a bracket as shown.
Knowing that both members are in compression and that the force is
20 kN in member A and 30 kN in member B, determine, using
trigonometry, the magnitude and direction of the resultant of the forces
applied to the bracket by members A and B.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines
We have:
Then:
γ = 180° − ( 45° + 25° ) = 110°
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2
2
= 1710.4 kN 2
R = 41.357 kN
and
30 kN
41.357 kN
=
sin α
sin110°
 30 kN 
sin α = 
 sin110°
 41.357 kN 
= 0.6816
α = 42.97°
Finally:
φ = α + 45° = 87.97°
R = 41.4 kN
20
88.0°
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
SOLUTION
20 kN Force:
Fx = + ( 20 kN ) cos 40°,
Fx = 15.32 kN
Fy = + ( 20 kN ) sin 40°,
Fy = 12.86 kN
Fx = − ( 30 kN ) cos 70°,
Fx = −10.26 kN
Fy = + ( 30 kN ) sin 70°,
Fy = 28.2 kN
Fx = − ( 42 kN ) cos 20°,
Fx = −39.5 kN
Fy = + ( 42 kN ) sin 20°,
Fy = 14.36 kN
30 kN Force:
42 kN Force:
21
PROBLEM 2.22
Determine the x and y components of each of the forces shown.
SOLUTION
40 lb Force:
Fx = − ( 40 lb ) sin 50°,
Fx = −30.6 lb
Fy = − ( 40 lb ) cos 50°,
Fy = −25.7 lb
Fx = + ( 60 lb ) cos 60°,
Fx = 30.0 lb
Fy = − ( 60 lb ) sin 60°,
Fy = −52.0 lb
Fx = + ( 80 lb ) cos 25°,
Fx = 72.5 lb
Fy = + ( 80 lb ) sin 25°,
Fy = 33.8 lb
60 lb Force:
80 lb Force:
22
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
We compute the following distances:
OA =
( 48)2 + ( 90 )2
= 102 in.
OB =
( 56 )2 + ( 90 )2
= 106 in.
OC =
(80 )2 + ( 60 )2
= 100 in.
Then:
204 lb Force:
Fx = − (102 lb )
48
,
102
Fx = −48.0 lb
Fy = + (102 lb )
90
,
102
Fy = 90.0 lb
Fx = + ( 212 lb )
56
,
106
Fx = 112.0 lb
Fy = + ( 212 lb )
90
,
106
Fy = 180.0 lb
Fx = − ( 400 lb )
80
,
100
Fx = −320 lb
Fy = − ( 400 lb )
60
,
100
Fy = −240 lb
212 lb Force:
400 lb Force:
23
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
We compute the following distances:
OA =
( 70 )2 + ( 240 )2
OB =
( 210 )2 + ( 200 )2
= 290 mm
OC =
(120 )2 + ( 225)2
= 255 mm
= 250 mm
500 N Force:
 70 
Fx = −500 N 

 250 
Fx = −140.0 N
 240 
Fy = +500 N 

 250 
Fy = 480 N
 210 
Fx = +435 N 

 290 
Fx = 315 N
 200 
Fy = +435 N 

 290 
Fy = 300 N
 120 
Fx = +510 N 

 255 
Fx = 240 N
 225 
Fy = −510 N 

 255 
Fy = −450 N
435 N Force:
510 N Force:
24
PROBLEM 2.25
While emptying a wheelbarrow, a gardener exerts on each handle AB a
force P directed along line CD. Knowing that P must have a 135-N
horizontal component, determine (a) the magnitude of the force P, (b) its
vertical component.
SOLUTION
(a)
P=
Px
cos 40°
=
135 N
cos 40°
or P = 176.2 N
(b)
Py = Px tan 40° = P sin 40°
= (135 N ) tan 40°
or Py = 113.3 N
25
PROBLEM 2.26
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 960-N vertical component, determine (a) the
magnitude of the force P, (b) its horizontal component.
SOLUTION
(a)
P=
=
Py
sin 35°
960 N
sin 35°
or P = 1674 N
(b)
Px =
=
Py
tan 35°
960 N
tan 35°
or Px = 1371 N
26
PROBLEM 2.27
Member CB of the vise shown exerts on block B a force P directed along
line CB. Knowing that P must have a 260-lb horizontal component,
determine (a) the magnitude of the force P, (b) its vertical component.
SOLUTION
We note:
CB exerts force P on B along CB, and the horizontal component of P is Px = 260 lb.
Then:
Px = P sin 50°
(a)
P=
Px
sin 50°
=
260 lb
sin50°
= 339.4 lb
(b)
P = 339 lb
Px = Py tan 50°
Py =
Px
tan 50°
=
260 lb
tan 50°
= 218.2 lb
27
Py = 218 lb
PROBLEM 2.28
Activator rod AB exerts on crank BCD a force P directed along line AB.
Knowing that P must have a 25-lb component perpendicular to arm BC of
the crank, determine (a) the magnitude of the force P, (b) its component
along line BC.
SOLUTION
Using the x and y axes shown.
Py = 25 lb
(a)
Then:
P=
=
Py
sin 75°
25 lb
sin 75°
or P = 25.9 lb
(b)
Px =
=
Py
tan 75°
25 lb
tan 75°
or Px = 6.70 lb
28
PROBLEM 2.29
The guy wire BD exerts on the telephone pole AC a force P directed
along BD. Knowing that P has a 450-N component along line AC,
determine (a) the magnitude of the force P, (b) its component in a
direction perpendicular to AC.
SOLUTION
Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC
is 450 N.
Then:
(a)
P=
450 N
= 549.3 N
cos 35°
P = 549 N
(b)
Px = ( 450 N ) tan 35°
= 315.1 N
Px = 315 N
29
PROBLEM 2.30
The guy wire BD exerts on the telephone pole AC a force P directed
along BD. Knowing that P has a 200-N perpendicular to the pole AC,
determine (a) the magnitude of the force P, (b) its component along
line AC.
SOLUTION
(a)
P=
Px
sin 38°
=
200 N
sin 38°
= 324.8 N
(b)
Py =
Px
tan 38°
=
200 N
tan 38°
or P = 325 N
= 255.98 N
or Py = 256 N
30
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.24.
Problem 2.24: Determine the x and y components of each of the forces
shown.
SOLUTION
From Problem 2.24:
F500 = − (140 N ) i + ( 480 N ) j
F425 = ( 315 N ) i + ( 300 N ) j
F510 = ( 240 N ) i − ( 450 N ) j
R = ΣF = ( 415 N ) i + ( 330 N ) j
Then:
α = tan −1
R=
330
= 38.5°
415
( 415 N )2 + ( 330 N )2
= 530.2 N
R = 530 N
Thus:
31
38.5°
PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.21.
Problem 2.21: Determine the x and y components of each of the forces
shown.
SOLUTION
From Problem 2.21:
F20 = (15.32 kN ) i + (12.86 kN ) j
F30 = − (10.26 kN ) i + ( 28.2 kN ) j
F42 = − ( 39.5 kN ) i + (14.36 kN ) j
R = ΣF = − ( 34.44 kN ) i + ( 55.42 kN ) j
Then:
α = tan −1
R=
55.42
= 58.1°
−34.44
( 55.42 kN )2 + ( −34.44 N )2
= 65.2 kN
R = 65.2 kN
32
58.2°
PROBLEM 2.33
Determine the resultant of the three forces of Problem 2.22.
Problem 2.22: Determine the x and y components of each of the forces
shown.
SOLUTION
The components of the forces were determined in 2.23.
Force
x comp. (lb)
y comp. (lb)
40 lb
−30.6
−25.7
60 lb
30
−51.96
80 lb
72.5
Rx = 71.9
33.8
Ry = −43.86
R = Rxi + Ry j
= ( 71.9 lb ) i − ( 43.86 lb ) j
tan α =
43.86
71.9
α = 31.38°
R=
( 71.9 lb )2 + ( −43.86 lb )2
= 84.23 lb
R = 84.2 lb
33
31.4°
PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.23.
Problem 2.23: Determine the x and y components of each of the forces
shown.
SOLUTION
The components of the forces were
determined in Problem 2.23.
F204 = − ( 48.0 lb ) i + ( 90.0 lb ) j
F212 = (112.0 lb ) i + (180.0 lb ) j
F400 = − ( 320 lb ) i − ( 240 lb ) j
Thus
R = Rx + R y
R = − ( 256 lb ) i + ( 30.0 lb ) j
Now:
tan α =
α = tan −1
30.0
256
30.0
= 6.68°
256
and
R=
( −256 lb )2 + ( 30.0 lb )2
= 257.75 lb
R = 258 lb
34
6.68°
PROBLEM 2.35
Knowing that α = 35°, determine the resultant of the three forces
shown.
SOLUTION
300-N Force:
Fx = ( 300 N ) cos 20° = 281.9 N
Fy = ( 300 N ) sin 20° = 102.6 N
400-N Force:
Fx = ( 400 N ) cos55° = 229.4 N
Fy = ( 400 N ) sin 55° = 327.7 N
600-N Force:
Fx = ( 600 N ) cos 35° = 491.5 N
Fy = − ( 600 N ) sin 35° = −344.1 N
and
Rx = ΣFx = 1002.8 N
Ry = ΣFy = 86.2 N
R=
(1002.8 N )2 + (86.2 N )2
= 1006.5 N
Further:
tan α =
α = tan −1
86.2
1002.8
86.2
= 4.91°
1002.8
R = 1007 N
35
4.91°
PROBLEM 2.36
Knowing that α = 65°, determine the resultant of the three forces
shown.
SOLUTION
300-N Force:
Fx = ( 300 N ) cos 20° = 281.9 N
Fy = ( 300 N ) sin 20° = 102.6 N
400-N Force:
Fx = ( 400 N ) cos85° = 34.9 N
Fy = ( 400 N ) sin 85° = 398.5 N
600-N Force:
Fx = ( 600 N ) cos 5° = 597.7 N
Fy = − ( 600 N ) sin 5° = −52.3 N
and
Rx = ΣFx = 914.5 N
Ry = ΣFy = 448.8 N
R=
( 914.5 N )2 + ( 448.8 N )2
= 1018.7 N
Further:
tan α =
α = tan −1
448.8
914.5
448.8
= 26.1°
914.5
R = 1019 N
36
26.1°
PROBLEM 2.37
Knowing that the tension in cable BC is 145 lb, determine the resultant of
the three forces exerted at point B of beam AB.
SOLUTION
Cable BC Force:
Fx = − (145 lb )
Fy = (145 lb )
84
= −105 lb
116
80
= 100 lb
116
100-lb Force:
Fx = − (100 lb )
3
= −60 lb
5
Fy = − (100 lb )
4
= −80 lb
5
156-lb Force:
Fx = (156 lb )
12
= 144 lb
13
Fy = − (156 lb )
5
= −60 lb
13
and
Rx = ΣFx = −21 lb,
R=
Ry = ΣFy = −40 lb
( −21 lb )2 + ( −40 lb )2
= 45.177 lb
Further:
tan α =
α = tan −1
40
21
40
= 62.3°
21
R = 45.2 lb
Thus:
37
62.3°
PROBLEM 2.38
Knowing that α = 50°, determine the resultant of the three forces
shown.
SOLUTION
The resultant force R has the x- and y-components:
Rx = ΣFx = (140 lb ) cos 50° + ( 60 lb ) cos85° − (160 lb ) cos 50°
Rx = −7.6264 lb
and
Ry = ΣFy = (140 lb ) sin 50° + ( 60 lb ) sin 85° + (160 lb ) sin 50°
Ry = 289.59 lb
Further:
tan α =
α = tan −1
290
7.6
290
= 88.5°
7.6
R = 290 lb
Thus:
38
88.5°
PROBLEM 2.39
Determine (a) the required value of α if the resultant of the three forces
shown is to be vertical, (b) the corresponding magnitude of the resultant.
SOLUTION
For an arbitrary angle α , we have:
Rx = ΣFx = (140 lb ) cos α + ( 60 lb ) cos (α + 35° ) − (160 lb ) cos α
(a) So, for R to be vertical:
Rx = ΣFx = (140 lb ) cos α + ( 60 lb ) cos (α + 35° ) − (160 lb ) cos α = 0
Expanding,
− cos α + 3 ( cos α cos 35° − sin α sin 35° ) = 0
Then:
tan α =
cos 35° −
sin 35°
1
3
or
 cos 35° −
α = tan −1 

sin 35°
1
3

 = 40.265°

α = 40.3°
(b) Now:
R = Ry = ΣFy = (140 lb ) sin 40.265° + ( 60 lb ) sin 75.265° + (160 lb ) sin 40.265°
R = R = 252 lb
39
PROBLEM 2.40
For the beam of Problem 2.37, determine (a) the required tension in cable
BC if the resultant of the three forces exerted at point B is to be vertical,
(b) the corresponding magnitude of the resultant.
Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine
the resultant of the three forces exerted at point B of beam AB.
SOLUTION
We have:
Rx = ΣFx = −
84
12
3
TBC + (156 lb ) − (100 lb )
116
13
5
Rx = −0.724TBC + 84 lb
or
and
Ry = ΣFy =
80
5
4
TBC − (156 lb ) − (100 lb )
116
13
5
Ry = 0.6897TBC − 140 lb
(a) So, for R to be vertical,
Rx = −0.724TBC + 84 lb = 0
TBC = 116.0 lb
(b) Using
TBC = 116.0 lb
R = Ry = 0.6897 (116.0 lb ) − 140 lb = −60 lb
R = R = 60.0 lb
40
PROBLEM 2.41
Boom AB is held in the position shown by three cables. Knowing that the
tensions in cables AC and AD are 4 kN and 5.2 kN, respectively,
determine (a) the tension in cable AE if the resultant of the tensions
exerted at point A of the boom must be directed along AB,
(b) the corresponding magnitude of the resultant.
SOLUTION
Choose x-axis along bar AB.
Then
(a) Require
Ry = ΣFy = 0:
( 4 kN ) cos 25° + ( 5.2 kN ) sin 35° − TAE sin 65° = 0
TAE = 7.2909 kN
or
TAE = 7.29 kN
(b)
R = ΣFx
= − ( 4 kN ) sin 25° − ( 5.2 kN ) cos 35° − ( 7.2909 kN ) cos 65°
= −9.03 kN
R = 9.03 kN
41
PROBLEM 2.42
For the block of Problems 2.35 and 2.36, determine (a) the required value
of α of the resultant of the three forces shown is to be parallel to the
incline, (b) the corresponding magnitude of the resultant.
Problem 2.35: Knowing that α = 35°, determine the resultant of the
three forces shown.
Problem 2.36: Knowing that α = 65°, determine the resultant of the
three forces shown.
SOLUTION
Selecting the x axis along aa′, we write
Rx = ΣFx = 300 N + ( 400 N ) cos α + ( 600 N ) sin α
(1)
Ry = ΣFy = ( 400 N ) sin α − ( 600 N ) cos α
(2)
(a) Setting Ry = 0 in Equation (2):
tan α =
Thus
600
= 1.5
400
α = 56.3°
(b) Substituting for α in Equation (1):
Rx = 300 N + ( 400 N ) cos 56.3° + ( 600 N ) sin 56.3°
Rx = 1021.1 N
R = Rx = 1021 N
42
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown. Determine the
tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
From the geometry, we calculate the distances:
AC =
(16 in.)2 + (12 in.)2
= 20 in.
BC =
( 20 in.)2 + ( 21 in.)2
= 29 in.
Then, from the Free Body Diagram of point C:
ΣFx = 0: −
or
and
or
TBC =
ΣFy = 0:
16
21
TAC +
TBC = 0
20
29
29 4
× TAC
21 5
12
20
TAC +
TBC − 600 lb = 0
20
29
12
20  29 4

TAC +
× TAC  − 600 lb = 0

20
29  21 5

TAC = 440.56 lb
Hence:
(a)
TAC = 441 lb
(b)
TBC = 487 lb
43
PROBLEM 2.44
Knowing that α = 25°, determine the tension (a) in cable AC, (b) in
rope BC.
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines:
TAC
T
5 kN
= BC =
sin115° sin 5° sin 60°
(a)
TAC =
5 kN
sin115° = 5.23 kN
sin 60°
TAC = 5.23 kN
(b)
TBC =
5 kN
sin 5° = 0.503 kN
sin 60°
TBC = 0.503 kN
44
PROBLEM 2.45
Knowing that α = 50° and that boom AC exerts on pin C a force
directed long line AC, determine (a) the magnitude of that force, (b) the
tension in cable BC.
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines:
FAC
TBC
400 lb
=
=
sin 25° sin 60° sin 95°
(a)
FAC =
400 lb
sin 25° = 169.69 lb
sin 95°
FAC = 169.7 lb
(b)
TBC =
400
sin 60° = 347.73 lb
sin 95°
TBC = 348 lb
45
PROBLEM 2.46
Two cables are tied together at C and are loaded as shown. Knowing that
α = 30°, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines:
TAC
TBC
2943 N
=
=
sin 60° sin 55°
sin 65°
(a)
TAC =
2943 N
sin 60° = 2812.19 N
sin 65°
TAC = 2.81 kN
(b)
TBC =
2943 N
sin 55° = 2659.98 N
sin 65°
TBC = 2.66 kN
46
PROBLEM 2.47
A chairlift has been stopped in the position shown. Knowing that each
chair weighs 300 N and that the skier in chair E weighs 890 N, determine
that weight of the skier in chair F.
SOLUTION
Free-Body Diagram Point B
In the free-body diagram of point B, the geometry gives:
θ AB = tan −1
9.9
= 30.51°
16.8
θ BC = tan −1
12
= 22.61°
28.8
Thus, in the force triangle, by the Law of Sines:
Force Triangle
TBC
1190 N
=
sin 59.49° sin 7.87°
TBC = 7468.6 N
Free-Body Diagram Point C
In the free-body diagram of point C (with W the sum of weights of chair
and skier) the geometry gives:
θ CD = tan −1
1.32
= 10.39°
7.2
Hence, in the force triangle, by the Law of Sines:
Force Triangle
W
7468.6 N
=
sin12.23° sin100.39°
W = 1608.5 N
Finally, the skier weight = 1608.5 N − 300 N = 1308.5 N
skier weight = 1309 N
47
PROBLEM 2.48
A chairlift has been stopped in the position shown. Knowing that each
chair weighs 300 N and that the skier in chair F weighs 800 N, determine
the weight of the skier in chair E.
SOLUTION
Free-Body Diagram Point F
In the free-body diagram of point F, the geometry gives:
θ EF = tan −1
12
= 22.62°
28.8
θ DF = tan −1
1.32
= 10.39°
7.2
Thus, in the force triangle, by the Law of Sines:
Force Triangle
TEF
1100 N
=
sin100.39° sin12.23°
TBC = 5107.5 N
Free-Body Diagram Point E
In the free-body diagram of point E (with W the sum of weights of chair
and skier) the geometry gives:
θ AE = tan −1
9.9
= 30.51°
16.8
Hence, in the force triangle, by the Law of Sines:
Force Triangle
W
5107.5 N
=
sin 7.89° sin 59.49°
W = 813.8 N
Finally, the skier weight = 813.8 N − 300 N = 513.8 N
skier weight = 514 N
48
PROBLEM 2.49
Four wooden members are joined with metal plate connectors and are in
equilibrium under the action of the four fences shown. Knowing that
FA = 510 lb and FB = 480 lb, determine the magnitudes of the other two
forces.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y components:
ΣFx = 0: FC + ( 510 lb ) sin15° − ( 480 lb ) cos15° = 0
or FC = 332 lb
ΣFy = 0: FD − ( 510 lb ) cos15° + ( 480 lb ) sin15° = 0
or FD = 368 lb
49
PROBLEM 2.50
Four wooden members are joined with metal plate connectors and are in
equilibrium under the action of the four fences shown. Knowing that
FA = 420 lb and FC = 540 lb, determine the magnitudes of the other two
forces.
SOLUTION
Resolving the forces into x and y components:
ΣFx = 0: − FB cos15° + ( 540 lb ) + ( 420 lb ) cos15° = 0
or
FB = 671.6 lb
FB = 672 lb
ΣFy = 0: FD − ( 420 lb ) cos15° + ( 671.6 lb ) sin15° = 0
or FD = 232 lb
50
PROBLEM 2.51
Two forces P and Q are applied as shown to an aircraft connection.
Knowing that the connection is in equilibrium and the P = 400 lb and
Q = 520 lb, determine the magnitudes of the forces exerted on the rods
A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y directions:
R = P + Q + FA + FB = 0
Substituting components:
R = − ( 400 lb ) j + ( 520 lb ) cos 55°  i − ( 520 lb ) sin 55° j
+ FBi − ( FA cos 55° ) i + ( FA sin 55° ) j = 0
In the y-direction (one unknown force)
−400 lb − ( 520 lb ) sin 55° + FA sin 55° = 0
Thus,
FA =
400 lb + ( 520 lb ) sin 55°
= 1008.3 lb
sin 55°
FA = 1008 lb
In the x-direction:
( 520 lb ) cos55° + FB − FA cos 55° = 0
Thus,
FB = FA cos 55° − ( 520 lb ) cos 55°
= (1008.3 lb ) cos 55° − ( 520 lb ) cos 55°
= 280.08 lb
FB = 280 lb
51
PROBLEM 2.52
Two forces P and Q are applied as shown to an aircraft connection.
Knowing that the connection is in equilibrium and that the magnitudes of
the forces exerted on rods A and B are FA = 600 lb and FB = 320 lb,
determine the magnitudes of P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y directions:
R = P + Q + FA + FB = 0
Substituting components:
R = ( 320 lb ) i − ( 600 lb ) cos 55° i + ( 600 lb ) sin 55° j
+ Pi + ( Q cos 55° ) i − ( Q sin 55° ) j = 0
In the x-direction (one unknown force)
320 lb − ( 600 lb ) cos 55° + Q cos 55° = 0
Thus,
Q=
−320 lb + ( 600 lb ) cos 55°
= 42.09 lb
cos 55°
Q = 42.1 lb
In the y-direction:
( 600 lb ) sin 55° − P − Q sin 55° = 0
Thus,
P = ( 600 lb ) sin 55° − Q sin 55° = 457.01 lb
P = 457 lb
52
PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that
W = 840 N, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
From geometry:
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.
The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.
Thus:
3
15
15
ΣFx = 0: − TCA + TCB − ( 680 N ) = 0
5
17
17
or
1
5
− TCA + TCB = 200 N
5
17
(1)
and
ΣFy = 0:
4
8
8
TCA + TCB − ( 680 N ) − 840 N = 0
5
17
17
or
1
2
TCA + TCB = 290 N
5
17
(2)
Solving Equations (1) and (2) simultaneously:
(a)
TCA = 750 N
(b)
TCB = 1190 N
53
PROBLEM 2.54
Two cables tied together at C are loaded as shown. Determine the range
of values of W for which the tension will not exceed 1050 N in either
cable.
SOLUTION
Free-Body Diagram
From geometry:
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.
The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.
Thus:
3
15
15
ΣFx = 0: − TCA + TCB − ( 680 N ) = 0
5
17
17
or
1
5
− TCA + TCB = 200 N
5
17
(1)
and
ΣFy = 0:
4
8
8
TCA + TCB − ( 680 N ) − W = 0
5
17
17
or
1
2
1
TCA + TCB = 80 N + W
5
17
4
(2)
Then, from Equations (1) and (2)
TCB = 680 N +
TCA =
17
W
28
25
W
28
Now, with T ≤ 1050 N
TCA : TCA = 1050 N =
25
W
28
W = 1176 N
or
and
TCB : TCB = 1050 N = 680 N +
or
W = 609 N
54
17
W
28
∴ 0 ≤ W ≤ 609 N
PROBLEM 2.55
The cabin of an aerial tramway is suspended from a set of wheels that can
roll freely on the support cable ACB and is being pulled at a constant
speed by cable DE. Knowing that α = 40° and β = 35°, that the
combined weight of the cabin, its support system, and its passengers is
24.8 kN, and assuming the tension in cable DF to be negligible,
determine the tension (a) in the support cable ACB, (b) in the traction
cable DE.
SOLUTION
Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If
considered as a rigid body (Chapter 4) it would be found that its center of
gravity should be located to the left of the centerline for the line CD to be
vertical.
Now
ΣFx = 0: TACB ( cos 35° − cos 40° ) − TDE cos 40° = 0
or
0.0531TACB − 0.766TDE = 0
(1)
and
ΣFy = 0: TACB ( sin 40° − sin 35° ) + TDE sin 40° − 24.8 kN = 0
or
0.0692TACB + 0.643TDE = 24.8 kN
(2)
From (1)
TACB = 14.426TDE
Then, from (2)
0.0692 (14.426TDE ) + 0.643TDE = 24.8 kN
and
(b) TDE = 15.1 kN
(a) TACB = 218 kN
55
PROBLEM 2.56
The cabin of an aerial tramway is suspended from a set of wheels that can
roll freely on the support cable ACB and is being pulled at a constant
speed by cable DE. Knowing that α = 42° and β = 32°, that the tension
in cable DE is 20 kN, and assuming the tension in cable DF to be
negligible, determine (a) the combined weight of the cabin, its support
system, and its passengers, (b) the tension in the support cable ACB.
SOLUTION
Free-Body Diagram
First, consider the sum of forces in the x-direction because there is only one unknown force:
ΣFx = 0: TACB ( cos 32° − cos 42° ) − ( 20 kN ) cos 42° = 0
or
0.1049TACB = 14.863 kN
(b) TACB = 141.7 kN
Now
ΣFy = 0: TACB ( sin 42° − sin 32° ) + ( 20 kN ) sin 42° − W = 0
or
(141.7 kN )( 0.1392 ) + ( 20 kN )( 0.6691) − W
=0
(a) W = 33.1 kN
56
PROBLEM 2.57
A block of weight W is suspended from a 500-mm long cord and two
springs of which the unstretched lengths are 450 mm. Knowing that the
constants of the springs are kAB = 1500 N/m and kAD = 500 N/m,
determine (a) the tension in the cord, (b) the weight of the block.
SOLUTION
Free-Body Diagram At A
First note from geometry:
The sides of the triangle with hypotenuse AD are in the ratio 8:15:17.
The sides of the triangle with hypotenuse AB are in the ratio 3:4:5.
The sides of the triangle with hypotenuse AC are in the ratio 7:24:25.
Then:
FAB = k AB ( LAB − Lo )
and
LAB =
( 0.44 m )2 + ( 0.33 m )2
= 0.55 m
So:
FAB = 1500 N/m ( 0.55 m − 0.45 m )
= 150 N
Similarly,
FAD = k AD ( LAD − Lo )
Then:
LAD =
( 0.66 m )2 + ( 0.32 m )2
= 0.68 m
FAD = 1500 N/m ( 0.68 m − 0.45 m )
= 115 N
(a)
ΣFx = 0: −
4
7
15
(150 N ) + TAC − (115 N ) = 0
5
25
17
or
TAC = 66.18 N
57
TAC = 66.2 N
PROBLEM 2.57 CONTINUED
(b) and
ΣFy = 0:
3
24
8
(150 N ) + ( 66.18 N ) + (115 N ) − W = 0
5
25
17
or W = 208 N
58
PROBLEM 2.58
A load of weight 400 N is suspended from a spring and two cords which
are attached to blocks of weights 3W and W as shown. Knowing that the
constant of the spring is 800 N/m, determine (a) the value of W, (b) the
unstretched length of the spring.
SOLUTION
Free-Body Diagram At A
First note from geometry:
The sides of the triangle with hypotenuse AD are in the ratio 12:35:37.
The sides of the triangle with hypotenuse AC are in the ratio 3:4:5.
The sides of the triangle with hypotenuse AB are also in the ratio
12:35:37.
Then:
ΣFx = 0: −
4
35
12
( 3W ) + (W ) + Fs = 0
5
37
37
or
Fs = 4.4833W
and
ΣFy = 0:
3
12
35
( 3W ) + (W ) + Fs − 400 N = 0
5
37
37
Then:
3
12
35
( 3W ) + (W ) + ( 4.4833W ) − 400 N = 0
5
37
37
or
W = 62.841 N
and
Fs = 281.74 N
or
W = 62.8 N
(a)
59
PROBLEM 2.58 CONTINUED
(b) Have spring force
Fs = k ( LAB − Lo )
Where
FAB = k AB ( LAB − Lo )
and
LAB =
( 0.360 m )2 + (1.050 m )2
= 1.110 m
So:
281.74 N = 800 N/m (1.110 − L0 ) m
or L0 = 758 mm
60
PROBLEM 2.59
For the cables and loading of Problem 2.46, determine (a) the value of α
for which the tension in cable BC is as small as possible, (b) the
corresponding value of the tension.
SOLUTION
The smallest TBC is when TBC is perpendicular to the direction of TAC
Free-Body Diagram At C
Force Triangle
α = 55.0°
(a)
(b)
TBC = ( 2943 N ) sin 55°
= 2410.8 N
TBC = 2.41 kN
61
PROBLEM 2.60
Knowing that portions AC and BC of cable ACB must be equal, determine
the shortest length of cable which can be used to support the load shown
if the tension in the cable is not to exceed 725 N.
SOLUTION
Free-Body Diagram: C
( For T = 725 N )
ΣFy = 0: 2Ty − 1000 N = 0
Ty = 500 N
Tx2 + Ty2 = T 2
Tx2 + ( 500 N ) = ( 725 N )
2
2
Tx = 525 N
By similar triangles:
BC 1.5 m
=
725
525
∴ BC = 2.07 m
L = 2 ( BC ) = 4.14 m
L = 4.14 m
62
PROBLEM 2.61
Two cables tied together at C are loaded as shown. Knowing that the
maximum allowable tension in each cable is 200 lb, determine (a) the
magnitude of the largest force P which may be applied at C, (b) the
corresponding value of α.
SOLUTION
Free-Body Diagram: C
Force Triangle
Force triangle is isoceles with
2β = 180° − 85°
β = 47.5°
P = 2 ( 200 lb ) cos 47.5° = 270 lb
(a)
P = 270 lb
Since P > 0, the solution is correct.
(b)
α = 180° − 55° − 47.5° = 77.5°
63
α = 77.5°
PROBLEM 2.62
Two cables tied together at C are loaded as shown. Knowing that the
maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC,
determine (a) the magnitude of the largest force P which may be applied
at C, (b) the corresponding value of α.
SOLUTION
Free-Body Diagram: C
Force Triangle
(a) Law of Cosines:
P 2 = ( 300 lb ) + (150 lb ) − 2 ( 300 lb )(150 lb ) cos85°
2
2
P = 323.5 lb
P = 324 lb
Since P > 300 lb, our solution is correct.
(b) Law of Sines:
sin β
sin 85°
=
300
323.5°
sin β = 0.9238
or
β = 67.49°
α = 180° − 55° − 67.49° = 57.5°
α = 57.5°
64
PROBLEM 2.63
For the structure and loading of Problem 2.45, determine (a) the value of
α for which the tension in cable BC is as small as possible, (b) the
corresponding value of the tension.
SOLUTION
TBC must be perpendicular to FAC to be as small as possible.
Free-Body Diagram: C
(a) We observe:
Force Triangle is
a right triangle
α = 55°
α = 55°
TBC = ( 400 lb ) sin 60°
(b)
or
TBC = 346.4 lb
65
TBC = 346 lb
PROBLEM 2.64
Boom AB is supported by cable BC and a hinge at A. Knowing that the
boom exerts on pin B a force directed along the boom and that the tension
in rope BD is 70 lb, determine (a) the value of α for which the tension in
cable BC is as small as possible, (b) the corresponding value of the
tension.
SOLUTION
Free-Body Diagram: B
TBD + FAB + TBC = 0
(a) Have:
where magnitude and direction of TBD are known, and the direction
of FAB is known.
Then, in a force triangle:
α = 90.0°
By observation, TBC is minimum when
(b) Have
TBC = ( 70 lb ) sin (180° − 70° − 30° )
= 68.93 lb
TBC = 68.9 lb
66
PROBLEM 2.65
Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless
vertical rod and is attached as shown to a spring. The constant of the
spring is 660 N/m, and the spring is unstretched when h = 300 mm.
Knowing that the system is in equilibrium when h = 400 mm, determine
the weight of the collar.
SOLUTION
Free-Body Diagram: Collar A
Fs = k ( L′AB − LAB )
Have:
where:
L′AB =
( 0.3 m )2 + ( 0.4 m )2
LAB = 0.3 2 m
= 0.5 m
(
)
Fs = 660 N/m 0.5 − 0.3 2 m
Then:
= 49.986 N
For the collar:
ΣFy = 0: − W +
4
( 49.986 N ) = 0
5
or W = 40.0 N
67
PROBLEM 2.66
The 40-N collar A can slide on a frictionless vertical rod and is attached
as shown to a spring. The spring is unstretched when h = 300 mm.
Knowing that the constant of the spring is 560 N/m, determine the value
of h for which the system is in equilibrium.
SOLUTION
ΣFy = 0: − W +
Free-Body Diagram: Collar A
or
Fs = 0
Fs = k ( L′AB − LAB )
Now..
Then:
( 0.3)2 + h2
hFs = 40 0.09 + h 2
or
where
h
L′AB =
h 560

(
( 0.3)2 + h2
LAB = 0.3 2 m
m
)
0.09 + h 2 − 0.3 2  = 40 0.09 + h 2

(14h − 1)
0.09 + h 2 = 4.2 2h
h∼m
Solving numerically,
h = 415 mm
68
PROBLEM 2.67
A 280-kg crate is supported by several rope-and-pulley arrangements as
shown. Determine for each arrangement the tension in the rope. (Hint:
The tension in the rope is the same on each side of a simple pulley. This
can be proved by the methods of Chapter 4.)
SOLUTION
(
Free-Body Diagram of pulley
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
(a)
T =
1
( 2746.8 N )
2
T = 1373 N
(
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
(b)
T =
1
( 2746.8 N )
2
T = 1373 N
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
(c)
T =
1
( 2746.8 N )
3
T = 916 N
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
(d)
T =
1
( 2746.8 N )
3
T = 916 N
(
)
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
(e)
T =
1
( 2746.8 N )
4
T = 687 N
69
PROBLEM 2.68
Solve parts b and d of Problem 2.67 assuming that the free end of the
rope is attached to the crate.
Problem 2.67: A 280-kg crate is supported by several rope-and-pulley
arrangements as shown. Determine for each arrangement the tension in
the rope. (Hint: The tension in the rope is the same on each side of a
simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram of pulley
and crate
(b)
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
3
T = 916 N
(d)
(
)
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
4
T = 687 N
70
PROBLEM 2.69
A 350-lb load is supported by the rope-and-pulley arrangement shown.
Knowing that β = 25°, determine the magnitude and direction of the
force P which should be exerted on the free end of the rope to maintain
equilibrium. (Hint: The tension in the rope is the same on each side of a
simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
ΣFx = 0: 2P sin 25° − P cos α = 0
and
cos α = 0.8452
or
α = ±32.3°
α = +32.3°
For
ΣFy = 0: 2P cos 25° + P sin 32.3° − 350 lb = 0
or P = 149.1 lb
32.3°
α = −32.3°
For
ΣFy = 0: 2P cos 25° + P sin − 32.3° − 350 lb = 0
or P = 274 lb
71
32.3°
PROBLEM 2.70
A 350-lb load is supported by the rope-and-pulley arrangement shown.
Knowing that α = 35°, determine (a) the angle β, (b) the magnitude of
the force P which should be exerted on the free end of the rope to
maintain equilibrium. (Hint: The tension in the rope is the same on each
side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
ΣFx = 0: 2 P sin β − P cos 25° = 0
Hence:
sin β =
(a)
1
cos 25°
2
or β = 24.2°
ΣFy = 0: 2P cos β + P sin 35° − 350 lb = 0
(b)
Hence:
2P cos 24.2° + P sin 35° − 350 lb = 0
or
P = 145.97 lb
72
P = 146.0 lb
PROBLEM 2.71
A load Q is applied to the pulley C, which can roll on the cable ACB. The
pulley is held in the position shown by a second cable CAD, which passes
over the pulley A and supports a load P. Knowing that P = 800 N,
determine (a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
ΣFx = 0: TACB ( cos 30° − cos 50° ) − ( 800 N ) cos 50° = 0
(a)
Hence
TACB = 2303.5 N
TACB = 2.30 kN
ΣFy = 0: TACB ( sin 30° + sin 50° ) + ( 800 N ) sin 50° − Q = 0
(b)
( 2303.5 N )( sin 30° + sin 50° ) + ( 800 N ) sin 50° − Q = 0
Q = 3529.2 N
or
73
Q = 3.53 kN
PROBLEM 2.72
A 2000-N load Q is applied to the pulley C, which can roll on the cable
ACB. The pulley is held in the position shown by a second cable CAD,
which passes over the pulley A and supports a load P. Determine (a) the
tension in the cable ACB, (b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
ΣFx = 0: TACB ( cos 30° − cos 50° ) − P cos 50° = 0
P = 0.3473TACB
or
(1)
ΣFy = 0: TACB ( sin 30° + sin 50° ) + P sin 50° − 2000 N = 0
1.266TACB + 0.766P = 2000 N
or
(2)
(a) Substitute Equation (1) into Equation (2):
1.266TACB + 0.766 ( 0.3473TACB ) = 2000 N
Hence:
TACB = 1305.5 N
TACB = 1306 N
(b) Using (1)
P = 0.3473 (1306 N ) = 453.57 N
P = 454 N
74
PROBLEM 2.73
Determine (a) the x, y, and z components of the 200-lb force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
(a)
Fx = ( 200 lb ) cos 30° cos 25° = 156.98 lb
Fx = +157.0 lb
Fy = ( 200 lb ) sin 30° = 100.0 lb
Fy = +100.0 lb
Fz = − ( 200 lb ) cos 30° sin 25° = −73.1996 lb
Fz = −73.2 lb
(b)
cosθ x =
156.98
200
or θ x = 38.3°
cosθ y =
100.0
200
or θ y = 60.0°
−73.1996
200
or θ z = 111.5°
cosθ z =
75
PROBLEM 2.74
Determine (a) the x, y, and z components of the 420-lb force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
(a)
Fx = − ( 420 lb ) sin 20° sin 70° = −134.985 lb
Fx = −135.0 lb
Fy = ( 420 lb ) cos 20° = 394.67 lb
Fy = +395 lb
Fz = ( 420 lb ) sin 20° cos 70° = 49.131 lb
Fz = +49.1 lb
(b)
cosθ x =
−134.985
420
θ x = 108.7°
cosθ y =
394.67
420
θ y = 20.0°
cosθ z =
49.131
420
θ z = 83.3°
76
PROBLEM 2.75
To stabilize a tree partially uprooted in a storm, cables AB and AC are
attached to the upper trunk of the tree and then are fastened to steel rods
anchored in the ground. Knowing that the tension in cable AB is 4.2 kN,
determine (a) the components of the force exerted by this cable on the
tree, (b) the angles θx, θy, and θz that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a)
Fx = ( 4.2 kN ) sin 50° cos 40° = 2.4647 kN
Fx = +2.46 kN
Fy = − ( 4.2 kN ) cos 50° = −2.6997 kN
Fy = −2.70 kN
Fz = ( 4.2 kN ) sin 50° sin 40° = 2.0681 kN
Fz = +2.07 kN
(b)
cosθ x =
2.4647
4.2
θ x = 54.1°
77
PROBLEM 2.75 CONTINUED
cosθ y =
−2.7
4.2
θ y = 130.0°
cosθ z =
2.0681
4.0
θ z = 60.5°
78
PROBLEM 2.76
To stabilize a tree partially uprooted in a storm, cables AB and AC are
attached to the upper trunk of the tree and then are fastened to steel rods
anchored in the ground. Knowing that the tension in cable AC is 3.6 kN,
determine (a) the components of the force exerted by this cable on the
tree, (b) the angles θx, θy, and θz that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a)
Fx = − ( 3.6 kN ) cos 45° sin 25° = −1.0758 kN
Fx = −1.076 kN
Fy = − ( 3.6 kN ) sin 45° = −2.546 kN
Fy = −2.55 kN
Fz = ( 3.6 kN ) cos 45° cos 25° = 2.3071 kN
Fz = +2.31 kN
(b)
cosθ x =
−1.0758
3.6
θ x = 107.4°
79
PROBLEM 2.76 CONTINUED
cosθ y =
−2.546
3.6
θ y = 135.0°
cosθ z =
2.3071
3.6
θ z = 50.1°
80
PROBLEM 2.77
A horizontal circular plate is suspended as shown from three wires which
are attached to a support at D and form 30° angles with the vertical.
Knowing that the x component of the force exerted by wire AD on the
plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles θx,
θy, and θz that the force exerted at A forms with the coordinate axes.
SOLUTION
(a)
Fx = F sin 30° sin 50° = 220.6 N (Given)
F =
220.6 N
= 575.95 N
sin30° sin50°
F = 576 N
(b)
cosθ x =
Fx
220.6
=
= 0.3830
F
575.95
θ x = 67.5°
Fy = F cos 30° = 498.79 N
cosθ y =
Fy
F
=
498.79
= 0.86605
575.95
θ y = 30.0°
Fz = − F sin 30° cos 50°
= − ( 575.95 N ) sin 30° cos 50°
= −185.107 N
cosθ z =
Fz
−185.107
=
= −0.32139
F
575.95
θ z = 108.7°
81
PROBLEM 2.78
A horizontal circular plate is suspended as shown from three wires which
are attached to a support at D and form 30° angles with the vertical.
Knowing that the z component of the force exerted by wire BD on the
plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles θx,
θy, and θz that the force exerted at B forms with the coordinate axes.
SOLUTION
(a)
Fz = − F sin 30° sin 40° = −64.28 N (Given)
F =
(b)
64.28 N
= 200.0 N
sin30° sin40°
F = 200 N
Fx = − F sin 30° cos 40°
= − ( 200.0 N ) sin 30° cos 40°
= −76.604 N
cosθ x =
Fx
−76.604
=
= −0.38302
F
200.0
θ x = 112.5°
Fy = F cos 30° = 173.2 N
cosθ y =
Fy
F
=
173.2
= 0.866
200
θ y = 30.0°
Fz = −64.28 N
cosθ z =
Fz
−64.28
=
= −0.3214
F
200
82
θ z = 108.7°
PROBLEM 2.79
A horizontal circular plate is suspended as shown from three wires which
are attached to a support at D and form 30° angles with the vertical.
Knowing that the tension in wire CD is 120 lb, determine (a) the
components of the force exerted by this wire on the plate, (b) the angles
θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
(a)
Fx = − (120 lb ) sin 30° cos 60° = −30 lb
Fx = −30.0 lb
Fy = (120 lb ) cos 30° = 103.92 lb
Fy = +103.9 lb
Fz = (120 lb ) sin 30° sin 60° = 51.96 lb
Fz = +52.0 lb
(b)
cosθ x =
Fx
−30.0
=
= −0.25
F
120
θ x = 104.5°
Fy
cosθ y =
F
=
103.92
= 0.866
120
θ y = 30.0°
cosθ z =
Fz
51.96
=
= 0.433
F
120
θ z = 64.3°
83
PROBLEM 2.80
A horizontal circular plate is suspended as shown from three wires which
are attached to a support at D and form 30° angles with the vertical.
Knowing that the x component of the forces exerted by wire CD on the
plate is –40 lb, determine (a) the tension in wire CD, (b) the angles θx, θy,
and θz that the force exerted at C forms with the coordinate axes.
SOLUTION
(a)
Fx = − F sin 30° cos 60° = −40 lb (Given)
F =
40 lb
= 160 lb
sin30° cos60°
F = 160.0 lb
(b)
Fx
−40
=
= −0.25
F
160
cosθ x =
θ x = 104.5°
Fy = (160 lb ) cos 30° = 103.92 lb
cosθ y =
Fy
F
=
103.92
= 0.866
160
θ y = 30.0°
Fz = (160 lb ) sin 30° sin 60° = 69.282 lb
cosθ z =
Fz
69.282
=
= 0.433
F
160
θ z = 64.3°
84
PROBLEM 2.81
Determine
the
magnitude
and
F = ( 800 lb ) i + ( 260 lb ) j − ( 320 lb ) k.
direction
of
the
force
SOLUTION
F =
Fx2 + Fy2 + Fz2 =
(800 lb )2 + ( 260 lb )2 + ( −320 lb )2
F = 900 lb
cosθ x =
Fx
800
=
= 0.8889
F
900
θ x = 27.3°
cosθ y =
Fy
θ y = 73.2°
cosθ z =
F
=
260
= 0.2889
900
Fz
−320
=
= −0.3555
F
900
85
θ z = 110.8°
PROBLEM 2.82
Determine
the
magnitude
and
direction
F = ( 400 N ) i − (1200 N ) j + ( 300 N ) k.
of
the
force
SOLUTION
F =
Fx2 + Fy2 + Fz2 =
( 400 N )2 + ( −1200 N )2 + ( 300 N )2
Fx
400
=
= 0.30769
F
1300
cosθ x =
cosθ y =
Fy
F
cosθ z =
=
−1200
= −0.92307
1300
Fz
300
=
= 0.23076
F
1300
86
F = 1300 N
θ x = 72.1°
θ y = 157.4°
θ z = 76.7°
PROBLEM 2.83
A force acts at the origin of a coordinate system in a direction defined by
the angles θx = 64.5° and θz = 55.9°. Knowing that the y component of
the force is –200 N, determine (a) the angle θy, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
( cosθ x )2 + ( cosθ y )
2
(
+ ( cosθ z ) = 1 ⇒ cosθ y
2
)
2
(
= 1 − cosθ y
2
) − ( cosθ z )2
Since Fy < 0 we must have cosθ y < 0
Thus, taking the negative square root, from above, we have:
cosθ y = − 1 − ( cos 64.5° ) − ( cos 55.9° ) = −0.70735
2
2
θ y = 135.0°
(b) Then:
F =
and
Fy
cosθ y
=
−200 N
= 282.73 N
−0.70735
Fx = F cosθ x = ( 282.73 N ) cos 64.5°
Fx = 121.7 N
Fz = F cosθ z = ( 282.73 N ) cos 55.9°
Fy = 158.5 N
F = 283 N
87
PROBLEM 2.84
A force acts at the origin of a coordinate system in a direction defined by
the angles θx = 75.4° and θy = 132.6°. Knowing that the z component of
the force is –60 N, determine (a) the angle θz, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
( cosθ x )2 + ( cosθ y )
2
(
+ ( cosθ z ) = 1 ⇒ cosθ y
2
)
2
(
= 1 − cosθ y
2
) − ( cosθ z )2
Since Fz < 0 we must have cosθ z < 0
Thus, taking the negative square root, from above, we have:
cosθ z = − 1 − ( cos 75.4° ) − ( cos132.6° ) = −0.69159
2
2
θ z = 133.8°
(b) Then:
F =
and
Fz
−60 N
=
= 86.757 N
cosθ z
−0.69159
F = 86.8 N
Fx = F cosθ x = ( 86.8 N ) cos 75.4°
Fx = 21.9 N
Fy = F cosθ y = ( 86.8 N ) cos132.6°
Fy = −58.8 N
88
PROBLEM 2.85
A force F of magnitude 400 N acts at the origin of a coordinate system.
Knowing that θx = 28.5°, Fy = –80 N, and Fz > 0, determine (a) the
components Fx and Fz, (b) the angles θy and θz.
SOLUTION
(a) Have
Fx = F cosθ x = ( 400 N ) cos 28.5°
Fx = 351.5 N
Then:
F 2 = Fx2 + Fy2 + Fz2
( 400 N )2
So:
= ( 352.5 N ) + ( −80 N ) + Fz2
2
2
Hence:
Fz = +
( 400 N )2 − ( 351.5 N )2 − ( −80 N )2
Fz = 173.3 N
(b)
cosθ y =
cosθ z =
Fy
F
=
−80
= −0.20
400
Fz 173.3
=
= 0.43325
F
400
89
θ y = 101.5°
θ z = 64.3°
PROBLEM 2.86
A force F of magnitude 600 lb acts at the origin of a coordinate system.
Knowing that Fx = 200 lb, θz = 136.8°, Fy < 0, determine (a) the
components Fy and Fz, (b) the angles θx and θy.
SOLUTION
Fz = F cosθ z = ( 600 lb ) cos136.8°
(a)
= −437.4 lb
Fz = −437 lb
Then:
F 2 = Fx2 + Fy2 + Fz2
So:
( 600 lb )2 = ( 200 lb )2 + ( Fy )
Hence:
Fy = −
2
+ ( −437.4 lb )
2
( 600 lb )2 − ( 200 lb )2 − ( −437.4 lb )2
= −358.7 lb
Fy = −359 lb
(b)
cosθ x =
cosθ y =
Fy
F
Fx
200
=
= 0.333
F
600
=
−358.7
= −0.59783
600
90
θ x = 70.5°
θ y = 126.7°
PROBLEM 2.87
A transmission tower is held by three guy wires anchored by bolts at B,
C, and D. If the tension in wire AB is 2100 N, determine the components
of the force exerted by the wire on the bolt at B.
SOLUTION
JJJG
BA = ( 4 m ) i + ( 20 m ) j − ( 5 m ) k
BA =
F = F λ BA
( 4 m )2 + ( 20 m )2 + ( −5 m )2
= 21 m
JJJG
BA 2100 N
( 4 m ) i + ( 20 m ) j − ( 5 m ) k 
= F
=
BA
21 m 
F = ( 400 N ) i + ( 2000 N ) j − ( 500 N ) k
Fx = +400 N, Fy = +2000 N, Fz = −500 N
91
PROBLEM 2.88
A transmission tower is held by three guy wires anchored by bolts at B,
C, and D. If the tension in wire AD is 1260 N, determine the components
of the force exerted by the wire on the bolt at D.
SOLUTION
JJJG
DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) k
DA =
F = F λ DA
( 4 m )2 + ( 20 m )2 + (14.8 m )2
= 25.2 m
JJJG
DA 1260 N
( 4 m ) i + ( 20 m ) j + (14.8 m ) k 
= F
=
DA
25.2 m 
F = ( 200 N ) i + (1000 N ) j + ( 740 N ) k
Fx = +200 N, Fy = +1000 N, Fz = +740 N
PROBLEM 2.89
A rectangular plate is supported by three cables as shown. Knowing that
the tension in cable AB is 204 lb, determine the components of the force
exerted on the plate at B.
SOLUTION
JJJG
BA = ( 32 in.) i + ( 48 in.) j − ( 36 in.) k
BA =
( 32 in.)2 + ( 48 in.)2 + ( −36 in.)2
F = F λ BA
= 68 in.
JJJG
BA 204 lb
( 32 in.) i + ( 48 in.) j − ( 36 in.) k 
= F
=
BA
68 in. 
F = ( 96 lb ) i + (144 lb ) j − (108 lb ) k
Fx = +96.0 lb, Fy = +144.0 lb, Fz = −108.0 lb
93
PROBLEM 2.90
A rectangular plate is supported by three cables as shown. Knowing that
the tension in cable AD is 195 lb, determine the components of the force
exerted on the plate at D.
SOLUTION
JJJG
DA = − ( 25 in.) i + ( 48 in.) j + ( 36 in.) k
DA =
( −25 in.)2 + ( 48 in.)2 + ( 36 in.)2
F = F λ DA
= 65 in.
JJJG
DA 195 lb
( −25 in.) i + ( 48 in.) j + ( 36 in.) k 
= F
=
DA
65 in. 
F = − ( 75 lb ) i + (144 lb ) j + (108 lb ) k
Fx = −75.0 lb, Fy = +144.0 lb, Fz = +108.0 lb
94
PROBLEM 2.91
A steel rod is bent into a semicircular ring of radius 0.96 m and is
supported in part by cables BD and BE which are attached to the ring at
B. Knowing that the tension in cable BD is 220 N, determine the
components of this force exerted by the cable on the support at D.
SOLUTION
JJJG
DB = ( 0.96 m ) i − (1.12 m ) j − ( 0.96 m ) k
DB =
TDB = T λ DB
( 0.96 m )2 + ( −1.12 m )2 + ( −0.96 m )2
= 1.76 m
JJJG
DB
220 N
( 0.96 m ) i − (1.12 m ) j − ( 0.96 m ) k 
=T
=
DB 1.76 m 
TDB = (120 N ) i − (140 N ) j − (120 N ) k
(TDB ) x
95
= +120.0 N, (TDB ) y = −140.0 N, (TDB ) z = −120.0 N
PROBLEM 2.92
A steel rod is bent into a semicircular ring of radius 0.96 m and is
supported in part by cables BD and BE which are attached to the ring at
B. Knowing that the tension in cable BE is 250 N, determine the
components of this force exerted by the cable on the support at E.
SOLUTION
JJJG
EB = ( 0.96 m ) i − (1.20 m ) j + (1.28 m ) k
EB =
TEB = T λ EB
( 0.96 m )2 + ( −1.20 m )2 + (1.28 m )2
= 2.00 m
JJJG
EB
250 N
( 0.96 m ) i − (1.20 m ) j + (1.28 m ) k 
=T
=
EB
2.00 m 
TEB = (120 N ) i − (150 N ) j + (160 N ) k
(TEB ) x
= +120.0 N, (TEB ) y = −150.0 N, (TEB ) z = +160.0 N
96
PROBLEM 2.93
Find the magnitude and direction of the resultant of the two forces shown
knowing that P = 500 N and Q = 600 N.
SOLUTION
P = ( 500 lb ) [ − cos 30° sin15°i + sin 30° j + cos 30° cos15°k ]
= ( 500 lb ) [ −0.2241i + 0.50 j + 0.8365k ]
= − (112.05 lb ) i + ( 250 lb ) j + ( 418.25 lb ) k
Q = ( 600 lb ) [ cos 40° cos 20°i + sin 40° j − cos 40° sin 20°k ]
= ( 600 lb ) [ 0.71985i + 0.64278j − 0.26201k ]
= ( 431.91 lb ) i + ( 385.67 lb ) j − (157.206 lb ) k
R = P + Q = ( 319.86 lb ) i + ( 635.67 lb ) j + ( 261.04 lb ) k
R=
( 319.86 lb )2 + ( 635.67 lb )2 + ( 261.04 lb )2
= 757.98 lb
R = 758 lb
cosθ x =
Rx
319.86 lb
=
= 0.42199
R
757.98 lb
θ x = 65.0°
cosθ y =
Ry
R
=
635.67 lb
= 0.83864
757.98 lb
θ y = 33.0°
cosθ z =
Rz
261.04 lb
=
= 0.34439
R
757.98 lb
θ z = 69.9°
97
PROBLEM 2.94
Find the magnitude and direction of the resultant of the two forces shown
knowing that P = 600 N and Q = 400 N.
SOLUTION
Using the results from 2.93:
P = ( 600 lb ) [ −0.2241i + 0.50 j + 0.8365k ]
= − (134.46 lb ) i + ( 300 lb ) j + ( 501.9 lb ) k
Q = ( 400 lb ) [ 0.71985i + 0.64278 j − 0.26201k ]
= ( 287.94 lb ) i + ( 257.11 lb ) j − (104.804 lb ) k
R = P + Q = (153.48 lb ) i + ( 557.11 lb ) j + ( 397.10 lb ) k
R=
(153.48 lb )2 + ( 557.11 lb )2 + ( 397.10 lb )2
= 701.15 lb
R = 701 lb
cosθ x =
Rx
153.48 lb
=
= 0.21890
R
701.15 lb
θ x = 77.4°
cosθ y =
Ry
R
=
557.11 lb
= 0.79457
701.15 lb
θ y = 37.4°
cosθ z =
Rz
397.10 lb
=
= 0.56637
R
701.15 lb
θ z = 55.5°
98
PROBLEM 2.95
Knowing that the tension is 850 N in cable AB and 1020 N in cable AC,
determine the magnitude and direction of the resultant of the forces
exerted at A by the two cables.
SOLUTION
JJJG
AB = ( 400 mm ) i − ( 450 mm ) j + ( 600 mm ) k
AB =
( 400 mm )2 + ( −450 mm )2 + ( 600 mm )2
= 850 mm
JJJG
AC = (1000 mm ) i − ( 450 mm ) j + ( 600 mm ) k
AC =
(1000 mm )2 + ( −450 mm )2 + ( 600 mm )2
TAB = TABλ AB = TAB
= 1250 mm
JJJG
 ( 400 mm ) i − ( 450 mm ) j + ( 600 mm ) k 
AB
= ( 850 N ) 

AB
850 mm


TAB = ( 400 N ) i − ( 450 N ) j + ( 600 N ) k
TAC = TAC λ AC = TAC
JJJG
 (1000 mm ) i − ( 450 mm ) j + ( 600 mm ) k 
AC
= (1020 N ) 

AC
1250 mm


TAC = ( 816 N ) i − ( 367.2 N ) j + ( 489.6 N ) k
R = TAB + TAC = (1216 N ) i − ( 817.2 N ) j + (1089.6 N ) k
R = 1825.8 N
Then:
and
cosθ x =
cosθ y =
cosθ z =
R = 1826 N
1216
= 0.66601
1825.8
θ x = 48.2°
−817.2
= −0.44758
1825.8
θ y = 116.6°
1089.6
= 0.59678
1825.8
θ z = 53.4°
99
PROBLEM 2.96
Assuming that in Problem 2.95 the tension is 1020 N in cable AB and
850 N in cable AC, determine the magnitude and direction of the resultant
of the forces exerted at A by the two cables.
SOLUTION
JJJG
AB = ( 400 mm ) i − ( 450 mm ) j + ( 600 mm ) k
AB =
( 400 mm )2 + ( −450 mm )2 + ( 600 mm )2
= 850 mm
JJJG
AC = (1000 mm ) i − ( 450 mm ) j + ( 600 mm ) k
AC =
(1000 mm )2 + ( −450 mm )2 + ( 600 mm )2
TAB = TABλ AB = TAB
= 1250 mm
JJJG
 ( 400 mm ) i − ( 450 mm ) j + ( 600 mm ) k 
AB
= (1020 N ) 

AB
850 mm


TAB = ( 480 N ) i − ( 540 N ) j + ( 720 N ) k
TAC = TAC λ AC = TAC
JJJG
 (1000 mm ) i − ( 450 mm ) j + ( 600 mm ) k 
AC
= ( 850 N ) 

AC
1250 mm


TAC = ( 680 N ) i − ( 306 N ) j + ( 408 N ) k
R = TAB + TAC = (1160 N ) i − ( 846 N ) j + (1128 N ) k
R = 1825.8 N
R = 1826 N
cosθ x =
1160
= 0.6353
1825.8
θ x = 50.6°
cosθ y =
−846
= −0.4634
1825.8
θ y = 117.6°
1128
= 0.6178
1825.8
θ z = 51.8°
Then:
and
cosθ z =
100
PROBLEM 2.97
For the semicircular ring of Problem 2.91, determine the magnitude and
direction of the resultant of the forces exerted by the cables at B knowing
that the tensions in cables BD and BE are 220 N and 250 N, respectively.
SOLUTION
For the solutions to Problems 2.91 and 2.92, we have
TBD = − (120 N ) i + (140 N ) j + (120 N ) k
TBE = − (120 N ) i + (150 N ) j − (160 N ) k
Then:
R B = TBD + TBE
= − ( 240 N ) i + ( 290 N ) j − ( 40 N ) k
and
R = 378.55 N
cosθ x = −
RB = 379 N
240
= −0.6340
378.55
θ x = 129.3°
cosθ y =
290
= −0.7661
378.55
θ y = 40.0°
cosθ z = −
40
= −0.1057
378.55
θ z = 96.1°
101
PROBLEM 2.98
To stabilize a tree partially uprooted in a storm, cables AB and AC are
attached to the upper trunk of the tree and then are fastened to steel rods
anchored in the ground. Knowing that the tension in AB is 920 lb and that
the resultant of the forces exerted at A by cables AB and AC lies in the yz
plane, determine (a) the tension in AC, (b) the magnitude and direction of
the resultant of the two forces.
SOLUTION
Have
TAB = ( 920 lb )( sin 50° cos 40°i − cos 50° j + sin 50° sin 40° j)
TAC = TAC ( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25° j)
(a)
R A = TAB + TAC
( RA ) x
∴
( RA ) x
= ΣFx = 0:
=0
( 920 lb ) sin 50° cos 40° − TAC cos 45° sin 25° = 0
or
TAC = 1806.60 lb
TAC = 1807 lb
(b)
( RA ) y
= ΣFy : − ( 920 lb ) cos 50° − (1806.60 lb ) sin 45°
( RA ) y
( RA ) z
= ΣFz :
= −1868.82 lb
( 920 lb ) sin 50° sin 40° + (1806.60 lb ) cos 45° cos 25°
( RA ) z
= 1610.78 lb
∴ RA = − (1868.82 lb ) j + (1610.78 lb ) k
Then:
RA = 2467.2 lb
RA = 2.47 kips
102
PROBLEM 2.98 CONTINUED
and
cosθ x =
cosθ y =
cosθ z =
0
=0
2467.2
θ x = 90.0°
−1868.82
= −0.7560
2467.2
θ y = 139.2°
1610.78
= 0.65288
2467.2
θ z = 49.2°
103
PROBLEM 2.99
To stabilize a tree partially uprooted in a storm, cables AB and AC are
attached to the upper trunk of the tree and then are fastened to steel rods
anchored in the ground. Knowing that the tension in AC is 850 lb and that
the resultant of the forces exerted at A by cables AB and AC lies in the yz
plane, determine (a) the tension in AB, (b) the magnitude and direction of
the resultant of the two forces.
SOLUTION
Have
TAB = TAB ( sin 50° cos 40°i − cos 50° j + sin 50° sin 40° j)
TAC = ( 850 lb )( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25° j)
(a)
( RA ) x
∴
( RA ) x
=0
= ΣFx = 0: TAB sin 50° cos 40° − ( 850 lb ) cos 45° sin 25° = 0
TAB = 432.86 lb
TAB = 433 lb
(b)
( RA ) y
= ΣFy : − ( 432.86 lb ) cos 50° − ( 850 lb ) sin 45°
( RA ) y
( RA ) z
= ΣFz :
= −879.28 lb
( 432.86 lb ) sin 50° sin 40° + (850 lb ) cos 45° cos 25°
( RA ) z
= 757.87 lb
∴ R A = − ( 879.28 lb ) j + ( 757.87 lb ) k
RA = 1160.82 lb
cosθ x =
cosθ y =
RA = 1.161 kips
0
=0
1160.82
θ x = 90.0°
−879.28
= −0.75746
1160.82
θ y = 139.2°
757.87
= 0.65287
1160.82
θ z = 49.2°
cosθ z =
104
PROBLEM 2.100
For the plate of Problem 2.89, determine the tension in cables AB and AD
knowing that the tension if cable AC is 27 lb and that the resultant of the
forces exerted by the three cables at A must be vertical.
SOLUTION
With:
JJJG
AC = ( 45 in.) i − ( 48 in.) j + ( 36 in.) k
( 45 in.)2 + ( −48 in.)2 + ( 36 in.)2
AC =
TAC = TAC λ AC = TAC
= 75 in.
JJJG
AC
27 lb
( 45 in.) i − ( 48 in.) j + ( 36 in.) k 
=
AC
75 in. 
TAC = (16.2 lb ) i − (17.28 lb ) j + (12.96 ) k
and
JJJG
AB = − ( 32 in.) i − ( 48 in.) j + ( 36 in.) k
( −32 in.)2 + ( −48 in.)2 + ( 36 in.)2
AB =
TAB = TABλ AB = TAB
= 68 in.
JJJG
AB
T
= AB ( −32 in.) i − ( 48 in.) j + ( 36 in.) k 
AB 68 in. 
TAB = TAB ( −0.4706i − 0.7059 j + 0.5294k )
and
JJJG
AD = ( 25 in.) i − ( 48 in.) j − ( 36 in.) k
AD =
( 25 in.)2 + ( −48 in.)2 + ( 36 in.)2
TAD = TADλ AD = TAD
= 65 in.
JJJG
AD
T
= AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k 
AD 65 in.
TAD = TAD ( 0.3846i − 0.7385 j − 0.5538k )
105
PROBLEM 2.100 CONTINUED
Now
R = TAB + TAD + TAD
= TAB ( −0.4706i − 0.7059 j + 0.5294k ) + (16.2 lb ) i − (17.28 lb ) j + (12.96 ) k 
+ TAD ( 0.3846i − 0.7385 j − 0.5538k )
Since R must be vertical, the i and k components of this sum must be zero.
Hence:
−0.4706TAB + 0.3846TAD + 16.2 lb = 0
(1)
0.5294TAB − 0.5538TAD + 12.96 lb = 0
(2)
Solving (1) and (2), we obtain:
TAB = 244.79 lb,
TAD = 257.41 lb
TAB = 245 lb
TAD = 257 lb
106
PROBLEM 2.101
The support assembly shown is bolted in place at B, C, and D and
supports a downward force P at A. Knowing that the forces in members
AB, AC, and AD are directed along the respective members and that the
force in member AB is 146 N, determine the magnitude of P.
SOLUTION
Note that AB, AC, and AD are in compression.
Have
and
d BA =
( −220 mm )2 + (192 mm )2 + ( 0 )2
d DA =
(192 mm )2 + (192 mm )2 + ( 96 mm )2
dCA =
( 0 )2 + (192 mm )2 + ( −144 mm )2
FBA = FBAλ BA =
= 292 mm
= 288 mm
= 240 mm
146 N
( −220 mm ) i + (192 mm ) j
292 mm 
= − (110 N ) i + ( 96 N ) j
FCA = FCAλ CA =
FCA
(192 mm ) j − (144 mm ) k 
240 mm 
= FCA ( 0.80j − 0.60k )
FDA = FDAλ DA =
FDA
(192 mm ) i + (192 mm ) j + ( 96 mm ) k 
288 mm 
= FDA [ 0.66667i + 0.66667 j + 0.33333k ]
P = − Pj
With
At A:
i-component:
ΣF = 0: FBA + FCA + FDA + P = 0
− (110 N ) + 0.66667 FDA = 0
or
FDA = 165 N
j-component:
96 N + 0.80 FCA + 0.66667 (165 N ) − P = 0
(1)
k-component:
−0.60FCA + 0.33333 (165 N ) = 0
(2)
Solving (2) for FCA and then using that result in (1), gives
107
P = 279 N
PROBLEM 2.102
The support assembly shown is bolted in place at B, C, and D and
supports a downward force P at A. Knowing that the forces in members
AB, AC, and AD are directed along the respective members and that
P = 200 N, determine the forces in the members.
SOLUTION
With the results of 2.101:
FBA = FBAλ BA =
FBA
( −220 mm ) i + (192 mm ) j
292 mm 
= FBA [ −0.75342i + 0.65753j] N
FCA = FCAλ CA =
FCA
(192 mm ) j − (144 mm ) k 
240 mm 
= FCA ( 0.80 j − 0.60k )
FDA = FDAλ DA =
FDA
(192 mm ) i + (192 mm ) j + ( 96 mm ) k 
288 mm 
= FDA [ 0.66667i + 0.66667 j + 0.33333k ]
P = − ( 200 N ) j
With:
ΣF = 0: FBA + FCA + FDA + P = 0
At A:
Hence, equating the three (i, j, k) components to 0 gives three equations
i-component:
−0.75342 FBA + 0.66667 FDA = 0
(1)
j-component:
0.65735FBA + 0.80FCA + 0.66667 FDA − 200 N = 0
(2)
k-component:
−0.60FCA + 0.33333FDA = 0
(3)
Solving (1), (2), and (3), gives
FBA = 104.5 N,
FCA = 65.6 N,
FDA = 118.1 N
FBA = 104.5 N
FCA = 65.6 N
FDA = 118.1 N
108
PROBLEM 2.103
Three cables are used to tether a balloon as shown. Determine the vertical
force P exerted by the balloon at A knowing that the tension in cable AB
is 60 lb.
SOLUTION
The forces applied at A are:
TAB , TAC , TAD and P
where P = Pj . To express the other forces in terms of the unit vectors
i, j, k, we write
JJJG
AB = − (12.6 ft ) i − (16.8 ft ) j
AB = 21 ft
JJJG
AC = ( 7.2 ft ) i − (16.8 ft ) j + (12.6 ft ) k
AC = 22.2 ft
JJJG
AD = − (16.8 ft ) j − ( 9.9 ft ) k
AD = 19.5 ft
JJJG
AB
= ( −0.6i − 0.8j) TAB
TAB = TABλ AB = TAB
and
AB
JJJG
AC
= ( 0.3242i − 0.75676 j + 0.56757k ) TAC
TAC = TAC λ AC = TAC
AC
JJJG
AD
= ( −0.8615 j − 0.50769k ) TAD
TAD = TADλ AD = TAD
AD
109
PROBLEM 2.103 CONTINUED
Equilibrium Condition
ΣF = 0: TAB + TAC + TAD + Pj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and
factoring i, j, and k:
( −0.6TAB + 0.3242TAC ) i + ( −0.8TAB − 0.75676TAC
− 0.8615TAD + P ) j
+ ( 0.56757TAC − 0.50769TAD ) k = 0
Equating to zero the coefficients of i, j, k:
−0.6TAB + 0.3242TAC = 0
(1)
−0.8TAB − 0.75676TAC − 0.8615TAD + P = 0
(2)
0.56757TAC − 0.50769TAD = 0
(3)
Setting TAB = 60 lb in (1) and (2), and solving the resulting set of
equations gives
TAC = 111 lb
TAD = 124.2 lb
P = 239 lb
110
PROBLEM 2.104
Three cables are used to tether a balloon as shown. Determine the vertical
force P exerted by the balloon at A knowing that the tension in cable AC
is 100 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
−0.6TAB + 0.3242TAC = 0
(1)
−0.8TAB − 0.75676TAC − 0.8615TAD + P = 0
(2)
0.56757TAC − 0.50769TAD = 0
(3)
Substituting TAC = 100 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives
TAB = 54 lb
TAD = 112 lb
P = 215 lb
111
PROBLEM 2.105
The crate shown in Figure P2.105 and P2.108 is supported by three
cables. Determine the weight of the crate knowing that the tension in
cable AB is 3 kN.
SOLUTION
The forces applied at A are:
TAB , TAC , TAD and P
where P = Pj . To express the other forces in terms of the unit vectors
i, j, k, we write
JJJG
AB = − ( 0.72 m ) i + (1.2 m ) j − ( 0.54 m ) k ,
AB = 1.5 m
JJJG
AC = (1.2 m ) j + ( 0.64 m ) k ,
AC = 1.36 m
JJJG
AD = ( 0.8 m ) i + (1.2 m ) j − ( 0.54 m ) k ,
AD = 1.54 m
JJJG
AB
TAB = TABλ AB = TAB
= ( −0.48i + 0.8 j − 0.36k ) TAB
and
AB
JJJG
AC
TAC = TAC λ AC = TAC
= ( 0.88235j + 0.47059k ) TAC
AC
JJJG
AD
TAD = TADλ AD = TAD
= ( 0.51948i + 0.77922 j − 0.35065k ) TAD
AD
Equilibrium Condition with W = −Wj
ΣF = 0: TAB + TAC + TAD − Wj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and
factoring i, j, and k:
( −0.48TAB + 0.51948TAD ) i + ( 0.8TAB + 0.88235TAC
+ 0.77922TAD − W ) j
+ ( −0.36TAB + 0.47059TAC − 0.35065TAD ) k = 0
112
PROBLEM 2.105 CONTINUED
Equating to zero the coefficients of i, j, k:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAB = 3 kN in Equations (1), (2) and (3) and solving the
resulting set of equations, using conventional algorithms for solving
linear algebraic equations, gives
TAC = 4.3605 kN
TAD = 2.7720 kN
W = 8.41 kN
113
PROBLEM 2.106
For the crate of Problem 2.105, determine the weight of the crate
knowing that the tension in cable AD is 2.8 kN.
Problem 2.105: The crate shown in Figure P2.105 and P2.108 is
supported by three cables. Determine the weight of the crate knowing that
the tension in cable AB is 3 kN.
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAD = 2.8 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives
TAB = 3.03 kN
TAC = 4.40 kN
W = 8.49 kN
114
PROBLEM 2.107
For the crate of Problem 2.105, determine the weight of the crate
knowing that the tension in cable AC is 2.4 kN.
Problem 2.105: The crate shown in Figure P2.105 and P2.108 is
supported by three cables. Determine the weight of the crate knowing that
the tension in cable AB is 3 kN.
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAC = 2.4 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives
TAB = 1.651 kN
TAD = 1.526 kN
W = 4.63 kN
115
PROBLEM 2.108
A 750-kg crate is supported by three cables as shown. Determine the
tension in each cable.
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
(
)
Substituting W = ( 750 kg ) 9.81 m/s 2 = 7.36 kN in Equations (1), (2), and (3) above, and solving the
resulting set of equations using conventional algorithms, gives
TAB = 2.63 kN
TAC = 3.82 kN
TAD = 2.43 kN
116
PROBLEM 2.109
A force P is applied as shown to a uniform cone which is supported by
three cords, where the lines of action of the cords pass through the vertex
A of the cone. Knowing that P = 0 and that the tension in cord BE is
0.2 lb, determine the weight W of the cone.
SOLUTION
Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all
have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the
generators of the cone.
Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB.
Hence:
It follows that:
λ AB = λ BE =
cos 45°i + 8j − sin 45°k
65
 cos 45°i + 8 j − sin 45°k 
TBE = TBE λ BE = TBE 

65


 cos 30°i + 8j + sin 30°k 
TCF = TCF λ CF = TCF 

65


 − cos15°i + 8 j − sin15°k 
TDG = TDG λ DG = TDG 

65


117
PROBLEM 2.109 CONTINUED
ΣF = 0: TBE + TCF + TDG + W + P = 0
At A:
Then, isolating the factors of i, j, and k, we obtain three algebraic equations:
i:
or
TBE
T
T
cos 45° + CF cos 30° − DG cos15° + P = 0
65
65
65
TBE cos 45° + TCF cos 30° − TDG cos15° + P 65 = 0
j: TBE
k: −
or
8
8
8
+ TCF
+ TDG
−W = 0
65
65
65
TBE + TCF + TDG − W
or
(1)
65
=0
8
(2)
TBE
T
T
sin 45° + CF sin 30° − DG sin15° = 0
65
65
65
−TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(3)
With P = 0 and the tension in cord BE = 0.2 lb:
Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination,
matrix methods or iteration – with MATLAB or Maple, for example), we obtain:
TCF = 0.669 lb
TDG = 0.746 lb
W = 1.603 lb
118
PROBLEM 2.110
A force P is applied as shown to a uniform cone which is supported by
three cords, where the lines of action of the cords pass through the vertex
A of the cone. Knowing that the cone weighs 1.6 lb, determine the range
of values of P for which cord CF is taut.
SOLUTION
See Problem 2.109 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
i : TBE cos 45° + TCF cos 30° − TDG cos15° + 65 P = 0
j: TBE + TCF + TDG − W
65
=0
8
k : − TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(1)
(2)
(3)
With W = 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1),
(2), and (3) for the tension TCF as a function of P and requiring it to be positive (> 0).
Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrix
methods or iteration – with MATLAB or Maple, for example), we obtain:
TCF = ( −1.729 P + 0.668 ) lb
Hence, for TCF > 0
or
−1.729 P + 0.668 > 0
P < 0.386 lb
∴ 0 < P < 0.386 lb
119
PROBLEM 2.111
A transmission tower is held by three guy wires attached to a pin at A and
anchored by bolts at B, C, and D. If the tension in wire AB is 3.6 kN,
determine the vertical force P exerted by the tower on the pin at A.
SOLUTION
The force in each cable can be written as the product of the magnitude of
the force and the unit vector along the cable. That is, with
JJJG
AC = (18 m ) i − ( 30 m ) j + ( 5.4 m ) k
AC =
(18 m )2 + ( −30 m )2 + ( 5.4 m )2
TAC = T λ AC = TAC
= 35.4 m
JJJG
AC
TAC
(18 m ) i − ( 30 m ) j + ( 5.4 m ) k 
=
35.4 m 
AC
TAC = TAC ( 0.5085i − 0.8475j + 0.1525k )
JJJG
AB = − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k
and
AB =
( −6 m )2 + ( −30 m )2 + ( 7.5 m )2
TAB = T λ AB = TAB
= 31.5 m
JJJG
AB
TAB
 − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k 
=
AB 31.5 m 
TAB = TAB ( −0.1905i − 0.9524 j + 0.2381k )
JJJG
AD = − ( 6 m ) i − ( 30 m ) j − ( 22.2 m ) k
Finally
AD =
( −6 m )2 + ( −30 m )2 + ( −22.2 m )2
TAD = T λ AD = TAD
= 37.8 m
JJJG
AD
TAD
 − ( 6 m ) i − ( 30 m ) j − ( 22.2 m ) k 
=
AD 37.8 m 
TAD = TAD ( −0.1587i − 0.7937 j − 0.5873k )
120
PROBLEM 2.111 CONTINUED
With P = Pj, at A:
ΣF = 0: TAB + TAC + TAD + Pj = 0
Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations:
i : − 0.1905TAB + 0.5085TAC − 0.1587TAD = 0
(1)
j: − 0.9524TAB − 0.8475TAC − 0.7937TAD + P = 0
(2)
k : 0.2381TAB + 0.1525TAC − 0.5873TAD = 0
(3)
In Equations (1), (2) and (3), set TAB = 3.6 kN, and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain:
TAC = 1.963 kN
TAD = 1.969 kN
P = 6.66 kN
121
PROBLEM 2.112
A transmission tower is held by three guy wires attached to a pin at A and
anchored by bolts at B, C, and D. If the tension in wire AC is 2.6 kN,
determine the vertical force P exerted by the tower on the pin at A.
SOLUTION
Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAC = 2.6 kN
and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations
(MATLAB or Maple, for example), to obtain
TAB = 4.77 kN
TAD = 2.61 kN
P = 8.81 kN
122
PROBLEM 2.113
A rectangular plate is supported by three cables as shown. Knowing that
the tension in cable AC is 15 lb, determine the weight of the plate.
SOLUTION
The (vector) force in each cable can be written as the product of the
(scalar) force and the unit vector along the cable. That is, with
JJJG
AB = ( 32 in.) i − ( 48 in.) j + ( 36 in.) k
AB =
( −32 in.)2 + ( −48 in.)2 + ( 36 in.)2
TAB = T λ AB = TAB
= 68 in.
JJJG
AB
T
= AB  − ( 32 in.) i − ( 48 in.) j + ( 36 in.) k 
AB 68 in.
TAB = TAB ( −0.4706i − 0.7059 j + 0.5294k )
JJJG
AC = ( 45 in.) i − ( 48 in.) j + ( 36 in.) k
and
AC =
( 45 in.)2 + ( −48 in.)2 + ( 36 in.)2
TAC = T λ AC = TAC
= 75 in.
JJJG
AC
T
= AC ( 45 in.) i − ( 48 in.) j + ( 36 in.) k 
75 in.
AC
TAC = TAC ( 0.60i − 0.64 j + 0.48k )
JJJG
AD = ( 25 in.) i − ( 48 in.) j − ( 36 in.) k
Finally,
AD =
123
( 25 in.)2 + ( −48 in.)2 + ( −36 in.)2
= 65 in.
PROBLEM 2.113 CONTINUED
TAD = T λ AD = TAD
JJJG
AD
T
= AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k 
AD 65 in. 
TAD = TAD ( 0.3846i − 0.7385 j − 0.5538k )
With W = Wj, at A we have:
ΣF = 0: TAB + TAC + TAD + Wj = 0
Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations:
i : − 0.4706TAB + 0.60TAC − 0.3846TAD = 0
(1)
j: − 0.7059TAB − 0.64TAC − 0.7385TAD + W = 0
(2)
k : 0.5294TAB + 0.48TAC − 0.5538TAD = 0
(3)
In Equations (1), (2) and (3), set TAC = 15 lb, and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain:
TAB = 136.0 lb
TAD = 143.0 lb
W = 211 lb
124
PROBLEM 2.114
A rectangular plate is supported by three cables as shown. Knowing that
the tension in cable AD is 120 lb, determine the weight of the plate.
SOLUTION
Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAD = 120 lb and
solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations
(MATLAB or Maple, for example), to obtain
TAC = 12.59 lb
TAB = 114.1 lb
W = 177.2 lb
125
PROBLEM 2.115
A horizontal circular plate having a mass of 28 kg is suspended as shown
from three wires which are attached to a support D and form 30° angles
with the vertical. Determine the tension in each wire.
SOLUTION
ΣFx = 0: − TAD sin 30° sin 50° + TBD sin 30° cos 40°
+ TCD sin 30° cos 60° = 0
Dividing through by the factor sin 30° and evaluating the trigonometric
functions gives
−0.7660TAD + 0.7660TBD + 0.50TCD = 0
(1)
Similarly,
ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40°
− TCD sin 30° sin 60° = 0
or
0.6428TAD + 0.6428TBD − 0.8660TCD = 0
(2)
TAD = TBD + 0.6527TCD
From (1)
Substituting this into (2):
TBD = 0.3573TCD
(3)
TAD = TCD
(4)
Using TAD from above:
Now,
ΣFy = 0: − TAD cos 30° − TBD cos 30° − TCD cos 30°
(
)
+ ( 28 kg ) 9.81 m/s 2 = 0
or
TAD + TBD + TCD = 317.2 N
126
PROBLEM 2.115 CONTINUED
Using (3) and (4), above:
TCD + 0.3573TCD + TCD = 317.2 N
TAD = 135.1 N
Then:
TBD = 46.9 N
TCD = 135.1 N
127
PROBLEM 2.119
A force P is applied as shown to a uniform cone which is supported by
three cords, where the lines of action of the cords pass through the vertex
A of the cone. Knowing that the cone weighs 2.4 lb and that P = 0,
determine the tension in each cord.
SOLUTION
Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all
have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the
generators of the cone.
Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB.
Hence:
λ AB = λBE =
cos 45°i + 8j − sin 45°k
65
It follows that:
 cos 45°i + 8 j − sin 45°k 
TBE = TBE λ BE = TBE 

65


 cos 30°i + 8j + sin 30°k 
TCF = TCF λ CF = TCF 

65


 − cos15°i + 8 j − sin15°k 
TDG = TDG λ DG = TDG 

65


At A:
ΣF = 0: TBE + TCF + TDG + W + P = 0
132
PROBLEM 2.119 CONTINUED
Then, isolating the factors if i, j, and k we obtain three algebraic equations:
i:
TBE
T
T
cos 45° + CF cos 30° − DG cos15° = 0
65
65
65
TBE cos 45° + TCF cos 30° − TDG cos15° = 0
or
j: TBE
k: −
or
8
8
8
+ TCF
+ TDG
−W = 0
65
65
65
TBE + TCF + TDG =
or
(1)
2.4
65 = 0.3 65
8
(2)
TBE
T
T
sin 45° + CF sin 30° − DG sin15° − P = 0
65
65
65
−TBE sin 45° + TCF sin 30° − TDG sin15° = P 65
(3)
With P = 0, the tension in the cords can be found by solving the resulting Equations (1), (2), and (3) using
conventional methods in Linear Algebra (elimination, matrix methods or iteration–with MATLAB or Maple,
for example). We obtain
TBE = 0.299 lb
TCF = 1.002 lb
TDG = 1.117 lb
133
PROBLEM 2.120
A force P is applied as shown to a uniform cone which is supported by
three cords, where the lines of action of the cords pass through the vertex
A of the cone. Knowing that the cone weighs 2.4 lb and that P = 0.1 lb,
determine the tension in each cord.
SOLUTION
See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
TBE cos 45° + TCF cos 30° − TDG cos15° = 0
(1)
TBE + TCF + TDG = 0.3 65
(2)
−TBE sin 45° + TCF sin 30° − TDG sin15° = P 65
(3)
With P = 0.1 lb, solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrix
methods or iteration–with MATLAB or Maple, for example), we obtain
TBE = 1.006 lb
TCF = 0.357 lb
TDG = 1.056 lb
134
PROBLEM 2.121
Using two ropes and a roller chute, two workers are unloading a 200-kg
cast-iron counterweight from a truck. Knowing that at the instant shown
the counterweight is kept from moving and that the positions of points A,
B, and C are, respectively, A(0, –0.5 m, 1 m), B(–0.6 m, 0.8 m, 0), and
C(0.7 m, 0.9 m, 0), and assuming that no friction exists between the
counterweight and the chute, determine the tension in each rope. (Hint:
Since there is no friction, the force exerted by the chute on the
counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute:
N=
N
( 2 j + k ) = N ( 0.8944 j + 0.4472k )
5
As in Problem 2.11, for example, the force in each rope can be written as
the product of the magnitude of the force and the unit vector along the
cable. Thus, with
JJJG
AB = − ( 0.6 m ) i + (1.3 m ) j + (1 m ) k
AB =
( −0.6 m )2 + (1.3 m )2 + (1 m )2
TAB = T λ AB = TAB
= 1.764 m
JJJG
AB
TAB
 − ( 0.6 m ) i + (1.3 m ) j + (1 m ) k 
=
AB 1.764 m 
TAB = TAB ( −0.3436i + 0.7444 j + 0.5726k )
JJJG
AC = ( 0.7 m ) i + (1.4 m ) j − (1 m ) k
and
AC =
( 0.7 m )2 + (1.4 m )2 + ( −1 m )2
TAC = T λ AC = TAC
= 1.8574 m
JJJG
AC
TAC
( 0.7 m ) i + (1.4 m ) j − (1 m ) k 
=
AC 1.764 m 
TAC = TAC ( 0.3769i + 0.7537 j − 0.5384k )
ΣF = 0: N + TAB + TAC + W = 0
Then:
135
PROBLEM 2.121 CONTINUED
With W = ( 200 kg )( 9.81 m/s ) = 1962 N, and equating the factors of i, j,
and k to zero, we obtain the linear algebraic equations:
i : − 0.3436TAB + 0.3769TAC = 0
(1)
j: 0.7444TAB + 0.7537TAC + 0.8944 N − 1962 = 0
(2)
k : − 0.5726TAB − 0.5384TAC + 0.4472 N = 0
(3)
Using conventional methods for solving Linear Algebraic Equations
(elimination, MATLAB or Maple, for example), we obtain
N = 1311 N
TAB = 551 N
TAC = 503 N
136
PROBLEM 2.122
Solve Problem 2.121 assuming that a third worker is exerting a force
P = −(180 N)i on the counterweight.
Problem 2.121: Using two ropes and a roller chute, two workers are
unloading a 200-kg cast-iron counterweight from a truck. Knowing that at
the instant shown the counterweight is kept from moving and that the
positions of points A, B, and C are, respectively, A(0, –0.5 m, 1 m),
B(–0.6 m, 0.8 m, 0), and C(0.7 m, 0.9 m, 0), and assuming that no friction
exists between the counterweight and the chute, determine the tension in
each rope. (Hint: Since there is no friction, the force exerted by the chute
on the counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute:
N=
N
( 2 j + k ) = N ( 0.8944 j + 0.4472k )
5
As in Problem 2.11, for example, the force in each rope can be written as
the product of the magnitude of the force and the unit vector along the
cable. Thus, with
JJJG
AB = − ( 0.6 m ) i + (1.3 m ) j + (1 m ) k
AB =
( −0.6 m )2 + (1.3 m )2 + (1 m )2
TAB = T λ AB = TAB
= 1.764 m
JJJG
AB
TAB
 − ( 0.6 m ) i + (1.3 m ) j + (1 m ) k 
=
AB 1.764 m 
TAB = TAB ( −0.3436i + 0.7444 j + 0.5726k )
JJJG
AC = ( 0.7 m ) i + (1.4 m ) j − (1 m ) k
and
AC =
( 0.7 m )2 + (1.4 m )2 + ( −1 m )2
TAC = T λ AC = TAC
= 1.8574 m
JJJG
AC
TAC
( 0.7 m ) i + (1.4 m ) j − (1 m ) k 
=
AC 1.764 m 
TAC = TAC ( 0.3769i + 0.7537 j − 0.5384k )
ΣF = 0: N + TAB + TAC + P + W = 0
Then:
137
PROBLEM 2.122 CONTINUED
P = − (180 N ) i
Where
and
(
)
W = − ( 200 kg ) 9.81 m/s 2  j


= − (1962 N ) j
Equating the factors of i, j, and k to zero, we obtain the linear equations:
i : − 0.3436TAB + 0.3769TAC − 180 = 0
j: 0.8944 N + 0.7444TAB + 0.7537TAC − 1962 = 0
k : 0.4472 N − 0.5726TAB − 0.5384TAC = 0
Using conventional methods for solving Linear Algebraic Equations
(elimination, MATLAB or Maple, for example), we obtain
N = 1302 N
TAB = 306 N
TAC = 756 N
138
PROBLEM 2.123
A piece of machinery of weight W is temporarily supported by cables AB,
AC, and ADE. Cable ADE is attached to the ring at A, passes over the
pulley at D and back through the ring, and is attached to the support at E.
Knowing that W = 320 lb, determine the tension in each cable. (Hint:
The tension is the same in all portions of cable ADE.)
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along
the cable. That is, with
JJJG
AB = − ( 9 ft ) i + ( 8 ft ) j − (12 ft ) k
AB =
( −9 ft )2 + (8 ft )2 + ( −12 ft )2
TAB = T λ AB = TAB
= 17 ft
JJJG
AB
T
= AB  − ( 9 ft ) i + ( 8 ft ) j − (12 ft ) k 
AB 17 ft 
TAB = TAB ( −0.5294i + 0.4706 j − 0.7059k )
and
JJJG
AC = ( 0 ) i + ( 8 ft ) j + ( 6 ft ) k
AC =
( 0 ft )2 + (8 ft )2 + ( 6 ft )2
TAC = T λ AC = TAC
= 10 ft
JJJG
AC
T
= AC ( 0 ft ) i + ( 8 ft ) j + ( 6 ft ) k 
AC 10 ft
TAC = TAC ( 0.8 j + 0.6k )
and
JJJG
AD = ( 4 ft ) i + ( 8 ft ) j − (1 ft ) k
AD =
( 4 ft )2 + (8 ft )2 + ( −1 ft )2
TAD = T λ AD = TADE
= 9 ft
JJJG
AD TADE
( 4 ft ) i + ( 8 ft ) j − (1 ft ) k 
=
9 ft 
AD
TAD = TADE ( 0.4444i + 0.8889 j − 0.1111k )
139
PROBLEM 2.123 CONTINUED
Finally,
JJJG
AE = ( −8 ft ) i + ( 8 ft ) j + ( 4 ft ) k
AE =
( −8 ft )2 + (8 ft )2 + ( 4 ft )2
TAE = T λ AE = TADE
= 12 ft
JJJG
AE TADE
( −8 ft ) i + ( 8 ft ) j + ( 4 ft ) k 
=
12 ft 
AE
TAE = TADE ( −0.6667i + 0.6667 j + 0.3333k )
With the weight of the machinery, W = −W j, at A, we have:
ΣF = 0: TAB + TAC + 2TAD − Wj = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
−0.5294TAB + 2 ( 0.4444TADE ) − 0.6667TADE = 0
(1)
0.4706TAB + 0.8TAC + 2 ( 0.8889TADE ) + 0.6667TADE − W = 0
(2)
−0.7059TAB + 0.6TAC − 2 ( 0.1111TADE ) + 0.3333TADE = 0
(3)
Knowing that W = 320 lb, we can solve Equations (1), (2) and (3) using conventional methods for solving
Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain
TAB = 46.5 lb
TAC = 34.2 lb
TADE = 110.8 lb
140
PROBLEM 2.124
A piece of machinery of weight W is temporarily supported by cables AB,
AC, and ADE. Cable ADE is attached to the ring at A, passes over the
pulley at D and back through the ring, and is attached to the support at E.
Knowing that the tension in cable AB is 68 lb, determine (a) the tension
in AC, (b) the tension in ADE, (c) the weight W. (Hint: The tension is the
same in all portions of cable ADE.)
SOLUTION
See Problem 2.123 for the analysis leading to the linear algebraic Equations (1), (2), and (3), below:
−0.5294TAB + 2 ( 0.4444TADE ) − 0.6667TADE = 0
(1)
0.4706TAB + 0.8TAC + 2 ( 0.8889TADE ) + 0.6667TADE − W = 0
(2)
−0.7059TAB + 0.6TAC − 2 ( 0.1111TADE ) + 0.3333TADE = 0
(3)
Knowing that the tension in cable AB is 68 lb, we can solve Equations (1), (2) and (3) using conventional
methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for
example) to obtain
(a) TAC = 50.0 lb
(b) TAE = 162.0 lb
(c)
141
W = 468 lb
PROBLEM 2.128
Solve Problem 2.127 assuming y = 550 mm.
Problem 2.127: Collars A and B are connected by a 1-m-long wire and
can slide freely on frictionless rods. If a force P = (680 N) j is applied at
A, determine (a) the tension in the wire when y = 300 mm, (b) the
magnitude of the force Q required to maintain the equilibrium of the
system.
SOLUTION
From the analysis of Problem 2.127, particularly the results:
y 2 + z 2 = 0.84 m 2
TAB =
Q=
680 N
y
680 N
z
y
With y = 550 mm = 0.55 m, we obtain:
z 2 = 0.84 m 2 − ( 0.55 m )
2
∴ z = 0.733 m
and
TAB =
(a)
680 N
= 1236.4 N
0.55
TAB = 1.236 kN
or
and
Q = 1236 ( 0.866 ) N = 906 N
(b)
Q = 0.906 kN
or
147
PROBLEM 2.129
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 300-lb horizontal component, determine
(a) the magnitude of the force P, (b) its vertical component.
SOLUTION
(a)
P sin 35° = 300 1b
P=
300 lb
sin 35°
P = 523 lb
(b) Vertical Component
Pv = P cos 35°
= ( 523 lb ) cos 35°
Pv = 428 lb
148
PROBLEM 2.130
A container of weight W is suspended from ring A, to which cables AC
and AE are attached. A force P is applied to the end F of a third cable
which passes over a pulley at B and through ring A and which is attached
to a support at D. Knowing that W = 1000 N, determine the magnitude
of P. (Hint: The tension is the same in all portions of cable FBAD.)
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along
the cable. That is, with
JJJG
AB = − ( 0.78 m ) i + (1.6 m ) j + ( 0 m ) k
AB =
( −0.78 m )2 + (1.6 m )2 + ( 0 )2
TAB = T λ AB = TAB
= 1.78 m
JJJG
AB
TAB
 − ( 0.78 m ) i + (1.6 m ) j + ( 0 m ) k 
=
AB 1.78 m 
TAB = TAB ( −0.4382i + 0.8989 j + 0k )
and
JJJG
AC = ( 0 ) i + (1.6 m ) j + (1.2 m ) k
AC =
( 0 m )2 + (1.6 m )2 + (1.2 m )2
TAC = T λ AC = TAC
= 2m
JJJG
AC TAC
( 0 ) i + (1.6 m ) j + (1.2 m ) k 
=
AC
2m
TAC = TAC ( 0.8 j + 0.6k )
and
JJJG
AD = (1.3 m ) i + (1.6 m ) j + ( 0.4 m ) k
AD =
(1.3 m )2 + (1.6 m )2 + ( 0.4 m )2
TAD = T λ AD = TAD
= 2.1 m
JJJG
AD
T
= AD (1.3 m ) i + (1.6 m ) j + ( 0.4 m ) k 
AD
2.1 m
TAD = TAD ( 0.6190i + 0.7619 j + 0.1905k )
149
PROBLEM 2.130 CONTINUED
Finally,
JJJG
AE = − ( 0.4 m ) i + (1.6 m ) j − ( 0.86 m ) k
AE =
( −0.4 m )2 + (1.6 m )2 + ( −0.86 m )2
TAE = T λ AE = TAE
= 1.86 m
JJJG
AE
TAE
 − ( 0.4 m ) i + (1.6 m ) j − ( 0.86 m ) k 
=
AE 1.86 m 
TAE = TAE ( −0.2151i + 0.8602 j − 0.4624k )
With the weight of the container W = −Wj, at A we have:
ΣF = 0: TAB + TAC + TAD − Wj = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
−0.4382TAB + 0.6190TAD − 0.2151TAE = 0
(1)
0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0
(2)
0.6TAC + 0.1905TAD − 0.4624TAE = 0
(3)
Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the
externally applied (unknown) force, we can solve the system of linear equations (1), (2) and (3) uniquely
for P.
P = 378 N
150
PROBLEM 2.131
A container of weight W is suspended from ring A, to which cables AC
and AE are attached. A force P is applied to the end F of a third cable
which passes over a pulley at B and through ring A and which is attached
to a support at D. Knowing that the tension in cable AC is 150 N,
determine (a) the magnitude of the force P, (b) the weight W of the
container. (Hint: The tension is the same in all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.130, the support of the container consists of the four cables AE, AC, AD, and AB, with
the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the
condition
TAB = TAD = P
and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain
(a)
P = 454 N
(b) W = 1202 N
151
PROBLEM 2.125
A container of weight W is suspended from ring A. Cable BAC passes
through the ring and is attached to fixed supports at B and C. Two forces
P = Pi and Q = Qk are applied to the ring to maintain the container is
the position shown. Knowing that W = 1200 N, determine P and Q.
(Hint: The tension is the same in both portions of cable BAC.)
SOLUTION
The (vector) force in each cable can be written as the product of the
(scalar) force and the unit vector along the cable. That is, with
JJJG
AB = − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k
AB =
( −0.48 m )2 + ( 0.72 m )2 + ( −0.16 m )2
TAB = T λ AB = TAB
= 0.88 m
JJJG
AB
TAB
 − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k 
=
AB 0.88 m 
TAB = TAB ( −0.5455i + 0.8182 j − 0.1818k )
and
JJJG
AC = ( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k
AC =
( 0.24 m )2 + ( 0.72 m )2 − ( 0.13 m )2
TAC = T λ AC = TAC
= 0.77 m
JJJG
AC
TAC
( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k 
=
0.77 m 
AC
TAC = TAC ( 0.3177i + 0.9351j − 0.1688k )
At A:
ΣF = 0: TAB + TAC + P + Q + W = 0
142
PROBLEM 2.125 CONTINUED
Noting that TAB = TAC because of the ring A, we equate the factors of
i, j, and k to zero to obtain the linear algebraic equations:
i:
( −0.5455 + 0.3177 ) T
+P=0
P = 0.2338T
or
j:
( 0.8182 + 0.9351) T
−W = 0
W = 1.7532T
or
k:
( −0.1818 − 0.1688) T
+Q =0
Q = 0.356T
or
With W = 1200 N:
T =
1200 N
= 684.5 N
1.7532
P = 160.0 N
Q = 240 N
143
PROBLEM 2.126
For the system of Problem 2.125, determine W and P knowing that
Q = 160 N.
Problem 2.125: A container of weight W is suspended from ring A.
Cable BAC passes through the ring and is attached to fixed supports at B
and C. Two forces P = Pi and Q = Qk are applied to the ring to
maintain the container is the position shown. Knowing that W = 1200 N,
determine P and Q. (Hint: The tension is the same in both portions of
cable BAC.)
SOLUTION
Based on the results of Problem 2.125, particularly the three equations relating P, Q, W, and T we substitute
Q = 160 N to obtain
T =
160 N
= 456.3 N
0.3506
W = 800 N
P = 107.0 N
144
PROBLEM 2.127
Collars A and B are connected by a 1-m-long wire and can slide freely on
frictionless rods. If a force P = (680 N) j is applied at A, determine
(a) the tension in the wire when y = 300 mm, (b) the magnitude of the
force Q required to maintain the equilibrium of the system.
SOLUTION
Free-Body Diagrams of collars
For both Problems 2.127 and 2.128:
( AB )2
(1 m )2
Here
= x2 + y 2 + z 2
= ( 0.40 m ) + y 2 + z 2
2
y 2 + z 2 = 0.84 m 2
or
Thus, with y given, z is determined.
Now
λ AB =
JJJG
AB
1
=
( 0.40i − yj + zk ) m = 0.4i − yk + zk
AB 1 m
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A:
ΣF = 0: N xi + N zk + Pj + TAB λ AB = 0
Setting the j coefficient to zero gives:
P − yTAB = 0
With P = 680 N,
TAB =
680 N
y
Now, from the free body diagram of collar B:
ΣF = 0: N xi + N y j + Qk − TABλ AB = 0
145
PROBLEM 2.127 CONTINUED
Setting the k coefficient to zero gives:
Q − TAB z = 0
And using the above result for TAB we have
Q = TAB z =
680 N
z
y
Then, from the specifications of the problem, y = 300 mm = 0.3 m
z 2 = 0.84 m 2 − ( 0.3 m )
2
∴ z = 0.866 m
and
TAB =
(a)
680 N
= 2266.7 N
0.30
TAB = 2.27 kN
or
and
Q = 2266.7 ( 0.866 ) = 1963.2 N
(b)
Q = 1.963 kN
or
146
PROBLEM 2.116
A transmission tower is held by three guy wires attached to a pin at A and
anchored by bolts at B, C, and D. Knowing that the tower exerts on the
pin at A an upward vertical force of 8 kN, determine the tension in each
wire.
SOLUTION
From the solutions of 2.111 and 2.112:
TAB = 0.5409 P
TAC = 0.295P
TAD = 0.2959 P
Using P = 8 kN:
TAB = 4.33 kN
TAC = 2.36 kN
TAD = 2.37 kN
128
PROBLEM 2.117
For the rectangular plate of Problems 2.113 and 2.114, determine the
tension in each of the three cables knowing that the weight of the plate is
180 lb.
SOLUTION
From the solutions of 2.113 and 2.114:
TAB = 0.6440 P
TAC = 0.0709 P
TAD = 0.6771P
Using P = 180 lb:
TAB = 115.9 lb
TAC = 12.76 lb
TAD = 121.9 lb
129
PROBLEM 2.118
For the cone of Problem 2.110, determine the range of values of P for
which cord DG is taut if P is directed in the –x direction.
SOLUTION
From the solutions to Problems 2.109 and 2.110, have
TBE + TCF + TDG = 0.2 65
−TBE sin 45° + TCF sin 30° − TDG sin15° = 0
TBE cos 45° + TCF cos 30° − TDG cos15° − P 65 = 0
(2′)
(3)
(1′ )
Applying the method of elimination to obtain a desired result:
Multiplying (2′) by sin 45° and adding the result to (3):
TCF ( sin 45° + sin 30° ) + TDG ( sin 45° − sin15° ) = 0.2 65 sin 45°
or
TCF = 0.9445 − 0.3714TDG
(4)
Multiplying (2′) by sin 30° and subtracting (3) from the result:
TBE ( sin 30° + sin 45° ) + TDG ( sin 30° + sin15° ) = 0.2 65 sin 30°
or
TBE = 0.6679 − 0.6286TDG
130
(5)
PROBLEM 2.118 CONTINUED
Substituting (4) and (5) into (1′) :
1.2903 − 1.7321TDG − P 65 = 0
∴ TDG is taut for P <
1.2903
lb
65
or 0 ≤ P < 0.1600 lb
131
PROBLEM 2.132
Two cables tied together at C are loaded as shown. Knowing that
Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
ΣFy = 0: TCA − Q cos 30° = 0
Q = 60 lb
With
TCA = ( 60 lb )( 0.866 )
(a)
TCA = 52.0 lb
ΣFx = 0: P − TCB − Q sin 30° = 0
(b)
P = 75 lb
With
TCB = 75 lb − ( 60 lb )( 0.50 )
or TCB = 45.0 lb
152
PROBLEM 2.133
Two cables tied together at C are loaded as shown. Determine the range
of values of Q for which the tension will not exceed 60 lb in either cable.
SOLUTION
ΣFx = 0: TCA − Q cos 30° = 0
Have
TCA = 0.8660 Q
or
TCA ≤ 60 lb
Then for
0.8660Q < 60 lb
Q ≤ 69.3 lb
or
ΣFy = 0: TCB = P − Q sin 30°
From
or
TCB = 75 lb − 0.50Q
For
TCB ≤ 60 lb
75 lb − 0.50Q ≤ 60 lb
0.50Q ≥ 15 lb
or
Q ≥ 30 lb
Thus,
30.0 ≤ Q ≤ 69.3 lb
Therefore,
153
PROBLEM 2.134
A welded connection is in equilibrium under the action of the four forces
shown. Knowing that FA = 8 kN and FB = 16 kN, determine the
magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of
Connection
ΣFx = 0:
3
3
FB − FC − FA = 0
5
5
FA = 8 kN, FB = 16 kN
With
FC =
4
4
(16 kN ) − (8 kN )
5
5
FC = 6.40 kN
ΣFy = 0: − FD +
3
3
FB − FA = 0
5
5
With FA and FB as above:
FD =
3
3
(16 kN ) − (8 kN )
5
5
FD = 4.80 kN
154
PROBLEM 2.135
A welded connection is in equilibrium under the action of the four forces
shown. Knowing that FA = 5 kN and FD = 6 kN, determine the
magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of
Connection
ΣFy = 0: − FD −
3
3
FA + FB = 0
5
5
FB = FD +
or
3
FA
5
FA = 5 kN, FD = 8 kN
With
FB =
5
3

6 kN + ( 5 kN ) 
3 
5

FB = 15.00 kN
ΣFx = 0: − FC +
FC =
=
4
4
FB − FA = 0
5
5
4
( FB − FA )
5
4
(15 kN − 5 kN )
5
FC = 8.00 kN
155
PROBLEM 2.136
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the magnitude of the force P
required to maintain the equilibrium of the collar when (a) x = 4.5 in.,
(b) x = 15 in.
SOLUTION
Free-Body Diagram of Collar
(a)
Triangle Proportions
ΣFx = 0: − P +
4.5
( 50 lb ) = 0
20.5
or P = 10.98 lb
(b)
Triangle Proportions
ΣFx = 0: − P +
15
( 50 lb ) = 0
25
or P = 30.0 lb
156
PROBLEM 2.137
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the collar
is in equilibrium when P = 48 lb.
SOLUTION
Free-Body Diagram of Collar
Triangle Proportions
Hence:
ΣFx = 0: − 48 +
xˆ =
or
50 xˆ
400 + xˆ 2
=0
48
400 + xˆ 2
50
(
xˆ 2 = 0.92 lb 400 + xˆ 2
)
xˆ 2 = 4737.7 in 2
xˆ = 68.6 in.
157
PROBLEM 2.138
A frame ABC is supported in part by cable DBE which passes through a
frictionless ring at B. Knowing that the tension in the cable is 385 N,
determine the components of the force exerted by the cable on the
support at D.
SOLUTION
The force in cable DB can be written as the product of the magnitude of the force and the unit vector along the
cable. That is, with
JJJG
DB = ( 480 mm ) i − ( 510 mm ) j + ( 320 mm ) k
DB =
F = F λ DB
( 480 )2 + ( 510 )2 + ( 320 )2
= 770 mm
JJJG
DB
385 N
( 480 mm ) i − ( 510 mm ) j + ( 320 mm ) k 
= F
=
DB
770 mm 
F = ( 240 N ) i − ( 255 N ) j + (160 N ) k
Fx = +240 N, Fy = −255 N, Fz = +160.0 N
158
PROBLEM 2.139
A frame ABC is supported in part by cable DBE which passes through a
frictionless ring at B. Determine the magnitude and direction of the
resultant of the forces exerted by the cable at B knowing that the tension
in the cable is 385 N.
SOLUTION
The force in each cable can be written as the product of the magnitude of the force and the unit vector along
the cable. That is, with
JJJG
BD = − ( 0.48 m ) i + ( 0.51 m ) j − ( 0.32 m ) k
BD =
( −0.48 m )2 + ( 0.51 m )2 + ( −0.32 m )2
TBD = T λ BD = TBD
= 0.77 m
JJJG
BD
TBD
 − ( 0.48 m ) i + ( 0.51 m ) j − ( 0.32 m ) k 
=
BD 0.77 m 
TBD = TBD ( −0.6234i + 0.6623j − 0.4156k )
and
JJJG
BE = − ( 0.27 m ) i + ( 0.40 m ) j − ( 0.6 m ) k
BE =
( −0.27 m )2 + ( 0.40 m )2 + ( −0.6 m )2
TBE = T λ BE = TBE
= 0.770 m
JJJG
BD
TBE
 − ( 0.26 m ) i + ( 0.40 m ) j − ( 0.6 m ) k 
=
BD 0.770 m 
TBE = TBE ( −0.3506i + 0.5195 j − 0.7792k )
Now, because of the frictionless ring at B, TBE = TBD = 385 N and the force on the support due to the two
cables is
F = 385 N ( −0.6234i + 0.6623j − 0.4156k − 0.3506i + 0.5195j − 0.7792k )
= − ( 375 N ) i + ( 455 N ) j − ( 460 N ) k
159
PROBLEM 2.139 CONTINUED
The magnitude of the resultant is
F =
Fx2 + Fy2 + Fz2 =
( −375 N )2 + ( 455 N )2 + ( −460 N )2
= 747.83 N
or F = 748 N
The direction of this force is:
θ x = cos −1
−375
747.83
or θ x = 120.1°
θ y = cos −1
455
747.83
or θ y = 52.5°
θ z = cos −1
−460
747.83
or θ z = 128.0°
160
PROBLEM 2.140
A steel tank is to be positioned in an excavation. Using trigonometry,
determine (a) the magnitude and direction of the smallest force P for
which the resultant R of the two forces applied at A is vertical, (b) the
corresponding magnitude of R.
SOLUTION
Force Triangle
(a) For minimum P it must be perpendicular to the vertical resultant R
∴ P = ( 425 lb ) cos 30°
or P = 368 lb
R = ( 425 lb ) sin 30°
(b)
or R = 213 lb
161
PROBLEM 3.1
A 13.2-N force P is applied to the lever which controls the auger of a
snowblower. Determine the moment of P about A when α is equal to 30°.
SOLUTION
First note
Px = P sin α = (13.2 N ) sin 30° = 6.60 N
Py = P cos α = (13.2 N ) cos 30° = 11.4315 N
Noting that the direction of the moment of each force component about A
is counterclockwise,
M A = xB/ A Py + yB/ A Px
= ( 0.086 m )(11.4315 N ) + ( 0.122 m )( 6.60 N )
= 1.78831 N ⋅ m
or M A = 1.788 N ⋅ m
W
PROBLEM 3.2
The force P is applied to the lever which controls the auger of a
snowblower. Determine the magnitude and the direction of the smallest
force P which has a 2.20- N ⋅ m counterclockwise moment about A.
SOLUTION
For P to be a minimum, it must be perpendicular to the line joining points
A and B.
rAB =
(86 mm )2 + (122 mm )2
 y
= 149.265 mm
 122 mm 
α = θ = tan −1   = tan −1 
 = 54.819°
x
 86 mm 
M A = rAB Pmin
Then
or
Pmin =
=
MA
rAB
2.20 N ⋅ m  1000 mm 


149.265 mm  1 m 
= 14.7389 N
∴ Pmin = 14.74 N
54.8°
or Pmin = 14.74 N
35.2° W
PROBLEM 3.3
A 13.1-N force P is applied to the lever which controls the auger of a
snowblower. Determine the value of α knowing that the moment of P
about A is counterclockwise and has a magnitude of 1.95 N ⋅ m.
SOLUTION
M A = rB/ A P sin θ
By definition
θ = φ + ( 90° − α )
where
 122 mm 
φ = tan −1 
 = 54.819°
 86 mm 
and
Also
Then
rB/ A =
(86 mm )2 + (122 mm )2
= 149.265 mm
1.95 N ⋅ m = ( 0.149265 m )(13.1 N ) sin ( 54.819° + 90° − α )
or
sin (144.819° − α ) = 0.99725
or
144.819° − α = 85.752°
and
144.819° − α = 94.248°
∴ α = 50.6°, 59.1° W
PROBLEM 3.4
A foot valve for a pneumatic system is hinged at B. Knowing that
α = 28°, determine the moment of the 4-lb force about point B by
resolving the force into horizontal and vertical components.
SOLUTION
Note that
θ = α − 20° = 28° − 20° = 8°
and
Fx = ( 4 lb ) cos8° = 3.9611 lb
Fy = ( 4 lb ) sin 8° = 0.55669 lb
Also
x = ( 6.5 in.) cos 20° = 6.1080 in.
y = ( 6.5 in.) sin 20° = 2.2231 in.
Noting that the direction of the moment of each force component about B
is counterclockwise,
M B = xFy + yFx
= ( 6.1080 in.)( 0.55669 lb ) + ( 2.2231 in.)( 3.9611 lb )
= 12.2062 lb ⋅ in.
or M B = 12.21 lb ⋅ in. W
PROBLEM 3.5
A foot valve for a pneumatic system is hinged at B. Knowing that
α = 28°, determine the moment of the 4-lb force about point B by
resolving the force into components along ABC and in a direction
perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
Q = ( 4.0 lb ) sin 28° = 1.87787 lb
Then
M B = rA/BQ
= ( 6.5 in.)(1.87787 lb )
= 12.2063 lb ⋅ in.
or M B = 12.21 lb ⋅ in. W
PROBLEM 3.6
It is known that a vertical force of 800 N is required to remove the nail at
C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of the
force P which creates the same moment about B if α = 10°, (c) the
smallest force P which creates the same moment about B.
SOLUTION
(a) Have
M B = rC/B FN
= ( 0.1 m )( 800 N )
= 80.0 N ⋅ m
or M B = 80.0 N ⋅ m
W
(b) By definition
M B = rA/B P sin θ
where
θ = 90° − ( 90° − 70° ) − α
= 90° − 20° − 10°
= 60°
∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60°
P = 205.28 N
or P = 205 N W
(c) For P to be minimum, it must be perpendicular to the line joining
points A and B. Thus, P must be directed as shown.
Thus
or
M B = dPmin = rA/B Pmin
80.0 N ⋅ m = ( 0.45 m ) Pmin
∴ Pmin = 177.778 N
or Pmin = 177.8 N
20° W
PROBLEM 3.7
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exert by the chain at B, (b) the smallest force applied at C which creates
the same moment about A.
SOLUTION
M A = rB/ A × TBF
(a) Have
Noting that the direction of the moment of each force component
about A is counterclockwise,
M A = xTBFy + yTBFx
= ( 6.5 ft )( 45 lb ) sin 60° + ( 4.4 ft − 3.1 ft )( 45 lb ) cos 60°
= 282.56 lb ⋅ ft
or M A = 283 lb ⋅ ft
(b) Have
W
M A = rC/ A × ( FC )min
For FC to be minimum, it must be perpendicular to the
line joining points A and C.
∴ M A = d ( FC )min
where
d = rC/ A =
( 6.5 ft )2 + ( 4.4 ft )2
= 7.8492 ft
∴ 282.56 lb ⋅ ft = ( 7.8492 ft ) ( FC )min
( FC )min
= 35.999 lb
 4.4 ft 
φ = tan −1 
 = 34.095°
 6.5 ft 
θ = 90° − φ = 90° − 34.095° = 55.905°
or
( FC )min
= 36.0 lb
55.9° W
PROBLEM 3.8
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exerted by the chain at B, (b) the magnitude and sense of the vertical
force applied at C which creates the same moment about A, (c) the
smallest force applied at B which creates the same moment about A.
SOLUTION
M A = rB/ A × TBF
(a) Have
Noting that the direction of the moment of each force component
about A is counterclockwise,
M A = xTBFy + yTBFx
= ( 6.5 ft )( 45 lb ) sin 60° + ( 4.4 ft − 3.1 ft )( 45 lb ) cos 60°
= 282.56 lb ⋅ ft
or M A = 283 lb ⋅ ft
W
M A = rC/ A × FC
(b) Have
M A = xFC
or
∴ FC =
MA
282.56 lb ⋅ ft
=
= 43.471 lb
x
6.5 ft
or FC = 43.5 lb W
M A = rB/ A × ( FB )min
(c) Have
For FB to be minimum, it must be perpendicular to the line joining
points A and B.
∴ M A = d ( FB )min
d =
where
∴
and
( 6.5 ft )2 + ( 4.4 ft
( FB )min
=
− 3.1 ft ) = 6.6287 ft
2
MA
282.56 lb ⋅ ft
=
= 42.627 lb
d
6.6287 ft

6.5 ft

θ = tan −1 
 = 78.690°
 4.4 ft − 3.1 ft 
or
( FB )min
= 42.6 lb
78.7° W
PROBLEM 3.9
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts
a 125-N force directed along its center line on the ball and socket at B,
determine the moment of the force about A.
SOLUTION
First note
dCB =
( 240 mm )2 + ( 46.6 mm )2
= 244.48 mm
Then
and
where
240 mm
244.48 mm
sin θ =
46.6 mm
244.48 mm
FCB = FCB cosθ i − FCB sin θ j
=
Now
cosθ =
125 N
( 240 mm ) i − ( 46.6 mm ) j
244.48 mm 
M A = rB/ A × FCB
rB/ A = ( 306 mm ) i − ( 240 mm + 46.6 mm ) j
= ( 306 mm ) i − ( 286.6 mm ) j
Then
125 N
M A = ( 306 mm ) i − ( 286.6 mm ) j ×
( 240i − 46.6 j)
244.48
= ( 27878 N ⋅ mm ) k = ( 27.878 N ⋅ m ) k
or M A = 27.9 N ⋅ m
W
PROBLEM 3.10
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts
a 125-N force directed along its center line on the ball and socket at B,
determine the moment of the force about A.
SOLUTION
First note
Then
and
dCB =
( 344 mm )2 + (152.4 mm )2
cosθ =
344 mm
376.25 mm
152.4 mm
376.25 mm
FCB = ( FCB cosθ ) i − ( FCB sin θ ) j
=
Now
sin θ =
= 376.25 mm
125 N
( 344 mm ) i + (152.4 mm ) j
376.25 mm 
M A = rB/ A × FCB
where
rB/ A = ( 410 mm ) i − ( 87.6 mm ) j
Then
125 N
M A = ( 410 mm ) i − ( 87.6 mm ) j ×
( 344i − 152.4 j)
376.25
= ( 30770 N ⋅ mm ) k
= ( 30.770 N ⋅ m ) k
or M A = 30.8 N ⋅ m
W
PROBLEM 3.11
A winch puller AB is used to straighten a fence post. Knowing that the
tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and
length d is 76 in., determine the moment about D of the force exerted by
the cable at C by resolving that force into horizontal and vertical
components applied (a) at point C, (b) at point E.
SOLUTION
Slope of line EC =
(a)
Then
and
Then
TABx =
35 in.
5
=
76 in. + 8 in. 12
12
(TAB )
13
=
12
( 260 lb ) = 240 lb
13
TABy =
5
( 260 lb ) = 100 lb
13
M D = TABx ( 35 in.) − TABy ( 8 in.)
= ( 240 lb )( 35 in.) − (100 lb )( 8 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.
(b) Have
M D = TABx ( y ) + TABy ( x )
= ( 240 lb )( 0 ) + (100 lb )( 76 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.
PROBLEM 3.12
It is known that a force with a moment of 7840 lb ⋅ in. about D is required
to straighten the fence post CD. If a = 8 in., b = 35 in., and d = 112
in., determine the tension that must be developed in the cable of winch
puller AB to create the required moment about point D.
SOLUTION
Slope of line EC =
35 in.
7
=
112 in. + 8 in. 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
Have
M D = TABx ( y ) + TABy ( x )
∴ 7840 lb ⋅ in. =
24
7
TAB ( 0 ) +
TAB (112 in.)
25
25
TAB = 250 lb
or TAB = 250 lb
PROBLEM 3.13
It is known that a force with a moment of 1152 N ⋅ m about D is required
to straighten the fence post CD. If the capacity of the winch puller AB is
2880 N, determine the minimum value of distance d to create the
specified moment about point D knowing that a = 0.24 m and
b = 1.05 m.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y ( d )
M D = 1152 N ⋅ m
where
(TAB max ) y
= TAB max sin θ = ( 2880 N ) sin θ
1.05 m
sin θ =
Now
(d

∴ 1152 N ⋅ m = 2880 N 


or
or
or
+ 0.24 ) + (1.05 ) m
2
1.05
( d + 0.24 )2 + (1.05)2
( d + 0.24 )2 + (1.05)2
(d
2

 (d )


= 2.625d
+ 0.24 ) + (1.05 ) = 6.8906d 2
2
2
5.8906d 2 − 0.48d − 1.1601 = 0
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 580 N is exerted on
the alternator B. Determine the moment of that force about bolt C if its
line of action passes through O.
SOLUTION
M C = rB/C × FB
Have
Noting the direction of the moment of each force component about C is
clockwise,
M C = xFBy + yFBx
where
x = 144 mm − 78 mm = 66 mm
y = 86 mm + 108 mm = 194 mm
and
FBx =
FBy =
78
( 78)
2
+ ( 86 )
2
86
( 78) + (86 )
2
2
( 580 N ) = 389.65 N
( 580 N ) = 429.62 N
∴ M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N )
= 103947 N ⋅ mm
= 103.947 N ⋅ m
or M C = 103.9 N ⋅ m
PROBLEM 3.15
Form the vector products B × C and B′ × C, where B = B′, and use the
results obtained to prove the identity
sin α cos β =
SOLUTION
1
sin
2
(α + β ) +
1
sin
2
(α − β ) .
B = B ( cos β i + sin β j)
First note
B′ = B ( cos β i − sin β j)
C = C ( cos α i + sin α j)
By definition
Now
B × C = BC sin (α − β )
(1)
B′ × C = BC sin (α + β )
(2)
B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α − sin β cos α ) k
(3)
B × C = B ( cos β i − sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α + sin β cos α ) k
Equating magnitudes of B × C from Equations (1) and (3),
(4)
(5)
sin (α − β ) = cos β sin α − sin β cos α
Similarly, equating magnitudes of B′ × C from Equations (2) and (4),
sin (α + β ) = cos β sin α + sin β cos α
(6)
Adding Equations (5) and (6)
sin (α − β ) + sin (α + β ) = 2 cos β sin α
∴ sin α cos β =
1
1
sin (α + β ) + sin (α − β )
2
2
PROBLEM 3.16
A line passes through the points (420 mm, −150 mm) and (−140 mm,
180 mm). Determine the perpendicular distance d from the line to the
origin O of the system of coordinates.
SOLUTION
d = λ AB × rO/ A
Have
λ AB =
where
rB/ A
rB/ A
rB/ A = ( −140 mm − 420 mm ) i + 180 mm − ( −150 mm )  j
and
= − ( 560 mm ) i + ( 330 mm ) j
rB/ A =
∴ λ AB =
( −560 )2 + ( 330 )2 mm = 650 mm
− ( 560 mm ) i + ( 330 mm ) j
1
=
( −56i + 33j)
650 mm
65
rO/ A = ( 0 − x A ) i + ( 0 − y A ) j = − ( 420 mm ) i + (150 mm ) j
∴ d =
1
( −56i − 33j) ×  − ( 420 mm ) i + (150 mm ) j = 84.0 mm
65
d = 84.0 mm
PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal
to the plane when A and B are equal to, respectively, (a) 4i − 2j + 3k and
−2i + 6j − 5k, (b) 7i + j − 4k and −6i − 3k + 2k.
SOLUTION
λ =
(a) Have
A×B
A×B
A = 4i − 2 j + 3k
where
B = −2i + 6 j − 5k
Then
i j k
A × B = 4 −2 3 = (10 − 18 ) i + ( −6 + 20 ) j + ( 24 − 4 ) k = 2 ( −4i + 7 j + 10k )
−2 6 −5
A×B = 2
and
∴ λ =
( −4 )2 + ( 7 )2 + (10 )2
2 ( −4i + 7 j + 10k )
or λ =
2 165
λ =
(b) Have
= 2 165
1
( −4i + 7 j + 10k )
165
A×B
A×B
A = 7i + j − 4k
where
B = −6i − 3j + 2k
Then
and
i j k
A × B = 7 1 −4 = ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k = 5 ( −2i + 2 j − 3k )
−6 −3 2
A×B = 5
∴ λ =
( −2 )2 + ( 2 )2 + ( −3)2
5 ( −2i + 2 j − 3k )
5 17
= 5 17
or λ =
1
( −2i + 2 j − 3k )
17
PROBLEM 3.18
The vectors P and Q are two adjacent sides of a parallelogram.
Determine the area of the parallelogram when (a) P = (8 in.)i + (2 in.)j −
(1 in.)k and Q = −(3 in.)i + (4 in.)j + (2 in.)k, (b) P = −(3 in.)i + (6 in.)j +
(4 in.)k and Q = (2 in.)i + (5 in.)j − (3 in.)k.
SOLUTION
A = P×Q
(a) Have
P = ( 8 in.) i + ( 2 in.) j − (1 in.) k
where
Q = − ( 3 in.) i + ( 4 in.) j + ( 2 in.) k
i j k
P × Q = 8 2 −1 in 2 = ( 4 + 4 ) i + ( 3 − 16 ) j + ( 32 + 6 ) k  in 2
−3 4 2
Then
(
) (
) (
)
= 8 in 2 i − 13 in 2 j + 38 in 2 k
∴ Α=
(8)2 + ( −13)2 + ( 38)2 in 2
= 40.951 in 2
or A = 41.0 in 2
A = P×Q
(b) Have
P = − ( 3 in.) i + ( 6 in.) j + ( 4 in.) k
where
Q = ( 2 in.) i + ( 5 in.) j − ( 3 in.) k
Then
i j k
P × Q = −3 6 4 in 2 = ( −18 − 20 ) i + ( 8 − 9 ) j + ( −15 − 12 ) k  in 2
2 5 −3
(
) (
) (
)
= − 38 in 2 i − 1 in 2 j − 27 in 2 k
∴ Α=
( −38)2 + ( −1)2 + ( −27 )2 in 2
= 46.626 in 2
or A = 46.6 in 2
PROBLEM 3.19
Determine the moment about the origin O of the force F = −(5 N)i − (2
N)j + (3 N)k which acts at a point A. Assume that the position vector of
A is (a) r = (4 m)i − (2 m)j − (1 m)k, (b) r = −(8 m)i + (3 m)j + (4 m)k,
(c) r = (7.5 m)i + (3 m)j − (4.5 m)k.
SOLUTION
MO = r × F
(a) Have
F = − (5 N ) i − ( 2 N ) j + (3 N ) k
where
r = ( 4 m ) i − ( 2 m ) j − (1 m ) k
∴ MO
i j k
= 4 −2 −1 N ⋅ m = ( −6 − 2 ) i + ( 5 − 12 ) j + ( −8 − 10 ) k  N ⋅ m
−5 −2 3
= ( −8i − 7 j − 18k ) N ⋅ m
or M O = − ( 8 N ⋅ m ) i − ( 7 N ⋅ m ) j − (18 N ⋅ m ) k
MO = r × F
(b) Have
F = − (5 N ) i − ( 2 N ) j + (3 N ) k
where
r = − (8 m ) i + ( 3 m ) j − ( 4 m ) k
∴ MO
i j k
= −8 3 4 N ⋅ m = ( 9 + 8 ) i + ( −20 + 24 ) j + (16 + 15 ) k  N ⋅ m
−5 −2 3
= (17i + 4 j + 31k ) N ⋅ m
or M O = (17 N ⋅ m ) i + ( 4 N ⋅ m ) j + ( 31 N ⋅ m ) k
(c) Have
where
MO = r × F
F = − (5 N ) i − ( 2 N ) j + (3 N ) k
r = ( 7.5 m ) i + ( 3 m ) j − ( 4.5 m ) k
PROBLEM 3.19 CONTINUED
∴ MO
i
j
k
= 7.5 3 −4.5 N ⋅ m = ( 9 − 9 ) i + ( 22.5 − 22.5 ) j + ( −15 + 15 ) k  N ⋅ m
−5 −2 3
or M O = 0
−2 

r  . Therefore, vector F has a line of action
This answer is expected since r and F are proportional  F =
3 

passing through the origin at O.
PROBLEM 3.20
Determine the moment about the origin O of the force F = −(1.5 lb)i +
(3 lb)j − (2 lb)k which acts at a point A. Assume that the position vector
of A is (a) r = (2.5 ft)i − (1 ft)j + (2 ft)k, (b) r = (4.5 ft)i − (9 ft)j +
(6 ft)k, (c) r = (4 ft)i − (1 ft)j + (7 ft)k.
SOLUTION
MO = r × F
(a) Have
F = − (1.5 lb ) i + ( 3 lb ) j + ( 2 lb ) k
where
r = ( 2.5 ft ) i − (1 ft ) j + ( 2 ft ) k
Then
MO
i
j k
= 2.5 −1 2 lb ⋅ ft = ( 2 − 6 ) i + ( −3 + 5 ) j + ( 7.5 − 1.5 ) k  lb ⋅ ft
−1.5 3 −2
or M O = − ( 4 lb ⋅ ft ) i + ( 2 lb ⋅ ft ) j + ( 6 lb ⋅ ft ) k
MO = r × F
(b) Have
F = − (1.5 lb ) i + ( 3 lb ) j − ( 2 lb ) k
where
r = ( 4.5 ft ) i − ( 9 ft ) j + ( 6 ft ) k
Then
MO
i
j k
= 4.5 −9 6 lb ⋅ ft = (18 − 18 ) i + ( −9 + 9 ) j + (13.5 − 13.5 ) k  lb ⋅ ft
−1.5 3 −2
or M O = 0
−1 

r .
This answer is expected since r and F are proportional  F =
3 

Therefore, vector F has a line of action passing through the origin at O.
MO = r × F
(c) Have
F = − (1.5 lb ) i − ( 3 lb ) j − ( 2 lb ) k
where
r = ( 4 ft ) i − (1 ft ) j + ( 7 ft ) k
Then
MO
i
j k
= 4 −1 7 lb ⋅ ft = ( 2 − 21) i + ( −10.5 + 8 ) j + (12 − 1.5) k  lb ⋅ ft
−1.5 3 −2
or M O = − (19 lb ⋅ ft ) i − ( 2.5 lb ⋅ ft ) j + (10.5 lb ⋅ ft ) k
PROBLEM 3.21
Before the trunk of a large tree is felled, cables AB and BC are attached as
shown. Knowing that the tension in cables AB and BC are 777 N and
990 N, respectively, determine the moment about O of the resultant force
exerted on the tree by the cables at B.
SOLUTION
M O = rB/O × FB
Have
rB/O = ( 8.4 m ) j
where
FB = TAB + TBC
TAB = λ BATAB =
− ( 0.9 m ) i − ( 8.4 m ) j + ( 7.2 m ) k
TBC = λ BCTBC =
( 0.9 )2 + ( 8.4 )2 + ( 7.2 )2
( 777 N )
m
( 5.1 m ) i − (8.4 m ) j + (1.2 m ) k 990 N
(
)
( 5.1)2 + (8.4 )2 + (1.2 )2 m
PROBLEM 3.21 CONTINUED
∴ FB =  − ( 63.0 N ) i − ( 588 N ) j + ( 504 N ) k  + ( 510 N ) i − ( 840 N ) j + (120 N ) k 
= ( 447 N ) i − (1428 N ) j + ( 624 N ) k
and
MO
i
j
k
= 0
8.4
0 N ⋅ m = ( 5241.6 N ⋅ m ) i − ( 3754.8 N ⋅ m ) k
447 −1428 624
or M O = ( 5.24 kN ⋅ m ) i − ( 3.75 kN ⋅ m ) k
PROBLEM 3.22
Before a telephone cable is strung, rope BAC is tied to a stake at B and is
passed over a pulley at A. Knowing that portion AC of the rope lies in a
plane parallel to the xy plane and that the tension T in the rope is 124 N,
determine the moment about O of the resultant force exerted on the
pulley by the rope.
SOLUTION
Have
where
M O = rA/O × R
rA/O = ( 0 m ) i + ( 9 m ) j + (1 m ) k
R = T1 + T2
T1 = − (124 N ) cos10°  i − (124 N ) sin10° j
= − (122.116 N ) i − ( 21.532 N ) j

(1.5 m ) i − ( 9 m ) j + (1.8 m ) k
T2 = λT2 = 

2
2
2
 (1.5 m ) + ( 9 m ) + (1.8 m )

 (124 N )


= ( 20 N ) i − (120 N ) j + ( 24 N ) k
∴ R = − (102.116 N ) i − (141.532 N ) j + ( 24 N ) k
MO
i
j
k
=
0
9
1 N⋅m
−102.116 −141.532 24
= ( 357.523 N ⋅ m ) i − (102.116 N ⋅ m ) j + ( 919.044 N ⋅ m ) k
or M O = ( 358 N ⋅ m ) i − (102.1 N ⋅ m ) j + ( 919 N ⋅ m ) k
PROBLEM 3.23
An 8-lb force is applied to a wrench to tighten a showerhead. Knowing
that the centerline of the wrench is parallel to the x axis, determine the
moment of the force about A.
SOLUTION
Have
M A = rC/ A × F
where
rC/ A = ( 8.5 in.) i − ( 2.0 in.) j + ( 5.5 in.) k
Fx = − ( 8cos 45° sin12° ) lb
Fy = − ( 8sin 45° ) lb
Fz = − ( 8cos 45° cos12° ) lb
∴ F = − (1.17613 lb ) i − ( 5.6569 lb ) j − ( 5.5332 lb ) k
and
i
j
k
MA =
8.5
−2.0
5.5 lb ⋅ in.
−1.17613 −5.6569 −5.5332
= ( 42.179 lb ⋅ in.) i + ( 40.563 lb ⋅ in.) j − ( 50.436 lb ⋅ in.) k
or M A = ( 42.2 lb ⋅ in.) i + ( 40.6 lb ⋅ in.) j − ( 50.4 lb ⋅ in.) k
PROBLEM 3.24
A wooden board AB, which is used as a temporary prop to support a
small roof, exerts at point A of the roof a 228 N force directed along BA.
Determine the moment about C of that force.
SOLUTION
Have
M C = rA/C × FBA
where
rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k
and
FBA = λ BA FBA


− ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k 
228 N )
=

(
2
2
2
( 0.1) + (1.8) + ( 0.6 ) m 

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
∴ MC
i
j
k
= 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k
or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k
PROBLEM 3.25
The ramp ABCD is supported by cables at corners C and D. The tension
in each of the cables is 360 lb. Determine the moment about A of the
force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
M A = rE/ A × TDE
(a) Have
rE/ A = ( 92 in.) j
where
TDE = λ DETDE
=
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 0 92
0 lb ⋅ in. = − ( 22, 080 lb ⋅ in.) i − ( 4416 lb ⋅ in ) k
48 264 −240
or M A = − (1840 lb ⋅ ft ) i − ( 368 lb ⋅ ft ) k
M A = rG/ A × TCG
(b) Have
rG/ A = (108 in.) i + ( 92 in.) j
where
TCG = λ CGTCG =
− ( 24 in.) i + (132 in.) j − (120 in.) k
( 24 )2 + (132 )2 + (120 )2 in.
( 360 lb )
= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 108 92
0 lb ⋅ in.
−48 264 −240
= − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k
or M A = − (1840 lb ⋅ ft ) i + ( 2160 lb ⋅ ft ) j + ( 2740 lb ⋅ ft ) k
PROBLEM 3.26
The arms AB and BC of a desk lamp lie in a vertical plane that forms an
o
angle of 30 with the xy plane. To reposition the light, a force of
magnitude 8 N is applied at C as shown. Determine the moment of the
force about O knowing that AB = 450 mm, BC = 325 mm, and line
CD is parallel to the z axis.
SOLUTION
Have
M O = rC/O × FC
where
( rC/O ) x = ( ABxz + BCxz ) cos 30°
ABxz = ( 0.450 m ) sin 45° = 0.31820 m
BC xz = ( 0.325 m ) sin 50° = 0.24896 m
( rC/O ) y = (OAy + ABy − BC y ) = 0.150 m + ( 0.450 m ) cos 45°
− ( 0.325 m ) cos 50° = 0.25929 m
( rC/O ) z = ( ABxz + BCxz ) sin 30°
= ( 0.31820 m + 0.24896 m ) sin 30° = 0.28358 m
rC/O = ( 0.49118 m ) i + ( 0.25929 m ) j + ( 0.28358 m ) k
or
or
( FC ) x
= − ( 8 N ) cos 45° sin 20° = −1.93476 N
( FC ) y
= − ( 8 N ) sin 45° = −5.6569 N
( FC ) z
= ( 8 N ) cos 45° cos 20° = 5.3157 N
FC = − (1.93476 N ) i − ( 5.6569 N ) j + ( 5.3157 N ) k
∴ MO
i
j
k
= 0.49118 0.25929 0.28358 N ⋅ m
−1.93476 −5.6569 5.3157
= ( 2.9825 N ⋅ m ) i − ( 3.1596 N ⋅ m ) j − ( 2.2769 N ⋅ m ) k
or M O = ( 2.98 N ⋅ m ) i − ( 3.16 N ⋅ m ) j − ( 2.28 N ⋅ m ) k
PROBLEM 3.27
In Problem 3.21, determine the perpendicular distance from point O to
cable AB.
Problem 3.21: Before the trunk of a large tree is felled, cables AB and
BC are attached as shown. Knowing that the tension in cables AB and BC
are 777 N and 990 N, respectively, determine the moment about O of the
resultant force exerted on the tree by the cables at B.
SOLUTION
Have
| M O | = TBAd
where
d = perpendicular distance from O to line AB.
Now
Μ O = rB/O × TBA
and
rB/O = ( 8.4 m ) j
TBA = λ BATAB =
− ( 0.9 m ) i − ( 8.4 m ) j + ( 7.2 m ) k
( 0.9 ) + (8.4 ) + ( 7.2 ) m
2
2
2
( 777 N )
= − ( 63.0 N ) i − ( 588 N ) j + ( 504 N ) k
∴ MO =
and
i
j
k
0
8.4
0 N ⋅ m = ( 4233.6 N ⋅ m ) i + ( 529.2 N ⋅ m ) k
−63.0 −588 504
| MO | =
( 4233.6 )2 + ( 529.2 )2
= 4266.5 N ⋅ m
∴ 4266.5 N ⋅ m = ( 777 N ) d
or
d = 5.4911 m
or d = 5.49 m
PROBLEM 3.28
In Problem 3.21, determine the perpendicular distance from point O to
cable BC.
Problem 3.21: Before the trunk of a large tree is felled, cables AB and
BC are attached as shown. Knowing that the tension in cables AB and BC
are 777 N and 990 N, respectively, determine the moment about O of the
resultant force exerted on the tree by the cables at B.
SOLUTION
Have
| M O | = TBC d
where
d = perpendicular distance from O to line BC.
M O = rB/O × TBC
rB/O = 8.4 m j
TBC = λ BCTBC =
( 5.1 m ) i − (8.4 m ) j + (1.2 m ) k 990 N
(
)
( 5.1)2 + (8.4 )2 + (1.2 )2 m
= ( 510 N ) i − ( 840 N ) j + (120 N ) k
∴ MO
and
i
j
k
= 0
8.4
0 = (1008 N ⋅ m ) i − ( 4284 N ⋅ m ) k
510 −840 120
| MO | =
(1008)2 + ( 4284 )2
= 4401.0 N ⋅ m
∴ 4401.0 N ⋅ m = ( 990 N ) d
d = 4.4454 m
or d = 4.45 m
PROBLEM 3.29
In Problem 3.24, determine the perpendicular distance from point D to a
line drawn through points A and B.
Problem 3.24: A wooden board AB, which is used as a temporary prop to
support a small roof, exerts at point A of the roof a 228 N force directed
along BA. Determine the moment about C of that force.
SOLUTION
Have
| M D | = FBAd
where
d = perpendicular distance from D to line AB.
M D = rA/D × FBA
rA/D = − ( 0.12 m ) j + ( 0.72 m ) k
FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2 m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
∴ MD
i
j
k
= 0
−0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k
and
|MD | =
(146.88)2 + (8.64 )2 + (1.44 )2
= 147.141 N ⋅ m
∴ 147.141 N ⋅ m = ( 228 N ) d
d = 0.64536 m
or d = 0.645 m
PROBLEM 3.30
In Problem 3.24, determine the perpendicular distance from point C to a
line drawn through points A and B.
Problem 3.24: A wooden board AB, which is used as a temporary prop to
support a small roof, exerts at point A of the roof a 228 N force directed
along BA. Determine the moment about C of that force.
SOLUTION
Have
| M C | = FBAd
where
d = perpendicular distance from C to line AB.
M C = rA/C × FBA
rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k
FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2
m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
i
j
k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k
and
| MC | =
(146.88)2 + ( 60.48)2 + ( 205.92 )2
= 260.07 N ⋅ m
∴ 260.07 N ⋅ m = ( 228 N ) d
d = 1.14064 m
or d = 1.141 m
PROBLEM 3.31
In Problem 3.25, determine the perpendicular distance from point A to
portion DE of cable DEF.
Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 lb. Determine the moment
about A of the force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
Have
M A = TDE d
where
d = perpendicular distance from A to line DE.
M A = rE/ A × TDE
rE/ A = ( 92 in.) j
TDE = λ DETDE =
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 0 92
0 N⋅m
48 264 −240
= − ( 22, 080 lb ⋅ in.) i − ( 4416 lb ⋅ in.) k
PROBLEM 3.31 CONTINUED
and
MA =
( 22, 080 )2 + ( 4416 )2
= 22,517 lb ⋅ in.
∴ 22,517 lb ⋅ in. = ( 360 lb ) d
d = 62.548 in.
or d = 5.21 ft W
PROBLEM 3.32
In Problem 3.25, determine the perpendicular distance from point A to a
line drawn through points C and G.
Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 lb. Determine the moment
about A of the force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
Have
M A = TCG d
where
d = perpendicular distance from A to line CG.
M A = rG/ A × TCG
rG/ A = (108 in.) i + ( 92 in.) j
TCG = λ CGTCG
=
− ( 24 in.) i + (132 in.) j − (120 in.) k
( 24 )
2
+ (132 ) + (120 ) in.
2
2
( 360 lb )
= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 108 92
0 lb ⋅ in.
− 48 264 −240
= − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k
and
MA =
( 22, 080 )2 + ( 25,920 )2 + ( 32,928)2
= 47,367 lb ⋅ in.
∴ 47,367 lb ⋅ in. = ( 360 lb ) d
d = 131.575 in.
or d = 10.96 ft W
PROBLEM 3.33
In Problem 3.25, determine the perpendicular distance from point B to a
line drawn through points D and E.
Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 lb. Determine the moment
about A of the force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
M B = TDE d
Have
where
d = perpendicular distance from B to line DE.
M B = rE/B × TDE
rE/B = − (108 in.) i + ( 92 in.) j
TDE = λ DETDE =
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
2
2
2
24
132
120
in.
+
+
( ) ( ) ( )
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M B = −108 92
0 lb ⋅ in.
48 264 −240
= − ( 22, 080 lb ⋅ in.) i − ( 25,920 lb ⋅ in.) j − ( 32,928 lb ⋅ in.) k
and
MB =
( 22, 080 )2 + ( 25,920 )2 + ( 32,928 )2
= 47,367 lb ⋅ in.
∴ 47,367 lb ⋅ in. = ( 360 lb ) d
d = 131.575 in.
or d = 10.96 ft W
PROBLEM 3.34
Determine the value of a which minimizes the perpendicular distance
from point C to a section of pipeline that passes through points A and B.
SOLUTION
Assuming a force F acts along AB,
M C = rA/C × F = F ( d )
where
d = perpendicular distance from C to line AB
F = λ AB F =
(8 m ) i + ( 7 m ) j − ( 9 m ) k F
( 8 )2 + ( 7 )2 + ( 9 )2 m
= F ( 0.57437 ) i + ( 0.50257 ) j − ( 0.64616 ) k
rA/C = (1 m ) i − ( 2.8 m ) j − ( a − 3 m ) k
i
j
k
∴ MC =
1
−2.8
3−a F
0.57437 0.50257 −0.64616
= ( 0.30154 + 0.50257a ) i + ( 2.3693 − 0.57437a ) j
+ 2.1108k ] F
Since
MC =
rA/C × F 2
or
rA/C × F 2 = ( dF )
2
∴ ( 0.30154 + 0.50257a ) + ( 2.3693 − 0.57437a ) + ( 2.1108 ) = d 2
2
Setting
2
2
( )
d
d 2 = 0 to find a to minimize d
da
2 ( 0.50257 )( 0.30154 + 0.50257a )
+ 2 ( −0.57437 )( 2.3693 − 0.57437a ) = 0
Solving
a = 2.0761 m
or a = 2.08 m W
PROBLEM 3.35
Given the vectors P = 7i − 2j + 5k, Q = −3i − 4j + 6k, and S = 8i +
j − 9k, compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S.
SOLUTION
P ⋅ Q = ( 7i − 2 j + 5k ) ⋅ ( −3i − 4 j + 6k )
= ( 7 )( −3) + ( −2 )( −4 ) + ( 5 )( 6 )
= 17
or P ⋅ Q = 17 W
P ⋅ S = ( 7i − 2 j + 5k ) ⋅ ( 8i + j − 9k )
= ( 7 )( 8 ) + ( −2 )(1) + ( 5 )( −9 )
=9
or P ⋅ S = 9 W
Q ⋅ S = ( −3i − 4 j + 6k ) ⋅ ( 8i + j − 9k )
= ( −3)( 8 ) + ( −4 )(1) + ( 6 )( −9 )
= −82
or Q ⋅ S = −82 W
PROBLEM 3.36
Form the scalar products B ⋅ C and B′ ⋅ C, where B = B′, and use the
results obtained to prove the identity
cos α cos β =
1
2
cos (α + β ) + 12 cos (α − β ) .
SOLUTION
By definition
B ⋅ C = BC cos (α − β )
where
B = B ( cos β ) i + ( sin β ) j
C = C ( cos α ) i + ( sin α ) j
∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β )
cos β cos α + sin β sin α = cos (α − β )
or
(1)
By definition
B′ ⋅ C = BC cos (α + β )
where
B′ = ( cos β ) i − ( sin β ) j
∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β )
or
cos β cos α − sin β sin α = cos (α + β )
(2)
Adding Equations (1) and (2),
2 cos β cos α = cos (α − β ) + cos (α + β )
or cos α cos β =
1
1
cos (α + β ) + cos (α − β ) W
2
2
PROBLEM 3.37
Consider the volleyball net shown. Determine the angle formed by guy
wires AB and AC.
SOLUTION
First note
AB = rB/ A =
( −1.95 m )2 + ( −2.4 m )2 + ( 0.6 m )2
= 3.15 m
AC = rC/ A =
( 0 m )2 + ( −2.4 m )2 + (1.8 m )2
= 3.0 m
and
rB/ A = − (1.95 m ) i − ( 2.40 m ) j + ( 0.6 m ) k
rC/ A = − ( 2.40 m ) j + (1.80 m ) k
By definition
rB/ A ⋅ rC/ A = rB/ A rC/ A cosθ
or
( −1.95i − 2.40 j + 0.6k ) ⋅ ( −2.40 j + 1.80k ) = ( 3.15)( 3.0 ) cosθ
( −1.95)( 0 ) + ( −2.40 )( −2.40 ) + ( 0.6 )(1.8) = 9.45cosθ
∴ cosθ = 0.72381
and
θ = 43.630°
or θ = 43.6° W
PROBLEM 3.38
Consider the volleyball net shown. Determine the angle formed by guy
wires AC and AD.
SOLUTION
First note
AC = rC/ A =
AD = rD/ A =
and
( −2.4 )2 + (1.8)2
m = 3m
(1.2 )2 + ( −2.4 )2 + ( 0.3)2
m = 2.7 m
rC/ A = − ( 2.4 m ) j + (1.8 m ) k
rD/ A = (1.2 m ) i − ( 2.4 m ) j + ( 0.3 m ) k
By definition
rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ
or
( −2.4 j + 1.8k ) ⋅ (1.2i − 2.4 j + 0.3k ) = ( 3)( 2.7 ) cosθ
( 0 )(1.2 ) + ( −2.4 )( −2.4 ) + (1.8)( 0.3) = 8.1cosθ
and
cosθ =
6.3
= 0.77778
8.1
θ = 38.942°
or θ = 38.9° W
PROBLEM 3.39
Steel framing members AB, BC, and CD are joined at B and C and are
braced using cables EF and EG. Knowing that E is at the midpoint of BC
and that the tension in cable EF is 330 N, determine (a) the angle
between EF and member BC, (b) the projection on BC of the force
exerted by cable EF at point E.
SOLUTION
λ BC ⋅ λ EF = (1)(1) cosθ
(a) By definition
where
λ BC =
λ EF =
∴
(16 m ) i − ( 4.5 m ) j − (12 m ) k
(16 )2 + ( 4.5)2 + (12 )2 m
− (7 m) i − (6 m) j + (6 m)k
( 7 ) 2 + ( 6 )2 + ( 6 ) 2 m
=
=
1
( −7i − 6 j + 6k )
11.0
(16i − 4.5j − 12k ) ⋅ ( −7i − 6 j + 6k )
20.5
1
(16i − 4.5j − 12k )
20.5
11.0
= cosθ
(16 )( −7 ) + ( −4.5)( −6 ) + ( −12 )( 6 ) = ( 20.5)(11.0 ) cosθ
and
 −157 
θ = cos −1 
 = 134.125°
 225.5 
or θ = 134.1° W
(b) By definition
(TEF )BC
= TEF cosθ
= ( 330 N ) cos134.125°
= −229.26 N
or (TEF ) BC = −230 N W
PROBLEM 3.40
Steel framing members AB, BC, and CD are joined at B and C and are
braced using cables EF and EG. Knowing that E is at the midpoint of BC
and that the tension in cable EG is 445 N, determine (a) the angle
between EG and member BC, (b) the projection on BC of the force
exerted by cable EG at point E.
SOLUTION
λ BC ⋅ λ EG = (1)(1) cosθ
(a) By definition
where
λ BC =
(16 m ) i − ( 4.5 m ) j − (12 m ) k
(16 m )2 + ( 4.5)2 + (12 )2 m
=
16i − 4.5j − 12k
20.5
= 0.78049i − 0.21951j − 0.58537k
λ EG =
(8 m ) i − ( 6 m ) j + ( 4.875 m ) k
(8)2 + ( 6 )2 + ( 4.875)2 m
=
8i − 6 j + 4.875k
11.125
= 0.71910i − 0.53933j + 0.43820k
∴ λ BC ⋅ λ EG =
and
16 ( 8 ) + ( −4.5 )( −6 ) + ( −12 )( 4.875 )
= cosθ
( 20.5)(11.25)
 96.5 
θ = cos −1 
 = 64.967°
 228.06 
or θ = 65.0° W
(b) By definition
(TEG )BC
= TEG cosθ
= ( 445 N ) cos 64.967°
= 188.295 N
or (TEG ) BC = 188.3 N W
PROBLEM 3.41
Slider P can move along rod OA. An elastic cord PC is attached to the
slider and to the vertical member BC. Knowing that the distance from O
to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle
between the elastic cord and the rod OA, (b) the projection on OA of the
force exerted by cord PC at point P.
SOLUTION
λ OA ⋅ λ PC = (1)(1) cosθ
(a) By definition
λ OA =
where
=
( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k
( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m
2
2
1
i+ j− k
3
3
3
Knowing that | rA/O | = LOA = 0.36 m and that P is located 0.12 m from O, it follows that the coordinates
of P are
1
the coordinates of A.
3
∴ P ( 0.08 m, 0.08 m, − 0.040 m )
Then
λ PC =
( 0.10 m ) i + ( 0.22 m ) j + ( 0.28 m ) k
( 0.10 )2 + ( 0.22 )2 + ( 0.28)2 m
= 0.27037i + 0.59481j + 0.75703k
2
1 
2
∴  i + j − k  ⋅ ( 0.27037i + 0.59481j + 0.75703k ) = cosθ
3
3 
3
θ = cos −1 ( 0.32445) = 71.068°
and
or θ = 71.1° W
(b)
(TPC )OA
= TPC cosθ = ( 30 N ) cos 71.068°
(TPC )OA
= 9.7334 N
or (TPC )OA = 9.73 N W
PROBLEM 3.42
Slider P can move along rod OA. An elastic cord PC is attached to the
slider and to the vertical member BC. Determine the distance from O to P
for which cord PC and rod OA are perpendicular.
SOLUTION
The requirement that member OA and the elastic cord PC be perpendicular implies that
λ OA ⋅ λ PC = 0
where
λ OA =
=
or
λ OA ⋅ rC/P = 0
( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k
( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m
2
2
1
i+ j− k
3
3
3
Letting the coordinates of P be P ( x, y, z ) , we have
rC/P = ( 0.18 − x ) i + ( 0.30 − y ) j + ( 0.24 − z ) k  m
2
1 
2
∴  i + j − k  ⋅ ( 0.18 − x ) i + ( 0.30 − y ) j + ( 0.24 − z ) k  = 0
3
3
3 

Since
Then
rP/O = λ OAdOP =
x=
2
dOP ,
3
y =
(1)
dOP
( 2i + 2 j − k ) ,
3
2
dOP,
3
z =
−1
dOP
3
(2)
Substituting the expressions for x, y, and z from Equation (2) into Equation (1),


1
2
2
1
( 2i + 2 j − k ) ⋅  0.18 − dOP  i +  0.30 − dOP  j +  0.24 + dOP  k  = 0
3
3
3
3




 

or
3dOP = 0.36 + 0.60 − 0.24 = 0.72
∴ dOP = 0.24 m
or dOP = 240 mm W
PROBLEM 3.43
Determine the volume of the parallelepiped of Figure 3.25 when
(a) P = −(7 in.)i − (1 in.)j + (2 in.)k, Q = (3 in.)i − (2 in.)j + (4 in.)k, and
S = −(5 in.)i + (6 in.)j − (1 in.)k, (b) P = (1 in.)i + (2 in.)j − (1 in.)k,
Q = −(8 in.)i − (1 in.)j + (9 in.)k, and S = (2 in.)i + (3 in.)j + (1 in.)k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(a)
Vol = P ⋅ ( Q × S )
−7 −1 2
= 3 −2 4 in 3 = ( −14 + 168 + 20 − 3 + 36 − 20 ) in 3
−5 6 −1
= 187 in 3
or Volume = 187 in 3 W
(b)
Vol = P ⋅ ( Q × S )
1 2 −1
= −8 −1 9 in 3 = ( −1 − 27 + 36 + 16 + 24 − 2 ) in 3
2 3 1
= 46 in 3
or Volume = 46 in 3 W
PROBLEM 3.44
Given the vectors P = 4i − 2j + Pzk, Q = i + 3j − 5k, and S = −6i +
2j − k, determine the value of Pz for which the three vectors are coplanar.
SOLUTION
For the vectors to all be in the same plane, the mixed triple product is zero.
P ⋅ (Q × S ) = 0
4 −2 Pz
∴ O = 1 3 −5 = −12 + 40 − 60 − 2 + Pz ( 2 + 18 )
−6 2 −1
so that
Pz =
34
= 1.70
20
or Pz = 1.700 W
PROBLEM 3.45
The 0.732 × 1.2-m lid ABCD of a storage bin is hinged along side AB
and is held open by looping cord DEC over a frictionless hook at E. If the
tension in the cord is 54 N, determine the moment about each of the
coordinate axes of the force exerted by the cord at D.
SOLUTION
First note
z =
( 0.732 )2 − ( 0.132 )2
m
= 0.720 m
Then
d DE =
( 0.360 )2 + ( 0.720 )2 + ( 0.720 )2
m
= 1.08 m
and
Have
rE/D = ( 0.360 m ) i + ( 0.720 m ) j − ( 0.720 m ) k
TDE =
=
(
TOE
rE/D
d DE
)
54 N
( 0.360i + 0.720 j − 0.720k )
1.08
= (18.0 N ) i + ( 36.0 N ) j − ( 36.0 N ) k
Now
M A = rD/ A × TDE
where
rD/ A = ( 0.132 m ) j + ( 0.720 m ) k
Then
i
j
k
M A = 0 0.132 0.720 N ⋅ m
18.0 36.0 −36.0
PROBLEM 3.45 CONTINUED
{
∴ M A = ( 0.132 )( −36.0 ) − ( 0.720 )( 36.0 )  i + ( 0.720 )(18.0 ) − 0  j
}
+ 0 − ( 0.132 )(18.0 )  k N ⋅ m
or
M A = − ( 30.7 N ⋅ m ) i + (12.96 N ⋅ m ) j − ( 2.38 N ⋅ m ) k
∴ M x = −30.7 N ⋅ m, M y = 12.96 N ⋅ m, M z = −2.38 N ⋅ m W
PROBLEM 3.46
The 0.732 × 1.2-m -m lid ABCD of a storage bin is hinged along side AB
and is held open by looping cord DEC over a frictionless hook at E. If the
tension in the cord is 54 N, determine the moment about each of the
coordinate axes of the force exerted by the cord at C.
SOLUTION
z =
First note
( 0.732 )2 − ( 0.132 )2
m
= 0.720 m
Then
dCE =
( 0.840 )2 + ( 0.720 )2 + ( 0.720 )2
m
= 1.32 m
and
TCE =
=
rE/C
dCE
(TCE )
− ( 0.840 m ) i + ( 0.720 m ) j − ( 0.720 m ) k
( 54 N )
1.32 m
= − ( 36.363 N ) i + ( 29.454 N ) j − ( 29.454 N ) k
Now
M A = rE/ A × TCE
where
rE/ A = ( 0.360 m ) i + ( 0.852 m ) j
Then
i
j
k
M A = 0.360 0.852
0
N⋅m
−34.363 29.454 −29.454
= − ( 25.095 N ⋅ m ) i + (10.6034 N ⋅ m ) j + ( 39.881 N ⋅ m ) k
∴ M x = −25.1 N ⋅ m, M y = 10.60 N ⋅ m, M z = 39.9 N ⋅ m W
PROBLEM 3.47
A fence consists of wooden posts and a steel cable fastened to each post
and anchored in the ground at A and D. Knowing that the sum of the
moments about the z axis of the forces exerted by the cable on the posts
at B and C is −66 N · m, determine the magnitude TCD when TBA = 56 N.
SOLUTION
Based on
| M z | = k ⋅ ( rB ) y × TBA  + k ⋅ ( rC ) y × TCD 




where
M z = − ( 66 N ⋅ m ) k
( rB ) y
= ( rC ) y = (1 m ) j
TBA = λ BATBA
=
(1.5 m ) i − (1 m ) j + ( 3 m ) k
3.5 m
( 56 N )
= ( 24 N ) i − (16 N ) j + ( 48 N ) k
TCD = λ CDTCD
=
( 2 m ) i − (1 m ) j − ( 2 m ) k T
=
1
TCD ( 2i − j − 2k )
3
3.0 m
CD
{
}
∴ − ( 66 N ⋅ m ) = k ⋅ (1 m ) j × ( 24 N ) i − (16 N ) j + ( 48 N ) k 

1

+ k ⋅ (1 m ) j ×  TCD ( 2i − j − 2k )  
3


or
−66 = −24 −
∴ TCD =
2
TCD
3
3
( 66 − 24 ) N
2
or TCD = 63.0 N W
PROBLEM 3.48
A fence consists of wooden posts and a steel cable fastened to each post
and anchored in the ground at A and D. Knowing that the sum of the
moments about the y axis of the forces exerted by the cable on the posts
at B and C is 212 N · m, determine the magnitude of TBA when
TCD = 33 N.
SOLUTION
Based on
| M y | = j ⋅ ( rB ) z × TBA + ( rC ) z × TCD 
where
M y = ( 212 N ⋅ m ) j
( rB ) z
= (8 m ) k
( rC ) z
= (2 m) k
TBA = λ BATBA
=
(1.5 m ) i − (1 m ) j − ( 3 m ) k T
=
TBA
(1.5i − j + 3k )
3.5
BA
3.5 m
TCD = λ CDTCD
=
( 2 m ) i − (1 m ) j − ( 2 m ) k
3.0 m
( 33 N )
= ( 22i − 11j − 22k ) N

T

∴ ( 212 N ⋅ m ) = j ⋅ ( 8 m ) k ×  BA (1.5i − j + 3k )  
 3.5


+ j ⋅ ( 2 m ) k × ( 22 i − 11j − 22k ) N 
or
212 =
8 (1.5 )
TBA + 2 ( 22 )
3.5
∴ TBA =
168
18.6667
or TBA = 49.0 N W
PROBLEM 3.49
To lift a heavy crate, a man uses a block and tackle attached to the bottom
of an I-beam at hook B. Knowing that the moments about the y and z axes
of the force exerted at B by portion AB of the rope are, respectively,
100 lb ⋅ ft and −400 lb ⋅ ft , determine the distance a.
SOLUTION
M O = rA/O × TBA
Based on
where
M O = M xi + M y j + M zk
= M xi + (100 lb ⋅ ft ) j − ( 400 lb ⋅ ft ) k
rA/O = ( 6 ft ) i + ( 4 ft ) j
TBA = λ BATBA
=
( 6 ft ) i − (12 ft ) j − ( a ) k T
BA
d BA
i j k
T
∴ M xi + 100 j − 400k = 6 4 0 BA
d
6 −12 −a BA
=
TBA
 − ( 4a ) i + ( 6a ) j − ( 96 ) k 
d BA 
100
d BA
6a
From j-coefficient:
100d AB = 6aTBA or TBA =
From k -coefficient:
−400d AB = −96TBA or TBA =
Equating Equations (1) and (2) yields
a=
400
d BA
96
(1)
(2)
100 ( 96 )
6 ( 400 )
or a = 4.00 ft W
PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the bottom
of an I-beam at hook B. Knowing that the man applies a 200-lb force to
end A of the rope and that the moment of that force about the y axis is
175 lb ⋅ ft , determine the distance a.
SOLUTION
(
| M y | = j ⋅ rA/O × TBA
Based on
)
rA/O = ( 6 ft ) i + ( 4 ft ) j
where
TBA = λ BATBA =
=
=
rA/B
d BA
TBA
( 6 ft ) i − (12 ft ) j − ( a ) k
d BA
( 200 lb )
200
( 6i − 12 j − ak )
d BA
0
1 0
200
∴ 175 lb ⋅ ft = 6
4 0
d
6 −12 −a BA
200
175 = 0 − 6 ( −a ) 
d
BA
where
d BA =
( 6 )2 + (12 )2 + ( a )2
ft
= 180 + a 2 ft
∴ 175 180 + a 2 = 1200a
or
180 + a 2 = 6.8571a
Squaring each side
180 + a 2 = 47.020a 2
Solving
a = 1.97771 ft
or a = 1.978 ft W
PROBLEM 3.51
A force P is applied to the lever of an arbor press. Knowing that P lies in
a plane parallel to the yz plane and that M x = 230 lb ⋅ in.,
M y = −200 lb ⋅ in., and M z = −35 lb ⋅ in., determine the magnitude of P
and the values of φ and θ.
SOLUTION
Based on
M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 
M y = − ( P cos φ )( 5 in.)
(2)
M z = − ( P sin φ )( 5 in.)
(3)
− ( P sin φ ) (5)
Equation (3) M z
:
=
Equation (2) M y
− ( P cos φ ) (5)
Then
or
(1)
tan φ =
−35
= 0.175
−200
φ = 9.9262°
or φ = 9.93° W
Substituting φ into Equation (2)
−200 lb ⋅ in. = − ( P cos 9.9262° ) (5 in.)
P = 40.608 lb
or P = 40.6 lb W
Then, from Equation (1)
230 lb ⋅ in. = ( 40.608 lb ) cos 9.9262° ( 9 in.) sin θ 
− ( 40.608 lb ) sin 9.9262°  ( 9 in.) cosθ 
or
0.98503sin θ − 0.172380cosθ = 0.62932
Solving numerically,
θ = 48.9° W
PROBLEM 3.52
A force P is applied to the lever of an arbor press. Knowing that P lies in
a plane parallel to the yz plane and that M y = −180 lb ⋅ in. and
M z = −30 lb ⋅ in., determine the moment M x of P about the x axis when
θ = 60°.
SOLUTION
Based on
M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 
(1)
M y = − ( P cos φ )( 5 in.)
(2)
M z = − ( P sin φ )( 5 in.)
(3)
− ( P sin φ )( 5 )
Equation (3) M z
=
:
− ( P cos φ )( 5 )
Equation (2) M y
Then
−30
= tan φ
−180
or
∴ φ = 9.4623°
From Equation (3),
−30 lb ⋅ in. = − ( P sin 9.4623° )( 5 in.)
∴ P = 36.497 lb
From Equation (1),
M x = ( 36.497 lb )( 9 in.)( cos 9.4623° sin 60° − sin 9.4623° cos 60° )
= 253.60 lb ⋅ in.
or M x = 254 lb ⋅ in. W
PROBLEM 3.53
The triangular plate ABC is supported by ball-and-socket joints at B and
D and is held in the position shown by cables AE and CF. If the force
exerted by cable AE at A is 220 lb, determine the moment of that force
about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rA/D × TAE
Have
λ DB =
where
( 48 in.) i − (14 in.) j
50 in.
)
= 0.96i − 0.28 j
rA/D = − ( 4 in.) j + ( 8 in.) k
( 36 in.) i − ( 24 in.) j + ( 8 in.) k 
TAE = λ AETAE = 
( 220 lb )
44 in.
= (180 lb ) i − (120 lb ) j + ( 40 lb ) k
∴ M DB
0.960 −0.280 0
8 lb ⋅ in.
= 0
−4
180 −120 40
= ( 0.960 ) ( −4 )( 40 ) − ( 8 )( −120 )  + ( −0.280 ) 8 (180 ) − 0 
= 364.8 lb ⋅ in.
or M DB = 365 lb ⋅ in. W
PROBLEM 3.54
The triangular plate ABC is supported by ball-and-socket joints at B and
D and is held in the position shown by cables AE and CF. If the force
exerted by cable CF at C is 132 lb, determine the moment of that force
about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rC/D × TCF
Have
where
λ DB =
( 48 in.) i − (14 in.) j
50 in.
)
= 0.96i − 0.28 j
rC/D = ( 8 in.) j − (16 in.) k
TCF = λ CFTCF =
( 24 in.) i − ( 36 in.) j − (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k
∴ M DB
0.96 −0.28 0
8
= 0
−16 lb ⋅ in.
72 −108 −24
= 0.96 ( 8 )( −24 ) − ( −16 )( −108 )  + ( −0.28 ) ( −16 )( 72 ) − 0 
= −1520.64 lb ⋅ in.
or M DB = −1521 lb ⋅ in. W
PROBLEM 3.55
A mast is mounted on the roof of a house using bracket ABCD and is
guyed by cables EF, EG, and EH. Knowing that the force exerted by
cable EF at E is 66 N, determine the moment of that force about the line
joining points D and I.
SOLUTION
M DI = λ DI ⋅ rF /I × TEF 
Have
where
λ DI =
(1.6 m ) i − ( 0.4 m ) j
(1.6 )2 + ( 0.4 )2 m
=
1
( 4i − j)
17
rF /I = ( 4.6 m + 0.8 m ) k = ( 5.4 m ) k
TEF = λ EF TEF
=
(1.2 m ) i − ( 3.6 m ) j + ( 5.4 m ) k
6.6 m
( 66 N )
= (12 N ) i − ( 36 N ) j + ( 54 N ) k
= 6 ( 2 N ) i − ( 6 N ) j + ( 9 N ) k 
−1 0
0 1
2 −6 9
4
∴ M DI =
( 6 N )( 5.4 m ) 0
17
= 7.8582 ( 0 + 24 ) + ( −2 − 0 ) 
= 172.879 N ⋅ m
or M DI = 172.9 N ⋅ m W
PROBLEM 3.56
A mast is mounted on the roof of a house using bracket ABCD and is
guyed by cables EF, EG, and EH. Knowing that the force exerted by
cable EG at E is 61.5 N, determine the moment of that force about the
line joining points D and I.
SOLUTION
Have
M DI = λ DI ⋅ rG/I × TEG 
where
λ DI =
=
(1.6 m ) i − ( 0.4 m ) j
0.4 17 m
1
( 4i − j)
17
rG/I = − (10.9 m + 0.8 m ) k = − (11.7 m ) k
TEG = λ EGTEG
=
(1.2 m ) i − ( 3.6 m ) j − (11.7 m ) k
12.3 m
( 61.5 N )
= 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k 
∴ M DI =
5 N (11.7 m )
17
4 −1
0
−1
0
0
1.2 −3.6 −11.7
{
}
= (14.1883 N ⋅ m )  0 − ( 4 )( −1)( −3.6 )  + ( −1)( −1)(1.2 ) − 0
= −187.286 N ⋅ m
or M DI = −187.3 N ⋅ m
PROBLEM 3.57
A rectangular tetrahedron has six edges of length a. A force P is directed
as shown along edge BC. Determine the moment of P about edge OA.
SOLUTION
(
M OA = λ OA ⋅ rC/O × P
Have
)
where
From triangle OBC
a
2
( OA) x
=
( OA) z
= ( OA ) x tan 30° =
( OA)2
Since
a 1 
a
=


2 3  2 3
= ( OA) x + ( OA) y + ( OAz )
2
2
2
 a 
2
a
a 2 =   + ( OA) y + 

2
 
2 3
or
∴
( OA) y
=
a2 −
2
2
a2 a2
2
−
=a
4
12
3
Then
rA/O =
a
2
a
i +a j+
k
2
3
2 3
and
λ OA =
1
i+
2
2
1
j+
k
3
2 3
P = λ BC P
=
=
( a sin 30° ) i − ( a cos30° ) k
a
(
P
i − 3k
2
)
rC/O = ai
( P)
PROBLEM 3.57 CONTINUED
∴ M OA
1
2
=
1
1
2
1
3 2 3
P
( a )  
0
0
2
0 − 3
=
aP  2 
−
 (1) − 3
2  3 
=
aP
2
(
)
M OA =
aP
2
PROBLEM 3.58
A rectangular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each other. (b)
Use this property and the result obtained in Problem 3.57 to determine the
perpendicular distance between edges OA and BC.
SOLUTION
(a) For edge OA to be perpendicular to edge BC,
JJJG JJJG
OA ⋅ BC = 0
where
From triangle OBC
and
Then
or
so that
a
2
( OA) x
=
( OA) z
= ( OA ) x tan 30° =
a 1 
a
=
2  3  2 3
JJJG  a 
 a 
∴ OA =   i + ( OA)y j + 
k
2
2 3
JJJG
BC = ( a sin 30° ) i − ( a cos 30° ) k
=
a
a 3
i−
k
2
2
=
a
i − 3k
2
(
)
a
 a  
a
=0
 i + ( OA ) y j + 
 k  ⋅ i − 3k
2
2
2
3

 

(
)
a2
a2
+ ( OA)y ( 0 ) −
=0
4
4
JJJG JJJG
∴ OA ⋅ BC = 0
JJJG
JJJG
OA is perpendicular to BC.
PROBLEM 3.58 CONTINUED
(b) Have M OA = Pd , with P acting along BC and d the
JJJG JJJG
perpendicular distance from OA to BC.
From the results of Problem 3.57,
M OA =
∴
or
Pa
2
Pa
= Pd
2
d =
a
2
PROBLEM 3.59
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is
partially supported by members EF and GH. Knowing that the
compressive force exerted by member EF on the walkway at F is 5400 lb,
determine the moment of that force about edge AD.
SOLUTION
(
M AD = λ AD ⋅ rE/ A × TEF
Having
λ AD =
where
( 24 ft ) i + ( 3 ft ) j
( 24 )2 + ( 3)2 ft
)
1
(8i + j)
65
=
rE/ A = ( 7 ft ) i − ( 3 ft ) j
TEF = λ EFTEF =
(8 ft − 7 ft ) i + 3 ft + ( 248 ) ( 3 ft ) j + (8 ft ) k
(1)2 + ( 4 )2 + (8)2
( 5400 lb )
ft
= 600 (1 lb ) i + ( 4 lb ) j + ( 8 lb ) k 
∴ M AD =
8 1 0
600
600
7 −3 0 lb ⋅ ft =
( −192 − 56 ) lb ⋅ ft
65
65
1 4 8
= −18,456.4 lb ⋅ ft
or M AD = −18.46 kip ⋅ ft
PROBLEM 3.60
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is
partially supported by members EF and GH. Knowing that the
compressive force exerted by member GH on the walkway at H is
4800 lb, determine the moment of that force about edge AD.
SOLUTION
(
M AD = λ AD ⋅ rG/ A × TGH
Having
λ AD =
where
( 24 ft ) i + ( 3 ft ) j
( 24 )2 + ( 3)2 ft
=
)
1
(8i + j)
65
rG/ A = ( 20 ft ) i − ( 6 ft ) j = 2 (10 ft ) i − ( 3 ft ) j
TGH = λ GH TGH =
(16 ft − 20 ft ) i + 6 ft + ( 1624 ) ( 3 ft ) j + (8 ft ) k
( 4 )2 + ( 8 )2 + ( 8 )2
( 4800 lb )
ft
= 1600  − (1 lb ) i + ( 2 lb ) j + ( 2 lb ) k 
8
∴ M AD =
1 0
3200 lb ⋅ ft
−3 0 =
( −48 − 20 )
65
−1 2 2
(1600 lb )( 2 ft ) 10
65
= −26,989 lb ⋅ ft
or M AD = −27.0 kip ⋅ ft
PROBLEM 3.61
Two forces F1 and F2 in space have the same magnitude F. Prove that the
moment of F1 about the line of action of F2 is equal to the moment of F2
about the line of action of F1.
SOLUTION
F1 = F1λ1
First note that
F2 = F2λ 2
and
Let M1 = moment of F2 about the line of action of M1
and M 2 = moment of F1 about the line of action of M 2
Now, by definition
(
)
(
)
(
)
(
)
M1 = λ1 ⋅ rB/ A × F2 = λ1 ⋅ rB/ A × λ 2 F2
M 2 = λ 2 ⋅ rA/B × F1 = λ 2 ⋅ rA/B × λ1 F1
F1 = F2 = F
Since
and
(
rA/B = −rB/ A
)
M1 = λ1 ⋅ rB/ A × λ 2 F
(
)
M 2 = λ 2 ⋅ −rB/ A × λ1 F
Using Equation (3.39)
(
)
(
λ1 ⋅ rB/ A × λ 2 = λ 2 ⋅ −rB/ A × λ1
so that
(
)
)
M 2 = λ1 ⋅ rB/ A × λ 2 F
∴ M12 = M 21
PROBLEM 3.62
In Problem 3.53, determine the perpendicular distance between cable AE
and the line joining points D and B.
Problem 3.53: The triangular plate ABC is supported by ball-and-socket
joints at B and D and is held in the position shown by cables AE and CF.
If the force exerted by cable AE at A is 220 lb, determine the moment of
that force about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rA/D × TAE
Have
where
λ DB =
( 48 in.) i − (14 in.) j
50 in.
)
= 0.96i − 0.28 j
rA/D = − ( 4 in.) j + ( 8 in.) k
TAE = λ AETAE
=
( 36 in.) i − ( 24 in.) j + (8 in.) k
44 in.
( 220 lb )
= (180 lb ) i − (120 lb ) j + ( 40 lb ) k
∴ M DB
0.96 −0.28 0
= 0
−4
8 lb ⋅ in.
180 −120 40
= 364.8 lb ⋅ in.
Only the perpendicular component of TAE contributes to the moment of
TAE about line DB. The parallel component of TAE will be used to find
the perpendicular component.
PROBLEM 3.62 CONTINUED
Have
(TAE )parallel
= λ DB ⋅ TAE
= ( 0.96i − 0.28 j) ⋅ (180 lb ) i − (120 lb ) j + ( 40 lb ) k 
= ( 0.96 )(180 ) + ( −0.28 )( −120 ) + ( 0 )( 40 )  lb
= (172.8 + 33.6 ) lb
= 206.4 lb
Since TAE = ( TAE )perpendicular + ( TAE )parallel
∴
(TAE )perpendicular
=
(TAE )2 − (TAE )2parallel
=
( 220 )2 − ( 206.41)2
= 76.151 lb
Then
M DB = (TAE )perpendicular ( d )
364.8 lb ⋅ in. = ( 76.151 lb ) d
d = 4.7905 in.
or d = 4.79 in.
PROBLEM 3.63
In Problem 3.54, determine the perpendicular distance between cable CF
and the line joining points D and B.
Problem 3.54: The triangular plate ABC is supported by ball-and-socket
joints at B and D and is held in the position shown by cables AE and CF.
If the force exerted by cable CF at C is 132 lb, determine the moment of
that force about the line joining points D and B.
SOLUTION
( M DB ) = λ DB ⋅ ( rC/D × TCF )
Have
λ DB =
where
( 48 in.) i − (14 in.) j
50 in.
= 0.96i − 0.28j
rC/D = ( 8 in.) j − (16 in.) k
TCF = λ CF TCF
=
( 24 in.) i − ( 36 in.) j − (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k
∴ M DB
0.96 −0.28 0
= 0
−16 lb ⋅ in
8
72 −108 −24
= −1520.64 lb ⋅ in.
Only the perpendicular component of TCF contributes to the moment of
TCF about line DB. The parallel component of TCF will be used to
obtain the perpendicular component.
PROBLEM 3.63 CONTINUED
Have
(TCF )parallel
= λ DB ⋅ TCF
= ( 0.96i − 0.28 j) ⋅ ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k 
= ( 0.96 )( 72 ) + ( −0.28 )( −108 ) + ( 0 )( −24 )  lb
= 99.36 lb
Since TCF = ( TCF )perp. + ( TCF )parallel
∴
(TCF )perp.
=
2
(TCF )2 − (TCF )parallel
=
(132 )2 − ( 99.36 )2
= 86.900 lb
Then
M DB = (TCF )perp. ( d )
−1520.64 lb ⋅ in. = ( 86.900 lb ) d
d = 17.4988 in.
or d = 17.50 in.
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between cable EF
and the line joining points D and I.
Problem 3.55: A mast is mounted on the roof of a house using bracket
ABCD and is guyed by cables EF, EG, and EH. Knowing that the force
exerted by cable EF at E is 66 N, determine the moment of that force
about the line joining points D and I.
SOLUTION
(
M DI = λ DI ⋅ rF /I × TEF
Have
λ DI =
where
(1.6 m ) i − ( 0.4 m ) j
0.4 17 m
=
)
1
( 4i − j)
17
rF /I = ( 5.4 m ) k
TEF = λ EFTEF =
(1.2 m ) i − ( 3.6 m ) j + ( 5.4 m ) k
6.6 m
( 66 N )
= 6 ( 2 N ) i − ( 6 N ) j + ( 9 N ) k 
∴ M DI =
4 −1 0
0 1 = 172.879 N ⋅ m
2 −6 9
( 6 N )( 5.4 m ) 0
17
Only the perpendicular component of TEF contributes to the moment of
TEF about line DI. The parallel component of TEF will be used to find
the perpendicular component.
Have
(TEF )parallel
= λ DI ⋅ TEF
=
1
( 4i − j) ⋅ (12 N ) i − ( 36 N ) j + ( 54 N ) k 
17
=
1
( 48 + 36 ) N
17
=
84
N
17
PROBLEM 3.64 CONTINUED
Since TEF = ( TEF )perp. + ( TEF )parallel
∴
(TEF )perp.
=
(TEF )2 − (TEF )2parallel
=
( 66 )2 − 
 84 

 17 
2
= 62.777 N
Then
M DI = (TEF )perp. ( d )
172.879 N ⋅ m = ( 62.777 N )( d )
d = 2.7539 m
or d = 2.75 m
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance between cable EG
and the line joining points D and I.
Problem 3.56: A mast is mounted on the roof of a house using bracket
ABCD and is guyed by cables EF, EG, and EH. Knowing that the force
exerted by cable EG at E is 61.5 N, determine the moment of that force
about the line joining points D and I.
SOLUTION
M DI = λ DI ⋅ rG/I × TEG 
Have
λ DI =
where
(1.6 m ) i − ( 0.4 m ) j
0.4 17 m
=
1
( 4i − j)
17
rG/I = − (10.9 m + 0.8 m ) k = − (11.7 m ) k
TEG = λ EGTEG =
(1.2 m ) i − ( 3.6 m ) j − (11.7 m ) k
12.3 m
( 61.5 N )
= 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k 
∴ M DI =
( 5 N )(11.7 m )
17
4 −1
0
−1
0
0
1.2 −3.6 −11.7
= −187.286 N ⋅ m
Only the perpendicular component of TEG contributes to the moment of
TEG about line DI. The parallel component of TEG will be used to find
the perpendicular component.
Have
TEG ( parallel ) = λ DI ⋅ TEG
=
1
( 4i − j) ⋅ 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k 
17
=
5
( 4.8 + 3.6 ) N
17
=
42
N
17
PROBLEM 3.65 CONTINUED
Since TEF = ( TEG )perp. + ( TEG )parallel
∴
(TEG )perp.
=
2
(TEG )2 − (TEG )parallel
=
( 61.5)
2
 42 
−

 17 
2
= 60.651 N
Then
M DI = (TEG )perp. ( d )
187.286 N ⋅ m = ( 60.651 N )( d )
d = 3.0880 m
or d = 3.09 m
PROBLEM 3.66
In Problem 3.41, determine the perpendicular distance between post BC
and the line connecting points O and A.
Problem 3.41: Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC. Knowing that the
distance from O to P is 0.12 m and the tension in the cord is 30 N,
determine (a) the angle between the elastic cord and the rod OA, (b) the
projection on OA of the force exerted by cord PC at point P.
SOLUTION
Assume post BC is represented by a force of magnitude FBC
where
FBC = FBC j
Have
M OA = λ OA ⋅ rB/O × FBC
where
(
λ OA =
)
( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k
0.36 m
2
2
1
i+ j− k
3
3
3
=
rB/O = ( 0.18 m ) i + ( 0.24 m ) k
∴ M OA
2 2 −1
1
F
= FBC 0.18 0 0.24 = BC ( −0.48 − 0.18 ) = −0.22 FBC
3
3
0 1 0
Only the perpendicular component of FBC contributes to the moment of FBC about line OA. The parallel
component will be found first so that the perpendicular component of FBC can be determined.
2
1 
2
FBC ( parallel ) = λ OA ⋅ FBC =  i + j − k  ⋅ FBC j
3
3 
3
=
2
FBC
3
FBC = ( FBC )parallel + ( FBC )perp.
Since
( FBC )perp.
=
( FBC )2 − ( FBC )2parallel
=
( FBC )2 − 
2 FBC 

 3 
2
= 0.74536 FBC
Then
M OA = ( FBC )perp. ( d )
0.22 FBC = ( 0.74536 FBC ) d
d = 0.29516 m
or d = 295 mm
PROBLEM 3.67
In Problem 3.45, determine the perpendicular distance between cord DE
and the y axis.
Problem 3.45: The 0.732 × 1.2 -m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over a frictionless
hook at E. If the tension in the cord is 54 N, determine the moment about
each of the coordinate axes of the force exerted by the cord at D.
SOLUTION
First note
z =
( 0.732 )2 − ( 0.132 )2
m
= 0.720 m
(
)
Have
M y = j ⋅ rD/ A × TDE
where
rD/ A = ( 0.132 j + 0.720k ) m
TDE = λ DETDE
=
( 0.360 m ) i + ( 0.732 m ) j − ( 0.720 m ) k
1.08 m
( 54 N )
= (18 N ) i + ( 36 N ) j − ( 36 N ) k
0
1
0
∴ M y = 0 0.132 0.720 = 12.96 N ⋅ m
−36
18 36
Only the perpendicular component of TDE contributes to the moment of
TDE about the y-axis. The parallel component will be found first so that
the perpendicular component of TDE can be determined.
TDE ( parallel ) = j ⋅ TDE = 36 N
PROBLEM 3.67 CONTINUED
( TDE ) = ( TDE )parallel + ( TDE )perp.
Since
(TDE )perp.
Then
=
2
(TDE )2 − (TDE )parallel
=
( 54 )2 − ( 36 )2
= 40.249 N
M y = (TDE )perp. (d )
12.96 N ⋅ m = ( 40.249 N )( d )
d = 0.32199 m
or d = 322 mm
PROBLEM 3.68
A plate in the shape of a parallelogram is acted upon by two couples.
Determine (a) the moment of the couple formed by the two 21-N forces,
(b) the perpendicular distance between the 12-N forces if the resultant of
the two couples is zero, (c) the value of α if the resultant couple is
1.8 N ⋅ m clockwise and d is 1.05 m.
SOLUTION
(a) Have
where
M1 = d1F1
d1 = 0.4 m
F1 = 21 N
∴ M1 = ( 0.4 m )( 21 N ) = 8.4 N ⋅ m
or M1 = 8.40 N ⋅ m
(b) Have
or
M1 + M 2 = 0
8.40 N ⋅ m − d 2 (12 N ) = 0
∴ d 2 = 0.700 m
(c) Have
or
M total = M1 + M 2
1.8 N ⋅ m = 8.40 N ⋅ m − (1.05 m )( sin α )(12 N )
∴ sin α = 0.52381
and
α = 31.588°
or α = 31.6°
PROBLEM 3.69
A couple M of magnitude 10 lb ⋅ ft is applied to the handle of a
screwdriver to tighten a screw into a block of wood. Determine the
magnitudes of the two smallest horizontal forces that are equivalent to M
if they are applied (a) at corners A and D, (b) at corners B and C,
(c) anywhere on the block.
SOLUTION
M = Pd
(a) Have
 1 ft 
10 lb ⋅ ft = P (10 in.) 

 12 in. 
or
∴ P = 12 lb
d BC =
(b)
=
or Pmin = 12.00 lb
( BE )2 + ( EC )2
(10 in.)2 + ( 6 in.)2
= 11.6619 in.
M = Pd
Have
 1 ft 
10 lb ⋅ ft = P (11.6619 in.) 

 12 in. 
P = 10.2899 lb
d AC =
(c)
=
or P = 10.29 lb
( AD )2 + ( DC )2
(10 in.)2 + (16 in.)2
= 2 89 in.
M = Pd AC
Have
(
)
 1 ft 
10 lb ⋅ ft = P 2 89 in. 

 12 in. 
P = 6.3600 lb
or P = 6.36 lb
PROBLEM 3.70
Two 60-mm-diameter pegs are mounted on a steel plate at A and C, and
two rods are attached to the plate at B and D. A cord is passed around the
pegs and pulled as shown, while the rods exert on the plate 10-N forces as
indicated. (a) Determine the resulting couple acting on the plate when T =
36 N. (b) If only the cord is used, in what direction should it be pulled to
create the same couple with the minimum tension in the cord?
(c) Determine the value of that minimum tension.
SOLUTION
M = Σ ( Fd )
(a) Have
= ( 36 N )( 0.345 m ) − (10 N )( 0.380 m )
= 8.62 N ⋅ m
M = 8.62 N ⋅ m
(b)
M = Td = 8.62 N ⋅ m
Have
For T to be minimum, d must be maximum.
∴ Tmin must be perpendicular to line AC
tan θ =
0.380 m
= 1.33333
0.285 m
θ = 53.130°
and
or θ = 53.1°
M = Tmin d max
(c) Have
M = 8.62 N ⋅ m
where

d max = 

( 0.380 )2 + ( 0.285)2

+ 2 ( 0.030 )  m = 0.535 m

∴ 8.62 N ⋅ m = Tmin ( 0.535 m )
Tmin = 16.1121 N
or Tmin = 16.11 N
PROBLEM 3.71
The steel plate shown will support six 50-mm-diameter idler rollers
mounted on the plate as shown. Two flat belts pass around the rollers, and
rollers A and D will be adjusted so that the tension in each belt is 45 N.
Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the
value of a so that the resultant couple acting on the plate is 54 N ⋅ m
clockwise.
SOLUTION
(a) Note when a = 0.2 m, rC/F is perpendicular to the inclined 45 N
forces.
Have
M = Σ ( Fd )
= − ( 45 N )  a + 0.2 m + 2 ( 0.025 m ) 
− ( 45 N )  2a 2 + 2 ( 0.025 m ) 


For a = 0.2 m,
M = − ( 45 N )( 0.450 m + 0.61569 m )
= −47.956 N ⋅ m
or M = 48.0 N ⋅ m
M = 54.0 N ⋅ m
(b)
M = Moment of couple due to horizontal forces at A and D
+ Moment of force-couple systems at C and F about C.
−54.0 N ⋅ m = −45 N  a + 0.2 m + 2 ( 0.025 m ) 
+  M C + M F + Fx ( a + 0.2 m ) + Fy ( 2a ) 
where
M C = − ( 45 N )( 0.025 m ) = −1.125 N ⋅ m
M F = M C = −1.125 N ⋅ m
PROBLEM 3.71 CONTINUED
∴
Fx =
−45
N
2
Fy =
−45
N
2
− 54.0 N ⋅ m = −45 N ( a + 0.25 m ) − 1.125 N ⋅ m − 1.125 N ⋅ m
−45 N
45 N
( a + 0.2 m ) −
( 2a )
2
2
1.20 = a + 0.25 + 0.025 + 0.025 +
a
0.20 2a
+
+
2
2
2
3.1213a = 0.75858
a = 0.24303 m
or a = 243 mm
PROBLEM 3.72
The shafts of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
SOLUTION
M = M1 + M 2
Based on
where
M1 = − ( 8 N ⋅ m ) j
M2 = − (6 N⋅m)k
∴ M = − (8 N ⋅ m ) j − ( 6 N ⋅ m ) k
M =
and
( 8 )2 + ( 6 ) 2
= 10 N ⋅ m
or M = 10.00 N ⋅ m
λ =
or
− (8 N ⋅ m ) j − ( 6 N ⋅ m ) k
M
=
= −0.8j − 0.6k
10 N ⋅ m
M
M = M λ = (10 N ⋅ m )( −0.8j − 0.6k )
cosθ x = 0
∴ θ x = 90°
cosθ y = −0.8
∴ θ y = 143.130°
cosθ z = −0.6
∴ θ z = 126.870°
or θ x = 90.0°, θ y = 143.1°, θ z = 126.9°
PROBLEM 3.73
Knowing that P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Have
M = M1 + M 2
where
M1 = rC/B × P1C
rC/B = ( 0.96 m ) i − ( 0.40 m ) j
P1C = − (100 N ) k
i
j
k
∴ M1 = 0.96 −0.40 0 = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j
0
0
−100
M 2 = rD/ A × P2 E
Also,
rD/ A = ( 0.20 m ) j − ( 0.55 m ) k
P2 E = λ ED P2 E
=
− ( 0.48 m ) i + ( 0.55 m ) k
( 0.48)2 + ( 0.55)2
(146 N )
m
= − ( 96 N ) i + (110 N ) k
i
j
k
∴ M 2 = 0 0.20 −0.55 N ⋅ m
−96 0
110
= ( 22.0 N ⋅ m ) i + ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k
PROBLEM 3.73 CONTINUED
M = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j + ( 22.0 N ⋅ m ) i
and
+ ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k 
= ( 62.0 N ⋅ m ) i + (148.8 N ⋅ m ) j + (19.2 N ⋅ m ) k
M =
M x2 + M y2 + M z2 =
( 62.0 )2 + (148.8)2 + (19.2 )2
= 162.339 N ⋅ m
or M = 162.3 N ⋅ m
λ =
M
62.0i + 148.8j + 19.2k
=
M
162.339
= 0.38192i + 0.91660 j + 0.118271k
cosθ x = 0.38192
∴ θ x = 67.547°
or θ x = 67.5°
cosθ y = 0.91660
∴ θ y = 23.566°
or θ y = 23.6°
cosθ z = 0.118271
∴ θ z = 83.208°
or θ z = 83.2°
PROBLEM 3.74
Knowing that P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Have
M = M4 + M7
where
M 4 = rG/C × F4G
rG/C = − (10 in.) i
F4G = ( 4 lb ) k
∴ M 4 = − (10 in.) i × ( 4 lb ) k = ( 40 lb ⋅ in.) j
M 7 = rD/F × F7 D
Also,
rD/F = − ( 5 in.) i + ( 3 in.) j
F7 D = λ ED F7 D
=
=
− ( 5 in.) i + ( 3 in.) j + ( 7 in.) k
( 5 )2 + ( 3)2 + ( 7 )2
( 7 lb )
in.
7 lb
( −5i + 3j + 7k )
83
i
7 lb ⋅ in.
∴ M7 =
−5
83
−5
j k
7 lb ⋅ in.
3 0 =
( 21i + 35j + 0k )
83
3 7
= 0.76835 ( 21i + 35 j) lb ⋅ in.
PROBLEM 3.74 CONTINUED
M = ( 40 lb ⋅ in.) j + 0.76835 ( 21i + 35j) lb ⋅ in.
and
= (16.1353 lb ⋅ in.) i + ( 66.892 lb ⋅ in.) j
M =
( M x )2 + ( M y )
2
=
(16.1353)2 + ( 66.892 )2
= 68.811 lb ⋅ in.
or M = 68.8 lb ⋅ in.
λ =
(16.1353 lb ⋅in.) i + ( 66.892 lb ⋅in.) j
M
=
68.811 lb ⋅ in.
M
= 0.23449i + 0.97212 j
cosθ x = 0.23449
∴ θ x = 76.438°
or θ x = 76.4°
cosθ y = 0.97212
∴ θ y = 13.5615°
or θ y = 13.56°
cosθ z = 0.0
∴ θ z = 90°
or θ z = 90.0°
PROBLEM 3.75
Knowing that P = 5 lb, replace the three couples with a single equivalent
couple, specifying its magnitude and the direction of its axis.
SOLUTION
M = M 4 + M 7 + M5
Have
where
M 4 = rG/C × F4G
M 7 = rD/F × F7 D
i j k
= −10 0 0 lb ⋅ in. = ( 40 lb ⋅ in.) j
0 0 4
i j k
 7 
= −5 3 0 
 lb ⋅ in. = 0.76835 ( 21i + 35j) lb ⋅ in.
 83 
−5 3 7
(See Solution to Problem 3.74.)
M 5 = rC/ A × F5C
i
j k
= 10 −6 7 lb ⋅ in. = − ( 35 lb ⋅ in.) i + ( 50 lb ⋅ in.) k
0 5 0
∴ M = (16.1353 − 35 ) i + ( 40 + 26.892 ) j + ( 50 ) k  lb ⋅ in.
= − (18.8647 lb ⋅ in.) i + ( 66.892 lb ⋅ in.) j + ( 50 lb ⋅ in.) k
M =
M x2 + M y2 + M z2 =
(18.8647 )2 + ( 66.892 )2 + ( 50 )2
= 85.618 lb ⋅ in.
or M = 85.6 lb ⋅ in.
λ =
M
−18.8647i + 66.892 j + 50k
=
= −0.22034i + 0.78129 j + 0.58399k
M
85.618
cosθ x = −0.22034
∴ θ x = 102.729°
or θ x = 102.7°
cosθ y = 0.78129
∴ θ y = 38.621°
or θ y = 38.6°
cosθ z = 0.58399
∴ θ z = 54.268°
or θ z = 54.3°
PROBLEM 3.76
Knowing that P = 210 N, replace the three couples with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
M = M1 + M 2 + M P
Have
M1 = rC/B × P1C
where
M 2 = rD/ A × P2 E
i
j
k
= 0.96 −0.40 0 = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j
0
0
−100
i
j
k
= 0 0.20 −0.55 = ( 22.0 N ⋅ m ) i + ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k
−96 0
110
(See Solution to Problem 3.73.)
M P = rE/ A
i
j
k
× PE = 0.48 0.20 −1.10 = ( 231 N ⋅ m ) i + (100.8 N ⋅ m ) k
0 210
0
∴ M = ( 40 + 22 + 231) i + ( 96 + 52.8 ) j + (19.2 + 100.8 ) k  N ⋅ m
= ( 293 N ⋅ m ) i + (148.8 N ⋅ m ) j + (120 N ⋅ m ) k
M =
M x2 + M y2 + M z2 =
( 293)2 + (148.8)2 + (120 )2
= 349.84 N ⋅ m
or M = 350 N ⋅ m
λ =
M
293i + 148.8 j + 120k
=
= 0.83752i + 0.42533j + 0.34301k
M
349.84
cosθ x = 0.83752
∴ θ x = 33.121°
or θ x = 33.1°
cosθ y = 0.42533
∴ θ y = 64.828°
or θ y = 64.8°
cosθ z = 0.34301
∴ θ z = 69.940°
or θ z = 69.9°
PROBLEM 3.77
In a manufacturing operation, three holes are drilled simultaneously in a
workpiece. Knowing that the holes are perpendicular to the surfaces of
the workpiece, replace the couples applied to the drills with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
M = M1 + M 2 + M 3
Have
M1 = − (1.1 lb ⋅ ft )( cos 25° j + sin 25°k )
where
M 2 = − (1.1 lb ⋅ ft ) j
M 3 = − (1.3 lb ⋅ ft )( cos 20° j − sin 20°k )
∴ M = ( −0.99694 − 1.1 − 1.22160 ) j + ( −0.46488 + 0.44463) k
= − ( 3.3185 lb ⋅ ft ) j − ( 0.020254 lb ⋅ ft ) k
and
M =
M x2 + M y2 + M z2 =
( 0 )2 + ( 3.3185)2 + ( 0.020254 )2
= 3.3186 lb ⋅ ft
or M = 3.32 lb ⋅ ft
λ =
( 0 ) i − 3.3185j − 0.020254k
M
=
3.3186
M
= −0.99997 j − 0.0061032k
cosθ x = 0
∴ θ x = 90°
cosθ y = −0.99997
∴ θ y = 179.555°
cosθ z = −0.0061032
∴ θ z = 90.349°
or θ x = 90.0°
or θ y = 179.6°
or θ z = 90.3°
PROBLEM 3.78
The tension in the cable attached to the end C of an adjustable boom ABC
is 1000 N. Replace the force exerted by the cable at C with an equivalent
force-couple system (a) at A, (b) at B.
SOLUTION
(a) Based on
ΣF : FA = T = 1000 N
or FA = 1000 N
20°
ΣM A : M A = (T sin 50° )( dA )
= (1000 N ) sin 50° ( 2.25 m )
= 1723.60 N ⋅ m
or M A = 1724 N ⋅ m
(b) Based on
ΣF : FB = T = 1000 N
or FB = 1000 N
20°
ΣMB : M B = (T sin 50° )( d B )
= (1000 N ) sin 50° (1.25 m )
= 957.56 N ⋅ m
or M B = 958 N ⋅ m
PROBLEM 3.79
The 20-lb horizontal force P acts on a bell crank as shown. (a) Replace P
with an equivalent force-couple system at B. (b) Find the two vertical
forces at C and D which are equivalent to the couple found in part a.
SOLUTION
(a) Based on
ΣF : PB = P = 20 lb
or PB = 20 lb
ΣM : MB = Pd B
= 20 lb ( 5 in.)
= 100 lb ⋅ in.
or M B = 100 lb ⋅ in.
(b) If the two vertical forces are to be equivalent to MB , they must be a
couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then, with PC and PD acting as shown,
ΣM : M D = PC d
100 lb ⋅ in. = PC ( 4 in.)
∴ PC = 25 lb
or PC = 25 lb
ΣFy : 0 = PD − PC
∴ PD = 25 lb
or PD = 25 lb
PROBLEM 3.80
A 700-N force P is applied at point A of a structural member. Replace P
with (a) an equivalent force-couple system at C, (b) an equivalent system
consisting of a vertical force at B and a second force at D.
SOLUTION
ΣF : PC = P = 700 N
(a) Based on
or PC = 700 N
60°
ΣM C : M C = − Px dCy + Py dCx
Px = ( 700 N ) cos60° = 350 N
where
Py = ( 700 N ) sin 60° = 606.22 N
dCx = 1.6 m
dCy = 1.1 m
∴ M C = − ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m )
= −385 N ⋅ m + 969.95 N ⋅ m
= 584.95 N ⋅ m
or M C = 585 N ⋅ m
ΣFx : PDx = P cos 60°
(b) Based on
= ( 700 N ) cos 60°
= 350 N
ΣM D :
( P cos 60° )( d DA ) =
PB ( d DB )
( 700 N ) cos 60° ( 0.6 m ) = PB ( 2.4 m )
PB = 87.5 N
or PB = 87.5 N
PROBLEM 3.80 CONTINUED
ΣFy : P sin 60° = PB + PDy
( 700 N ) sin 60° = 87.5 N + PDy
PDy = 518.72 N
PD =
=
( PDx )2 + ( PDy )
2
( 350 )2 + ( 518.72 )2
= 625.76 N
 PDy 
−1  518.72 
 = tan 
 = 55.991°
 350 
 PDx 
θ = tan −1 
or PD = 626 N
56.0°
PROBLEM 3.81
A landscaper tries to plumb a tree by applying a 240-N force as shown.
Two helpers then attempt to plumb the same tree, with one pulling at B
and the other pushing with a parallel force at C. Determine these two
forces so that they are equivalent to the single 240-N force shown in the
figure.
SOLUTION
Based on
ΣFx : − ( 240 N ) cos30° = − FB cosα − FC cosα
− ( FB + FC ) cos α = − ( 240 N ) cos 30°
or
ΣFy :
( 240 N ) sin 30° =
( FB + FC ) sin α
or
(1)
FB sin α + FC sin α
= ( 240 N ) sin 30°
(2)
From
Equation (2)
: tan α = tan 30°
Equation (1)
∴ α = 30°
Based on
ΣM C : ( 240 N ) cos ( 30° − 20° )  ( 0.25 m ) = ( FB cos10° )( 0.60 m )
∴ FB = 100 N
or FB = 100.0 N
From Equation (1),
30°
− (100 N + FC ) cos30° = −240cos30°
FC = 140 N
or FC = 140.0 N
30°
PROBLEM 3.82
A landscaper tries to plumb a tree by applying a 240-N force as shown.
(a) Replace that force with an equivalent force-couple system at C. (b)
Two helpers attempt to plumb the same tree, with one applying a
horizontal force at C and the other pulling at B. Determine these two
forces if they are to be equivalent to the single force of part a.
SOLUTION
ΣFx : − ( 240 N ) cos30° = − FC cos30°
(a) Based on
∴ FC = 240 N
or FC = 240 N
ΣM C : ( 240 N ) cos10° ( d A ) = M C
30°
d A = 0.25 m
∴ M C = 59.088 N ⋅ m
or M C = 59.1 N ⋅ m
ΣFy :
(b) Based on
( 240 N ) sin 30° =
FB sin α
FB sin α = 120
or
(1)
ΣM B : 59.088 N ⋅ m − ( 240 N ) cos10°  ( dC ) = − FC ( dC cos 20° )
59.088 N ⋅ m − ( 240 N ) cos10°  ( 0.60 m ) = − FC ( 0.60 m ) cos 20°
0.56382 FC = 82.724
FC = 146.722 N
or FC = 146.7 N
and
ΣFx : − ( 240 N ) cos30° = −146.722 N − FB cosα
FB cosα = 61.124
(2)
From
Equation (1)
:
Equation (2)
tan α =
α = 63.007°
From Equation (1),
FB =
120
= 1.96323
61.124
or α = 63.0°
120
= 134.670 N
sin 63.007°
or FB = 134.7 N
63.0°
PROBLEM 3.83
A dirigible is tethered by a cable attached to its cabin at B. If the tension
in the cable is 250 lb, replace the force exerted by the cable at B with an
equivalent system formed by two parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an angle
of α with the vertical.
Then for equivalence,
ΣFx :
( 250 lb ) sin 30° =
FA sin α + FB sin α
ΣFy : − ( 250 lb ) cos 30° = − FA cos α − FB cos α
(1)
(2)
Dividing Equation (1) by Equation (2),
( 250 lb ) sin 30°
− ( 250 lb ) cos 30°
=
( FA + FB ) sin α
− ( FA + FB ) cos α
Simplifying yields α = 30°
Based on
ΣM C : ( 250 lb ) cos 30° (12 ft ) = ( FA cos 30° )( 32 ft )
∴ FA = 93.75 lb
or FA = 93.8 lb
60°
Based on
ΣM A : − ( 250 lb ) cos 30° ( 20 ft ) = ( FC cos 30° ) ( 32 ft )
∴ FC = 156.25 lb
or FC = 156.3 lb
60°
PROBLEM 3.84
Three workers trying to move a 3 × 3 × 4-ft crate apply to the crate the
three horizontal forces shown. (a) If P = 60 lb, replace the three forces
with an equivalent force-couple system at A. (b) Replace the force-couple
system of part a with a single force, and determine where it should be
applied to side AB. (c) Determine the magnitude of P so that the three
forces can be replaced with a single equivalent force applied at B.
SOLUTION
(a) Based on
ΣFz : − 50 lb + 50 lb + 60 lb = FA
FA = 60 lb
or FA = ( 60.0 lb ) k
Based on
ΣM A :
( 50 lb )( 2 ft ) − ( 50 lb )( 0.6 ft ) = M A
M A = 70 lb ⋅ ft
(a)
or M A = ( 70.0 lb ⋅ ft ) j
(b) Based on
ΣFz : − 50 lb + 50 lb + 60 lb = F
F = 60 lb
or F = ( 60.0 lb ) k
Based on
ΣM A : 70 lb ⋅ ft = 60 lb (x )
(b)
x = 1.16667 ft
or x = 1.167 ft from A along AB
(c) Based on
ΣMB : − ( 50 lb ) (1 ft ) + ( 50 lb ) (2.4 ft ) − P (3 ft ) = R (0 )
P=
70
= 23.333 lb
3
or P = 23.3 lb
(c)
PROBLEM 3.85
A force and a couple are applied to a beam. (a) Replace this system with
a single force F applied at point G, and determine the distance d.
(b) Solve part a assuming that the directions of the two 600-N forces are
reversed.
SOLUTION
(a)
ΣFy : FC + FD + FE = F
Have
F = −800 N + 600 N − 600 N
F = −800 N
Have
or F = 800 N
ΣM G : FC ( d − 1.5 m ) − FD ( 2 m ) = 0
(800 N )( d
− 1.5 m ) − ( 600 N )( 2 m ) = 0
d =
1200 + 1200
800
d = 3m
or d = 3.00 m
(b)
Changing directions of the two 600 N forces only changes sign of the couple.
∴ F = −800 N
and
or F = 800 N
ΣM G : FC ( d − 1.5 m ) + FD ( 2 m ) = 0
(800 N )( d
d =
− 1.5 m ) + ( 600 N )( 2 m )
1200 − 1200
=0
800
or d = 0
PROBLEM 3.86
Three cables attached to a disk exert on it the forces shown. (a) Replace
the three forces with an equivalent force-couple system at A.
(b) Determine the single force which is equivalent to the force-couple
system obtained in part a, and specify its point of application on a line
drawn through points A and D.
SOLUTION
(a) Have
ΣF : FB + FC + FD = FA
FB = −FD
Since
∴ FA = FC = 110 N
20°
or FA = 110.0 N
Have
20.0°
ΣM A : − FBT ( r ) − FCT ( r ) + FDT ( r ) = M A
− (140 N ) sin15° ( 0.2 m ) − (110 N ) sin 25° ( 0.2 m ) + (140 N ) sin 45° ( 0.2 m ) = M A
M A = 3.2545 N ⋅ m
or M A = 3.25 N ⋅ m
(b) Have
ΣF : FA = FE
or FE = 110.0 N
20.0°
ΣM : M A = [ FE cos 20°] ( a )
∴ 3.2545 N ⋅ m = (110 N ) cos 20° ( a )
a = 0.031485 m
or a = 31.5 mm below A
PROBLEM 3.87
While tapping a hole, a machinist applies the horizontal forces shown to
the handle of the tap wrench. Show that these forces are equivalent to a
single force, and specify, if possible, the point of application of the single
force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
Have FB = 26.5 N + 2.5 N, where the 26.5 N force be part of the couple. Combining the two parallel forces,
M couple = ( 26.5 N ) ( 0.080 m + 0.070 m ) cos 25°
= 3.60 N ⋅ m
and, M couple = 3.60 N ⋅ m
A single equivalent force will be located in the negative z-direction.
Based on
ΣMB : − 3.60 N ⋅ m = (2.5 N )cos 25°  ( a )
a = −1.590 m
F′ = ( 2.5 N )( cos 25°i + sin 25° j)
and is applied on an extension of handle BD
at a distance of 1.590 m to the right of B
PROBLEM 3.88
A rectangular plate is acted upon by the force and couple shown. This
system is to be replaced with a single equivalent force. (a) For α = 40°,
specify the magnitude and the line of action of the equivalent force.
(b) Specify the value of α if the line of action of the equivalent force is
to intersect line CD 12 in. to the right of D.
SOLUTION
ΣFx : − ( 3 lb ) sin 40° + ( 3 lb ) sin 40° = Fx
(a) Have
∴ Fx = 0
ΣFy : − ( 3 lb ) cos 40° − 10 lb + ( 3 lb ) cos 40° = Fy
Have
∴ Fy = −10 lb
or F = 10.00 lb
Note: The two 3-lb forces form a couple
ΣM A : rC/ A × PC + rB/ A × PB = rX / A × F
and
i
j
k
i j k
i j k
3 16
−10 0 + 160 1 0 0 = 10 d 0 0
sin 40° cos 40° 0
0 −1 0
0 −1 0
k : 3 (16 ) cos 40° − ( −10 ) 3sin 40° − 160 = −10d
36.770 + 19.2836 − 160 = −10d
∴ d = 10.3946 in.
or F = 10.00 lb
(b) From part (a),
Have
at 10.39 in. right of A or at 5.61 in. left of B
F = 10.00 lb
ΣM A : rC/ A × PC + rB/ A × PB = (12 in.) i × F
i
j k
i j k
i j k
3 16 −10 0 + 160 1 0 0 = 120 1 0 0
sin α cos α 0
0 −1 0
0 −1 0
k : 48cos α + 30sin α − 160 = −120
24cosα = 20 − 15sin α
PROBLEM 3.88 CONTINUED
Squaring both sides of the equation, and
using the identity cos 2 α = 1 − sin 2 α , results in
sin 2 α − 0.74906sin α − 0.21973 = 0
Using quadratic formula
sin α = 0.97453
sinα = − 0.22547
so that
α = 77.0°
and
α = −13.03°
PROBLEM 3.89
A hexagonal plate is acted upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at E.
SOLUTION
Since the minimum value of P acting at B is realized when Pmin is
perpendicular to a line connecting B and E, α = 30°
Then,
ΣM E : rB/E × Pmin + rD/ A × PD = 0
where
rB/E = − ( 0.30 m ) i +  2 ( 0.30 m ) cos 30° j
= − ( 0.30 m ) i + ( 0.51962 m ) j
rD/ A = 0.30 m + 2 ( 0.3 m ) sin 30°  i
= ( 0.60 m ) i
PD = ( 450 N ) j
Pmin = Pmin ( cos 30° ) i + ( sin 30° ) j
∴ Pmin
i
j
k
i
j k
−0.30 0.51962 0 + 0.60 0 0 N ⋅ m = 0
0.86603 0.50 0
0 450 0
Pmin ( −0.15 m − 0.45 m ) k + ( 270 N ⋅ m ) k = 0
∴ Pmin = 450 N
or Pmin = 450 N
30°
PROBLEM 3.90
An eccentric, compressive 270-lb force P is applied to the end of a
cantilever beam. Replace P with an equivalent force-couple system at G.
SOLUTION
Have
ΣF : − ( 270 lb ) i = F
∴ F = − ( 270 lb ) i
Also, have
ΣM G : rA/G × P = M
i j
k
270 0 −4 −2.4 lb ⋅ in. = M
−1 0
0
∴ M = ( 270 lb ⋅ in.) ( −2.4 )( −1) j − ( −4 )( −1) k 
or M = ( 648 lb ⋅ in.) j − (1080 lb ⋅ in.) k
PROBLEM 3.91
Two workers use blocks and tackles attached to the bottom of an I-beam
to lift a large cylindrical tank. Knowing that the tension in rope AB is
324 N, replace the force exerted at A by rope AB with an equivalent
force-couple system at E.
SOLUTION
ΣF : TAB = F
Have
where
TAB = λ ABTAB
=
( 0.75 m ) i − ( 6.0 m ) j + ( 3.0 m ) k
6.75 m
( 324 N )
∴ TAB = 36 N ( i − 8j + 4k )
so that
F = ( 36.0 N ) i − ( 288 N ) j + (144.0 N ) k
Have
ΣM E : rA/E × TAB = M
or
i j k
( 7.5 m )( 36 N ) 0 1 0 = M
1 −8 4
∴ M = ( 270 N ⋅ m )( 4i − k )
or M = (1080 N ⋅ m ) i − ( 270 N ⋅ m ) k
PROBLEM 3.92
Two workers use blocks and tackles attached to the bottom of an I-beam
to lift a large cylindrical tank. Knowing that the tension in rope CD is
366 N, replace the force exerted at C by rope CD with an equivalent
force-couple system at O.
SOLUTION
ΣF : TCD = F
Have
where
TCD = λ CDTCD
=
− ( 0.3 m ) i − ( 5.6 m ) j + ( 2.4 m ) k
( 366 N )
6.1 m
∴ TCD = ( 6.0 N )( −3i − 56 j + 24k )
so that
F = − (18.00 N ) i − ( 336 N ) j + (144.0 N ) k
Have
ΣM O : rC/O × TCD = M
or
i
j k
( 7.5 m )( 6 N ) 0 1 0 = M
−3 −56 24
∴ M = ( 45 N ⋅ m )( 24i + 3k )
or M = (1080 N ⋅ m ) i + (135.0 N ⋅ m ) k
PROBLEM 3.93
To keep a door closed, a wooden stick is wedged between the floor and
the doorknob. The stick exerts at B a 45-lb force directed along line AB.
Replace that force with an equivalent force-couple system at C.
SOLUTION
Have
ΣF : PAB = FC
where
PAB = λ AB PAB
=
( 2.0 in.) i + ( 38 in.) j − ( 24 in.) k
44.989 in.
( 45 lb )
or FC = ( 2.00 lb ) i + ( 38.0 lb ) j − ( 24.0 lb ) k
Have
ΣM C : rB/C × PAB = M C
MC
i
j
k
= 2 29.5 −33
0 lb ⋅ in.
1 19 −12
= ( 2 lb ⋅ in.) {( −33)( −12 ) i − ( 29.5 )( −12 ) j
+ ( 29.5 )(19 ) − ( −33)(1)  k
}
or M C = ( 792 lb ⋅ in.) i + ( 708 lb ⋅ in.) j + (1187 lb ⋅ in.) k
PROBLEM 3.94
A 25-lb force acting in a vertical plane parallel to the yz plane is applied
to the 8-in.-long horizontal handle AB of a socket wrench. Replace the
force with an equivalent force-couple system at the origin O of the
coordinate system.
SOLUTION
Have
ΣF : PB = F
where
PB = 25 lb  − ( sin 20° ) j + ( cos 20° ) k 
= − ( 8.5505 lb ) j + ( 23.492 lb ) k
or F = − ( 8.55 lb ) j + ( 23.5 lb ) k
Have
ΣM O : rB/O × PB = M O
where
rB/O = ( 8cos 30° ) i + (15 ) j − ( 8sin 30° ) k  in.
= ( 6.9282 in.) i + (15 in.) j − ( 4 in.) k
∴
i
j
k
6.9282
15
−4 lb ⋅ in. = M O
0
−8.5505 23.492
M O = ( 318.18 ) i − (162.757 ) j − ( 59.240 ) k  lb ⋅ in.
or M O = ( 318 lb ⋅ in.) i − (162.8 lb ⋅ in.) j − ( 59.2 lb ⋅ in.) k
PROBLEM 3.95
A 315-N force F and 70-N · m couple M are applied to corner A of the
block shown. Replace the given force-couple system with an equivalent
force-couple system at corner D.
SOLUTION
Have
ΣF : F = FD
= λ AI F
=
( 0.360 m ) i − ( 0.120 m ) j + ( 0.180 m ) k
0.420 m
( 315 N )
= ( 750 N )( 0.360i − 0.120 j + 0.180k )
or FD = ( 270 N ) i − ( 90.0 N ) j + (135.0 N ) k
Have
ΣM D : M + rI /D × F = M D
where
M = λ AC M
=
( 0.240 m ) i − ( 0.180 m ) k
0.300 m
( 70.0 N ⋅ m )
= ( 70.0 N ⋅ m )( 0.800i − 0.600k )
rI /D = ( 0.360 m ) k
∴ MD
i
j
k
= ( 70.0 N ⋅ m )( 0.8i − 0.6k ) + 0
0
0.36 ( 750 N ⋅ m )
0.36 −0.12 0.18
= ( 56.0 N ⋅ m ) i − ( 42.0 N ⋅ m ) k + ( 32.4 N ⋅ m ) i + ( 97.2 N ⋅ m ) j
or M D = ( 88.4 N ⋅ m ) i + ( 97.2 N ⋅ m ) j − ( 42.0 N ⋅ m ) k
PROBLEM 3.96
The handpiece of a miniature industrial grinder weighs 2.4 N, and its
center of gravity is located on the y axis. The head of the handpiece is
offset in the xz plane in such a way that line BC forms an angle of 25°
with the x direction. Show that the weight of the handpiece and the two
couples M1 and M 2 can be replaced with a single equivalent force.
Further assuming that M1 = 0.068 N ⋅ m and M 2 = 0.065 N ⋅ m,
determine (a) the magnitude and the direction of the equivalent force,
(b) the point where its line of action intersects the xz plane.
SOLUTION
First assume that the given force W and couples M1 and M 2 act at the
origin.
Now
W = −Wj
and
M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k
Note that since W and M are perpendicular, it follows that they can be
replaced with a single equivalent force.
F =W
(a) Have
or
F = −Wj = − ( 2.4 N ) j
or F = − ( 2.40 N ) j
(b) Assume that the line of action of F passes through point P (x, 0, z).
Then for equivalence
M = rP/O × F
where
∴
rP/O = xi + zk
− ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k
i j k
= x 0 z = (Wz ) i − (Wx ) k
0 −W 0
PROBLEM 3.96 CONTINUED
Equating the i and k coefficients,
z =
(b) For
−M z cos 25°
W
and
 M − M 2 sin 25° 
x = − 1

W


W = 2.4 N, M1 = 0.068 N ⋅ m, M 2 = 0.065 N ⋅ m
x=
0.068 − 0.065sin 25°
= −0.0168874 m
−2.4
or x = −16.89 mm
z =
−0.065cos 25°
= −0.024546 m
2.4
or z = −24.5 mm
PROBLEM 3.97
A 20-lb force F1 and a 40- lb ⋅ ft couple M1 are applied to corner E of the
bent plate shown. If F1 and M1 are to be replaced with an equivalent
force-couple system ( F2 , M 2 ) at corner B and if ( M 2 ) z = 0, determine
(a) the distance d, (b) F2 and M 2.
SOLUTION
ΣM Bz : M 2 z = 0
(a) Have
(
)
k ⋅ rH /B × F1 + M1z = 0
where
(1)
rH /B = ( 31 in.) i − ( 2 in.) j
F1 = λ EH F1
=
=
( 6 in.) i + ( 6 in.) j − ( 7 in.) k
11.0 in.
( 20 lb )
20 lb
( 6i + 6 j − 7k )
11.0
M1z = k ⋅ M1
M1 = λ EJ M1
=
−di + ( 3 in.) j − ( 7 in.) k
d 2 + 58 in.
Then from Equation (1),
0 0 1
20 lb ⋅ in. ( −7 )( 480 lb ⋅ in.)
31 −2 0
+
=0
11.0
d 2 + 58
6 6 −7
( 480 lb ⋅in.)
PROBLEM 3.97 CONTINUED
Solving for d, Equation (1) reduces to
20 lb ⋅ in.
3360 lb ⋅ in.
=0
(186 + 12 ) − 2
11.0
d + 58
d = 5.3955 in.
From which
or d = 5.40 in.
(b)
F2 = F1 =
20 lb
( 6i + 6 j − 7k )
11.0
= (10.9091i + 10.9091j − 12.7273k ) lb
or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k
M 2 = rH /B × F1 + M1
i j k
20 lb ⋅ in. ( −5.3955 ) i + 3j − 7k
= 31 −2 0
+
( 480 lb ⋅ in.)
11.0
9.3333
6 6 −7
= ( 25.455i + 394.55 j + 360k ) lb ⋅ in.
+ ( −277.48i + 154.285 j − 360k ) lb ⋅ in.
M 2 = − ( 252.03 lb ⋅ in.) i + ( 548.84 lb ⋅ in.) j
or M 2 = − ( 21.0 lb ⋅ ft ) i + ( 45.7 lb ⋅ ft ) j
PROBLEM 3.98
A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each
loading with an equivalent force-couple system at end A of the beam.
(b) Which of the loadings are equivalent?
SOLUTION
(a)
ΣFy : − 200 lb − 100 lb = Ra
(a) Have
or R a = 300 lb
and
ΣM A : 900 lb ⋅ ft − (100 lb )( 4 ft ) = M a
or M a = 500 lb ⋅ ft
ΣFy : − 300 lb = Rb
(b) Have
or R b = 300 lb
ΣM A : − 450 lb ⋅ ft = M b
and
or M b = 450 lb ⋅ ft
ΣFy : 150 lb − 450 lb = Rc
(c) Have
or R c = 300 lb
and
ΣM A : 2250 lb ⋅ ft − ( 450 lb )( 4 ft ) = M c
or M c = 450 lb ⋅ ft
ΣFy : − 200 lb + 400 lb = Rd
(d) Have
or R d = 200 lb
and
ΣM A :
( 400 lb )( 4 ft ) − 1150 lb ⋅ ft
= Md
or M d = 450 lb ⋅ ft
(e) Have
ΣFy : − 200 lb − 100 lb = Re
or R e = 300 lb
and
ΣM A : 100 lb ⋅ ft + 200 lb ⋅ ft − (100 lb )( 4 ft ) = M e
or M e = 100 lb ⋅ ft
PROBLEM 3.98 CONTINUED
(f) Have
ΣFy : − 400 lb + 100 lb = R f
or R f = 300 lb
and
ΣM A : −150 lb ⋅ ft + 150 lb ⋅ ft + (100 lb )( 4 ft ) = M f
or M f = 400 lb ⋅ ft
(g) Have
ΣFy : −100 lb − 400 lb = Rg
or R g = 500 lb
and
ΣM A : 100 lb ⋅ ft + 2000 lb ⋅ ft − ( 400 lb )( 4 ft ) = M g
or M g = 500 lb ⋅ ft
(h) Have
ΣFy : −150 lb − 150 lb = Rh
or R h = 300 lb
and
ΣM A : 1200 lb ⋅ ft − 150 lb ⋅ ft − (150 lb )(4 ft ) = M h
or M h = 450 lb ⋅ ft
(b)
Therefore, loadings (c) and (h) are equivalent
PROBLEM 3.99
A 4-ft-long beam is loaded as shown. Determine the loading of Problem
3.98 which is equivalent to this loading.
SOLUTION
ΣFy : −100 lb − 200 lb = R
Have
or R = 300 lb
and
ΣM A : − 200 lb ⋅ ft + 1400 lb ⋅ ft − ( 200 lb )( 4 ft ) = M
or M = 400 lb ⋅ ft
Equivalent to case (f) of Problem 3.98
Problem 3.98 Equivalent force-couples at A
case
R
(a)
300 lb
500 lb ⋅ ft
(b)
300 lb
450 lb ⋅ ft
(c)
300 lb
450 lb ⋅ ft
(d )
200 lb
450 lb ⋅ ft
(e)
300 lb
100 lb ⋅ ft
(f )
300 lb
400 lb ⋅ ft
(g )
500 lb
500 lb ⋅ ft
(h)
300 lb
450 lb ⋅ ft
M
PROBLEM 3.100
Determine the single equivalent force and the distance from point A to its
line of action for the beam and loading of (a) Problem 3.98b,
(b) Problem 3.98d, (c) Problem 3.98e.
Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings.
(a) Replace each loading with an equivalent force-couple system at end A
of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
For equivalent single force at distance d from A
ΣFy : − 300 lb = R
Have
or R = 300 lb
ΣM C :
and
( 300 lb )( d ) − 450 lb ⋅ ft
=0
or d = 1.500 ft
(b)
ΣFy : − 200 lb + 400 lb = R
Have
or R = 200 lb
and
ΣM C :
( 200 lb )( d ) + ( 400 lb )( 4 − d ) − 1150 lb ⋅ ft
=0
or d = 2.25 ft
(c)
Have
ΣFy : − 200 lb − 100 lb = R
or R = 300 lb
and
ΣM C : 100 lb ⋅ ft + ( 200 lb )( d ) − (100 lb )( 4 − d ) + 200 lb ⋅ ft = 0
or d = 0.333 ft
PROBLEM 3.101
Five separate force-couple systems act at the corners of a metal block,
which has been machined into the shape shown. Determine which of
these systems is equivalent to a force F = (10 N ) j and a couple of
moment M = ( 6 N ⋅ m ) i + ( 4 N ⋅ m ) k located at point A.
SOLUTION
The equivalent force-couple system at A for each of the five force-couple
systems will be determined. Each will then be compared to the given
force-couple system to determine if they are equivalent.
Force-couple system at B
ΣF :
Have
F = (10 N ) j
or
and
(10 N ) j = F
(
)
ΣM A : ΣM B + rB/ A × F = M
( 4 N ⋅ m ) i + ( 2 N ⋅ m ) k + ( 0.2 m ) i × (10 N ) j = M
M = (4 N⋅m)i + ( 4 N⋅m)k
Comparing to given force-couple system at A,
Is Not Equivalent
Force-couple system at C
Have
or
and
ΣF :
(10 N ) j = F
F = (10 N ) j
(
)
ΣM A : M C + rC/ A × F = M
(8.5 N ⋅ m ) i + ( 0.2 m ) i + ( 0.25 m ) k  × (10 N ) j = M
M = ( 6 N ⋅ m ) i + ( 2.0 N ⋅ m ) k
Comparing to given force-couple system at A,
Is Not Equivalent
PROBLEM 3.101 CONTINUED
Force-couple system at E
ΣF :
Have
(10 N ) j = F
F = (10 N ) j
or
(
)
ΣM A : M E + rE/ A × F = M
and
( 6 N ⋅ m ) i + ( 0.4 m ) i − ( 0.08 m ) j × (10 N ) j = M
M = (6 N⋅m)i + ( 4 N⋅m)k
Comparing to given force-couple system at A,
Is Equivalent
Force-couple system at G
ΣF :
Have
(10 N ) i + (10 N ) j = F
F = (10 N ) i + (10 N ) j
or
F has two force components
∴ force-couple system at G
Is Not Equivalent
Force-couple system at I
(10 N ) j = F
Have
ΣF :
or
F = (10 N ) j
and
(
)
ΣM A : ΣM I + rI / A × F = M
(10 N ⋅ m ) i − ( 2 N ⋅ m ) k
+ ( 0.4 m ) i − ( 0.2 m ) j + ( 0.4 m ) k  × (10 N ) j = M
or
M = (6 N ⋅ m) i + (2 N ⋅m)k
Comparing to given force-couple system at A,
Is Not Equivalent
PROBLEM 3.102
The masses of two children sitting at ends A and B of a seesaw are 38 kg
and 29 kg, respectively. Where should a third child sit so that the
resultant of the weights of the three children will pass through C if she
has a mass of (a) 27 kg, (b) 24 kg.
SOLUTION
First
WA = mA g = ( 38 kg ) g
WB = mB g = ( 29 kg ) g
(a)
WC = mC g = ( 27 kg ) g
For resultant weight to act at C,
ΣM C = 0
Then ( 38 kg ) g  ( 2 m ) − ( 27 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0
∴ d =
76 − 58
= 0.66667 m
27
or d = 0.667 m
(b)
WC = mC g = ( 24 kg ) g
For resultant weight to act at C,
ΣM C = 0
Then ( 38 kg ) g  ( 2 m ) − ( 24 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0
∴ d =
76 − 58
= 0.75 m
24
or d = 0.750 m
PROBLEM 3.103
Three stage lights are mounted on a pipe as shown. The mass of each
light is mA = mB = 1.8 kg and mC = 1.6 kg . (a) If d = 0.75 m,
determine the distance from D to the line of action of the resultant of the
weights of the three lights. (b) Determine the value of d so that the
resultant of the weights passes through the midpoint of the pipe.
SOLUTION
WA = WB = m A g = (1.8 kg ) g
First
WC = mC g = (1.6 kg ) g
d = 0.75 m
(a)
Have
R = WA + WB + WC
R = (1.8 + 1.8 + 1.6 ) kg  g
R = ( 5.2 g ) N
or
Have
ΣM D : −1.8g ( 0.3 m ) − 1.8g (1.3 m ) − 1.6 g ( 2.05 m ) = −5.2 g ( D )
∴ D = 1.18462 m
or D = 1.185 m
D=
(b)
L
= 1.25 m
2
Have
ΣM D : − (1.8 g )( 0.3 m ) − (1.8g )(1.3 m ) − (1.6 g )(1.3 m + d )
= − ( 5.2 g )(1.25 m )
∴ d = 0.9625 m
or d = 0.963 m
PROBLEM 3.104
Three hikers are shown crossing a footbridge. Knowing that the weights
of the hikers at points C, D, and E are 800 N, 700 N, and 540 N,
respectively, determine (a) the horizontal distance from A to the line of
action of the resultant of the three weights when a = 1.1 m, (b) the value
of a so that the loads on the bridge supports at A and B are equal.
SOLUTION
a = 1.1 m
(a)
ΣF : −WC − WD − WE = R
Have
∴ R = −800 N − 700 N − 540 N
R = 2040 N
(a)
R = 2040 N
or
Have
ΣM A : − (800 N )(1.5 m ) − (700 N )(2.6 m ) − (540 N )(4.25 m )
= −R ( d )
∴
− 5315 N ⋅ m = − ( 2040 N ) d
d = 2.6054 m
and
or d = 2.61 m to the right of A
(b) For equal reaction forces at A and B, the resultant, R, must act at the
center of the span.
(b)
L
ΣM A = − R  
2
From
∴
− ( 800 N )(1.5 m ) − ( 700 N )(1.5 m + a ) − ( 540 N )(1.5 m + 2.5a )
= − ( 2040 N )( 3 m )
3060 + 2050a = 6120
and
a = 1.49268 m
or a = 1.493 m
PROBLEM 3.105
Gear C is rigidly attached to arm AB. If the forces and couple shown can
be reduced to a single equivalent force at A, determine the equivalent
force and the magnitude of the couple M.
SOLUTION
For equivalence
ΣFx : − ( 90 N ) sin 30° + (125 N ) cos 40° = Rx
or Rx = 50.756 N
ΣFy : − ( 90 N ) cos30° − 200 N − (125 N ) sin 40° = Ry
or Ry = −358.29 N
and
( 50.756 )2 + ( −358.29 )2
R=
Then
tan θ =
Ry
Rx
=
−358.29
= −7.0591
50.756
= 361.87 N
∴ θ = −81.937°
or R = 362 N
81.9°
Also
ΣM A : M − ( 90 N ) sin 35° ( 0.6 m ) − ( 200 N ) cos 25°  ( 0.85 m ) − (125 N ) sin 65°  (1.25 m ) = 0
∴ M = 326.66 N ⋅ m
or M = 327 N ⋅ m
PROBLEM 3.106
To test the strength of a 25 × 20-in. suitcase, forces are applied as shown.
If P = 18 lb, (a) determine the resultant of the applied forces, (b) locate
the two points where the line of action of the resultant intersects the edge
of the suitcase.
SOLUTION
(a) P = 18 lb
Have
ΣF : − ( 20 lb ) i +
42 lb
( −3i + 2 j) + (18 lb ) j + ( 36 lb ) i = Rxi + Ry j
13
∴ − (18.9461 lb ) i + ( 41.297 lb ) j = Rxi + Ry j
R = − (18.95 lb ) i + ( 41.3 lb ) j
or
R=
Rx2 + Ry2 =
(18.9461)2 + ( 41.297 )2
= 45.436 lb
 Ry 
−1  41.297 
 = tan 
 = −65.355°
 −18.9461 
 Rx 
θ x = tan −1 
or R = 45.4 lb
(b) Have
65.4°
ΣM B = M B
 42 lb

M B = ( 4 in.) j × ( −20 lb ) i + ( 21 in.) i × 
( −3i + 2 j) + (12 in.) j × ( 36 lb ) i + ( 3 in.) i × (18 lb ) j
13


∴ M B = (191.246 lb ⋅ in.) k
PROBLEM 3.106 CONTINUED
M B = rB × R
Since
i
j
k
∴ (191.246 lb ⋅ in.) k =
x
y
0 = ( 41.297 x + 18.9461y ) k
−18.9461 41.297 0
For
y = 0,
x=
191.246
= 4.6310 in.
41.297
or x = 4.63 in.
For
x = 0,
y =
191.246
= 10.0942 in.
18.9461
or y = 10.09 in.
PROBLEM 3.107
Solve Problem 3.106 assuming that P = 28 lb.
Problem 3.106: To test the strength of a 25 × 20-in. suitcase, forces are
applied as shown. If P = 18 lb, (a) determine the resultant of the applied
forces, (b) locate the two points where the line of action of the resultant
intersects the edge of the suitcase.
SOLUTION
(a) P = 28 lb
Have
ΣF : − ( 20 lb ) i +
42
( −3i + 2 j) + ( 28 lb ) j + ( 36 lb ) i = Rxi + Ry j
13
∴ − (18.9461 lb ) i + ( 51.297 lb ) j = Rxi + Ry j
R = − (18.95 lb ) i + ( 51.3 lb ) j
or
R=
Rx2 + Ry2 =
(18.9461)2 + ( 51.297 )2
= 54.684 lb
 Ry 
−1  51.297 
 = tan 
 = −69.729°
 −18.9461 
 Rx 
θ x = tan −1 
or R = 54.7 lb
(b) Have
69.7°
ΣM B = M B
 42 lb

M B = ( 4 in.) j × ( −20 lb ) i + ( 21 in.) i × 
( −3i + 2j) + (12 in.) j × ( 36 lb ) i + ( 3 in.) i × ( 28 lb ) j
13


∴ M B = ( 221.246 lb ⋅ in.) k
PROBLEM 3.107 CONTINUED
M B = rB × R
Since
i
j
k
∴ ( 221.246 lb ⋅ in.) k =
x
y
0 = ( 51.297 x + 18.9461y ) k
−18.9461 51.297 0
For
y = 0,
x=
221.246
= 4.3130 in.
51.297
or x = 4.31 in.
For
x = 0,
y =
221.246
= 11.6776 in.
18.9461
or y = 11.68 in.
PROBLEM 3.108
As four holes are punched simultaneously in a piece of aluminum sheet
metal, the punches exert on the piece the forces shown. Knowing that the
forces are perpendicular to the surfaces of the piece, determine (a) the
resultant of the applied forces when α = 45° and the point of
intersection of the line of action of that resultant with a line drawn
through points A and B, (b) the value of α so that the line of action of the
resultant passes through fold EF.
SOLUTION
Position the origin for the coordinate system along the centerline of the
sheet metal at the intersection with line EF.
ΣF = R
(a) Have
R = −  2.6 j − 5.25 j − 10.5 ( cos 45°i + sin 45° j) − 3.2i  kN
∴ R = − (10.6246 kN ) i − (15.2746 kN ) j
R=
Rx2 + Ry2 =
(10.6246 )2 + (15.2746 )2
= 18.6064 kN
 Ry 
−1  −15.2746 
 = tan 
 = 55.179°
 −10.6246 
 Rx 
θ = tan −1 
or R = 18.61 kN
55.2°
M EF = ΣM EF
Have
where
MEF = (2.6 kN )(90 mm ) + (5.25 kN )(40 mm )
− (10.5 kN )(20 mm ) − ( 3.2 kN ) ( 40 mm ) sin 45° + 40 mm 
∴ MEF = 15.4903 N ⋅ m
To obtain distance d left of EF,
M EF = dRy = d ( −15.2746 kN )
Have
∴d =
15.4903 N ⋅ m
= −1.01412 × 10−3 m
−3
−15.2746 × 10 N
or d = 1.014 mm left of EF
PROBLEM 3.108 CONTINUED
(b) Have
M EF = 0
M EF = 0 = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm )
− (10.5 kN )( 20 mm )
− ( 3.2 kN ) ( 40 mm ) sin α + 40 mm 
∴ (128 N ⋅ m ) sin α = 106 N ⋅ m
sin α = 0.828125
α = 55.907°
or α = 55.9°
PROBLEM 3.109
As four holes are punched simultaneously in a piece of aluminum sheet
metal, the punches exert on the piece the forces shown. Knowing that the
forces are perpendicular to the surfaces of the piece, determine (a) the
value of α so that the resultant of the applied forces is parallel to the
10.5 N force, (b) the corresponding resultant of the applied forces and the
point of intersection of its line of action with a line drawn through points
A and B.
SOLUTION
(a) For the resultant force, R, to be parallel to the 10.5 kN force,
α =φ
∴ tan α = tan φ =
Ry
Rx
where
Rx = −3.2 kN − (10.5 kN ) sin α
Ry = −2.6 kN − 5.25 kN − (10.5 kN ) cos α
∴ tan α =
3.2 + 10.5 sinα
7.85 + 10.5cos α
tan α =
and
3.2
= 0.40764
7.85
α = 22.178°
or α = 22.2°
α = 22.178°
(b) From
Rx = −3.2 kN − (10.5 kN ) sin 22.178° = −7.1636 kN
Ry = −7.85 kN − (10.5 kN ) cos 22.178° = −17.5732 kN
R=
Rx2 + Ry2 =
( 7.1636 )2 + (17.5732 )2
= 18.9770 kN
R = 18.98 kN
or
67.8°
Then
M EF = ΣM EF
where
M EF = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm ) − (10.5 kN )( 20 mm )
− ( 3.2 kN ) ( 40 mm ) sin 22.178° + 40 mm 
= 57.682 N ⋅ m
PROBLEM 3.109 CONTINUED
To obtain distance d left of EF,
M EF = dRy = d ( −17.5732 )
Have
∴d =
57.682 N ⋅ m
= −3.2824 × 10−3 m
−17.5732 × 103 N
or d = 3.28 mm left of EF
PROBLEM 3.110
A truss supports the loading shown. Determine the equivalent force
acting on the truss and the point of intersection of its line of action with a
line through points A and G.
SOLUTION
R = ΣF
Have
R = ( 240 N )( cos 70°i − sin 70° j) − (160 N ) j
+ ( 300 N )( − cos 40°i − sin 40° j) − (180 N ) j
∴ R = − (147.728 N ) i − ( 758.36 N ) j
R=
Rx2 + Ry2 =
(147.728)2 + ( 758.36 )2
= 772.62 N
 Ry 
−1  −758.36 
 = tan 
 = 78.977°
 −147.728 
 Rx 
θ = tan −1 
or R = 773 N
79.0°
ΣM A = dRy
Have
where
ΣM A = − [ 240 N cos 70°] ( 6 m ) − [ 240 N sin 70°] ( 4 m )
− (160 N )(12 m ) +
[300 N cos 40°] ( 6 m )
− [300 N sin 40°] ( 20 m ) − (180 N )( 8 m )
= −7232.5 N ⋅ m
∴d =
−7232.5 N ⋅ m
= 9.5370 m
−758.36 N
or d = 9.54 m to the right of A
PROBLEM 3.111
Three forces and a couple act on crank ABC. For P = 5 lb and α = 40°,
(a) determine the resultant of the given system of forces, (b) locate the
point where the line of action of the resultant intersects a line drawn
through points B and C, (c) locate the point where the line of action of the
resultant intersects a line drawn through points A and B.
SOLUTION
P = 5 lb,
(a)
α = 40°
R = ΣF
Have
= ( 5 lb )( cos 40°i + sin 40° j) − ( 3 lb ) i − ( 2 lb ) j
∴ R = ( 0.83022 lb ) i + (1.21394 lb ) j
R=
Rx2 + Ry2 =
( 0.83022 )2 + (1.21394 )2
= 1.47069 lb
 Ry 
−1  1.21394 
 = tan 
 = 55.632°
 0.83022 
 Rx 
θ = tan −1 
or R = 1.471 lb
55.6°
MB = ΣMB = dRy
(b) From
where
MB = − (5 lb )cos 40°  (15 in. )sin 50°  − (5 lb )sin 40° 
× (15 in. )sin 50°  + ( 3 lb ) ( 6 in.) sin 50°
− ( 2 lb )( 6 in.) + 50 lb ⋅ in.
∴ M B = −23.211 lb ⋅ in.
and
d =
MB
−23.211 lb ⋅ in.
=
= −19.1205 in.
Ry
1.21394 lb
or d = 19.12 in. to the left of B
PROBLEM 3.111 CONTINUED
M B = rD/B × R
(c) From
− ( 23.211 lb ⋅ in.) k = ( −d1 cos 50°i + d1 sin 50° j)
× ( −0.83022 lb ) i + (1.21394 lb ) j
− ( 23.211 lb ⋅ in.) k = ( −0.78028d1 − 0.63599d1 ) k
∴ d1 =
or
23.211
= 16.3889 in.
1.41627
d1 = 16.39 in. from B along line AB
or 1.389 in. above and to the left of A
PROBLEM 3.112
Three forces and a couple act on crank ABC. Determine the value of d so
that the given system of forces is equivalent to zero at (a) point B, (b)
point D.
SOLUTION
ΣFx = 0
Based on
P cos α − 3 lb = 0
∴ P cos α = 3 lb
(1)
ΣFy = 0
and
P sin α − 2 lb = 0
∴ P sin α = 2 lb
(2)
Dividing Equation (2) by Equation (1),
tan α =
2
3
∴ α = 33.690°
Substituting into Equation (1),
P=
3 lb
= 3.6056 lb
cos 33.690°
P = 3.61 lb
or
(a) Based on
33.7°
ΣM B = 0
− ( 3.6056 lb ) cos 33.690° ( d + 6 in.) sin 50° 
− ( 3.6056 lb ) sin 33.690° ( d + 6 in.) cos 50°
+ ( 3 lb ) ( 6 in.) sin 50° − ( 2 lb )( 6 in.) + 50 lb ⋅ in. = 0
−3.5838d = −30.286
∴ d = 8.4509 in.
or d = 8.45 in.
PROBLEM 3.112 CONTINUED
(b) Based on
ΣM D = 0
− ( 3.6056 lb ) cos 33.690°  ( d + 6 in.) sin 50° 
− ( 3.6056 lb ) sin 33.690°  ( d + 6 in.) cos 50° + 6 in.
+ ( 3 lb ) ( 6 in.) sin 50° + 50 lb ⋅ in. = 0
−3.5838d = −30.286
∴ d = 8.4509 in.
or d = 8.45 in.
This result is expected, since R = 0 and M RB = 0 for
d = 8.45 in. implies that R = 0 and M = 0 at any other point for
the value of d found in part a.
PROBLEM 3.113
Pulleys A and B are mounted on bracket CDEF. The tension on each side
of the two belts is as shown. Replace the four forces with a single
equivalent force, and determine where its line of action intersects the
bottom edge of the bracket.
SOLUTION
Equivalent force-couple at A due to belts on pulley A
ΣF : −120 N − 160 N = RA
Have
∴ R A = 280 N
ΣM A : − 40 N ( 0.02 m ) = M A
Have
∴ M A = 0.8 N ⋅ m
Equivalent force-couple at B due to belts on pulley B
ΣF :
Have
( 210 N + 150 N )
∴ R B = 360 N
25° = R B
25°
ΣM B : − 60 N ( 0.015 m ) = M B
Have
∴ M B = 0.9 N ⋅ m
Equivalent force-couple at F
Have
ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j)
= ( 326.27 N ) i − (127.857 N ) j
R = RF =
2
2
RFx
+ RFy
=
( 326.27 )2 + (127.857 )2
= 350.43 N
 RFy 
−1  −127.857 
 = tan 
 = −21.399°
 326.27 
 RFx 
θ = tan −1 
or R F = R = 350 N
21.4°
PROBLEM 3.113 CONTINUED
Have
ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m
− ( 360 N ) cos 25° ( 0.010 m )
+ ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m
M F = − ( 3.5056 N ⋅ m ) k
To determine where a single resultant force will intersect line FE,
M F = dR y
∴ d =
MF
−3.5056 N ⋅ m
=
= 0.027418 m = 27.418 mm
Ry
−127.857 N
or d = 27.4 mm
PROBLEM 3.114
As follower AB rolls along the surface of member C, it exerts a constant
force F perpendicular to the surface. (a) Replace F with an equivalent
force-couple system at the point D obtained by drawing the perpendicular
from the point of contact to the x axis (b) For a = 1 m and b = 2 m,
determine the value of x for which the moment of the equivalent forcecouple system at D is maximum.
SOLUTION
(a) The slope of any tangent to the surface of member C is
dy
d  
x 2   −2b
=
b 1 − 2   = 2 x
dx
dx  
a  
a
Since the force F is perpendicular to the surface,
 dy 
tan α = −  
 dx 
−1
=
a2  1 
 
2b  x 
For equivalence
ΣF : F = R
ΣM D :
( F cosα )( y A ) = M D
where
cos α =
2bx
(a )
2
2
+ ( 2bx )
∴ MD =
,
2

x2 
y A = b 1 − 2 
a 


x3 
2 Fb 2  x − 2 
a 

a 4 + 4b 2 x 2
Therefore, the equivalent force-couple system at D is
R = F
 a2 
tan −1 

 2bx 

x3 
2Fb 2  x − 2 
a 

M =
a 4 + 4b 2 x 2
PROBLEM 3.114 CONTINUED
dM
=0
dx
(b) To maximize M, the value of x must satisfy
where, for a = 1 m, b = 2 m
M =
∴
dM
= 8F
dx
)
1 + 16 x 2
1
1 + 16 x 2 1 − 3x 2 − x − x3  ( 32 x ) 1 + 16 x 2
2

(
) (
)
(
(1 + 16x )
(1 + 16x )(1 − 3x ) − 16x ( x − x ) = 0
2
2
2
)
−
1
2
 = 0
3
32 x 4 + 3x 2 − 1 = 0
or
x2 =
(
8F x − x 3
−3 ± 9 − 4 ( 32 )( −1)
2 ( 32 )
= 0.136011 m 2 and − 0.22976 m 2
Using the positive value of x 2 ,
x = 0.36880 m
or x = 369 mm
PROBLEM 3.115
As plastic bushings are inserted into a 3-in.-diameter cylindrical sheet
metal container, the insertion tool exerts the forces shown on the
enclosure. Each of the forces is parallel to one of the coordinate axes.
Replace these forces with an equivalent force-couple system at C.
SOLUTION
For equivalence
Σ F: FA + FB + FC + FD = R C
R C = − ( 5 lb ) j − ( 3 lb ) j − ( 4 lb ) k − ( 7 lb ) i
∴ R C = ( −7 lb ) i − ( 8 lb ) j − ( 4 lb ) k
Also for equivalence
ΣM C : rA′/C × FA + rB′/C × FB + rD′/C × FD = M C
or
MC
i j
k
i
j
k
i
j
k
= 0 0 −1.5 in. + 1 in. 0 −1.5 in. + 0 1.5 in. 1.5 in.
0 5 lb
0
0 −3 lb
0
−7 lb
0
0
= ( −7.50 lb ⋅ in. − 0 ) i  + ( 0 − 4.50 lb ⋅ in.) i + ( −3.0 lb ⋅ in. − 0 ) k 
+ (10.5 lb ⋅ in. − 0 ) j + ( 0 + 10.5 lb ⋅ in.) k 
or M C = − (12.0 lb ⋅ in.) i + (10.5 lb ⋅ in.) j + ( 7.5 lb ⋅ in.) k
PROBLEM 3.116
Two 300-mm-diameter pulleys are mounted on line shaft AD. The belts B
and C lie in vertical planes parallel to the yz plane. Replace the belt forces
shown with an equivalent force-couple system at A.
SOLUTION
Equivalent force-couple at each pulley
Pulley B
R B = ( 290 N )( − cos 20° j + sin 20°k ) − 430 Nj
= − ( 702.51 N ) j + ( 99.186 N ) k
M B = − ( 430 N − 290 N )( 0.15 m ) i
= − ( 21 N ⋅ m ) i
Pulley C
R C = ( 310 N + 480 N )( − sin10° j − cos10°k )
= − (137.182 N ) j − ( 778.00 N ) k
M C = ( 480 N − 310 N )( 0.15 m ) i
= ( 25.5 N ⋅ m ) i
Then
R = R B + R C = − ( 839.69 N ) j − ( 678.81 N ) k
or R = − ( 840 N ) j − ( 679 N ) k
M A = M B + M C + rB/ A × R B + rC/ A × R C
i
j
k
= − ( 21 N ⋅ m ) i + ( 25.5 N ⋅ m ) i + 0.45
0
0 N⋅m
0 −702.51 99.186
i
j
k
+ 0.90
0
0
N⋅m
0 −137.182 −778.00
= ( 4.5 N ⋅ m ) i + ( 655.57 N ⋅ m ) j − ( 439.59 N ⋅ m ) k
or M A = ( 4.50 N ⋅ m ) i + ( 656 N ⋅ m ) j − ( 440 N ⋅ m ) k
PROBLEM 3.117
A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic
holds the socket wrench handle at points A and B and applies forces at
these points. Knowing that these forces are equivalent to a force-couple
system at C consisting of the force C = − ( 40 N ) i + ( 20 N ) k and the
couple M C = ( 40 N ⋅ m ) i , determine the forces applied at A and B when
Az = 10 N.
SOLUTION
ΣF : A + B = C
Have
Fx : Ax + Bx = −40 N
or
∴ Bx = − ( Ax + 40 N )
(1)
ΣFy : Ay + By = 0
Ay = − By
or
(2)
ΣFz : 10 N + Bz = 20 N
Bz = 10 N
or
ΣM C : rB/C × B + rA/C × A = M C
Have
∴
or
(3)
i
j
k
i
j k
0.2 0 −0.05 + 0.2 0 0.2 N ⋅ m = ( 40 N ⋅ m ) i
Bx By 10
Ax Ay 10
( 0.05By − 0.2 Ax ) i + ( −0.05Bx − 2 + 0.2 Ax − 2) j
(
)
+ 0.2By + 0.2 Ay k = ( 40 N ⋅ m ) i
From
i - coefficient
0.05By − 0.2 Ay = 40 N ⋅ m
(4)
j - coefficient
− 0.05Bx + 0.2 Ax = 4 N ⋅ m
(5)
k - coefficient
0.2 By + 0.2 Ay = 0
(6)
PROBLEM 3.117 CONTINUED
From Equations (2) and (4):
(
)
0.05By − 0.2 − By = 40
By = 160 N, Ay = −160 N
From Equations (1) and (5):
−0.05 ( − Ax − 40 ) + 0.2 Ax = 4
Ax = 8 N
From Equation (1):
Bx = − ( 8 + 40 ) = −48 N
∴ A = ( 8 N ) i − (160 N ) j + (10 N ) k
B = − ( 48 N ) i + (160 N ) j + (10 N ) k
PROBLEM 3.118
While using a pencil sharpener, a student applies the forces and couple
shown. (a) Determine the forces exerted at B and C knowing that these
forces and the couple are equivalent to a force-couple system at A
consisting of the force R = ( 3.9 lb ) i + Ry j − (1.1 lb ) k and the couple
M RA = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k. . (b) Find the corresponding
values of Ry and M x .
SOLUTION
ΣF : B + C = R
Have
ΣFx : Bx + Cx = 3.9 lb
or
Bx = 3.9 lb − Cx
(1)
ΣFy : C y = Ry
(2)
ΣFz : C z = −1.1 lb
(3)
ΣM A : rB/ A × B + rC/ A × C + M B = M RA
Have
i j k
i
j
k
1
1
4 0 2.0 + ( 2 lb ⋅ ft ) i = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k
∴
x 0 4.5 +
12
12
Bx 0 0
C x C y −1.1
( 2 − 0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k
= M xi + (1.5 ) j − (1.1) k
From
i - coefficient
2 − 0.166667C y = M x
(4)
j - coefficient
0.375Bx + 0.166667Cx + 0.36667 = 1.5
(5)
k - coefficient
0.33333C y = −1.1
(6)
or
C y = −3.3 lb
(a) From Equations (1) and (5):
0.375 ( 3.9 − Cx ) + 0.166667Cx = 1.13333
Cx =
From Equation (1):
0.32917
= 1.58000 lb
0.20833
Bx = 3.9 − 1.58000 = 2.32 lb
∴ B = ( 2.32 lb ) i
C = (1.580 lb ) i − ( 3.30 lb ) j − (1.1 lb ) k
(b) From Equation (2):
From Equation (4):
Ry = C y = −3.30 lb
or R y = − ( 3.30 lb ) j
M x = −0.166667 ( −3.30 ) + 2.0 = 2.5500 lb ⋅ ft
or M x = ( 2.55 lb ⋅ ft ) i
PROBLEM 3.119
A portion of the flue for a furnace is attached to the ceiling at A. While
supporting the free end of the flue at F, a worker pushes in at E and pulls
out at F to align end E with the furnace. Knowing that the 10-lb force at
F lies in a plane parallel to the yz plane, determine (a) the angle α the
force at F should form with the horizontal if duct AB is not to tend to
rotate about the vertical, (b) the force-couple system at B equivalent to
the given force system when this condition is satisfied.
SOLUTION
(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:
(
)
R
M By
= j ⋅ ΣM RB = j ⋅ rF /B × FF + rE/B × FE = 0
where
rF /B = ( 45 in.) i − ( 23 in.) j + ( 28 in.) k
rE/B = ( 54 in.) i − ( 34 in.) j + ( 28 in.) k
FF = 10 lb ( sin α ) j + ( cos α ) k 
FE = − ( 5 lb ) k
∴
ΣM RB
i
j
k
i
j k
= (10 lb ) 45 in. −23 in. 28 in. + ( 5 lb )( 2 in.) 27 −17 14
0
sin α cos α
0 0 −1
= ( −230 cos α − 280sin α + 170 ) i − ( 450 cos α − 270 ) j + ( 450sin α ) k  lb ⋅ in.
Thus,
R
M By
= −450 cos α + 270 = 0
cos α = 0.60
α = 53.130°
or α = 53.1°
PROBLEM 3.119 CONTINUED
(b) R = FE + FF
where
FE = − ( 5 lb ) k
FF = (10 lb )( sin 53.130° j + cos 53.130°k ) = ( 8 lb ) j + ( 6 lb ) k
∴ R = ( 8 lb ) j + (1 lb ) k
and
M = ΣM RB = −  230 ( 0.6 ) + 280 ( 0.8 ) − 170  i −  450 ( 0.6 ) − 270 j +  450 ( 0.8 ) k
= − (192 lb ⋅ in.) i − ( 0 ) j + ( 360 lb ⋅ in.) k
or M = − (192 lb ⋅ in.) i + ( 360 lb ⋅ in.) k
PROBLEM 3.120
A portion of the flue for a furnace is attached to the ceiling at A. While
supporting the free end of the flue at F, a worker pushes in at E and pulls
out at F to align end E with the furnace. Knowing that the 10-lb force at
F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the
given force system with an equivalent force-couple system at C, (b)
determine whether duct CD will tend to rotate clockwise or
counterclockwise relative to elbow C, as viewed from D to C.
SOLUTION
R = ΣF = FF + FE
(a) Have
FF = 10 lb ( sin 60° ) j + ( cos 60° ) k  = (8.6603 lb ) j + ( 5.0 lb ) k
where
FE = − ( 5 lb ) k
∴ R = ( 8.6603 lb ) j
or R = ( 8.66 lb ) j
Have
M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE
where
rF /C = ( 9 in.) i − ( 2 in.) j
rE/C = (18 in.) i − (13 in.) j
∴
M CR
i
j
k
i
j k
= 9
−2
0 lb ⋅ in. + 18 −13 0 lb ⋅ in.
0 8.6603 5.0
0 0 −5
= ( 55 lb ⋅ in.) i + ( 45 lb ⋅ in.) j + ( 77.942 lb ⋅ in.) k
or M CR = ( 55.0 lb ⋅ in.) i + ( 45.0 lb ⋅ in.) j + ( 77.9 lb ⋅ in.) k
(b) To determine which direction duct section CD has a tendency to turn, have
R
M CD
= λ DC ⋅ M CR
where
λ DC =
Then
− (18 in.) i + ( 4 in.) j
2 85 in.
R
M CD
=
=
1
( −9i + 2 j)
85
1
( −9i + 2 j) ⋅ ( 55i + 45j + 77.942k ) lb ⋅ in.
85
= ( −53.690 + 9.7619 ) lb ⋅ in.
= −43.928 lb ⋅ in.
Since λ DC ⋅ M CR < 0, duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.
PROBLEM 3.121
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the chuck
and bit parallel to the y axis. The assembly was then rotated 25o about
the y axis and 20o about the centerline of the horizontal arm AB, bringing
it into the position shown. The drilling process was started by switching
on the motor and rotating the handle to bring the bit into contact with the
workpiece. Replace the force and couple exerted by the drill press with an
equivalent force-couple system at the center O of the base of the vertical
column.
SOLUTION
R =F
Have
= ( 44 N ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 
= (13.6389 N ) i − ( 41.346 N ) j − ( 6.3599 N ) k
or R = (13.64 N ) i − ( 41.3 N ) j − ( 6.36 N ) k
M O = rB/O × F + M C
Have
where
rB/O = ( 0.280 m ) sin 25°  i + ( 0.300 m ) j + ( 0.280 m ) cos 25°  k
= ( 0.118333 m ) i + ( 0.300 m ) j + ( 0.25377 m ) k
M C = ( 7.2 N ⋅ m ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 
= ( 2.2318 N ⋅ m ) i − ( 6.7658 N ⋅ m ) j − (1.04072 N ⋅ m ) k
∴ MO
i
j
k
= 0.118333 0.300 0.25377 N ⋅ m
13.6389 −41.346 −6.3599
+ ( 2.2318i − 6.7658 j − 1.04072k ) N ⋅ m
= (10.8162 N ⋅ m ) i − ( 2.5521 N ⋅ m ) j − (10.0250 N ⋅ m ) k
or M O = (10.82 N ⋅ m ) i − ( 2.55 N ⋅ m ) j − (10.03 N ⋅ m ) k
PROBLEM 3.122
While a sagging porch is leveled and repaired, a screw jack is used to
support the front of the porch. As the jack is expanded, it exerts on the
porch the force-couple system shown, where R = 300 N and
M = 37.5 N ⋅ m. Replace this force-couple system with an equivalent
force-couple system at C.
SOLUTION
 − ( 0.2 m ) i + (1.4 m ) j − ( 0.5 m ) k 
R C = R = ( 300 N ) λ AB = 300 N 

1.50 m


From
R C = − ( 40.0 N ) i + ( 280 N ) j − (100 N ) k
M C = rA/C × R + M
From
where
rA/C = ( 2.6 m ) i + ( 0.5 m ) k
 ( 0.2 m ) i − (1.4 m ) j + ( 0.5 m ) k 
M = ( 37.5 N ⋅ m ) λ BA = ( 37.5 N ⋅ m ) 

1.50 m


= ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k
∴ MC
i
j k
= (10 N ⋅ m ) 2.6 0 0.5 + ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k
−4 28 −10
= ( −140 + 5 ) N ⋅ m  i + ( −20 + 260 − 35 ) N ⋅ m  j + ( 728 + 12.5 ) N ⋅ m  k
or M C = − (135.0 N ⋅ m ) i + ( 205 N ⋅ m ) j + ( 741 N ⋅ m ) k
PROBLEM 3.123
Three children are standing on a 15 × 15-ft raft. If the weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively,
determine the magnitude and the point of application of the resultant of
the three weights.
SOLUTION
Have
ΣF : FA + FB + FC = R
− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j = R
− ( 235 lb ) j = R
Have
or R = 235 lb
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )
(85 lb)( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( zD )
∴ z D = 9.0957 ft
Have
or z D = 9.10 ft
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( xD )
∴ xD = 7.6915 ft
or xD = 7.69 ft
PROBLEM 3.124
Three children are standing on a 15 × 15-ft raft. The weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a
fourth child of weight 95 lb climbs onto the raft, determine where she
should stand if the other children remain in the positions shown and the
line of action of the resultant of the four weights is to pass through the
center of the raft.
SOLUTION
Have
ΣF : FA + FB + FC + FD = R
− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R
∴ R = − ( 330 lb ) j
Have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft )
∴ z D = 3.5523 ft
Have
or z D = 3.55 ft
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft )
∴ xD = 7.0263 ft
or xD = 7.03 ft
PROBLEM 3.125
The forces shown are the resultant downward loads on sections of the flat
roof of a building because of accumulated snow. Determine the
magnitude and the point of application of the resultant of these four
loads.
SOLUTION
Have
ΣF : FA + FB + FC + FD = R
− ( 580 kN ) j − ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j = R
∴ R = − ( 3400 kN ) j
Have
R = 3400 kN
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )
( 580 kN )(8 m ) + ( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) = ( 3400 kN )( zE )
∴ z E = 14.3853 m
Have
or z E = 14.39 m
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )
( 580 kN )(10 m ) + ( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) = ( 3400 kN )( xE )
∴ xE = 30.382 m
or xE = 30.4 m
PROBLEM 3.126
The forces shown are the resultant downward loads on sections of the flat
roof of a building because of accumulated snow. If the snow represented
by the 580-kN force is shoveled so that the this load acts at E, determine
a and b knowing that the point of application of the resultant of the four
loads is then at B.
SOLUTION
Have
ΣF : FB + FC + FD + FE = R
− ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j − ( 580 kN ) j = R
∴ R = − ( 3400 kN ) j
Have
ΣM x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )
( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) + ( 580 kN )( b ) = ( 3400 kN )(16 m )
∴ b = 17.4655 m
Have
or b = 17.47 m
ΣM z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )
( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) + ( 580 kN )( a ) = ( 3400 kN )( 32 m )
∴ a = 19.4828 m
or a = 19.48 m
PROBLEM 3.127
A group of students loads a 2 × 4-m flatbed trailer with two
0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R
Have
∴ R = − ( 780 N ) j
Have
ΣM z :
( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x )
∴ x = 1.34103 m
Have
ΣM x :
( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z )
∴ z = 0.93846 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
( 0.3 m ≤
x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m )
PROBLEM 3.127 CONTINUED
Let x = 0.3 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0
∴ W = 380 N
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0
∴ z = 3.5684 m < 3.7 m
Let z = 3.7 m,
∴ acceptable
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0
∴ W = 395.29 N > 380 N
Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the
smallest weight W.
or W = 380 N at
( 0.3 m, 0, 3.57 m )
PROBLEM 3.128
Solve Problem 3.127 if the students want to place as much weight as
possible in the fourth box and that at least one side of the box must
coincide with a side of the trailer.
Problem 3.127: A group of students loads a 2 × 4-m flatbed trailer with
two 0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)
SOLUTION
For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of
the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box
touching the z-axis, satisfies this condition.
Let x = 0.6 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0
∴ W = 665 N
and
ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0
∴ z = 2.8105 m
(2 m <
z < 4 m)
∴ acceptable
or W = 665 N at
( 0.6 m, 0, 2.81 m )
PROBLEM 3.129
A block of wood is acted upon by three forces of the same magnitude P
and having the directions shown. Replace the three forces with an
equivalent wrench and determine (a) the magnitude and direction of the
resultant R, (b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the xy plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
Have
ΣF : Pi − Pi − Pk = R
∴ R = − Pk
Have
ΣM O : − P ( 3a ) k − P ( 3a ) j + P ( −ai + 3aj) = M OR
∴ M OR = Pa ( −i − 3k )
Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same.
R = − Pk
(a)
or Magnitude of R = P
Direction of R : θ x = 90°, θ y = 90°, θ z = −180°
(b) Have
M1 = λ R ⋅ M OR
= −k ⋅  Pa ( −i − 3k ) 
= 3Pa
and pitch
p=
M1 3Pa
=
= 3a
R
P
or p = 3a
PROBLEM 3.129 CONTINUED
(c) Have
M OR = M1 + M 2
∴ M 2 = M OR − M1 = Pa ( −i − 3k ) − ( −3Pak ) = − Pai
Require
M 2 = rQ/O × R
− Pai = ( xi + yj) × ( − P ) k = Pxj − Pyi
From
i : − Pa = − Py
or
y =a
j: x = 0
∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at x = 0, y = a
PROBLEM 3.130
A piece of sheet metal is bent into the shape shown and is acted upon by
three forces. Replace the three forces with an equivalent wrench and
determine (a) the magnitude and direction of the resultant R, (b) the pitch
of the wrench, (c) the point where the axis of the wrench intersects the
yz plane.
SOLUTION
First, reduce the given force system to a force-couple system at the origin.
( 2P ) i − ( P ) j + ( P ) j = R
ΣF :
Have
∴ R = ( 2P ) i
ΣM O : Σ ( rO × F ) = M OR
Have
M OR
i j k
i j k
= Pa 2 2 2.5 + 0 0 4 = Pa ( −1.5i + 5j − 6k )
2 −1 0
0 1 0
R = 2 Pi
(a)
or Magnitude of R = 2 P
Direction of R : θ x = 0°, θ y = −90°, θ z = 90°
(b) Have
M1 = λ R ⋅ M OR
λR =
R
R
= i ⋅ ( −1.5Pai + 5Paj − 6 Pak )
= −1.5Pa
and pitch
p=
M1
−1.5Pa
=
= −0.75a
R
2P
or p = −0.75a
PROBLEM 3.130 CONTINUED
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( 5Pa ) j − ( 6Pa ) k
Require
M 2 = rQ/O × R
( 5Pa ) j − ( 6Pa ) k = ( yj + zk ) × ( 2Pi ) = − ( 2Py ) k + ( 2Pz ) j
From
i : 5Pa = 2 Pz
∴ z = 2.5a
From
k : − 6 Pa = −2 Py
∴ y = 3a
∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at y = 3a, z = 2.5a
PROBLEM 3.131
The forces and couples shown are applied to two screws as a piece of
sheet metal is fastened to a block of wood. Reduce the forces and the
couples to an equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the wrench
intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
ΣF : − (10 N ) j − (11 N ) j = R
Have
∴ R = − ( 21 N ) j
ΣM O : Σ ( rO × F ) + ΣM C = M OR
Have
M OR
i j
k
i j
k
= 0 0 0.5 N ⋅ m + 0 0 −0.375 N ⋅ m − (12 N ⋅ m ) j
0 −10 0
0 −11
0
= ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
R = − ( 21 N ) j
(a)
(b) Have
M1 = λ R ⋅ M OR
λR =
or R = − ( 21 N ) j
R
R
= ( − j) ⋅ ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
= 12 N ⋅ m
and pitch
p=
and
M1 = − (12 N ⋅ m ) j
M1 12 N ⋅ m
=
= 0.57143 m
R
21 N
or p = 0.571 m
PROBLEM 3.131 CONTINUED
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( 0.875 N ⋅ m ) i
M 2 = rQ/O × R
Require
∴
( 0.875 N ⋅ m ) i = ( xi + zk ) × − ( 21 N ) j
0.875i = − ( 21x ) k + ( 21z ) i
From i:
0.875 = 21z
∴ z = 0.041667 m
From k:
0 = −21x
∴ z =0
∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 41.7 mm
PROBLEM 3.132
The forces and couples shown are applied to two screws as a piece of
sheet metal is fastened to a block of wood. Reduce the forces and the
couples to an equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the wrench
intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple system.
Have
Have
ΣF : − ( 6 lb ) i − ( 4.5 lb ) j = R
R = 7.5 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
M OR = −6 lb ( 8 in.) j − (160 lb ⋅ in.) i − ( 72 lb ⋅ in.) j
= − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j
M OR = 200 lb ⋅ in.
R = − ( 6 lb ) i − ( 4.5 lb ) j
(a)
(b) Have
M1 = λ R ⋅ M OR
λ =
R
R
= ( −0.8i − 0.6 j) ⋅  − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j
= 200 lb ⋅ in.
M1 = 200 lb ⋅ in. ( −0.8i − 0.6j)
and
Pitch
p=
200 lb ⋅ in.
M1
=
= 26.667 in.
7.50 lb
R
or p = 26.7 in.
(c) From above note that
M1 = M OR
Therefore, the axis of the wrench goes through the origin. The line
of action of the wrench lies in the xy plane with a slope of
dy
3
=
dx
4
PROBLEM 3.133
Two bolts A and B are tightened by applying the forces and couple
shown. Replace the two wrenches with a single equivalent wrench and
determine (a) the resultant R, (b) the pitch of the single equivalent
wrench, (c) the point where the axis of the wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
Have
ΣF : − ( 20 lb ) k − ( 21 lb ) j = − ( 21 lb ) j − ( 20 lb ) k = R
R = 29 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
and
i j k
i j k
20 lb ( 4 in.) 4 3 0 + 21 lb ( 4 in.) 6 0 1 + ( −300 j − 320k ) lb ⋅ in. = M OR
0 0 −1
0 −1 0
∴ M OR = − (156 lb ⋅ in.) i + ( 20 lb ⋅ in.) j − ( 824 lb ⋅ in.) k
R = − ( 21 lb ) j − ( 20 lb ) k
(a)
(b) Have
M1 = λ R ⋅ M OR
=−
λR =
R
R
−21j − 20k
⋅  − (156 lb ⋅ in.) i + ( 20 lb ⋅ in.) j − ( 824 lb ⋅ in.) k 
29
= 553.80 lb ⋅ in.
PROBLEM 3.133 CONTINUED
M1 = M1λ R = − ( 401.03 lb ⋅ in.) j − ( 381.93 lb ⋅ in.) k
and
Then pitch
p=
M1 553.80 lb ⋅ in.
=
= 19.0964 in.
29 lb
R
or p = 19.10 in.
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( −156i + 20 j − 824k ) − ( −401.03j − 381.93k )  lb ⋅ in.
= − (156.0 lb ⋅ in.) i + ( 421.03 lb ⋅ in.) j − ( 442.07 lb ⋅ in.) k
M 2 = rQ/O × R
Require
( −156i + 421.03j − 442.07k ) = ( xi + zk ) × ( −21j − 20k )
= ( 21z ) i + ( 20 x ) j − ( 21x ) k
From i:
−156 = 21z
∴ z = −7.4286 in.
or
From k:
z = −7.43 in.
−442.07 = −21x
∴ x = 21.051 in.
or
x = 21.1 in.
∴ The axis of the wrench intersects the xz-plane at
x = 21.1 in., z = −7.43 in.
PROBLEM 3.134
Two bolts A and B are tightened by applying the forces and couple
shown. Replace the two wrenches with a single equivalent wrench and
determine (a) the resultant R, (b) the pitch of the single equivalent
wrench, (c) the point where the axis of the wrench intersects the xz plane.
SOLUTION
First reduce the given force system to a force-couple at the origin at B.
15 
 8
j = R
ΣF : − ( 79.2 lb ) k − ( 51 lb )  i +
17 
 17
(a) Have
∴ R = − ( 24.0 lb ) i − ( 45.0 lb ) j − ( 79.2 lb ) k
R = 94.2 lb
and
ΣM B : rA/B × FA + M A + M B = M RB
Have
M RB
i j
k
15 
 8
= 0 −20
j  = 1584i − 660k − 42 ( 8i + 15 j)
0 − 660k − 714  i +
17 
 17
0 0 −79.2
∴ M RB = (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k
(b) Have
M1 = λ R ⋅ M OR
=
λR =
R
R
−24.0i − 45.0 j − 79.2k
⋅ (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k 
94.2
= 537.89 lb ⋅ in.
PROBLEM 3.134 CONTINUED
M1 = M1λ R
and
= − (137.044 lb ⋅ in.) i − ( 256.96 lb ⋅ in.) j − ( 452.24 lb ⋅ in.) k
Then pitch
p=
M1 537.89 lb ⋅ in.
=
= 5.7101 in.
94.2 lb
R
or p = 5.71 in.
M RB = M1 + M 2
(c) Have
∴ M 2 = M RB − M1 = (1248i − 630 j − 660k ) − ( −137.044i − 256.96 j − 452.24k )
= (1385.04 lb ⋅ in.) i − ( 373.04 lb ⋅ in.) j − ( 207.76 lb ⋅ in.) k
M 2 = rQ/B × R
Require
i
j
k
1385.04i − 373.04 j − 207.76k = x
0
z
−24 −45 −79.2
= ( 45 z ) i − ( 24 z ) j + ( 79.2 x ) j − ( 45 x ) k
From i:
From k:
1385.04 = 45 z
−207.76 = −45x
∴ z = 30.779 in.
∴ x = 4.6169 in.
∴ The axis of the wrench intersects the xz-plane at
x = 4.62 in., z = 30.8 in.
PROBLEM 3.135
A flagpole is guyed by three cables. If the tensions in the cables have the
same magnitude P, replace the forces exerted on the pole with an
equivalent wrench and determine (a) the resultant force R, (b) the pitch of
the wrench, (c) the point where the axis of the wrench intersects the xz
plane.
SOLUTION
(a) First reduce the given force system to a force-couple at the origin.
ΣF : Pλ BA + Pλ DC + Pλ DE = R
Have
 4
3  3
4   −9
4
12  
R = P  j − k  +  i − j  +  i − j +
k
5
5
5
5
25
5
25  
 
 

∴ R =
R=
3P
25
( 2 )2 + ( 20 )2 + (1)2
=
3P
( 2i − 20 j − k )
25
27 5
P
25
ΣM : Σ ( rO × P ) = M OR
Have
−4 P
3P 
4P 
4P
12 P 
 3P
 −9 P
j−
k  + ( 20a ) j × 
i−
j  + ( 20a ) j × 
i−
j+
k  = M OR
5 
5 
5
25 
 5
 5
 25
( 24a ) j × 
∴ M OR =
M1 = λ R ⋅ M OR
(b) Have
where
24 Pa
( −i − k )
5
λR =
3P
25
1
R
=
=
( 2i − 20 j − k )
( 2i − 20 j − k )
R
25
27 5 P 9 5
PROBLEM 3.135 CONTINUED
Then
M1 =
and pitch
p=
M1 = M 1λ R =
(c)
Then
1
9 5
( 2i − 20 j − k ) ⋅
24 Pa
−8Pa
( −i − k ) =
5
15 5
M1
−8Pa  25  −8a
=

=
R
81
15 5  27 5 P 
or p = −0.0988a
−8Pa  1 
8Pa
( −2i + 20 j + k )

 ( 2i − 20 j − k ) =
675
15 5  9 5 
M 2 = M OR − M1 =
24Pa
8Pa
8Pa
( −i − k ) −
( −2i + 20 j + k ) =
( −403i − 20 j − 406k )
5
675
675
M 2 = rQ/O × R
Require
 8Pa 
 3P 

 ( −403i − 20 j − 406k ) = ( xi + zk ) × 
 ( 2i − 20 j − k )
675


 25 
 3P 
=
  20 zi + ( x + 2 z ) j − 20 xk 
 25 
From i:
8 ( −403)
Pa
 3P 
= 20 z 

675
 25 
∴ z = −1.99012a
From k:
8 ( −406 )
Pa
 3P 
= −20 x 

675
 25 
∴ x = 2.0049a
∴ The axis of the wrench intersects the xz-plane at
x = 2.00a, z = −1.990a
PROBLEM 3.136
Determine whether the force-and-couple system shown can be reduced to
a single equivalent force R. If it can, determine R and the point where the
line of action of R intersects the yz plane. If it cannot be so reduced,
replace the given system with an equivalent wrench and determine its
resultant, its pitch, and the point where its axis intersects the yz plane.
SOLUTION
First, reduce the given force system to a force-couple at D.
Have
ΣF : FDA + FED = FDAλ DA + FEDλ ED = R
where
 − ( 0.300 m ) i + ( 0.225 m ) j + ( 0.200 m ) k 
FDA = 136 N 

0.425 m


= − ( 96 N ) i + ( 72 N ) j + ( 64 N ) k
 − ( 0.150 m ) i − ( 0.200 m ) k 
FED = 120 N 
 = − ( 72 N ) i − ( 96 N ) k
0.250 m


∴ R = − (168 N ) i + ( 72 N ) j − ( 32 N ) k
Have
or
R
ΣM D : M A = M D
 − ( 0.150 m ) i − ( 0.150 m ) j + ( 0.450 m ) k  16 N ⋅ m
M RD = (16 N ⋅ m ) 
( −i − j + 3k )
 =
0.150 11 m
11


PROBLEM 3.136 CONTINUED
The force-couple at D can be replaced by a single force if R is perpendicular to M RD . To be perpendicular,
R ⋅ M RD = 0.
Have
R ⋅ M RD = ( −168i + 72 j − 32k ) ⋅
=
16
( −i − j + 3k )
11
128
( 21 − 9 − 12 )
11
=0
∴ Force-couple can be reduced to a single equivalent force.
To determine the coordinates where the equivalent single force intersects the yz-plane, M RD = rQ/D × R
where
rQ/D = ( 0 − 0.300 ) m  i + ( y − 0.075 ) m  j + ( z − 0 ) m  k
i
16 N ⋅ m
∴
( −i − j + 3k ) = (8 N ) −0.3
11
−21
j
k
( y − 0.075) z m
9
−4
or
16 N ⋅ m
( −i − j + 3k ) = (8 N ) −4 ( y − 0.075) − 9 z  i + ( −21z − 1.2 ) j + −2.7 + 21( y − 0.075) k m
11
{
From j:
From k:
−16
= 8 ( −21z − 1.2 )
11
48
= 8  −2.7 + 21( y − 0.075 ) 
11
}
∴ z = −0.028427 m = −28.4 mm
∴ y = 0.28972 m = 290 mm
∴ line of action of R intersects the yz-plane at
y = 290 mm, z = −28.4 mm
PROBLEM 3.137
Determine whether the force-and-couple system shown can be reduced to
a single equivalent force R. If it can, determine R and the point where the
line of action of R intersects the yz plane. If it cannot be so reduced,
replace the given system with an equivalent wrench and determine its
resultant, its pitch, and the point where its axis intersects the yz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
ΣF : FA + FG = R
Have
 ( 4 in.) i + ( 6 in.) j − (12 in.) k 
∴ R = (10 lb ) k + 14 lb 
 = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k
14 in.


R=
and
Have
56 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
{
}
M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k 
 (16 in.) i − (12 in.) j 
 ( 4 in.) i − (12 in.) j + ( 6 in.) k 
+ ( 84 lb ⋅ in.) 
 + ( 120 lb ⋅ in.) 

20 in.
14 in.




∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k
= (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k
PROBLEM 3.137 CONTINUED
The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR .
To be perpendicular R ⋅ M OR = 0
Have
R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0?
= 73.829 + 19.3716 − 24.572
≠0
∴ System cannot be reduced to a single equivalent force.
To reduce to an equivalent wrench, the moment component along the line of action of P is found.
M1 = λ R ⋅ M OR
λR =
R
R
 ( 4i + 6 j − 2k ) 
=
 ⋅ (18.4572i + 3.2286 j + 12.2858k )
56


= 9.1709 lb ⋅ ft
M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k )
and
And pitch
p=
M1 9.1709 lb ⋅ ft
=
= 1.22551 ft
R
56 lb
or p = 1.226 ft
Have
M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k )
= (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k
Require
M 2 = rQ/O × R
(13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6j − 2k )
= − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k
From j:
−4.1244 = 4z
From k:
14.7368 = −4 y
or
or
z = −1.0311 ft
y = −3.6842 ft
∴ line of action of the wrench intersects the yz plane at
y = −3.68 ft, z = 1.031 ft
PROBLEM 3.138
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y axis and applied respectively at A and B.
SOLUTION
Express the forces at A and B as
A = Axi + Az k
B = Bxi + Bzk
Then, for equivalence to the given force system
ΣFx : Ax + Bx = 0
(1)
ΣFz : Az + Bz = R
(2)
ΣM x : Az ( a ) + Bz ( a + b ) = 0
(3)
ΣM z : − Ax ( a ) − Bx ( a + b ) = M
(4)
Bx = − Ax
From Equation (1),
Substitute into Equation (4)
− Ax ( a ) + Ax ( a + b ) = M
∴ Ax =
From Equation (2),
and Equation (3),
M
b
and
Bx = −
M
b
Bz = R − Az
Az a + ( R − Az )( a + b ) = 0
a

∴ Az = R 1 + 
b

PROBLEM 3.138 CONTINUED
and
a

Bz = R − R 1 + 
b

a
∴ Bz = − R
b
Then
M
A=
 b
a


 i + R 1 +  k
b


M
B = −
 b

a 
i −  Rk

b 
PROBLEM 3.139
Show that, in general, a wrench can be replaced with two forces chosen in
such a way that one force passes through a given point while the other
force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rxi + Ry j + Rzk
M = M xi + M y j + M zk
and
while the unknown forces A and B can be expressed as
A = Axi + Ay j + Azk
and
B = Bxi + Bzk
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position
vector rP are also known.
Then, for equivalence of the two systems
ΣFx : Rx = Ax + Bx
(1)
ΣFy : Ry = Ay
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
PROBLEM 3.139 CONTINUED
(
)
Based on the above six independent equations for the six unknowns Ax , Ay , Az , Bx , Bz , b , there exists a
unique solution for A and B.
Ay = Ry
From Equation (2)
Equation (6)
1
Ax =   xRy − M z
 y
)
Equation (1)
1
Bx = Rx −   xRy − M z
 y
(
)
Equation (4)
1
Az =   M x + zRy
 y
)
Equation (3)
1
Bz = Rz −   M x + zRy
 y
)
Equation (5)
(
(
(
b=
( xM x + yM y + zM z )
( M x − yRz + zRy )
PROBLEM 3.140
Show that a wrench can be replaced with two perpendicular forces, one
of which is applied at a given point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
Have
R = Rj
and
M = Mj
and are known.
The unknown forces A and B can be expressed as
A = Axi + Ay j + Azk
and
B = Bxi + By j + Bzk
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence
∑ Fx : 0 = Ax + Bx
(1)
∑ Fy : R = Ay + By
(2)
∑ Fz : 0 = Az + Bz
(3)
∑ M x : 0 = − zBy
(4)
∑ M y : M = −aAz − xBz + zBx
(5)
∑ M z : 0 = aAy + xBy
(6)
Since A and B are made perpendicular,
A⋅B = 0
There are eight unknowns:
or
Ax Bx + Ay By + Az Bz = 0
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
(7)
PROBLEM 3.140 CONTINUED
0 = − zBy
Next consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3) this equation becomes
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
By ≠ 0, so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become.
0 = Bx
Then
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = −aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0
(7)′
Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,
 R − By 
M = −aAz −  −a
 ( − Az )

By 

or
Az = −
M
By
aR
Substituting into Equation (7)′,
M
 M

By 
By  = 0
( R − By ) By +  − aR
 aR 
(8)
PROBLEM 3.140 CONTINUED
By =
or
a 2 R3
a R2 + M 2
2
Then from Equations (2), (8), and (3)
Ay = R −
Az = −
Bz =
a 2 R3
RM 2
=
a2R2 + M 2
a2R2 + M 2

M 
a 2 R3
aR 2 M
=
−
 2 2

aR  a R + M 2 
a2R2 + M 2
aR 2 M
a2R2 + M 2
In summary
A=
RM
( Mj − aRk )
a R2 + M 2
B=
aR 2
( aRj + Mk )
a2R2 + M 2
2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a
given point.
Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0.
From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane
parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to
the left of the origin, as shown in the figure below.
PROBLEM 3.141
Show that a wrench can be replaced with two forces, one of which has a
prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action ( AA′ ) . Note that it has been assumed that the line of
action of force B intersects the xz plane at point P ( x, 0, z ) . Denoting the known direction of line AA′ by
λ A = λxi + λ y j + λzk
it follows that force A can be expressed as
(
A = Aλ A = A λxi + λ y j + λz k
)
Force B can be expressed as
B = Bxi + By j + Bzk
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows
that the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
(
ΣFx : 0 = Aλx + Bx
(1)
ΣFy : R = Aλ y + By
(2)
ΣFz : 0 = Aλz + Bz
(3)
ΣM x : 0 = − zBy
(4)
ΣM y : M = −aAλz + zBx − xBz
(5)
ΣM z : 0 = aAλ y + xBy
(6)
)
Since there are six unknowns A, Bx , By , Bz , x, z and six independent equations, it will be possible to
obtain a solution.
PROBLEM 3.141 CONTINUED
Case 1: Let z = 0 to satisfy Equation (4)
Aλ y = R − By
Now Equation (2)
Bz = − Aλz
Equation (3)
x=−
Equation (6)
aAλ y
By
 a 
= −
 R − By
 By 


(
)
Substitution into Equation (5)
  a 

M = −aAλz −  − 
 R − By ( − Aλz ) 
  By 

(
∴ A=−
1 M 
λz  aR 
)
By
Substitution into Equation (2)
R=−
1 M 
B λ + By
λz  aR  y y
∴ By =
Then
A=−
λz aR 2
λz aR − λ y M
MR
R
=
aR
λz aR − λ y M
λy −
λz
M
Bx = − Aλx =
λx MR
λz aR − λ y M
Bz = − Aλz =
λz MR
λz aR − λ y M
In summary
A=
B=
and
P
λA
aR
λy −
λz
M
R
( λ Mi + λz aRj + λz Mk )
λz aR − λ y M x


 λz aR − λ y M
R 
x = a 1 −
 = a 1 − R 

By 
λz aR 2




 
 
or x =
Note that for this case, the lines of action of both A and B intersect the x axis.
λy M
λz R
PROBLEM 3.141 CONTINUED
Case 2: Let By = 0 to satisfy Equation (4)
A=
Now Equation (2)
R
λy
Equation (1)
λ
Bx = − R  x
 λy





Equation (3)
λ
Bz = − R  z
 λy





aAλ y = 0
Equation (6)
which requires a = 0
Substitution into Equation (5)
 λ
M = z −R  x
  λ y
  λ 

 − x −R  z 

  λ y  

or
This last expression is the equation for the line of action of force B.
In summary
 R
A=
 λy

 R
B=
 λy


λ A



 ( −λ x i − λ z k )


Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below.
Note that the component of A in the xz plane is parallel to B.
M 
λz x − λx z =   λ y
 R
PROBLEM 3.142
A worker tries to move a rock by applying a 360-N force to a steel bar as
shown. (a) Replace that force with an equivalent force-couple system at
D. (b) Two workers attempt to move the same rock by applying a vertical
force at A and another force at D. Determine these two forces if they are
to be equivalent to the single force of part a.
SOLUTION
(a) Have
ΣF : 360 N ( − sin 40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j = F
or F = 360 N
ΣM D : rB/D × R = M
Have
where
50°
rB/D = − ( 0.65 m ) cos 30°  i + ( 0.65 m ) sin 30°  j
= − ( 0.56292 m ) i + ( 0.32500 m ) j
i
j
k
∴ M = −0.56292 0.32500 0 N ⋅ m = (155.240 + 75.206 ) N ⋅ m  k
−231.40 −275.78 0
= ( 230.45 N ⋅ m ) k
(b) Have
where
or M = 230 N ⋅ m
ΣM D : M = rA/D × FA
rA/D = − (1.05 m ) cos 30° i + (1.05 m ) sin 30°  j
= − ( 0.90933 m ) i + ( 0.52500 m ) j
PROBLEM 3.142 CONTINUED
i
j
k
∴ FA −0.90933 0.52500 0 N ⋅ m = [ 230.45 N ⋅ m ] k
0
−1
0
( 0.90933FA ) k
or
= 230.45k
∴ FA = 253.42 N
or FA = 253 N
ΣF : F = FA + FD
Have
− ( 231.40 N ) i − ( 275.78 N ) j = − ( 253.42 N ) j + FD ( − cosθ i − sin θ j)
From
i : 231.40 N = FD cosθ
(1)
j: 22.36 N = FD sin θ
(2)
Equation (2) divided by Equation (1)
tan θ = 0.096629
∴ θ = 5.5193°
or
θ = 5.52°
Substitution into Equation (1)
FD =
231.40
= 232.48 N
cos5.5193°
or FD = 232 N
5.52°
PROBLEM 3.143
A worker tries to move a rock by applying a 360-N force to a steel bar as
shown. If two workers attempt to move the same rock by applying a force
at A and a parallel force at C, determine these two forces so that they will
be equivalent to the single 360-N force shown in the figure.
SOLUTION
ΣF : R = FA + FC
Have
− ( 360 N ) sin 40° i − ( 360 N ) cos 40° j = − ( FA + FC ) sin θ  i − ( FA + FC ) cosθ  j
From
i:
( 360 N ) sin 40° = ( FA + FC ) sin θ
(1)
j:
( 360 N ) cos 40° = ( FA + FC ) cosθ
(2)
Dividing Equation (1) by Equation (2),
tan 40° = tan θ
∴ θ = 40°
Substituting θ = 40° into Equation (1),
FA + FC = 360 N
Have
where
ΣM C : rB/C × R = rA/C × FA
rB/C = ( 0.35 m )( −cos30°i + sin 30° j) = − ( 0.30311 m ) i + ( 0.175 m ) j
(3)
PROBLEM 3.143 CONTINUED
R = ( 360 N )( −sin40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j
rA/C = ( 0.75 m )( −cos30°i + sin 30 j) = − ( 0.64952 m ) i + ( 0.375 m ) j
FA = FA ( − sin 40°i − cos 40° j) = FA ( −0.64279i − 0.76604 j)
∴
i
j
k
i
j
k
−0.30311 0.175 0 N ⋅ m = FA −0.64952 0.375 0 N ⋅ m
−231.40 −275.78 0
−0.64279 −0.76604 0
83.592 + 40.495 = ( 0.49756 + 0.24105 ) FA
∴ FA = 168.002 N
or
FA = 168.0 N
Substituting into Equation (3),
FC = 360 − 168.002 = 191.998 N
or
FC = 192.0 N
or FA = 168.0 N
50°
FC = 192.0 N
50°
PROBLEM 3.144
A force and a couple are applied as shown to the end of a cantilever
beam. (a) Replace this system with a single force F applied at point C,
and determine the distance d from C to a line drawn through points D and
E. (b) Solve part a if the directions of the two 360-N forces are reversed.
SOLUTION
(a)
(a) Have
ΣF : F = ( 360 N ) j − ( 360 N ) j − ( 600 N ) k
or F = − ( 600 N ) k
and
ΣM D :
( 360 N )( 0.15 m ) = ( 600 N )( d )
∴ d = 0.09 m
or d = 90.0 mm below ED
F = − ( 600 N ) k
(b) Have from part a
(b)
and
ΣM D : − ( 360 N )( 0.15 m ) = − ( 600 N )( d )
∴ d = 0.09 m
or d = 90.0 mm above ED
PROBLEM 3.145
A crate of mass 80 kg is held in the position shown. Determine (a) the
moment produced by the weight W of the crate about E, (b) the smallest
force applied at B which creates a moment of equal magnitude and
opposite sense about E.
SOLUTION
(
)
W = mg = 80 kg 9.81 m/s 2 = 784.8 N
(a) By definition
ΣM E : M E = ( 784.8 N )( 0.25 m )
Have
∴ M E = 196.2 N ⋅ m
(b) For the force at B to be the smallest, resulting in a moment ( M E )
about E, the line of action of force FB must be perpendicular to the
line connecting E to B. The sense of FB must be such that the force
produces a counterclockwise moment about E.
Note:
Have
d =
( 0.85 m )2 + ( 0.5 m )2
= 0.98615 m
ΣM E : 196.2 N ⋅ m = FB ( 0.98615 m )
∴ FB = 198.954 N
and
 0.85 m 
θ = tan −1 
 = 59.534°
 0.5 m 
or FB = 199.0 N
59.5°
PROBLEM 3.146
A crate of mass 80 kg is held in the position shown. Determine (a) the
moment produced by the weight W of the crate about E, (b) the smallest
force applied at A which creates a moment of equal magnitude and
opposite sense about E, (c) the magnitude, sense, and point of application
on the bottom of the crate of the smallest vertical force which creates a
moment of equal magnitude and opposite sense about E.
SOLUTION
(
)
W = mg = 80 kg 9.81 m/s 2 = 784.8 N
(a) By definition
ΣM E : M E = ( 784.8 N )( 0.25 m )
Have
∴ M E = 196.2 N ⋅ m
(b) For the force at A to be the smallest, resulting in a moment about E,
the line of action of force FA must be perpendicular to the line
connecting E to A. The sense of FA must be such that the force
produces a counterclockwise moment about E.
Note:
Have
d =
( 0.35 m )2 + ( 0.5 m )2
= 0.61033 m
ΣM E : 196.2 N ⋅ m = FA ( 0.61033 m )
∴ FA = 321.47 N
and
 0.35 m 
θ = tan −1 
 = 34.992°
 0.5 m 
or FA = 321 N
35.0°
(c) The smallest force acting on the bottom of the crate resulting in a
moment about E will be located at the point on the bottom of the
crate farthest from E and acting perpendicular to line CED. The
sense of the force will be such as to produce a counterclockwise
moment about E. A force acting vertically upward at D satisfies
these conditions.
PROBLEM 3.146 CONTINUED
Have
ΣM E : M E = rD/E × FD
(196.2 N ⋅ m ) k = ( 0.85 m ) i × ( FD ) j
(196.2 N ⋅ m ) k = ( 0.85FD ) k
∴ FD = 230.82 N
or FD = 231 N
PROBLEM 3.147
A farmer uses cables and winch pullers B and E to plumb one side of a
small barn. Knowing that the sum of the moments about the x axis of the
forces exerted by the cables on the barn at points A and D is equal to
4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found
using the z-components of each force acting at their intersection with the
xy-plane (A and D). The x-components of the forces are parallel to the xaxis, and the y-components of the forces intersect the x-axis. Therefore,
neither the x or y components produce a moment about the x-axis.
ΣM x :
Have
(TAB ) z
where
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
= k ⋅ TAB = k ⋅ (TABλ AB )

 −i − 12 j + 12k  
= k ⋅  255 lb 
  = 180 lb
17



(TDE ) z
= k ⋅ TDE = k ⋅ (TDE λ DE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 
  = 0.64865TDE
18.5



y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
∴
and
(180 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ⋅ ft
TDE = 282.79 lb
or TDE = 283 lb
PROBLEM 3.148
Solve Problem 3.147 when the tension in cable AB is 306 lb.
Problem 3.147: A farmer uses cables and winch pullers B and E to
plumb one side of a small barn. Knowing that the sum of the moments
about the x axis of the forces exerted by the cables on the barn at points A
and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when
TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found
using the z components of each force acting at the intersection with the xy
plane (A and D). The x components of the forces are parallel to the x axis,
and the y components of the forces intersect the x axis. Therefore, neither
the x or y components produce a moment about the x axis.
ΣM x :
Have
(TAB ) z
where
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
= k ⋅ TAB = k ⋅ (TABλ AB )

 −i − 12 j + 12k  
= k ⋅ 306 lb 
  = 216 lb
17



(TDE ) z
= k ⋅ TDE = k ⋅ (TDE λ DE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 
  = 0.64865TDE
18.5



y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
∴
and
( 216 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ⋅ ft
TDE = 235.21 lb
or TDE = 235 lb
PROBLEM 3.149
As an adjustable brace BC is used to bring a wall into plumb, the forcecouple system shown is exerted on the wall. Replace this force-couple
system with an equivalent force-couple system at A knowing that
R = 21.2 lb and M = 13.25 lb ⋅ ft.
SOLUTION
ΣF : R = R A = Rλ BC
Have
λ BC =
where
∴ RA =
( 42 in.) i − ( 96 in.) j − (16 in.) k
106 in.
21.2 lb
( 42i − 96 j − 16k )
106
or R A = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k
ΣM A : rC/ A × R + M = M A
Have
where
rC/ A = ( 42 in.) i + ( 48 in.) k =
1
( 42i + 48k ) ft
12
= ( 3.5 ft ) i + ( 4.0 ft ) k
R = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k
M = −λ BC M
=
−42i + 96 j + 16k
(13.25 lb ⋅ ft )
106
= − ( 5.25 lb ⋅ ft ) i + (12 lb ⋅ ft ) j + ( 2 lb ⋅ ft ) k
PROBLEM 3.149 CONTINUED
Then
i
j
k
3.5
0
4.0 lb ⋅ ft + ( −5.25i + 12 j + 2k ) lb ⋅ ft = M A
8.40 −19.20 −3.20
∴ M A = ( 71.55 lb ⋅ ft ) i + ( 56.80 lb ⋅ ft ) j − ( 65.20 lb ⋅ ft ) k
or M A = ( 71.6 lb ⋅ ft ) i + ( 56.8 lb ⋅ ft ) j − ( 65.2 lb ⋅ ft ) k
PROBLEM 3.150
Two parallel 60-N forces are applied to a lever as shown. Determine the
moment of the couple formed by the two forces (a) by resolving each
force into horizontal and vertical components and adding the moments of
the two resulting couples, (b) by using the perpendicular distance
between the two forces, (c) by summing the moments of the two forces
about point A.
SOLUTION
(a) Have
where
ΣM B : − d1Cx + d 2C y = M
d1 = ( 0.360 m ) sin 55° = 0.29489 m
d 2 = ( 0.360 m ) cos 55° = 0.20649 m
Cx = ( 60 N ) cos 20° = 56.382 N
C y = ( 60 N ) sin 20° = 20.521 N
∴ M = − ( 0.29489 m )( 56.382 N ) k + ( 0.20649 m )( 20.521 N ) k = − (12.3893 N ⋅ m ) k
or M = 12.39 N ⋅ m
(b) Have
M = Fd ( −k ) = 60 N ( 0.360 m ) sin ( 55° − 20° )  ( −k )
= − (12.3893 N ⋅ m ) k
or M = 12.39 N ⋅ m
PROBLEM 3.150 CONTINUED
(c) Have
ΣM A : Σ ( rA × F ) = rB/ A × FB + rC/ A × FC = M
i
j
k
i
j
k
∴ M = ( 0.520 m )( 60 N ) cos 55° sin 55° 0 + ( 0.880 m )( 60 N ) cos 55° sin 55° 0
− cos 20° − sin 20° 0
cos 20° sin 20° 0
= (17.8956 N ⋅ m − 30.285 N ⋅ m ) k = − (12.3892 N ⋅ m ) k
or M = 12.39 N ⋅ m
PROBLEM 3.151
A 32-lb motor is mounted on the floor. Find the resultant of the weight
and the forces exerted on the belt, and determine where the line of action
of the resultant intersects the floor.
SOLUTION
ΣF :
Have
( 60 lb ) i − ( 32 lb ) j + (140 lb )( cos 30°i + sin 30°j) = R
∴ R = (181.244 lb ) i + ( 38.0 lb ) j
or R = 185.2 lb
11.84°
ΣM O : ΣM O = xRy
Have
∴
− (140 lb ) cos 30° ( 4 + 2 cos 30° ) in. − (140 lb ) sin 30° ( 2 in.) sin 30°
− ( 60 lb )( 2 in.) = x ( 38.0 lb )
x=
and
1
( −694.97 − 70.0 − 120 ) in.
38.0
x = −23.289 in.
Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor.
PROBLEM 3.152
To loosen a frozen valve, a force F of magnitude 70 lb is applied to the
handle of the valve. Knowing that θ = 25°, M x = −61 lb ⋅ ft, and
M z = −43 lb ⋅ ft, determine θ and d.
SOLUTION
Have
ΣM O : rA/O × F = M O
where
rA/O = − ( 4 in.) i + (11 in.) j − ( d ) k
F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k )
F = 70 lb, θ = 25°
For
F = ( 70 lb ) ( 0.90631cos φ ) i − 0.42262 j + ( 0.90631sin φ ) k 
∴ MO
i
j
k
= ( 70 lb )
−4
11
−d
in.
−0.90631cos φ −0.42262 0.90631sin φ
= ( 70 lb ) ( 9.9694sin φ − 0.42262d ) i + ( −0.90631d cos φ + 3.6252sin φ ) j
+ (1.69048 − 9.9694 cos φ ) k  in.
and
M x = ( 70 lb )( 9.9694sin φ − 0.42262d ) in. = − ( 61 lb ⋅ ft )(12 in./ft )
(1)
M y = ( 70 lb )( −0.90631d cos φ + 3.6252sin φ ) in.
(2)
M z = ( 70 lb )(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft (12 in./ft )
(3)
PROBLEM 3.152 CONTINUED
From Equation (3)
 634.33 
φ = cos −1 
 = 24.636°
 697.86 
or φ = 24.6°
From Equation (1)
 1022.90 
d =
 = 34.577 in.
 29.583 
or d = 34.6 in.
PROBLEM 3.153
When a force F is applied to the handle of the valve shown, its moments
about the x and z axes are, respectively, M x = −77 lb ⋅ ft and
M z = −81 lb ⋅ ft. For d = 27 in., determine the moment M y of F about
the y axis.
SOLUTION
Have
ΣM O : rA/O × F = M O
where
rA/O = − ( 4 in.) i + (11 in.) j − ( 27 in.) k
F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k )
∴ MO
i
j
k
= F
−4
11
−27 lb ⋅ in.
cosθ cos φ − sin θ cosθ sin φ
= F (11cosθ sin φ − 27sin θ ) i + ( −27 cosθ cos φ + 4cosθ sin φ ) j
+ ( 4sin θ − 11cosθ cos φ ) k  ( lb ⋅ in.)
and
M x = F (11cosθ sin φ − 27sin θ )( lb ⋅ in.)
(1)
M y = F ( −27 cosθ cos φ + 4cosθ sin φ )( lb ⋅ in.)
(2)
M z = F ( 4sin θ − 11cosθ cos φ )( lb ⋅ in.)
(3)
Now, Equation (1)
cosθ sin φ =
1  Mx

+ 27sin θ 

11  F

(4)
and
cosθ cos φ =
1
Mz 
 4sin θ −

F 
11 
(5)
Equation (3)
Substituting Equations (4) and (5) into Equation (2),
1
1 M
M 

  
M y = F −27   4sin θ − z   + 4   x + 27sin θ   
F 
  
11 
 11  F

or
My =
1
( 27M z + 4M x )
11
PROBLEM 3.153 CONTINUED
Noting that the ratios
27
4
and
are the ratios of lengths, have
11
11
My =
27
4
( −81 lb ⋅ ft ) + ( −77 lb ⋅ ft ) = 226.82 lb ⋅ ft
11
11
or M y = −227 lb ⋅ ft
PROBLEM 4.1
The boom on a 4300-kg truck is used to unload a pallet of shingles of
mass 1600 kg. Determine the reaction at each of the two (a) rear
wheels B, (b) front wheels C.
SOLUTION
(
)
(
)
WA = mA g = (1600 kg ) 9.81 m/s 2
= 15696 N
WA = 15.696 kN
or
WG = mG g = ( 4300 kg ) 9.81 m/s 2
= 42 183 N
WG = 42.183 kN
or
(a) From f.b.d. of truck with boom
ΣM C = 0:
(15.696 kN ) ( 0.5 + 0.4 + 6 cos15° ) m  − 2FB ( 0.5 + 0.4 + 4.3) m 
+ ( 42.183 kN )( 0.5 m ) = 0
∴ 2 FB =
126.185
= 24.266 kN
5.2
or FB = 12.13 kN
(b) From f.b.d. of truck with boom
ΣM B = 0:
(15.696 kN ) ( 6 cos15° − 4.3) m  − ( 42.183 kN ) ( 4.3 + 0.4 ) m 
+ 2 FC ( 4.3 + 0.9 ) m  = 0
∴ 2 FC =
174.786
= 33.613 kN
5.2
or FC = 16.81 kN
Check:
ΣFy = 0:
( 33.613 − 42.183 + 24.266 − 15.696 ) kN = 0?
( 57.879 − 57.879 ) kN = 0
ok
PROBLEM 4.2
Two children are standing on a diving board of mass 65 kg. Knowing that
the masses of the children at C and D are 28 kg and 40 kg, respectively,
determine (a) the reaction at A, (b) the reaction at B.
SOLUTION
(
)
(
)
(
)
WG = mG g = ( 65 kg ) 9.81 m/s 2 = 637.65 N
WC = mC g = ( 28 kg ) 9.81 m/s 2 = 274.68 N
WD = mD g = ( 40 kg ) 9.81 m/s 2 = 392.4 N
(a) From f.b.d. of diving board
ΣM B = 0: − Ay (1.2 m ) − ( 637.65 N )( 0.48 m ) − ( 274.68 N )(1.08 m ) − ( 392.4 N )( 2.08 m ) = 0
∴ Ay = −
1418.92
= −1182.43 N
1.2
or A y = 1.182 kN
(b) From f.b.d. of diving board
ΣM A = 0: By (1.2 m ) − 637.65 N (1.68 m ) − 274.68 N ( 2.28 m ) − 392.4 N ( 3.28 m ) = 0
∴ By =
2984.6
= 2487.2 N
1.2
or B y = 2.49 kN
Check:
ΣFy = 0:
( −1182.43 + 2487.2 − 637.65 − 274.68 − 392.4 ) N = 0?
( 2487.2 − 2487.2 ) N = 0
ok
PROBLEM 4.3
Two crates, each weighing 250 lb, are placed as shown in the bed of a
3000-lb pickup truck. Determine the reactions at each of the two
(a) rear wheels A, (b) front wheels B.
SOLUTION
(a) From f.b.d. of truck
ΣM B = 0:
( 250 lb )(12.1 ft ) + ( 250 lb )( 6.5 ft ) + ( 3000 lb )( 3.9 ft ) − ( 2FA )( 9.8 ft ) = 0
∴ 2 FA =
16350
= 1668.37 lb
9.8
∴ FA = 834 lb
(b) From f.b.d. of truck
ΣM A = 0:
( 2FB )( 9.8 ft ) − ( 3000 lb )( 5.9 ft ) − ( 250 lb )( 3.3 ft ) + ( 250 lb )( 2.3 ft ) = 0
∴ 2 FB =
17950
= 1831.63 lb
9.8
∴ FB = 916 lb
Check:
ΣFy = 0:
( −250 + 1668.37 − 250 − 3000 + 1831.63) lb = 0?
( 3500 − 3500 ) lb = 0
ok
PROBLEM 4.4
Solve Problem 4.3 assuming that crate D is removed and that the position
of crate C is unchanged.
P4.3 The boom on a 4300-kg truck is used to unload a pallet of
shingles of mass 1600 kg. Determine the reaction at each of the two
(a) rear wheels B, (b) front wheels C
SOLUTION
(a) From f.b.d. of truck
ΣM B = 0:
( 3000 lb )( 3.9 ft ) − ( 2FA )( 9.8 ft ) + ( 250 lb )(12.1 ft ) = 0
∴ 2 FA =
14725
= 1502.55 lb
9.8
or FA = 751 lb
(b) From f.b.d. of truck
ΣM A = 0:
( 2FB )( 9.8 ft ) − ( 3000 lb )( 5.9 ft ) + ( 250 lb )( 2.3 ft ) = 0
∴ 2 FB =
17125
= 1747.45 lb
9.8
or FB = 874 lb
Check:
ΣFy = 0:  2 ( 751 + 874 ) − 3000 − 250  lb = 0?
( 3250 − 3250 ) lb = 0
ok
PROBLEM 4.5
A T-shaped bracket supports the four loads shown. Determine the reactions
at A and B if (a) a = 100 mm, (b) a = 70 mm.
SOLUTION
(a)
From f.b.d. of bracket
ΣM B = 0: − (10 N )( 0.18 m ) − ( 30 N )( 0.1 m ) + ( 40 N )( 0.06 m ) + A ( 0.12 m ) = 0
∴ A=
2.400
= 20 N
0.12
or A = 20.0 N
ΣM A = 0: B ( 0.12 m ) − ( 40 N )( 0.06 m ) − ( 50 N )( 0.12 m ) − ( 30 N )( 0.22 m ) − (10 N )( 0.3 m ) = 0
∴ B=
18.000
= 150 N
0.12
or B = 150.0 N
(b)
From f.b.d. of bracket
ΣM B = 0: − (10 N )( 0.15 m ) − ( 30 N )( 0.07 m ) + ( 40 N )( 0.06 m ) + A ( 0.12 m ) = 0
∴ A=
1.200
= 10 N
0.12
or A = 10.00 N
ΣM A = 0: B ( 0.12 m ) − ( 40 N )( 0.06 m ) − ( 50 N )( 0.12 m ) − ( 30 N )( 0.19 m )
− (10 N )( 0.27 m ) = 0
∴ B=
16.800
= 140 N
0.12
or B = 140.0 N
PROBLEM 4.6
For the bracket and loading of Problem 4.5, determine the smallest
distance a if the bracket is not to move.
P4.5 A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B if (a) a = 100 mm, (b) a = 70 mm.
SOLUTION
The amin value will be based on A = 0
From f.b.d. of bracket
ΣM B = 0:
( 40 N )( 60 mm ) − ( 30 N )( a ) − (10 N )( a + 80 mm ) = 0
∴a=
1600
= 40 mm
40
or amin = 40.0 mm
PROBLEM 4.7
A hand truck is used to move two barrels, each weighing 80 lb.
Neglecting the weight of the hand truck, determine (a) the vertical force
P which should be applied to the handle to maintain equilibrium when
α = 35o , (b) the corresponding reaction at each of the two wheels.
SOLUTION
a1 = ( 20 in.) sin α − ( 8 in.) cos α
a2 = ( 32 in.) cos α − ( 20 in.) sin α
b = ( 64 in.) cos α
From f.b.d. of hand truck
ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0
ΣFy = 0: P − 2w + 2 B = 0
(2)
α = 35°
For
a1 = 20sin 35° − 8cos 35° = 4.9183 in.
a2 = 32cos 35° − 20sin 35° = 14.7413 in.
b = 64cos 35° = 52.426 in.
(a)
From Equation (1)
P ( 52.426 in.) − 80 lb (14.7413 in.) + 80 lb ( 4.9183 in.) = 0
∴ P = 14.9896 lb
(b)
or P = 14.99 lb
From Equation (2)
14.9896 lb − 2 ( 80 lb ) + 2 B = 0
∴ B = 72.505 lb
(1)
or B = 72.5 lb
PROBLEM 4.8
Solve Problem 4.7 when α = 40o.
P4.7 A hand truck is used to move two barrels, each weighing 80 lb.
Neglecting the weight of the hand truck, determine (a) the vertical force P
which should be applied to the handle to maintain equilibrium when
α = 35o , (b) the corresponding reaction at each of the two wheels.
SOLUTION
a1 = ( 20 in.) sin α − ( 8 in.) cos α
a2 = ( 32 in.) cos α − ( 20 in.) sin α
b = ( 64 in.) cos α
From f.b.d. of hand truck
ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0
ΣFy = 0: P − 2w + 2B = 0
(2)
α = 40°
For
a1 = 20sin 40° − 8cos 40° = 6.7274 in.
a2 = 32cos 40° − 20sin 40° = 11.6577 in.
b = 64cos 40° = 49.027 in.
(a)
From Equation (1)
P ( 49.027 in.) − 80 lb (11.6577 in.) + 80 lb ( 6.7274 in.) = 0
∴ P = 8.0450 lb
or P = 8.05 lb
(b)
(1)
From Equation (2)
8.0450 lb − 2 ( 80 lb ) + 2B = 0
∴ B = 75.9775 lb
or B = 76.0 lb
PROBLEM 4.9
Four boxes are placed on a uniform 14-kg wooden plank which rests
on two sawhorses. Knowing that the masses of boxes B and D are
4.5 kg and 45 kg, respectively, determine the range of values of the
mass of box A so that the plank remains in equilibrium when box C is
removed.
SOLUTION
WA = m A g
WB = mB g = 4.5 g
WD = mD g = 45 g
WG = mG g = 14 g
For ( m A )min, E = 0
ΣM F = 0:
( mA g )( 2.5 m ) + ( 4.5g )(1.6 m )
+ (14 g )(1 m ) − ( 45g )( 0.6 m ) = 0
∴ m A = 2.32 kg
For ( m A )max, F = 0:
ΣM E = 0: mA g ( 0.5 m ) − ( 4.5g )( 0.4 m ) − (14 g )(1 m )
− ( 45g )( 2.6 m ) = 0
∴ mA = 265.6 kg
or 2.32 kg ≤ mA ≤ 266 kg
PROBLEM 4.10
A control rod is attached to a crank at A and cords are attached at B and
C. For the given force in the rod, determine the range of values of the
tension in the cord at C knowing that the cords must remain taut and that
the maximum allowed tension in a cord is 180 N.
SOLUTION
(TC )max,
For
ΣM O = 0:
TB = 0
(TC )max ( 0.120 m ) − ( 400 N )( 0.060 m ) = 0
(TC )max
= 200 N > Tmax = 180 N
∴ (TC )max = 180.0 N
(TC )min ,
For
ΣM O = 0:
TB = Tmax = 180 N
(TC )min ( 0.120 m ) + (180 N )( 0.040 m )
− ( 400 N )( 0.060 m ) = 0
∴ (TC )min = 140.0 N
Therefore,
140.0 N ≤ TC ≤ 180.0 N
PROBLEM 4.11
The maximum allowable value of each of the reactions is 360 N.
Neglecting the weight of the beam, determine the range of values of the
distance d for which the beam is safe.
SOLUTION
From f.b.d. of beam
ΣFx = 0: Bx = 0
so that
B = By
ΣFy = 0: A + B − (100 + 200 + 300 ) N = 0
A + B = 600 N
or
Therefore, if either A or B has a magnitude of the maximum of 360 N,
the other support reaction will be < 360 N ( 600 N − 360 N = 240 N ) .
ΣM A = 0:
(100 N )( d ) − ( 200 N )( 0.9 − d ) − ( 300 N )(1.8 − d )
+ B (1.8 − d ) = 0
d =
or
720 − 1.8B
600 − B
Since B ≤ 360 N,
d =
720 − 1.8 ( 360 )
= 0.300 m
600 − 360
ΣM B = 0:
d ≥ 300 mm
(100 N )(1.8) − A (1.8 − d ) + ( 200 N )( 0.9 ) = 0
d =
or
or
1.8 A − 360
A
Since A ≤ 360 N,
d =
1.8 ( 360 ) − 360
= 0.800 m
360
or
d ≤ 800 mm
or 300 mm ≤ d ≤ 800 mm
PROBLEM 4.12
Solve Problem 4.11 assuming that the 100-N load is replaced by a 160-N
load.
P4.11 The maximum allowable value of each of the reactions is 360 N.
Neglecting the weight of the beam, determine the range of values of the
distance d for which the beam is safe.
SOLUTION
From f.b.d of beam
ΣFx = 0: Bx = 0
so that
B = By
ΣFy = 0: A + B − (160 + 200 + 300 ) N = 0
A + B = 660 N
or
Therefore, if either A or B has a magnitude of the maximum of 360 N,
the other support reaction will be < 360 N ( 660 − 360 = 300 N ) .
ΣM A = 0: 160 N ( d ) − 200 N ( 0.9 − d ) − 300 N (1.8 − d )
+ B (1.8 − d ) = 0
d =
or
720 − 1.8B
660 − B
Since B ≤ 360 N,
d =
720 − 1.8 ( 360 )
= 0.240 m
660 − 360
or
d ≥ 240 mm
ΣM B = 0: 160 N (1.8 ) − A (1.8 − d ) + 200 N ( 0.9 ) = 0
d =
or
1.8 A − 468
A
Since A ≤ 360 N,
d =
1.8 ( 360 ) − 468
= 0.500 m
360
or
d ≥ 500 mm
or 240 mm ≤ d ≤ 500 mm
PROBLEM 4.13
For the beam of Sample Problem 4.2, determine the range of values of
P for which the beam will be safe knowing that the maximum
allowable value of each of the reactions is 45 kips and that the
reaction at A must be directed upward.
SOLUTION
For the force of P to be a minimum, A = 0.
With A = 0,
ΣM B = 0: Pmin ( 6 ft ) − ( 6 kips )( 2 ft ) − ( 6 kips )( 4 ft ) = 0
∴ Pmin = 6.00 kips
For the force P to be a maximum, A = A max = 45 kips
With A = 45 kips,
ΣM B = 0: − ( 45 kips )( 9 ft ) + Pmax ( 6 ft ) − ( 6 kips )( 2 ft ) − ( 6 kips )( 4 ft ) = 0
∴ Pmax = 73.5 kips
A check must be made to verify the assumption that the maximum value of P is based on the reaction force at
A. This is done by making sure the corresponding value of B is < 45 kips.
ΣFy = 0: 45 kips − 73.5 kips + B − 6 kips − 6 kips = 0
∴ B = 40.5 kips < 45 kips
∴ ok
or Pmax = 73.5 kips
and 6.00 kips ≤ P ≤ 73.5 kips
PROBLEM 4.14
For the beam and loading shown, determine the range of values of the
distance a for which the reaction at B does not exceed 50 lb
downward or 100 lb upward.
SOLUTION
To determine amax the two 150-lb forces need to be as close to B without
having the vertical upward force at B exceed 100 lb.
From f.b.d. of beam with B = 100 lb
ΣM D = 0: − (150 lb ) ( amax − 4 in.) − (150 lb ) ( amax − 1 in.)
− ( 25 lb )( 2 in.) + (100 lb )( 8 in.) = 0
amax = 5.00 in.
or
To determine amin the two 150-lb forces need to be as close to A without
having the vertical downward force at B exceed 50 lb.
From f.b.d. of beam with B = 50 lb
ΣM D = 0:
(150 lb )( 4 in. − amin ) − (150 lb )( amin
− 1 in.)
− ( 25 lb )( 2 in.) − ( 50 lb )( 8 in.) = 0
or
Therefore,
amin = 1.00 in.
or 1.00 in. ≤ a ≤ 5.00 in.
PROBLEM 4.15
A follower ABCD is held against a circular cam by a stretched spring,
which exerts a force of 21 N for the position shown. Knowing that the
tension in rod BE is 14 N, determine (a) the force exerted on the roller
at A, (b) the reaction at bearing C.
SOLUTION
Note: From f.b.d. of ABCD
Ax = A cos 60° =
Ay = A sin 60° = A
A
2
3
2
(a) From f.b.d. of ABCD
 A
ΣM C = 0:   ( 40 mm ) − 21 N ( 40 mm )
2
+ 14 N ( 20 mm ) = 0
∴ A = 28 N
or A = 28.0 N
60°
(b) From f.b.d. of ABCD
ΣFx = 0: C x + 14 N + ( 28 N ) cos 60° = 0
∴ C x = −28 N
ΣFy = 0:
C y − 21 N + ( 28 N ) sin 60° = 0
∴ C y = −3.2487 N
Then
and
C =
C x = 28.0 N
or
C x2 + C y2 =
or
C y = 3.25 N
( 28)2 + ( 3.2487 )2
= 28.188 N
 Cy 
−1  −3.2487 
 = tan 
 = 6.6182°
 −28 
 Cx 
θ = tan −1 
or C = 28.2 N
6.62°
PROBLEM 4.16
A 6-m-long pole AB is placed in a hole and is guyed by three cables.
Knowing that the tensions in cables BD and BE are 442 N and 322 N,
respectively, determine (a) the tension in cable CD, (b) the reaction
at A.
SOLUTION
Note:
DB =
( 2.8)2 + ( 5.25)2
= 5.95 m
DC =
( 2.8)2 + ( 2.10 )2
= 3.50 m
(a) From f.b.d. of pole
 2.8 m 

ΣM A = 0: − ( 322 N )( 6 m ) + 
 ( 442 N )  ( 6 m )
5.95
m



 2.8 m 

+ 
 TCD  ( 2.85 m ) = 0
 3.50 m 

∴ TCD = 300 N
or TCD = 300 N
(b) From f.b.d. of pole
 2.8 m 
ΣFx = 0: 322 N − 
 ( 442 N )
 5.95 m 
 2.8 m 
−
 ( 300 N ) + Ax = 0
 3.50 m 
∴ Ax = 126 N
or
A x = 126 N
 5.25 m 
ΣFy = 0: Ay − 
 ( 442 N )
 5.95 m 
Then
and
 2.10 m 
−
 ( 300 N ) = 0
 3.50 m 
∴ Ay = 570 N
or
A=
(126 )2 + ( 570 )2
Ax2 + Ay2 =
A y = 570 N
= 583.76 N
 570 N 
θ = tan −1 
 = 77.535°
 126 N 
or A = 584 N
77.5°
PROBLEM 4.17
Determine the reactions at A and C when (a) α = 0, (b) α = 30o.
SOLUTION
(a)
(a) α = 0°
From f.b.d. of member ABC
(80 lb )(10 in.) + (80 lb )( 20 in.) − A ( 40 in.) = 0
ΣM C = 0:
∴ A = 60 lb
or A = 60.0 lb
ΣFy = 0: C y + 60 lb = 0
∴ C y = −60 lb
or
C y = 60 lb
ΣFx = 0: 80 lb + 80 lb + Cx = 0
∴ C x = −160 lb
C =
Then
C x2 + C y2 =
or
C x = 160 lb
(160 )2 + ( 60 )2
= 170.880 lb
 Cy 
−1  −60 
 = tan 
 = 20.556°
 −160 
 Cx 
θ = tan −1 
and
or C = 170.9 lb
(b)
20.6°
(b) α = 30°
From f.b.d. of member ABC
ΣM C = 0:
(80 lb )(10 in.) + ( 80 lb )( 20 in.) − ( A cos 30° )( 40 in.)
+ ( A sin 30° )( 20 in.) = 0
∴ A = 97.399 lb
or A = 97.4 lb
60°
PROBLEM 4.17 CONTINUED
ΣFx = 0: 80 lb + 80 lb + ( 97.399 lb ) sin 30° + Cx = 0
∴ C x = −208.70 lb
C x = 209 lb
or
ΣFy = 0: C y + ( 97.399 lb ) cos 30° = 0
∴ C y = −84.350 lb
Then
and
C =
C x2 + C y2 =
or
C y = 84.4 lb
( 208.70 )2 + (84.350 )2
= 225.10 lb
 Cy 
−1  −84.350 
 = tan 
 = 22.007°
C
 −208.70 
 x
θ = tan −1 
or C = 225 lb
22.0°
PROBLEM 4.18
Determine the reactions at A and B when (a) h = 0, (b) h = 8 in.
SOLUTION
(a)
(a) h = 0
From f.b.d. of plate
ΣM A = 0:
( B sin 30° )( 20 in.) − ( 40 lb )(10 in.) = 0
∴ B = 40 lb
or B = 40.0 lb
30°
ΣFx = 0: Ax − ( 40 lb ) cos 30° = 0
∴ Ax = 34.641 lb
or
A x = 34.6 lb
ΣFy = 0: Ay − 40 lb + ( 40 lb ) sin 30° = 0
∴ Ay = 20 lb
A=
Then
Ax2 + Ay2 =
A y = 20.0 lb
or
( 34.641)2 + ( 20 )2
= 39.999 lb
 Ay 
20 
−1 
 = tan 
 = 30.001°
 34.641 
 Ax 
θ = tan −1 
and
or A = 40.0 lb
(b)
30°
(b) h = 8 in.
From f.b.d. of plate
ΣM A = 0:
( B sin 30° )( 20 in.) − ( B cos 30° )(8 in.)
− ( 40 lb )(10 in.) = 0
∴ B = 130.217 lb
or B = 130.2 lb
30.0°
PROBLEM 4.18 CONTINUED
ΣFx = 0: Ax − (130.217 lb ) cos30° = 0
∴ Ax = 112.771 lb
or
A x = 112.8 lb
ΣFy = 0: Ay − 40 lb + (130.217 lb ) sin 30° = 0
∴ Ay = −25.108 lb
Then
and
A=
Ax2 + Ay2 =
or
A y = 25.1 lb
(112.771)2 + ( 25.108)2
= 115.532 lb
 Ay 
−1  −25.108 
 = tan 
 = −12.5519°
A
 112.771 
 x
θ = tan −1 
or A = 115.5 lb
12.55°
PROBLEM 4.19
The lever BCD is hinged at C and is attached to a control rod at B. If
P = 200 N, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
(a) From f.b.d. of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
∴ TAB = 300 N
(b) From f.b.d. of lever BCD
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
∴ C x = −380 N
C x = 380 N
or
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
∴ C y = −240 N
Then
and
C =
C x2 + C y2 =
C y = 240 N
or
( 380 )2 + ( 240 )2
= 449.44 N
 Cy 
−1  −240 
 = tan 
 = 32.276°
 −380 
 Cx 
θ = tan −1 
or C = 449 N
32.3°
PROBLEM 4.20
The lever BCD is hinged at C and is attached to a control rod at B.
Determine the maximum force P which can be safely applied at D if the
maximum allowable value of the reaction at C is 500 N.
SOLUTION
From f.b.d. of lever BCD
ΣM C = 0: TAB ( 50 mm ) − P ( 75 mm ) = 0
∴ TAB = 1.5P
(1)
ΣFx = 0: 0.6TAB + P − C x = 0
∴ C x = P + 0.6TAB
(2)
Cx = P + 0.6 (1.5P ) = 1.9 P
From Equation (1)
ΣFy = 0: 0.8TAB − C y = 0
∴ C y = 0.8TAB
(3)
C y = 0.8 (1.5P ) = 1.2P
From Equation (1)
From Equations (2) and (3)
C =
C x2 + C y2 =
(1.9P )2 + (1.2P )2
= 2.2472 P
Since Cmax = 500 N,
∴ 500 N = 2.2472Pmax
or
Pmax = 222.49 lb
or P = 222 lb
PROBLEM 4.21
The required tension in cable AB is 800 N. Determine (a) the vertical
force P which must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
(a) From f.b.d. of pedal
ΣM C = 0: P ( 0.4 m ) − ( 800 N ) ( 0.18 m ) sin 60° = 0
∴ P = 311.77 N
or P = 312 N
(b) From f.b.d. of pedal
ΣFx = 0: Cx − 800 N = 0
∴ Cx = 800 N
C x = 800 N
or
ΣFy = 0: C y − 311.77 N = 0
∴ C y = 311.77 N
C y = 311.77 N
or
Then
and
C =
C x2 + C y2 =
(800 )2 + ( 311.77 )2
= 858.60 N
 Cy 
−1  311.77 
 = tan 
 = 21.291°
C
 800 
 x
θ = tan −1 
or C = 859 N
21.3°
PROBLEM 4.22
Determine the maximum tension which can be developed in cable AB
if the maximum allowable value of the reaction at C is 1000 N.
SOLUTION
Cmax = 1000 N
Have
C 2 = C x2 + C y2
Now
∴ Cy =
(1000 )2 − Cx2
(1)
From f.b.d. of pedal
ΣFx = 0: C x − Tmax = 0
∴ C x = Tmax
(2)
ΣM D = 0: C y ( 0.4 m ) − Tmax ( 0.18 m ) sin 60° = 0
∴ C y = 0.38971Tmax
(3)
Equating the expressions for C y in Equations (1) and (3), with Cx = Tmax
from Equation (2)
2
(1000 )2 − Tmax
= 0.389711Tmax
2
∴ Tmax
= 868,150
and
Tmax = 931.75 N
or Tmax = 932 N
PROBLEM 4.23
A steel rod is bent to form a mounting bracket. For each of the mounting
brackets and loadings shown, determine the reactions at A and B.
SOLUTION
(a) From f.b.d. of mounting bracket
ΣM E = 0: A ( 8 in.) − 80 lb ⋅ in. − (10 lb )( 6 in.)
(a)
− ( 20 lb )(12 in.) = 0
∴ A = 47.5 lb
or A = 47.5 lb
ΣFx = 0: Bx − 10 lb + 47.5 lb = 0
∴ Bx = −37.5 lb
B x = 37.5 lb
or
ΣFy = 0: By − 20 lb = 0
∴ By = 20 lb
B y = 20.0 lb
or
Then
and
B=
Bx2 + By2 =
( 37.5)2 + ( 20.0 )2
= 42.5 lb
 By 
−1  20 
 = tan 
 = −28.072°
 −37.5 
 Bx 
θ = tan −1 
or B = 42.5 lb
28.1°
(b) From f.b.d. of mounting bracket
(b)
ΣM B = 0:
( A cos 45° )(8 in.) − 80 lb ⋅ in.
− (10 lb )( 6 in.) − ( 20 lb )(12 in.) = 0
∴ A = 67.175 lb
or A = 67.2 lb
ΣFx = 0: Bx − 10 lb + 67.175cos 45° = 0
∴ Bx = −37.500 lb
or
B x = 37.5 lb
45°
PROBLEM 4.23 CONTINUED
ΣFy = 0: By − 20 lb + 67.175sin 45° = 0
∴ By = −27.500 lb
B y = 27.5 lb
or
Then
and
B=
Bx2 + By2 =
( 37.5)2 + ( 27.5)2
= 46.503 lb
 By 
−1  −27.5 
 = tan 
 = 36.254°
 −37.5 
 Bx 
θ = tan −1 
or B = 46.5 lb
36.3°
PROBLEM 4.24
A steel rod is bent to form a mounting bracket. For each of the
mounting brackets and loadings shown, determine the reactions at A
and B.
SOLUTION
(a)
(a) From f.b.d. of mounting bracket
ΣM A = 0: − B ( 8 in.) − ( 20 lb )(12 in.)
+ (10 lb )( 2 in.) − 80 lb ⋅ in. = 0
∴ B = −37.5 lb
or B = 37.5 lb
ΣFx = 0: − 37.5 lb − 10 lb + Ax = 0
∴ Ax = 47.5 lb
A x = 47.5 lb
or
ΣFy = 0: − 20 lb + Ay = 0
∴ Ay = 20 lb
A y = 20.0 lb
or
Then
A=
Ax2 + Ay2 =
and
θ = tan −1 
( 47.5)2 + ( 20 )2
= 51.539 lb
 Ay 
−1  20 
 = tan 
 = 22.834°
 47.5 
 Ax 
or A = 51.5 lb
(b)
22.8°
(b) From f.b.d. of mounting bracket
ΣM A = 0: − ( B cos 45° )( 8 in.) − ( 20 lb )( 2 in.)
−80 lb ⋅ in. + (10 lb )( 2 in.) = 0
∴ B = −53.033 lb
or B = 53.0 lb
ΣFx = 0: Ax + ( −53.033 lb ) cos 45° − 10 = 0
∴ Ax = 47.500 lb
or
A x = 47.5 lb
45°
PROBLEM 4.24 CONTINUED
ΣFy = 0: Ay − ( 53.033 lb ) sin 45° − 20 = 0
∴ Ay = −17.500 lb
A y = 17.50 lb
or
Then
and
A=
Ax2 + Ay2 =
( 47.5)2 + (17.5)2
= 50.621 lb
 Ay 
−1  −17.5 
 = tan 
 = −20.225°
 47.5 
 Ax 
θ = tan −1 
or A = 50.6 lb
20.2°
PROBLEM 4.25
A sign is hung by two chains from mast AB. The mast is hinged at A and
is supported by cable BC. Knowing that the tensions in chains DE and
FH are 50 lb and 30 lb, respectively, and that d = 1.3 ft, determine
(a) the tension in cable BC, (b) the reaction at A.
SOLUTION
(8.4 )2 + (1.3)2
BC =
First note
= 8.5 ft
(a) From f.b.d. of mast AB
 8.4 

ΣM A = 0: 
 TBC  ( 2.5 ft ) − ( 30 lb )( 7.2 ft )
 8.5 

−50 lb ( 2.2 ft ) = 0
∴ TBC = 131.952 lb
or TBC = 132.0 lb
(b) From f.b.d. of mast AB
 8.4 
ΣFx = 0: Ax − 
 (131.952 lb ) = 0
 8.5 
∴ Ax = 130.400 lb
A x = 130.4 lb
or
 1.3 
ΣFy = 0: Ay + 
 (131.952 lb ) − 30 lb − 50 lb = 0
 8.5 
∴ Ay = 59.819 lb
A y = 59.819 lb
or
Then
and
A=
Ax2 + Ay2 =
(130.4 )2 + ( 59.819 )2
= 143.466 lb
 Ay 
−1  59.819 
 = tan 
 = 24.643°
A
 130.4 
 x
θ = tan −1 
or A = 143.5 lb
24.6°
PROBLEM 4.26
A sign is hung by two chains from mast AB. The mast is hinged at A
and is supported by cable BC. Knowing that the tensions in chains DE
and FH are 30 lb and 20 lb, respectively, and that d = 1.54 ft,
determine (a) the tension in cable BC, (b) the reaction at A.
SOLUTION
BC =
First note
(8.4 )2 + (1.54 )2
= 8.54 ft
(a) From f.b.d. of mast AB
 8.4 

ΣM A = 0: 
 TBC  ( 2.5 ft ) − 20 lb ( 7.2 ft )
 8.54 

− 30 lb ( 2.2 ft ) = 0
∴ TBC = 85.401 lb
or TBC = 85.4 lb
(b) From f.b.d. of mast AB
 8.4 
ΣFx = 0: Ax − 
 ( 85.401 lb ) = 0
 8.54 
∴ Ax = 84.001 lb
A x = 84.001 lb
or
 1.54 
ΣFy = 0: Ay + 
 ( 85.401 lb ) − 20 lb − 30 lb = 0
 8.54 
∴ Ay = 34.600 lb
A y = 34.600 lb
or
Then
and
A=
Ax2 + Ay2 =
(84.001)2 + ( 34.600 )2
= 90.848 lb
 Ay 
−1  34.6 
 = tan 
 = 22.387°
 84.001 
 Ax 
θ = tan −1 
or A = 90.8 lb
22.4°
PROBLEM 4.27
For the frame and loading shown, determine the reactions at A and E
when (a) α = 30o , (b) α = 45o.
SOLUTION
(a) Given α = 30°
(a)
From f.b.d. of frame
ΣM A = 0: − ( 90 N )( 0.2 m ) − ( 90 N )( 0.06 m )
+ ( E cos 60° )( 0.160 m ) + ( E sin 60° )( 0.100 m ) = 0
∴ E = 140.454 N
or E = 140.5 N
60°
ΣFx = 0: Ax − 90 N + (140.454 N ) cos 60° = 0
∴ Ax = 19.7730 N
A x = 19.7730 N
or
ΣFy = 0: Ay − 90 N + (140.454 N ) sin 60° = 0
∴ Ay = −31.637 N
A y = 31.6 N
or
Then
A=
Ax2 + Ay2 =
(19.7730 )2 + ( 31.637 )2
= 37.308 lb
and
 Ay 
−1  −31.637 
 = tan 

 19.7730 
 Ax 
θ = tan −1 
= −57.995°
or A = 37.3 N
58.0°
PROBLEM 4.27 CONTINUED
(b)
(b) Given α = 45°
From f.b.d. of frame
ΣM A = 0: − ( 90 N )( 0.2 m ) − ( 90 N )( 0.06 m )
+ ( E cos 45° )( 0.160 m ) + ( E sin 45° )( 0.100 m ) = 0
∴ E = 127.279 N
or E = 127.3 N
45°
ΣFx = 0: Ax − 90 + (127.279 N ) cos 45° = 0
∴ Ax = 0
ΣFy = 0: Ay − 90 + (127.279 N ) sin 45° = 0
∴ Ay = 0
or A = 0
PROBLEM 4.28
A lever AB is hinged at C and is attached to a control cable at A. If the
lever is subjected to a 300-N vertical force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
First
x AC = ( 0.200 m ) cos 20° = 0.187 939 m
y AC = ( 0.200 m ) sin 20° = 0.068 404 m
Then
yDA = 0.240 m − y AC
= 0.240 m − 0.068404 m
= 0.171596 m
tan α =
and
yDA
0.171 596
=
x AC
0.187 939
∴ α = 42.397°
β = 90° − 20° − 42.397° = 27.603°
and
(a) From f.b.d. of lever AB
ΣM C = 0: T cos 27.603° ( 0.2 m )
− 300 N ( 0.3 m ) cos 20° = 0
∴ T = 477.17 N
or T = 477 N
(b) From f.b.d. of lever AB
ΣFx = 0: C x + ( 477.17 N ) cos 42.397° = 0
∴ C x = −352.39 N
or
C x = 352.39 N
ΣFy = 0: C y − 300 N − ( 477.17 N ) sin 42.397° = 0
∴ C y = 621.74 N
or
C y = 621.74 N
PROBLEM 4.28 CONTINUED
Then
and
C =
C x2 + C y2 =
( 352.39 )2 + ( 621.74 )2
= 714.66 N
 Cy 
−1  621.74 
 = tan 
 = −60.456°
C
 −352.39 
 x
θ = tan −1 
or C = 715 N
60.5°
PROBLEM 4.29
Neglecting friction and the radius of the pulley, determine the tension
in cable BCD and the reaction at support A when d = 80 mm.
SOLUTION
First
 60 
 = 12.0948°
 280 
α = tan −1 
 60 
β = tan −1   = 36.870°
 80 
From f.b.d. of object BAD
ΣM A = 0:
( 40 N )( 0.18 m ) + (T cosα )( 0.08 m )
+ (T sin α )( 0.18 m ) − (T cos β )( 0.08 m )
− (T sin β )( 0.18 m ) = 0
 7.2 N ⋅ m 
∴ T =
 = 128.433 N
 0.056061 
or T = 128.4 N
ΣFx = 0:
(128.433 N )( cos β
− cos α ) + Ax = 0
∴ Ax = 22.836 N
A x = 22.836 N
or
ΣFy = 0: Ay + (128.433 N )( sin β + sin α ) + 40 N = 0
∴ Ay = −143.970 N
A y = 143.970 N
or
Then
and
A=
Ax2 + Ay2 =
( 22.836 )2 + (143.970 )2
= 145.770 N
 Ay 
−1  −143.970 
 = tan 
 = −80.987°
 22.836 
 Ax 
θ = tan −1 
or A = 145.8 N
81.0°
PROBLEM 4.30
Neglecting friction and the radius of the pulley, determine the tension in
cable BCD and the reaction at support A when d = 144 mm.
SOLUTION
First note
 60 
α = tan −1 
 = 15.5241°
 216 
 60 
β = tan −1 
 = 22.620°
 144 
From f.b.d. of member BAD
ΣM A = 0:
( 40 N )( 0.18 m ) + (T cosα )( 0.08 m )
+ (T sin α )( 0.18 m ) − (T cos β )( 0.08 m )
− (T sin β )( 0.18 m ) = 0
 7.2 N ⋅ m 
∴ T =
 = 404.04 N
 0.0178199 m 
or T = 404 N
ΣFx = 0: Ax + ( 404.04 N )( cos β − cos α ) = 0
∴ Ax = 16.3402 N
A x = 16.3402 N
or
ΣFy = 0: Ay + ( 404.04 N )( sin β + sin α ) + 40 N = 0
∴ Ay = −303.54 N
A y = 303.54 N
or
Then
and
A=
Ax2 + Ay2 =
(16.3402 )2 + ( 303.54 )2
= 303.98 N
 Ay 
−1  −303.54 
 = tan 
 = −86.919°
A
 16.3402 
 x
θ = tan −1 
or A = 304 N
86.9°
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
SOLUTION
From f.b.d. of inverted T-member
ΣM C = 0: T ( 25 in.) − T (10 in.) − ( 30 lb )(10 in.) = 0
∴ T = 20 lb
or T = 20.0 lb W
ΣFx = 0: Cx − 20 lb = 0
∴ C x = 20 lb
C x = 20.0 lb
or
ΣFy = 0: C y + 20 lb − 30 lb = 0
∴ C y = 10 lb
C y = 10.00 lb
or
Then
and
or
C =
C x2 + C y2 =
( 20 )2 + (10 )2
= 22.361 lb
 Cy 
−1  10 
 = tan   = 26.565°
 20 
 Cx 
θ = tan −1 
C = 22.4 lb
26.6° W
PROBLEM 4.32
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
that θ = 35o , determine the reaction (a) at B, (b) at C.
SOLUTION
For θ = 35°
(a) From the f.b.d. of rod ABC
ΣM D = 0: Cx( R ) − P( R ) = 0
∴ Cx = P
Cx = P
or
ΣFx = 0: P − B sin 35° = 0
∴ B=
P
= 1.74345P
sin 35°
or B = 1.743P
55.0° W
(b) From the f.b.d. of rod ABC
ΣFy = 0: C y + (1.74345P ) cos 35° − P = 0
∴ C y = −0.42815P
C y = 0.42815P
or
Then
and
C =
C x2 + C y2 =
( P )2 + ( 0.42815P )2
= 1.08780 P
 Cy 
−1  −0.42815P 
 = tan 
 = −23.178°
P


 Cx 
θ = tan −1 
or C = 1.088P
23.2° W
PROBLEM 4.33
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
that θ = 50o , determine the reaction (a) at B, (b) at C.
SOLUTION
For θ = 50°
(a) From the f.b.d. of rod ABC
ΣM D = 0: Cx ( R ) − P ( R ) = 0
∴ Cx = P
Cx = P
or
ΣFx = 0: P − B sin 50° = 0
∴ B=
P
= 1.30541P
sin 50°
or B = 1.305P
40.0° W
(b) From the f.b.d. of rod ABC
ΣFy = 0: C y − P + (1.30541P ) cos 50° = 0
∴ C y = 0.160900P
C y = 0.1609 P
or
Then
C =
C x2 + C y2 =
and
θ = tan −1 
( P )2 + ( 0.1609P )2
= 1.01286 P
 Cy 
−1  0.1609 P 
 = tan 
 = 9.1405°
C
P


 x
or C = 1.013P
9.14° W
PROBLEM 4.34
Neglecting friction and the radius of the pulley, determine (a) the
tension in cable ABD, (b) the reaction at C.
SOLUTION
First note
 15 
α = tan −1   = 22.620°
 36 
 15 
β = tan −1   = 36.870°
 20 
(a) From f.b.d. of member ABC
ΣM C = 0:
( 30 lb )( 28 in.) − (T sin 22.620° )( 36 in.)
− (T sin 36.870° )( 20 in.) = 0
∴ T = 32.500 lb
or T = 32.5 lb W
(b) From f.b.d. of member ABC
ΣFx = 0: Cx + ( 32.500 lb )( cos 22.620° + cos 36.870° ) = 0
∴ C x = −56.000 lb
C x = 56.000 lb
or
ΣFy = 0: C y − 30 lb + ( 32.500 lb )( sin 22.620° + sin 36.870° ) = 0
∴ C y = −2.0001 lb
C y = 2.0001 lb
or
Then
and
C =
C x2 + C y2 =
( 56.0 )2 + ( 2.001)2
= 56.036 lb
 Cy 
−1  −2.0 
 = tan 
 = 2.0454°
 −56.0 
 Cx 
θ = tan −1 
or C = 56.0 lb
2.05° W
PROBLEM 4.35
Neglecting friction, determine the tension in cable ABD and the
reaction at C when θ = 60o.
SOLUTION
From f.b.d. of bent ACD
ΣM C = 0:
(T cos30° )( 2a sin 60° ) + (T sin 30° )( a + 2a cos 60° )
−T (a) − P (a) = 0
∴ T =
P
1.5
or T =
2P
W
3
or C = 0.577P
W
 2P 
ΣFx = 0: C x − 
 cos 30° = 0
 3 
∴ Cx =
3
P = 0.57735P
3
C x = 0.577 P
or
ΣFy = 0: C y +
2
 2P 
P−P+
 cos 60° = 0
3
 3 
∴ Cy = 0
PROBLEM 4.36
Neglecting friction, determine the tension in cable ABD and the
reaction at C when θ = 30o.
SOLUTION
From f.b.d. of bent ACD
ΣM C = 0:
(T cos 60° )( 2a sin 30° ) + T sin 60° ( a + 2a cos 30° )
− P (a) − T (a) = 0
∴ T =
P
= 0.53590P
1.86603
or T = 0.536 P W
ΣFx = 0: C x − ( 0.53590P ) cos 60° = 0
∴ Cx = 0.26795P
or
C x = 0.268P
ΣFy = 0: C y + 0.53590 P − P + ( 0.53590 P ) sin 60° = 0
∴ Cy = 0
or C = 0.268P
W
PROBLEM 4.37
Determine the tension in each cable and the reaction at D.
SOLUTION
First note
BE =
( 20 )2 + (8)2
in. = 21.541 in.
CF =
(10 )2 + (8)2
in. = 12.8062 in.
From f.b.d. of member ABCD
ΣM C = 0:


8 
 TBE  (10 in.) = 0
 21.541 

(120 lb )( 20 in.) − 
∴ TBE = 646.24 lb
or TBE = 646 lb W
8
 8 


ΣFy = 0: − 120 lb + 
 ( 646.24 lb ) − 
 TCF = 0
21.541
12.8062




∴ TCF = 192.099 lb
or TCF = 192.1 lb W
 20 
 10 
ΣFx = 0: 
 ( 646.24 lb ) + 
 (192.099 lb ) − D = 0
 21.541 
 12.8062 
∴ D = 750.01 lb
or D = 750 lb
W
PROBLEM 4.38
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and
rests against frictionless surfaces at A and D. Knowing that the collar
at B can move freely on the rod and that θ = 45o. determine (a) the
tension in cord OB, (b) the reactions at A and D.
SOLUTION
(a) From f.b.d. of rod ABCD
ΣM E = 0:
( 25 N ) cos 60° ( dOE ) − (T cos 45° ) ( dOE ) = 0
∴ T = 17.6777 N
or T = 17.68 N W
(b) From f.b.d. of rod ABCD
ΣFx = 0: − (17.6777 N ) cos 45° + ( 25 N ) cos 60°
+ N D cos 45° − N A cos 45° = 0
∴ N A − ND = 0
or
ND = N A
(1)
ΣFy = 0: N A sin 45° + N D sin 45° − (17.6777 N ) sin 45°
− ( 25 N ) sin 60° = 0
∴ N A + N D = 48.296 N
(2)
Substituting Equation (1) into Equation (2),
2 N A = 48.296 N
N A = 24.148 N
or N A = 24.1 N
45.0° W
and N D = 24.1 N
45.0° W
PROBLEM 4.39
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and
rests against frictionless surfaces at A and D. Knowing that the collar
at B can move freely on the rod, determine (a) the value of θ for
which the tension in cord OB is as small as possible, (b) the
corresponding value of the tension, (c) the reactions at A and D.
SOLUTION
(a) From f.b.d. of rod ABCD
ΣM E = 0:
( 25 N ) cos 60° ( dOE ) − (T cosθ ) ( dOE ) = 0
T =
or
12.5 N
cosθ
(1)
∴ T is minimum when cosθ is maximum,
or θ = 0° W
(b) From Equation (1)
T =
12.5 N
= 12.5 N
cos 0
or Tmin = 12.50 N W
ΣFx = 0: − N A cos 45° + N D cos 45° + 12.5 N
(c)
− ( 25 N ) cos 60° = 0
∴ ND − N A = 0
or
ND = N A
(2)
ΣFy = 0: N A sin 45° + N D sin 45° − ( 25 N ) sin 60° = 0
∴ N D + N A = 30.619 N
(3)
Substituting Equation (2) into Equation (3),
2 N A = 30.619
N A = 15.3095 N
or N A = 15.31 N
45.0° W
and N D = 15.31 N
45.0° W
PROBLEM 4.40
Bar AC supports two 100-lb loads as shown. Rollers A and C rest against
frictionless surfaces and a cable BD is attached at B. Determine (a) the
tension in cable BD, (b) the reaction at A, (c) the reaction at C.
SOLUTION
First note that from similar triangles
yDB
10
=
6
20
and
BD =
∴ yDB = 3 in.
( 3)2 + (14 )2
in. = 14.3178 in.
Tx =
14
T = 0.97780T
14.3178
Ty =
3
T = 0.20953T
14.3178
(a) From f.b.d. of bar AC
ΣM E = 0:
( 0.97780T )( 7 in.) − ( 0.20953T )( 6 in.)
− (100 lb )(16 in.) − (100 lb )( 4 in.) = 0
∴ T = 357.95 lb
or T = 358 lb W
(b) From f.b.d. of bar AC
ΣFy = 0: A − 100 − 0.20953 ( 357.95 ) − 100 = 0
∴ A = 275.00 lb
or A = 275 lb W
(c) From f.b.d of bar AC
ΣFx = 0: 0.97780 ( 357.95 ) − C = 0
∴ C = 350.00 lb
or C = 350 lb
W
PROBLEM 4.41
A parabolic slot has been cut in plate AD, and the plate has been
placed so that the slot fits two fixed, frictionless pins B and C. The
equation of the slot is y = x 2 /100, where x and y are expressed in mm.
Knowing that the input force P = 4 N, determine (a) the force each
pin exerts on the plate, (b) the output force Q.
SOLUTION
y =
The equation of the slot is
Now
x2
100
 dy 
  = slope of the slot at C
 dx C
 2x 
=
= 1.200

100  x = 60 mm
∴ α = tan −1 (1.200 ) = 50.194°
and
θ = 90° − α = 90° − 50.194° = 39.806°
Coordinates of C are
xC = 60 mm,
Also, the coordinates of D are
yC =
( 60 ) 2
100
= 36 mm
xD = 60 mm
yD = 46 mm + ( 40 mm ) sin β
where
 120 − 66 
 = 12.6804°
 240 
β = tan −1 
∴ yD = 46 mm + ( 40 mm ) tan12.6804°
= 55.000 mm
PROBLEM 4.41 CONTINUED
yED =
Also,
60 mm
60 mm
=
tan β
tan12.6804°
= 266.67 mm
From f.b.d. of plate AD
ΣM E = 0:
( NC cosθ )  yED − ( yD − yC ) + ( NC sin θ )( xC ) − ( 4 N )( yED −
yD ) = 0
( NC cos 39.806° )  266.67 − ( 55.0 − 36.0 ) mm + NC sin ( 39.806° )( 60 mm ) − ( 4 N )( 266.67 − 55.0 ) mm = 0
∴ NC = 3.7025 N
or
NC = 3.70 N
39.8°
ΣFx = 0: − 4 N + NC cosθ + Q sin β = 0
−4 N + ( 3.7025 N ) cos 39.806° + Q sin12.6804° = 0
∴ Q = 5.2649 N
or
Q = 5.26 N
77.3°
ΣFy = 0: N B + NC sin θ − Q cos β = 0
N B + ( 3.7025 N ) sin 39.806° − ( 5.2649 N ) cos12.6804° = 0
∴ N B = 2.7662 N
or
(a)
(b)
N B = 2.77 N
N B = 2.77 N , NC = 3.70 N
Q = 5.26 N
39.8°
77.3° ( output )
PROBLEM 4.42
A parabolic slot has been cut in plate AD, and the plate has been placed
so that the slot fits two fixed, frictionless pins B and C. The equation of
the slot is y = x 2 /100, where x and y are expressed in mm. Knowing that
the maximum allowable force exerted on the roller at D is 8.5 N,
determine (a) the corresponding magnitude of the input force P, (b) the
force each pin exerts on the plate.
SOLUTION
y =
The equation of the slot is,
x2
100
 dy 
  = slope of slot at C
 dx C
Now
 2x 
=
= 1.200

100  x = 60 mm
∴ α = tan −1 (1.200 ) = 50.194°
and
θ = 90° − α = 90° − 50.194° = 39.806°
Coordinates of C are
xC = 60 mm, yC =
( 60 )2
100
= 36 mm
Also, the coordinates of D are
xD = 60 mm
yD = 46 mm + ( 40 mm ) sin β
where
 120 − 66 
 = 12.6804°
 240 
β = tan −1 
∴ yD = 46 mm + ( 40 mm ) tan12.6804° = 55.000 mm
xE = 0
Note:
yE = yC + ( 60 mm ) tan θ
= 36 mm + ( 60 mm ) tan 39.806°
= 86.001 mm
(a) From f.b.d. of plate AD
ΣM E = 0: P ( yE ) − ( 8.5 N ) sin β  ( yE − yD )
− ( 8.5 N ) cos β  ( 60 mm ) = 0
PROBLEM 4.42 CONITNIUED
P ( 86.001 mm ) − ( 8.5 N ) sin12.6804° ( 31.001 mm )
− ( 8.5 N ) cos12.6804° ( 60 mm ) = 0
∴ P = 6.4581 N
or P = 6.46 N
(b)
ΣFx = 0: P − ( 8.5 N ) sin β − NC cosθ = 0
6.458 N − ( 8.5 N )( sin12.6804° ) − NC ( cos 39.806° ) = 0
∴ NC = 5.9778 N
or NC = 5.98 N
39.8°
ΣFy = 0: N B + NC sin θ − ( 8.5 N ) cos β = 0
N B + ( 5.9778 N ) sin 39.806° − ( 8.5 N ) cos12.6804° = 0
∴ N B = 4.4657 N
or N B = 4.47 N
PROBLEM 4.43
A movable bracket is held at rest by a cable attached at E and by frictionless
rollers. Knowing that the width of post FG is slightly less than the distance
between the rollers, determine the force exerted on the post by each roller
when α = 20o.
SOLUTION
From f.b.d. of bracket
ΣFy = 0: T sin 20° − 60 lb = 0
∴ T = 175.428 lb
Tx = (175.428 lb ) cos 20° = 164.849 lb
Ty = (175.428 lb ) sin 20° = 60 lb
Note: Ty and 60 lb force form a couple of
60 lb (10 in.) = 600 lb ⋅ in.
ΣM B = 0: 164.849 lb ( 5 in.) − 600 lb ⋅ in. + FCD ( 8 in.) = 0
∴ FCD = −28.030 lb
or
FCD = 28.0 lb
ΣFx = 0: FCD + FAB − Tx = 0
−28.030 lb + FAB − 164.849 lb = 0
∴ FAB = 192.879 lb
or
FAB = 192.9 lb
Rollers A and C can only apply a horizontal force to the right onto the
vertical post corresponding to the equal and opposite force to the left on
the bracket. Since FAB is directed to the right onto the bracket, roller B
will react FAB. Also, since FCD is acting to the left on the bracket, it will
act to the right on the post at roller C.
PROBLEM 4.43 CONTINUED
∴ A=D=0
B = 192.9 lb
C = 28.0 lb
Forces exerted on the post are
A=D=0
B = 192.9 lb
C = 28.0 lb
PROBLEM 4.44
Solve Problem 4.43 when α = 30o.
P4.43 A movable bracket is held at rest by a cable attached at E and by
frictionless rollers. Knowing that the width of post FG is slightly less
than the distance between the rollers, determine the force exerted on the
post by each roller when α = 20o.
SOLUTION
From f.b.d. of bracket
ΣFy = 0: T sin 30° − 60 lb = 0
∴ T = 120 lb
Tx = (120 lb ) cos 30° = 103.923 lb
Ty = (120 lb ) sin 30° = 60 lb
Note: Ty and 60 lb force form a couple of
( 60 lb )(10 in.) = 600 lb ⋅ in.
ΣM B = 0:
(103.923 lb )( 5 in.) − 600 lb ⋅ in. + FCD (8 in.) = 0
∴ FCD = 10.0481 lb
or
FCD = 10.05 lb
ΣFx = 0: FCD + FAB − Tx = 0
10.0481 lb + FAB − 103.923 lb = 0
∴ FAB = 93.875 lb
or
FAB = 93.9 lb
Rollers A and C can only apply a horizontal force to the right on the
vertical post corresponding to the equal and opposite force to the left on
the bracket. The opposite direction apply to roller B and D. Since both
FAB and FCD act to the right on the bracket, rollers B and D will react
these forces.
∴ A=C=0
B = 93.9 lb
D = 10.05 lb
Forces exerted on the post are
A=C=0
B = 93.9 lb
D = 10.05 lb
PROBLEM 4.45
A 20-lb weight can be supported in the three different ways shown.
Knowing that the pulleys have a 4-in. radius, determine the reaction at A
in each case.
SOLUTION
(a) From f.b.d. of AB
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 20 lb = 0
Ay = 20.0 lb
or
and A = 20.0 lb
ΣM A = 0: M A − ( 20 lb )(1.5 ft ) = 0
∴ M A = 30.0 lb ⋅ ft
or M A = 30.0 lb ⋅ ft
 1 ft 
4 in. 
 = 0.33333 ft
 12 in. 
(b) Note:
From f.b.d. of AB
ΣFx = 0: Ax − 20 lb = 0
Ax = 20.0 lb
or
ΣFy = 0: Ay − 20 lb = 0
Ay = 20.0 lb
or
Then
A=
Ax2 + Ay2 =
( 20.0 )2 + ( 20.0 )2
= 28.284 lb
∴ A = 28.3 lb
45°
ΣM A = 0: M A + ( 20 lb )( 0.33333 ft )
− ( 20 lb )(1.5 ft + 0.33333 ft ) = 0
∴ M A = 30.0 lb ⋅ ft
or M A = 30.0 lb ⋅ ft
PROBLEM 4.45 CONTINUED
(c) From f.b.d. of AB
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 20 lb − 20 lb = 0
or
Ay = 40.0 lb
and A = 40.0 lb
ΣM A = 0: M A − ( 20 lb )(1.5 ft − 0.33333 ft )
− ( 20 lb )(1.5 ft + 0.33333 ft ) = 0
∴ M A = 60.0 lb ⋅ ft
or M A = 60.0 lb ⋅ ft
PROBLEM 4.46
A belt passes over two 50-mm-diameter pulleys which are mounted on a
bracket as shown. Knowing that M = 0 and Ti = TO = 24 N, determine
the reaction at C.
SOLUTION
From f.b.d. of bracket
ΣFx = 0: Cx − 24 N = 0
∴ Cx = 24 N
ΣFy = 0: C y − 24 N = 0
∴ C y = 24 N
Then
C =
C x2 + C y2 =
( 24 )2 + ( 24 )2
= 33.941 N
∴ C = 33.9 N
45.0°
ΣM C = 0: M C − ( 24 N ) ( 45 − 25 ) mm 
+ ( 24 N ) ( 25 + 50 − 60 ) mm  = 0
∴ M C = 120 N ⋅ mm
or M C = 0.120 N ⋅ m
PROBLEM 4.47
A belt passes over two 50-mm-diameter pulleys which are mounted on a
bracket as shown. Knowing that M = 0.40 N ⋅ m m and that Ti and TO
are equal to 32 N and 16 N, respectively, determine the reaction at C.
SOLUTION
From f.b.d. of bracket
ΣFx = 0: C x − 32 N = 0
∴ C x = 32 N
ΣFy = 0: C y − 16 N = 0
∴ C y = 16 N
Then
and
C =
C x2 + C y2 =
( 32 )2 + (16 )2
= 35.777 N
 Cy 
−1  16 
 = tan   = 26.565°
 32 
 Cx 
θ = tan −1 
or C = 35.8 N
26.6°
ΣM C = 0: M C − ( 32 N )( 45 mm − 25 mm )
+ (16 N )( 25 mm + 50 mm − 60 mm ) − 400 N ⋅ mm = 0
∴ M C = 800 N ⋅ mm
or M C = 0.800 N ⋅ m
PROBLEM 4.48
A 350-lb utility pole is used to support at C the end of an electric wire.
The tension in the wire is 120 lb, and the wire forms an angle of 15°
with the horizontal at C. Determine the largest and smallest allowable
tensions in the guy cable BD if the magnitude of the couple at A may not
exceed 200 lb ⋅ ft.
SOLUTION
First note
LBD =
Tmax :
( 4.5)2 + (10 )2
= 10.9659 ft
From f.b.d. of utility pole with M A = 200 lb ⋅ ft
ΣM A = 0: − 200 lb ⋅ ft − (120 lb ) cos15° (14 ft )
 4.5 

+ 
 Tmax  (10 ft ) = 0
 10.9659 

∴ Tmax = 444.19 lb
or Tmax = 444 lb
Tmin :
From f.b.d. of utility pole with M A = 200 lb ⋅ ft
ΣM A = 0: 200 lb ⋅ ft − (120 lb ) cos15°  (14 ft )
 4.5 

+ 
 Tmin  (10 ft ) = 0
 10.9659 

∴ Tmin = 346.71 lb
or Tmin = 347 lb
PROBLEM 4.49
In a laboratory experiment, students hang the masses shown from a beam
of negligible mass. (a) Determine the reaction at the fixed support A
knowing that end D of the beam does not touch support E. (b) Determine
the reaction at the fixed support A knowing that the adjustable support E
exerts an upward force of 6 N on the beam.
SOLUTION
(
)
WB = mB g = (1 kg ) 9.81 m/s 2 = 9.81 N
(
)
WC = mC g = ( 0.5 kg ) 9.81 m/s 2 = 4.905 N
(a) From f.b.d. of beam ABCD
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − WB − WC = 0
Ay − 9.81 N − 4.905 N = 0
∴ Ay = 14.715 N
or A = 14.72 N
ΣM A = 0: M A − WB ( 0.2 m ) − WC ( 0.3 m ) = 0
M A − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) = 0
∴ M A = 3.4335 N ⋅ m
or M A = 3.43 N ⋅ m
(b) From f.b.d. of beam ABCD
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − WB − WC + 6 N = 0
Ay − 9.81 N − 4.905 N + 6 N = 0
∴ Ay = 8.715 N
or A = 8.72 N
ΣM A = 0: M A − WB ( 0.2 m ) − WC ( 0.3 m ) + ( 6 N )( 0.4 m ) = 0
M A − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + ( 6 N )( 0.4 m ) = 0
∴ M A = 1.03350 N ⋅ m
or M A = 1.034 N ⋅ m
PROBLEM 4.50
In a laboratory experiment, students hang the masses shown from a beam
of negligible mass. Determine the range of values of the force exerted on
the beam by the adjustable support E for which the magnitude of the
couple at A does not exceed 2.5 N ⋅ m.
SOLUTION
(
)
WB = mB g = 1 kg 9.81 m/s 2 = 9.81 N
(
)
WC = mC g = 0.5 kg 9.81 m/s 2 = 4.905 N
Maximum M A value is 2.5 N ⋅ m
Fmin :
From f.b.d. of beam ABCD with M A = 2.5 N ⋅ m
ΣM A = 0: 2.5 N ⋅ m − WB ( 0.2 m ) − WC ( 0.3 m )
+ Fmin ( 0.4 m ) = 0
2.5 N ⋅ m − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + Fmin ( 0.4 m ) = 0
∴ Fmin = 2.3338 N
or
Fmax :
Fmin = 2.33 N
From f.b.d. of beam ABCD with M A = 2.5 N ⋅ m
ΣM A = 0: − 2.5 N ⋅ m − WB ( 0.2 m ) − WC ( 0.3 m )
+ Fmax ( 0.4 m ) = 0
−2.5 N ⋅ m − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + Fmax ( 0.4 m ) = 0
∴ Fmax = 14.8338 N
or
Fmax = 14.83 N
or 2.33 N ≤ FE ≤ 14.83 N
PROBLEM 4.51
Knowing that the tension in wire BD is 300 lb, determine the reaction at
fixed support C for the frame shown.
SOLUTION
From f.b.d. of frame with T = 300 lb
 5
ΣFx = 0: C x − 100 lb +   300 lb = 0
 13 
∴ C x = −15.3846 lb
or
C x = 15.3846 lb
 12 
ΣFy = 0: C y − 180 lb −   300 lb = 0
 13 
∴ C y = 456.92 lb
Then
and
C =
C x2 + C y2 =
or
C y = 456.92 lb
(15.3846 )2 + ( 456.92 )2
= 457.18 lb
 Cy 
−1  456.92 
 = tan 
 = −88.072°
C
 −15.3846 
 x
θ = tan −1 
or C = 457 lb
88.1°
 12 

ΣM C = 0: M C + (180 lb )( 20 in.) + (100 lb )(16 in.) −   300 lb  (16 in.) = 0
 13 

∴ M C = −769.23 lb ⋅ in.
or M C = 769 lb ⋅ in.
PROBLEM 4.52
Determine the range of allowable values of the tension in wire BD if the
magnitude of the couple at the fixed support C is not to exceed 75 lb ⋅ ft.
SOLUTION
Tmax
From f.b.d. of frame with M C = 75 lb ⋅ ft
= 900 lb ⋅ in.
 12 

ΣM C = 0: 900 lb ⋅ in. + (180 lb )( 20 in.) + (100 lb )(16 in.) −   Tmax  (16 in.) = 0
 13 

∴ Tmax = 413.02 lb
Tmin
From f.b.d. of frame with
M C = 75 lb ⋅ ft
= 900 lb ⋅ in.
 12 

ΣM C = 0: − 900 lb ⋅ in. + (180 lb )( 20 in.) + (100 lb )(16 in.) −   Tmin  (16 in.) = 0
 13 

∴ Tmin = 291.15 lb
∴ 291 lb ≤ T ≤ 413 lb
PROBLEM 4.53
Uniform rod AB of length l and weight W lies in a vertical plane and is
acted upon by a couple M. The ends of the rod are connected to small
rollers which rest against frictionless surfaces. (a) Express the angle θ
corresponding to equilibrium in terms of M, W, and l. (b) Determine the
value of θ corresponding to equilibrium when M = 1.5 lb ⋅ ft,
W = 4 lb, and l = 2 ft.
SOLUTION
(a) From f.b.d. of uniform rod AB
ΣFx = 0: − A cos 45° + B cos 45° = 0
∴ −A + B = 0
or
B= A
(1)
ΣFy = 0: A sin 45° + B sin 45° − W = 0
∴ A+B =
2W
(2)
From Equations (1) and (2)
2A =
2W
∴ A=
1
W
2
From f.b.d. of uniform rod AB
 l 

ΣM B = 0: W   cosθ  + M
 2 

 1

−
W  l cos ( 45° − θ )  = 0
2


From trigonometric identity
cos (α − β ) = cos α cos β + sin α sin β
Equation (3) becomes
 Wl 
 Wl 
  cosθ + M −   ( cosθ + sin θ ) = 0
 2 
 2 
(3)
PROBLEM 4.53 CONTINUED
or
 Wl 
 Wl 
 Wl 
  cosθ + M −   cosθ −   sin θ = 0
 2 
 2 
 2 
∴ sin θ =
2M
Wl
 2M 
or θ = sin −1 

 Wl 
(b)
 2 (1.5 lb ⋅ ft ) 
 = 22.024°
 ( 4 lb )( 2 ft ) 
θ = sin −1 
or θ = 22.0°
PROBLEM 4.54
A slender rod AB, of weight W, is attached to blocks A and B, which
move freely in the guides shown. The blocks are connected by an elastic
cord which passes over a pulley at C. (a) Express the tension in the cord
in terms of W and θ . (b) Determine the value of θ for which the tension
in the cord is equal to 3W.
SOLUTION
(a) From f.b.d. of rod AB
 l 

ΣM C = 0: T ( l sin θ ) + W   cosθ  − T ( l cosθ ) = 0
 2 

∴T =
W cosθ
2 ( cosθ − sin θ )
Dividing both numerator and denominator by cosθ ,
T =
W
1



2  1 − tan θ 
W 
 
 2
or T =
(1 − tan θ )
(b) For T = 3W ,
W 
 
 2
3W =
(1 − tan θ )
∴ 1 − tan θ =
or
1
6
5
θ = tan −1   = 39.806°
6
or θ = 39.8°
PROBLEM 4.55
A thin, uniform ring of mass m and radius R is attached by a frictionless
pin to a collar at A and rests against a small roller at B. The ring lies in a
vertical plane, and the collar can move freely on a horizontal rod and is
acted upon by a horizontal force P. (a) Express the angle θ
corresponding to equilibrium in terms of m and P. (b) Determine the
value of θ corresponding to equilibrium when m = 500 g and P = 5 N.
SOLUTION
(a) From f.b.d. of ring
ΣM C = 0: P ( R cosθ + R cosθ ) − W ( R sin θ ) = 0
2P = W tan θ where W = mg
∴ tan θ =
2P
mg
 2P 
or θ = tan −1 

 mg 
(b) Have
m = 500 g = 0.500 kg and P = 5 N

2 (5 N )
∴ θ = tan −1 
 ( 0.500 kg ) 9.81 m/s 2

(
)




= 63.872°
or θ = 63.9°
PROBLEM 4.56
Rod AB is acted upon by a couple M and two forces, each of magnitude
P. (a) Derive an equation in θ , P, M, and l which must be satisfied when
the rod is in equilibrium. (b) Determine the value of θ corresponding to
equilibrium when M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.
SOLUTION
(a) From f.b.d. of rod AB
ΣM C = 0: P ( l cosθ ) + P ( l sin θ ) − M = 0
or sin θ + cosθ =
(b) For
M
Pl
M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.
sin θ + cosθ =
150 lb ⋅ in.
5
= = 1.25
( 20 lb )( 6 in.) 4
sin 2 θ + cos 2 θ = 1
Using identity
(
sin θ + 1 − sin 2 θ
(
1 − sin 2 θ
)
1
2
)
1
2
= 1.25
= 1.25 − sin θ
1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ
2sin 2 θ − 2.5sin θ + 0.5625 = 0
Using quadratic formula
sin θ =
=
or
− ( −2.5 ) ±
( 6.25) − 4 ( 2 )( 0.5625)
2 ( 2)
2.5 ± 1.75
4
sin θ = 0.95572
∴ θ = 72.886°
and
and
sin θ = 0.29428
θ = 17.1144°
or θ = 17.11° and θ = 72.9°
PROBLEM 4.57
A vertical load P is applied at end B of rod BC. The constant of the spring
is k, and the spring is unstretched when θ = 90o. (a) Neglecting the
weight of the rod, express the angle θ corresponding to equilibrium in
terms of P, k, and l. (b) Determine the value of θ corresponding to
1
equilibrium when P = kl.
4
SOLUTION
First note
T = tension in spring = ks
s = elongation of spring
where
( )θ − ( AB )θ
= AB
= 90°
θ 
 90° 
= 2l sin   − 2l sin 

2
 2 
  θ   1 
= 2l sin   − 

  2   2 
  θ   1 
∴ T = 2kl sin   − 

  2   2 
(a) From f.b.d. of rod BC

 θ 
ΣM C = 0: T l cos    − P ( l sin θ ) = 0
 2 

Substituting T From Equation (1)
  θ   1  
 θ 
2kl sin   − 
  l cos  2   − P ( l sin θ ) = 0
2
 
    2  
  θ   1 

θ 
θ 
 θ 
2kl 2 sin   − 
  cos  2  − Pl  2sin  2  cos  2   = 0
2
 
 
 

    2 
Factoring out
θ 
2l cos   , leaves
2
(1)
PROBLEM 4.57 CONTINUED
  θ   1 
θ 
kl sin   − 
 − P sin   = 0

2
  2   2 
or
1  kl 
θ 
sin   =


2  kl − P 
2


kl
∴ θ = 2sin −1 

 2 ( kl − P ) 
(b) P =
kl
4

kl
2 kl −

θ = 2sin −1 
(
kl
4
)

 kl  4  
−1  4 
 = 2sin −1 

  = 2sin 

3 2 
 2  3 kl  

= 2sin −1 ( 0.94281)
= 141.058°
or θ = 141.1°
PROBLEM 4.58
Solve Sample Problem 4.5 assuming that the spring is unstretched when
θ = 90o.
SOLUTION
First note
T = tension in spring = ks
s = deformation of spring
where
= rβ
∴ F = kr β
From f.b.d. of assembly
ΣM 0 = 0: W ( l cos β ) − F ( r ) = 0
Wl cos β − kr 2 β = 0
or
∴ cos β =
For
kr 2
β
Wl
k = 250 lb/in., r = 3 in., l = 8 in., W = 400 lb
cos β =
or
( 250 lb/in.)( 3 in.)2 β
( 400 lb )(8 in.)
cos β = 0.703125β
Solving numerically,
β = 0.89245 rad
or
Then
β = 51.134°
θ = 90° + 51.134° = 141.134°
or θ = 141.1°
PROBLEM 4.59
A collar B of weight W can move freely along the vertical rod shown. The
constant of the spring is k, and the spring is unstretched when θ = 0.
(a) Derive an equation in θ , W, k, and l which must be satisfied when the
collar is in equilibrium. (b) Knowing that W = 3 lb, l = 6 in., and
k = 8 lb/ft, determine the value of θ corresponding to equilibrium.
SOLUTION
T = ks
First note
k = spring constant
where
s = elongation of spring
=
l
l
−l =
(1 − cosθ )
cosθ
cosθ
∴ T =
kl
(1 − cosθ )
cosθ
(a) From f.b.d. of collar B
ΣFy = 0: T sin θ − W = 0
kl
(1 − cosθ ) sin θ − W = 0
cosθ
or
or tan θ − sin θ =
W
kl
(b) For W = 3 lb, l = 6 in., k = 8 lb/ft
l =
6 in.
= 0.5 ft
12 in./ft
tan θ − sin θ =
3 lb
= 0.75
(8 lb/ft )( 0.5 ft )
Solving Numerically,
θ = 57.957°
or θ = 58.0°
PROBLEM 4.60
A slender rod AB, of mass m, is attached to blocks A and B which move
freely in the guides shown. The constant of the spring is k, and the spring
is unstretched when θ = 0 . (a) Neglecting the mass of the blocks, derive
an equation in m, g, k, l, and θ which must be satisfied when the rod is in
equilibrium. (b) Determine the value of θ when m = 2 kg, l = 750
mm, and k = 30 N/m.
SOLUTION
First note
Fs = spring force = ks
k = spring constant
where
s = spring deformation
= l − l cosθ
= l (1 − cosθ )
∴ Fs = kl (1 − cosθ )
(a) From f.b.d. of assembly
l

ΣM D = 0: Fs ( l sin θ ) − W  cosθ  = 0
2

l

kl (1 − cosθ )( l sin θ ) − W  cosθ  = 0
2

W
kl ( sin θ − cosθ sin θ ) − 
 2

 cosθ = 0

Dividing by cosθ
kl ( tan θ − sin θ ) =
W
2
∴ tan θ − sin θ =
W
2kl
or tan θ − sin θ =
(b) For m = 2 kg, l = 750 mm, k = 30 N/m
l = 750 mm = 0.750 m
mg
2kl
PROBLEM 4.60 CONTINUED
Then
tan θ − sin θ =
( 2 kg ) ( 9.81 m/s2 )
2 ( 30 N/m )( 0.750 m )
= 0.436
Solving Numerically,
θ = 50.328°
or θ = 50.3°
PROBLEM 4.61
The bracket ABC can be supported in the eight different ways shown. All
connections consist of smooth pins, rollers, or short links. In each case,
determine whether (a) the plate is completely, partially, or improperly
constrained, (b) the reactions are statically determinate or indeterminate,
(c) the equilibrium of the plate is maintained in the position shown. Also,
wherever possible, compute the reactions assuming that the magnitude of
the force P is 100 N.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket:
ΣM A = 0: B (1 m ) − (100 N )( 0.6 m ) = 0
∴ B = 60.0 N
ΣFx = 0: Ax − 60 N = 0
∴ A x = 60.0 N
ΣFy = 0: Ay − 100 N = 0
∴ A y = 100 N
Then
A=
( 60.0 )2 + (100 )2
= 116.619 N
 100 
θ = tan −1 
 = 59.036°
 60.0 
and
∴ A = 116.6 N
59.0°
2. Four concurrent reactions through A
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
3. Two reactions
(a)
Partially constrained
(b)
Determinate
(c)
Equilibrium
PROBLEM 4.61 CONTINUED
From f.b.d. of bracket
ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0
∴ C = 50.0 N
ΣFy = 0: A − 100 N + 50 N = 0
∴ A = 50.0 N
4. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket
 1.0 
θ = tan −1   = 39.8°
 1.2 
BC =
(1.2 )2 + (1.0 )2
= 1.56205 m
 1.2  
ΣM A = 0: 
 B  (1 m ) − (100 N )( 0.6 m ) = 0
 1.56205  
∴ B = 78.1 N
39.8°
ΣFx = 0: C − ( 78.102 N ) cos 39.806° = 0
∴ C = 60.0 N
ΣFy = 0: A + ( 78.102 N ) sin 39.806° − 100 N = 0
∴ A = 50.0 N
5. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of bracket
ΣM C = 0:
(100 N )( 0.6 m ) − Ay (1.2 m ) = 0
∴ Ay = 50 N
or A y = 50.0 N
6. Four non-concurrent non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
PROBLEM 4.61 CONTINUED
From f.b.d. of bracket
ΣM A = 0: − Bx (1 m ) − (100 N )( 0.6 m ) = 0
∴ Bx = −60.0 N
or B x = 60.0 N
ΣFx = 0: − 60 + Ax = 0
∴ Ax = 60.0 N
or A x = 60.0 N
7. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket
ΣFx = 0: Ax = 0
ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0
∴ C = 50.0 N
or C = 50.0 N
ΣFy = 0: Ay − 100 N + 50.0 N = 0
∴ Ay = 50.0 N
∴ A = 50.0 N
8. Three concurrent, non-parallel reactions
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
PROBLEM 4.62
Eight identical 20 × 30-in. rectangular plates, each weighing 50 lb, are
held in a vertical plane as shown. All connections consist of frictionless
pins, rollers, or short links. For each case, answer the questions listed in
Problem 4.61, and, wherever possible, compute the reactions.
P6.1 The bracket ABC can be supported in the eight different ways
shown. All connections consist of smooth pins, rollers, or short links. In
each case, determine whether (a) the plate is completely, partially, or
improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the
position shown. Also, wherever possible, compute the reactions assuming
that the magnitude of the force P is 100 N.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
ΣM A = 0: C ( 30 in.) − 50 lb (15 in.) = 0
C = 25.0 lb
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 50 lb + 25 lb = 0
Ay = 25 lb
A = 25.0 lb
2. Three non-current, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
ΣFx = 0:
ΣM B = 0:
B=0
( 50 lb )(15 in.) − D ( 30 in.) = 0
D = 25.0 lb
ΣFy = 0: 25.0 lb − 50 lb + C = 0
C = 25.0 lb
PROBLEM 4.62 CONTINUED
3. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of plate
ΣM D = 0: Ax ( 20 in.) − ( 50 lb )(15 in.)
∴ A x = 37.5 lb
ΣFx = 0: Dx + 37.5 lb = 0
∴ D x = 37.5 lb
4. Three concurrent reactions
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
5. Two parallel reactions
(a)
Partial constraint
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
ΣM D = 0: C ( 30 in.) − ( 50 lb )(15 in.) = 0
C = 25.0 lb
ΣFy = 0: D − 50 lb + 25 lb = 0
D = 25.0 lb
6. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
ΣM D = 0: B ( 20 in.) − ( 50 lb )(15 in.) = 0
B = 37.5 lb
ΣFx = 0: Dx + 37.5 lb = 0
ΣFy = 0: Dy − 50 lb = 0
D x = 37.5 lb
D y = 50.0 lb
or D = 62.5 lb
53.1°
PROBLEM 4.62 CONTINUED
7. Two parallel reactions
(a)
Improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
8. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of plate
ΣM D = 0: B ( 30 in.) − ( 50 lb )(15 in.) = 0
B = 25.0 lb
ΣFy = 0: Dy − 50 lb + 25.0 lb = 0
D y = 25.0 lb
ΣFx = 0: Dx + C = 0
PROBLEM 4.63
Horizontal and vertical links are hinged to a wheel, and forces are applied
to the links as shown. Knowing that a = 3.0 in., determine the value of P
and the reaction at A.
SOLUTION
As shown on the f.b.d., the wheel is a three-force body. Let point D be
the intersection of the three forces.
From force triangle
A P
21 lb
=
=
5
4
3
∴ P=
4
( 21 lb ) = 28 lb
3
or P = 28.0 lb
and
A=
5
( 21 lb ) = 35 lb
3
3
 
θ = tan −1   = 36.870°
4
∴ A = 35.0 lb
36.9°
PROBLEM 4.64
Horizontal and vertical links are hinged to a wheel, and forces are applied
to the links as shown. Determine the range of values of the distance a for
which the magnitude of the reaction at A does not exceed 42 lb.
SOLUTION
Let D be the intersection of the three forces acting on the wheel.
From the force triangle
21 lb
=
a
or
16 + a 2
A = 21
16
+1
a2
A = 42 lb
For
21 lb
=
a
a2 =
or
or
A
a=
42 lb
16 + a 2
16 + a 2
4
16
= 2.3094 in.
3
or a ≥ 2.31 in.
Since
as a increases, A decreases
A = 21
16
+1
a2
PROBLEM 4.65
Using the method of Section 4.7, solve Problem 4.21.
P4.21 The required tension in cable AB is 800 N. Determine (a) the
vertical force P which must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
Let E be the intersection of the three forces acting on the pedal device.
First note
 (180 mm ) sin 60° 
 = 21.291°
400 mm


α = tan −1 
From force triangle
(a)
P = ( 800 N ) tan 21.291°
= 311.76 N
or P = 312 N
(b)
C =
800 N
cos 21.291°
= 858.60 N
or C = 859 N
21.3°
PROBLEM 4.66
Using the method of Section 4.7, solve Problem 4.22.
P4.22 Determine the maximum tension which can be developed in cable
AB if the maximum allowable value of the reaction at C is 1000 N.
SOLUTION
Let E be the intersection of the three forces acting on the pedal device.
First note
 (180 mm ) sin 60° 
 = 21.291°
400 mm


α = tan −1 
From force triangle
Tmax = (1000 N ) cos 21.291°
= 931.75 N
or Tmax = 932 N
PROBLEM 4.67
To remove a nail, a small block of wood is placed under a crowbar, and a
horizontal force P is applied as shown. Knowing that l = 3.5 in. and
P = 30 lb, determine the vertical force exerted on the nail and the
reaction at B.
SOLUTION
Let D be the intersection of the three forces acting on the crowbar.
First note
 ( 36 in.) sin 50° 
 = 82.767°
3.5 in.


θ = tan −1 
From force triangle
FN = P tan θ = ( 30 lb ) tan 82.767°
= 236.381 lb
∴ on nail FN = 236 lb
RB =
P
30 lb
=
= 238.28 lb
cosθ
cos82.767°
or R B = 238 lb
82.8°
PROBLEM 4.68
To remove a nail, a small block of wood is placed under a crowbar, and a
horizontal force P is applied as shown. Knowing that the maximum
vertical force needed to extract the nail is 600 lb and that the horizontal
force P is not to exceed 65 lb, determine the largest acceptable value of
distance l.
SOLUTION
Let D be the intersection of the three forces acting on the crowbar.
From force diagram
tan θ =
FN
600 lb
=
= 9.2308
P
65 lb
∴ θ = 83.817°
From f.b.d.
tan θ =
∴ l =
( 36 in.) sin 50°
l
( 36 in.) sin 50°
tan 83.817°
= 2.9876 in.
or l = 2.99 in.
PROBLEM 4.69
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since member BD is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and
be opposite in direction for BD to be in equilibrium. The force B acting at B of member ABC will be equal in
magnitude but opposite in direction to force B acting on member BD. Member ABC is a three-force body with
member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate the above conditions. The
force triangle for member ABC is also shown. The angles α and β are found from the member dimensions:
 0.5 m 
 = 26.565°
 1.0 m 
α = tan −1 
 1.5 m 
β = tan −1 
 = 56.310°
 1.0 m 
Applying the law of sines to the force triangle for member ABC,
150 N
C
B
=
=
sin ( β − α ) sin ( 90° + α ) sin ( 90° − β )
or
150 N
C
B
=
=
sin 29.745° sin116.565° sin 33.690°
∴ C =
and
(150 N ) sin116.565°
D= B=
sin 29.745°
= 270.42 N
(150 N ) sin 33.690°
sin 29.745°
or C = 270 N
56.3°
or D = 167.7 N
26.6°
= 167.704 N
PROBLEM 4.70
For the frame and loading shown, determine the reactions at A and C.
SOLUTION
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with
member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle β is found from the member dimensions:
 60 m 
β = tan −1 
 = 30.964°
 100 m 
Applying of the law of sines to the force triangle for member BCD,
130 N
B
C
=
=
sin ( 45° − β ) sin β
sin135°
or
130 N
B
C
=
=
sin14.036° sin 30.964° sin135°
∴ A= B=
and
C =
(130 N ) sin 30.964°
sin14.036°
(130 N ) sin135°
sin14.036°
= 275.78 N
or A = 276 N
45.0°
or C = 379 N
59.0°
= 379.02 N
PROBLEM 4.71
To remove the lid from a 5-gallon pail, the tool shown is used to apply an
upward and radially outward force to the bottom inside rim of the lid.
Assuming that the rim rests against the tool at A and that a 100-N force is
applied as indicated to the handle, determine the force acting on the rim.
SOLUTION
The three-force member ABC has forces that intersect at D, where

α = tan −1 
 yDC

90 mm

− yBC − 45 mm 
and
yDC =
( 360 mm ) cos 35°
xBC
=
tan 20°
tan 20°
= 810.22 mm
yBC = ( 360 mm ) sin 35°
= 206.49 mm
 90 
∴ α = tan −1 
 = 9.1506°
 558.73 
Based on the force triangle, the law of sines gives
100 N
A
=
sin α
sin 20°
∴A=
or
(100 N ) sin 20°
sin 9.1506°
A = 215 N
= 215.07 N
80.8° on tool
and A = 215 N
80.8° on rim of can
PROBLEM 4.72
To remove the lid from a 5-gallon pail, the tool shown is used to apply
an upward and radially outward force to the bottom inside rim of the
lid. Assuming that the top and the rim of the lid rest against the tool at
A and B, respectively, and that a 60-N force is applied as indicated to
the handle, determine the force acting on the rim.
SOLUTION
The three-force member ABC has forces that intersect at point D, where,
from the law of sines ( ∆CDE )
150 mm + (19 mm ) tan 35°
L
=
sin 95°
sin 30°
∴ L = 325.37 mm
Then
 45 mm 

 yBD 
α = tan −1 
where
yBD = L − y AE − 22 mm
= 325.37 mm −
19 mm
− 22 mm
cos 35°
= 280.18 mm
 45 mm 
∴ α = tan −1 
 = 9.1246°
 280.18 mm 
Applying the law of sines to the force triangle,
B
60 N
=
sin150° sin 9.1246°
∴ B = 189.177 N
Or, on member
and, on lid
B = 189.2 N
80.9°
B = 189.2 N
80.9°
PROBLEM 4.73
A 200-lb crate is attached to the trolley-beam system shown. Knowing
that a = 1.5 ft, determine (a) the tension in cable CD, (b) the reaction
at B.
SOLUTION
From geometry of forces
 y

β = tan −1  BE 
 1.5 ft 
where
yBE = 2.0 − yDE
= 2.0 − 1.5 tan 35°
= 0.94969 ft
 0.94969 
∴ β = tan −1 
 = 32.339°
 1.5 
and
α = 90° − β = 90° − 32.339° = 57.661°
θ = β + 35° = 32.339° + 35° = 67.339°
Applying the law of sines to the force triangle,
200 lb
T
B
=
=
sin θ
sin α
sin 55°
( 200 lb )
or
sin 67.339°
(a)
T =
=
T
B
=
sin 57.661° sin 55°
( 200 lb )( sin 57.661° )
sin 67.339°
= 183.116 lb
or T = 183.1 lb
(b)
B=
( 200 lb )( sin 55° )
sin 67.339°
= 177.536 lb
or B = 177.5 lb
32.3°
PROBLEM 4.74
Solve Problem 4.73 assuming that a = 3 ft.
P4.73 A 200-lb crate is attached to the trolley-beam system shown.
Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (b) the
reaction
at B.
SOLUTION
From geometry of forces
y

β = tan −1  BE 
 3 ft 
where
yBE = yDE − 2.0 ft
= 3tan 35° − 2.0
= 0.100623 ft
 0.100623 
∴ β = tan −1 
 = 1.92103°
3


and
α = 90° + β = 90° + 1.92103° = 91.921°
θ = 35° − β = 35° − 1.92103° = 33.079°
Applying the law of sines to the force triangle,
200 lb
T
B
=
=
sin θ
sin α
sin 55°
or
200 lb
T
B
=
=
sin 33.079° sin 91.921° sin 55°
(a)
T =
( 200 lb )( sin 91.921° )
sin 33.079°
= 366.23 lb
or T = 366 lb
(b)
B=
( 200 lb )( sin 55° )
sin 33.079°
= 300.17 lb
or B = 300 lb
1.921°
PROBLEM 4.75
A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 8 mm, determine the force P required to move the roller
onto the tiles if the roller is pushed to the left.
SOLUTION
Based on the roller having impending motion to the left, the only contact
between the roller and floor will be at the edge of the tile.
First note
(
)
W = mg = ( 20 kg ) 9.81 m/s 2 = 196.2 N
From the geometry of the three forces acting on the roller
 92 mm 
 = 23.074°
 100 mm 
α = cos −1 
and
θ = 90° − 30° − α
= 60° − 23.074
= 36.926°
Applying the law of sines to the force triangle,
W
P
=
sin θ
sin α
or
196.2 N
P
=
sin 36.926° sin 23.074°
∴ P = 127.991 N
or P = 128.0 N
30°
PROBLEM 4.76
A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 8 mm, determine the force P required to move the roller
onto the tiles if the roller is pulled to the right.
SOLUTION
Based on the roller having impending motion to the right, the only
contact between the roller and floor will be at the edge of the tile.
First note
(
W = mg = ( 20 kg ) 9.81 m/s 2
)
= 196.2 N
From the geometry of the three forces acting on the roller
 92 mm 
α = cos −1 
 = 23.074°
 100 mm 
and
θ = 90° + 30° − α
= 120° − 23.074°
= 96.926°
Applying the law of sines to the force triangle,
W
P
=
sin θ
sin α
or
196.2 N
P
=
sin 96.926° sin 23.074
∴ P = 77.460 N
or P = 77.5 N
30°
PROBLEM 4.77
A small hoist is mounted on the back of a pickup truck and is used to lift
a 120-kg crate. Determine (a) the force exerted on the hoist by the
hydraulic cylinder BC, (b) the reaction at A.
SOLUTION
First note
(
)
W = mg = (120 kg ) 9.81 m/s 2 = 1177.2 N
From the geometry of the three forces acting on the small hoist
x AD = (1.2 m ) cos 30° = 1.03923 m
y AD = (1.2 m ) sin 30° = 0.6 m
and
Then
yBE = x AD tan 75° = (1.03923 m ) tan 75° = 3.8785 m
 yBE − 0.4 m 
−1  3.4785 
 = tan 
 = 73.366°
x AD
 1.03923 


α = tan −1 
β = 75° − α = 75° − 73.366° = 1.63412°
θ = 180° − 15° − β = 165° − 1.63412° = 163.366°
Applying the law of sines to the force triangle,
W
B
A
=
=
sin β
sin θ
sin15°
or
(a)
1177.2 N
B
A
=
=
sin1.63412° sin163.366° sin15°
B = 11 816.9 N
or B = 11.82 kN
(b)
75.0°
A = 10 684.2 N
or A = 10.68 kN
73.4°
PROBLEM 4.78
The clamp shown is used to hold the rough workpiece C. Knowing that
the maximum allowable compressive force on the workpiece is 200 N
and neglecting the effect of friction at A, determine the corresponding
(a) reaction at B, (b) reaction at A, (c) tension in the bolt.
SOLUTION
From the geometry of the three forces acting on the clamp
y AD = (105 mm ) tan 78° = 493.99 mm
yBD = y AD − 70 mm = ( 493.99 − 70 ) mm = 423.99 mm
Then

y

 423.99 
−1
BD
θ = tan −1 
 = tan 
 = 65.301°
 195 mm 
 195 
α = 90° − θ − 12° = 78° − 65.301° = 12.6987°
(a) Based on the maximum allowable compressive force on the
workpiece of 200 N,
( RB ) y
= 200 N
RB sin θ = 200 N
or
∴ RB =
200 N
= 220.14 N
sin 65.301°
or R B = 220 N
65.3°
Applying the law of sines to the force triangle,
RB
NA
T
=
=
sin12° sin α
sin ( 90° + θ )
or
(b)
220.14 N
NA
T
=
=
sin12°
sin12.6987° sin155.301°
N A = 232.75 N
or N A = 233 N
(c)
T = 442.43 N
or T = 442 N
PROBLEM 4.79
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg.
Knowing that θ = 45° and that the force exerted at C by the worker is
perpendicular to the handle of the peavey, determine (a) the force exerted
at C, (b) the reaction at A.
SOLUTION
First note
(
)
W = mg = ( 36 kg ) 9.81 m/s 2 = 353.16 N
From the geometry of the three forces acting on the modified peavey

1.1 m

β = tan −1 
 = 40.236°
 1.1 m + 0.2 m 
α = 45° − β = 45° − 40.236° = 4.7636°
Applying the law of sines to the force triangle,
W
C
A
=
=
sin β
sin α
sin135°
or
(a)
(b)
353.16 N
C
A
=
=
sin 40.236° sin 4.7636 sin135°
C = 45.404 N
or C = 45.4 N
45.0°
or A = 387 N
85.2°
A = 386.60 N
PROBLEM 4.80
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg.
Knowing that θ = 60° and that the force exerted at C by the worker is
perpendicular to the handle of the peavey, determine (a) the force exerted
at C, (b) the reaction at A.
SOLUTION
First note
(
)
W = mg = ( 36 kg ) 9.81 m/s 2 = 353.16 N
From the geometry of the three forces acting on the modified peavey


1.1 m
β = tan −1 

 DC + 0.2 m 
where
DC = (1.1 m + a ) tan 30°
 R 
a=
−R
 tan 30° 
 0.1 m 
=
 − 0.1 m
 tan 30° 
= 0.073205 m
∴ DC = (1.173205 ) tan 30°
= 0.67735 m
and

1.1

β = tan −1 
 = 51.424°
 0.87735 
α = 60° − β = 60° − 51.424° = 8.5756°
Applying the law of sines to the force triangle,
W
C
A
=
=
sin β
sin α
sin120°
or
(a)
353.16 N
C
A
=
=
sin 51.424° sin 8.5756° sin120°
C = 67.360 N
or C = 67.4 N
(b)
30°
A = 391.22 N
or A = 391 N
81.4°
PROBLEM 4.81
Member ABC is supported by a pin and bracket at B and by an
inextensible cord at A and C and passing over a frictionless pulley at D.
The tension may be assumed to be the same in portion AD and CD of the
cord. For the loading shown and neglecting the size of the pulley,
determine the tension in the cord and the reaction at B.
SOLUTION
From the f.b.d. of member ABC, it is seen that the member can be treated
as a three-force body.
From the force triangle
T − 300 3
=
T
4
3T = 4T − 1200
∴ T = 1200 lb
B
5
=
T
4
Also,
∴ B=
5
5
T = (1200 lb ) = 1500 lb
4
4
3
θ = tan −1   = 36.870°
4
and B = 1500 lb
36.9°
PROBLEM 4.82
Member ABCD is supported by a pin and bracket at C and by an
inextensible cord attached at A and D and passing over frictionless
pulleys at B and E. Neglecting the size of the pulleys, determine the
tension in the cord and the reaction at C.
SOLUTION
From the geometry of the forces acting on member ABCD
 200 
β = tan −1 
 = 33.690°
 300 
 375 
 = 61.928°
 200 
α = tan −1 
α − β = 61.928° − 33.690° = 28.237°
180° − α = 180° − 61.928° = 118.072°
Applying the law of sines to the force triangle,
T − 80 N
T
C
=
=
sin (α − β ) sin β
sin (180° − α )
or
T − 80 N
T
C
=
=
sin 28.237° sin 33.690° sin118.072°
Then
(T
− 80 N ) sin 33.690° = T sin 28.237°
∴ T = 543.96 N
or T = 544 N
and
( 543.96 N ) sin118.072 = C sin 33.690°
∴ C = 865.27 N
or C = 865 N
33.7°
PROBLEM 4.83
Using the method of Section 4.7, solve Problem 4.18.
P4.18 Determine the reactions at A and B when (a) h = 0 , (b) h = 8 in.
SOLUTION
(a) Based on symmetry
α = 30°
From force triangle
A = B = 40 lb
or A = 40.0 lb
30°
and B = 40.0 lb
30°
(b) From geometry of forces
 8 in. − (10 in.) tan 30° 
 = 12.5521°
10 in.


α = tan −1 
Also,
30° − α = 30° − 12.5521° = 17.4479°
90° + α = 90° + 12.5521° = 102.5521°
Applying law of sines to the force triangle,
40 lb
A
B
=
=
sin ( 30° − α ) sin 60° sin ( 90° + α )
or
40 lb
A
B
=
=
sin17.4479° sin 60° sin102.5521
A = 115.533 lb
or A = 115.5 lb
12.55°
B = 130.217 lb
or B = 130.2 lb
30.0°
PROBLEM 4.84
Using the method of Section 4.7, solve Problem 4.28.
P4.28 A lever is hinged at C and is attached to a control cable at A. If the
lever is subjected to a 300-N vertical force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
From geometry of forces acting on lever
 yDA 

 xDA 
α = tan −1 
where
yDA = 0.24 m − y AC = 0.24 m − ( 0.2 m ) sin 20°
= 0.171596 m
xDA = ( 0.2 m ) cos 20°
= 0.187939 m
 0.171596 
∴ α = tan −1 
 = 42.397°
 0.187939 
 y AC + yEA 

xCE


β = 90° − tan −1 
where
xCE = ( 0.3 m ) cos 20° = 0.28191 m
y AC = ( 0.2 m ) sin 20° = 0.068404 m
yEA = ( xDA + xCE ) tan α
= ( 0.187939 + 0.28191) tan 42.397°
= 0.42898 m
 0.49739 
∴ β = 90° − tan −1 
 = 29.544°
 0.28191 
Also,
90° − (α + β ) = 90° − 71.941° = 18.0593°
90° + α = 90° + 42.397° = 132.397°
PROBLEM 4.84 CONTINUED
Applying the law of sines to the force triangle,
300 N
T
C
=
=
sin ( 90° + α )
sin 90° − (α + β )  sin β
or
300 N
T
C
=
=
sin18.0593° sin 29.544° sin132.397°
(a) T = 477.18 N
(b) C = 714.67 N
or T = 477 N
or C = 715 N
60.5°
PROBLEM 4.85
Knowing that θ = 35o , determine the reaction (a) at B, (b) at C.
SOLUTION
From the geometry of the three forces applied to the member ABC
y

α = tan −1  CD 
 R 
where
yCD = R tan 55° − R = 0.42815R
∴ α = tan −1 ( 0.42815 ) = 23.178°
Then
55° − α = 55° − 23.178° = 31.822°
90° + α = 90° + 23.178° = 113.178°
Applying the law of sines to the force triangle,
P
B
C
=
=
sin ( 55° − α ) sin ( 90° + α ) sin 35°
or
(a)
P
B
C
=
=
sin 31.822° sin113.178° sin 35°
B = 1.74344P
or B = 1.743P
(b)
55.0°
C = 1.08780P
or C = 1.088P
23.2°
PROBLEM 4.86
Knowing that θ = 50o , determine the reaction (a) at B, (b) at C.
SOLUTION
From the geometry of the three forces acting on member ABC
y

α = tan −1  DC 
 R 
where
yDC = R − y AD = R 1 − tan ( 90° − 50° ) 
= 0.160900 R
∴ α = tan −1 ( 0.160900 ) = 9.1406°
Then
90° − α = 90° − 9.1406° = 80.859°
40° + α = 40° + 9.1406° = 49.141°
Applying the law of sines to the force triangle,
P
B
C
=
=
sin ( 40° + α ) sin ( 90° − α ) sin 50°
or
(a)
(b)
P
B
C
=
=
sin 49.141° sin ( 80.859° ) sin 50°
B = 1.30540P
or B = 1.305P
40.0°
or C = 1.013P
9.14°
C = 1.01286P
PROBLEM 4.87
A slender rod of length L and weight W is held in equilibrium as shown,
with one end against a frictionless wall and the other end attached to a
cord of length S. Derive an expression for the distance h in terms of L and
S. Show that this position of equilibrium does not exist if S > 2L.
SOLUTION
From the f.b.d of the three-force member AB, forces must intersect at D.
Since the force T intersects point D, directly above G,
yBE = h
For triangle ACE:
S 2 = ( AE ) + ( 2h )
2
2
(1)
For triangle ABE:
L2 = ( AE ) + ( h )
2
2
(2)
Subtracting Equation (2) from Equation (1)
S 2 − L2 = 3h 2
(3)
or h =
S 2 − L2
3
As length S increases relative to length L, angle θ increases until rod AB
is vertical. At this vertical position:
h+L=S
or
h≥S−L
Therefore, for all positions of AB
S 2 − L2
≥S−L
3
or
or
or
and
For
h=S−L
(
)
S 2 − L2 ≥ 3 ( S − L ) = 3 S 2 − 2SL + L2 = 3S 2 − 6SL + 3L2
2
0 ≥ 2S 2 − 6SL + 4 L2
0 ≥ S 2 − 3SL + 2 L2 = ( S − L )( S − 2L )
S−L=0
S = L
∴ Minimum value of S is L
For
S − 2L = 0
S = 2L
∴ Maximum value of S is 2L
Therefore, equilibrium does not exist if S > 2L
(4)
PROBLEM 4.88
A slender rod of length L = 200 mm is held in equilibrium as shown,
with one end against a frictionless wall and the other end attached to a
cord of length S = 300 mm. Knowing that the mass of the rod is 1.5 kg,
determine (a) the distance h, (b) the tension in the cord, (c) the reaction
at B.
SOLUTION
From the f.b.d of the three-force member AB, forces must intersect at D.
Since the force T intersects point D, directly above G,
yBE = h
For triangle ACE:
S 2 = ( AE ) + ( 2h )
2
2
(1)
For triangle ABE:
L2 = ( AE ) + ( h )
2
2
(2)
Subtracting Equation (2) from Equation (1)
S 2 − L2 = 3h 2
or h =
(a) For L = 200 mm and
h=
S 2 − L2
3
S = 300 mm
( 300 )2 − ( 200 )2
3
= 129.099 mm
or h = 129.1 mm
(b) Have
(
)
W = mg = (1.5 kg ) 9.81 m/s 2 = 14.715 N
 2h 
 2 (129.099 ) 

300

θ = sin −1   = sin −1 
 s 

and
θ = 59.391°
From the force triangle
T =
W
14.715 N
=
= 17.0973 N
sin θ
sin 59.391°
or T = 17.10 N
(c)
B=
W
14.715 N
=
= 8.7055 N
tan θ
tan 59.391°
or B = 8.71 N
PROBLEM 4.89
A slender rod of length L and weight W is attached to collars which can
slide freely along the guides shown. Knowing that the rod is in
equilibrium, derive an expression for the angle θ in terms of the
angle β .
SOLUTION
As shown in the f.b.d of the slender rod AB, the three forces intersect at
C. From the force geometry
tan β =
xGB
y AB
where
y AB = L cosθ
xGB =
and
∴ tan β =
1
2
1
L sin θ
2
L sin θ
L cosθ
=
1
tan θ
2
or tan θ = 2 tan β
PROBLEM 4.90
A 10-kg slender rod of length L is attached to collars which can slide
freely along the guides shown. Knowing that the rod is in equilibrium and
that β = 25°, determine (a) the angle θ that the rod forms with the
vertical, (b) the reactions at A and B.
SOLUTION
(a) As shown in the f.b.d. of the slender rod AB, the three forces
intersect at C. From the geometry of the forces
xCB
yBC
tan β =
where
xCB =
1
L sin θ
2
yBC = L cosθ
and
∴ tan β =
1
tan θ
2
or
tan θ = 2 tan β
For
β = 25°
tan θ = 2 tan 25° = 0.93262
∴ θ = 43.003°
or θ = 43.0°
(
)
W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N
(b)
From force triangle
A = W tan β
= ( 98.1 N ) tan 25°
= 45.745 N
or A = 45.7 N
and
B=
W
98.1 N
=
= 108.241 N
cos β
cos 25°
or B = 108.2 N
65.0°
PROBLEM 4.91
A uniform slender rod of mass 5 g and length 250 mm is balanced on a
glass of inner diameter 70 mm. Neglecting friction, determine the angle θ
corresponding to equilibrium.
SOLUTION
From the geometry of the forces acting on the three-force member AB
Triangle ACF
yCF = d tan θ
xFE = yCF tan θ = d tan 2 θ
Triangle CEF
Triangle AGE
cosθ =
=
d + xFE
d + d tan 2 θ
=
L
L
 
 
2
2
(
2d
1 + tan 2 θ
L
Now
(1 + tan θ ) = sec θ
Then
cosθ =
2
2
)
1
cosθ
2d
2d  1 
sec 2 θ =


L
L  cos 2 θ 
∴ cos3 θ =
For
secθ =
and
d = 70 mm and
cos3 θ =
2d
L
L = 250 mm
2 ( 70 )
= 0.56
250
∴ cosθ = 0.82426
and
θ = 34.487°
or θ = 34.5°
PROBLEM 4.92
Rod AB is bent into the shape of a circular arc and is lodged between two
pegs D and E. It supports a load P at end B. Neglecting friction and the
weight of the rod, determine the distance c corresponding to equilibrium
when a = 1 in. and R = 5 in.
SOLUTION
yED = xED = a,
Since
Slope of ED is
∴ slope of HC is
45°
45°
DE =
Also
and
2a
a
1
DH = HE =   DE =
2
2
For triangles DHC and EHC
sin β =
a = 1 in. and
sin β =
R = 5 in.
1 in.
= 0.141421
2 ( 5 in.)
∴ β = 8.1301°
and
a
2R
c = R sin ( 45° − β )
Now
For
a/ 2
=
R
or β = 8.13°
c = ( 5 in.) sin ( 45° − 8.1301° ) = 3.00 in.
or c = 3.00 in.
PROBLEM 4.93
A uniform rod AB of weight W and length 2R rests inside a hemispherical
bowl of radius R as shown. Neglecting friction determine the angle θ
corresponding to equilibrium.
SOLUTION
Based on the f.b.d., the uniform rod AB is a three-force body. Point E is
the point of intersection of the three forces. Since force A passes through
O, the center of the circle, and since force C is perpendicular to the rod,
triangle ACE is a right triangle inscribed in the circle. Thus, E is a point
on the circle.
Note that the angle α of triangle DOA is the central angle corresponding
to the inscribed angle θ of triangle DCA.
∴ α = 2θ
The horizontal projections of AE , ( x AE ) , and AG, ( x AG ) , are equal.
∴ x AE = x AG = x A
or
( AE ) cos 2θ
= ( AG ) cosθ
and
( 2R ) cos 2θ
Now
cos 2θ = 2cos 2 θ − 1
then
4cos 2 θ − 2 = cosθ
or
= R cosθ
4cos 2 θ − cosθ − 2 = 0
Applying the quadratic equation
cosθ = 0.84307
∴ θ = 32.534°
and
cosθ = −0.59307
and θ = 126.375°(Discard)
or θ = 32.5°
PROBLEM 4.94
A uniform slender rod of mass m and length 4r rests on the surface shown
and is held in the given equilibrium position by the force P. Neglecting
the effect of friction at A and C, (a) determine the angle θ, (b) derive an
expression for P in terms of m.
SOLUTION
The forces acting on the three-force member intersect at D.
(a) From triangle ACO
 r 
−1  1 
 = tan   = 18.4349°
 3r 
 3
θ = tan −1 
tan θ =
(b) From triangle DCG
∴ DC =
and
or θ = 18.43°
r
DC
r
r
=
= 3r
tan θ
tan18.4349°
DO = DC + r = 3r + r = 4r
 yDO 

 x AG 
α = tan −1 
where
yDO = ( DO ) cosθ = ( 4r ) cos18.4349°
= 3.4947r
and
x AG = ( 2r ) cosθ = ( 2r ) cos18.4349°
= 1.89737r
 3.4947r 
∴ α = tan −1 
 = 63.435°
 1.89737r 
where
90° + (α − θ ) = 90° + 45° = 135.00°
Applying the law of sines to the force triangle,
mg
R
= A


sin 90° + (α − θ )  sin θ
∴ RA = ( 0.44721) mg
Finally,
P = RA cos α
= ( 0.44721mg ) cos 63.435°
= 0.20000mg
or P =
mg
5
PROBLEM 4.95
A uniform slender rod of length 2L and mass m rests against a roller at D
and is held in the equilibrium position shown by a cord of length a.
Knowing that L = 200 mm, determine (a) the angle θ, (b) the length a.
SOLUTION
(a) The forces acting on the three-force member AB intersect at E. Since
triangle DBC is isosceles, DB = a.
From triangle BDE
ED = DB tan 2θ = a tan 2θ
From triangle GED
ED =
∴ a tan 2θ =
L−a
tan θ
tan θ
a ( tan θ tan 2θ + 1) = L
or
a=
From triangle BCD
( L − a)
1 (1.25L )
2
cosθ
or
L
= 1.6cosθ
a
Substituting Equation (2) into Equation (1) yields
1.6cosθ = 1 + tan θ tan 2θ
Now
tan θ tan 2θ =
sin θ sin 2θ
cosθ cos 2θ
=
sin θ 2sin θ cosθ
cosθ 2 cos 2 θ − 1
=
2 (1 − cos 2 θ )
2cos 2 θ − 1
2 (1 − cos 2 θ )
2cos 2 θ − 1
Then
1.6cosθ = 1 +
or
3.2cos3 θ − 1.6cosθ − 1 = 0
Solving numerically
θ = 23.515° or θ = 23.5°
(b) From Equation (2) for L = 200 mm and θ = 23.5°
a=
5 ( 200 mm )
= 136.321 mm
8 cos 23.515°
or a = 136.3 mm
(1)
(2)
PROBLEM 4.96
Gears A and B are attached to a shaft supported by bearings at C and D.
The diameters of gears A and B are 150 mm and 75 mm, respectively, and
the tangential and radial forces acting on the gears are as shown.
Knowing that the system rotates at a constant rate, determine the
reactions at C and D. Assume that the bearing at C does not exert any
axial force, and neglect the weights of the gears and the shaft.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of
shaft
ΣFx = 0: ∴ Dx = 0
ΣM D( z -axis ) = 0: − C y (175 mm ) + ( 482 N ) ( 75 mm )
+ ( 2650 N ) ( 50 mm ) = 0
∴ C y = 963.71 N
or
C y = ( 964 N ) j
ΣM D( y -axis ) = 0: Cz (175 mm ) + (1325 N ) ( 75 mm )
+ ( 964 N ) ( 50 mm ) = 0
∴ C z = −843.29 N
or
C z = ( 843 N ) k
and C = ( 964 N ) j − ( 843 N ) k
ΣM C ( z -axis ) = 0: − ( 482 N ) (100 mm ) + Dy (175 mm )
+ ( 2650 N ) ( 225 mm ) = 0
∴ D y = −3131.7 N
or
D y = − ( 3130 N ) j
ΣM C ( y -axis ) = 0: − (1325 N ) (100 mm ) − Dz (175 mm )
+ ( 964 N ) ( 225 mm ) = 0
∴ Dz = 482.29 N
or
D z = ( 482 N ) k
and D = − ( 3130 N ) j + ( 482 N ) k
PROBLEM 4.97
Solve Problem 4.96 assuming that for gear A the tangential and radial
forces are acting at E, so that FA = (1325 N)j + (482 N)k.
P4.96 Gears A and B are attached to a shaft supported by bearings at C and
D. The diameters of gears A and B are 150 mm and 75 mm, respectively,
and the tangential and radial forces acting on the gears are as shown.
Knowing that the system rotates at a constant rate, determine the reactions at
C and D. Assume that the bearing at C does not exert any axial force, and
neglect the weights of the gears and the shaft.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of
shaft
ΣFx = 0: ∴ Dx = 0
ΣM D( z -axis ) = 0: − C y (175 mm ) − (1325 N ) ( 75 mm )
+ ( 2650 N ) ( 50 mm ) = 0
∴ C y = 189.286 N
C y = (189.3 N ) j
or
ΣM D( y -axis ) = 0: C z (175 mm ) + ( 482 N ) ( 75 mm )
+ ( 964 N ) ( 50 mm ) = 0
∴ C z = −482.00 N
C z = − ( 482 N ) k
or
and C = (189.3 N ) j − ( 482 N ) k
ΣM C ( z -axis ) = 0:
(1325 N ) (100 mm ) + Dy (175 mm )
+ ( 2650 N ) ( 225 mm ) = 0
∴ Dy = − 4164.3 N
D y = − ( 4160 N ) j
or
ΣM C ( y -axis ) = 0:
− ( 482 N )(100 mm ) − Dz (175 mm )
+ ( 964 N )( 225 mm ) = 0
∴ Dz = 964.00 N
or
D z = ( 964 N ) k
and D = − ( 4160 N ) j + ( 964 N ) k
PROBLEM 4.98
Two transmission belts pass over sheaves welded to an axle supported by
bearings at B and D. The sheave at A has a radius of 50 mm, and the
sheave at C has a radius of 40 mm. Knowing that the system rotates with
a constant rate, determine (a) the tension T, (b) the reactions at B and D.
Assume that the bearing at D does not exert any axial thrust and neglect
the weights of the sheaves and the axle.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft
(a)
ΣM x-axis = 0:
( 240 N − 180 N )( 50 mm ) + ( 300 N − T )( 40 mm ) = 0
∴ T = 375 N
(b)
ΣFx = 0: Bx = 0
ΣM D( z -axis ) = 0:
( 300 N + 375 N )(120 mm ) − By ( 240 mm ) = 0
∴ By = 337.5 N
ΣM D( y -axis ) = 0:
( 240 N + 180 N )( 400 mm ) + Bz ( 240 mm ) = 0
∴ Bz = −700 N
or B = ( 338 N ) j − ( 700 N ) k
ΣM B( z -axis ) = 0: − ( 300 N + 375 N )(120 mm ) + D y ( 240 mm ) = 0
∴ D y = 337.5 N
ΣM B( y -axis ) = 0:
( 240 N + 180 N )(160 mm ) + Dz ( 240 mm ) = 0
∴ Dz = −280 N
or D = ( 338 N ) j − ( 280 N ) k
PROBLEM 4.99
For the portion of a machine shown, the 4-in.-diameter pulley A and
wheel B are fixed to a shaft supported by bearings at C and D. The spring
of constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does
not exert any axial force. Knowing that θ = 180° and that the machine is
at rest and in equilibrium, determine (a) the tension T, (b) the reactions at
C and D. Neglect the weights of the shaft, pulley, and wheel.
SOLUTION
First, determine the spring force, FE , at θ = 180°.
FE = ks x
ks = 2 lb/in.
where
x = ( yE )final − ( yE )initial = (12 in. + 3.5 in.) − (12 in. − 3.5 in.) = 7.0 in.
∴ FE = ( 2 lb/in.)( 7.0 in.) = 14.0 lb
(a) From f.b.d. of machine part
ΣM x = 0:
( 34 lb )( 2 in.) − T ( 2 in.) = 0
or T = 34.0 lb
∴ T = 34 lb
(b)
ΣM D ( z -axis ) = 0: − C y (10 in.) − FE ( 2 in. + 1 in.) = 0
− C y (10 in.) − 14.0 lb ( 3 in.) = 0
∴ C y = −4.2 lb
or
C y = − ( 4.20 lb ) j
ΣM D( y -axis ) = 0: C z (10 in.) + 34 lb ( 4 in.) + 34 lb ( 4 in.) = 0
∴ C z = −27.2 lb
or
C z = − ( 27.2 lb ) k
and C = − ( 4.20 lb ) j − ( 27.2 lb ) k
PROBLEM 4.99 CONTINUED
ΣFx = 0: Dx = 0
ΣM C ( z -axis ) = 0: Dy (10 in.) − FE (12 in. + 1 in.) = 0
Dy (10 in.) − 14.0 (13 in.) = 0
or
∴ Dy = 18.2 lb
or
D y = (18.20 lb ) j
ΣM C ( y -axis ) = 0: − 2 ( 34 lb ) ( 6 in.) − Dz (10 in.) = 0
∴ Dz = −40.8 lb
or
D z = − ( 40.8 lb ) k
and D = (18.20 lb ) j − ( 40.8 lb ) k
PROBLEM 4.100
Solve Problem 4.99 for θ = 90°.
P4.99 For the portion of a machine shown, the 4-in.-diameter pulley A and
wheel B are fixed to a shaft supported by bearings at C and D. The spring of
constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does not
exert any axial force. Knowing that θ = 180° and that the machine is at rest
and in equilibrium, determine (a) the tension T, (b) the reactions at C and D.
Neglect the weights of the shaft, pulley, and wheel.
SOLUTION
First, determine the spring force, FE , at θ = 90°.
FE = ks x
ks = 2 lb/in.
where
and

x = Lfinal − Linitial = 

( 3.5)2 + (12 )2  − (12 − 3.5) = 12.5 − 8.5 = 4.0 in.

∴ FE = ( 2 lb/in.)( 4.0 in.) = 8.0 lb
Then
FE =
−12.0
3.5
(8.0 lb ) j +
(8.0 lb ) k = − ( 7.68 lb ) j + ( 2.24 lb ) k
12.5
12.5
(a) From f.b.d. of machine part
ΣM x = 0:
( 34 lb ) ( 2 in.) − T ( 2 in.) − ( 7.68 lb )( 3.5 in.) = 0
or T = 20.6 lb
∴ T = 20.56 lb
(b)
ΣM D ( z -axis ) = 0: − C y (10 in.) − ( 7.68 lb ) ( 3.0 in.) = 0
∴ C y = −2.304 lb
or
C y = − ( 2.30 lb ) j
ΣM D ( y -axis ) = 0: Cz (10 in.) + ( 34 lb ) ( 4.0 in.) + ( 20.56 lb ) ( 4.0 in.) − ( 2.24 lb ) ( 3 in.) = 0
∴ C z = −21.152 lb
or
C z = − ( 21.2 lb ) k
and C = − ( 2.30 lb ) j − ( 21.2 lb ) k
PROBLEM 4.100 CONTINUED
ΣFx = 0: Dx = 0
ΣM C ( z -axis ) = 0: Dy (10 in.) − ( 7.68 lb )(13 in.) = 0
∴ D y = 9.984 lb
or
D y = ( 9.98 lb ) j
ΣM C ( y -axis ) = 0: − ( 34 lb )( 6 in.) − ( 20.56 lb )( 6 in.) − Dz (10 in.) − ( 2.24 lb )(13 in.) = 0
∴ Dz = −35.648 lb
or
D z = − ( 35.6 lb ) k
and D = ( 9.98 lb ) j − ( 35.6 lb ) k
PROBLEM 4.101
A 1.2 × 2.4-m sheet of plywood having a mass of 17 kg has been
temporarily placed among three pipe supports. The lower edge of the
sheet rests on small collars A and B and its upper edge leans against pipe
C. Neglecting friction at all surfaces, determine the reactions at A, B,
and C.
SOLUTION
(
)
W = mg = (17 kg ) 9.81 m/s 2 = 166.77 N
First note
h=
(1.2 )2 − (1.125)2
= 0.41758 m
From f.b.d. of plywood sheet
 (1.125 m ) 
ΣM z = 0: C ( h ) − W 
 =0
2


C ( 0.41758 m ) − (166.77 N ) ( 0.5625 m ) = 0
∴ C = 224.65 N
C = − ( 225 N ) i
or
ΣM B( y -axis ) = 0: − ( 224.65 N ) ( 0.6 m ) + Ax (1.2 m ) = 0
∴ Ax = 112.324 N
or
A x = (112.3 N ) i
ΣM B( x-axis ) = 0: (166.77 N ) ( 0.3 m ) − Ay (1.2 m ) = 0
∴ Ay = 41.693 N
or
A y = ( 41.7 N ) j
ΣM A( y -axis ) = 0: ( 224.65 N ) ( 0.6 m ) − Bx (1.2 m ) = 0
∴ Bx = 112.325 N
or
B x = (112.3 N ) i
PROBLEM 4.101 CONTINUED
ΣM A( x-axis ) = 0: By (1.2 m ) − (166.77 N ) ( 0.9 m ) = 0
∴ By = 125.078 N
or
B y = (125.1 N ) j
∴ A = (112.3 N ) i + ( 41.7 N ) j
B = (112.3 N ) i + (125.1 N ) j
C = − ( 225 N ) i
PROBLEM 4.102
The 200 × 200-mm square plate shown has a mass of 25 kg and is
supported by three vertical wires. Determine the tension in each wire.
SOLUTION
(
)
W = mg = ( 25 kg ) 9.81 m/s 2 = 245.25 N
First note
From f.b.d. of plate
ΣM x = 0: ( 245.25 N ) (100 mm ) − TA (100 mm ) − TC ( 200 mm ) = 0
∴ TA + 2TC = 245.25 N
(1)
ΣM z = 0: TB (160 mm ) + TC (160 mm ) − ( 245.25 N ) (100 mm ) = 0
∴ TB + TC = 153.281 N
(2)
ΣFy = 0: TA + TB + TC − 245.25 N = 0
∴ TB + TC = 245.25 − TA
(3)
TA = 245.25 N − 153.281 N = 91.969 N
(4)
Equating Equations (2) and (3) yields
TA = 92.0 N
or
Substituting the value of TA into Equation (1)
TC =
( 245.25 N − 91.969 N )
2
= 76.641 N
(5)
TC = 76.6 N
or
Substituting the value of TC into Equation (2)
TB = 153.281 N − 76.641 N = 76.639 N
or
TB = 76.6 N
TA = 92.0 N
TB = 76.6 N
TC = 76.6 N
PROBLEM 4.103
The 200 × 200-mm square plate shown has a mass of 25 kg and is
supported by three vertical wires. Determine the mass and location of the
lightest block which should be placed on the plate if the tensions in the
three cables are to be equal.
SOLUTION
First note
(
)
WG = m p1g = ( 25 kg ) 9.81 m/s 2 = 245.25 N
(
)
W1 = mg = m 9.81 m/s 2 = ( 9.81m ) N
From f.b.d. of plate
ΣFy = 0: 3T − WG − W1 = 0 (1)
ΣM x = 0: WG (100 mm ) + W1 ( z ) − T (100 mm ) − T ( 200 mm ) = 0
or − 300T + 100WG + W1z = 0
(2)
ΣM z = 0: 2T (160 mm ) − WG (100 mm ) − W1 ( x ) = 0
or 320T − 100WG − W1x = 0
Eliminate T by forming 100 ×  Eq. (1) + Eq. ( 2 ) 
−100W1 + W1z = 0
∴ z = 100 mm
0 ≤ z ≤ 200 mm, ∴ okay
Now, 3 ×  Eq. ( 3)  − 320 ×  Eq. (1) yields
3 ( 320T ) − 3 (100 )WG − 3W1x − 320 ( 3T ) + 320WG + 320W1 = 0
(3)
PROBLEM 4.103 CONTINUED
or
20WG + ( 320 − 3x )W1 = 0
or
W1
20
=
WG
3
x
−
320 )
(
The smallest value of
W1
will result in the smallest value of W1 since WG is given.
WG
∴ Use x = xmax = 200 mm
W1
20
1
=
=
WG
3 ( 200 ) − 320 14
and then
∴ W1 =
and
WG
245.25 N
=
= 17.5179 N ( minimum )
14
14
m=
W1 17.5179 N
=
= 1.78571 kg
g
9.81 m/s 2
or m = 1.786 kg
at x = 200 mm, z = 100 mm
PROBLEM 4.104
A camera of mass 240 g is mounted on a small tripod of mass 200 g.
Assuming that the mass of the camera is uniformly distributed and that
the line of action of the weight of the tripod passes through D, determine
(a) the vertical components of the reactions at A, B, and C when θ = 0,
(b) the maximum value of θ if the tripod is not to tip over.
SOLUTION
First note
Wtp = mtp
For θ = 0
(
)
g = ( 0.20 kg ) ( 9.81 m/s ) = 1.9620 N
WC = mC g = ( 0.24 kg ) 9.81 m/s 2 = 2.3544 N
2
xC = − ( 60 mm − 24 mm ) = −36 mm
zC = 0
(a) From f.b.d. of camera and tripod as projected onto plane ABCD
ΣFy = 0: Ay + By + C y − WC − Wtp = 0
∴ Ay + By + C y = 2.3544 N + 1.9620 N = 4.3164 N
ΣM x = 0: C y ( 38 mm ) − By ( 38 mm ) = 0
∴ C y = By
(1)
(2)
ΣM z = 0: By ( 35 mm ) + C y ( 35 mm ) + ( 2.3544 N ) ( 36 mm ) − Ay ( 45 mm ) = 0
∴ 9 Ay − 7By − 7C y = 16.9517
(3)
Substitute C y with By from Equation (2) into Equations (1) and (3), and solve by elimination
(
7 Ay + 2 By = 4.3164
)
9 Ay − 14By = 16.9517
16 Ay
= 47.166
PROBLEM 4.104 CONTINUED
∴ Ay = 2.9479 N
or A y = 2.95 N
Substituting Ay = 2.9479 N into Equation (1)
2.9479 N + 2 By = 4.3164
∴ By = 0.68425 N
C y = 0.68425 N
or B y = C y = 0.684 N
(b) By = 0 for impending tipping
From f.b.d. of camera and tripod as projected onto plane ABCD
ΣFy = 0: Ay + C y − WC − Wtp = 0
∴ Ay + C y = 4.3164 N
(1)
ΣM x = 0: C y ( 38 mm ) − ( 2.3544 N ) ( 36 mm ) sin θ  = 0
∴ C y = 2.2305sin θ
(2)
ΣM z = 0: C y ( 35 mm ) − Ay ( 45 mm ) + ( 2.3544 N ) ( 36 mm ) cosθ  = 0
∴ 9 Ay − 7C y = (16.9517 N ) cosθ
(3)
Forming 7 ×  Eq. (1)  +  Eq. ( 3)  yields
16 Ay = 30.215 N + (16.9517 N ) cosθ
(4)
PROBLEM 4.104 CONTINUED
Substituting Equation (2) into Equation (3)
9 Ay − (15.6134 N ) sin θ = (16.9517 N ) cosθ
(5)
Forming 9 ×  Eq. ( 4 )  − 16 ×  Eq. ( 5 )  yields
( 249.81 N ) sin θ
or
= 271.93 N − (118.662 N ) cosθ
cos 2 θ =  2.2916 N − ( 2.1053 N ) sin θ 
2
cos 2 θ = 1 − sin 2 θ
Now
∴ 5.4323sin 2 θ − 9.6490sin θ + 4.2514 = 0
Using quadratic formula to solve,
sin θ = 0.80981 and sin θ = 0.96641
∴ θ = 54.078° and θ = 75.108°
or θ max = 54.1° before tipping
PROBLEM 4.105
Two steel pipes AB and BC, each having a weight per unit length of
5 lb/ft, are welded together at B and are supported by three wires.
Knowing that a = 1.25 ft, determine the tension in each wire.
SOLUTION
WAB = ( 5 lb/ft )( 2 ft ) = 10 lb
First note
WBC = ( 5 lb/ft )( 4 ft ) = 20 lb
W = WAB + WBC = 30 lb
To locate the equivalent force of the pipe assembly weight
rG/B × W = Σ ( ri × Wi ) = rG ( AB ) × WAB + rG ( BC ) × WBC
( xGi + zGk ) × ( −30 lb ) j = (1 ft ) k × ( −10 lb ) j + ( 2 ft ) i × ( −20 lb ) j
or
∴
From i-coefficient
k-coefficient
− ( 30 lb ) xGk + ( 30 lb ) zG i = (10 lb ⋅ ft ) i − ( 40 lb ⋅ ft ) k
zG =
10 lb ⋅ ft 1
= ft
30 lb
3
xG =
40 lb ⋅ ft
1
= 1 ft
30 lb
3
From f.b.d. of piping
ΣM x = 0: W ( zG ) − TA ( 2 ft ) = 0
1 
1 
∴ TA =  ft  30 lb  ft  = 5 lb
2 
3 
or
TA = 5.00 lb
ΣFy = 0: 5 lb + TD + TC − 30 lb = 0
∴ TD + TC = 25 lb
(1)
PROBLEM 4.105 CONTINUED
4 
ΣM z = 0: TD (1.25 ft ) + TC ( 4 ft ) − 30 lb  ft  = 0
3 
∴ 1.25TD + 4TC = 40 lb ⋅ ft
−4  Equation (1) 
−4TD − 4TC = −100
Equation (2) + Equation (3)
Results:
(3)
−2.75TD = −60
∴ TD = 21.818 lb
From Equation (1)
(2)
or
TD = 21.8 lb
TC = 25 − 21.818 = 3.1818 lb
or
TC = 3.18 lb
TA = 5.00 lb
TC = 3.18 lb
TD = 21.8 lb
PROBLEM 4.106
For the pile assembly of Problem 4.105, determine (a) the largest
permissible value of a if the assembly is not to tip, (b) the corresponding
tension in each wire.
P4.105 Two steel pipes AB and BC, each having a weight per unit length
of 5 lb/ft, are welded together at B and are supported by three wires.
Knowing that a = 1.25 ft, determine the tension in each wire.
SOLUTION
First note
WAB = ( 5 lb/ft )( 2 ft ) = 10 lb
WBC = ( 5 lb/ft )( 4 ft ) = 20 lb
From f.b.d. of pipe assembly
ΣFy = 0: TA + TC + TD − 10 lb − 20 lb = 0
∴ TA + TC + TD = 30 lb
(1)
ΣM x = 0: (10 lb )(1 ft ) − TA ( 2 ft ) = 0
or
TA = 5.00 lb
(2)
TC + TD = 25 lb
From Equations (1) and (2)
(3)
ΣM z = 0: TC ( 4 ft ) + TD ( amax ) − 20 lb ( 2 ft ) = 0
or
( 4 ft ) TC
+ TD amax = 40 lb ⋅ ft
(4)
PROBLEM 4.106 CONTINUED
Using Equation (3) to eliminate TC
4 ( 25 − TD ) + TD amax = 40
amax = 4 −
or
60
TD
By observation, a is maximum when TD is maximum. From Equation (3), (TD )max occurs when TC = 0.
Therefore, (TD )max = 25 lb and
amax = 4 −
60
25
= 1.600 ft
Results: (a)
amax = 1.600 ft
(b)
TA = 5.00 lb
TC = 0
TD = 25.0 lb
PROBLEM 4.107
A uniform aluminum rod of weight W is bent into a circular ring of radius
R and is supported by three wires as shown. Determine the tension in
each wire.
SOLUTION
From f.b.d. of ring
ΣFy = 0: TA + TB + TC − W = 0
∴ TA + TB + TC = W
(1)
ΣM x = 0: TA ( R ) − TC ( R sin 30° ) = 0
∴ TA = 0.5TC
(2)
ΣM z = 0: TC ( R cos 30° ) − TB ( R ) = 0
∴ TB = 0.86603TC
(3)
Substituting TA and TB from Equations (2) and (3) into Equation (1)
0.5TC + 0.86603TC + TC = W
∴ TC = 0.42265W
From Equation (2)
TA = 0.5 ( 0.42265W ) = 0.21132W
From Equation (3)
TB = 0.86603 ( 0.42265W ) = 0.36603W
or TA = 0.211W
TB = 0.366W
TC = 0.423W
PROBLEM 4.108
A uniform aluminum rod of weight W is bent into a circular ring of radius
R and is supported by three wires as shown. A small collar of weight W ′
is then placed on the ring and positioned so that the tensions in the three
wires are equal. Determine (a) the position of the collar, (b) the value of
W ′, (c) the tension in the wires.
SOLUTION
Let θ = angle from x-axis to small collar of weight W ′
From f.b.d. of ring
ΣFy = 0: 3T − W − W ′ = 0
(1)
ΣM x = 0: T ( R ) − T ( R sin 30° ) + W ′ ( R sin θ ) = 0
or
1
W ′ sin θ = − T
2
(2)
ΣM z = 0: T ( R cos 30° ) − W ′ ( R cosθ ) − T ( R ) = 0
or

3
W ′ cosθ = − 1 −
T

2 

(3)
Dividing Equation (2) by Equation (3)
 1    3 
tan θ =   1 − 

 2    2  
∴ θ = 75.000°
and
−1
= 3.7321
θ = 255.00°
Based on Equations (2) and (3), θ = 75.000° will give a negative value
for W ′, which is not acceptable.
(a)
∴ W ′ is located at θ = 255° from the x-axis or 15° from A
towards B.
(b) From Equation (1) and Equation (2)
W ′ = 3 ( −2W ′ )( sin 255° ) − W
∴ W ′ = 0.20853W
or W ′ = 0.209W
(c) From Equation (1)
T = −2 ( 0.20853W ) sin 255°
= 0.40285W
or T = 0.403W
PROBLEM 4.109
An opening in a floor is covered by a 3 × 4-ft sheet of plywood weighing
12 lb. The sheet is hinged at A and B and is maintained in a position
slightly above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
From f.b.d. of plywood sheet
ΣM x = 0:
(12 lb )( 2 ft ) − C y ( 3.5 ft ) = 0
∴ C y = 6.8571 lb
ΣM B( z -axis ) = 0:
or
C y = 6.86 lb
(12 lb )(1 ft ) + ( 6.8571 lb )( 0.5 ft ) − Ay ( 2 ft ) = 0
∴ Ay = 7.7143 lb
or
Ay = 7.71 lb
ΣM A( z -axis ) = 0: − (12 lb )(1 ft ) + By ( 2 ft ) + ( 6.8571 lb )( 2.5 ft ) = 0
∴ By = 2.5714 lb
or
By = 2.57 lb
(a)
Ay = 7.71 lb
(b)
By = 2.57 lb
(c)
C y = 6.86 lb
PROBLEM 4.110
Solve Problem 4.109 assuming that the small block C is moved and
placed under edge DE at a point 0.5 ft from corner E.
SOLUTION
rB/ A = ( 2 ft ) i
First,
rC/ A = ( 2 ft ) i + ( 4 ft ) k
rG/ A = (1 ft ) i + ( 2 ft ) k
From f.b.d. of plywood sheet
ΣM A = 0: rB/ A × ( B y j + Bzk ) + rC/ A × C y j + rG/ A × ( −Wj) = 0
( 2 ft ) i × By j + ( 2 ft ) i × Bzk + [( 2 ft ) i + ( 4 ft ) k ] × C y j
+ [(1 ft ) i + ( 2 ft ) k ] × ( −12 lb ) j = 0
2Byk − 2 Bz j + 2C yk − 4C y i − 12k + 24i = 0
i-coeff.
−4C y + 24 = 0
∴ C y = 6.00 lb
j-coeff.
−2 Bz = 0
∴ Bz = 0
k-coeff.
2By + 2C y − 12 = 0
or
2By + 2 ( 6 ) − 12 = 0
∴ By = 0
PROBLEM 4.110 CONTINUED
ΣF = 0: Ay j + Azk + By j + Bzk + C y j − Wj = 0
Ay j + Az k + 0 j + 0k + 6 j − 12 j = 0
j-coeff.
Ay + 6 − 12 = 0
k-coeff.
Az = 0
∴ Ay = 6.00 lb
Az = 0
∴ a) Ay = 6.00 lb
b) By = 0
c) C y = 6.00 lb
PROBLEM 4.111
The 10-kg square plate shown is supported by three vertical wires.
Determine (a) the tension in each wire when a = 100 mm, (b) the value
of a for which tensions in the three wires are equal.
SOLUTION
First note
(a)
(
)
W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N
(a) From f.b.d. of plate
ΣFy = 0: TA + TB + TC − W = 0
∴ TA + TB + TC = 98.1 N
(1)
ΣM x = 0: W (150 mm ) − TB( 300 mm ) − TC (100 mm ) = 0
∴ 6TB + 2TC = 294.3
(2)
ΣM z = 0: TB(100 mm ) + TC ( 300 mm ) − ( 98.1 N )(150 mm ) = 0
∴
− 6TB − 18TC = −882.9
(3)
Equation (2) + Equation (3)
−16TC = −588.6
∴ TC = 36.788 N
TC = 36.8 N W
or
Substitution into Equation (2)
6TB + 2 ( 36.788 N ) = 294.3 N
∴ TB = 36.788 N
or
TB = 36.8 N W
From Equation (1)
TA + 36.788 + 36.788 = 98.1 N
∴ TA = 24.525 N
or
TA = 24.5 N W
PROBLEM 4.111 CONTINUED
(b)
(b) From f.b.d. of plate
ΣFy = 0: 3T − W = 0
∴ T =
1
W
3
(1)
ΣM x = 0: W (150 mm ) − T ( a ) − T ( 300 mm ) = 0
∴ T =
150W
a + 300
(2)
Equating Equation (1) to Equation (2)
1
150W
W =
3
a + 300
or
a + 300 = 3 (150 )
or a = 150.0 mm W
PROBLEM 4.112
The 3-m flagpole AC forms an angle of 30o with the z axis. It is held by
a ball-and-socket joint at C and by two thin braces BD and BE. Knowing
that the distance BC is 0.9 m, determine the tension in each brace and the
reaction at C.
SOLUTION
TBE can be found from ΣM about line CE
From f.b.d. of flagpole
(
)
ΣM CE = 0: λ CE ⋅ rB/C × TBD + λ CE ⋅ ( rA/C × FA ) = 0
where
λ CE =
( 0.9 m ) i + ( 0.9 m ) j
( 0.9 )2 + ( 0.9 )2 m
=
1
( i + j)
2
rB/C = ( 0.9 m ) sin 30° j + ( 0.9 m ) cos 30°  k
= ( 0.45 m ) j + ( 0.77942 m ) k
 − ( 0.9 m ) i + 0.9 m − ( 0.9 m ) sin 30°  j − ( 0.9 m ) cos 30° k 



 T
TBD = λ BDTBD = 
 BD
2
2
2
( 0.9 ) + ( 0.45 ) + ( 0.77942 ) m


T
=  − ( 0.9 m ) i + ( 0.45 m ) j − ( 0.77942 m ) k  BD
1.62
= ( −0.70711i + 0.35355 j − 0.61237k ) TBD
rA/C = ( 3 m ) sin 30° j + ( 3 m ) cos30°k = (1.5 m ) j + ( 2.5981 m ) k
FA = − ( 300 N ) j
∴
1
1
0
1 1
0
 TBD 
 1 
0
0.45
0.77942 
 + 0 1.5 2.5981 
=0
 2
 2
0 −300
0
−0.70711 0.35355 −0.61237
PROBLEM 4.112 CONTINUED
−1.10227TBD + 779.43 = 0
or
∴ TBD = 707.12 N
or TBD = 707 N W
TBE = TBD = 707.12 N
Based on symmetry with yz-plane,
or TBE = 707 N W
The reaction forces at C are found from ΣF = 0
ΣFx = 0: − (TBD ) x + (TBE ) x + C x = 0
ΣFy = 0:
or
Cx = 0
(TBD ) y + (TBE ) y + C y − 300 N = 0
C y = 300 N − 2 ( 0.35355)( 707.12 N )
∴ C y = −200.00 N
ΣFz = 0: Cz − (TBD ) z − (TBE ) z = 0
Cz = 2 ( 0.61237 )( 707.12 N )
∴ C z = 866.04 N
or C = − ( 200 N ) j + ( 866 N ) k W
PROBLEM 4.113
A 3-m boom is acted upon by the 4-kN force shown. Determine the
tension in each cable and the reaction at the ball-and-socket joint at A.
SOLUTION
From f.b.d. of boom
(
)
(
)
ΣM AE = 0: λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0
where
λ AE =
( 2.1 m ) j − (1.8 m ) k
( 2.1)2 + (1.8)2 m
= 0.27451j − 0.23529k
rB/ A = (1.8 m ) i
TBD = λ BDTBD =
( −1.8 m ) i + ( 2.1 m ) j + (1.8 m ) k T
BD
(1.8)2 + ( 2.1)2 + (1.8)2 m
= ( −0.54545i + 0.63636 j + 0.54545k ) TBD
rC/ A = ( 3.0 m ) i
FC = − ( 4 kN ) j
PROBLEM 4.113 CONTINUED
∴
0
0.27451 −0.23529
0 0.27451 −0.23529
1.8
0
0
0
0
=0
TBD + 3
−0.54545 0.63636 0.54545
0
−4
0
( −0.149731 − 0.149729 )1.8TBD + 2.82348 = 0
∴ TBD = 5.2381 kN
Based on symmetry,
or TBD = 5.24 kN W
TBE = TBD = 5.2381 kN
ΣFz = 0: Az + (TBD ) z − (TBE ) z = 0
or TBE = 5.24 kN W
Az = 0
ΣFy = 0: Ay + (TBD ) y + (TBD ) y − 4 kN = 0
Ay + 2 ( 0.63636 )( 5.2381 kN ) − 4 kN = 0
∴ Ay = −2.6666 kN
ΣFx = 0: Ax − (TBD ) x − (TBE ) x = 0
Ax − 2 ( 0.54545 )( 5.2381 kN ) = 0
∴ Ax = 5.7142 kN
and A = ( 5.71 N ) i − ( 2.67 N ) j W
PROBLEM 4.114
An 8-ft-long boom is held by a ball-and-socket joint at C and by two
cables AD and BE. Determine the tension in each cable and the reaction
at C.
SOLUTION
From f.b.d. of boom
(
)
(
)
ΣM CE = 0: λ CE ⋅ rA/C × TAD + λ CE ⋅ rA/C × FA = 0
where
λ CE =
( 2 ft ) j − ( 3 ft ) k
( 2 )2 + ( 3)2 ft
=
1
( 2 j − 3k )
13
rA/C = ( 8 ft ) i
TAD = λ ADTAD =
− ( 8 ft ) i + (1 ft ) j + ( 4 ft ) k
(8) + (1) + ( 4 ) ft
2
2
2
TAD
1
=   TAD ( −8i + j + 4k )
9
FA = − (198 lb ) j
∴
0 2 −3
0 2 −3
 TAD 
 198 
8 0 0 
+ 8 0 0 
=0
 9 13 
 13 
0 −1 0
−8 1 4
PROBLEM 4.114 CONTINUED
( −64 − 24 )
TAD
198
+ ( 24 )
=0
9 13
13
∴ TAD = 486.00 lb
or TAD = 486 lb W
(
)
(
ΣM CD = 0: λ CD ⋅ rB/C × TBE + λ CD ⋅ rA/C × FA
where
λ CD =
(1 ft ) j + ( 4 ft ) k
17 ft
=
)
1
(1j + 4k )
17
rB/C = ( 6 ft ) i
TBE = λ BETBE =
− ( 6 ft ) i + ( 2 ft ) j − ( 3 ft ) k
(6)
∴
2
+ ( 2 ) + ( 3) ft
2
2
1
TBE =   TBE ( −6i + 2 j − 3k )
7
0 1 4
0 1 4
198
TBE
6 0 0
+ 8 0 0
=0
7 17
17
0 −1 0
−6 2 −3
(18 + 48)
TBE
+ ( −32 )198 = 0
7
∴ TBE = 672.00 lb
or TBE = 672 lb W
ΣFx = 0: C x − (TAD ) x − (TBE ) x = 0
8
6
Cx −   486 −   672 = 0
9
 
7
∴ C x = 1008 lb
ΣFy = 0: C y + (TAD ) y + (TBE ) y − 198 lb = 0
1
2
C y +   486 +   672 − 198 lb = 0
9
7
∴ C y = −48.0 lb
ΣFz = 0: Cz + (TAD ) z − (TBE ) z = 0
4
3
Cz +   486 −   ( 672 ) = 0
9
 
7
∴ Cz = 72.0 lb
or C = (1008 lb ) i − ( 48.0 lb ) j + ( 72.0 lb ) k W
PROBLEM 4.115
Solve Problem 4.114 assuming that the given 198-lb load is replaced
with two 99-lb loads applied at A and B.
P4.114 An 8-ft-long boom is held by a ball-and-socket joint at C and by
two cables AD and BE. Determine the tension in each cable and the
reaction at C.
SOLUTION
From f.b.d. of boom
(
)
(
)
(
)
ΣM CE = 0: λ CE ⋅ rA/C × TAD + λ CE ⋅ rA/C × FA + λ CE ⋅ rB/C × FB = 0
λ CE =
where
( 2 ft ) j − ( 3 ft ) k
( 2 )2 + ( 3)2 ft
=
1
( 2 j − 3k )
13
rA/C = ( 8 ft ) i
rB/C = ( 6 ft ) i
TAD = λ ADTAD =
− ( 8 ft ) i + (1 ft ) j + ( 4 ft ) k
(8) + (1) + ( 4 ) ft
2
2
2
TAD
1
=   TAD ( −8i + j + 4k )
9
FA = − ( 99 lb ) j
FB = − ( 99 lb ) j
∴
0 2 −3
0 2 −3
0 2 −3
99
99
TAD
8 0 0
+ 8 0 0
+ 6 0 0
=0
9 13
13
13
0 −1 0
0 −1 0
−8 1 4
PROBLEM 4.115 CONTINUED
( −64 − 24 )
TAD
99
+ ( 24 + 18 )
=0
9 13
13
TAD = 425.25 lb
or
or TAD = 425 lb W
(
)
(
)
(
)
ΣM CD = 0: λ CD ⋅ rB/C × TBE + λ CD ⋅ rA/C × FA + λ CD ⋅ rB/C × FB = 0
where
λ CD =
(1 ft ) j + ( 4 ft ) k
17
=
1
( j + 4k )
17
rB/C = ( 6 ft ) i
rA/C = ( 8 ft ) j
TBE = λ BETBE =
∴
− ( 6 ft ) i + ( 2 ft ) j − ( 3 ft ) k
(6)
0 1 4
 T
6 0 0  BE
 7 17
−6 2 −3
2
+ ( 2 ) + ( 3) ft
2
2
TBE =
TBE
( −6i + 2 j − 3k )
7
0 1 4
0 1 4

 99 
 99 
+ 8 0 0
+ 6 0 0
=0

 17 
 17 
0 −1 0
0 −1 0
(18 + 48) 
TBE 
 99 
 + ( −32 − 24 ) 
=0
 7 17 
 17 
or
TBE = 588.00 lb
or TBE = 588 lb W
ΣFx = 0: C x − (TAD ) x − (TBE ) x = 0
8
6
Cx −   425.25 −   588.00 = 0
9
7
∴ C x = 882 lb
ΣFy = 0: C y + (TAD ) y + (TBE ) y − 99 − 99 = 0
1
2
C y +   425.25 +   588.00 − 198 = 0
9
 
7
∴ C y = −17.25 lb
ΣFz = 0: Cz + (TAD ) z − (TBE ) z = 0
4
3
Cz +   425.25 −   588.00 = 0
9
7
∴ Cz = 63.0 lb
or C = ( 882 lb ) i − (17.25 lb ) j + ( 63.0 lb ) k W
PROBLEM 4.116
The 18-ft pole ABC is acted upon by a 210-lb force as shown. The pole is
held by a ball-and-socket joint at A and by two cables BD and BE. For
a = 9 ft, determine the tension in each cable and the reaction at A.
SOLUTION
From f.b.d. of pole ABC
(
)
(
)
ΣM AE = 0: λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0
where
λ AE =
( 4.5 ft ) i + ( 9 ft ) k
( 4.5 )2 + ( 9 )2 ft
=
1
( 4.5i + 9k )
101.25
rB/ A = ( 9 ft ) j
rC/ A = (18 ft ) j
TBD = λ BDTBD =
( 4.5 ft ) i − ( 9 ft ) j − ( 9 ft ) k T
BD
( 4.5)2 + ( 9 )2 + ( 9 )2 ft
T 
=  BD  ( 4.5i − 9 j − 9k )
 13.5 
FC = λ CF ( 210 lb ) =
∴
−9i − 18 j + 6k
( 9 )2 + (18)2 + ( 6 )2
( 210 lb ) = 10 lb ( −9i − 18j + 6k )
4.5 0 9
4.5 0 9
TBD


 10 lb 
0 9 0 
 + 0 18 0 
=0
 13.5 101.25 
 101.25 
−9 −18 6
4.5 −9 −9
PROBLEM 4.116 CONTINUED
( −364.5 − 364.5) T
BD
13.5 101.25
+
( 486 + 1458) (10 lb ) = 0
101.25
TBD = 360.00 lb
and
or TBD = 360 lb
(
)
(
)
ΣM AD = 0: λ AD ⋅ rB/ A × TBE + λ AD ⋅ rC/ A × FC = 0
where
λ AD =
( 4.5 ft ) i − ( 9 ft ) k
( 4.5 )2 + ( 9 )2 ft
1
( 4.5i − 9k )
101.25
=
rB/ A = ( 9 ft ) j
rC/ A = (18 ft ) j
TBE = λ BETBE =
∴
( 4.5 ft ) i − ( 9 ft ) j + ( 9 ft ) k T
2
2
=
TBE
( 4.5i − 9 j + 9k )
13.5
4.5 0 −9
4.5 0 −9
TBE


 10 lb 
0 9 0 
 + 0 18 0 
=0
 13.5 101.25 
 101.25 
4.5 −9 9
−9 −18 6
( 364.5 + 364.5 ) T
13.5 101.25
or
BE
( 4.5 ) + ( 9 ) + ( 9 ) ft
2
BE
+
( 486 − 1458 )10 lb
101.25
=0
TBE = 180.0 lb
or TBE = 180.0 lb
ΣFx = 0: Ax + (TBD ) x + (TBE ) x − ( FC ) x = 0
 4.5 
 4.5 
 9 
Ax + 
 360 + 
180 −   210 = 0
 13.5 
 13.5 
 21 
∴ Ax = −90.0 lb
ΣFy = 0: Ay − (TBD ) y − (TBE ) y − ( FC ) y = 0
 9 
 9 
 18 
Ay − 
 360 − 
180 −   210 = 0
 13.5 
 13.5 
 21 
∴ Ay = 540 lb
ΣFz = 0: Az − (TBD ) z + (TBE ) z + ( FC ) z = 0
 9 
 9 
 6 
Az − 
 360 + 
180 +   210 = 0
13.5
13.5




 21 
∴ Az = 60.0 lb
or A = − ( 90.0 lb ) i + ( 540 lb ) j + ( 60.0 lb ) k
PROBLEM 4.117
Solve Problem 4.116 for a = 4.5 ft.
P4.116 The 18-ft pole ABC is acted upon by a 210-lb force as shown. The
pole is held by a ball-and-socket joint at A and by two cables BD and BE.
For a = 9 ft, determine the tension in each cable and the reaction at A.
SOLUTION
From f.b.d. of pole ABC
(
)
(
)
ΣM AE = 0 : λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0
λ AE =
where
( 4.5 ft ) i + ( 9 ft ) k
( 4.5 )2 + ( 9 )2 ft
=
1
( 4.5i + 9k )
101.25
rB/ A = ( 9 ft ) j
rC/ A = (18 ft ) j
TBD = λ BDTBD =
( 4.5 ft ) i − ( 9 ft ) j − ( 9 ft ) k T
BD
( 4.5)2 + ( 9 )2 + ( 9 )2 ft
T 
=  BD  ( 4.5i − 9 j − 9k )
 13.5 
FC = λ CF ( 210 lb ) =
−4.5i − 18 j + 6k
( 4.5)2 + (18)2 + ( 6 )2
( 210 lb )
 210 lb 
=
 ( −4.5i − 18 j + 6k )
 19.5 
∴
4.5 0 9
4.5 0 9
TBD


 210 lb 
0 9 0 
18 0 
+ 0
=0
 13.5 101.25 
 19.5 101.25 
4.5 −9 −9
−4.5 −18 6
PROBLEM 4.117 CONTINUED
( −364.5 − 364.5) T
+
BD
13.5 101.25
( 486 + 729 )
19.5 101.25
( 210 lb ) = 0
TBD = 242.31 lb
or
(
or TBD = 242 lb
)
(
)
ΣM AD = 0: λ AD ⋅ rB/ A × TBE + λ AD ⋅ rC/ A × FC = 0
where
λ AD =
( 4.5 ft ) i − ( 9 ft ) k
( 4.5 )2 + ( 9 )2 ft
1
( 4.5i − 9k ) ,
101.25
=
rB/ A = ( 9 ft ) j
rC/ A = (18 ft ) j
TBE = λ BETBE =
∴
( 4.5 ft ) i − ( 9 ft ) j + ( 9 ft ) k T
2
2
=
TBE
( 4.5i − 9 j + 9k )
13.5
4.5 0 −9
4.5 0 −9
TBE


 210 lb 
0 9 0 
18 0 
+ 0
=0
 13.5 101.25 
 19.5 101.25 
4.5 −9 9
−4.5 −18 6
( 364.5 + 364.5 ) T
13.5 101.25
or
BE
( 4.5 ) + ( 9 ) + ( 9 ) ft
2
BE
+
( 486 − 729 )( 210 lb )
19.5 101.25
=0
TBE = 48.462 lb
or TBE = 48.5 lb
ΣFx = 0: Ax + (TBD ) x + (TBE ) x − ( FC ) x = 0
 4.5 
 4.5 
 4.5 
Ax + 
 242.31 + 
 48.462 − 
 210 = 0
 13.5 
 13.5 
 19.5 
∴ Ax = −48.459 lb
ΣFy = 0: Ay − (TBD ) y − (TBE ) y − ( FC ) y = 0
 9 
 9 
 18 
Ay − 
 242.31 − 
 48.462 − 
 210 =
 13.5 
 13.5 
 19.5 
∴ Ay = 387.69 lb
ΣFz = 0: Az − (TBD ) z + (TBE ) z + ( FC ) z = 0
 9 
 9 
 6 
Az − 
 242.31 + 
 48.462 + 
2
 13.5 
 13.5 
 19.5 
∴ Az = 64.591 lb
or A = − ( 48.5 lb ) i + ( 388 lb ) j + ( 64.6 lb ) k
PROBLEM 4.118
Two steel pipes ABCD and EBF are welded together at B to form the
boom shown. The boom is held by a ball-and-socket joint at D and by
two cables EG and ICFH; cable ICFH passes around frictionless pulleys
at C and F. For the loading shown, determine the tension in each cable
and the reaction at D.
SOLUTION
From f.b.d. of boom
(
)
(
)
ΣM z = 0: k ⋅ rC/D × TCI + k ⋅ rA/D × FA = 0
where
rC/D = (1.8 m ) i
TCI = λ CI TCI =
− (1.8 m ) i + (1.12 m ) j
(1.8) + (1.12 ) m
2
2
TCI
 T 
=  CI  ( −1.8i + 1.12 j)
 2.12 
rA/D = ( 3.5 m ) i
FA = − ( 560 N ) j
0
0 1
0 0 1
 TCI 
0 0
∴ ΣM z = 1.8
 + 3.5 0 0 ( 560 N ) = 0
 2.12 
−1.8 1.12 0
0 −1 0
( 2.016 )
or
TCI
+ ( −3.5 ) 560 = 0
2.12
TCI = TFH = 2061.1 N
TICFH = 2.06 kN
PROBLEM 4.118 CONTINUED
(
)
(
)
ΣM y = 0: j ⋅ rG/D × TEG + j ⋅ rH /D × TFH = 0
where
rG/D = ( 3.4 m ) k
rH /D = − ( 2.5 m ) k
TEG =
− ( 3.0 m ) i + ( 3.15 m ) k
( 3)2 + ( 3.15)2
TFH = λ FH TFH =
∴
m
− ( 3.0 m ) i − ( 2.25 m ) k
( 3)
2
+ ( 2.25 ) m
2
( 2061.1 N ) =
2061.1 N
( −3i − 2.25k )
3.75
0 1 0
0 1
0
 TEG 
 2061.1 N 
0 0 3.4 
 + 0 0 −2.5 
=0
 4.35 
 3.75 
−3 0 3.15
−3 0 −2.25
− (10.2 )
or
T 
TEG =  EG  ( −3i + 3.15k )
 4.35 
TEG
2061.1 N
+ ( 7.5 )
=0
4.35
3.75
TEG = 1758.00 N
TEG = 1.758 kN
ΣFx = 0: Dx − (TCI ) x − (TFH ) x − (TEG ) x = 0
 1.8  (
 3.0  (
 3 (
Dx − 
 2061.1 N ) − 
 2061.1 N ) − 
 1758 N ) = 0
 2.12 
 3.75 
 4.35 
∴ Dx = 4611.3 N
ΣFy = 0: Dy + (TCI ) y − 560 N = 0
 1.12  (
Dy + 
 2061.1 N ) − 560 N = 0
 2.12 
∴ D y = −528.88 N
ΣFz = 0: Dz + (TEG ) z − (TFH ) z = 0
 3.15  (
 2.25  (
Dz + 
 1758 N ) − 
 2061.1 N ) = 0
 4.35 
 3.75 
∴ Dz = −36.374 N
and D = ( 4610 N ) i − ( 529 N ) j − ( 36.4 N ) k
PROBLEM 4.119
Solve Problem 4.118 assuming that the 560-N load is applied at B.
P4.118 Two steel pipes ABCD and EBF are welded together at B to form
the boom shown. The boom is held by a ball-and-socket joint at D and by
two cables EG and ICFH; cable ICFH passes around frictionless pulleys
at C and F. For the loading shown, determine the tension in each cable
and the reaction at D.
SOLUTION
From f.b.d. of boom
(
)
(
)
ΣM z = 0: k ⋅ rC/D × TCI + k ⋅ rB/D × FB = 0
rC/D = (1.8 m ) i
where
TCI = λ CI TCI =
− (1.8 m ) i + (1.12 m ) j
(1.8)2 + (1.12 )2
TCI
m
 T 
=  CI  ( −1.8i + 1.12 j)
 2.12 
rB/D = ( 3.0 m ) i
FB = − ( 560 N ) j
∴
0
0 1
0 0 1
 TCI 
1.8
0 0
 + 3 0 0 ( 560 N ) = 0
 2.12 
−1.8 1.12 0
0 −1 0
( 2.016 )
or
TCI
+ ( −3) 560 = 0
2.12
TCI = TFH = 1766.67 N
TICFH = 1.767 kN
PROBLEM 4.119 CONTINUED
(
)
(
)
ΣM y = 0: j ⋅ rG/D × TEG + j ⋅ rH /D × TFH = 0
where
rG/D = ( 3.4 m ) k
rH /D = − ( 2.5 m ) k
TEG = λ EGTEG =
TFH = λ FH TFH =
∴
− ( 3.0 m ) i + ( 3.15 m ) k
( 3)2 + ( 3.15)2
( 3)
TEG
( −3i + 3.15k )
4.35
TFH =
1766.67 N
( −3i − 2.25k )
3.75
m
− ( 3.0 m ) i − ( 2.25 m ) k
2
TEG =
+ ( 2.25 ) m
2
0 1 0
0 1
0
 TEG 
 1766.67 
0 0 3.4 
 + 0 0 −2.5 
=0
 4.35 
 3.75 
−3 0 3.15
−3 0 −2.25
− (10.2 )
TEG
1766.67
+ ( 7.5 )
=0
4.35
3.75
TEG = 1506.86 N
or
TEG = 1.507 kN
ΣFx = 0: Dx − (TCI ) x − (TFH ) x − (TEG ) x = 0
 1.8  (
 3 (
 3 (
Dx − 
 1766.67 N ) − 
 1766.67 N ) − 
 1506.86 N ) = 0
 2.12 
 3.75 
 4.35 
∴ Dx = 3952.5 N
ΣFy = 0: Dy + (TCI ) y − 560 N = 0
 1.12  (
Dy + 
 1766.67 N ) − 560 N = 0
 2.12 
∴ D y = −373.34 N
ΣFz = 0: Dz + (TEG ) z − (TFH ) z = 0
 3.15  (
 2.25  (
Dz + 
 1506.86 N ) − 
 1766.67 N ) = 0
 4.35 
 3.75 
∴ Dz = −31.172 N
D = ( 3950 N ) i − ( 373 N ) j − ( 31.2 N ) k
PROBLEM 4.120
The lever AB is welded to the bent rod BCD which is supported by
bearings at E and F and by cable DG. Knowing that the bearing at E does
not exert any axial thrust, determine (a) the tension in cable DG, (b) the
reactions at E and F.
SOLUTION
(a) From f.b.d. of assembly
 − ( 0.12 m ) j − ( 0.225 m ) k 
TDG
TDG = λ DGTDG = 
 = 0.255  − ( 0.12 ) j − ( 0.225 ) k 
2
2
 ( 0.12 ) + ( 0.225 ) m 

 0.225  
ΣM y = 0: − ( 220 N )( 0.24 m ) + TDG 
  ( 0.16 m ) = 0
 0.255  

∴ TDG = 374.00 N
or TDG = 374 N
(b) From f.b.d. of assembly
ΣM F ( z -axis ) = 0:

0.120  
  ( 0.16 m ) = 0
 0.255  
( 220 N )( 0.19 m ) − Ex ( 0.13 m ) − 374 N 

∴ Ex = 104.923 N
ΣFx = 0: Fx + 104.923 N − 220 N = 0
∴ Fx = 115.077 N

 0.225  
ΣM F ( x-axis ) = 0: Ez ( 0.13 m ) + 374 N 
  ( 0.06 m ) = 0
 0.255  

∴ Ez = −152.308 N
PROBLEM 4.120 CONTINUED
 0.225 
ΣFz = 0: Fz − 152.308 N − ( 374 N ) 
=0
 0.255 
∴ Fz = 482.31 N
 0.12 
ΣFy = 0: Fy − ( 374 N ) 
=0
 0.255 
∴ Fy = 176.0 N
E = (104.9 N ) i − (152.3 N ) k
F = (115.1 N ) i + (176.0 N ) j + ( 482 N ) k
PROBLEM 4.121
A 30-kg cover for a roof opening is hinged at corners A and B. The roof
forms an angle of 30o with the horizontal, and the cover is maintained in
a horizontal position by the brace CE. Determine (a) the magnitude of the
force exerted by the brace, (b) the reactions at the hinges. Assume that the
hinge at A does not exert any axial thrust.
SOLUTION
(
)
W = mg = ( 30 kg ) 9.81 m/s2 = 294.3 N
First note
FEC = λ EC FEC = ( sin15° ) i + ( cos15° ) j FEC
From f.b.d. of cover
ΣM z = 0:
(a)
or
( FEC cos15° ) (1.0 m ) − W ( 0.5 m ) = 0
FEC cos15° (1.0 m ) − ( 294.3 N )( 0.5 m ) = 0
∴ FEC = 152.341 N
or FEC = 152.3 N
ΣM x = 0: W ( 0.4 m ) − Ay ( 0.8 m ) − ( FEC cos15° ) ( 0.8 m ) = 0
(b)
or
( 294.3 N )( 0.4 m ) − Ay ( 0.8 m ) − (152.341 N ) cos15° ( 0.8 m ) = 0
∴ Ay = 0
ΣM y = 0: Ax ( 0.8 m ) + ( FEC sin15° ) ( 0.8 m ) = 0
or
Ax ( 0.8 m ) + (152.341 N ) sin15° ( 0.8 m ) = 0
∴ Ax = −39.429 N
ΣFx = 0: Ax + Bx + FEC sin15° = 0
−39.429 N + Bx + (152.341 N ) sin15° = 0
∴ Bx = 0
PROBLEM 4.121 CONTINUED
ΣFy = 0: FEC cos15° − W + By = 0
or
(152.341 N ) cos15° − 294.3 N + By
=0
∴ By = 147.180 N
or A = − ( 39.4 N ) i
B = (147.2 N ) j
PROBLEM 4.122
The rectangular plate shown has a mass of 15 kg and is held in the
position shown by hinges A and B and cable EF. Assuming that the hinge
at B does not exert any axial thrust, determine (a) the tension in the cable,
(b) the reactions at A and B.
SOLUTION
(
)
W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N
First note
 ( 0.08 m ) i + ( 0.25 m ) j − ( 0.2 m ) k 
TEF
TEF = λ EFTEF = 
 TEF = 0.33 ( 0.08i + 0.25j − 0.2k )
2
2
2
( 0.08 ) + ( 0.25) + ( 0.2 ) m 

From f.b.d. of rectangular plate
ΣM x = 0:
or
or
(147.15 N )( 0.1 m ) − (TEF ) y ( 0.2 m ) = 0
 0.25 

14.715 N ⋅ m − 
 TEF  ( 0.2 m ) = 0
 0.33 

TEF = 97.119 N
or TEF = 97.1 N
ΣFx = 0: Ax + (TEF ) x = 0
 0.08 
Ax + 
 ( 97.119 N ) = 0
 0.33 
∴ Ax = −23.544 N
PROBLEM 4.122 CONTINUED
ΣM B( z -axis ) = 0: − Ay ( 0.3 m ) − (TEF ) y ( 0.04 m ) + W ( 0.15 m ) = 0
or
 0.25 

− Ay ( 0.3 m ) − 
 97.119 N  ( 0.04 m ) + 147.15 N ( 0.15 m ) = 0
 0.33 

∴ Ay = 63.765 N
ΣM B( y -axis ) = 0: Az ( 0.3 m ) + (TEF ) x ( 0.2 m ) + (TEF ) z ( 0.04 m ) = 0
 0.08 

 0.2 

Az ( 0.3 m ) + 
 TEF  ( 0.2 m ) − 
 TEF  ( 0.04 m ) = 0
 0.33 

 0.33 

∴ Az = −7.848 N
and A = − ( 23.5 N ) i + ( 63.8 N ) j − ( 7.85 N ) k
ΣFy = 0: Ay − W + (TEF ) y + By = 0
 0.25 
63.765 N − 147.15 N + 
 ( 97.119 N ) + By = 0
 0.33 
∴ By = 9.81 N
ΣFz = 0: Az − (TEF ) z + Bz = 0
 0.2 
−7.848 N − 
 ( 97.119 N ) + Bz = 0
 0.33 
∴ Bz = 66.708 N
and B = ( 9.81 N ) j + ( 66.7 N ) k
PROBLEM 4.123
Solve Problem 4.122 assuming that cable EF is replaced by a cable
attached at points E and H.
P4.122 The rectangular plate shown has a mass of 15 kg and is held in
the position shown by hinges A and B and cable EF. Assuming that the
hinge at B does not exert any axial thrust, determine (a) the tension in the
cable, (b) the reactions at A and B.
SOLUTION
(
)
W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N
First note
TEH = λ EH TEH


− ( 0.3 m ) i + ( 0.12 m ) j − ( 0.2 m ) k 
T

T = EH  − ( 0.3) i + ( 0.12 ) j − ( 0.2 ) k 
=

 EH
2
2
2
0.38
( 0.3) + ( 0.12 ) + ( 0.2 ) m 

From f.b.d. of rectangular plate
ΣM x = 0:
or
or
(147.15 N )( 0.1 m ) − (TEH ) y ( 0.2 m ) = 0
 0.12 

 TEH  ( 0.2 m ) = 0
 0.38 

(147.15 N )( 0.1 m ) − 
TEH = 232.99 N
or TEH = 233 N
ΣFx = 0: Ax + (TEH ) x = 0
 0.3 
Ax − 
 ( 232.99 N ) = 0
 0.38 
∴ Ax = 183.938 N
PROBLEM 4.123 CONTINUED
ΣM B( z -axis ) = 0: − Ay ( 0.3 m ) − (TEH ) y ( 0.04 m ) + W ( 0.15 m ) = 0
or
 0.12
− Ay ( 0.3 m ) − 
( 232.99 N ) ( 0.04 m ) + (147.15 N )( 0.15 m ) = 0
 0.38

∴ Ay = 63.765 N
ΣM B( y -axis ) = 0: Az ( 0.3 m ) + (TEH ) x ( 0.2 m ) + (TEH ) z ( 0.04 m ) = 0
or
 0.3 

 0.2 

Az ( 0.3 m ) − 
 ( 232.99 N )  ( 0.2 m ) − 
 ( 232.99 )  ( 0.04 m ) = 0
 0.38 

 0.38 

∴ Az = 138.976 N
and A = (183.9 N ) i + ( 63.8 N ) j + (139.0 N ) k
ΣFy = 0: Ay + By − W + (TEH ) y = 0
 0.12 
63.765 N + By − 147.15 N + 
 ( 232.99 N ) = 0
 0.38 
∴ By = 9.8092 N
ΣFz = 0: Az + Bz − (TEH ) z = 0
 0.2 
138.976 N + Bz − 
 ( 232.99 N ) = 0
 0.38 
∴ Bz = −16.3497 N
and B = ( 9.81 N ) j − (16.35 N ) k
PROBLEM 4.124
A small door weighing 16 lb is attached by hinges A and B to a wall and
is held in the horizontal position shown by rope EFH. The rope passes
around a small, frictionless pulley at F and is tied to a fixed cleat at H.
Assuming that the hinge at A does not exert any axial thrust, determine
(a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
T = λ EFT =
First note
=
(12 in.) i + ( 54 in.) j − ( 28 in.) k T
(12 )2 + ( 54 )2 + ( 28)2 in.
T
T
(12i + 54 j − 28k ) = ( 6i + 27 j − 14k )
62
31
W = − (16 lb ) j at G
From f.b.d. of door ABCD
(a)
ΣM x = 0: Ty ( 28 in.) − W (14 in.) = 0
  27  
T  31   ( 28 in.) − (16 lb )(14 in.) = 0
  
∴ T = 9.1852 lb
or T = 9.19 lb
(b)
ΣM B( z -axis ) = 0: − Ay ( 30 in.) + W (15 in.) − Ty ( 4 in.) = 0

 27  
− Ay ( 30 in.) + (16 lb )(15 in.) − ( 9.1852 lb )    ( 4 in.) = 0
 31  

∴ Ay = 6.9333 lb
PROBLEM 4.124 CONTINUED
ΣM B( y -axis ) = 0: Az ( 30 in.) + Tx ( 28 in.) − Tz ( 4 in.) = 0


 6 
 14  
Az ( 30 in.) + ( 9.1852 lb )    ( 28 in.) − ( 9.1852 lb )    ( 4 in.) = 0
 31  
 31  


∴ Az = −1.10617 lb
or A = ( 6.93 lb ) j − (1.106 lb ) k
 6
ΣFx = 0: Bx + Tx = Bx + ( 9.1852 lb )   = 0
 31 
∴ Bx = −1.77778 lb
ΣFy = 0: By + Ty − W + Ay = 0
 27 
By + ( 9.1852 lb )   − 16 lb + 6.9333 lb = 0
 31 
∴ By = 1.06666 lb
ΣFz = 0: Az − Tz + Bz = 0
 14 
−1.10617 lb − ( 9.1852 lb )   + Bz = 0
 31 
∴ Bz = 5.2543 lb
or B = − (1.778 lb ) i + (1.067 lb ) j + ( 5.25 lb ) k
PROBLEM 4.125
Solve Problem 4.124 assuming that the rope is attached to the door at I.
P4.124 A small door weighing 16 lb is attached by hinges A and B to a
wall and is held in the horizontal position shown by rope EFH. The rope
passes around a small, frictionless pulley at F and is tied to a fixed cleat
at H. Assuming that the hinge at A does not exert any axial thrust,
determine (a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
T = λ IF T =
First note
=
( 3 in.) i + ( 54 in.) j − (10 in.) k T
( 3)2 + ( 54 )2 + (10 )2 in.
T
( 3i + 54 j − 10k )
55
W = − (16 lb ) j
From f.b.d. of door ABCD
(a)
ΣM x = 0: W (14 in.) − Ty (10 in.) = 0
(16 lb )(14 in.) − 
54 
 T (10 in.) = 0
 55 
∴ T = 22.815 lb
or T = 22.8 lb
(b)
ΣM B( z -axis ) = 0: − Ay ( 30 in.) + W (15 in.) + Ty ( 5 in.) = 0
 54 
− Ay ( 30 in.) + (16 lb )(15 in.) + ( 22.815 lb )   ( 5 in.) = 0
 55 
∴ Ay = 11.7334 lb
PROBLEM 4.125 CONTINUED
ΣM B( y -axis ) = 0: Az ( 30 in.) + Tx (10 in.) + Tz ( 5 in.) = 0


 3 
 10  
Az ( 30 in.) + ( 22.815 lb )    (10 in.) + ( 22.815 lb )    ( 5 in.) = 0
 55  
 55  


∴ Az = −1.10618 lb
or A = (11.73 lb ) j − (1.106 lb ) k
ΣFx = 0: Bx + Tx = 0
 3 
Bx +   ( 22.815 lb ) = 0
 55 
∴ Bx = −1.24444 lb
ΣFy = 0: Ay − W + Ty + By = 0
 54 
11.7334 lb − 16 lb + ( 22.815 lb )   + By = 0
 55 
∴ By = −18.1336 lb
ΣFz = 0: Az − Tz + Bz = 0
 10 
−1.10618 lb − ( 22.815 lb )   + Bz = 0
 55 
∴ Bz = 5.2544 lb
or B = − (1.244 lb ) i − (18.13 lb ) j + ( 5.25 lb ) k
PROBLEM 4.126
A 285-lb uniform rectangular plate is supported in the position shown by
hinges A and B and by cable DCE, which passes over a frictionless hook
at C. Assuming that the tension is the same in both parts of the cable,
determine (a) the tension in the cable, (b) the reactions at A and B.
Assume that the hinge at B does not exert any axial thrust.
SOLUTION
λ CD =
First note
=
λ CE =
=
− ( 23 in.) i + ( 22.5 in.) j − (15 in.) k
35.5 in.
1
( −23i + 22.5 j − 15k )
35.5
( 9 in.) i + ( 22.5 in.) j − (15 in.) k
28.5 in.
1
( 9i + 22.5 j − 15k )
28.5
W = − ( 285 lb ) j
From f.b.d. of plate
(a)
ΣM x = 0:
 22.5  
 22.5  
 T  (15 in.) − 
 T  (15 in.) = 0
 35.5  
 28.5  
( 285 lb )( 7.5 in.) − 
∴ T = 100.121 lb
or T = 100.1 lb
PROBLEM 4.126 CONTINUED
 23 
 9 
ΣFx = 0: Ax − T 
 +T
=0
 35.5 
 28.5 
(b)
 23  (
 9 
Ax − (100.121 lb ) 
 + 100.121 lb ) 
=0
 35.5 
 28.5 
∴ Ax = 33.250 lb
  22.5  
ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T 
  ( 6 in.) −
  35.5  
or
  22.5  
T  28.5   ( 6 in.) = 0

 

 22.5  
− Ay ( 26 in.) + ( 285 lb )(13 in.) − (100.121 lb ) 
  ( 6 in.)
 35.5  


 22.5  
− (100.121 lb ) 
  ( 6 in.) = 0
 28.5  

∴ Ay = 109.615 lb
  15  
ΣM B( y -axis ) = 0: Az ( 26 in.) − T 
  ( 6 in.) −
  35.5  
  15  
− T 
  ( 6 in.) +
  28.5  
or
  23  
T  35.5   (15 in.)

 
  9 
T  28.5   (15 in.) = 0

 
1
 −1
Az ( 26 in.) + 
( 90 + 345) −
( 90 − 135) (100.121 lb ) = 0
35.5
28.5


∴ Az = 41.106 lb
or A = ( 33.3 lb ) i + (109.6 lb ) j + ( 41.1 lb ) k
 22.5 
 22.5 
ΣFy = 0: By − W + T 
 +T
 + Ay = 0
 35.5 
 28.5 
 22.5 22.5 
By − 285 lb + (100.121 lb ) 
+
 + 109.615 lb = 0
 35.5 28.5 
∴ By = 32.885 lb
 15 
 15 
ΣFz = 0: Bz + Az − T 
 −T
=0
 35.5 
 28.5 
15 
 15
Bz + 41.106 lb − (100.121 lb ) 
+
=0
 35.5 28.5 
∴ Bz = 53.894 lb
or B = ( 32.9 lb ) j + ( 53.9 lb ) k
PROBLEM 4.127
Solve Problem 4.126 assuming that cable DCE is replaced by a cable
attached to point E and hook C.
P4.126 A 285-lb uniform rectangular plate is supported in the position
shown by hinges A and B and by cable DCE, which passes over a
frictionless hook at C. Assuming that the tension is the same in both parts
of the cable, determine (a) the tension in the cable, (b) the reactions at A
and B. Assume that the hinge at B does not exert any axial thrust.
SOLUTION
First note
λ CE =
=
( 9 in.) i + ( 22.5 in.) j − (15 in.) k
28.5 in.
1
( 9i + 22.5 j − 15k )
28.5
W = − ( 285 lb ) j
From f.b.d. of plate
(a)
ΣM x = 0:
 22.5  
 T  (15 in.) = 0
 28.5  
( 285 lb )( 7.5 in.) − 
∴ T = 180.500 lb
or T = 180.5 lb
(b)
 9 
ΣFx = 0: Ax + T 
=0
 28.5 
 9 
Ax + 180.5 lb 
=0
 28.5 
∴ Ax = −57.000 lb
PROBLEM 4.127 CONTINUED
  22.5  
ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T 
  ( 6 in.) = 0
  28.5  

 22.5  
− Ay ( 26 in.) + ( 285 lb )(13 in.) − (180.5 lb ) 
  ( 6 in.) = 0
 28.5  

∴ Ay = 109.615 lb
  15  
ΣM B( y -axis ) = 0: Az ( 26 in.) − T 
  ( 6 in.) +
  28.5  
  9 
T  28.5   (15 in.) = 0

 
 45 
Az ( 26 in.) + (180.5 lb ) 
=0
 28.5 
∴ Az = −10.9615 lb
or A = − ( 57.0 lb ) i + (109.6 lb ) j − (10.96 lb ) k
 22.5 
ΣFy = 0: By − W + T 
 + Ay = 0
 28.5 
 22.5 
By − 285 lb + (180.5 lb ) 
 − 109.615 lb = 0
 28.5 
∴ By = 32.885 lb
 15 
ΣFz = 0: Bz + Az − T 
=0
 28.5 
 15 
Bz − 10.9615 lb − 180.5 lb 
=0
 28.5 
∴ Bz = 105.962 lb
or B = ( 32.9 lb ) j + (106.0 lb ) k
PROBLEM 4.128
The tensioning mechanism of a belt drive consists of frictionless pulley
A, mounting plate B, and spring C. Attached below the mounting plate is
slider block D which is free to move in the frictionless slot of bracket E.
Knowing that the pulley and the belt lie in a horizontal plane, with
portion F of the belt parallel to the x axis and portion G forming an angle
of 30° with the x axis, determine (a) the force in the spring, (b) the
reaction at D.
SOLUTION
From f.b.d. of plate B
(a)
ΣFx = 0: 12 N + (12 N ) cos 30° − T = 0
∴ T = 22.392 N
(b)
or T = 22.4 N
ΣFy = 0: Dy = 0
ΣFz = 0: Dz − (12 N ) sin 30° = 0
∴ Dz = 6 N
or D = ( 6.00 N ) k
ΣM x = 0: M Dx − (12 N ) sin 30°  ( 22 mm ) = 0
∴ M Dx = 132.0 N ⋅ mm
ΣM D( y -axis ) = 0: M Dy + ( 22.392 N )( 30 mm ) − (12 N )( 75 mm ) − (12 N ) cos 30° ( 75 mm ) = 0
∴ M Dy = 1007.66 N ⋅ mm
ΣM D( z -axis ) = 0: M Dz + ( 22.392 N )(18 mm ) − (12 N )( 22 mm ) − (12 N ) cos 30° ( 22 mm ) = 0
∴ M Dz = 89.575 N ⋅ mm
or M D = ( 0.1320 N ⋅ m ) i + (1.008 N ⋅ m ) j + ( 0.0896 N ⋅ m ) k
PROBLEM 4.129
The assembly shown is welded to collar A which fits on the vertical pin
shown. The pin can exert couples about the x and z axes but does not
prevent motion about or along the y axis. For the loading shown,
determine the tension in each cable and the reaction at A.
SOLUTION
TCF = λ CFTCF =
First note
− ( 0.16 m ) i + ( 0.12 m ) j
( 0.16 ) + ( 0.12 ) m
2
2
TCF
= TCF ( −0.8i + 0.6 j)
TDE = λ DETDE =
( 0.24 m ) j − ( 0.18 m ) k T
DE
( 0.24 )2 + ( 0.18)2 m
= TDE ( 0.8j − 0.6k )
(a) From f.b.d. of assembly
ΣFy = 0: 0.6TCF + 0.8TDE − 800 N = 0
or
0.6TCF + 0.8TDE = 800 N
(1)
ΣM y = 0: − ( 0.8TCF ) ( 0.27 m ) + ( 0.6TDE )( 0.16 m ) = 0
or
TDE = 2.25TCF
(2)
PROBLEM 4.129 CONTINUED
Substituting Equation (2) into Equation (1)
0.6TCF + 0.8 ( 2.25 ) TCF  = 800 N
∴ TCF = 333.33 N
and from Equation (2)
or TCF = 333 N
TDE = 2.25 ( 333.33 N ) = 750.00 N
or TDE = 750 N
(b) From f.b.d. of assembly
ΣFz = 0: Az − ( 0.6 )( 750.00 N ) = 0
∴ Az = 450.00 N
ΣFx = 0: Ax − ( 0.8 )( 333.33 N ) = 0
∴ Ax = 266.67 N
or A = ( 267 N ) i + ( 450 N ) k
ΣM x = 0: M Ax + ( 800 N )( 0.27 m ) − ( 333.33 N )( 0.6 )  ( 0.27 m ) − ( 750 N )( 0.8 )  ( 0.18 m ) = 0
∴ M Ax = −54.001 N ⋅ m
ΣM z = 0: M Az − ( 800 N )( 0.16 m ) + ( 333.33 N )( 0.6 )  ( 0.16 m ) + ( 750 N )( 0.8 )  ( 0.16 m ) = 0
∴ M Az = 0
or M A = − ( 54.0 N ⋅ m ) i
PROBLEM 4.130
The lever AB is welded to the bent rod BCD which is supported by
bearing E and by cable DG. Assuming that the bearing can exert an axial
thrust and couples about axes parallel to the x and z axes, determine
(a) the tension in cable DG, (b) the reaction at E.
SOLUTION
TDG = λ DGTDG =
First note
=
− ( 0.12 m ) j − ( 0.225 m ) k
( 0.12 ) + ( 0.225) m
2
2
TDG
TDG
( −0.12 j − 0.225k )
0.255
(a) From f.b.d. of weldment
 0.225 

ΣM y = 0: 
 TDG  ( 0.16 m ) − ( 220 N )( 0.24 m ) = 0
 0.255 

∴ TDG = 374.00 N
(b) From f.b.d. of weldment
ΣFx = 0: Ex − 220 N = 0
∴ Ex = 220.00 N
 0.12 
ΣFy = 0: E y − ( 374.00 N ) 
=0
 0.255 
∴ E y = 176.000 N
or TDG = 374 N
PROBLEM 4.130 CONTINUED
 0.225 
ΣFz = 0: Ez − ( 374.00 N ) 
=0
 0.255 
∴ Ez = 330.00 N
or E = ( 220 N ) i + (176.0 N ) j + ( 330 N ) k
ΣM x = 0: M Ex + ( 330.00 N )( 0.19 m ) = 0
∴ M Ex = −62.700 N ⋅ m
ΣM z = 0:
( 220 N )( 0.06 m ) + M Ez

 0.12  
− ( 374.00 N ) 
  ( 0.16 m ) = 0
 0.255  

∴ M Ez = −14.9600 N ⋅ m
or M E = − ( 62.7 N ⋅ m ) i − (14.96 N ⋅ m ) k
PROBLEM 4.131
Solve Problem 4.124 assuming that the hinge at A is removed and that the
hinge at B can exert couples about the y and z axes.
P4.124 A small door weighing 16 lb is attached by hinges A and B to a
wall and is held in the horizontal position shown by rope EFH. The rope
passes around a small, frictionless pulley at F and is tied to a fixed cleat
at H. Assuming that the hinge at A does not exert any axial thrust,
determine (a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
From f.b.d. of door
ΣM B = 0: rG/B × W + rE/B × TEF + M B = 0
(a)
where
W = − (16 lb ) j
M B = M By j + M Bz k
TEF = λ EFTEF =
=
(12 in.) i + ( 54 in.) j − ( 28 in.) k T
EF
(12 )2 + ( 54 )2 + ( 28)2 in.
TEF
( 6i + 27 j − 14k )
31
rG/B = − (15 in.) i + (14 in.) k
rE/B = − ( 4 in.) i + ( 28 in.) k
∴
i
j k
i j k
T 
−15 0 14 (16 lb ) + −4 0 28  EF  + M By j + M Bz k = 0
 31 
0 −1 0
6 27 −14
(
or
( 224 − 24.387TEF ) i + ( 3.6129TEF
(
)
)
+ M By j
)
+ 240 − 3.4839TEF + M Bz k = 0
From i-coefficient
224 − 24.387TEF = 0
∴ TEF = 9.1852 lb
or TEF = 9.19 lb W
(b) From j-coefficient
3.6129 ( 9.1852 ) + M By = 0
∴ M By = −33.185 lb ⋅ in.
PROBLEM 4.131 CONTINUED
From k-coefficient
240 − 3.4839 ( 9.1852 ) + M Bz = 0
∴ M Bz = −208.00 lb ⋅ in.
or M B = − ( 33.2 lb ⋅ in.) j − ( 208 lb ⋅ in.) k W
ΣFx = 0: Bx +
6
( 9.1852 lb ) = 0
31
∴ Bx = −1.77778 lb
ΣFy = 0: By − 16 lb +
27
( 9.1852 lb ) = 0
31
∴ By = 8.0000 lb
ΣFz = 0: Bz −
14
( 9.1852 lb ) = 0
31
∴ Bz = 4.1482 lb
or B = − (1.778 lb ) i + ( 8.00 lb ) j + ( 4.15 lb ) k W
PROBLEM 4.132
The frame shown is supported by three cables and a ball-and-socket joint
at A. For P = 0, determine the tension in each cable and the reaction at A.
SOLUTION
First note
TDI = λ DI TDI =
=
TEH = λ EH TEH =
=
TFG = λ FGTFG =
=
− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k
( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m
TDI
TDI
( −0.65i + 0.2 j − 0.44k )
0.81
− ( 0.45 m ) i + ( 0.24 m ) j
( 0.45)2 + ( 0.24 )2 m
TEH
TEH
( −0.45i ) + ( 0.24 j)
0.51
− ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k
( 0.45
)
2
+ ( 0.2 ) + ( 0.36 ) m
2
2
TFG
TFG
( −0.45i + 0.2 j + 0.36k )
0.61
From f.b.d. of frame
ΣM A = 0: rD/ A × TDI + rC/ A × ( −280 N ) j + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j = 0
or
i
j
k
i
j k
i
j k
i
j
k
 TDI 
 TEH 
 TFG 
0.65 0.2
0 
0.32 0 
 + 0.65 0 0 ( 280 N ) + 0
 + 0.45 0 0.06 

 0.81 
 0.51 
 0.61 
−0.65 0.2 −0.44
−0.45 0.24 0
−0.45 0.2 0.36
0 −1 0
i
j
k
+ 0.45 0 0.06 ( 360 N ) = 0
0 −1 0
or
( −0.088i + 0.286 j + 0.26k )
TDI
T
+ ( −0.65k ) 280 N + ( 0.144k ) EH
0.81
0.51
+ ( −0.012i − 0.189 j + 0.09k )
TFG
+ ( 0.06i − 0.45k )( 360 N ) = 0
0.61
PROBLEM 4.132 CONTINUED
From i-coefficient
T 
T 
−0.088  DI  − 0.012  FG  + 0.06 ( 360 N ) = 0
 0.81 
 0.61 
∴ 0.108642TDI + 0.0196721TFG = 21.6
From j-coefficient
(1)
T 
T 
0.286  DI  − 0.189  FG  = 0
0.81


 0.61 
∴ TFG = 1.13959TDI
(2)
From k-coefficient
T 
T 
T 
0.26  DI  − 0.65 ( 280 N ) + 0.144  EH  + 0.09  FG 
 0.81 
 0.51 
 0.61 
− 0.45 ( 360 N ) = 0
∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 N
(3)
Substitution of Equation (2) into Equation (1)
0.108642TDI + 0.0196721(1.13959TDI ) = 21.6
∴ TDI = 164.810 N
TDI = 164.8 N W
or
Then from Equation (2)
TFG = 1.13959 (164.810 N ) = 187.816 N
TFG = 187.8 N W
or
And from Equation (3)
0.32099 (164.810 N ) + 0.28235TEH + 0.147541(187.816 N ) = 344 N
∴ TEH = 932.84 N
TEH = 933 N W
or
The vector forms of the cable forces are:
TDI =
164.810 N
( −0.65i + 0.2 j − 0.44k )
0.81
= − (132.25 N ) i + ( 40.694 N ) j − ( 89.526 N ) k
TEH =
932.84 N
( −0.45i + 0.24 j) = − (823.09 N ) i + ( 438.98 N ) j
0.51
TFG =
187.816 N
( −0.45i + 0.2 j + 0.36k )
0.61
= − (138.553 N ) i + ( 61.579 N ) j + (110.842 N ) k
PROBLEM 4.132 CONTINUED
Then, from f.b.d. of frame
ΣFx = 0: Ax − 132.25 − 823.09 − 138.553 = 0
∴ Ax = 1093.89 N
ΣFy = 0: Ay + 40.694 + 438.98 + 61.579 − 360 − 280 = 0
∴ Ay = 98.747 N
ΣFz = 0: Az − 89.526 + 110.842 = 0
∴ Az = −21.316 N
or
A = (1094 N ) i + ( 98.7 N ) j − ( 21.3 N ) k W
PROBLEM 4.133
The frame shown is supported by three cables and a ball-and-socket joint
at A. For P = 50 N, determine the tension in each cable and the reaction
at A.
SOLUTION
First note
TDI = λ DI TDI =
=
TEH = λ EH TEH =
=
TFG = λ FGTFG =
=
− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k
( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m
TDI
TDI
( −65i + 20 j − 44k )
81
− ( 0.45 m ) i + ( 0.24 m ) j
( 0.45)2 + ( 0.24 )2 m
TEH
TEH
( −15i + 8j)
17
− ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k
( 0.45 ) + ( 0.2 ) + ( 0.36 ) m
2
2
2
TFG
TFG
( −45i + 20 j + 36k )
61
From f.b.d. of frame
ΣM A = 0: rD/ A × TDI + rC/ A ×  − ( 280 N ) j + ( 50 N ) k 
+ rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j
or
i
j
k
i
j
k
i
j k
 TDI 
T 
0.65 0.2 0 
0 + 0 0.32 0  EH 
 + 0.65 0
 81 
 17 
−65 20 −44
−15 8 0
0 −280 50
i
j
k
i
j
k
 TFG 
+ 0.45 0 0.06 
 + 0.45 0 0.06 ( 360 N ) = 0
 61 
−45 20 36
0 −1 0
and
( −8.8i + 28.6 j + 26k ) 
TDI 
 TEH 
 + ( −32.5 j − 182k ) + ( 4.8k ) 

 81 
 17 
T 
+ ( −1.2i − 18.9 j + 9.0k )  FG  + ( 0.06i − 0.45k ) ( 360 ) = 0
 61 
PROBLEM 4.133 CONTINUED
T 
T
−8.8  DI  − 1.2  FG
 81 
 61
From i-coefficient

 + 0.06 ( 360 ) = 0

∴ 0.108642TDI + 0.0196721TFG = 21.6
(1)
T 
T 
From j-coefficient 28.6  DI  − 32.5 − 18.9  FG  = 0
 81 
 61 
∴ 0.35309TDI − 0.30984TFG = 32.5
(2)
From k-coefficient
T 
T 
T 
26  DI  − 182 + 4.8  EH  + 9.0  FG  − 0.45 ( 360 ) = 0
 81 
 17 
 61 
∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344
−3.25 × Equation (1)
Add Equation (2)
(3)
−0.35309TDI − 0.063935TFG = −70.201
0.35309TDI − 0.30984TFG =
−0.37378TFG
32.5
= −37.701
∴ TFG = 100.864 N
TFG = 100.9 N W
or
Then from Equation (1)
0.108642TDI + 0.0196721(100.864 ) = 21.6
∴ TDI = 180.554 N
TDI = 180.6 N W
or
and from Equation (3)
0.32099 (180.554 ) + 0.28235TEH + 0.147541(100.864 ) = 344
∴ TEH = 960.38 N
TEH = 960 N W
or
The vector forms of the cable forces are:
TDI =
180.554 N
( −65i + 20 j − 44k )
81
= − (144.889 N ) i + ( 44.581 N ) j − ( 98.079 N ) k
TEH =
960.38 N
( −15i + 8j) = − (847.39 N ) i + ( 451.94 N ) j
17
TFG =
100.864 N
( −45i + 20 j + 36k )
61
= − ( 74.409 N ) i + ( 33.070 N ) j + ( 59.527 N ) k
PROBLEM 4.133 CONTINUED
Then from f.b.d. of frame
ΣFx = 0: Ax − 144.889 − 847.39 − 74.409 = 0
∴ Ax = 1066.69 N
ΣFy = 0: Ay + 44.581 + 451.94 + 33.070 − 360 − 280 = 0
∴ Ay = 110.409 N
ΣFz = 0: Az − 98.079 + 59.527 + 50 = 0
∴ Az = −11.448 N
Therefore,
A = (1067 N ) i + (110.4 N ) j − (11.45 N ) k W
PROBLEM 4.134
The rigid L-shaped member ABF is supported by a ball-and-socket joint
at A and by three cables. For the loading shown, determine the tension in
each cable and the reaction at A.
SOLUTION
First note
TBG = λ BGTBG =
− (18 in.) i + (13.5 in.) k
(18)
2
+ (13.5 ) in.
2
TBG
= TBG ( −0.8i + 0.6k )
TDH = λ DH TDH =
− (18 in.) i + ( 24 in.) j
(18)2 + ( 24 )2 in.
TDH
= TDH ( −0.6i + 0.8 j)
Since λ FJ = λ DH ,
TFJ = TFJ ( −0.6i + 0.8 j)
From f.b.d. of member ABF
ΣM A( x-axis ) = 0:
( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) − (120 lb )(12 in.) = 0
∴ 3.2TFJ + 1.6TDH = 480
ΣM A( z -axis ) = 0:
( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) − (120 lb )(18 in.) = 0
∴
− 3.2TFJ − 3.2TDH = −960
Equation (1) + Equation (2)
Substituting in Equation (1)
ΣM A( y -axis ) = 0:
(1)
(2)
TDH = 300 lb W
TFJ = 0 W
( 0.6TFJ ) ( 48 in.) + 0.6 ( 300 lb ) ( 24 in.) − ( 0.6TBG ) (18 in.) = 0
∴ TBG = 400 lb W
PROBLEM 4.134 CONTINUED
ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0
−0.6 ( 300 lb ) − 0.8 ( 400 lb ) + Ax = 0
∴ Ax = 500 lb
ΣFy = 0: 0.8TFJ + 0.8TDH − 240 lb + Ay = 0
0.8 ( 300 lb ) − 240 + Ay = 0
∴ Ay = 0
ΣFz = 0: 0.6TBG + Az = 0
0.6 ( 400 lb ) + Az = 0
∴ Az = −240 lb
Therefore,
A = ( 500 lb ) i − ( 240 lb ) k W
PROBLEM 4.135
Solve Problem 4.134 assuming that the load at C has been removed.
P4.134 The rigid L-shaped member ABF is supported by a ball-andsocket joint at A and by three cables. For the loading shown, determine
the tension in each cable and the reaction at A.
SOLUTION
First
TBG = λ BGTBG =
− (18 in.) i + (13.5 in.) k
(18)
2
+ (13.5 ) in.
2
TBG
= TBG ( −0.8i + 0.6k )
TDH = λ DH TDH =
− (18 in.) i + ( 24 in.) j
(18)2 + ( 24 )2 in.
TDH
= TDH ( −0.6i + 0.8 j)
λ FJ = λ DH
Since
TFJ = TFJ ( −0.6i + 0.8 j)
From f.b.d. of member ABF
ΣM A( x-axis ) = 0:
( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) = 0
∴ 3.2TFJ + 1.6TDH = 360
ΣM A( z -axis ) = 0:
(1)
( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) = 0
∴
− 3.2TFJ − 3.2TDH = −480
(2)
Equation (1) + Equation (2)
TDH = 75.0 lb W
Substituting into Equation (2)
TFJ = 75.0 lb W
ΣM A( y -axis ) = 0:
or
( 0.6TFJ ) ( 48 in.) + ( 0.6TDH )( 24 in.) − ( 0.6TBG ) (18 in.) = 0
( 75.0 lb )( 48 in.) + ( 75.0 lb )( 24 in.) = TBG (18 in.)
TBG = 300 lb W
PROBLEM 4.135 CONTINUED
ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0
−0.6 ( 75.0 + 75.0 ) − 0.8 ( 300 ) + Ax = 0
∴ Ax = 330 lb
ΣFy = 0: 0.8TFJ + 0.8TDH − 120 lb + Ay = 0
0.8 (150 lb ) − 120 lb + Ay = 0
∴ Ay = 0
ΣFz = 0: 0.6TBG + Az = 0
0.6 ( 300 lb ) + Az = 0
∴ Az = −180 lb
Therefore
A = ( 330 lb ) i − (180 lb ) k W
PROBLEM 4.136
In order to clean the clogged drainpipe AE, a plumber has disconnected
both ends of the pipe and inserted a power snake through the opening at
A. The cutting head of the snake is connected by a heavy cable to an
electric motor which rotates at a constant speed as the plumber forces the
cable into the pipe. The forces exerted by the plumber and the motor on
the end of the cable can be represented by the wrench F = − ( 60 N ) k ,
M = − (108 N ⋅ m ) k. Determine the additional reactions at B, C, and D
caused by the cleaning operation. Assume that the reaction at each
support consists of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
ΣFx = 0: Bx = 0
ΣM D( x-axis ) = 0:
( 60 N )( 2.5 m ) − Bz ( 2 m ) = 0
∴ Bz = 75.0 N
and B = ( 75.0 N ) k
ΣM D( z -axis ) = 0: C y ( 3 m ) − 108 N ⋅ m = 0
∴ C y = 36.0 N
ΣM D( y -axis ) = 0: − Cz ( 3 m ) − ( 75 N )( 4 m ) + ( 60 N )( 4 m ) = 0
∴ C z = −20.0 N
and C = ( 36.0 N ) j − ( 20.0 N ) k
ΣFy = 0: Dy + 36.0 = 0
∴ D y = −36.0 N
ΣFz = 0: Dz − 20.0 N + 75.0 N − 60 N = 0
∴ Dz = 5.00 N
and D = − ( 36.0 N ) j + ( 5.00 N ) k
PROBLEM 4.137
Solve Problem 4.136 assuming that the plumber exerts a force
F = − ( 60 N ) k and that the motor is turned off ( M = 0 ) .
P4.136 In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power snake through
the opening at A. The cutting head of the snake is connected by a heavy
cable to an electric motor which rotates at a constant speed as the
plumber forces the cable into the pipe. The forces exerted by the plumber
and the motor on the end of the cable can be represented by the wrench
F = − ( 60 N ) k ,
M = − (108 N ⋅ m ) k. Determine the additional
reactions at B, C, and D caused by the cleaning operation. Assume that
the reaction at each support consists of two force components
perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
ΣFx = 0: Bx = 0
ΣM D( x-axis ) = 0:
( 60 N )( 2.5 m ) − Bz ( 2 m ) = 0
∴ Bz = 75.0 N
and B = ( 75.0 N ) k
ΣM D( z -axis ) = 0: C y ( 3 m ) − Bx ( 2 m ) = 0
∴ Cy = 0
ΣM D( y -axis ) = 0: Cz ( 3 m ) − ( 75.0 N )( 4 m ) + ( 60 N )( 4 m ) = 0
∴ C z = −20 N
and C = − ( 20.0 N ) k
ΣFy = 0: Dy + C y = 0
∴ Dy = 0
ΣFz = 0: Dz + Bz + C z − F = 0
Dz + 75 N − 20 N − 60 N = 0
∴ Dz = 5.00 N
and D = ( 5.00 N ) k
PROBLEM 4.138
Three rods are welded together to form a “corner” which is supported by
three eyebolts. Neglecting friction, determine the reactions at A, B, and C
when P = 240 N, a = 120 mm, b = 80 mm, and c = 100 mm.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0
i
j k
i j k
i
j k
120 0 0 + 0 80 0 + 0 0 100 = 0
Bx 0 Bz
Cx Cy 0
0 Ay Az
( −120 Az j + 120 Ayk ) + (80Bzi − 80Bxk ) + ( −100C yi + 100Cx j) = 0
From i-coefficient
80Bz − 100C y = 0
Bz = 1.25C y
or
j-coefficient
−120 Az + 100Cx = 0
Cx = 1.2 Az
or
k-coefficient
(1)
(2)
120 Ay − 80Bx = 0
Bx = 1.5 Ay
or
(3)
ΣF = 0: A + B + C − P = 0
( Bx + Cx ) i + ( Ay + C y − 240 N ) j + ( Az + Bz ) k = 0
or
Bx + Cx = 0
From i-coefficient
Cx = − Bx
or
j-coefficient
Ay + C y − 240 N = 0
Ay + C y = 240 N
or
(5)
Az + Bz = 0
k-coefficient
or
(4)
Az = − Bz
(6)
PROBLEM 4.138 CONTINUED
Substituting Cx from Equation (4) into Equation (2)
− Bz = 1.2 Az
(7)
Using Equations (1), (6), and (7)
Cy =
Bz
− Az
1  Bx  Bx
=
=

=
1.25 1.25 1.25  1.2  1.5
(8)
From Equations (3) and (8)
Cy =
1.5 Ay
1.5
C y = Ay
or
and substituting into Equation (5)
2 Ay = 240 N
∴ Ay = C y = 120 N
(9)
Using Equation (1) and Equation (9)
Bz = 1.25 (120 N ) = 150.0 N
Using Equation (3) and Equation (9)
Bx = 1.5 (120 N ) = 180.0 N
From Equation (4)
Cx = −180.0 N
From Equation (6)
Az = −150.0 N
Therefore
A = (120.0 N ) j − (150.0 N ) k
B = (180.0 N ) i + (150.0 N ) k
C = − (180.0 N ) i + (120.0 N ) j
PROBLEM 4.139
Solve Problem 4.138 assuming that the force P is removed and is
replaced by a couple M = + ( 6 N ⋅ m ) j acting at B.
P4.138 Three rods are welded together to form a “corner” which is
supported by three eyebolts. Neglecting friction, determine the reactions
at A, B, and C when P = 240 N, a = 120 mm, b = 80 mm, and
c = 100 mm.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0
i
j k
i
j
k
i
j k
0.12 0 0 + 0 0.08 0 + 0 0 0.1 + ( 6 N ⋅ m ) j = 0
Bx 0 Bz
Cx Cy 0
0 Ay Az
( −0.12 Az j + 0.12 Ayk ) + ( 0.08Bz j − 0.08Bxk )
(
)
+ −0.1C y i + 0.1Cx j + ( 6 N ⋅ m ) j = 0
From i-coefficient
0.08Bz − 0.1C y = 0
C y = 0.8Bz
or
j-coefficient
(1)
−0.12 Az + 0.1Cx + 6 = 0
Cx = 1.2 Az − 60
or
k-coefficient
(2)
0.12 Ay − 0.08Bx = 0
Bx = 1.5 Ay
or
(3)
ΣF = 0: A + B + C = 0
( Bx + Cx ) i + ( Ay + C y ) j + ( Az
+ Bz ) k = 0
From i-coefficient
Cx = − Bx
(4)
j-coefficient
C y = − Ay
(5)
k-coefficient
Az = − Bz
(6)
Substituting Cx from Equation (4) into Equation (2)
B 
Az = 50 −  x 
 1.2 
(7)
PROBLEM 4.139 CONTINUED
Using Equations (1), (6), and (7)
2
C y = 0.8Bz = −0.8 Az =   Bx − 40
3
(8)
From Equations (3) and (8)
C y = Ay − 40
2 Ay = 40
Substituting into Equation (5)
∴ Ay = 20.0 N
From Equation (5)
C y = −20.0 N
Equation (1)
Bz = −25.0 N
Equation (3)
Bx = 30.0 N
Equation (4)
Cx = −30.0 N
Equation (6)
Az = 25.0 N
Therefore
A = ( 20.0 N ) j + ( 25.0 N ) k
B = ( 30.0 N ) i − ( 25.0 N ) k
C = − ( 30.0 N ) i − ( 20.0 N ) j
PROBLEM 4.140
The uniform 10-lb rod AB is supported by a ball-and-socket joint at A and
leans against both the rod CD and the vertical wall. Neglecting the effects
of friction, determine (a) the force which rod CD exerts on AB, (b) the
reactions at A and B. (Hint: The force exerted by CD on AB must be
perpendicular to both rods.)
SOLUTION
(a) The force acting at E on the f.b.d. of rod AB is perpendicular to AB
and CD. Letting λ E = direction cosines for force E,
λE =
rB/ A × k
rB/ A × k
 − ( 32 in.) i + ( 24 in.) j − ( 40 in.) k  × k
=
2
2
( 32 ) + ( 24 ) in.
= 0.6i + 0.8 j
Also,
W = − (10 lb ) j
B = Bk
E = E ( 0.6i + 0.8 j)
From f.b.d. of rod AB
ΣM A = 0: rG/ A × W + rE/ A × E + rB/ A × B = 0
i
j
k
i
j
k
i
j
k
∴ −16 12 −20 (10 lb ) + −24 18 −30 E + −32 24 −40 B = 0
0 −1 0
0.6 0.8 0
0
0
1
( −20i + 16k )(10 lb ) + ( 24i − 18 j − 30k ) E + ( 24i + 32 j) B = 0
From k-coefficient
160 − 30 E = 0
∴ E = 5.3333 lb
and
E = 5.3333 lb ( 0.6i + 0.8 j)
E = ( 3.20 lb ) i + ( 4.27 lb ) j
or
(b) From j-coefficient
−18 ( 5.3333 lb ) + 32 B = 0
∴ B = 3.00 lb
or
B = ( 3.00 lb ) k
PROBLEM 4.140 CONTINUED
From f.b.d. of rod AB
ΣF = 0: A + W + E + B = 0
Ax i + Ay j + Az k − (10 lb ) j + ( 3.20 lb ) i + ( 4.27 lb ) j + ( 3.00 lb ) k = 0
From i-coefficient
Ax + 3.20 lb = 0
∴ Ax = −3.20 lb
j-coefficient
Ay − 10 lb + 4.27 lb = 0
∴ Ay = 5.73 lb
k-coefficient
Az + 3.00 lb = 0
∴ Az = −3.00 lb
Therefore
A = − ( 3.20 lb ) i + ( 5.73 lb ) j − ( 3.00 lb ) k
PROBLEM 4.141
A 21-in.-long uniform rod AB weighs 6.4 lb and is attached to a ball-andsocket joint at A. The rod rests against an inclined frictionless surface and
is held in the position shown by cord BC. Knowing that the cord is 21 in.
long, determine (a) the tension in the cord, (b) the reactions at A and B.
SOLUTION
First note
W = − ( 6.4 lb ) j
N B = N B ( 0.8j + 0.6k )
LAB = 21 in.
=
( xB )2 + (13 + 3)2 + ( 4 )2
=
xB2 + (16 ) + ( 4 )
2
2
∴ xB = 13 in.
TBC = λ BCTBC =
=
(13 in.) i + (16 in.) j − ( 4 in.) k T
21 in.
BC
TBC
(13i + 16 j − 4k )
21
From f.b.d. of rod AB
ΣM A = 0: rG/ A × W + rB/ A × N B + rC/ A × TBC = 0
i
j k
i
j
k
i j k
26TBC
6.5 −8 2 + 13 −16 4 N B + 1 0 0
=0
21
0 −6.4 0
0 0.8 0.6
13 16 −4
(12.8i − 41.6k ) + ( −12.8i − 7.8j + 10.4k ) N B + ( 4 j + 16k )
26TBC
=0
21
PROBLEM 4.141 CONTINUED
12.8 − 12.8N B = 0
From i-coeff.
∴ N B = 1.00 lb
N B = ( 0.800 lb ) j + ( 0.600 lb ) k
or
 26 
−7.8 N B + 4   TBC = 0
 21 
From j-coeff.
∴ TBC = 1.575 lb
From f.b.d. of rod AB
ΣF = 0: A + W + N B + TBC = 0

( Axi + Ay j + Azk ) − ( 6.4 lb ) j + ( 0.800 lb ) j + ( 0.600 lb ) k +  1.575
 (13i + 16 j − 4k ) = 0
21 
From i-coefficient
Ax = −0.975 lb
j-coefficient
Ay = 4.40 lb
k-coefficient
Az = −0.3 lb
∴ (a)
TBC = 1.575 lb
(b)
A = − ( 0.975 lb ) i + ( 4.40 lb ) j − ( 0.300 lb ) k
N B = ( 0.800 lb ) j + ( 0.600 lb ) k
PROBLEM 4.142
While being installed, the 56-lb chute ABCD is attached to a wall with
brackets E and F and is braced with props GH and IJ. Assuming that the
weight of the chute is uniformly distributed, determine the magnitude of
the force exerted on the chute by prop GH if prop IJ is removed.
SOLUTION
First note
 42 in. 
θ = tan −1 
 = 16.2602°
 144 in. 
xG = ( 50 in.) cos16.2602° = 48 in.
yG = 78 in. − ( 50 in.) sin16.2602° = 64 in.
λ BA =
− (144 in.) i + ( 42 in.) j
(144 )
2
+ ( 42 ) in.
2
=
1
( −24i + 7 j)
25
rK / A = ( 72 in.) i − ( 21 in.) j + ( 9 in.) k
rG/ A = ( 48 in.) i − ( 78 in. − 64 in.) j + (18 in.) k = ( 48 in.) i − (14 in.) j + (18 in.) k
W = − ( 56 lb ) j
PHG = λ HG PHG
=
=
− ( 2 in.) i + ( 64 in.) j − (16 in.) k
( 2 ) + ( 64 ) + (16 ) in.
2
2
PHG
( −i + 32 j − 8k )
33
2
PHG
PROBLEM 4.142 CONTINUED
From the f.b.d. of the chute
ΣM BA = 0: λ BA ⋅ ( rK / A × W ) + λ BA ⋅ ( rG/ A × PHG ) = 0
−24
−24
7 0
7 0
 56 
 P

72 −21 9   + 48 −14 18  HG  = 0
(
)
 25 
 33 25 
−1 32 −8
0 −1 0
−216 ( 56 )
P
+ [ −24 ( −14 ) ( −8 ) − ( −24 ) (18 )( 32 ) + 7 (18 ) ( −1) − ( 7 )( 48 )( −8 )] HG = 0
25
33 ( 25 )
∴ PHG = 29.141 lb
or PHG = 29.1 lb
PROBLEM 4.143
While being installed, the 56-lb chute ABCD is attached to a wall with
brackets E and F and is braced with props GH and IJ. Assuming that the
weight of the chute is uniformly distributed, determine the magnitude of
the force exerted on the chute by prop IJ if prop GH is removed.
SOLUTION
First note
 42 in. 
θ = tan −1 
 = 16.2602°
 144 in. 
xI = (100 in.) cos16.2602° = 96 in.
yI = 78 in. − (100 in.) sin16.2602° = 50 in.
λ BA =
− (144 in.) i + ( 42 in.) j
(144 )
2
+ ( 42 ) in.
2
=
1
( −24i + 7 j)
25
rK / A = ( 72 in.) i − ( 21 in.) j + ( 9 in.) k
rI / A = ( 96 in.) i − ( 78 in. − 50 in.) j + (18 in.) k = ( 96 in.) i − ( 28 in.) j + (18 in.) k
W = − ( 56 lb ) j
PJI = λ JI PJI
=
=
− (1 in.) i + ( 50 in.) j − (10 in.) k
(1)
2
+ ( 50 ) + (10 ) in.
2
PJI
( −i + 50 j − 10k )
51
2
PJI
PROBLEM 4.143 CONTINUED
From the f.b.d. of the chute
ΣM BA = 0: λ BA ⋅ ( rK / A × W ) + λ BA ⋅ ( rI / A × PJI ) = 0
−24
−24 7
7 0
0
 56 
 P 
72 −21 9   + 96 −28 18  JI  = 0
 25 
 51( 25 ) 
−1 50 −10
0 −1 0
−216 ( 56 )
P
+ [ −24 ( −28 )( −10 ) − ( −24 ) (18 )( 50 ) + 7 (18 ) ( −1) − ( 7 )( 96 )( −10 )] JI = 0
25
51( 25 )
∴ PJI = 28.728 lb
or PJI = 28.7 lb
PROBLEM 4.144
To water seedlings, a gardener joins three lengths of pipe, AB, BC,
and CD, fitted with spray nozzles and suspends the assembly using
hinged supports at A and D and cable EF. Knowing that the pipe
weighs 0.85 lb/ft, determine the tension in the cable.
SOLUTION
First note rG/ A = (1.5 ft ) i
WAB = − ( 0.85 lb/ft )( 3 ft ) j = − ( 2.55 lb ) j
rF / A = ( 2 ft ) i
T = λ FET =
− ( 2 ft ) i + ( 3 ft ) j − ( 4.5 ft ) k
( 2 )2 + ( 3)2 + ( 4.5)2
T
ft
 T 
=
 ( −2i + 3j − 4.5k )
 33.25 
rB/ A = ( 3 ft ) i
WBC = − ( 0.85 lb/ft )(1 ft ) j = − ( 0.85 lb ) j
rH / A = ( 3 ft ) i − ( 2.25 ft ) k
WCD = − ( 0.85 lb/ft )( 4.5 ft ) j = − ( 3.825 lb ) j
λ AD =
( 3 ft ) i − (1 ft ) j − ( 4.5 ft ) k
( 3)2 + (1)2 + ( 4.5)2 ft
=
1
( 3i − j − 4.5k )
5.5
PROBLEM 4.144 CONTINUED
From f.b.d. of the pipe assembly
ΣM AD = 0:
λ AD ⋅ ( rG/ A × WAB ) + λ AD ⋅ ( rF / A × T )
+ λ AD ⋅ ( rB/ A × WBC ) + λ AD ⋅ ( rH / A × WCD ) = 0
3
−1 −4.5
3 −1 −4.5
T
 1 


0
0 
0 
∴ 1.5
+ 2 0

 5.5 
 5.5 33.25 
0 −2.55 0
−2 3 −4.5
3 −1 −4.5
3
−1
−4.5
 1 
 1 
+ 3 0
+
−
0 
3
0
2.25 

=0
 5.5 
 5.5 
0 −0.85 0
0 −3.825
0

T 
 + (11.475 ) + ( 25.819 ) = 0
33.25


(17.2125) + ( −36 ) 
∴ T = 8.7306 lb
or T = 8.73 lb
PROBLEM 4.145
Solve Problem 4.144 assuming that cable EF is replaced by a cable
connecting E and C.
P4.144 To water seedlings, a gardener joins three lengths of pipe, AB,
BC, and CD, fitted with spray nozzles and suspends the assembly using
hinged supports at A and D and cable EF. Knowing that the pipe weighs
0.85 lb/ft, determine the tension in the cable.
SOLUTION
First note
rG/ A = (1.5 ft ) i
WAB = − ( 0.85 lb/ft )( 3 ft ) j = − ( 2.55 lb ) j
rC/ A = ( 3 ft ) i − (1 ft ) j
T = λ CET =
− ( 3 ft ) i + ( 4 ft ) j − ( 4.5 ft ) k
( 3)2 + ( 4 )2 + ( 4.5)2
T
ft
 T

=
 ( −3i + 4 j − 4.5k )
 45.25 
rB/ A = ( 3 ft ) i
WBC = − ( 0.85 lb/ft )(1 ft ) j = − ( 0.85 lb ) j
rH / A = ( 3 ft ) i − ( 2.25 ft ) k
WCD = − ( 0.85 lb/ft )(1 ft ) j = − ( 3.825 lb ) j
λ AD =
( 3 ft ) i − (1 ft ) j − ( 4.5 ft ) k
( 3)2 + (1)2 + ( 4.5)2 ft
=
1
( 3i − j − 4.5k )
5.5
PROBLEM 4.145 CONTINUED
From f.b.d. of the pipe assembly
ΣM AD = 0:
λ AD ⋅ ( rG/ A × WAB ) + λ AD ⋅ ( rC/ A × T )
+ λ AD ⋅ ( rB/ A × WBC ) + λ AD ⋅ ( rH / A × WCD ) = 0
−1 −4.5
3
3 −1 −4.5
T
 1 


∴ 1.5
0
0 
 + 3 −1 0 

 5.5 
 5.5 45.25 
0 −2.55 0
−3 4 −4.5
3 −1 −4.5
3
−1
−4.5
 1 
 1 
+3
−2.25 
0
0 
0
+ 3
=0
 5.5 
 5.5 
0 −0.85 0
0 −3.825
0

T

 + (11.475 ) + ( 25.819 ) = 0
 45.25 
(17.2125) + ( −40.5) 
∴ T = 9.0536 lb
or T = 9.05 lb
PROBLEM 4.146
The bent rod ABDE is supported by ball-and-socket joints at A and E and
by the cable DF. If a 600-N load is applied at C as shown, determine the
tension in the cable.
SOLUTION
First note
λ AE =
− ( 70 mm ) i + ( 240 mm ) k
( 70 )
2
=
+ ( 240 ) mm
2
1
( −7i + 24k )
25
rC/ A = ( 90 mm ) i + (100 mm ) k
FC = − ( 600 N ) j
rD/ A = ( 90 mm ) i + ( 240 mm ) k
T = λ DFT =
=
− (160 mm ) i + (110 mm ) j − ( 80 mm ) k
(160 )2 + (110 )2 + (80 )2
T
mm
T
( −16 i + 11j − 8k )
21
From the f.b.d. of the bend rod
(
)
(
)
ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rD/ A × T = 0
∴
−7 0 24
−7 0 24
 600 
 T 
90 0 100 
=0
 + 90 0 240 
25 ( 21) 
 25 

0 −1 0
−16 11 −8
( −700 − 2160 ) 
600 
 T 
 + (18 480 + 23 760 ) 
 =0
25


 25 ( 21) 
∴ T = 853.13 N
or T = 853 N
PROBLEM 4.147
Solve Problem 4.146 assuming that cable DF is replaced by a cable
connecting B and F.
P4.146 The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 600-N load is applied at C as shown,
determine the tension in the cable.
SOLUTION
First note
λ AE =
− ( 70 mm ) i + ( 240 mm ) k
( 70 )
2
=
+ ( 240 ) mm
2
1
( −7i + 24k )
25
rC/ A = ( 90 mm ) i + (100 mm ) k
FC = − ( 600 N ) j
rB/ A = ( 90 mm ) i
T = λ BFT =
=
− (160 mm ) i + (110 mm ) j + (160 mm ) k
(160 )2 + (110 )2 + (160 )2
T
mm
1
( −160 i + 110 j + 160k )
251.59
From the f.b.d. of the bend rod
(
)
(
)
ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rB/ A × T = 0
−7 0 24
−7
0
24
T
 600 


∴ 90 0 100 
0
0 
=0
 + 90
25 ( 251.59 ) 
 25 

0 −1 0
−160 110 160
( −700 − 2160 ) 
T
600 


=0
 + ( 237 600 ) 

25
251.59
 25 
)
 (
∴ T = 1817.04 N
or T = 1817 N
PROBLEM 4.148
Two rectangular plates are welded together to form the assembly shown.
The assembly is supported by ball-and-socket joints at B and D and by a
ball on a horizontal surface at C. For the loading shown, determine the
reaction at C.
SOLUTION
λ BD =
First note
=
− ( 80 mm ) i − ( 90 mm ) j + (120 mm ) k
(80 )2 + ( 90 )2 + (120 )2
mm
1
( −8i − 9 j + 12k )
17
rA/B = − ( 60 mm ) i
P = ( 200 N ) k
rC/D = ( 80 mm ) i
C = (C ) j
From the f.b.d. of the plates
(
)
(
)
ΣM BD = 0: λ BD ⋅ rA/B × P + λ BD ⋅ rC/D × C = 0
∴
−8 −9 12
−8 −9 12
 60 ( 200 ) 
 C ( 80 ) 
1 0 0 
−1 0 0 
+

 =0
 17 
 17 
0 0 1
0 1 0
( −9 )( 60 )( 200 ) + (12 )(80 ) C = 0
∴ C = 112.5 N
or C = (112.5 N ) j
PROBLEM 4.149
Two 1 × 2-m plywood panels, each of mass 15 kg, are nailed together as
shown. The panels are supported by ball-and-socket joints at A and F and
by the wire BH. Determine (a) the location of H in the xy plane if the
tension in the wire is to be minimum, (b) the corresponding minimum
tension.
SOLUTION
(
)
W1 = W2 = − ( mg ) j = − (15 kg ) 9.81 m/s 2 j
Let
= − (147.15 N ) j
From the f.b.d. of the panels
(
)
(
)
(
=
1
( 2i − j − 2k )
3
)
ΣM AF = 0: λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rT / A × W2 = 0
where
λ AF =
( 2 m ) i − (1 m ) j − ( 2 m ) k
( 2 )2 + (1)2 + ( 2 )2 m
rG/ A = (1 m ) i
rB/ A = ( 2 m ) i
rI / A = ( 2 m ) i − (1 m ) k
PROBLEM 4.149 CONTINUED
λ BH =
( x − 2) i + ( y ) j − ( 2) k
( x − 2 ) 2 + y 2 + ( 2 )2
T = λ BH T =
∴
( x − 2) i + ( y ) j − ( 2) k
( x − 2 ) 2 + y 2 + ( 2 )2
2 −1 −2
2
−1 −2
T
 147.15 

1 0 0 
0 0 
+ 2
2
2
2
 3 

0 −1 0
x − 2 y −2  3 ( x − 2 ) + y + ( 2 )
2 −1 −2

 147.15 
 + 2 0 −1  3  = 0

 0 −1 0
2 (147.15 )
147.15
T
+ ( −4 − 4 y )
+ ( −2 + 4 )
=0
2
2
3
3
3 ( x − 2) + y 2 + ( 2)
T =
or
For x − 2 m, T = Tmin
147.15
( x − 2 ) 2 + y 2 + ( 2 )2
1+ y
∴ Tmin =
1
147.15 2
( y + 4) 2
(1 + y )
1
(1 + y ) ( y 2 + 4 ) 2 ( 2 y ) − ( y 2 + 4 ) 2 (1)
−1
The y-value for Tmin is found from
 dT 
 dy  = 0:


Setting the numerator equal to zero,
(1 + y ) y
2
1
(1 + y )2
=0
= y2 + 4
y = 4m
Then
T min =
147.15 (
2
2
2
2 − 2 ) + ( 4 ) + ( 2 ) = 131.615 N
(1 + 4 )
∴ (a)
x = 2.00 m, y = 4.00 m
(b)
Tmin = 131.6 N
PROBLEM 4.150
Solve Problem 4.149 subject to the restriction that H must lie on the
y axis.
P4.149 Two 1 × 2-m plywood panels, each of mass 15 kg, are nailed
together as shown. The panels are supported by ball-and-socket joints at
A and F and by the wire BH. Determine (a) the location of H in the xy
plane if the tension in the wire is to be minimum, (b) the corresponding
minimum tension.
SOLUTION
(
)
W1 = W2 = − ( mg ) j = − (15 kg ) 9.81 m/s 2 j = − (147.15 N ) j
Let
From the f.b.d. of the panels
(
)
(
)
(
=
1
( 2i − j − 2k )
3
)
ΣM AF = 0: λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0
where
λ AF =
( 2 m ) i − (1 m ) j − ( 2 m ) k
( 2 )2 + (1)2 + ( 2 )2 m
rG/ A = (1 m ) i
rB/ A = ( 2 m ) i
rI / A = ( 2 m ) i − (1 m ) k
T = λ BH T =
=
− (2 m) i + ( y) j − (2 m) k
( 2 )2 + ( y )2 + ( 2 )2
T
8 + y2
( −2i +
m
yj − 2k )
T
PROBLEM 4.150 CONTINUED
∴
2 −1 −2
2 −1 −2
T
 147.15 

1 0 0
+ 2 0 0 
2

 3 
−2 y −2  3 8 + y
0 −1 0
2 −1 −2

 147.15 
 + 2 0 −1  3  = 0
 0 −1 0
)
(
2 (147.15 ) + ( −4 − 4 y ) T 8 + y 2 + ( 2 )147.15 = 0
∴ T =
For Tmin ,
 dT 

=0
 dy 
∴
(147.15) 8 +
(1 + y )
(1 + y ) 12 (8 +
y2
) ( 2 y ) − (8 +
(1 + y )2
y2
− 12
y2
)
1
2
(1)
=0
Setting the numerator equal to zero,
(1 + y ) y
= 8 + y2
∴ y = 8.00 m
and
Tmin =
(147.15) 8 + (8)2
(1 + 8)
= 138.734 N
∴ (a)
x = 0, y = 8.00 m
(b)
Tmin = 138.7 N
PROBLEM 4.151
A uniform 20 × 30-in. steel plate ABCD weighs 85 lb and is attached to
ball-and-socket joints at A and B. Knowing that the plate leans against a
frictionless vertical wall at D, determine (a) the location of D, (b) the
reaction at D.
SOLUTION
(a) Since rD/ A is perpendicular to rB/ A ,
rD/ A ⋅ rB/ A = 0
where coordinates of D are ( 0, y, z ) , and
rD/ A = − ( 4 in.) i + ( y ) j + ( z − 28 in.) k
rB/ A = (12 in.) i − (16 in.) k
∴ rD/ A ⋅ rB/ A = −48 − 16 z + 448 = 0
or
z = 25 in.
Since
LAD = 30 in.
30 =
( 4 )2 + ( y )2 + ( 25 − 28)2
900 = 16 + y 2 + 9
y =
or
875 in. = 29.580 in.
∴ Coordinates of D :
x = 0, y = 29.6 in., z = 25.0 in.
(b) From f.b.d. of steel plate ABCD
ΣM AB = 0:
where
λ AB =
λ AB ⋅ ( rD/ A × N D ) + λ AB ⋅ ( rG/B × W ) = 0
(12 in.) i − (16 in.) k
(12 )2 + (16 )2 in.
=
1
( 3i − 4k )
5
rD/ A = − ( 4 in.) i + ( 29.580 in.) j − ( 3 in.) k
N D = N Di
PROBLEM 4.151 CONTINUED
rG/B =
1
1
rD/B =  − (16 in.) i + ( 29.580 in.) j + ( 25 in. − 12 in.) k 
2
2
W = − ( 85 lb ) j
3
0
3
0
−4
−4
 ND 
 85 
∴ −4 29.580 −3 
=0
 + −16 29.580 13 
2 ( 5 ) 
 5 

1
0
0
0
0
−1
118.32 N D + ( 39 − 64 ) 42.5 = 0
∴ N D = 8.9799 lb
or N D = ( 8.98 lb ) i
PROBLEM 4.152
Beam AD carries the two 40-lb loads shown. The beam is held by a fixed
support at D and by the cable BE which is attached to the counter-weight
W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb.
SOLUTION
(a) W = 100 lb
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: Dy − 40 lb − 40 lb + 100 lb = 0
∴ Dy = −20.0 lb
or D = 20.0 lb
M D − (100 lb )( 5 ft ) + ( 40 lb )( 8 ft )
ΣM D = 0:
+ ( 40 lb )( 4 ft ) = 0
∴ M D = 20.0 lb ⋅ ft
or M D = 20.0 lb ⋅ ft
(b) W = 90 lb
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: Dy + 90 lb − 40 lb − 40 lb = 0
∴ Dy = −10.00 lb
or D = 10.00 lb
ΣM D = 0:
M D − ( 90 lb )( 5 ft ) + ( 40 lb )( 8 ft )
+ ( 40 lb )( 4 ft ) = 0
∴ M D = −30.0 lb ⋅ ft
or M D = 30.0 lb ⋅ ft
PROBLEM 4.153
For the beam and loading shown, determine the range of values of W for
which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.
SOLUTION
For Wmin ,
M D = −40 lb ⋅ ft
From f.b.d. of beam AD
ΣM D = 0:
( 40 lb )(8 ft ) − Wmin ( 5 ft ) + ( 40 lb )( 4 ft ) − 40 lb ⋅ ft
=0
∴ Wmin = 88.0 lb
For Wmax ,
M D = 40 lb ⋅ ft
From f.b.d. of beam AD
ΣM D = 0:
( 40 lb )(8 ft ) − Wmax ( 5 ft ) + ( 40 lb )( 4 ft ) + 40 lb ⋅ ft
=0
∴ Wmax = 104.0 lb
or 88.0 lb ≤ W ≤ 104.0 lb
PROBLEM 4.154
Determine the reactions at A and D when β = 30°.
SOLUTION
From f.b.d. of frame ABCD
ΣM D = 0: − A ( 0.18 m ) + (150 N ) sin 30°  ( 0.10 m )
+ (150 N ) cos 30° ( 0.28 m ) = 0
∴ A = 243.74 N
or A = 244 N
ΣFx = 0:
( 243.74 N ) + (150 N ) sin 30° + Dx
=0
∴ Dx = −318.74 N
ΣFy = 0: Dy − (150 N ) cos 30° = 0
∴ Dy = 129.904 N
Then
and
D=
( Dx )2 + Dx2
=
( 318.74 )2 + (129.904 )2
= 344.19 N
 Dy 
−1  129.904 
 = tan 
 = −22.174°
D
 −318.74 
 x
θ = tan −1 
or D = 344 N
22.2°
PROBLEM 4.155
Determine the reactions at A and D when β = 60°.
SOLUTION
From f.b.d. of frame ABCD
ΣM D = 0: − A ( 0.18 m ) + (150 N ) sin 60° ( 0.10 m )
+ (150 N ) cos 60° ( 0.28 m ) = 0
∴ A = 188.835 N
or A = 188.8 N
(188.835 N ) + (150 N ) sin 60° + Dx
ΣFx = 0:
=0
∴ Dx = −318.74 N
ΣFy = 0: Dy − (150 N ) cos 60° = 0
∴ Dy = 75.0 N
Then
and
D=
( Dx )2 + ( Dy )
2
=
( 318.74 )2 + ( 75.0 )2
= 327.44 N
 Dy 
−1  75.0 
 = tan 
 = −13.2409°
 −318.74 
 Dx 
θ = tan −1 
or D = 327 N
13.24°
PROBLEM 4.156
A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at
each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
(a) From f.b.d. of tractor
ΣM B = 0:
( 2100 lb )( 40 in.) − ( 2 A)( 60 in.) − ( 900 lb )( 50 in.) = 0
∴ A = 325 lb
or A = 325 lb
(b) From f.b.d. of tractor
ΣM A = 0:
( 2B )( 60 in.) − ( 2100 lb )( 20 in.) − ( 900 lb )(110 in.) = 0
∴ B = 1175 lb
or B = 1175 lb
PROBLEM 4.157
A tension of 5 lb is maintained in a tape as it passes the support system
shown. Knowing that the radius of each pulley is 0.4 in., determine the
reaction at C.
SOLUTION
From f.b.d. of system
ΣFx = 0: C x + ( 5 lb ) = 0
∴ Cx = −5 lb
ΣFy = 0: C y − ( 5 lb ) = 0
∴ C y = 5 lb
Then
and
C =
( Cx )2 + ( C y )
2
=
( 5 )2 + ( 5 )2
= 7.0711 lb
 +5 
θ = tan −1   = −45°
 −5 
or C = 7.07 lb
45.0°
ΣM C = 0: M C + ( 5 lb )( 6.4 in.) + ( 5 lb )( 2.2 in.) = 0
∴ M C = −43.0 lb ⋅ in
or M C = 43.0 lb ⋅ in.
PROBLEM 4.158
Solve Problem 4.157 assuming that 0.6-in.-radius pulleys are used.
P4.157 A tension of 5 lb is maintained in a tape as it passes the support
system shown. Knowing that the radius of each pulley is 0.4 in., determine
the reaction at C.
SOLUTION
From f.b.d of system
ΣFx = 0: C x + ( 5 lb ) = 0
∴ Cx = −5 lb
ΣFy = 0: C y − ( 5 lb ) = 0
∴ C y = 5 lb
Then
and
C =
( C x )2 + ( C y )
2
=
( 5 )2 + ( 5 )2
= 7.0711 lb
 5 
θ = tan −1   = −45.0°
 −5 
or C = 7.07 lb
45.0°
ΣM C = 0: M C + ( 5 lb )( 6.6 in.) + ( 5 lb )( 2.4 in.) = 0
∴ M C = −45.0 lb ⋅ in.
or M C = 45.0 lb ⋅ in.
PROBLEM 4.159
The bent rod ABEF is supported by bearings at C and D and by wire AH.
Knowing that portion AB of the rod is 250 mm long, determine (a) the
tension in wire AH, (b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.
SOLUTION
(a) From f.b.d. of bent rod
(
)
(
)
ΣM CD = 0: λ CD ⋅ rH /B × T + λ CD ⋅ rF /E × F = 0
where
λ CD = i
rH /B = ( 0.25 m ) j
T = λ AH T
=
( y AH ) j − ( z AH ) k T
( y AH )2 + ( z AH )2
y AH = ( 0.25 m ) − ( 0.25 m ) sin 30°
= 0.125 m
z AH = ( 0.25 m ) cos 30°
= 0.21651 m
∴T=
T
( 0.125j − 0.21651k )
0.25
rF /E = ( 0.25 m ) k
F = −400 N j
1
0
0
1 0 0
 T 
1
0
0.25
∴ 0
+
( )
 0 0 1 ( 0.25 )( 400 N ) = 0
 0.25 
0 0.125 −0.21651
0 −1 0
−0.21651T + 0.25 ( 400 N ) = 0
∴ T = 461.88 N
or T = 462 N
PROBLEM 4.159 CONTINUED
(b) From f.b.d. of bent rod
ΣFx = 0: C x = 0
ΣM D( z -axis ) = 0: − ( 461.88 N ) sin 30°  ( 0.35 m ) − C y ( 0.3 m )
− ( 400 N )( 0.05 m ) = 0
∴ C y = −336.10 N
ΣM D( y -axis ) = 0: Cz ( 0.3 m ) − ( 461.88 N ) cos 30°  ( 0.35 m ) = 0
∴ Cz = 466.67 N
or C = − ( 336 N ) j + ( 467 N ) k
ΣFy = 0: Dy − 336.10 N + ( 461.88 N ) sin 30° − 400 N = 0
∴ Dy = 505.16 N
ΣFz = 0: Dz + 466.67 N − ( 461.88 N ) cos30° = 0
∴ Dz = −66.670 N
or D = ( 505 N ) j − ( 66.7 N ) k
PROBLEM 4.160
For the beam and loading shown, determine (a) the reaction at A, (b) the
tension in cable BC.
SOLUTION
(a) From f.b.d of beam
ΣFx = 0: Ax = 0
ΣM B = 0:
(15 lb )( 28 in.) + ( 20 lb )( 22 in.) + ( 35 lb )(14 in.)
+ ( 20 lb )( 6 in.) − Ay ( 6 in.) = 0
∴ Ay = 245 lb
or A = 245 lb
(b) From f.b.d of beam
ΣM A = 0:
(15 lb )( 22 in.) + ( 20 lb )(16 in.) + ( 35 lb )(8 in.)
− (15 lb )( 6 in.) − TB ( 6 in.) = 0
∴ TB = 140.0 lb
or TB = 140.0 lb
Check:
ΣFy = 0: −15 lb − 20 lb − 35 lb − 20 lb
− 15 lb − 140 lb + 245 lb = 0?
245 lb − 245 lb = 0 ok
PROBLEM 4.161
Frame ABCD is supported by a ball-and-socket joint at A and by three
cables. For a = 150 mm, determine the tension in each cable and the
reaction at A.
SOLUTION
First note
TDG = λ DGTDG =
− ( 0.48 m ) i + ( 0.14 m ) j
( 0.48)2 + ( 0.14 )2
=
−0.48i + 0.14 j
TDG
0.50
=
TDG
( 24i + 7 j)
25
TBE = λ BETBE =
− ( 0.48 m ) i + ( 0.2 m ) k
( 0.48)2 + ( 0.2 )2 m
=
−0.48i + 0.2k
TBE
0.52
=
TBE
( −12 j + 5k )
13
TDG
m
TBE
From f.b.d. of frame ABCD
 7

ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.15 m ) = 0
 25

or TDG = 625 N
 24

 5

ΣM y = 0: 
× 625 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0
13
 25



or TBE = 975 N
 7

ΣM z = 0: TCF ( 0.14 m ) + 
× 625 N  ( 0.48 m )
 25

− ( 350 N )( 0.48 m ) = 0
or TCF = 600 N
PROBLEM 4.161 CONTINUED
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
 12
  24

Ax − 600 N −  × 975 N  − 
× 625 N  = 0
 13
  25

∴ Ax = 2100 N
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
 7

Ay + 
× 625 N  − 350 N = 0
25


∴ Ay = 175.0 N
ΣFz = 0: Az + (TBE ) z = 0
5

Az +  × 975 N  = 0
 13

∴ Az = −375 N
Therefore
A = ( 2100 N ) i + (175.0 N ) j − ( 375 N ) k
PROBLEM 4.162
Frame ABCD is supported by a ball-and-socket joint at A and by three
cables. Knowing that the 350-N load is applied at D (a = 300 mm),
determine the tension in each cable and the reaction at A.
SOLUTION
First note
TDG = λ DGTDG =
− ( 0.48 m ) i + ( 0.14 m ) j
( 0.48)2 + ( 0.14 )2 m
=
−0.48i + 0.14 j
TDG
0.50
=
TDG
( 24i + 7 j)
25
TBE = λ BETBE =
− ( 0.48 m ) i + ( 0.2 m ) k
( 0.48)2 + ( 0.2 )2 m
=
−0.48i + 0.2k
TBE
0.52
=
TBE
( −12i + 5k )
13
TDG
TBE
From f.b.d of frame ABCD
 7

ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.3 m ) = 0
 25

or TDG = 1250 N
 24

 5

ΣM y = 0: 
× 1250 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0
13
 25



or TBE = 1950 N
 7

ΣM z = 0: TCF ( 0.14 m ) + 
× 1250 N  ( 0.48 m )
 25

− ( 350 N )( 0.48 m ) = 0
or TCF = 0
PROBLEM 4.162 CONTINUED
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
 12
  24

Ax + 0 −  × 1950 N  − 
× 1250 N  = 0
 13
  25

∴ Ax = 3000 N
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
 7

Ay + 
× 1250 N  − 350 N = 0
25


∴ Ay = 0
ΣFz = 0: Az + (TBE ) z = 0
5

Az +  × 1950 N  = 0
 13

∴ Az = −750 N
Therefore
A = ( 3000 N ) i − ( 750 N ) k
PROBLEM 4.163
In the problems listed below, the rigid bodies considered were completely
constrained and the reactions were statically determinate. For each of
these rigid bodies it is possible to create an improper set of constraints by
changing a dimension of the body. In each of the following problems
determine the value of a which results in improper constraints.
(a) Problem 4.81, (b) Problem 4.82.
SOLUTION
(a)
ΣM B = 0:
(a)
( 300 lb )(16 in.) − T (16 in.) + T ( a ) = 0
T =
or
( 300 lb )(16 in.)
(16 − a ) in.
∴ T becomes infinite when
16 − a = 0
or a = 16.00 in.
ΣM C = 0:
(b)
(b)
(T
 8 
− 80 N )( 0.2 m ) −  T  ( 0.175 m )
 17 
 15 
−  T  ( 0.4 m − a ) = 0
 17 
0.2T − 16.0 − 0.82353T − 0.35294T + 0.88235Ta = 0
or
T =
16.0
0.88235a − 0.23529
∴ T becomes infinite when
0.88235a − 0.23529 = 0
a = 0.26666 m
or a = 267 mm
PROBLEM 5.1
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
8 × 6 = 48
−4
9
−192
432
2
16 × 12 = 192
8
6
1536
1152
Σ
240
1344
1584
Then
X =
Σ xA 1344 in 3
=
ΣA
240 in 2
or X = 5.60 in. W
and
Y =
Σ yA 1584 in 3
=
ΣA
240 in 2
or Y = 6.60 in. W
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
1
× 60 × 75 = 2250
2
40
25
90 000
56 250
2
105 × 75 = 7875
112.5
37.5
885 900
295 300
Σ
10 125
975 900
351 600
Then
X =
ΣxA 975 900 mm3
=
ΣA
10 125 mm 2
or X = 96.4 mm W
and
Y =
Σ yA 351 600 mm3
=
ΣA
10 125 mm 2
or Y = 34.7 mm W
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
For the area as a whole, it can be concluded by observation that
Y =
Then
2
( 24 in.)
3
or Y = 16.00 in. W
A, in 2
x , in.
xA, in 3
1
1
× 24 × 10 = 120
2
2
(10 ) = 6.667
3
800
2
1
× 24 × 16 = 192
2
Σ
312
X =
10 +
1
(16 ) = 15.333
3
2944
3744
Σ xA 3744 in 3
=
ΣA
312 in 2
or X = 12.00 in. W
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
1
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
21 × 22 = 462
1.5
11
693
5082
2
−
1
( 6 )( 9 ) = −27
2
−6
2
162
−54
3
−
1
( 6 )(12 ) = −36
2
8
2
−288
−72
567
4956
Σ
399
Then
X =
Σ xA 567 mm3
=
ΣA
399 mm 2
or X = 1.421 mm W
and
Y =
Σ yA 4956 mm3
=
ΣA
399 mm 2
or Y = 12.42 mm W
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
120 × 200 = 24 000
60
120
1 440 000
2 880 000
2
−
94.5
120
−534 600
−678 600
905 400
2 201 400
Σ
π ( 60 )
2
2
= −5654.9
18 345
Then
X =
Σ xA 905 400 mm3
=
ΣA
18 345 mm 2
or X = 49.4 mm W
and
Y =
Σ yA 2 201 400 mm3
=
ΣA
18 345 mm 2
or Y = 93.8 mm W
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
1
π (9)
4
2
= 63.617
2
1
(15)( 9 ) = 67.5
2
Σ
131.1
x , in.
y , in.
x A, in 3
y A, in 3
−4 ( 9 )
= −3.8917
( 3π )
3.8917
−243
243
5
3
337.5
202.5
94.5
445.5
Then
X =
Σ xA
94.5 in 3
=
ΣA
131.1 in 2
or X = 0.721 in. W
and
Y =
Σ yA 445.5 in 3
=
ΣA
131.1 in 2
or Y = 3.40 in. W
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
First note that symmetry implies X = Y
A, mm 2
40 × 40 = 1600
1
−
2
Σ
Then
π (40) 2
4
= −1257
343
X =
Σ xA 10 667 mm3
=
ΣA
343 mm 2
x , mm
xA, mm3
20
32 000
16.98
−21 330
10 667
or X = 31.1 mm W
and Y = X = 31.1 mm W
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
X =0 W
First note that symmetry implies
y , in.
yA, in 3
= −25.13
1.6977
−42.67
= 56.55
2.546
A, in 2
1
2
−
π ( 4)
2
π ( 6)
2
Σ
Then
2
2
31.42
Y =
Σ yA 101.33 in 3
=
ΣA
31.42 in 2
144
101.33
or Y = 3.23 in. W
PROBLEM 5.9
For the area of Problem 5.8, determine the ratio r2 /r1 so that y = 3r1/4.
SOLUTION
A
y
yA
π
4r1
3π
2
− r13
3
4r2
3π
2 3
r2
3
− r12
2
1
π
2
Σ
Then
or
2
π
(r
2
2
2
r22
(
)
Y ΣA = Σy A
3
π 2
2 3
r1 ×
r2 − r12 =
r2 − r13
4
2
3
(
)
(
)
2
  r 3
9π  r2 
  − 1 =  2  − 1
16  r1 
  r1 


Let
p=
r2
r1
9π
( p + 1)( p − 1)] = ( p − 1)( p 2 + p + 1)
[
16
or
)
2 3
r2 − r13
3
− r12
16 p 2 + (16 − 9π) p + (16 − 9π) = 0
PROBLEM 5.9 CONTINUED
Then
or
Taking the positive root
p=
−(16 − 9π) ± (16 − 9π) 2 − 4(16)(16 − 9π)
2(16)
p = −0.5726
p = 1.3397
r2
= 1.340 W
r1
PROBLEM 5.10
Show that as r1 approaches r2 , the location of the centroid approaches that
of a circular arc of radius ( r1 + r2 ) / 2.
SOLUTION
First, determine the location of the centroid.
y2 =
From Fig. 5.8A:
=
y1 =
Similarly
Then
(
)
2
cos α
r2 π
3
−α
2
(
2
cos α
r1
3 π2 − α
(
)
A2 =
)
A1 =
)
(
)
(
)
)
( π2 − α ) r12
(
(
Now
( π2 − α ) r22
2
cosα  π
2
cosα  π
r2 π
− α r22  − r1 π
− α r12 
2


 2

3
3
α
α
−
−
2
2
2 3
r2 − r13 cosα
=
3
π

π

Σ A =  − α  r22 −  − α  r12
2

2

π
 2
2
=  − α  r2 − r1
2

Y Σ A = Σ yA
 π
 2 3

Y  − α  r22 − r12  =
r2 − r13 cos α

 2
 3
2 r23 − r13 cos α
Y =
3 r22 − r12 π2 − α
Σ yA =
(
and
(π
2 sin 2 − α
r2 π
3
−α
2
(
)
)
)
(
)
(
)
PROBLEM 5.10 CONTINUED
Using Figure 5.8B, Y of an arc of radius
1
( r1 + r2 ) is
2
Y =
=
Now
Let
Then
(π
sin − α
1
( r1 + r2 ) π 2
2
−α
2
(
)
)
1
cos α
(r1 + r2 ) π
2
−α
2
(
(
)
( r2 − r1 ) r22 + r1 r2 + r12
r23 − r13
=
r22 − r12
( r2 − r1 )( r2 + r1 )
2
r + r1 r2 + r12
= 2
r2 + r1
(1)
)
r2 = r + ∆
r1 = r − ∆
1
r = ( r1 + r2 )
2
( r + ∆ ) + ( r + ∆ )( r − ∆ ) + ( r − ∆ )
r23 − r13
and
=
2
2
r2 − r1
(r + ∆) + (r − ∆)
3r 2 + ∆ 2
=
2r
In the limit as ∆ → 0 (i.e., r1 = r2 ), then
2
so that
Which agrees with Eq. (1).
r23 − r13
3
= r
2
2
2
r2 − r1
3 1
= × (r1 + r2 )
2 2
2 3
cos α
Y = × ( r1 + r2 ) π
−α
3 4
2
2
or Y =
1
cos α
W
( r1 + r2 ) π
−α
2
2
PROBLEM 5.11
Locate the centroid of the plane area shown.
SOLUTION
X =0 W
First note that symmetry implies
r2 = 2 2 in., α = 45°
y2′ =
1
2
3
Σ
( ) ( ) = 1.6977 in.
2r sin α 2 2 2 sin
=
3α
3 π4
( )
−
4
A, in 2
y , in.
y A, in 3
1
( 4)(3) = 6
2
1
6
2 − y ′ = 0.3024
1.8997
0.6667
−2.667
(2 2 )
4
π
π
2
= 6.283
1
( 4)( 2) = −4
2
8.283
5.2330
Y Σ A = Σ yA
Then
(
)
Y 8.283 in 2 = 5.2330 in 3
or Y = 0.632 in. W
PROBLEM 5.12
Locate the centroid of the plane area shown.
SOLUTION
1
2
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
(40)(90) = 3600
−15
20
−54 000
72 000
10
−15
6750
−10 125
−25.47
−19.099
−54 000
−40 500
−101 250
21 375
π ( 40)( 60)
4
= 2121
3
1
(30)( 45) = 675
2
Σ
6396
XA = Σ xA
Then
(
)
X 6396 mm 2 = −101 250 mm3
or X = −15.83 mm W
YA = Σ yA
and
(
)
Y 6396 mm 2 = 21 375 mm3
or Y = 3.34 mm W
PROBLEM 5.13
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
−
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
2
( 40)(80) = 2133
3
48
15
102 400
32 000
53.33
13.333
−85 330
−21 330
17 067
10 667
1
( 40)(80) = −1600
2
533.3
X Σ A = Σ XA
Then
(
)
X 533.3 mm 2 = 17 067 mm3
or X = 32.0 mm W
Y Σ A = Σ yA
and
(
)
Y 533.3 mm 2 = 10 667 mm3
or Y = 20.0 mm W
PROBLEM 5.14
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
2
(150 )( 240 ) = 24 000
3
56.25
96
1 350 000
2 304 000
2
−
50
40
−450 000
−360 000
900 000
1 944 000
Σ
1
(150)(120) = −9000
2
15 000
X Σ A = Σ xA
Then
(
)
X 15 000 mm 2 = 900 000 mm3
or X = 60.0 mm W
Y Σ A = Σ yA
and
(
)
Y 15 000 mm 2 = 1 944 000
or Y = 129.6 mm W
PROBLEM 5.15
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
(10)(15) = 50
3
4.5
7.5
225
375
6.366
16.366
1125
2892
1350
3267
π
4
(15)2
= 176.71
226.71
X Σ A = Σx A
Then
(
)
X 226.71 in 2 = 1350 in 3
or X = 5.95 in. W
Y ΣA = Σy A
and
(
)
Y 226.71 in 2 = 3267 in 3
or Y = 14.41 in. W
PROBLEM 5.16
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
−
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
2
(8)(8) = 42.67
3
3
2.8
128
119.47
1.5
−0.8
−8
4.267
120
123.73
2
( 4)( 2) = −5.333
3
37.33
X Σ A = Σx A
Then
(
)
X 37.33 in 2 = 120 in 3
or X = 3.21 in. W
Y ΣA = Σy A
and
(
)
Y 37.33 in 2 = 123.73 in 3
or Y = 3.31 in. W
PROBLEM 5.17
The horizontal x axis is drawn through the centroid C of the area shown
and divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x axis, and
explain the results obtained.
SOLUTION
Qx = Σ yA
Note that
Then
and
(Qx )1 =  53
( Qx )2
1

m  × 6 × 5 m 2
2

or ( Qx )1 = 25.0 × 103 mm3 W
 2
 1

 1
 1

=  − × 2.5 m  × 9 × 2.5  m 2 +  − × 2.5 m  × 6 × 2.5  m 2
 3
 2

 3
 2

or ( Qx ) 2 = −25.0 × 103 mm3 W
Now
Qx = ( Qx )1 + ( Qx ) 2 = 0
This result is expected since x is a centroidal axis ( thus y = 0 )
and
Qx = Σ y A = Y Σ A
(y
= 0 ⇒ Qx = 0 )
PROBLEM 5.18
The horizontal x axis is drawn through the centroid C of the area shown
and divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x axis, and
explain the results obtained.
SOLUTION
First, locate the position y of the figure.
A, mm 2
y , mm
yA, mm3
1
160 × 300 = 48 000
150
7 200 000
2
−150 × 80 = −16 000
160
−2 560 000
Σ
32 000
Y ΣA = Σy A
Then
(
)
Y 32 000 mm 2 = 4 640 000 mm3
or
Y = 145.0 mm
4 640 000
PROBLEM 5.18 CONTINUED
A I: Q I = Σ yA
155
115
 − ( 80 × 115) 
(160 × 155) +
=
2
2 
6
3
= 1.393 × 10 mm
W
A II : Q II = Σ yA
145
=−
(160 × 145) −
2
= −1.393 × 106 mm3
∴
( Qarea ) x
 85

 − 2 ( 80 × 85 ) 


W
= QI + QII = 0
Which is expected since Qx = Σ yA = yA and y = 0 ,
since x is a centroidal axis.
PROBLEM 5.19
The first moment of the shaded area with respect to the x axis is denoted
by Qx . (a) Express Qx in terms of r and θ . (b) For what value of θ is
Qx maximum, and what is the maximum value?
SOLUTION
(a) With Qx = Σ yA and using Fig. 5.8 A,
(
)
 2 r sin π − θ 
  r 2 π2 − θ  −
Qx =  3 π 2

−
θ


2


2
= r 3 cos θ − cos θ sin 2 θ
3
(
) ( 32 r sin θ )  12 × 2r cos θ × r sin θ 
(
)
or Qx =
(b) By observation, Qx is maximum when
and then
2 3
r cos3 θ W
3
θ =0W
Qx =
2 3
r W
3
PROBLEM 5.20
A composite beam is constructed by bolting four plates to four
2 × 2 × 3/8-in. angles as shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load. As proved in mechanics of
materials, the shearing forces exerted on the bolts at A and B are
proportional to the first moments with respect to the centroidal x axis of
the red shaded areas shown, respectively, in parts a and b of the figure.
Knowing that the force exerted on the bolt at A is 70 lb, determine the
force exerted on the bolt at B.
SOLUTION
From the problem statement: F ∝ Qx
FA
FB
so that
=
(Qx ) A (Qx ) B
FB =
and
Qx = ∑ yA
Now
So
and
(Qx ) B
F
(Qx ) A A
( Qx ) A
( Qx )B
0.375


=  7.5 in. +
in.  10 in. × ( 0.375 in.)  = 28.82 in 3
2


0.375


= ( Qx ) A + 2  7.5 in. −
in.  (1.625 in.)( 0.375 in.) 
2



+ 2 ( 7.5 in. − 1 in.) ( 2 in.)( 0.375 in.) 
= 28.82 in 3 + 8.921 in 3 + 9.75 in 3
= 47.49 in 3
Then
FB =
47.49 in 3
( 70 lb) = 115.3 lb W
28.82 in 3
PROBLEM 5.21
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
First note that because the wire is homogeneous, its center of gravity will
coincide with the centroid of the corresponding line.
Then
L, in.
x , in.
y , in.
xL, in 2
yL, in 2
1
16
8
0
128
0
2
12
16
6
102
72
3
24
4
12
96
288
4
6
−8
9
−48
54
5
8
−4
6
−32
48
6
6
0
3
0
18
Σ
72
336
480
X ΣL = Σ x L
X ( 72 in.) = 336 in 2
and
or X = 4.67 in.
Y ΣL = Σ y L
Y (72 in.) = 480 in 2
or Y = 6.67 in.
PROBLEM 5.22
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
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
165
82.5
0
13 612
0
2
75
165
37.5
12 375
2812
3
105
112.5
75
11 812
7875
30
37.5
2881
3602
40 680
14 289
4
Σ
Then
602 + 752 = 96.05
441.05
X ΣL = Σx L
X (441.05 mm) = 40 680 mm 2
and
or X = 92.2 mm
Y ΣL = Σ y L
Y (441.05 mm) = 14 289 mm 2
Y = 32.4 mm
PROBLEM 5.23
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
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
L, mm
1
y , mm
xL, mm 2
yL, mm 2
6
3
80.50
40.25
2
16
12
14
192
224
3
21
1.5
22
31.50
462
4
16
−9
14
−144
224
−4.5
3
−48.67
32.45
111.32
982.7
5
Σ
Then
122 + 62 = 13.416
x , mm
62 + 92 = 10.817
77.233
X ΣL = Σx L
X (77.233 mm) = 111.32 mm 2
and
or X = 1.441 mm
Y ΣL = Σ y L
Y (77.233 mm) = 982.7 mm 2
or Y = 12.72 mm
PROBLEM 5.24
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
First note that because the wire is homogeneous, its center of gravity will
coincide with the centroid of the corresponding line.
X =0
By symmetry
Then
L, in.
y , in.
yL, in 2
1
2
0
0
2
π ( 6)
= 3.820
72
3
2
0
0
4
π ( 4)
= 2.546
32
Σ
35.416
2 ( 6)
π
2 ( 4)
π
104
Y ΣL = Σ y L
Y (35.416 in.) = 104 in 2
or Y = 2.94 in.
PROBLEM 5.25
A 750 = g uniform steel rod is bent into a circular arc of radius 500 mm
as shown. The rod is supported by a pin at A and the cord BC. Determine
(a) the tension in the cord, (b) the reaction at A.
SOLUTION
First note, from Figure 5.8B: X =
( 0.5 m ) sin 30°
π/6
=
1.5
m
π
W = mg
= ( 0.75 kg ) 9.81 m/s 2
= 7.358 N
Then
(
)
Also note that ∆ ABD is an equilateral triangle.
Equilibrium then requires
(a) ΣM A = 0:


 1.5 
m  cos 30° ( 7.358 N ) − ( 0.5 m ) sin 60° TBC = 0
0.5 m − 
 π



or
TBC = 1.4698 N
or TBC = 1.470 N
(b) ΣFx = 0: Ax + (1.4698 N ) cos 60° = 0
or
Ax = −0.7349 N
ΣFy = 0: Ay − 7.358 N + (1.4698 N ) sin 60° = 0
or
Ay = 6.085 N
thus A = 6.13 N
83.1°
PROBLEM 5.26
The homogeneous wire ABCD is bent as shown and is supported by a pin
at B. Knowing that l = 8 in., determine the angle θ for which portion
BC of the wire is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie
on a vertical line through B. Further, because the wire is homogeneous,
its center of gravity will coincide with the centroid of the corresponding
line.
Thus ΣM B = 0, which implies that x = 0 or ΣxL = 0
Hence
−
2(6 in.)
π
(π × 6 in.) + 
8 in. 
 ( 8 in.)
 2 
6 in.


cosθ  ( 6 in.) = 0
+  8 in. −
2


Then
cosθ =
4
9
or θ = 63.6°
PROBLEM 5.27
The homogeneous wire ABCD is bent as shown and is supported by a pin
at B. Knowing that θ = 30°, determine the length l for which portion CD
of the wire is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on
a vertical line through B. Further, because the wire is homogeneous, its
center of gravity will coincide with the centroid of the corresponding line.
Thus ΣM B = 0, which implies that x = 0
Hence
or
Σxi Li = 0
 2 ( 6 in.)

−
cos 30° + ( 6 in.) sin 30°  (π × 6 in.)
 π

 ( l in.)

+
cos 30° ( l in.)
 2

6 in. 

+ ( l in.) cos 30° −
( 6 in.) = 0
2 

or
l 2 + 12.0l − 316.16 = 0
with roots l1 = 12.77 and −24.77.
Taking the positive root
l = 12.77 in.
PROBLEM 5.28
The homogeneous wire ABCD is bent as shown and is attached to a hinge
at C. Determine the length L for which the portion BCD of the wire is
horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on
a vertical line through C. Further, because the wire is homogeneous, its
center of gravity will coincide with the centroid of the corresponding line.
Thus ΣM C = 0, which implies that x = 0
or
Hence
or
Σ xi Li = 0
L
( L ) + ( −4 in.)(8 in.) + ( −4 in.)(10 in.) = 0
2
L2 = 144 in 2
or L = 12.00 in.
PROBLEM 5.29
Determine the distance h so that the centroid of the shaded area is as close
to line BB′ as possible when (a) k = 0.2, (b) k = 0.6.
SOLUTION
y =
Then
ΣyA
ΣA
 (a + h) 
a
( ab ) − 
  kb ( a − h ) 
2
2 

y =
ba − kb ( a − h )
or
=
2
2
1 a (1 − k ) + kh
2 a(1 − k ) + kh
c =1− k
Let
y =
Then
ζ =
and
h
a
a c + kζ 2
2 c + kζ
(1)
Now find a value of ζ (or h) for which y is minimum:
(
)
2
dy
a 2kζ ( c + kζ ) − k c + kζ
=
=0
2
dζ
( c + kζ ) 2
or
(
)
2ζ ( c + kζ ) − c + kζ 2 = 0
(2)
PROBLEM 5.29 CONTINUED
2cζ + 2ζ
Expanding (2)
2
− c − kζ
ζ=
Then
2
=0
or
kζ
2
+ 2cζ − c = 0
( 2c) 2 − 4 ( k ) ( c )
−2c ±
2k
Taking the positive root, since h > 0 (hence ζ > 0 )
−2 (1 − k ) + 4 (1 − k ) + 4k (1 − k )
2
h=a
(a) k = 0.2:
(b) k = 0.6:
h=a
h=a
2k
−2 (1 − 0.2 ) +
4 (1 − 0.2 ) + 4 ( 0.2 )(1 − 0.2 )
−2 (1 − 0.6 ) +
4 (1 − 0.6 ) + 4 ( 0.6 )(1 − 0.6 )
2
2 ( 0.2 )
or h = 0.472a
2
2 ( 0.6 )
or h = 0.387a
PROBLEM 5.30
Show when the distance h is selected to minimize the distance y from
line BB′ to the centroid of the shaded area that y = h.
SOLUTION
From Problem 5.29, note that Eq. (2) yields the value of ζ that minimizes h.
Then from Eq. (2)
We see
c + kζ 2
c + kζ
2ζ =
(3)
Then, replacing the right-hand side of (1) by 2ζ , from Eq. (3)
We obtain
But
So
y =
a
( 2ζ)
2
ζ=
h
a
y =h
Q.E.D.
PROBLEM 5.31
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
y =
For the element of area (EL) shown
h
x
a
dA = ( h − y ) dx
and
x

= h 1 −  dx
a

xEL = x
Then
1
(h +
2
h
= 1 +
2
yEL =
Then area
and
a 
h 1
0 

A = ∫ dA = ∫
x

a
a

x
x2 
1
−  dx = h  x −
= ah

2a 
2
a

0
a
 x 2 x3 
x 
1 2
−  dx  = h 
−
 = a h
a 
3a 
6
 
 2
0
a  
x h 1
0  
∫ xEL dA = ∫
a
∫ yEL dA = ∫0
h
x  
x   h2 a 
x2 
h
dx
1
1
1
+
−
=
−
∫  a 2 dx

 
 
a  
a 
2
2 0 

h2 
x3 
x
=
−


2 
3a 2 
Hence
y)
a
=
0
1 2
ah
3
xA = ∫ xEL dA
1  1
x  ah  = a 2h
2  6
x =
1
a
3
yA = ∫ yEL dA
1  1
y  ah  = ah 2
2  3
y =
2
h
3
PROBLEM 5.32
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
x = a, y = h : h = ka3
or
x=
Now
dA = xdy
xEL =
a 1/3
y dy
h1/3
1
1 a 1/3
x=
y , yEL = y
2
2 h1/ 3
h
A = ∫ dA =∫0
Then
h
a3
a 1/3
y
h1/3
Then
=
k =
( )
a 1/3
3 a
y dy =
y 4/3
1/3
4 h1/3
h
h
=
0
3
ah
4
h
h
and ∫ xEL dA = ∫0
1 a 1/3  a 1/3  1 a  3 5/3 
3 2
y  1/3 y dy  =
y  =
a h
2/3 
2 h1/3
2
5
10
h 
h

0
h
a  3 7/3 
3 2
h  a

1/3
∫ yEL dA = ∫0 y  h1/3 y dy  = h1/3  7 y  = 7 ah



0
Hence
3 2
3 
xA = ∫ xEL dA : x  ah  =
a h
 4  10
3  3
yA = ∫ yEL dA: y  ah  = ah 2
4  7
x =
2
a
5
y =
4
h
7
PROBLEM 5.33
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
x = a, y = h: h = k1a3
At
a = k 2 h3
k2 =
or
k1 =
or
h
a3
a
h3
Hence, on line 1
y =
h 3
x
a3
and on line 2
h 1/3
x
a1/3
y =
Then
h 
 h
dA =  1/3 x1/3 − 3 x3  dx
a

a
and
yEL =
1  h 1/3
h 3
 1/3 x + 3 x 
2 a
a
a
h 
1
1
a h
 3

∴ A = ∫ dA = ∫0  1/3 x1/3 − 3 x3  dx = h  1/3 x 4/3 − 3 x 4  = ah
2
4
4
a
a
a
a



0
a
a 
x
0 
∫ xELdA = ∫
h 1/3
h 
1
8 2
 3

x − 3 x3  dx = h  1/3 x7/3 − 3 x5  =
a h
1/3
5a
a
a

 7a
 0 35
a 
1/3
3 
1/3
3
∫ yEL dA = ∫0 2  a1/3 x + a3 x  a1/3 x − a3 x  dx



1
h
h
h
h
a
h 2 a  x 2/3 x 6 
h 2  3 x5/3 1 x6 
8 2
=
−
dx
=
−
=
ah



∫
6
5/3
6
0  2/3

2 a
2 5 a
7a 
35
a 
0
From
8 2
 ah 
xA = ∫ xEL dA: x   =
a h
2
35
 
or x =
16
a
35
and
8 2
 ah 
yA = ∫ yEL dA: y   =
ah
 2  35
or y =
16
h
35
PROBLEM 5.34
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0
First note that symmetry implies
For the element (EL) shown
yEL =
2r
(Figure 5.8B)
π
dA = π rd r
r2
Then
and
r2
π rd r
r1
A = ∫ dA = ∫
r
2
∫ yEL dA = ∫r1
 r2 
π 2
r2 − r12
= π   =
2
2
  r1
1
(

r2
)
(
( π rd r ) = 2  r 3  = r23 − r13
π
3
3
r
2r
2
)
1
So
π
 2 3
yA = ∫ yEL dA: y  r22 − r12  =
r2 − r13
2
 3
4 r23 − r13
or y =
3π r22 − r12
(
)
(
)
PROBLEM 5.35
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0
First note that symmetry implies
For the element (EL) shown
y = R cos θ, x = R sin θ
dx = R cos θ d θ
dA = ydx = R 2 cos 2θ dθ
Hence
α
1 2
α
 θ sin 2θ 
A = ∫ dA = 2∫0 R 2 cos 2 θ dθ = 2 R 2  +
 = R ( 2α sin 2α )
2
4
2

0
(
α
)
R
2

2
2
31
2
∫ yEL dA = 2∫0 2 cosθ R cos θ dθ = R  3 cos θ sin θ + 3 sin θ 

0
α
=
(
R3
cos 2 α sin α + 2sin α
3
But yA = ∫ yEL dA so
or
Alternatively,
)
(
R3
cos 2 α sin α + 2sin α
3
y =
R2
( 2α + sin 2α )
2
(
)
)
cos 2 α + 2
2
y = R sin α
3
( 2α + sin 2α )
y =
2
3 − sin 2 α
R sin α
3
2α + sin 2α
PROBLEM 5.36
Determine by direct integration the centroid of the area shown.
SOLUTION
For the element (EL) shown
y =
b 2
a − x2
a
dA = ( b − y ) dx
and
)
1
b
= x; y = ( y + b ) =
a+ a −x )
2
2a (
b
A = ∫ dA = ∫ ( a − a − x ) dx
a
=
xEL
(
b
a − a 2 − x 2 dx
a
2
a
0
Then
2
EL
2
2
To integrate, let x = a sin θ : a 2 − x 2 = a cosθ , dx = a cosθ dθ
Then
π /2 b
A = ∫0
a
( a − a cosθ )( a cosθ dθ )
π /2
b
2θ  
θ
=  a 2 sin θ − a 2  + sin

a
4  0
2
π

= ab  1 − 
4

)
(
1
a b
 b  a
and ∫ xEL dA = ∫0 x  a − a 2 − x 2 dx  =  x 2 + a 2 − x 2
a
a
2
3



=
1 3
ab
6
(
) (
(
3/2  
)

 0
)
a b

2
2 b
2
2
∫ yEL dA = ∫0 2a a + a − x  a a − a − x dx 


b2 a 2
b 2  x3 
x dx =
=
 
2 ∫0
2a
2a 2  3 
( )
a
=
0
π /2
1 2
ab
6
xA = ∫ xEL dA:
 
π  1
x  ab 1 −   = a 2b
4  6
 
or x =
2a
3 ( 4 − π)
yA = ∫ yEL dA:
 
π  1
y  ab 1 −   = ab 2
4  6
 
or y =
2b
3 ( 4 − π)
PROBLEM 5.37
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For the element (EL) shown on line 1 at
x = a, b = k2a 2
∴ y =
∴ y =
b
a2
b 2
x
a2
x = a, −2b = k1a3
On line 2 at
k2 =
or
or
k2 =
−2b
a3
−2b 3
x
a3
2b 
 b
dA =  2 x 2 + 3 x3  dx
a

a
Then
b 
2 x3 
b  x3 2 x 4 
=
+
A = ∫ dA = ∫ 2  x 2 +
dx
x 
4a 
a 
a 2  3
a
a
0
0
 1 1 5
= ab  +  = ab
 3 2 6
a
b 2 2b 3 
b  x 4 2 x5 
2
2 1
x
x
dx
+
=
+
 = a b  + 

2
3
2

5a 
a
a  4
a

4 5
0
a 
x
0 
and ∫ xEL dA = ∫
13 2
ab
20
2b 3   b 2 2b 3  
a1 b 2
∫ yEL dA = ∫0 2  a 2 x − a3 x   a 2 x + a3 x  dx 

 
 
=
1  b

 2b 
= ∫  2 x 2  −  3 x 3 
2  a

a

2
a
0
a
2

b 2  x5
2
− 2 x7 
 dx =

4
2a  5
7a

0
2
13
 1
= b 2a5 
−  = − ab 2
70
 10 7 
Then
xA = ∫ xEL dA:
yA = ∫ yEL dA:
 5  13 2
x  ab  =
ab
 6  20
 5  13 2
y  ab  −
ab
 6  70
or x =
39
a
50
or y = −
39
b
175
PROBLEM 5.38
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
x = 0, y = b
At
b = k (0 − a)
y =
Then
and
a
A = ∫ dA = ∫0
b
a2
b
( x − a )2
a2
dA = ydx =
and
k =
or
y
b
2
=
x − a)
2(
2
2a
xEL = x, yEL =
Now
Then
2
b
( x − a )2 dx
a2
a
b
b
1
2
3
x − a ) dx = 2 ( x − a )  = ab
2(
0 3
3a 
a
2
a 
a 3

2
2
∫ xEL dA = ∫0 x  a 2 ( x − a ) dx  = a 2 ∫0 ( x − 2ax + a x )dx


b
=
b
b  x4 2 3 a2 2 
1 2
− ax +
x  =
ab
2

3
2
a  4
 12
a
2
2
5
a


∫ yEL dA = ∫0 2a 2 ( x − a )  a 2 ( x − a ) dx  = 2a 4  5 ( x − a ) 



0
b
=
Hence
b
b2
1
1 2
ab
10
1 2
1 
xA = ∫ xEL dA: x  ab  =
ab
3
12


1 2
1 
yA = ∫ yEL dA: y  ab  =
ab
 3  10
x =
y =
1
a
4
3
b
10
PROBLEM 5.39
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x
Have
yEL =
1
a
x x2 
y = 1 − + 2 
2
2
L L 

x x2 
dA = ydx = a 1 − + 2  dx
L L 

Then
2L
2L 
a 1
0
A = ∫ dA = ∫

x x2 
x2
x3 
8
− + 2  dx = a  x −
+ 2  = aL
L
L
2
3
L
L
3



0
2L
and
 x2
x x2  
x3
x4 
2L  
∫ xEL dA = ∫0 x  a 1 − L + L2  dx  = a  2 − 3L + 4L2 
 
 

0
10 2
aL
=
3
2L
∫ yEL dA = ∫0
Hence
a
x x2   
x x2  
1
1
−
+
a
−
+


 
 dx 
2 
L L2   
L L2  
=
a 2 EL 
x
x2
x3 x 4 
1
2
3
2
−
+
−
+

 dx
∫
2 0 
L
L2
L3 L4 
=
a2
2
2L

x5 
11
x 2 x3
x4
+ 2 − 3 + 4  = a2L
x −
5
L
2L
5L  0
L

 8  10 2
xA = ∫ xEL dA: x  aL  =
aL
3
3 
 1  11
yA = ∫ yEL dA: y  a  = a 2
5
8 
x =
y =
5
L
4
33
a
40
PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
y1 at x = a, y = 2b
For
y1 =
Then
2b = ka 2
or k =
2b
a2
2b 2
x
a2
b
x
( x + 2b) = b  2 − 
a
a
By observation
y2 = −
Now
xEL = x
and for 0 ≤ x ≤ a :
1
b
y1 = 2 x 2
2
a
and
dA = y1dx =
1
b
x
y2 =  2 − 
2
2
a
and
x

dA = y2dx = b  2 −  dx
a

yEL =
2b 2
x dx
a2
For a ≤ x ≤ 2a :
yEL =
Then
a
A = ∫ dA = ∫0
2b 2
x
2a 
x dx + ∫a b  2 − dx
2
a
a

2a
a
2
 a
2b  x3 
x 
7
= 2   + b  −  2 −   = ab
a  
6
a  3 0
 2 
0
and
x 
a  2b 2
2a  

∫ xEL dA = ∫0 x  a 2 x dx  + ∫a x b  2 − a  dx 


 


2a
a
=
 2 x3 
2b  x 4 
b
+


x −

3a  0
a2  4 0

=
1 2
1  2
2
2
3 
a b + b ( 2a ) − ( a )  +
2a − ( a )  

 3a 

2
=
7 2
ab
6
{
( )
PROBLEM 5.40 CONTINUED
x 
x 
a b 2  2b 2 
2a b 
∫ yEL dA = ∫0 a 2 x  a 2 x dx  + ∫0 2  2 − a  b  2 − a  dx 



 
 
2b 2
= 4
a
=
Hence
2a
a
3
 x5 
b2  a 
x 
+
−
−
2

 

 
a  
2  3 
 5 0
a
17 2
ab
30
7  7
xA = ∫ xEL dA: x  ab  = a 2b
6  6
 7  17 2
yA = ∫ yEL dA: y  ab  =
ab
 6  30
x =a
y =
17
b
35
PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For y2
x = a, y = b : a = kb 2
at
y2 =
Now
xEL = x
For
0≤ x≤
k =
a
b2
b 1/2
x
a
Then
and for
or
a
y
b x1/2
x1/2
dx
: yEL = 2 =
, dA = y2dx = b
2
2
2 a
a
a
1
b  x 1 x1/2 
≤ x ≤ a : yEL = ( y1 + y2 ) =  − +

2
2
2a 2
a 
 x1/2 x 1 
dA = ( y2 − y1 ) dx = b 
− +  dx
 a a 2
Then
a/2
A = ∫ dA = ∫0 b
b
=
a
=
 x1/2 x 1 
x1/2
a
− +  dx
dx + ∫a/2 b 
a
 a a 2
a
a/2
 2 x3/2
x2 1 
 2 3/2 
x
b
+
−
+ x

3

2a 2  a/2

0
3 a
3/2
3/2
2 b  a 
3/2
a 
  + ( a ) −   
3 a  2 
 2  
2

 1 
a  1
 a  
2
+ b  −  a −    + ( a ) −    
2a 
 2   2 
 2  


( )
=
13
ab
24
PROBLEM 5.41 CONTINUED
and
1/2
  x1/2 x 1  

a/2  x
a
=
+
− +  dx
x
dA
x
b
dx
x


∫ EL
∫0 
∫ b 
a  a/2   a a 2  

a
a/2
 2 x5/2 x3 x 4 
b  2 5/2 
=
x
+
b
−
+



4  a/2
a  5
0
 5 a 3a
=
5/2
5/2
2 b  a 
5/2
a 
  + ( a ) −   
5 a  2 
 2  
 1  3  a 3  1  2  a 2  
+ b  − ( a ) −    +  ( a ) −    
 2   
 2   4 
 3a 

=
71 2
ab
240
a/2 b
x1/2  x1/2

dx 
b
∫ yEL dA = ∫0 2
a a 
1 x1/2    x1/2 x 1  
a b x
+ ∫a/2  − +
− +  dx 
 b 
2 a 2
a    a a 2  
a
3
a/2
1  x 1  
b2  1 2 
b 2  x 2


=
+
−
x
 −  
2a  2  0
2  2a 3a  a 2   

 a/2
=
b  a 
2
a
  + ( a ) −  
a
4  2 
2
=
11 2
ab
48
2
Hence
2
b2  a 1 
−
 − 
 6a  2 2 
71 2
 13 
xA = ∫ xEL dA: x  ab  =
ab
 24  240
 13  11 2
yA = ∫ yEL dA: y  ab  =
ab
 24  48
x =
3
17
a = 0.546a
130
y =
11
b = 0.423b
26
PROBLEM 5.42
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid. Express your answer in terms
of a.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Have at
x = a, y = a : a = ka 2
or
1 2
x
a
dy =
y =
Thus
and
2
k =
1
a
2
xdx
a
2
 dy 
2 
dL = 1 +   dx = 1 +  x  dx
 dx 
a 
Then
a
0
∴ L = ∫ dL = ∫
=
x
4
4x2
a 2
4x2
1 + 2 x 2 dx =  1 + 2 + ln  x + 1 + 2
4 a
2
a
a
a


(
)
a
a
5 + ln 2 + 5 = 1.4789a
2
4
a

3/2
4
4x2   2  a2  
 
1 + 2 dx  =    1 + 2 x 2  
  3  8 
a
a
 

 
0
a
x
0 
∫ xELdL = ∫
Then
(
)
a 2 3/2
5 − 1 = 0.8484a 2
12
xL = ∫ xEL dL: x (1.4789a ) = 0.8484a 2
=
x = 0.574a
a



0
PROBLEM 5.43
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Now
Then
and
xEL = r cos θ
and
7π /4
dL = rd θ
7π /4
L = ∫ dL = ∫π /4 rdθ = r [θ ]π /4 =
3
πr
2
7π /4
∫ xELdL = ∫π /4 r cosθ ( rdθ )
1 
 1
7π /4
2
= r 2 [sin θ ]π /4 = r 2  −
−
 = −r 2
2
2

Thus
3 
xL = ∫ xdL : x  π r  = −r 2 2
2 
x =−
2 2
r
3π
PROBLEM 5.44
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
xEL = a cos3 θ
Now
dL =
and
dx 2 + dy 2
x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ
Where
y = a sin 3 θ : dy = 3a sin 2 θ cosθ dθ
Then
(

dL =  −3a cos 2 θ sin θ dθ

1/2
) + (3a sin θ cosθ dθ ) 
2
2
2
(
= 3a cosθ sin θ cos 2 θ + sin 2 θ
)
1/2
dθ
= 3a cosθ sin θ dθ
π /2
∴ L = ∫ dL = ∫0
=
and
π /2
1

3a cosθ sin θ dθ = 3a  sin 2 θ 
2

0
3
a
2
π /2
3
∫ xEL dL = ∫0 a cos θ ( 3a cosθ sin θ dθ )
π /2
 1

= 3a 2  − cos5 θ 
 5
0
Hence
=
3 2
a
5
3  3
xL = ∫ xEL dL : x  a  = a 2
2  5
x =
2
a
5
PROBLEM 5.44 CONTINUED
Alternative solution
 x
x = a cos3 θ ⇒ cos 2 θ =  
 a
 y
y = a sin θ ⇒ sin θ =  
 a
3
 x
∴  
 a
2/3
2/3
 y
+ 
 a
=1
(
y = a 2/3 − x 2/3
or
(
dy
= a 2/3 − x 2/3
dx
Now
xEL = x
and
2/3
2
Then
 dy 
dL = 1 +  
 dx 
2/3
3/2
) (−x )
1/2
−1/3


dx = 1 +  a 2/3 − x 2/3


2
)
(
) (−x )
1/2
−1/3


1/2
2

 dx

a
Then
and
Hence
a
L = ∫ dL = ∫0
a1/3
3
3

dx = a1/3  x 2/3  = a
1/ 3
2
2
x

0
1/3
a a
x

1/3
0
∫ xELdL = ∫
x
a

3
1/3  3 5/3 
dx = a  x  = a 2

5
0 5
3  3
xL = ∫ xEL dL : x  a  = a 2
2  5
x =
2
a
5
PROBLEM 5.45
Determine by direct integration the centroid of the area shown.
SOLUTION
Have
yEL
and
2
2
r cosθ = aeθ cosθ
3
3
2
2 θ
= r sin θ = ae sin θ
3
3
xEL =
dA =
1
1
( r )( rdθ ) = a 2e2θ dθ
2
2
Then
π
(
)
1 1
1
π 1

A = ∫ dA = ∫0 a 2e2θ dθ = a 2  e2θ  = a 2 e2π − 1 = 133.623a 2
2
2 2
4
0
and
π
 2 2θ 
θ
3 π 3θ
∫ xELdA = ∫0 3ae cosθ  2 a e dθ  = 3 a ∫0 e cosθ dθ


2
1
1
To proceed, use integration by parts, with
u = e3θ
du = 3e3θ dθ
and
dv = cosθ dθ
Then
Now let
and
3θ
3θ
3θ
∫ e cosθ dθ = e sin θ − ∫ sin θ ( 3e dθ )
u = e3θ
So that
du = 3e3θ dθ
then
dv = sin θ dθ ,
Then
v = sin θ
then
v = − cosθ
3θ
3θ
−3θ
3θ
∫ e sin θ dθ = e sin θ − 3  −e cosθ − ∫ ( − cosθ ) ( 3e dθ ) 
e3θ
3θ
∫ e cosθ dθ = 10 ( sin θ + 3cosθ )
π

1  e3θ
a3
∴ ∫ xEL dA = a3 
( sin θ + 3cos θ)  = −3e3π − 3 = −1239.26a3
3  10
0 30
Also
(
)
π
 2 2θ 
θ
3 π 3θ
∫ yEL dA = ∫0 3 ae sin θ  2 a e dθ  = 3 a ∫0 e sin θ dθ


2
1
1
PROBLEM 5.45 CONTINUED
Using integration by parts, as above, with
u = e3θ
dv = ∫ sin θ dθ
Then
du = 3e3θ dθ
and
and
v = − cosθ
3θ
3θ
3θ
∫ e sin θd θ = −e cos θ − ∫ ( − cos θ) ( 3e d θ)
So that ∫ e3θ sin θd θ =
e3θ
( − cos θ + 3sin θ)
10
π

1  e3θ
a3
∴ ∫ yEL dA = a3 
( − cos θ + 3sin θ)  = e3π + 1 = 413.09a3
3  10
0 30
(
)
y (133.623a ) = 413.09a
(
Hence xA = ∫ xEL dA: x 133.623a 2 = −1239.26a3
yA = ∫ yEL dA:
2
3
)
or x = −9.27a
or y = 3.09a
PROBLEM 5.46
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x,
Have
yEL =
dA = ydx
and
L/2
A = ∫ dA = ∫0 x sin
and
1
πx
x sin
2
L
L/2
 L2
πx L
πx
− x cos 
dx =  2 sin
π
L
L
L 0
π
πx
L2
π2
πx 
L/2 
x = ∫ xEL dA = ∫0 x  x sin
dx 
L


L/2
 2 L2
 π x  2 L3
 π x  L 2  π x 
=  2 x sin 
 + 3 cos 
 − x sin 

 L  π
 L  π
 L 0
π
Also
=
L/2 1
y = ∫ yEL dA = ∫0
2
x sin
=
πx
πx 
dx 
 x sin
L 
L

L/2
1  2 L2
πx  L
2 L3 
π x
=  2 x sin
−  x − 3  cos 
2  π
L π
L 
π 
0
=

1  1  L3 
L2  L 
L3
1
6 + π2
−
=
   −

(
)


2
2  6  8  4π 2  2 
 96π
(
)
L3
π2
−2
L3
π3
PROBLEM 5.46 CONTINUED
Hence
 L2 
z 
 1
xA = ∫ xEL dA: x  2  = L3  2 − 3 
π 
π
π 
or x = 0.363L
 L2 
L3  1
2 
yA = ∫ yEL dA: y  2  =
− 3
2 2
π 
 π  96π  π
or y = 0.1653L
PROBLEM 5.47
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem. 5.2 about (a) the x axis, (b) the line
x = 165 mm.
SOLUTION
From the solution to Problem 5.2:
A = 10 125 mm 2 , X area = 96.4 mm, Yarea = 34.7 mm
( Area )
From the solution to Problem 5.22:
L = 441.05 mm X line = 92.2 mm, Yline = 32.4 mm
( Line )
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x axis:
Area = 2π Yline L = 2π ( 32.4 mm )( 441.05 mm ) = 89.786 × 103 mm 2
A = 89.8 × 103 mm 2
Volume = 2π Yarea A = 2π ( 34.7 mm )(10 125 mm ) = 2.2075 × 106 mm3
V = 2.21 × 106 mm3
(b) Rotation about x = 165 mm:
(
)
Area = 2π 165 − X line L = 2π (165 − 92.2 ) mm  ( 441.05 mm ) = 2.01774 × 105 mm 2
A = 0.202 × 106 mm 2
(
)
Volume = 2π 165 − X area A = 2π (165 − 96.4 ) mm  (10 125 mm ) = 4.3641 × 106 mm3
V = 4.36 × 106 mm3
PROBLEM 5.48
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem 5.4 about (a) the line y = 22 mm, (b) the
line x = 12 mm.
SOLUTION
From the solution to Problem 5.4:
A = 399 mm 2 , X area = 1.421 mm, Yarea = 12.42 mm
(Area)
From the solution to Problem 5.23:
L = 77.233 mm, X line = 1.441 mm, Yline = 12.72 mm
(Line)
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the line y = 22 mm:
(
)
Area = 2π 22 − Yline L = 2π ( 22 − 12.72 ) mm  ( 77.233 mm ) = 4503 mm 2
A = 4.50 × 103 mm 2
(
)
(
)
Volume = 2π 22 − Yarea A = 2π ( 22 − 12.42 ) mm  399 mm 2 = 24 016.97 mm3
V = 24.0 × 103 mm3
(b) Rotation about line x = 12 mm:
(
)
Area = 2π 12 − X line L = 2π (12 − 1.441) mm  ( 77.233 mm ) = 5124.45 mm 2
A = 5.12 × 103 mm 2
(
)
Volume = 2π (12 − 1.421) A = 2π (12 − 1.421) mm  399 mm 2 = 26 521.46 mm3
V = 26.5 × 103 mm3
PROBLEM 5.49
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem 5.1 about (a) the x axis, (b) the line
x = 16 in.
SOLUTION
From the solution to Problem 5.1:
A = 240 in 2 , X area = 5.60 in., Yarea = 6.60 in.
(Area)
From the solution to Problem 5.21:
L = 72 in., X line = 4.67 in., Yline = 6.67 in.
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x axis:
Ax = 2π Yline L = 2π ( 6.67 in.)( 72 in.) = 3017.4 in 2
A = 3020 in 2
(
)
Vx = 2π Yarea A = 2π ( 6.60 in.) 240 in 2 = 9952.6 in 3
V = 9950 in 3
(b) Rotation about x = 16 in.:
(
)
Ax =16 = 2π 16 − X line L = 2π (16 − 4.67 ) in. ( 72 in.) = 5125.6 in 2
Ax =16 = 5130 in 2
(
)
(
)
Vx =16 = 2π 16 − X area A = 2π (16 − 5.60 ) in. 240 in 2 = 15 682.8 in 3
Vx =16 = 15.68 × 103 in 3
PROBLEM 5.50
Determine the volume of the solid generated by rotating the semielliptical
area shown about (a) the axis AA′, (b) the axis BB′, (c) the y axis.
SOLUTION
Applying the second theorem of Pappus-Guldinus, we have
(a) Rotation about axis AA′:
 π ab 
2 2
Volume = 2π yA = 2π ( a ) 
=π ab
 2 
V = π 2a 2b
(b) Rotation about axis BB′:
 π ab 
2 2
Volume = 2π yA = 2π ( 2a ) 
 = 2π a b
 2 
V = 2π 2a 2b
 4a  π ab  2 2
Volume = 2π yA = 2π  
 = πa b
 3π  2  3
V =
(c) Rotation about y-axis:
2 2
πa b
3
PROBLEM 5.51
Determine the volume and the surface area of the chain link shown,
which is made from a 2-in.-diameter bar, if R = 3 in. and L = 10 in.
SOLUTION
First note that the area A and the circumference C of the cross section of the bar are
A=
π
4
d2
and
C = πd
Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Then, applying the theorems of Pappus-Guldinus, we have
Volume = 2 (Vside ) + 2 (Vend ) = 2 ( AL ) + 2 (π RA) = 2 ( L + π R ) A
2
π
= 2 10 in. + π ( 3 in.)   ( 2 in.) 
4

= 122.049 in 3
V = 122.0 in 3
Area = 2 ( Aside ) + 2 ( Aend ) = 2 ( CL ) + 2 (π RC ) = 2 ( L + π R ) C
= 2 10 in. + π ( 3 in.)  π ( 4 in.) 
= 488.198 in 2
A = 488 in 2
PROBLEM 5.52
Verify that the expressions for the volumes of the first four shapes in
Figure 5.21 on page 261 are correct.
SOLUTION
Following the second theorem of Pappus-Guldinus, in each case a
specific generating area A will be rotated about the x axis to produce the
given shape. Values of y are from Fig. 5.8A.
(1) Hemisphere: the generating area is a quarter circle
Have
 4a  π 
V = 2π yA = 2π   a 2 
 3π  4 
or V =
2 3
πa
3
(2) Semiellipsoid of revolution: the generating area is a quarter ellipse
Have
 4a  π 
V = 2π yA = 2π   ha 
 3π  4 
or V =
2 2
πa h
3
(3) Paraboloid of revolution: the generating area is a quarter parabola
Have
 3  2 
V = 2π yA = 2π  a  ah 
 8  3 
or V =
1 2
πa h
2
or V =
1 2
πa h
3
(4) Cone: the generating area is a triangle
Have
 a  1 
V = 2π yA = 2π   ha 
 3  2 
PROBLEM 5.53
A 15-mm-diameter hole is drilled in a piece of 20-mm-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
 5
 1


V = 2π xA = 2π  + 7.5  mm  ×  × 5 mm × 5 mm 
3
2



 
or V = 720 mm3
PROBLEM 5.54
Three different drive belt profiles are to be studied. If at any given time
each belt makes contact with one-half of the circumference of its pulley,
determine the contact area between the belt and the pulley for each
design.
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area AC of a
belt is given by
AC = π yL = πΣ yL
Where the individual lengths are the “Lengths” of the belt cross section
that are in contact with the pulley
Have
AC = π  2 ( y1L1 ) + y2 L2 
  2.5 mm 
2.5 
 
= π 2  60 −
 mm  

2 
 
  cos 20° 

+ ( 60 − 2.5 ) mm  (12.5 mm ) 

or AC = 3.24 × 103 mm 2
Have
AC = π  2 ( y1L1 ) 

  7.5 mm 
7.5 
= 2π  60 − 1.6 −
 mm  × 
D 
2 

  cos 20 
or AC = 2.74 × 103 mm 2
Have


2× 5
AC = π ( y1L1 ) = π  60 −
 mm  (π × 5 mm )
π 


or AC = 2.80 × 103 mm 2
PROBLEM 5.55
Determine the capacity, in gallons, of the punch bowl shown if R = 12 in.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the
second theorem of Pappus-Guldinus and using Fig. 5.8A, we have
V = 2π xA = 2πΣxA = 2π ( x1 A1 + x2 A2 )




 1 1   1 1


3   2 R sin 30D
π


= 2π  × R   × R ×
cos30D   R 2  
R  + 
2   3× π
 3 2   2 2
  6  


6



 R3
R3  3 3
= 2π 
+
π R3
 =
8
 16 3 2 3 
=
Since
3 3
3
π (12 in.) = 3526.03 in 3
8
1 gal = 231 in 3
V =
3526.03 in 3
= 15.26 gal
231 in 3/gal
V = 15.26 gal
PROBLEM 5.56
The aluminum shade for a small high-intensity lamp has a uniform
thickness of 3/32 in. Knowing that the specific weight of aluminum is
0.101 lb/in 3 , determine the weight of the shade.
SOLUTION
The weight of the lamp shade is given by
W = γ V = γ At
where A is the surface area of the shade. This area can be generated by rotating the line shown about the
x axis. Applying the first theorem of Pappus-Guldinus, we have
A = 2π yL = 2πΣyL = 2π ( y1L1 + y2 L2 + y3 L3 + y4 L4 )
 0.6 mm
0.60 + 0.75 
= 2π 
( 0.6 mm ) + 
 mm ×
2


 2
 0.75 + 1.25 
+
 mm ×
2


 1.25 + 1.5 
+
 mm ×
2


= 22.5607 in
Then
( 0.15 mm )2 + (1.5 mm )2
( 0.50 mm )2 + ( 0.40 mm )2

( 0.25 mm )2 + (1.25 mm )2 

2
W = 0.101 lb/in 3 × 22.5607 in 2 ×
3
in. = 0.21362 lb
32
W = 0.214 lb W
PROBLEM 5.57
The top of a round wooden table has the edge profile shown. Knowing
that the diameter of the top is 1100 mm before shaping and that the
density of the wood is 690 kg/m3 , determine the weight of the waste
wood resulting from the production of 5000 tops.
SOLUTION
All dimensions are in mm
Have
Vwaste = Vblank − Vtop
Vblank = π ( 550 mm ) × ( 30 mm ) = 9.075π × 106 mm3
2
Vtop = V1 + V2 + V3 + V4
Applying the second theorem of Pappus-Guldinus to parts 3 and 4
2
2
Vtop = π ( 529 mm ) × (18 mm )  + π ( 535 mm ) × (12 mm ) 

 

 
 π
4 × 12 
2

+ 2π   535 +
 mm  × (12 mm ) 
3
π
4



 

 
 π
4 × 18 
2

+ 2π   529 +
 mm  × (18 mm ) 
3
π
4



 

= π ( 5.0371 + 3.347 + 0.1222 + 0.2731) × 106 mm3
= 8.8671π × 106 mm3
∴ Vwaste = ( 9.0750 − 8.8671) π × 106 mm3
= 0.2079π × 10−3 m3
Finally
Wwaste = ρ wood Vwaste g N tops
(
)
= 690 kg/m3 × 0.2079π × 10−3 m3 × 9.81 m/s 2 × 5000 ( tops )
or Wwaste = 2.21 kN W
PROBLEM 5.58
The top of a round wooden table has the shape shown. Determine how
many liters of lacquer are required to finish 5000 tops knowing that each
top is given three coats of lacquer and that 1 liter of lacquer covers 12 m2.
SOLUTION
Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have
Asurface = Atop circle + Abottom circle + Aedge
2
2
= π ( 535 mm )  + π ( 529 mm ) 

 

 π
2 × 12 
 

+ 2π  535 +
 mm  × (12 mm ) 
π 
 

 2
 π
2 × 18 
 

+ 2π  529 +
 mm  × (18 mm ) 
π 
 
 2

= 617.115π × 103 mm 2
Then
# liters = Asurface × Coverage × N tops × N coats
= 617.115π × 10−3 m 2 ×
1 liter
× 5000 × 3
12 m 2
or # liters = 2424 L W
PROBLEM 5.59
The escutcheon (a decorative plate placed on a pipe where the pipe exits
from a wall) shown is cast from yellow brass. Knowing that the specific
3
weight of yellow brass is 0.306 lb/in . determine the weight of the
escutcheon.
SOLUTION
The weight of the escutcheon is given by
W = (specific weight)V
where V is the volume of the plate. V can be generated by rotating the area A about the x axis.
Have
and
Then
a = 3.0755 in. − 2.958 in. = 0.1175 in.
sin φ =
0.5
⇒ φ = 0.16745 R = 9.5941°
3
2α = 26D − 9.5941D = 16.4059D
or
α = 8.20295D = 0.143169 rad
The area A can be obtained by combining the following four areas, as indicated.
Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have
V = 2π yA = 2πΣ yA
PROBLEM 5.59 CONTINUED
A, in 2
y , in.
yA, in 3
1
1
( 3.0755)(1.5) = 2.3066
2
1
(1.5) = 0.5
3
1.1533
2
−α ( 3) = −1.28851
2
1
( 2.958)( 0.5) = −0.7395
2
3
−
4
− ( 0.1755 )( 0.5 ) = −0.05875
2 ( 3) sin α
× sin (α + φ ) = 0.60921
3α
1
( 0.5 ) = 0.16667
3
1
( 0.5) = 0.25
2
–0.78497
–0.12325
–0.14688
Σ yA = 0.44296 in 3
Then
so that
(
)
V = 2π 0.44296 in 3 = 1.4476 in 3
(
)
W = 1.4476 in 3 0.306 lb/in 3 = 0.44296 lb
W = 0.443 lb W
PROBLEM 5.60
The reflector of a small flashlight has the parabolic shape shown.
Determine the surface area of the inside of the reflector.
SOLUTION
First note that the required surface area A can be generated by rotating
the parabolic cross section through 2π radians about the x axis. Applying
the first theorem of Pappus-Guldinus, we have
A = 2π yL
Now, since
x = ky ,
2
a = 56.25 k
or
At
2
a + 15 = 156.25k
Eq. (2) a + 15 156.25k
:
=
Eq. (1)
56.25k
a
and
(2)
or a = 8.4375 mm
Eq. (1) ⇒ k = 0.15
∴ x = 0.15 y 2
2
(1)
x = ( a + 15 ) mm: a + 15 = k (12.5 )
or
Then
x = a : a = k ( 7.5 )
at
1
mm
dx
= 0.3 y
dy
2
Now
So
 dx 
dL = 1 +   dy = 1 + 0.09 y 2 dy
 dy 
A = 2π yL
yL = ∫ ydL
and
∴ A = 2π ∫7.5 y 1 + 0.09 y 2 dy
12.5
12.5
3/2 
2  1 
2
= 2π  
 1 + 0.09 y
3
0.18

 
)
= 1013 mm 2
or A = 1013 mm 2 W
(

 7.5
PROBLEM 5.61
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
Resultant
R = R1 + R2
(a) Have
R1 = ( 40 lb/ft )(18 ft ) = 720 lb
R2 =
1
(120 lb/ft )(18 ft ) = 1080 lb
2
R = 1800 lb
or
The resultant is located at the centroid C of the distributed load x
Have
or
ΣM A:
(1800 lb ) x
= ( 40 lb/ft )(18 ft )( 9 ft ) +
x = 10.80 ft
1
(120 lb/ft )(18 ft )(12 ft )
2
R = 1800 lb W
x = 10.80 ft
(b)
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1800 lb = 0, Ay = 1800 lb
∴ A = 1800 lb
W
ΣM A = 0: M A − (1800 lb )(10.8 ft ) = 0
M A = 19.444 lb ⋅ ft
or M A = 19.44 kip ⋅ ft
W
PROBLEM 5.62
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
(a) Have
RI = ( 300 N/m )( 6 m ) = 1800 N
RII =
ΣFy : − R = − RI − RII
Then
or
1
( 6 m )( 900 N/m ) = 1800 N
3
R = 1800 N + 1800 N = 3600 N
ΣM A : − x ( 3600 N ) = − ( 3 m )(1800 N ) − ( 4.5 m )(1800 N )
x = 3.75 m
or
R = 3600 N W
x = 3.75 m
(b) Reactions
ΣFx = 0: Ax = 0
ΣM A = 0:
or
( 6 m ) By − ( 3600 N )( 3.75 m ) = 0
By = 2250 N
B = 2250 N
W
A = 1350 N
W
ΣFy = 0: Ay + 2250 N = 3600 N
or
Ay = 1350 N
PROBLEM 5.63
Determine the reactions at the beam supports for the given loading.
SOLUTION
RI = (100 lb/ft )( 4 ft ) = 400 lb
Have
1
( 200 lb/ft )( 6 ft ) = 600 lb
2
= ( 200 lb/ft )( 4 ft ) = 800 lb
RII =
RIII
ΣFx = 0: Ax = 0
Then
ΣM A = 0:
or
( 2 ft )( 400 lb ) − ( 4 ft )( 600 lb ) − (12 ft )(800 lb ) + (10 ft ) By
By = 800 lb
=0
B = 800 lb
W
A = 1000 lb
W
ΣFy = 0: Ay + 800 lb − 400 lb − 600 lb − 800 = 0
or
Ay = 1000 lb
PROBLEM 5.64
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI = ( 9 ft )( 200 lb/ft ) = 1800 lb
RII =
Then
1
( 3 ft )( 200 lb/ft ) = 300 lb
2
ΣFx = 0: Ax = 0
ΣM A = 0: − ( 4.5 ft )(1800 lb ) − (10 ft )( 300 lb ) + ( 9 ft ) By = 0
or
By = 1233.3 lb
B = 1233 lb
W
A = 867 lb
W
ΣFy = 0: Ay − 1800 lb − 300 lb + 1233.3 lb = 0
or
Ay = 866.7 lb
PROBLEM 5.65
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI =
1
( 200 N/m )( 0.12 m ) = 12 N
2
RII = ( 200 N/m )( 0.2 m ) = 40 N
Then
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 18 N − 12 N − 40 N = 0
or
Ay = 34 N
A = 34.0 N
W
ΣM A = 0: M A − ( 0.8 m )(12 N ) − ( 0.22 m )( 40 N ) + ( 0.38 m )(18 N )
or
M A = 2.92 N ⋅ m
M A = 2.92 N ⋅ m
W
PROBLEM 5.66
Determine the reactions at the beam supports for the given loading.
SOLUTION
First, replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a linear relation between load and distance, and the values at the end points are the same.
Have
RI = (1.8 m )( 2000 N/m ) = 3600 N
RII =
Then
1
(1.8 m )( 4500 N/m ) = 4050 N
2
ΣFx = 0: Ax = 0
ΣM B = 0: − ( 3 m ) Ay − ( 2.1 m )( 3600 N ) + ( 2.4 m )( 4050 N )
or
Ay = 270 N
A = 270 N
W
B = 720 N
W
ΣFy = 0: 270 N − 3600 N + 4050 N − By = 0
or
By = 720 N
PROBLEM 5.67
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
Then
RI =
1
( 4 m )( 2000 kN/m ) = 2667 N
3
RII =
1
( 2 m )(1000 kN/m ) = 666.7 N
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2667 N − 666.7 N = 0
or
Ay = 3334 N
A = 3.33 kN
W
ΣM A = 0: M A − (1 m )( 2667 N ) − ( 5.5 m )( 666.7 N )
or
M A = 6334 N ⋅ m
M A = 6.33 kN ⋅ m
W
PROBLEM 5.68
Determine the reactions at the beam supports for the given loading.
SOLUTION
First, replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a parabolic relation between load and distance, and the values at end points are the same.
Have
RI = ( 8 ft )(100 lb/ft ) = 800 lb
RII =
Then
2
(8 ft )( 600 lb/ft ) = 3200 lb
3
ΣFx = 0: Ax = 0
ΣM A = 0: 11B + ( 5 ft )( 800 lb ) − ( 4 ft )( 3200 ) lb = 0
or B = 800 lb
W
or A = 1600 lb
W
ΣFy = 0: Ay − 3200 lb + 800 lb + 800 lb = 0
PROBLEM 5.69
Determine (a) the distance a so that the vertical reactions at supports A
and B are equal, (b) the corresponding reactions at the supports.
SOLUTION
(a) Have
1
( a ft )(120 lb/ft ) = ( 60a ) lb
2
RII =
1
(12 − a )( 40 lb/ft ) = ( 240 − 20a ) lb
2
ΣFy = 0: Ay − 60a − ( 240 − 2a ) + By = 0
Then
Ay + By = 240 + 40a
or
Ay = By ⇒ Ay = By = 120 + 20a
Now
Also
RI =
(1)

a    1

ΣM B = 0: − (12 m ) Ay + ( 60a ) lb  12 −  ft  +  (12 − a ) ft   ( 240 − 20a ) lb  = 0
3    3


or
Ay = 80 −
140
10 2
a−
a
3
9
(2)
Equating Eqs. (1) and (2)
120 + 20a = 80 −
or
Then
Now
(b) Have
Eq. (1)
140
10
a − a2
3
9
40 2
a − 320a + 480 = 0
3
a = 1.6077 ft,
a = 22.392
a ≤ 12 ft
a = 1.608 ft W
ΣFx = 0: Ax = 0
Ay = By = 120 + 20 (1.61) = 152.2 lb
A = B = 152.2 lb
W
PROBLEM 5.70
Determine (a) the distance a so that the vertical reaction at support B is
minimum, (b) the corresponding reactions at the supports.
SOLUTION
RI =
1
( a ft )(120 lb/ft ) = 60a lb
2
RII =
1
(12 − a ) ft  ( 40 lb/ft ) = ( 240 − 20a ) lb
2
(a) Have
Then
or
Then
(b) Eq. (1)

a 
a 
ΣM A = 0: −  ft  ( 60a lb ) − ( 240 − 20a ) lb   8 +  ft  + (12 ft ) By = 0
3 
3 

By =
dBy
da
By =
10 2 20
a −
a + 160
9
3
=
20
20
a−
=0
9
3
or a = 3.00 ft W
10
20
( 3.00 )2 − ( 3.00 ) + 160
9
3
= 150 lb
and
(1)
B = 150.0 lb
W
A = 210 lb
W
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 60 ( 3.00 )  lb −  240 − 20 ( 3.00 )  lb + 150 lb = 0
or
Ay = 210 lb
PROBLEM 5.71
Determine the reactions at the beam supports for the given loading when
w0 = 1.5 kN/m.
SOLUTION
Have
RI =
1
( 9 m )( 2 kN/m ) = 9 kN
2
RII = ( 9 m )(1.5 kN/m ) = 13.5 kN
Then
ΣFx = 0: Cx = 0
ΣM B = 0: − 50 kN ⋅ m − (1 m )( 9 kN ) − ( 2.5 m )(13.5 kN ) + ( 6 m ) C y = 0
or
C y = 15.4583 kN
C = 15.46 kN
W
B = 7.04 kN
W
ΣFy = 0: By − 9 kN − 13.5 kN + 15.4583 = 0
or
By = 7.0417 kN
PROBLEM 5.72
Determine (a) the distributed load w0 at the end D of the beam ABCD
for which the reaction at B is zero, (b) the corresponding reactions at C.
SOLUTION
Have
RI =
1
( 9 m ) ( 3.5 − w0 ) kN/m  = 4.5 ( 3.5 − w0 ) kN
2
RII = ( 9 m ) ( w0 kN/m ) = 9w0 kN
(a) Then
or
so
ΣM C = 0: − 50 kN ⋅ m + ( 5 m )  4.5 ( 3.5 − w0 ) kN  + ( 3.5 m ) ( 9w0 kN ) = 0
9w0 + 28.75 = 0
w0 = −3.1944 kN/m
w0 = 3.19 kN/m
W
C = 1.375 kN
W
Note: the negative sign means that the distributed force w0 is upward.
ΣFx = 0: Cx = 0
(b)
ΣFy = 0: − 4.5 ( 3.5 + 3.19 ) kN + 9 ( 3.19 ) kN + C y = 0
or
C y = 1.375 kN
PROBLEM 5.73
A grade beam AB supports three concentrated loads and rests on soil and
the top of a large rock. The soil exerts an upward distributed load, and
the rock exerts a concentrated load RR as shown. Knowing that P = 4 kN
and wB = 12 wA , determine the values of wA and RR corresponding to
equilibrium.
SOLUTION
Have
RI = (1.2 m )( wA kN/m ) = 1.2 wA kN
RII =
1
1
(1.8 m )  wA kN/m  = 0.45 wA kN
2
2

1

RIII = (1.8 m )  wA kN/m  = 0.9 wA kN
2


Then
ΣM C = 0: − ( 0.6 m ) (1.2 wA ) kN  + ( 0.6 m ) ( 0.45 wA ) kN/m 
+ ( 0.9 m ) ( 0.9 wA ) kN/m  − (1.2 m )( 4 kN/m )
− ( 0.8 m )(18 kN/m ) + ( 0.7 m )( 24 kN/m ) = 0
or
and
wA = 6.667 kN/m
wA = 6.67 kN/m W
ΣFy = 0: RR + (1.2 m )( 6.67 kN/m ) + ( 0.45 m )( 6.67 kN/m )
+ ( 0.9 m )( 6.67 kN/m ) − 24 kN − 18 kN − 4 kN
or
RR = 29.0 kN
RR = 29.0 kN W
PROBLEM 5.74
A grade beam AB supports three concentrated loads and rests on soil and
the top of a large rock. The soil exerts an upward distributed load, and
the rock exerts a concentrated load RR as shown. Knowing that wB =
0.4wA, determine (a) the largest value of P for which the beam is in
equilibrium, (b) the corresponding value of wA.
In the following problems, use γ = 62.4 lb/ft3 for the specific weight of
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S.
customary units are used. With SI units, use ρ = 103 kg/m3 for the
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of
concrete. (See the footnote on page 222 for how to determine the specific
weight of a material given its density.)
j
SOLUTION
Have
RI = (1.2 m )( wA kN/m ) = 1.2 wA kN
RII =
1
(1.8 m )( 0.6 wA kN/m ) = 0.54 wA kN
2
RIII = (1.8 m )( 0.4 wA kN/m ) = 0.72 wA kN
(a) Then
ΣM A = 0:
( 0.6 m ) (1.2 wA ) kN  + (1.2 m ) RR + (1.8 m ) ( 0.54 wA ) kN 
+ ( 2.1 m ) ( 0.72 wA ) kN  − ( 0.5 m )( 24 kN )
− ( 2.0 m )(18 kN ) + ( 2.4 m ) P = 0
or
and
or
3.204 wA + 1.2 RR − 2.4 P = 48
(1)
ΣFy = 0: RR + 1.2 WA + 0.54 WA + 0.72 WA − 24 − 18 − P = 0
RR + 2.46 WA − P = 42
(2)
Now combine Eqs. (1) and (2) to eliminate wA :
( 3.204 ) Eq. 2 − ( 2.46 ) Eq. 1 ⇒
0.252 RR = 16.488 − 2.7 P
Since RR must be ≥ 0, the maximum acceptable value of P is that for which R = 0,
or
(b) Then, from Eq. (2):
P = 6.1067 kN
2.46 WA − 6.1067 = 42
P = 6.11 kN W
or WA = 19.56 kN/m W
PROBLEM 5.75
The cross section of a concrete dam is as shown. For a dam section of
unit width, determine (a) the reaction forces exerted by the ground on the
base AB of the dam, (b) the point of application of the resultant of the
reaction forces of part a, (c) the resultant of the pressure forces exerted
by the water on the face BC of the dam.
In the following problems, use γ = 62.4 lb/ft3 for the specific weight of
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S.
customary units are used. With SI units, use ρ = 103 kg/m3 for the
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of
concrete. (See the footnote on page 222 for how to determine the specific
weight of a material given its density.)
SOLUTION
The free body shown consists of a 1-m thick section of the dam and the
triangular section BCD of the water behind the dam.
X1 = 6 m
Note:
X 2 = ( 9 + 3) m = 12 m
X 3 = (15 + 2 ) m = 17 m
X 4 = (15 + 4 ) m = 19 m
W = ρ gV
(a) Now
(
)(
)
(
)(
)
(
)(
)
(
)(
)
so that
1

W1 = 2400 kg/m3 9.81 m/s 2  ( 9 m )(15 m )(1 m )  = 1589 kN
2


W2 = 2400 kg/m3 9.81 m/s 2 ( 6 m )(18 m )(1 m )  = 2543 kN
1

W3 = 2400 kg/m3 9.81 m/s 2  ( 6 m )(18 m )(1 m )  = 1271 kN
2

1

W4 = 2400 kg/m3 9.81 m/s 2  ( 6 m )(18 m )(1 m )  = 529.7 kN
2

Also
P=
(
)(
)
1
1
Ap = (18 m )(1 m )   103 kg/m3 9.81 m/s 2 (18 m ) 


2
2
= 1589 kN
Then
or
ΣFx = 0: H − 1589 kN = 0
H = 1589 kN
H = 1589 kN
W
ΣFy = 0: V − 1589 kN − 2543 kN − 1271 kN − 529.7 kN
or
V = 5933 kN
V = 5.93 MN
W
PROBLEM 5.75 CONTINUED
(b) Have
ΣM A = 0: X ( 5933 kN ) + ( 6 m )(1589 kN )
− ( 6 m )(1589 kN ) − (12 m )( 2543 kN )
− (17 m )(1271 kN ) − (19 m )( 529.7 ) = 0
X = 10.48 m
or
X = 10.48 m W
to the right of A
(c) Consider water section BCD as the free body.
ΣF = 0
Have
−R = 1675 kN
Then
18.43°
or R = 1675 kN
18.43° W
Alternative solution to part (c)
Consider the face BC of the dam.
Have
and
BC =
62 + 182 = 18.9737 m
tan θ =
6
18
(
θ = 18.43°
)(
)
p = ( ρ g ) h = 1000 kg/m3 9.81 m/s 2 (18 m )
= 176.6 kN/m 2
Then
R=
(
1
1
Ap = (18.97 m )(1 m )  176.6 kN/m 2
2
2
= 1675 kN
∴ R = 1675 kN
18.43°
)
PROBLEM 5.76
The cross section of a concrete dam is as shown. For a dam section of
unit width, determine (a) the reaction forces exerted by the ground on the
base AB of the dam, (b) the point of application of the resultant of the
reaction forces of part a, (c) the resultant of the pressure forces exerted
by the water on the face BC of the dam.
SOLUTION
The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (and
behind) the dam.
Note
x1 =
5
(16 ft ) = 10 ft
8
1


x2 = 16 + ( 6 )  ft = 19 ft
2


1


x3 =  22 + (12 )  ft = 25 ft
4


5


x4 =  22 + (12 )  ft = 29.5 ft
8


PROBLEM 5.76 CONTINUED
W = γV
Now
(
)
(
)
(
)
2

W1 = 150 lb/ft 3  (16 ft )( 24 ft ) × (1 ft )  = 38, 400 lb
3


W2 = 150 lb/ft 3 ( 6 ft )( 24 ft ) × (1 ft )  = 21, 600 lb
1

W3 = 150 lb/ft 3  (12 ft )(18 ft ) × (1 ft )  = 10,800 lb
3

(
)
2

W4 = 62.4 lb/ft 3  (12 ft )(18 ft ) × (1 ft )  = 8985.6 lb
3

Also
P=
(
)
1
1
Ap = (18 × 1) ft 2  × 62.4 lb/ft 3 × 18 ft = 10,108.8 lb
2
2
ΣFx = 0: H − 10,108.8 lb = 0
(a) Then
or H = 10.11 kips
W
ΣFy = 0: V − 38, 400 lb − 21, 600 lb − 10,800 lb − 8995.6 lb = 0
V = 79,785.6
or
V = 79.8 kips
W
ΣM A = 0: X ( 79, 785.6 lb ) − ( 6 ft )( 38, 400 lb ) − (19 ft )( 21,600 lb ) − ( 25 ft )(10,800 lb )
(b)
− ( 29.5 ft )( 8985.6 lb ) + ( 6 ft )(10,108.8 lb ) = 0
or
X = 15.90 ft
The point of application of the resultant is 15.90 ft to the right of A W
(c) Consider the water section BCD as the free body.
Have
ΣF = 0
∴ R = 13.53 kips
θ = 41.6°
On the face BD of the dam
R = 13.53 kips
41.6° W
PROBLEM 5.77
The 9 × 12-ft side AB of a tank is hinged at its bottom A and is held in
place by a thin rod BC. The maximum tensile force the rod can withstand
without breaking is 40 kips, and the design specifications require the
force in the rod not exceed 20 percent of this value. If the tank is slowly
filled with water, determine the maximum allowable depth of water d in
the tank.
SOLUTION
Consider the free-body diagram of the side.
Have
Now
P=
1
1
Ap = A (γ d )
2
2
ΣM A = 0:
( 9 ft ) T
−
d
P=0
3
Then, for d max:
( 9 ft ) ( 0.2 ) ( 40 × 103 lb ) −
or
or
(
)
d max  1

3
 (12 ft ) ( d max )  62.4 lb/ft d max  = 0
3 2

3
216 × 103 ft 3 = 374.4 d max
3
d max
= 576.92 ft 3
d max = 8.32 ft W
PROBLEM 5.78
The 9 × 12-ft side of an open tank is hinged at its bottom A and is held
in place by a thin rod. The tank is filled with glycerine, whose specific
weight is 80 lb/ft 3. Determine the force T in the rod and the reactions at
the hinge after the tank is filled to a depth of 8 ft.
SOLUTION
Consider the free-body diagram of the side.
Have
P=
=
(
)
1
( 8 ft )(12 ft )  80 lb/ft 3 ( 8 ft ) = 30, 720 lb
2
ΣFy = 0: Ay = 0
Then
ΣM A = 0:
or
1
1
Ap = A (γ d )
2
2
( 9 ft ) T
8 
−  ft  ( 30, 720 lb ) = 0
3 
T = 9102.22 lb
T = 9.10 kips
W
A = 21.6 kips
W
ΣFx = 0: Ax + 30,720 lb − 9102.22 lb = 0
or
A = −21, 618 lb
PROBLEM 5.79
The friction force between a 2 × 2-m square sluice gate AB and its
guides is equal to 10 percent of the resultant of the pressure forces
exerted by the water on the face of the gate. Determine the initial force
needed to lift the gate that its mass is 500 kg.
SOLUTION
Consider the free-body diagram of the gate.
PI =
Now
(
)(
)
1
1
ApI = ( 2 × 2 ) m 2   103 kg/m3 9.81 m/s 2 ( 3 m ) 


2
2
= 58.86 kN
PII =
(
)(
)
1
1
ApII = ( 2 × 2 ) m 2   103 kg/m3 9.81 m/s 2 ( 5 m ) 


2
2
= 98.10 kN
Then
F = 0.1P = 0.1( PI + PII )
= 0.1( 58.86 + 98.10 ) kN
= 15.696 kN
Finally
(
)
ΣFy = 0: T − 15.696 kN − ( 500 kg ) 9.81 m/s 2 = 0
or T = 20.6 kN W
PROBLEM 5.80
The dam for a lake is designed to withstand the additional force caused
by silt which has settled on the lake bottom. Assuming that silt is
equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 and considering
a 1-m-wide section of dam, determine the percentage increase in the
force acting on the dam face for a silt accumulation of depth 1.5 m.
SOLUTION
First, determine the force on the dam face without the silt.
Pw =
Have
=
1
1
Apw = A ( ρ gh )
2
2
(
)(
)
1
( 6 m )(1 m )   103 kg/m3 9.81 m/s 2 ( 6 m ) 


2
= 176.58 kN
Next, determine the force on the dam face with silt.
Have
Pw′ =
(
)(
)
1
( 4.5 m )(1m )   103 kg/m3 9.81 m/s 2 ( 4.5 m ) 


2
= 99.326 kN
( Ps )I
(
)(
)
= (1.5 m )(1 m )   103 kg/m3 9.81 m/s 2 ( 4.5 m ) 


= 66.218 kN
( Ps )II
=
(
)(
)
1
(1.5 m )(1 m )   1.76 × 103 kg/m3 9.81 m/s 2 (1.5 m ) 


2
= 19.424 kN
Then
P′ = Pw′ + ( Ps )I + ( Ps )II = 184.97 kN
The percentage increase, % inc., is then given by
% inc. =
(184.97 − 176.58) × 100% = 4.7503%
P′ − Pw
× 100% =
176.58
Pw
% inc. = 4.75% W
PROBLEM 5.81
The base of a dam for a lake is designed to resist up to 150 percent of the
horizontal force of the water. After construction, it is found that silt
(which is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is
settling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide
section of dam, determine the number of years until the dam becomes
unsafe.
SOLUTION
From Problem 5.80, the force on the dam face before the silt is deposited, is Pw = 176.58 kN. The maximum
allowable force Pallow on the dam is then:
Pallow = 1.5Pw = (1.5 )(176.58 kN ) = 264.87 kN
Next determine the force P′ on the dam face after a depth d of silt has settled.
Have
Pw′ =
(
)(
)(
)
)
1
( 6 − d ) m × (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m 


2
= 4.905 ( 6 − d ) kN
2
( Ps )I
(
=  d (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m 


(
)
= 9.81 6d − d 2 kN
( Ps )II
=
(
)(
)
1
 d (1 m )   1.76 × 103 kg/m3 9.81 m/s 2 ( d ) m 


2
= 8.6328d 2 kN
(
)
(
)
P′ = Pw′ + ( Ps )I + ( Ps )II =  4.905 36 − 12d + d 2 + 9.81 6d − d 2 + 8.6328d 2  kN


= 3.7278d 2 + 176.58 kN
PROBLEM 5.81 CONTINUED
Now required that P′ = Pallow to determine the maximum value of d.
∴
or
Finally
( 3.7278d
2
)
+ 176.58 kN = 264.87 kN
d = 4.8667 m
4.8667 m = 20 × 10−3
m
×N
year
or N = 243 years W
PROBLEM 5.82
The square gate AB is held in the position shown by hinges along its top
edge A and by a shear pin at B. For a depth of water d = 3.5 m,
determine the force exerted on the gate by the shear pin.
SOLUTION
First consider the force of the water on the gate.
Have
Then
PI =
P=
1
1
Ap = A ( ρ gh )
2
2
(
)(
)
(
)(
)
1
(18 m )2 103 kg/m3 9.81 m/s2 (1.7 m )
2
= 26.99 kN
PII =
1
(18 m )2 103 kg/m3 9.81 m/s2 (1.7 × 1.8cos 30° ) m
2
= 51.74 kN
Now
or
or
ΣM A = 0:
1
2
( LAB ) PI + ( LAB ) PII − LAB FB = 0
3
3
1
2
( 26.99 kN ) + ( 51.74 kN ) − FB = 0
3
3
FB = 43.49 kN
FB = 4.35 kN
30.0° W
PROBLEMS 5.83 AND 5.84
Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh
water channel by nailing two boards to pilings located at the sides of the
channel and propping a third board AB against the pilings and the floor of
the channel. Neglecting friction, determine the reactions at A and B when
rope BC is slack.
Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh
water channel by nailing two boards to pilings located at the sides of the
channel and propping a third board AB against the pilings and the floor of
the channel. Neglecting friction, determine the magnitude and direction
of the minimum tension required in rope BC to move board AB.
SOLUTION
First, consider the force of the water on the gate.
P=
Have
So that
PI =
1
1
Ap = A (γ h )
2
2
(
)
(
)
1
(1.5 ft )( 5 ft )   62.4 lb/ft 3 (1.8 ft ) 


2
= 421.2 lb
PII =
1
(1.5 ft )( 5 ft )   62.4 lb/ft 3 ( 3 ft ) 


2
= 702 lb
5.83
Find the reactions at A and B when rope is slack.
ΣM A = 0: − ( 0.9 ft ) B + ( 0.5 ft )( 421.2 lb ) + (1.0 ft )( 702 lb ) = 0
B = 1014 lb
or
B = 1014 lb
ΣFx = 0: 2 Ax +
W
4
4
( 421.2 lb ) + ( 702 lb ) = 0
5
5
Ax = −449.28 lb
or
Note that the factor 2 (2 Ax ) is included since Ax is the horizontal force
exerted by the board on each piling.
ΣFy = 0: 1014 lb −
or
3
3
( 421.2 lb ) − ( 702 lb ) + Ay = 0
5
5
Ay = −340.08 lb
∴ A = 563 lb
37.1° W
PROBLEMS 5.83 AND 5.84 CONTINUED
5.84
Note that there are two ways to move the board:
1. Pull upward on the rope fastened at B so that the board
rotates about A. For this case B → 0 and TBC is
perpendicular to AB for minimum tension.
2. Pull horizontally at B so that the edge B of the board moves
to the left. For this case Ay → 0 and the board remains
against the pilings because of the force of the water.
Case (1)
ΣM A = 0: −1.5TBC + ( 0.5 ft )( 421.2 lb )
+ (1.0 ft )( 702 lb ) = 0
or
Case (2)
TBC = 608.4 lb
ΣM B = 0: − (1.2 ft ) ( 2 Ax ) − ( 0.5 ft )( 702 lb )
− (1.0 ft )( 421.2 lb ) = 0
or
2 Ax = − 643.5 lb
ΣFx = 0: − TBC − 643.5 lb +
+
or
4
( 421.2 lb )
5
4
( 702 lb ) = 0
5
TBC = 255.06 lb
∴
( TBC )min
= 255 lb
W
PROBLEMS 5.85 AND 5.86
Problem 5.85: A 2 × 3-m gate is hinged at A and is held in position by
rod CD. End D rests against a spring whose constant is 12 kN/m. The
spring is undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will move
to the end of the cylindrical portion of the floor.
Problem 5.86: Solve Problem 5.85 if the mass of the gate is 500 kg.
SOLUTION
First, determine the forces exerted on the gate by the spring and the
water when B is at the end of the cylindrical portion of the floor.
sin θ =
Have
1
2
∴ θ = 30°
Then
xsp = (1.5 m ) tan 30°
and
Fsp = kxsp
= (12 kN/m )(1.5 m ) tan 30°
= 10.39 kN
d ≥ 2m
Assume
P=
Have
PI =
Then
1
1
Ap = A ( ρ g ) h
2
2
(
)(
)
1
( 2 m )( 3 m )   103 kg/m3 9.81 m/s 2 ( d − 2 ) m 


2
= 29.43 ( d − 2 ) kN
PII =
(
)(
)
1
( 2 m )( 3 m )   103 kg/m3 9.81 m/s 2 ( d − 2 + 2cos 30° ) m 


2
= 29.43 ( d − 0.2679 ) kN
PROBLEMS 5.85 AND 5.86 CONTINUED
5.85
Find d min so that gate opens, W = 0.
Using the above free-body diagrams of the gate, we have
2 
ΣM A = 0:  m   29.43 ( d − 2 ) kN 
3 
4 
+  m   29.43 ( d − 0.2679 ) kN 
3 
− (1.5 m )(10.39 kN ) = 0
or
19.62 ( d − 2 ) + 39.24 ( d − 0.2679 ) = 15.585
58.86d = 65.3374
5.86
d = 1.110 m W
d = 1.1105 m
or
Find d min so that the gate opens.
(
)
W = 9.81 m/s 2 ( 500 kg ) = 4.905 kN
Using the above free-body diagrams of the gate, we have
2 
ΣM A = 0:  m   29.43 ( d − 2 ) kN 
3 
4 
+  m   29.43 ( d − 0.2679 ) kN 
3 
− (1.5 m )(10.39 kN ) +
− ( 0.5 m )( 4.905 kN ) = 0
or
or
19.62 ( d − 2 ) + 39.24 ( d − 0.2679 ) = 18.0375
d = 1.15171 m
d = 1.152 m W
PROBLEMS 5.87 AND 5.88
Problem 5.87: The gate at the end of a 3-ft-wide fresh water channel is
fabricated from three 240-lb, rectangular steel plates. The gate is hinged
at A and rests against a frictionless support at D. Knowing that d = 2.5 ft,
determine the reactions at A and D.
Problem 5.88: The gate at the end of a 3-ft-wide fresh water channel is
fabricated from three 240-lb, rectangular steel plates. The gate is hinged
at A and rests against a frictionless support at D. Determine the depth of
water d for which the gate will open.
SOLUTION
5.87
Thus, at
Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged
surfaces ( p = γ h ) .
(
)
h = 0.5 ft, p0.5 = 62.4 lb/ft 3 ( 0.5 ) ft = 31.2 lb/ft 2
(
)
h = 2.5 ft, p2.5 = 6.24 lb/ft 3 ( 2.5 ) ft = 156.0 lb/ft 2
Then
(
)
1
( 0.5 ft )( 3 ft )  31.2 lb/ft 2 = 23.4 lb
2
P1 =
(
)
P2 = ( 2 ft )( 3 ft )  31.2 lb/ft 2 = 187.2 lb
and
or
(
)
(
)
P3 =
1
( 2 ft )( 3 ft )  31.2 lb/ft 2 = 93.6 lb
2
P4 =
1
( 2 ft )( 3 ft )  156 lb/ft 2 = 468 lb
2


1

ΣM A = 0: − 4 D + ( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) −  2 + × 0.5  ft ( 23.4 lb )  − (1 ft )(187.2 lb )
3



2
1
− ( 2 ft )( 93.6 lb ) − ( 2 ft )( 468 lb ) = 0
3
3
D = 11.325 lb
∴ D = 11.33 lb
W
PROBLEMS 5.87 AND 5.88 CONTINUED
ΣFx = 0: Ax + 11.32 + 23.4 + 93.6 + 468 = 0
Ax = −596.32 lb
or
ΣFy = 0: Ay − 240 − 240 − 240 + 187.2 = 0
Ay = 532.8 lb
or
5.88
∴ A = 800 lb
h = ( d − 2 ) ft, pd − 2 = γ ( d − 2 ) lb/ft 2
At
41.8° W
γ = 62.4 lb/ft 3
where
h = d ft, pd = (γ d ) lb/ft 2
P1 =
Then
1
1
3
2
A1 pd − 2 = ( d − 2 ) ft × ( 3 ft )  γ lb/ft 3 ( d − 2 ) ft  = γ ( d − 2 ) lb
2
2
2
(Note: For simplicity, the numerical value of the density γ will be substituted into the equilibrium equations
below, rather than at this level of the calculations.)
P2 = A2 pd − 2 = ( 2 ft )( 3 ft )  γ ( d − 2 ) ft  = 6γ ( d − 2 ) lb
P3 =
P4 =
1
1
A3 pd − 2 = ( 2 ft )( 3 ft )  γ ( d − 2 ) ft  = 3γ ( d − 2 ) lb
2
2
1
1
A4 pd = ( 2 ft )( 3 ft )  γ ( d ft ) = ( 3γ d ) lb = 3γ ( d − 2 ) + 6γ  lb
2
2
As the gate begins to open, D → 0
∴
ΣM A = 0:
( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) − 2 ft + ( d − 2 ) ft  

3
2 
γ ( d − 2 ) lb  +
2

1
3
2

− (1 ft ) 6γ ( d − 2 ) lb  −  ( 2 ft )  3γ ( d − 2 ) lb 
3

1

−  ( 2 ft )  3γ ( d − 2 ) lb + 6γ lb  = 0
3

or
1
720
−4
( d − 2 )3 + 3 ( d − 2 )2 + 12 ( d − 2 ) =
γ
2
=
720
−4
62.4
= 7.53846
Solving numerically yields
d = 2.55 ft W
PROBLEM 5.89
A rain gutter is supported from the roof of a house by hangers that are
spaced 0.6 m apart. After leaves clog the gutter’s drain, the gutter slowly
fills with rainwater. When the gutter is completely filled with water,
determine (a) the resultant of the pressure force exerted by the water on
the 0.6-m section of the curved surface of the gutter, (b) the force-couple
system exerted on a hanger where it is attached to the gutter.
SOLUTION
(a) Consider a 0.6 m long parabolic section of water.
Then
P=
=
1
1
Ap = A ( ρ gh )
2
2
(
)(
)
1
( 0.08 m )( 0.6 m )  103 kg/m3 9.81 m/s2 ( 0.08 m )
2
= 18.84 N
Ww = ρ gV
(
)(
)
2

= 103 kg/m3 9.81 m/s 2  ( 0.12 m )( 0.08 m )( 0.6 m ) 
3

= 37.67 N
ΣF = 0:
Now
So that R =
( −R ) + P + Ww
tan θ =
P 2 + Ww2 ,
=0
Ww
P
= 42.12 N, θ = 63.4°
R = 42.1 N
63.4° W
(b) Consider the free-body diagram of a 0.6 m long section of water and
gutter.
Then
ΣFx = 0: Bx = 0
ΣFy = 0: By − 37.67 N = 0
or
By = 37.67 N
ΣM B = 0: M B + ( 0.06 − 0.048) m  ( 37.67 N ) = 0
or
M B = −0.4520 N ⋅ m
The force-couple system exerted on the hanger is then
37.7 N , 0.452 N ⋅ m
W
PROBLEM 5.90
The composite body shown is formed by removing a hemisphere of
radius r from a cylinder of radius R and height 2R. Determine (a) the
y coordinate of the centroid when r = 3R/4, (b) the ratio r/R for which
y = −1.2 R.
SOLUTION
Note, for the axes shown
V
1
(π R ) ( 2R ) = 2π R
2
2
− π r3
3
Σ

r3 
2π  R3 − 
3

2
y
yV
−R
−2π R 4
3
− r
8
1 4
πr
4
3

r4 
−2π  R 4 − 
8 

1
R4 − r 4
Σ yV
8
Y =
=−
1
ΣV
3
R − r3
3
Then
1 r 
 
8 R 
4
1 r 
1−  
3 R 
3
1−
=
r =
(a )
3
R: y = −
4
1−
1 3
 
3 4
4
3
1 3
1−  
3 4
R
or y = −1.118R W
1 r 
 
8 R 
4
1 r 
1−  
3 R 
3
1−
y = −1.2 R : − 1.2 R = −
(b)
4
or
Solving numerically
R
3
r
r
  − 3.2   + 1.6 = 0
R
R
r
= 0.884 W
R
PROBLEM 5.91
Determine the y coordinate of the centroid of the body shown.
SOLUTION
First note that the values of Y will be the same for the given body and the body shown below. Then
V
y
yV
Cone
1 2
πa h
3
1
− h
4
1
− π a 2h 2
12
2
Cylinder
1
a
−π   b = − π a 2b
2
4
 
1
− b
2
1 2 2
πa b
8
Σ
Have
Then
π
12
a 2 ( 4h − 3b )
−
π
24
(
a 2 2h 2 − 3b 2
)
Y ΣV = ΣyV
π
π

Y  a 2 ( 4h − 3b )  = − a 2 2h 2 − 3b 2
24
12

(
)
or Y = −
2h 2 − 3b 2
W
2 ( 4h − 3b )
PROBLEM 5.92
Determine the z coordinate of the centroid of the body shown. (Hint:
Use the result of Sample Problem 5.13.)
SOLUTION
First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.
V
1 2
πa h
6
Half-Cone
Half-Cylinder
Σ
−
π a
2
−
π
2
  b=− ab
22
8
π
24
−
z
a
π
zV
1 3
− ah
6
*
4 a
2a
 =−
3π  2 
3π
a 2 ( 4h − 3b )
1 3
ab
12
−
1 3
a ( 2h − b )
12
From Sample Problem 5.13
Have
Then
Z ΣV = ΣzV
1
π

Z  a 2 ( 4h − 3b )  = − a3 ( 2h − b )
24
12


or Z = −
2a 2h − b
W
π 4h − 3b
PROBLEM 5.93
Consider the composite body shown. Determine (a) the value of x when
h = L/2 , (b) the ratio h/L for which x = L .
SOLUTION
x
1
L
2
1
L+ h
4
V
Rectangular prism
Lab
Pyramid
1 b
a h
3 2
Then
1 

ΣV = ab  L + h 
6 

ΣxV =
X ΣV = ΣxV
Now
xV
1 2
L ab
2
1
1 

abh  L + h 
6
4 

1  2
1 

ab 3L + h  L + h  
6 
4 

so that
 
1  1 
1 
X  ab  L + h   = ab  3L2 + hL + h 2 
6  6 
4 
 
1 h 1 
h 1 h2 

X 1 +

 = L  3 + +
6 L 6 
L 4 L2 

or
(1)
1
L
2
h
1
Substituting
=
into Eq. (1)
L
2
(a) X = ? when h =
2

1  1  1 
1 11 
X 1 +    = L 3 +   +   
6  2   6 
 2  4  2  

or
(b)
X =
57
L
104
X = 0.548L W
h
= ? when X = L
L
Substituting into Eq. (1)
or
or
1 h 1 
h 1 h2 

L 1 +
=
L
3
+
+



6 L  6 
L 4 L2 

1+
1h
1 1h
1 h2
= +
+
6L
2 6 L 24 L2
h2
= 12
L2
∴
h
= 2 3W
L
PROBLEMS 5.94 AND 5.95
Problem 5.94: For the machine element shown, determine the
x coordinate of the center of gravity.
Problem 5.95: For the machine element shown, determine the
y coordinate of the center of gravity.
SOLUTIONS
First, assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the corresponding volume.
x , in.
y , in.
xV , in 4
yV , in 4
I
II
V , in 3
(4)(3.6)(0.75) = 10.8
(2.4)(2.0)(0.6) = 2.88
2.0
3.7
0.375
1.95
21.6
10.656
4.05
5.616
III
π(0.45)2 (0.4) = 0.2545
4.2
2.15
1.0688
0.54711
IV
−π(0.5) 2 (0.75) = − 0.5890
1.2
0.375
−0.7068
−0.22089
Σ
13.3454
32.618
9.9922
5.94
X ΣV = Σ x V
Have
(
)
X 13.3454 in 3 = 32.618 in 4
or X = 2.44 in. W
5.95
Y ΣV = Σ yV
Have
(
)
Y 13.3454 in 3 = 9.9922 in 4
or Y = 0.749 in. W
PROBLEMS 5.96 AND 5.97
Problem 5.96: For the machine element shown, locate the x coordinate
of the center of gravity.
Problem 5.97: For the machine element shown, locate the y coordinate
of the center of gravity.
SOLUTIONS
First, assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
xV , mm 4
yV , mm 4
1
(160 )( 54 )(18) = 155 520
80
9
12 441 600
1 399 680
2
1
(120 )( 42 )( 54 ) = 136 080
2
40
32
5 443 200
4 354 560
9
3 534 114
185 508
9
−2 316 233
−130 288
19 102 681
5 809 460
3
π
2
(27) 2 (18) = 6561π
4
−π(16)2 (18) = −4608π
Σ
297 736
160 +
36
π
160
5.96
X ΣV = Σ x V
Have
(
)
X 297 736 mm3 = 19 102 681 mm 4
or X = 64.2 mm W
5.97
Y ΣV = Σ yV
Have
(
)
Y 297 736 mm3 = 5 809 460 mm 4
or Y = 19.51 mm W
PROBLEMS 5.98 AND 5.99
Problem 5.98: For the stop bracket shown, locate the x coordinate of the
center of gravity.
Problem 5.99: For the stop bracket shown, locate the z coordinate of the
center of gravity.
SOLUTIONS
First, assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the
corresponding volume.
Z II = 24 mm +
1
(90 + 86) mm = 112 mm
2
Z III = 24 mm +
1
(102) mm = 58 mm
3
X III = 68 mm +
1
( 20) mm = 78 mm
2
Have..
Z IV = 110 mm +
2
( 90) mm = 170 mm
3
X IV = 60 mm +
2
(132) mm = 156 mm
3
V , mm3
x , mm
z , mm
xV , mm 4
zV , mm 4
I
( 200 )(176 )( 24 ) = 844 800
100
12
84 480 000
1 013 760
II
( 200 )( 24 )(176 ) = 844 800
100
112
84 480 000
94 617 600
78
58
9 865 440
733 840
156
170
−22 239 360
−24 235 200
156 586 080
8 785 584
III
IV
Σ
1
( 20 )(124 )(102 ) = 126 480
2
1
− ( 90 )(132 )( 24 ) = −142 560
2
1 673 520
5.98
X ΣV = ΣxV
Have
(
)
X 1 673 520 mm3 = 156 586 080 mm 4
or X = 93.6 mm W
5.99
Z ΣV = ΣzV
Have
(
)
Z 1 673 520 mm3 = 8 785 584 mm 4
or Z = 52.5 mm W
PROBLEM 5.100
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity coincides with the centroid of
the corresponding area.
A, mm 2
1
2
3
1
( 90 )( 60 )
2
= 2700
( 90 )( 200 )
= 18 000
− ( 45 )(100 )
= −4500
π
( 45)2
4
2
= 1012.5π
Σ
Have
19 380.9
x , mm
30
y , mm
z , mm
xA, mm3
yA, mm3
zA, mm3
0
81 000
378 000
0
120 + 20
= 140
45
60
80
810 000
1 080 000
1 440 000
22.5
30
120
−101 250
−135 000
−540 000
143 139
0
569 688
932 889
1 323 000
1 469 688
45
0
160 +
( 4 )( 45)
3π
= 179.1
X ΣA = ΣxA:
(
)
X 19 380.9 mm 2 = 932 889 mm3
= 48.1 mm
or
X = 48.1 mm W
Y ΣA = Σ yA
(
)
Y 19 380.9 mm 2 = 1 323 000 mm3
Y = 68.3 mm
or
Y = 68.3 mm W
Z ΣA = Σ zA
(
)
Z 19 380.9 mm 2 = 1 469 688 mm3
or
Z = 75.8 mm
Z = 75.8 mm W
PROBLEM 5.101
A mounting bracket for electronic components is formed from sheet
metal of uniform thickness. Locate the center of gravity of the bracket.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with
the centroid of the corresponding area. Then (see diagram)
zV = 22.5 −
4 ( 6.25 )
3π
= 19.85 mm
π
AV = −
2
( 6.25)2
= −61.36 mm 2
A, mm 2
IV
( 25)( 60) = 1500
(12.5)( 60 ) = 750
( 7.5)( 60 ) = 450
− (12.5 )( 30 ) = −375
V
Σ
−61.36
2263.64
I
II
III
x , mm
y , mm
z , mm
xA, mm3
yA, mm3
zA, mm3
12.5
0
30
18 750
0
45 000
25
−6.25
30
18 750
−4687.5
22 500
28.75
−12.5
30
12 937.5
−5625
13 500
10
0
37.5
−3750
0
−14 062.5
10
0
19.85
−613.6
46 074
0
−10 313
−1218.0
65 720
X ΣA = ΣxA
Have
(
)
X 2263.64 mm 2 = 46 074 mm3
or X = 20.4 mm W
Y ΣA = Σ yA
(
)
Y 2263.64 mm 2 = −10 313 mm3
or Y = −4.55 mm W
Z ΣA = Σ zA
(
)
Z 2263.64 mm 2 = 65 720 mm3
or Z = 29.0 mm W
PROBLEM 5.102
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the
centroid of the corresponding area.
yI = 6 +
zI =
4
= 7.333 in.
3
1
( 6) = 2 in.
3
1
xII = yII =
xIV = 11 −
I
II
III
IV
Σ
A, in 2
12
56.55
30
−3.534
95.01
x , in.
y , in.
z , in.
0
3.8197
8.5
10.363
7.333
3.8197
0
0
2
3
3
3
π
( 2)( 6) = 3.8197 in.
1
( 4)(1.5) = 10.363 in.
3π
x A, in 3
0
216
255
−36.62
434.4
y A, in 3
88
216
0
0
304
z A, in 3
24
169.65
90
−10.603
273.0
X ΣA = Σ x A
Have
(
)
X 95.01 in 2 = 434 in 3
or X = 4.57 in. W
Y ΣA = Σ y A
(
)
Y 95.01 in 2 = 304.0 in 3
or Y = 3.20 in. W
Z ΣA = Σ z A
(
)
Z 95.01 in 2 = 273.0 in 3
or Z = 2.87 in. W
PROBLEM 5.103
An enclosure for an electronic device is formed from sheet metal of
uniform thickness. Locate the center of gravity of the enclosure.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the
centroid of the corresponding area.
Consider the division of the back, sides, and top into eight segments according to the sketch.
Z = 6.00 in. W
Note that symmetry implies
and
A8 = A2
A7 = A 3
A6 = A5
Thus
A, in 2
1
2
3
4
5
6
7
8
Σ
Have
(12 )( 9 ) = 108
(11.3)( 9 ) = 101.7
1
(11.3)( 2.4 ) = 13.56
2
(12 )(11.454 ) = 137.45
1
(1.5)(11.454 ) = 8.591
2
8.591
13.56
101.7
493.2
x , in.
y , in.
xA, in 3
yA, in 3
0
4.5
0
486
5.6
4.5
569.5
457.6
7.467
9.8
101.25
132.89
5.6
10.2
769.72
1402.0
7.467
10.6
64.15
91.06
7.467
7.467
5.6
10.6
9.8
4.5
64.15
101.25
569.5
2239.5
91.06
132.89
457.6
3251.1
(
)
Y ΣA = Σ yA: Y ( 493.2 in ) = 3251.1 in
X ΣA = ΣxA: X 493.2 in 2 = 2239.5 in 3
2
3
or X = 4.54 in. W
or Y = 6.59 in. W
PROBLEM 5.104
A 200-mm-diameter cylindrical duct and a 100 × 200-mm rectangular
duct are to be joined as indicated. Knowing that the ducts are fabricated
from the same sheet metal, which is of uniform thickness, locate the center
of gravity of the assembly.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also note that symmetry implies
Z = 0W
1
2
3
4
5
6
x, m
y, m
xA, m3
yA, m3
π ( 0.2 )( 0.3) = 0.1885
0
0.15
0
0.028274
= 0.06366
0.25
−0.02000
−0.007854
−4 ( 0.1)
= −0.04244
3π
0.30
−0.000667
0.004712
0.15
0.30
0.00900
0.001800
0.15
0.20
0.00900
0.001200
4 ( 0.1)
3π
0.20
−0.000667
−0.003142
0.15
0.25
0.004500
0.007500
0.15
0.25
0.004500
0.023667
0.007500
0.066991
−
π
2
( 0.2 )( 0.1) = −0.0314
2 ( 0.1)
π
π
2
( 0.1) = 0.01571
2
( 0.3)( 0.2 ) = 0.060
( 0.3)( 0.2 ) = 0.060
−
π
2
( 0.1)2
= −0.1571
8
( 0.3)( 0.1) = 0.030
( 0.3)( 0.1) = 0.030
Σ
0.337080
7
Have
A, m 2
X ΣA = Σ xA: X (0.337080 mm 2 ) = 0.023667 mm3
X = 0.0702 m
or
(
X = 70.2 mm W
)
Y ΣA = Σ yA: Y 0.337080 mm 2 = 0.066991 mm3
or
Y = 0.19874 m
Y = 198.7 mm W
PROBLEM 5.105
An elbow for the duct of a ventilating system is made of sheet metal of
uniform thickness. Locate the center of gravity of the elbow.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also, note that the shape of the duct implies
Y = 1.5 in. W
xI = zI = 16 in. −
Note that
xII = 16 in. −
zII = 12 in. −
2
π
2
π
2
π
(16 in.) = 5.81408 in.
(8 in.) = 10.9070 in.
(8 in.) = 6.9070 in.
xIV = zIV = 16 in. −
4
(16 in.) = 9.2094 in.
3π
4
(8 in.) = 12.6047 in.
3π
4
zV = 12 in. −
(8 in.) = 8.6047 in.
3π
xV = 16 in. −
Also note that the corresponding top and bottom areas will contribute equally when determining x and z .
Thus
A, in 2
I
II
Have
π
2
(16 )( 3) = 75.3982
π
2
(8)( 3) = 37.6991
xA, in 3
zA, in 3
x , in.
z , in.
5.81408
5.81408
438.37
438.37
10.9070
6.9070
411.18
260.39
III
4 ( 3) = 12
8
14
96.0
168.0
IV
 π
2
2   (16) = 402.1239
 4
9.2094
9.2094
3703.32
3703.32
V
π 2
−2   ( 8 ) = −100.5309
4
12.6047
8.6047
−1267.16
−865.04
VI
−2 ( 4 )( 8 ) = −64
12
14
−768.0
−896.0
Σ
362.69
2613.71
2809.04
(
)
X ΣA = Σ xA: X 362.69 in 2 = 2613.71 in 3
or X = 7.21 in. W
(
)
Z ΣA = Σ zA: Z 362.69 in 2 = 2809.04 in 3
or Z = 7.74 in. W
PROBLEM 5.106
A window awning is fabricated from sheet metal of uniform thickness.
Locate the center of gravity of the awning.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with
the centroid of the corresponding area.
yII = yVI = 80 +
zII = zVI =
yIV = 80 +
zIV =
π
π
2
π
4
= 292.2 mm
= 212.2 mm
= 398.3 mm
= 318.3 mm
( 500 )2
= 196 350 mm 2
( 500 )( 680 ) = 534 071 mm2
yA, mm3
zA, mm3
A, mm 2
(80)(500) = 40 000
y , mm
40
250
1.6 × 106
10 × 106
292.2
212.2
57.4 × 106
41.67 × 106
III
196 350
(80)(680) = 54 400
40
500
0.2176 × 106
27.2 × 106
IV
534 071
398.3
318.3
212.7 × 106
170 × 106
V
(80)(500) = 40 000
40
250
1.6 × 106
10 × 106
VI
196 350
292.2
212.2
57.4 × 106
Σ
1.061 × 106
I
II
Now, symmetry implies
and
( 4 )( 500 )
( 2 )( 500 )
π
3π
3π
( 2 )( 500 )
AII = AVI =
AIV =
( 4 )( 500 )
z , mm
332.9 × 106
(
Y ΣA = Σ yA: Y 1.061 × 10 mm
6
2
) = 332.9 × 10
41.67 × 106
300.5 × 106
X = 340 mm W
6
mm
3
or Y = 314 mm W
(
)
Z ΣA = Σ zA: Z 1.061 × 106 mm 2 = 300.5 × 106 mm3
or Z = 283 mm W
PROBLEM 5.107
The thin, plastic front cover of a wall clock is of uniform thickness.
Locate the center of gravity of the cover.
SOLUTION
First, assume that the plastic is homogeneous so that the center of gravity of the cover coincides with the
centroid of the corresponding area.
X = 150.0 mm
Next, note that symmetry implies
A, mm 2
1
2
3
4
5
6
7
Σ
Have
( 300 )( 280 )
= 84 000
( 280 )( 50 )
= 14 000
( 300 )( 50 )
= 15 000
( 280 )( 50 )
= 14 000
−π (100 )
y A, mm3
z , mm3
y , mm
z , mm
0
140
0
11 760 000
25
140
350 000
1 960 000
25
0
375 000
0
25
140
350 000
1 960 000
0
130
0
−4 084 070
= 247.29
0
−174 783
= 247.29
0
−174 783
1 075 000
11 246 363
2
= −31 416
−π
( 30 )2
4
= −706.86
−π
( 30 )2
4
= −706.86
0
260 −
0
260 −
( 4 )( 30 )
3π
( 4 )( 30 )
3π
94 170
(
)
Y Σ A = Σ yA: Y 94 170 mm 2 = 1 075 000 mm3
or Y = 11.42 mm W
(
)
Z Σ A = Σ zA: Z 94 170 mm 2 = 11 246 363 mm3
or Z = 119.4 mm W
PROBLEM 5.108
A thin steel wire of uniform cross section is bent into the shape shown,
where arc BC is a quarter circle of radius R. Locate its center of gravity.
SOLUTION
First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the
corresponding line.
L, in 2
x , in.
y , in.
z ,in.
x L, in 2
y L, in 2
z L, in 2
1
15
0
7.5
0
0
112.5
0
2
14
7
0
0
98
0
0
13
 5
9 
 12 
0
6
149.5
0
78
135.0
225.0
180.0
382.5
337.5
258.0
3
π
4
Σ
Have
(15)
2
= 23.56
= 11.5
3  2 × 15 


5 π 
30
24
π
π
= 5.73
= 9.549
= 7.639
65.56
X ΣL = Σ x L: X ( 65.56 in.) = 382.5 in 2
or X = 5.83 in. W
Y ΣL = Σ y L: Y ( 65.56 in.) = 337.5 in 2
or Y = 5.15 in. W
Z ΣL = Σ z L: Z ( 65.56 in.) = 258.0 in 2
or Z = 3.94 in. W
PROBLEM 5.109
A thin steel wire of uniform cross section is bent into the shape shown,
where arc BC is a quarter circle of radius R. Locate its center of gravity.
SOLUTION
First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the
corresponding line
x2 = z2 =
Have
L1 =
( 2 )(8)
π
=
16
π
ft
82 + 32 = 8.5440 ft
8π
L 2 = 2 = 4π ft
4
Have
x L, ft 2
34.176
y L, ft 2
12.816
z L, ft 2
0
π
64.0
0
64.0
4
0
0
0
98.176
0
4.5
17.316
32
0
96.0
L, ft
x , ft
y , ft
z , ft
1
8.5440
4
1.5
2
4π
16π
0
0
16
3
4
8
3
32.110
0
0
0
1.5
X ΣL = Σ x L: X ( 32.110 ft ) = 98.176 ft 2
or X = 3.06 ft W
Y ΣL = Σ y L: Y ( 32.110 ft ) = 17.316 ft 2
or Y = 0.539 ft W
Z ΣL = Σ z L: Z ( 32.110 ft ) = 96.0 ft 2
or Z = 2.99 ft W
PROBLEM 5.110
The frame of a greenhouse is constructed from uniform aluminum
channels. Locate the center of gravity of the portion of the frame shown.
SOLUTION
First, assume that the channels are homogeneous so that the center of gravity of the frame coincides with the
centroid of the corresponding line.
Note
x8 = x9 =
( 2 )( 0.9 )
π
y8 = y9 = 1.5 +
L 7 = L8 =
1
2
3
4
5
6
7
8
9
10
Σ
Have
L, m
x, m
y, m
z, m
0.6
0.9
1.5
1.5
2.4
0.6
0.9
1.4137
1.4137
0.6
11.827
0.9
0.45
0.9
0.9
0
0.9
0.45
0.573
0.573
0
0
0
0.75
0.75
1.2
1.5
1.5
2.073
2.073
2.4
0.3
0.6
0
0.6
0.6
0.3
0.6
0
0.6
0.3
xL, m 2
0.540
0.4050
1.350
1.350
0
0.540
0.4050
0.8100
0.8100
0
6.210
π
2
= 0.57296 m
( 2 )( 0.9 )
π
= 2.073 m
( 0.9 ) = 1.4137 m
yL, m 2
0
0
1.125
1.125
2.880
0.9
1.350
2.9306
2.9306
1.440
14.681
zL, m 2
0.18
0.54
0
0.9
1.44
0.18
0.54
0
0.8482
0.18
4.8082
X ΣL = Σ xL: X (11.827 m ) = 6.210 m 2
or X = 0.525 m W
Y ΣL = Σ yL: Y (11.827 m ) = 14.681 m 2
or Y = 1.241 m W
Z ΣL = Σ zL: Z (11.827 m ) = 4.8082 m 2
or Z = 0.406 m W
*
PROBLEM 5.111
The decorative metalwork at the entrance of a store is fabricated from
uniform steel structural tubing. Knowing that R = 1.2 m, locate the center
of gravity of the metalwork.
SOLUTION
First, assume that the tubes are homogeneous so that the center of gravity of the metalwork coincides with the
centroid of the corresponding line.
Z = 0W
Note that symmetry implies
Have
y, m
xL, m 2
yL, m 2
1.5
2.5456
4.5
1.5
2.5456
4.5
3.7639
0
14.1897
= 0.7639
3
2.88
11.3097
= 0.7639
3.7639
1.44
7.0949
9.4112
41.594
x, m
L, m
1
3
2
3
(1.2) cos 45° = 0.8485
(1.2) cos 45° = 0.8485
3
1.2π
0
4
1.2π
5
0.6π
Σ
15.425
( 2 )(1.2 )
π
( 2 )(1.2 )
π
X ΣL = Σ xL: X (15.425 m ) = 9.4112 m 2
or X = 0.610 m W
Y ΣL = Σ yL: Y (15.425 m ) = 41.594 m 2
or Y = 2.70 m W
PROBLEM 5.112
A scratch awl has a plastic handle and a steel blade and shank. Knowing
that the specific weight of plastic is 0.0374 lb/in 3 and of steel is
0.284 lb/in 3, locate the center of gravity of the awl.
SOLUTION
Y = Z = 0W
First, note that symmetry implies
xI =
5
2π
( 0.5 in.) = 0.3125 in., WI = 0.0374 lb/in 3   ( 0.5 in.)3 = 0.009791 lb
8
 3 
(
)
(
)
xII = 1.6 in. + 0.5 in. = 2.1 in. WII = 0.0374 lb/in 3 (π )( 0.5 in.) ( 3.2 in.) = 0.093996 lb
2
2
π 
xIII = 3.7 in. − 1 in. = 2.7 in., WIII = − 0.0374 lb/in 3   ( 0.12 in.) ( 2 in.) = −0.000846 lb
4
(
)
2
2
π 
xIV = 7.3 in. − 2.8 in. = 4.5 in., WIV = 0.284 lb/in 3   ( 0.12 in.) ( 5.6 in.) = 0.017987 lb
4
(
xV = 7.3 in. +
1
π
( 0.4 in.) = 7.4 in., WV = 0.284 lb/in 3   ( 0.06 in.)2 ( 0.4 in.) = 0.000428 lb
4
3
(
I
II
III
IV
V
Σ
Have
)
W , lb
0.009791
0.093996
−0.000846
0.017987
0.000428
0.12136
)
x , in.
0.3125
2.1
2.7
4.5
7.4
xW , in ⋅ lb
0.003060
0.197393
−0.002284
0.080942
0.003169
0.28228
X ΣW = ΣxW : X ( 0.12136 lb ) = 0.28228 in.⋅ lb
or X = 2.33 in. W
PROBLEM 5.113
A bronze bushing is mounted inside a steel sleeve. Knowing that the
density of bronze is 8800 kg/m3 and of steel is 7860 kg/m3 , determine
the center of gravity of the assembly.
SOLUTION
X = Z = 0W
First, note that symmetry implies
W = ( ρ g )V
Now
 π 

yI = 4 mm , WI = 7860 kg/m3 9.81 m/s2    0.0362 − 0.0152 m 2  ( 0.008 m ) 


 4 

(
)(
)
(
)
= 0.51887 N
 π 

yII = 18 mm, WII = 7860 kg/m3 9.81 m/s 2    0.02252 − 0.052 m 2  ( 0.02 m ) 

 4  

(
)(
)
(
)
= 0.34065 N
 π 

yIII = 14 mm, WIII = 8800 kg/m3 9.81 m/s2    0.152 − 0.102 m 2  ( 0.028 m ) 


 4 

(
)(
)
(
)
= 0.23731 N
Y ΣW = Σ yW
Have
Y =
( 4 mm)( 0.5189 N ) + (18 mm)( 0.3406 N ) + (14 mm )( 0.2373 N )
0.5189 N + 0.3406 N + 0.2373 N
or Y = 10.51 mm W
( above base )
PROBLEM 5.114
A marker for a garden path consists of a truncated regular pyramid carved
from stone of specific weight 160 lb/ft3. The pyramid is mounted on a
steel base of thickness h. Knowing that the specific weight of steel is
490 lb/ft3 and that steel plate is available in 14 in. increments, specify the
minimum thickness h for which the center of gravity of the marker is
approximately 12 in. above the top of the base.
SOLUTION
First, locate the center of gravity of the stone. Assume that the stone is
homogeneous so that the center of gravity coincides with the centroid of
the corresponding volume.
Have
y1 =
3
( 56 in.) = 42 in.,
4
V1 =
1
(12 in.)(12 in.)( 56 in.)
3
= 2688 in 3
y2 =
3
( 28 in.) = 21 in.,
4
V2 = −
1
( 6 in.)( 6 in.)( 28 in.)
3
= −366 in 3
Vstone = 2688 in 3 − 366 in 3
Then
= 2352 in 3
and
Y =
=
ΣyV
ΣV
( 42 in.) ( 2688 in 3 ) + ( 21 in.) ( −366 in 3 )
2352 in 3
= 45 in.
Therefore, the center of gravity of the stone is ( 45 − 28 ) in. = 17 in.
above the base.
Now
(
Wstone = γ stoneVstone = 160 lb/ft
3
)(
 1 ft 
2352 in 

 12 in. 
3
)
= 217.78 lb
Wsteel = γ steelVsteel
 1 ft 
= 490 lb/ft (12 in.)(12 in.) h  

 12 in. 
(
3
)
= ( 40.833h ) 1b
3
3
PROBLEM 5.114 CONTINUED
Then
Ymarker =
=
or
ΣyW
= 12 in.
ΣW
(17 in.)( 217.78 lb ) +  −
h 
in.  ( 40.833 h ) lb
 2 
( 217.78 + 40.833h ) lb
h 2 + 24h − 53.334 = 0
With positive solution h = 2.0476 in.
∴ specify h = 2 in. W
PROBLEM 5.115
The ends of the park bench shown are made of concrete, while the seat
and back are wooden boards. Each piece of wood is 36 × 120 ×
1180 mm. Knowing that the density of concrete is 2320 kg/m3 and of
wood is 470 kg/m3 , determine the x and y coordinates of the center of
gravity of the bench.
SOLUTION
First, note that we will account for the two concrete ends by counting twice the weights of components 1, 2,
and 3.
(
)(
)
W1 = ( ρc g )V1 = 2320 kg/m3 9.81 m/s 2 ( 0.480 m )( 0.408 m )( 0.072 m ) 
= 320.9 N
(
)(
)
W2 = − ( ρc g )V2 = − 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.048 m )( 0.072 m ) 
= −7.551 N
(
)(
)
W3 = ( ρc g )V3 = 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.384 m )( 0.072 m ) 
= 60.41 N
W4 = W5 = W6 = W7 = ρ wVboard
(
)(
)
= 470 kg/m3 9.81 m/s 2 ( 0.120 m )( 0.036 m )(1.180 m ) 
= 23.504 N
PROBLEM 5.115 CONTINUED
W, N
x , mm
y , mm
x W, mm ⋅ N
y W, mm ⋅ N
1
2 ( 320.4 ) = 641.83
312
−204
200 251.4
−130 933.6
2
2 ( −7.551) = −15.10
312
−384
−4711.8
5799.1
84
228
360
442
124.7
160.1
192
18
18
18
328.3
139.6
10 148.5
5358.8
8461.3
10 388.5
2930.9
3762.9
236 590
23 196.5
423.1
423.1
423.1
7716.2
3281.1
−89 671
3
4
5
6
7
8
Σ
Have
2 ( 60.41) = 120.82
23.504
23.504
23.504
23.504
23.504
865.06
X ΣW = Σ xW : X ( 865.06 N ) = 236 590 mm ⋅ N
or X = 274 mm W
Y ΣW = Σ yW : Y ( 865.06 N ) = −89 671 mm ⋅ N
or Y = −103.6 mm W
PROBLEM 5.116
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A hemisphere.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is x 2 + y 2 = a 2 so that
r 2 = a 2 − x 2 and then
(
)
dV = π a 2 − x 2 dx
Component 1
a/2
π
0
V1 = ∫
2
)
11 3
πa
24
=
and
(
a/2

x3 
a − x dx = π a 2 x − 
3 0

2
a/2
2
2
∫1 xEL dV = ∫0 x π ( a − x ) dx 
a/2
 x2 x4 
= π a 2
−

4 0
 2
=
Now
7
π a4
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a3  =
π a4
24
64


or x1 =
Component 2
a
π
a /2
V2 = ∫
(
a

x3 
a − x dx = π  a 2 x − 
3  a/2

2
2
)


a3 

a
= π a 2 ( a ) −  − a2   −

3

   2 

5
=
πa 3
24
3 
( a2 )
 
3 
 
21
aW
88
PROBLEM 5.116 CONTINUED
and
a
 2 x2 x4 
a
2
2
∫2 xELdV = ∫a/2 x π a − x dx  = π  a 2 − 4 

 a/2
(
)
2

2
4

a )   2 a2
(
 2 (a)


=π a
−
− a
−
2
4  
2
 




9
=
π a4
64
( )
Now

 
4 
 
( a2 )
4
9
 5

x2V2 = ∫2 xEL dV : x2  π a3  =
π a4
24
64


or x2 =
27
aW
40
PROBLEM 5.117
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A semiellipsoid of revolution.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is
r2 =
(
x2
y2
+
= 1 so that
h2 a 2
)
a2 2
h − x 2 and then
h2
dV = π
(
)
a2 2
h − x 2 dx
h2
Component 1
h/2
π
0
V1 = ∫
=
and
h/2
a2 2
a2  2
x3 
2
−
=
−
π
h
x
dx
h
x


3 0
h2
h2 
(
)
11 2
πa h
24
 a2

h/2
2
2
∫1 xEL dV = ∫0 x π h 2 ( h − x ) dx 


a2
=π 2
h
h/2
 2 x2 x4 
− 
h
4 0
 2
7
π a 2h 2
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a 2h  =
π a 2h 2
 24
 64
=
Now
or x1 =
21
hW
88
PROBLEM 5.117 CONTINUED
Component 2
h
V2 = ∫h/2 π
(
)
a2 2
a2
h − x 2 dx = π 2
2
h
h
h
 2
x3 
h x − 
3 h/2

()
3

3

h 
h)   2  h 
a 2  2
(
2 
− h   −
= π 2 h ( h ) −

3    2
3 
h 






5
=
πa 2 h
24
and
 a2

h
2
2
∫2 xELdV = ∫h/2 x π h2 ( h − x ) dx 


=π
a2
h2
h
 2 x2 x4 
−
h

4  h/2
 2
2

2
4

h )   2 h2
(
a2  2 ( h )
− h
= π 2 h
−
−
2
4  
2
h 




9
=
π a 2h 2
64
( )
Now

 
4 
 
( h2 )
4
9
 5

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h 2
24
64


or x2 =
27
hW
40
PROBLEM 5.118
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A paraboloid of revolution.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is x = h −
r2 =
h 2
y so that
a2
a2
( h − x ) and then
h
dV = π
a2
( h − x ) dx
h
V1 = ∫0 π
a2
( h − x ) dx
h
Component 1
h/2
=π
=
and
a2
h
h/2

x2 
 hx −

2 0

3 2
πa h
8
2

h/2  a
=
x
dV
x
π
∫1 EL
∫0  h ( h − x ) dx 


h/2
a 2  x 2 x3 
=π
− 
h
3 0
h  2
=
Now
1
π a 2h 2
12
1
3

x1V1 = ∫1 xEL dV : x1  π a 2h  = π a 2h 2
8
 12
or x1 =
2
hW
9
PROBLEM 5.118 CONTINUED
Component 2
h
V2 = ∫h/2 π
h
a2
=π
h
=
a2
a2 
x2 
( h − x ) dx = π  hx − 
2  h/2
h
h 

2

h)    h 
(



− h  −
 h ( h) −
2   2
 




 
2 
 
( h2 )
2
1 2
πa h
8
h
and
 a2

a 2  x 2 x3 
h
∫2 xELdV = ∫h/2 x π h ( h − x ) dx  = π h  h 2 − 3 



 h/2
2

2
3

h )   h2
(
a2  ( h )


=π
−
− h
−
 h
3   2
h  2





1
= π a 2h 2
12
( )
Now
3 
( h2 )
 
3 
 
1
1

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h2
8
12


or x2 =
2
hW
3
PROBLEM 5.119
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y = 0W
First note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x

x2 
Now r = b1 − 2  so that
a 

2
2
x2 
dV = πb 1 − 2  dx
a 

2
Then


x2 
2x2 x4 
a
a
V = ∫0 π b 2 1 − 2  dx = ∫0 π b 2  1 − 2 + 4  dx
a 
a
a 


a
2
x5 
2 x3
= π b  x − 2 + 4 
3a
5a 

0
2 1

= π ab 2 1 − + 
3 5

8
= π ab 2
15
and
2x2 x4 
a
2 
x
dV
π
b
x
1
=
−
+ 4  dx

∫ EL
∫0

a2
a 

2
x2 2x4
x6 
= π b 
− 2 + 4 
4a
6a 
 2
1 1 1
= π a 2b 2  − + 
2 2 6
=
1 2 2
πa b
6
a
0
PROBLEM 5.119 CONTINUED
Then
1
 8

xV = ∫ xEL dV : x  π ab 2  = π a 2b 2
 15
 16
or x =
15
aW
6
PROBLEM 5.120
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y = 0W
First, note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
Now r = 1 −
1
so that
x
2
1

dV = π 1 −  dx
x

2
1 

= π 1 − + 2  dx
x x 

3
Then
2
1 
1
3 

V = ∫1 π 1 − + 2  dx = π  x − 2 ln x − 
x x 
x 1



1 
1 
= π  3 − 2 ln3 −  − 1 − 2 ln 1 −  
3 
1 

= ( 0.46944π ) m3
and
3
 x2

2
1  

1
dx
π
−
+
=
 − 2 x + ln x 


2
x x  
 
2
1
3 
x π
1 
∫ x ELdV = ∫
  32
 13
 
= π   − 2 ( 3) + ln 3 −  − 2 (1) + ln1 
 2
 
  2
= (1.09861π ) m
Now
(
)
xV = ∫ x EL dV : X 0.46944π m3 = 1.09861π m 4
or x = 2.34 m W
PROBLEM 5.121
Locate the centroid of the volume obtained by rotating the shaded area
about the line x = h.
SOLUTION
x = hW
First, note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dy, yEL = y
Now x 2 =
Then
and
Let
Then
(
)
h2 2
h 2
a − y 2 so that r = h −
a − y2
2
a
a
2
h2
dV = π 2 a − a 2 − y 2 dy
a
)
(
V = ∫0 π
a
(
h2
a − a2 − y2
2
a
) dy
2
y = a sin θ ⇒ dy = a cos θ d θ
V =π
=π
(
h 2 π /2
2
2
2
∫ a − a − a sin θ
a2 0
) a cosθ dθ
2
(
)
h 2 π /2  2
2
2
2
∫ a − 2a ( a cosθ ) + a − a sin θ  a cosθ dθ
a2 0 
π /2
= π ah 2 ∫0
( 2 cosθ − 2 cos θ − sin θ cosθ ) dθ
2
2
π /2

 θ sin 2θ  1 3 
= π ah 2  2sin θ − 2  +
 − sin θ 
4  3
2

0

= π ah 2  2 −

 π  1
2 2  − 
2 3

 
= 0.095870π ah 2
PROBLEM 5.121 CONTINUED
and
 h2
(

2
)
(
=π
h2 a
2a 2 y − 2ay a 2 − y 2 − y 3 dy
2 ∫0
a
=π
h2
a2
=π
h2  2
1  2
2
a ( a ) − a4  −  a a2
2 
4  3
a 
=
Now
)
a
2
2
∫ y ELdV = ∫0 y π a 2 a − a − y dy 


(
 2 2 2
2
2
a y + 3 a a − y

)
3/2
a
−
1 4
y
4  0
3/2  
( )


1
π a 2h 2
12
(
)
yV = ∫ y EL dV : y 0.095870π ah 2 =
1
π a 2h 2
12
or y = 0.869a W
PROBLEM 5.122
Locate the centroid of the volume generated by revolving the portion of
the sine curve shown about the x axis.
SOLUTION
y = 0W
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
First, note that symmetry implies
dV = πr 2dx, xEL = x
πx
Now
r = b sin
so that
dV = πb 2 sin 2
Then
2a
πx
2a
dx
V = ∫a π b 2 sin 2
2a
πx
2a
dx
2a
 x sin π2ax 
= πb  −

2 πa 
 2
a
2
( ) ( a2 )
= π b 2  22a −

1
= π ab 2
2
and
πx
2a 

2
2
∫ x ELdV = ∫a x  π b sin 2a dx 


Use integration by parts with
u = x
du = dx
dV = sin 2
V =
x
−
2
πx
2a
sin π x
2π
a
a
PROBLEM 5.122 CONTINUED
Then
 
x sin πax
=
− 2π
π
x
dV
b
x



∫ EL

   2
a

2
2a

sin π x
2a  x
  − ∫a  − 2π a
a
  a
2
 
 dx 
 
2a

π x  
a2

 2a 
 a   1
= π b 2   2a   − a    −  x 2 +
cos
 
a a 
2π 2
 2 
 2   4
 

 3   1
1
a2
a 2  
2
2
= π b 2  a 2  −  ( 2a ) +
−
+
a
(
)

2π 2 4
2π 2  
 2   4
1 
3
= π a 2b 2  − 2 
4
π 

Now
= 0.64868π a 2b 2
1

xV = ∫ x EL dV : x  π ab 2  = 0.64868π a 2b 2
2

or x = 1.297a W
PROBLEM 5.123
Locate the centroid of the volume generated by revolving the portion of
the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell
of radius r and thickness dr as the element of volume.)
SOLUTION
x = 0W
First note that symmetry implies
z = 0W
Choose as the element of volume a cylindrical shell of radius r and
thickness dr.
1
dV = ( 2πr )( y )( dr ) , yEL = y
Then
2
Now
so that
Then
y = b sin
πr
2a
dV = 2πbr sin
2a
V = ∫a 2πbr sin
πr
2a
πr
2a
dr
dr
Use integration by parts with
u =r
dv = sin
du = dr
Then
v=−
2a
π
πr
2a
cos
dr
πr
2a
2a
  2a
πr  
πr  
2a  2a
V = 2πb ( r )  −
cos   − ∫a  cos  dr 
 π
 π
2a  a
2a  


2a
 2a
 4a 2
πr  

( 2a ) ( −1)  +  2 sin  
= 2πb −
2a a 
π
 π

2
2
 4a
4a 
V = 2πb 
− 2 
π 
 π
1

= 8a 2b 1 − 

π
= 5.4535 a 2b
Also
πr
πr
2a 


∫ y ELdV = ∫a  12 b sin 2a   2πbr sin 2a dr 
2a
= πb 2 ∫a r sin 2
πr
dr
2a
PROBLEM 5.123 CONTINUED
Use integration by parts with
u =r
dv = sin 2
du = dr
Then
v=
r
−
2
 
r sin πar
∫ y EL dV = π b  ( r )  2 − 2π
  
a

2
πr
dr
2a
sin πar
2π
a
2a

sin π r
2a  r
  − ∫a  − 2π a
a
  a
2
 
 dr 
 
2a

π r  
a2

 2a 
 a   r 2
= π b 2   ( 2a )   − ( a )    −  +
cos
 
a a 
2π 2
 2 
 2   4
 

3
 ( 2a ) 2
( a )2 + a 2  
a2

= π b2  a2 − 
+
−

4
2π 2
2π 2  
 4
 2

1 
3
= π a 2b 2  − 2 
4 π 
= 2.0379a 2b 2
Now
(
)
yV = ∫ y EL dV : y 5.4535a 2b = 2.0379a 2b2
or y = 0.374b W
PROBLEM 5.124
Show that for a regular pyramid of height h and n sides ( n = 3, 4, … ) the
centroid of the volume of the pyramid is located at a distance h / 4 above
the base.
SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any
number N of sides, the area of the base of the pyramid is given by
Abase = kb 2
where k = k ( N ) ; see note below. Using similar triangles, have
or
Then
and
s
h−y
=
b
h
b
s = (h − y)
h
b2
2
dV = Aslicedy = ks 2dy = k 2 ( h − y ) dy
h
h
V = ∫0 k
h
3
 1
− 3 ( h − y ) 

0
1 2
kb h
3
=
Also
b2
b2
2
−
=
h
y
dy
k
(
)
h2
h2
yEL = y
2

b2 h
h  b
2
so then ∫ y EL dV = ∫0 y k 2 ( h − y ) dy  = k 2 ∫0 h 2 y − 2hy 2 + y 3 dy
h
 h

(
)
h
=k
Now
b2  1 2 2 2 3 1 4 
1 2 2
h y − hy + y  =
kb h
2 
3
4 0 12
h 2
1 2 2
1

yV = ∫ y EL dV : y  kb 2h  =
kb h
3
 12
or y =
Note
1
Abase = N  × b ×
2
=
N
b2
4 tan Nπ
= k ( N ) b2
b
2
tan Nπ



1
h Q.E.D. W
4
PROBLEM 5.125
Determine by direct integration the location of the centroid of one-half of a
thin, uniform hemispherical shell of radius R.
SOLUTION
x = 0W
First note that symmetry implies
The element of area dA of the shell shown is obtained by cutting the shell
with two planes parallel to the xy plane. Now
dA = (π r )( Rdθ )
yEL = −
2r
π
r = R sin θ
dA = π R 2 sin θ dθ
where
so that
yEL = −
2R
π
sin θ
π
π
A = ∫02 π R 2 sin θ dθ = π R 2 [ − cosθ ]02
Then
= π R2
and


2
∫ y ELdA = ∫02  − π sin θ  (π R sin θ dθ )


π
2R
π
3 θ
sin 2θ  2
= −2 R  −
4  0
2
=−
Now
π
2
R3
(
)
yA = ∫ y EL dA: y π R 2 = −
π
2
R3
1
or y = − R W
2
Symmetry implies
z = y∴ z =−
1
RW
2
PROBLEM 5.126
The sides and the base of a punch bowl are of uniform thickness t. If
t << R and R = 350 mm, determine the location of the center of gravity
of (a) the bowl, (b) the punch.
SOLUTION
(a) Bowl
First note that symmetry implies
x = 0W
z = 0W
for the coordinate axes shown below. Now assume that the bowl may be
treated as a shell; the center of gravity of the bowl will coincide with the
centroid of the shell. For the walls of the bowl, an element of area is
obtained by rotating the arc ds about the y axis. Then
dAwall = ( 2π R sin θ )( Rdθ )
and ( yEL ) wall = − R cos θ
Then
and
π /2
π /2
Awall = ∫π /6 2π R 2 sin θ dθ = 2π R 2 [ − cosθ ]π /6
= π 3R 2
ywall Awall = ∫ ( yEL )wall dA
π /2
(
= ∫π /6 ( − R cosθ ) 2π R 2 sin θ dθ
= π R3 cos 2 θ 
)
π /2
π /6
3
= − π R3
4
By observation
Now
or
or
Abase =
π
4
R2,
ybase = −
3
R
2
y ΣA = ΣyA
π
3
π  3 


y  π 3R 2 + R 2  = − π R3 + R 2  −
R
4 
4
4  2 

y = −0.48763R
R = 350 mm
∴ y = −170.7 mm W
(b) Punch
First note that symmetry implies
x = 0W
z = 0W
and that because the punch is homogeneous, its center of gravity will
coincide with the centroid of the corresponding volume. Choose as the
element of volume a disk of radius x and thickness dy. Then
dV = π x 2dy, yEL = y
PROBLEM 5.126 CONTINUED
Now
Then
x2 + y 2 = R2
0
π
− 3/2 R
V =∫
(
(
)
dV = π R 2 − y 2 dy
so that
)
0
1 

R 2 − y 2 dy = π  R 2 y − y 3 
3 −

3/2 R
3
 
3  1
3   3
R −  −
R
= −π  R 2  −
= π 3R3
  2  3  2   8


and
0
∫ yEL dV = ∫−
(
0
)
1 
1
y π R 2 − y 2 dy  = π  R 2 y 2 − y 4 
3/2 R ( ) 

2
4 −

3/2 R
2
4

1 2
3 
1
3  
15

R −
R − −
R  = − π R4
= −π




2  2 
4 2  
64


Now
or
15
3

yV = ∫ yEL dV : y  π 3 R3  = − π R 4
8
64


y =−
5
8 3
R
R = 350 mm
∴ y = −126.3 mm W
PROBLEM 5.127
After grading a lot, a builder places four stakes to designate the corners of
the slab for a house. To provide a firm, level base for the slab, the builder
places a minimum of 60 mm of gravel beneath the slab. Determine the
volume of gravel needed and the x coordinate of the centroid of the
volume of the gravel. (Hint: The bottom of the gravel is an oblique plane,
which can be represented by the equation y = a + bx + cz.)
SOLUTION
First, determine the constants a, b, and c.
At
x = 0, z = 0 : y = −60 mm
∴ −60 mm = a; a = −60 mm
At
x = 7200 mm, z = 0 : y = −100 mm
∴ − 100 mm = −60 + b ( 7200)
1
180
x = 0 , z = 12 000 mm: y = −120 mm
b=−
At
∴ − 120 mm = −60 mm + c (12000 )
c=−
1
200
It follows that y = −60 −
1
1
x−
z where all dimensions
180
120
are in mm.
Choose as the element of volume a filament of base dx × dz and height
| y |. Then
dV = y dxdz, xEL = x
dV = −60 −
or
Then
12000 7200 
 60
0
V = ∫0
∫

+
1
1
x−
z dxdz
180
200
1
1 
x+
z  dxdz
180
200 
7200
12000 
= ∫0
1 2
1

60 x + 360 x + 200 xz 

0

12000

0
=∫

( 60 )( 7200 )
2
7200 )
(
+
360
+
dz
( 7200 ) z dz
200

12000
36 2 

= 576000 z +
z
2  0

= 9.504 × 109 mm3 = 9.50 m3
V = 9.50 m3 W
PROBLEM 5.127 CONTINUED
and
12000 7200 

∫ x ELdV = ∫0 ∫0 x  60 + 180 x + 200 z  dxdz


1
12000  60
= ∫0
12000
= ∫0
2

1
7200
x2 +
1 3
1 2 
x +
x z
540
400
0
( 2246.4 × 10
6
dz
)
+ 129 600 z dz
12000
129 600 2 

=  2246.4 × 106 z −
z 
2

0
= 2.695 × 1013 + 0.933 × 1013
= 3.63 × 1013 mm 4 = 36.3 m 4
Now
(
)
xV = ∫ x EL dV : x 9.50 m3 = 36.3 m 4
or x = 3.82 m W
PROBLEM 5.128
Determine by direct integration the location of the centroid of the volume
between the xz plane and the portion shown of the surface
16h ax − x 2 bz − z 2
y =
.
a 2b 2
(
)(
)
SOLUTION
a
2
b
z =
2
x =
First note that symmetry implies
Choose as the element of volume a filament of base dx × dz and height y. Then
dV = ydxdz, yEL =
(
dV =
Then
V = ∫0 ∫0
b a
=
and
)(
)
16h
ax − x 2 bz − z 2 dxdz
a 2b 2
or
V =
1
y
2
(
)(
)
16h
ax − x 2 bz − z 2 dxdz
2 2
ab
(
)
a
16h b
1 
a
bz − z 2  x 2 − x3  dz
2 2 ∫0
3 0
ab
z
b
( )
16h  a 2
1 3 b
1 
8ah  b 2 1 3  4
a − ( a )   z 2 − z 3  = 2  ( b ) − ( b )  = abh
2 2 
3
3 0 3b  2
3
a b 2
 2
 9
b a 

2
2 
2
2
∫ yELdV = ∫0 ∫0 2  a 2b2 ( ax − x )( bz − z )  a 2b2 ( ax − x )( bz − z ) dxdz 
1 16h

16h

(
2
)(

)
=
128h b a 2 2
a x − 2ax3 + x 4 b2 z 2 − 2bz 3 + z 4 dxdz
4 4 ∫0 ∫0
ab
=
2
128h 2 b 2 2
a 4 1 5
3
4 a
3
−
+
−
2
b
z
bz
z
x
x + x  dz

∫
2
5 0
a 2b 4 0
3
=
128h 2
a 4b 4
=

b
64ah 2  b3
1
32
abh 2
( b )3 − ( b )4 + ( b )5  =
4 
2
5
225
15b  3

(
)
a
b
 a2
1 5   b2 3 b 4 1 5 
a
3
4
 (a) − (a) + (a)   Z − Z + Z 
Z
5 0
2
5
3
 3
PROBLEM 5.128 CONTINUED
Now
32
4

yV = ∫ y EL dV : y  abh  =
abh 2
9
225


or y =
8
h
25
PROBLEM 5.129
Locate the centroid of the section shown, which was cut from a circular
cylinder by an inclined plane.
SOLUTION
x =0
First note that symmetry implies
Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then
1
dV = 2 xy dz, yEL = y, z EL = z
2
h
h
h
z
y = −
z = 1 − 
x = a2 − z 2
and
Now
2 2a
2
a
z

dV = h a 2 − z 2 1 −  dz

a
So
Then
a
h
0
V =∫
=
=
z
 1 

 z 1 2
a − z 1 −  dz = h   z a 2 − z 2 + a 2 sin −1    +
a − z2

 a   3a
a
 2 
2
2
1 2  −1
a h sin (1) − sin −1 ( −1) 
2
π
2
a 2h
(
)
3/2 

a

 − a
PROBLEM 5.129 CONTINUED
h
z  
a 1
2
2
∫ yEL dV = ∫−a  2 × 2 1 − a    h a − z 1 −

 


Then
z 
 dz
a  
=

h2 a
z
z2 
a 2 − z 2 1 − 2 + 2  dz
∫
−a
4
a a 

=
2 2
h 2  1 
2
2
2
−1  z  
a − z2
  z a − z + a sin    + 
4  2 
 a    3a
(
1
+ 2
a
=
 z 2
2
− a − z
4

(
)
)
3
2



a
3
2
a2z 2
a4
 z   
a − z2 +
sin −1    
+
8
8
 a   
−a
5h 2a 2  −1
sin (1) − sin −1 ( −1) 
32 
 π a 2  5h 2a 2
yV = ∫ yEL dV : y 
h  =
(π )
32
 2 
Then
or y =
and
5
h
16

z 
a
2
2
∫ zELdV = ∫−a z  h a − z 1 − a  dz 


 1
= h − a 2 − z 2
 3
(
=−
)
3
2


1 z
− − a 2 − z 2
a 4
(
)
3
2
a2 z 2
a4
 z   
a − z2 +
sin −1    
+
8
8
 a   
a
−a
a3h  −1
sin (1) − sin −1 ( −1) 

8
 π a 2h 
π a3h
zV = ∫ z EL dV : z 
 = −
8
 2 
or z = −
a
4
PROBLEM 5.130
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
14 × 20 = 280
7
10
1960
2800
2
−π ( 4 ) = −16π
6
12
−301.59
−603.19
Σ
229.73
1658.41
2196.8
2
X Σ A = ΣxA
Then
(
)
X 229.73 in 2 = 1658.41 in 3
or X = 7.22 in.
Y Σ A = ΣyA
and
(
)
Y 229.73 in 2 = 2196.8 in 3
or Y = 9.56 in.
PROBLEM 5.131
For the area shown, determine the ratio a/b for which x = y.
SOLUTION
A
1
2
Σ
Then
2
ab
3
1
− ab
2
1
ab
6
x
3
a
8
1
a
3
y
xA
yA
3
b
5
2
b
3
2
ab
4
a 2b
−
6
2ab 2
5
ab 2
−
3
a 2b
12
ab 2
15
X Σ A = ΣxA
 1  a 2b
X  ab =
6 
12
or
or
Now
1
a
2
Y Σ A = ΣyA
X =
 1  ab 2
Y  ab =
6 
15
2
Y = b
5
1
2
X =Y ⇒
a= b
2
5
or
a
4
=
b
5
PROBLEM 5.132
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
1
2
3
Σ
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
126 × 54 = 6804
1
× 126 × 30 = 1890
2
1
× 72 × = 1728
2
10 422
9
27
61 236
183 708
30
64
56 700
120 960
48
−16
82 944
−27 648
200 880
277 020
X ΣA = Σ xA
Then
(
)
X 10 422 m 2 = 200 880 mm 2
or X = 19.27 mm W
Y Σ A = Σ yA
and
(
)
Y 10 422 m 2 = 270 020 mm3
or Y = 26.6 mm W
PROBLEM 5.133
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
By observation
h
x+h
a
x

= h 1 − 

a
y1 = −
For y2:
At
x = 0, y = h: h = k (1 − 0)
At
(
x = a, y = 0: 0 = h 1 − ca 2
1
a2
yEL
 x x2 
= h  − 2  dx
a a 
1
h 
x 
x2  
= ( y1 − y2 ) = 1 −  + 1 − 2  
2
2 
a 
a  
=
h
x x2 
−
−
2
a a 2 
2 
a
a 
h
0 
A = ∫ dA = ∫
=
and
C =

x2  
x 
dA = ( y2 − y1 ) dx = h 1 − 2  − 1 −   dx
a  
a  

Now
Then
or

x2 
y2 = h 1 − 2 
a 

Then
xEL = x
)
k =h
or
1
ah
6
a  
x h
0  
∫ x ELdA = ∫
=
 x2
x x2 
x3 
− 2  dx = h 
− 2
a a 
 2a 3a  0
a
 x 3
x x2  
x4 
− 2  dx  = h 
− 2  
  a a  
 3a 4a   0
1 2
a h
12
PROBLEM 5.133 CONTINUED
a
∫ y EL dA = ∫0
=
h
x x2    x x2  
−
−
−
dx 
2
h
a a 2    a a 2  
2 
h2 a  x
x2
x4 
2 − 3 2 + 4  dx
∫

0
2  a
a
a 
a
h2  x2
x3
x5 
1 2
=
−
+
=
ah

2
4
2 a
a
5a 0 10
1 2
1 
xA = ∫ x EL dA: x  ah =
a h
 6  12
x =
1
aW
2
1 2
1 
yA = ∫ y EL dA: y  ah =
a h
 6  10
y =
3
hW
5
PROBLEM 5.134
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 l = 2 m, determine the distance d so that portion BCD of the
member is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the component
must lie on a vertical line through C. Further, because the tubing is
uniform, the center of gravity of the component will coincide with the
centroid of the corresponding line. Thus, X = 0
So that
Then
Σ xL = 0
0.75


− d −
cos 55°  m × ( 0.75 m )
2


+ ( 0.75 − d ) m × (1.5 m )

1

+ (1.5 − d ) m −  × 2 m × cos 55°   × ( 2 m ) = 0
2


or
( 0.75 + 1.5 + 2 ) d
2
1

=  ( 0.75 ) − 2  cos 55° + ( 0.75 )(1.5 ) + 3
2

or d = 0.739 m W
PROBLEM 5.135
A cylindrical hole is drilled through the center of a steel ball bearing
shown here in cross section. Denoting the length of the hole by L, show
that the volume of the steel remaining is equal to the volume of a sphere
of diameter L.
SOLUTION
Calculate volumes by rotating cross sections about a line and using Theorem II of Pappus-Guldinus
y AA =
For the sector:
2R sin α
3α
A = α R2
2
For the triangle:
h=
A=
=
1
L
R2 −   =
4R 2 − L2
2
2
y AA =
2
1
h=
4R 2 − L2 ,
3
3
1
( L )( h )
2
1
L 4R 2 − L2
4
Using Theorem II of Pappus-Guldinus
Vball = 2π ( y AA )1 A1 − 2π ( y AA )2 A2
 2 R sin α
1
 1

α R 2 −  4R 2 − L2  L 4R 2 − L2  
= 2π 
3
α
3
4




(
)
(
)
L
2

4 R 2 − L2 
= 2π  R3 sin α −
12
3

Now
Then
R sin α =
L
2
2  L 
1
1 3
V = 2π    R 2 − LR 2 +
L
3
12 
3  2 
=
Note Vsphere =
L
If r = , then
2
Therefore,
π
6
L3
4
πr where r is the radius
3
3
Vsphere
4 L
π
= π   = L3
3 2
6
Vball = Vsphere =
π
6
L3 Q.E.D.
PROBLEM 5.136
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
(a) Have
RI = ( 4 m )( 200 N/m ) = 800 N
2
( 4 m )( 600 N/m ) = 1600 N
3
ΣFy : −R = − RI − RII
RII =
Then
or
and
or
R = 800 + 1600 = 2400 N
ΣM A: − X ( 2400 ) = −2 ( 800 ) − 2.5 (1600 )
X =
7
m
3
∴ R = 2400 N , X = 2.33 m
(b) Reactions
ΣFx = 0: Ax = 0
7 
ΣM A = 0: ( 4 m ) By −  m ( 2400 N ) = 0
3 
or
By = 1400 N
ΣFy = 0: Ay + 1400 N − 2400 N = 0
or
Ay = 1000 N
∴ A = 1000 N , B = 1400 N
PROBLEM 5.137
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI =
1
( 3 ft )( 480 lb/ft ) = 720 lb
2
RII =
1
( 6 ft )( 600 lb/ft ) = 1800 lb
2
RIII = ( 2 ft )( 600 lb/ft ) = 1200 lb
ΣFx = 0: Bx = 0
Then
ΣM B = 0:
( 2 ft )( 720 lb ) − ( 4 ft )(1800 lb )
+ ( 6 ft ) Cy − ( 7 ft )(1200 lb ) = 0
or
C y = 2360 1b
C = 2360 lb
ΣFy = 0: − 720 lb + By − 1800 lb + 2360 lb − 1200 lb = 0
or
By = 1360 lb
B = 1360 lb
PROBLEM 5.138
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
yI = −
1
(1.2) = −0.4 m
3
1
(3.6) = 1.2 m
3
4 (1.8)
2.4
m
=−
=−
3π
π
zI =
xIII
A, m 2
I
II
III
Σ
Have
1
( 3.6) (1.2) = 2.16
2
(3.6)(1.7) = 6.12
π
2
(1.8) 2
= 5.0894
x, m
y, m
z, m
xA, m3
yA, m3
zA, m3
1.5
−0.4
1.2
3.24
−0.864
2.592
0.75
0.4
1.8
4.59
2.448
11.016
0.8
1.8
−3.888
4.0715
9.1609
3.942
5.6555
22.769
−
2.4
π
13.3694
(
)
X ΣV = ΣxV : X 13.3694 m 2 = 3.942 m3
(
)
or X = 0.295 m
Y ΣV = ΣyV : Y 13.3694 m 2 = 5.6555 m3
or Y = 0.423 m
(
)
Z ΣV = Σ zV : Z 13.3694 m 2 = 22.769 m3
or Z = 1.703 m
PROBLEM 5.139
The composite body shown is formed by removing a semiellipsoid of
revolution of semimajor axis h and semiminor axis a2 from a hemisphere of
radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b)
the ratio h/a for which y = −0.4a.
SOLUTION
V
y
yV
Hemisphere
2 3
πa
3
3
− a
8
1
− πa 4
4
2
Semiellipsoid
2 a
1
− π   h = − πa 2 h
3 2
6
3
− h
8
ΣV =
Then
π
6
a 2 ( 4a − h )
ΣyV = −
π
16
+
(
1
πa 2 h 2
16
a 2 4a 2 − h 2
)
Y ΣV = Σ yV
Now
so that
π
π

Y  a 2 ( 4a − h )  = − a 2 4a 2 − h 2
16
6

or
2
h
3 

 h 
Y  4 −  = − a 4 −   

 a 
a
8 

(
(a) Y = ? when h =
Substituting
)
(1)
a
2
h 1
= into Eq. (1)
a
2
2
1
3 

1 
Y  4 −  = − a 4 −   
2
8 

 2  
or
Y =−
45
a
112
Y = −0.402a
PROBLEM 5.139 CONTINUED
(b)
h
= ? when Y = −0.4a
a
Substituting into Eq. (1)
( −0.4a )  4 −

2
h
3 
h 
a
4
=
−
−


  
a
8 
 a  
2
or
Then
 h
 h
3   − 3.2   + 0.8 = 0
 a
 a
2
h 3.2 ± ( −3.2 ) − 4 ( 3)( 0.8 )
=
a
2 ( 3)
=
3.2 ± 0.8
6
or
h
2
h
2
=
and
=
a
5
a
3
PROBLEM 5.140
A thin steel wire of uniform cross section is bent into the shape shown.
Locate its center of gravity.
SOLUTION
First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the
corresponding line.
x1 = 0.3sin 60° = 0.15 3 m
z1 = 0.3cos 60° = 0.15 m
 0.6sin 30° 
x2 = 
 sin 30°
π

6


0.9
=
m
π
 0.6sin 30° 
z2 = 
 cos 30°
π

6


0.9
=
3m
π
π 
L2 =   ( 0.6 ) = ( 0.2π ) m
3
Have
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
zL, m 2
1
1.0
0.15 3
0.4
0.15
0.25981
0.4
0.15
2
0.2π
0.9
π
0
π
0.18
0
0.31177
3
4
∑
0.8
0.6
3.0283
0
0
0.4
0.8
0.6
0.3
0
0
0.43981
0.32
0.48
1.20
0.48
0.18
1.12177
0.9 3
X Σ L = Σ x L: X ( 3.0283 m ) = 0.43981 m 2
Y Σ L = Σ yL: Y ( 3.0283 m ) = 1.20 m 2
Z Σ L = Σ z L: Z ( 3.0283 m ) = 1.12177 m 2
or X = 0.1452 m
or Y = 0.396 m
or Z = 0.370 m
PROBLEM 5.141
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y =0
First note that symmetry implies
and z = 0
y = k ( X − h)
Have
x = 0, y = a: a = k ( −h )
At
k =
or
2
2
a
h2
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = π r 2dx, X EL = x
a
2
r = 2 ( x − h)
h
Now
dV = π
so that
Then
V = ∫0 π
h
=
and
a2
( x − h)4 dx
h4
h
a2
π a2 
4
5
−
=
x
h
dx
x − h ) 
(
)
(
4
4 
0
5h 
h
1 2
πa h
5
 a2

4
h
∫ x EL dV = ∫0 x π h 4 ( x − h ) dx 


(
)
=π
a2 h 5
x − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h 4 x dx
4 ∫0
h
=π
a2  1 6 4 5 3 2 4 4 3 3 1 4 2 
x − hx + h x − h x + h x 
5
2
3
2
h 4  6
0
h
=
Now
1
π a 2h 2
30
π
 π 2 2
xV = ∫ x EL dV : x  a 2h  =
a h
5
 30
or x =
1
h
6
PROBLEM 6.1
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM B = 0: ( 6.25 m ) C y − ( 4 m )( 315 N ) = 0
ΣFy = 0: By − 315 N + C y = 0
ΣFx = 0:
C y = 240 N
B y = 75 N
Bx = 0
Joint FBDs:
Joint B:
FAB
F
75 N
= BC =
5
4
3
N
Joint C:
FBC = 100.0 N T W
By inspection:
N
FAB = 125.0 N C W
FAC = 260 N C W
PROBLEM 6.2
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM A = 0: (14 ft ) Cx − ( 7.5 ft )( 5.6 kips ) = 0
ΣFx = 0: − Ax + Cx = 0
ΣFy = 0: Ay − 5.6 kips = 0
C x = 3 kips
A x = 3 kips
A y = 5.6 kips
Joint FBDs:
Joint C:
FBC
F
3 kips
= AC =
5
4
3
FBC = 5.00 kips C W
FAC = 4.00 kips T W
Joint A:
FAB 1.6 kips
=
8.5
4
FAB = 3.40 kips T W
PROBLEM 6.3
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.
SOLUTION
FBD Truss:
ΣM B = 0: ( 6 ft )( 6 kips ) − ( 9 ft ) C y = 0
ΣFy = 0: By − 6 kips − C y = 0
C y = 4 kips
B y = 10 kips
ΣFx = 0: C x = 0
Joint FBDs:
Joint C:
FAC
F
4 kips
= BC =
17
15
8
FAC = 8.50 kips T W
FBC = 7.50 kips C W
Joint B:
By inspection:
FAB 10 kips
=
5
4
FAB = 12.50 kips C W
PROBLEM 6.4
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM B = 0: (1.5 m ) C y + ( 2 m )(1.8 kN ) − 3.6 m ( 2.4 kN ) = 0
C y = 3.36 kN
ΣFy = 0: By + 3.36 kN − 2.4 kN = 0
B y = 0.96 kN
Joint FBDs:
Joint D:
ΣFy = 0:
2
FAD − 2.4 kN = 0
2.9
ΣFx = 0: FCD −
FCD =
FAD = 3.48 kN T W
2.1
FAD = 0
2.9
2.1
(3.48 kN)
2.9
FCD = 2.52 kN C W
Joint C:
By inspection:
FAC = 3.36 kN C W
FBC = 2.52 kN C W
Joint B:
ΣFy = 0:
4
FAB − 0.9 kN = 0
5
FAB = 1.200 kN T W
PROBLEM 6.5
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
ΣFx = 0 : C x = 0
FBD Truss:
By symmetry: C y = D y = 6 kN
Joint FBDs:
Joint B:
ΣFy = 0: − 3 kN +
1
FAB = 0
5
FAB = 3 5 = 6.71 kN T W
ΣFx = 0:
Joint C:
2
FAB − FBC = 0
5
ΣFy = 0: 6 kN −
ΣFx = 0: 6 kN −
FBC = 6.00 kN C W
3
FAC = 0
5
FAC = 10.00 kN C W
4
FAC + FCD = 0
5
FCD = 2.00 kN T W
Joint A:
 1

3

ΣFy = 0: − 2 
3 5 kN  + 2  10 kN  − 6 kN = 0 check
5
5




By symmetry:
FAE = FAB = 6.71 kN T W
FAD = FAC = 10.00 kN C W
FDE = FBC = 6.00 kN C W
PROBLEM 6.6
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM A = 0: ( 25.5 ft ) C y + ( 6 ft )( 3 kips ) − ( 8 ft )( 9.9 kips ) = 0
C y = 2.4 kips
ΣFy = 0: Ay + 2.4 kips − 9.9 kips = 0
A y = 7.4 kips
ΣFx = 0: − Ax + 3 kips = 0
A x = 3 kips
2.4 kips
F
F
= CD = BC
12
18.5 18.5
Joint FBDs:
Joint C:
FCD = 3.70 kips T W
FBC = 3.70 kips C W
or: ΣFx = 0: FBC = FCD
ΣFy = 0: 2.4 kips − 2
6
FBC = 0
18.5
same answers
Joint D:
ΣFx = 0: 3 kips +
17.5
4
( 3.70 kips ) − FAD = 0
18.5
5
FAD = 8.125 kips
ΣFy = 0:
FAD = 8.13 kips T W
6
3
( 3.7 kips ) + (8.125 kips ) − FBD = 0
18.5
5
FBD = 6.075 kips
FBD = 6.08 kips C W
PROBLEM 6.6 CONTINUED
Joint A:
ΣFx = 0: −3 kips +
FAB = 4.375 kips
4
4
(8.125 kips ) − FAB = 0
5
5
FAB = 4.38 kips C W
PROBLEM 6.7
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣFy = 0: Ay − 480 N = 0
A y = 480 N
ΣM A = 0: ( 6 m ) Dx = 0
ΣFx = 0: − Ax = 0
Dx = 0
Ax = 0
Joint FBDs:
Joint A:
480 N
F
F
= AB = AC
6
2.5
6.5
FAB = 200 N C W
FAC = 520 N T W
Joint B:
200 N
F
F
= BE = BC
2.5
6
6.5
FBE = 480 N C W
FBC = 520 N T W
PROBLEM 6.7 CONTINUED
Joint C:
FCD = FCE = 520 N T W
By inspection:
ΣFx = 0:
Joint D:
2.5
( 520 N ) − FDE = 0
6.5
FDE = 200 N C W
PROBLEM 6.8
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM E = 0: ( 9 ft ) Fy − ( 6.75 ft )( 4 kips ) − (13.5 ft )( 4 kips ) = 0
Fy = 9 kips
ΣFy = 0: − E y + 9 kips = 0
E y = 9 kips
ΣFx = 0: − Ex + 4 kips + 4 kips = 0
E x = 8 kips
FEC = 9.00 kips T W
By inspection of joint E:
FEF = 8.00 kips T W
FAB = 0 W
By inspection of joint B:
FBD = 0 W
Joint FBDs:
Joint F:
ΣFx = 0:
4
FCF − 8 kips = 0
5
FCF = 10.00 kips C W
3
(10 kips) = 0
5
FDF = 6.00 kips T W
ΣFy = 0: FDF −
Joint C:
ΣFx = 0: 4 kips −
4
(10 kips ) + FCD = 0
5
FCD = 4.00 kips T W
ΣFy = 0: FAC − 9 kips +
3
(10 kips ) = 0
5
FAC = 3.00 kips T W
PROBLEM 6.8 CONTINUED
Joint A:
ΣFx = 0: 4 kips −
4
FAD = 0
5
FAD = 5.00 kips C W
PROBLEM 6.9
Determine the force in each member of the Gambrel roof truss shown.
State whether each member is in tension or compression.
SOLUTION
ΣFx = 0: H x = 0
FBD Truss:
By symmetry: A = H y = 12 kN
FCE = FAC and FBC = 0 W
By inspection of joints C and G,
FEG = FGH and FFG = 0 W
Joint A:
ΣFy = 0: 12 kN − 3 kN −
ΣFx = 0: FAC −
ΣFx = 0:
Joint B:
ΣFy = 0:
3
FAB = 0
5
4
(15 kN ) = 0
5
FAB = 15.00 kN C W
FAC = 12.00 kN T W
4
10
4
FBD − FBE = 0
(15 kN ) −
5
10.44
5
3
3
3
FBD + FBE = 0
(15 kN ) − 6 kN −
5
10.44
5
FBD = 11.93 kN C W
Solving yields
FBE = 0.714 kN C W
ΣFx = 0: by symmetry
Joint D:
ΣFy = 0: − FDE − 6 kN + 2
FDF = 11.93 kN C W
3
(11.93 kN ) = 0
10.44
FDE = 0.856 kN T W
By symmetry:
From above
FEF = FBE
so
FEF = 0.714 kN C W
FFH = FAB
FFH = 15.00 kN C W
FGH = FAC
FGH = 12.00 kN T W
FCE = FAC
FCE = 12.00 kN T W
FEG = FGH
FEG = 12.00 kN T W
PROBLEM 6.10
Determine the force in each member of the Howe roof truss shown.
State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣFx = 0: A x = 0
By symmetry:
A y = H y = 12 kN
FFH = FAB ; FGH = FAC ; FFG = FBC
and
FDF = FBD ; FEF = FBE ; FEG = FCE
9 kN
F
F
= AC = AB
3
4
5
Joint FBDs:
FAC = 12.00 kN T W
FAB = 15.00 kN C W
so FFH = 15.00 kN C W
FGH = 12.00 kN T W
By inspection:
FBC = 0; FCE = 12 kN
Joint C:
FBC = 0 = FFG W
FCE = 12.00 kN T W
FEG = 12.00 kN T W
Note:
Joint B:
α = tan −1
4
3
β = 2 tan −1
so
3
4
so
sin α = 0.8
sin β = 0.96
ΣFy 1 = 0: − 6 kN sin α + FBE sin β = 0
FBE = 5.00 kN C W
so FEF = 5.00 kN C W
ΣFy = 0: − 6 kN −
3
3
3
FBD + FBE + (15 kN ) = 0
5
5
5
PROBLEM 6.10 CONTINUED
Joint D:
3
3
3
FBD = ( 5 kN ) + (15 kN ) − 6 kN
5
5
5
FBD = 10.00 kN C W
so FDF = 10.00 kN C W
3

ΣFy = 0: −6 kN + 2  10 kN  − FDE = 0
5


FDE = 6.00 kN T W
PROBLEM 6.11
Determine the force in each member of the Fink roof truss shown. State
whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣFx = 0 : A x = 0
By symmetry: A y = G y = 24 kips
FFG = FAB ; FEG = FAC ; FEF = FBC
also;
Joint FBDs:
Joint A:
FDF = FBD ; FDE = FCD
18 kips
F
F
= AC = AB ;
4
9
97
FAC = 40.5 kips T W
FAB = 44.32 kips
FAB = 44.3 kips C W
so FFG = 44.3 kips C W
FEG = 40.5 kips T W
ΣFx = 0:
Joint B:
ΣFy = 0:
Solving:
9
3
( 44.32 kips − FBD ) − FBC = 0
5
97
4
4
( 44.32 kips − FBD ) + FBC − 12 = 0
5
97
FBC = 11.25 kips C W
FBD = 36.9 kips C W
so FEF = 11.25 kips C W
FDF = 36.9 kips C W
PROBLEM 6.11 CONTINUED
ΣFy = 0:
Joint C:
4
4
(11.25 kips ) − FCD = 0
5
5
FCD = 11.25 kips T W
FDE = 11.25 kips T W
3

ΣFx = 0: FCE + 2  (11.25 kips )  − 40.5 kips = 0
5

FCE = 27.0 kips T W
PROBLEM 6.12
Determine the force in each member of the fan roof truss shown. State
whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣFx = 0 :
Ax = 0
By symmetry: A y = I y = 12 kN
and
FAB = FHI ; FAC = FGI ; FBC = FGH
FBD = FFH ; FDC = FFG ; FDE = FEF
FCE = FEG
Joint FBDs:
10 kN
F
F
= AC = AB
4
9
97
Joint A:
FAC = 22.5 kN T W
so FGI = 22.5 kN T W
FAB = 2.5 97 kN
FAB = 24.6 kN C W
so FHI = 24.6 kN C W
Joint B:
ΣFx = 0: 22.5 kN −
ΣFy = 0: 10 kN − 4 kN +
9
( FBD + FBC ) = 0
97
4
( FBC − FBD ) = 0
97
FBD = 19.70 kN C W
Solving:
so FFH = 19.70 kN C W
and FBC = 4.92 kN C W
so FGH = 4.92 kN C W
Joint D:
By inspection:
FDE = 19.70 kN C W
so FEF = 19.70 kN C W
and FCD = 4.00 kN C W
so FFG = 4.00 kN C W
PROBLEM 6.12 CONTINUED
ΣFx = 0: − 22.5 kN +
Joint C:
ΣFy = 0: − 4 kN −
Solving:
9
3
( 4.92 kN ) + FCE + FCG = 0
5
97
4
4
( 4.92 kN ) + FCE = 0
5
97
FCE = 7.50 kN T W
so FEG = 7.50 kN T W
and FCG = 13.50 kN T W
PROBLEM 6.13
Determine the force in each member of the roof truss shown. State
whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM A = 0: (16 ft ) H y − (16 ft )( 200 lb ) − (12 ft )( 300 lb ) − ( 8 ft )( 400 lb ) − ( 4 ft )( 400 lb ) = 0
H y = 725 lb
ΣFy = 0: Ay − 200 lb − 400 lb − 400 lb − 300 lb − 200 lb + 725 lb = 0
A y = 775 lb
ΣFx = 0 :
Ax = 0
Joint FBDs:
Joint A:
575 lb
F
F
= AC = AB
1
26
29
FAB = 3096.5 lb; FAB = 3.10 kips C W
FAC = 2931.9 lb; FAC = 2.93 kips T W
PROBLEM 6.13 CONTINUED
Joint C:
By inspection:
FBC = 0 W
FCE = 2.93 kips T W
Joint B:
ΣFy = 0:
2
( 3097 lb ) − 400 lb −
29
ΣFx = 0:
2
FBD = 0
29
FBD = 2020.0 lb; FBD = 2.02 kips C W
5
( 3097 − 2020 ) lb − FBE = 0 FBE = 1000.0 lb; FBE = 1.000 kip C W
29
Joint D:
ΣFx = 0:
ΣFy = 0: FDE + 2
5
( 2.02 kips − FDF ) = 0
29
2
( 2.02 kips ) − 0.4 kips = 0
29
FDF = 2.02 kips C W
FDE = 1.100 kips C W
Joint H:
525 lb
F
F
= FH = GH
1
29
26
FFH = 2827 lb; FFH = 2.83 kips C W
FGH = 2677 lb; FGH = 2.68 kips T W
Joint G:
By inspection:
FEG = 2.68 kips T W
FFG = 0 W
PROBLEM 6.13 CONTINUED
Joint F:
ΣFy = 0:
2
( 2.83 kips − FDF ) − 0.3 kips = 0
29
ΣFx = 0: FEF +
5
( 2.02 kips − 2.827 kips ) = 0
29
FDF = 2.02 kips
FEF = 0.750 kips C W
PROBLEM 6.14
Determine the force in each member of the Gambrel roof truss
shown. State whether each member is in tension or compression.
SOLUTION
ΣFx = 0
FBD Truss:
By symmetry:
Ax = 0
A y = J y = 5 kN
FAB = FHJ ; FAC = FIJ ; FBD = FGH
FCD = FGI ; FDE = FEG ; FDF = FFG
and
FBC = FHI
FEF = 0 W
By inspection of joint F:
Joint FBDs:
ΣFy = 0: 5 kN − 1 kN −
Joint A:
8
FAB = 0
89
FAB =
89
kN
2
FAB = 4.72 kN C W
ΣFx = 0: FAC −
5
89
89
kN = 0
2
FAC = 2.50 kN T W
so FHJ = 4.72 kN C W
FIJ = 2.50 kN T W
Joint B:
ΣFx = 0:
ΣFy = 0:

5  89
kN − FBD − FBC  = 0


89  2


8  89
kN − FBD + FBC  − 1 kN = 0


89  2

Solving: FBD = 4.127 kN
FAB = 0.5896 kN
so FBD = 4.13 kN C W
and FBC = 0.590 kN C W
so FGH = 4.13 kN C W
and FHI = 0.590 kN C W
PROBLEM 6.14 CONTINUED
Joint C:
5
(.59 kN ) − 2.5 kN = 0;
89
ΣFx = 0: FCI +
FCI = 2.187 kN
FCI = 2.19 kN T W
ΣFy = 0: FCD −
8
(.59 kN ) = 0
89
FCD = 0.500 kN T W
so FGI = 0.500 kN T W
ΣFy = 0:
8
( 4.127 kN ) − 2.5 kN −
89
5
FDE = 0
281
FDE = 3.352 kN
Joint D:
so FDE = 3.35 kN C W
and FEG = 3.35 kN C W
ΣFx =
5
16
( 4.127 kN ) −
( 3.352 kN ) + FDF = 0
89
281
FDF = 1.012 kN T W
FFG = 1.012 kN T W
PROBLEM 6.15
Determine the force in each member of the Pratt bridge truss shown.
State whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = H y = 9 kN
FAB = FFH ; FAC = FGH
and
FBC = FFG ; FBD = FDF
FBE = FEF ; FCE = FEG
FDE = 0 W
By inspection of joint D:
FBDs Joints:
ΣFy = 0: 9 kN −
4
FAB = 0
5
FAB = 11.25 kN C W
ΣFx = 0: FAC −
3
FAB = 0
5
FAC = 6.75 kN T W
ΣFx = 0: FCE − 6.75 kN = 0
FCE = 6.75 kN T W
ΣFy = 0: FBC − 6 kN = 0
FBC = 6.00 kN T W
ΣFy = 0:
4
4
(11.25 kN ) − 6 kN + FBE = 0
5
5
FBE = 3.75 kN C W
ΣFx = 0: FBD −
3
3
(11.25 kN ) − ( 3.75 kN ) = 0
5
5
FBD = 9.00 kN T W
From symmetry conditions above
FFH = 11.25 kN C W
FGH = 6.75 kN T W
FEG = 6.75 kN T W
FFG = 6.00 kN T W
FEF = 3.75 kN C W
FDF = 9.00 kN T W
PROBLEM 6.16
Determine the force in each member of the Pratt bridge truss shown.
State whether each member is in tension or compression. Assume
that the load at G has been removed.
SOLUTION
ΣFx = 0: A x = 0
FBD Truss:
ΣM A = 0: (12 m ) H y − ( 6 m )( 6 kN ) − ( 3 m )( 6 kN ) = 0
H y = 4.5 kN
ΣFy = 0: Ay − 6 kN − 6 kN + 4.5 kN = 0
A y = 7.5 kN
4.5 kN
F
F
= GH = FH
4
3
5
Joint FBDs:
Joint H:
FGH = 3.375 kN
FGH = 3.38 kN T W
FFH = 5.625 kN
FFH = 5.63 kN C W
FFG = 0 W
By inspection of joint G:
FEG = FGH = 3.38 kN T W
Joint F:
5.625 kN
F
F
= EF = DF
5
5
6
FEF = 5.63 kN T W
FDF = 6.75 kN C W
FDE = 0 W
By inspection of joint D :
FBD = FDF = 6.75 kN C W
Joint B:
By inspection of joint C: FAC = FCE
and FBC = 6.00 kN T W
ΣFx = 0:
3
( FAB + FBE ) − 6.75 kN = 0
5
ΣFy = 0:
4
( FAB − FBE ) − 6 kN = 0
5
ΣFy = 0:
4
( FAB − FBE ) − 6 kN = 0
5
PROBLEM 6.16 CONTINUED
FAB = 9.375 kN
Solving:
FBE = 1.875 kN
Joint A:
FAC
7.5 kN 9.375 kN
=
=
3
4
5
so
FAB = 9.38 kN C W
FBE = 1.875 kN T W
FAC = 5.625 kN
FAC = 5.63 kN T W
From above
FCE = 5.63 kN T W
PROBLEM 6.17
Determine the force in member DE and in each of the members
located to the left of DE for the inverted Howe roof truss shown.
State whether each member is in tension or compression.
SOLUTION
ΣFx = 0: A x = 0
FBD Truss:
By load symmetry A y = H y = 2400 lb
θ = tan −1
Note:
10.08
= 16.26°
15.81 + 18.75
β = 90 − 2θ = 57.48°; α = 180 − β = 32.52°
Joint FBDs:
Joint A:
ΣFy′ = 0: ( 2400 lb − 600 lb ) cosθ − FAC sin θ = 0
FAC =
(1800 lb )
tan16.26°
= 6171.5 lb;
FAC = 6.17 kips T W
ΣFx = 0: (6171.5 lb) cos 2θ − FAB cosθ = 0
FAB = 6171.5
Joint B:
cos 32.52°
= 5420.7 lb;
cos16.26°
ΣFx′ = 0: 5420.7 lb + (1200 lb ) sin θ − FBD = 0
FBD = 5420.7 + 1200sin16.26° = 5756.7 lb
Joint C:
FAB = 5.42 kips C W
ΣFy′ = 0: FBC − (1200 lb ) cos θ = 0
FBD = 5.76 kips C W
FBC = 1152 lb
FBC = 1.152 kips C W
ΣFy′′ = 0: FCD sin 2θ − (1152 lb ) cosθ = 0
FCD = 1152
cos16.26°
= 2057.2 lb
sin 32.52°
FCD = 2.06 kips T W
ΣFx′′ = 0: FCE + (1152 lb ) sin θ + ( 2057.2 lb ) cos 2θ − 6171.5 lb = 0
PROBLEM 6.17 CONTINUED
FCE = 6171.5 − 1152sin16.26° − 2057.2cos 32.52°
= 4114.3 lb
Joint E:
FCE = 4.11 kips T W
ΣFy = 0: ( 4114.3 lb ) sin 2θ − FDE cosθ = 0
FDE = 4114.3
sin 32.52°
= 2304.0 lb
cos16.26°
FDE = 2.30 kips C W
PROBLEM 6.18
Determine the force in each of the members located to the right of
DE for the inverted Howe roof truss shown. State whether each
member is in tension or compression.
SOLUTION
ΣFx = 0: A x = 0
FBD Truss:
By symmetry of loads A y = H y = 2400 lb
θ = tan −1
Note:
10.08
= 16.26°
15.81 + 18.75
β = 90 − 2θ = 57.48°
Joint FBDs:
Joint H:
ΣFy = 0: 2400 lb − 600 lb − FFH sin θ = 0
FFH =
1800 lb
= 6428.7 lb
sin16.26°
FFH = 6.43 kips C W
ΣFx = 0: ( 6428.7 lb ) cosθ − FGH = 0
FGH = 6428.7 cos16.32° = 6171.5 lb
FGH = 6.17 kips T W
Joint F:
ΣFy′ = 0: FFG − (1200 lb ) cosθ = 0
FFG = 1152.0 lb
FFG = 1.152 kips C W
ΣFx′ = 0: FDF + (1200 lb ) sin θ − 6428.7 lb = 0
FDF = 6428.7 − 1200sin16.26° = 6092.7 lb
FDF = 6.09 kips C W
PROBLEM 6.18 CONTINUED
Joint G:
ΣFy = 0: FDG sin 2θ − (1152 lb) cosθ = 0
FDG = 1152
cos16.26°
= 2057.2 lb
sin 32.52°
FDG = 2.06 kips T W
ΣFx = 0: 6171.5 lb − 2057.2cos 2θ − FEG − 1152 lb sin θ = 0
FEG = 6171.5 − 2057.2cos 32.52° − (1152lb ) sin 16.26° = 4114.3 lb
FEG = 4.11 kips T W
PROBLEM 6.19
Determine the force in each of the members located to the left of
member FG for the scissor roof truss shown. State whether each
member is in tension or compression.
SOLUTION
FBD Truss:
ΣM A = 0: ( 6 ft ) 6 Ly − 3 (.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0
L y = 0.75 kip
ΣFy = 0: Ay − 0.5 kip − 1 kip − 1 kip − 0.5 kip + 0.75 kip = 0
A y = 2.25 kips
ΣFx = 0: A x = 0
Joint FBDs:
Joint A:
1.75 kips
F
F
= AC = AB
1
5
8
FAB = 4.95 kips C W
FAC = 3.913 kips
FAC = 3.91 kips T W
FBC = 0 W
By inspection of joint C:
FCE = FAC
and
Joint B:
ΣFy = 0:
so
4.95 kips
F
− 1 kip − BD = 0
2
2
FBD = 3.536 kips
ΣFx = 0: ( 4.95 kips − 3.536 kips )
FBE = 1.000 kip
Joint E:
ΣFx = 0:
FBD = 3.54 kips C W
1
− FBE = 0
2
FBE = 1.000 kip C W
2
[ FEG − 3.913 kips] + 1 kip = 0
5
FEG = 2.795 kips
ΣFy = 0:
FCE = 3.91 kips T W
FEG = 2.80 kips T W
1
( 2.795 kips − 3.913 kips ) + FDE = 0
5
FDE = 0.500 kip T W
PROBLEM 6.19 CONTINUED
Joint D:
ΣFx = 0:
ΣFy = 0:
Solving:
1
2
( 3.536 kips ) − ( FDF + FDG ) = 0
2
5
1
1
( 3.536 kips ) − 1.5 kips + ( FDG − FDF ) = 0
2
5
FDF = 2.516 kips
FDG = 0.280 kip
so
FDF = 2.52 kips C W
FDG = 0.280 kip C W
PROBLEM 6.20
Determine the force in member FG and in each of the members
located to the right of member FG for the scissor roof truss shown.
State whether each member is in tension or compression.
SOLUTION
ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0
FBD Truss:
L y = 0.75 kip
FJK = 0 W
Inspection of joints K , J , and I , in order, shows that
FIJ = 0 W
FHI = 0 W
FIK = FKL ; FHJ = FJL and FGI = FIK
and that
Joint FBDs:
0.75
F
F
= JL = KL
1
8
5
FJL = 2.1213 kips
FKL = 1.6771 kips
FJL = 2.12 kips C W
FKL = 1.677 kips T W
FHJ = 2.12 kips C W
and, from above:
FGI = FIK = 1.677 kips T W
ΣFx = 0:
ΣFy = 0:
Solving:
2
1
( FFH + FGH ) − ( 2.1213 kips ) = 0
5
2
1
1
( FGH + FFH ) + ( 2.1213 kips ) = 0
5
2
FFH = 2.516 kips
FGH = − 0.8383 kips
FFH = 2.52 kips C W
FGH = 0.838 kips T W
PROBLEM 6.20 CONTINUED
ΣFx = 0:
2
( FDF − 2.516 kips ) = 0
5
ΣFy = 0: FFG − 0.5 kip +
FDF = 2.52 kips C
1
( 2 )( 2.516 kips ) = 0
5
FFG = 1.750 kips T W
PROBLEM 6.21
The portion of truss shown represents the upper part of a power
transmission line tower. For the given loading, determine the force in
each of the members located above HJ. State whether each member is
in tension or compression.
SOLUTION
Joint FBDs:
FDF
F
1.5 kN
= EF =
2.29 2.29
1.2
FDF = FEF = 2.8625 kN
FDF = 2.86 kN T W
FEF = 2.86 kN C W
FBD
F
2.8625 kN
= DE =
= 1.25 kN
2.21
0.6
2.29
FBD = 2.7625 kN
FBD = 2.76 kN T W
FDE = 0.750 kN C W
FAB = 2.86 kN T W
By symmetry of joint A vs. joint F
FAC = 2.86 kN T W
ΣFx = 0: 2.7625 kN −
2.21
4
( 2.8625 kN ) + FBE = 0
2.29
5
FBE = 0 W
ΣFy = 0: FBC −
0.6
( 2.8625 kN ) = 0;
2.29
FBC = 0.750 kN C W
ΣFx = 0:
2.21
( 2.8625 kN ) − FCE = 0
2.29
FCE = 2.7625 kN
FCE = 2.76 kN C W
ΣFy = 0: FCH − 0.75 kN −
0.6
( 2.8625 kN ) = 0
2.21
FCH = 1.500 kN C W
PROBLEM 6.21 CONTINUED
ΣFx = 0: 2.7625 kN −
2.21
4
( 2.8625 kN ) − FHE = 0
2.29
5
FHE = 0 W
ΣFy = 0: FEJ − 0.75 kN −
0.6
( 2.8625 kN ) = 0
2.29
FEJ = 1.500 kN W
PROBLEM 6.22
For the tower and loading of Prob. 6.21 and knowing that
FCH = FEJ = 1.5 kN C and FEH = 0 , determine the force in member
HJ and in each of the members located between HJ and NO. State
whether each member is in tension or compression.
SOLUTION
1.5 kN
F
F
= JL = KL
1.2
3.03 3.03
Joint FBDs:
FJL = FKL = 3.7875 kN
FJL = 3.79 kN T W
FKL = 3.79 kN C W
FGH = 3.79 kN T W
By symmetry:
FGI = 3.79 kN C W
ΣFx = 0:
2.97
( 3.7875 kN ) − FHJ = 0
3.03
FHJ = 3.7125 kN
ΣFy = 0: FJK −
FHJ = 3.71 kN T W
0.6
( 3.7875 kN ) − 1.5 kN = 0
3.03
FJK = 2.25 kN C W
Knowing FHE = 0; by symmetry
FHK = 0 W
FHI = 2.25 kN C W
ΣFx = 0:
2.97
( 3.7875 kN ) − FIK = 0
3.03
FIK = 3.7125
FIK = 3.71 kN C W
ΣFy = 0: FIN − 2.25 kN −
0.6
( 3.7875 kN ) = 0
3.03
FIN = 3.00 kN C W
Knowing that FHK = 0, by symmetry
FKO = 3.00 kN C W
FKN = 0 W
PROBLEM 6.23
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM F = 0: (10 m ) G y − ( 7.5 m )( 80 kN ) − ( 8 m )( 30 kN ) = 0
G y = 84 kN
ΣFx = 0: − Fx + 30 kN = 0
Fx = 30 kN
ΣFy = 0: Fy + 84 kN − 80 kN = 0
Fy = 4 kN
FEG = 0 W
By inspection of joint G:
FCG = 84 kN C W
84 kN
F
F
= CE = AC = 10.5 kN
8
61
29
Joint FBDs:
FCE = 82.0 kN T W
FAC = 56.5 kN C W
ΣFy = 0:
2
6
FAE +
(82.0 kN ) − 80 kN = 0
5
61
FAE = 19.01312
ΣFx = 0: − FDE −
1
(19.013 kN ) +
5
FDE = 43.99 kN
ΣFx = 0:
FAE = 19.01 kN T W
5
(82.0 kN ) = 0
61
FDE = 44.0 kN T W
1
FDF − 30 kN = 0
5
FDF = 67.082 kN
FDF = 67.1 kN T W
PROBLEM 6.23 CONTINUED
ΣFy = 0:
2
( 67.082 kN ) − FBF − 4 kN = 0
5
FBF = 56.00 kN
ΣFx = 0: 30 kN +
ΣFy = 0: 56 kN −
5
FBD −
61
6
FBD −
61
FBD = 42.956 kN
Solving:
FAB = 61.929 kN
ΣFy = 0:
FBF = 56.0 kN C W
5
FAB = 0
29
3
FAB = 0
29
FBD = 43.0 kN T W
FAB = 61.9 kN C W
6
2
( 42.956 N ) + ( FAD − 67.082 N ) = 0
61
5
FAD = 30.157 kN
FAD = 30.2 N T W
PROBLEM 6.24
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Joint FBDs:
9 kN
F
F
= AC = AB
9
40
41
FAB = 41.0 kN T W
FAC = 40.0 kN C W
FBD = 41.0 kN T W
By inspection of joint B:
FBC = 3.00 kN C W
Note: to determine slope of CD: DE =
ΣFy = 0:
ΣFx = 0:
1
FCD − 3 kN = 0
5
(
4.6
( 2.25 m ) = 1.035 m
10.0
FCD = 3 5 kN = 6.71 kN T W
)
2
3 5 kN + 40 kN − FCE = 0
5
ΣFx = 0:
(
)
40
2
3 5 kN = 0
( FDG − 41 kN ) −
41
5
FDG = 47.15 kN
ΣFy = 0: FDE +
FCE = 46.0 kN C W
FDG = 47.2 kN T W
(
)
9
1
3 5 kN = 0
( 47.15 kN − 41 kN ) −
41
5
FDE = 1.650 kN C W
ΣFy = 0:
5
FEG − 1.65 kN = 0
13
ΣFx = 0: 46 kN +
12
( 4.29 kN ) − FEF = 0
13
FEG = 4.29 kN T W
FEF = 49.96 kN
FEF = 50.0 kN C W
PROBLEM 6.24 CONTINUED
49.96 kN
F
F
= FG = FH
5
4
3
FFG = 40.0 kN C W
FFH = 30.0 kN C W
PROBLEM 6.25
For the roof truss shown in Fig. P6.25 and P6.26, determine the force
in each of the members located to the left of member GH. State
whether each member is in tension or compression.
SOLUTION
FBD Truss:
ΣM M = 0: ( 3 ft )( 300 lb ) + ( 9 ft )( 300 lb ) + (15 ft )( 300 lb )
+ (18 ft )( 300 lb ) + ( 24 ft )( 300 lb ) + ( 30 ft )( 200 lb )
( )
− ( 30 ft ) Ay = 0
A y = 890 lb
ΣFx = 0: A x = 0
690 lb
F
F
= AC = AB
5
12
13
FAB = 1794 lb C W
FAC = 1656 lb T W
Joint FBDs:
FCF = 0 W
By inspection of joint F:
FEF = 0 W
1656 lb
F
F
= CE = BC
12
13
5
FCE = 1794 lb T W
FBC = 890 lb C W
ΣFy = 0:
5
(1794 lb − FBD ) + 690 lb − 300 lb = 0
13
FBD = 2808 lb
ΣFx = 0: FBE +
FBD = 2.81 kips C W
12
(1794 lb − 2808 lb ) = 0
13
ΣFx = 0:
6
12
FEH − 936 lb − 1794 lb = 0
13
37
FEH = 432 37 lb
ΣFy = 0: FDE −
FBE = 936 lb T W
5
(1794 lb ) −
13
FEH = 2.63 kips T W
(
)
1
432 37 lb = 0
37
FDE = 1122 lb T W
PROBLEM 6.25 CONTINUED
ΣFx = 0:
ΣFy = 0:
12
1
( 2808 lb + FDG ) − FDH = 0
13
2
5
1
( 2808 lb + FDG ) + FDH − 300 lb − 1122 lb = 0
13
2
Solving:
FDG = 1721 lb T W
FDH = 1419 lb C W
PROBLEM 6.26
Determine the force in member GH and in each of the members located
to the right of GH for the roof truss shown. State whether each member
is in tension or compression.
SOLUTION
ΣM A = 0: ( 30 ft ) M y − ( 30 ft )( 200 lb ) − ( 27 ft )( 300 lb )
FBD Truss:
− ( 21 ft )( 300 lb ) − (15 ft )( 300 lb )
− (12 ft )( 300 lb ) − ( 6 ft )( 300 lb ) = 0
M y = 1010 lb
810 lb
F
F
= LM = KM
5
12
13
Joint FBDs:
FKM = 2106 lb
FKM = 2.11 kips C W
FLM = 1944 lb
FLM = 1.944 kips T W
FJL
F
1944 lb
= KL =
1
6
37
FKL = 324 lb C W
FJL = 1970.8 lb
ΣFx = 0:
ΣFy = 0:
FJL = 1.971 kips T W
12
( FIK − 2106 lb ) +
13
5
( − FIK + 2106 lb ) −
13
1
FJK + 24 lb = 0
577
FIK = 2162.7 lb
Solving:
24
FJK = 0
577
FIK = 2.16 kips C W
FJK = 52.4 lb T W
ΣFx = 0:
6
(1970.8 lb − FHJ ) +
37
FHJ = 2024 lb
ΣFy = 0:
1
( 2024 lb − 1970.8 lb ) +
37
24
( 52.4 lb ) = 0
577
FHJ = 2.02 kips T W
1
( 52.4 lb ) − FIJ = 0
577
FIJ = 10.90 lb C W
PROBLEM 6.26 CONTINUED
ΣFy′ = 0: FHI sin 38.88° + (10.9 lb − 300 lb ) cos 22.62° = 0
FHJ = 425.1 lb
FHI = 425 lb C W
ΣFx′ = 0: FGI + FHI cos 38.88° + ( 300 lb − 10.9 lb ) sin 22.62°
−2162.7 lb = 0
FGI = 1720.5 lb = 1.721 kips C W
By symmetry FDG = 1720.5 lb
ΣFy = 0:
5
2 (1720.5 lb ) − 300 lb − FGH = 0
13
FGH = 1023.46 lb
FGH = 1.023 kips T W
PROBLEM 6.27
Determine whether the trusses of Probs. 6.13, 6.14, and 6.25 are
simple trusses.
SOLUTION
Truss of 6.13:
Start with ∆ABC and add, in order, joints E, D, F , G, H
This is a simple truss. W
Truss of 6.14:
ABDC, DEGF, and GHJI are all individually simple trusses, but no
simple extension (one joint, two members at a time) will produce the
given truss:
∴ Not a simple truss. W
Truss of 6.25:
Starting with ∆ABC , add, in order, joints E, F , D, H , G, I , J , K , L, M
This is a simple truss. W
PROBLEM 6.28
Determine whether the trusses of Probs. 6.19, 6.21, and 6.23 are
simple trusses.
SOLUTION
Truss of 6.19:
Start with ∆ABC , and add, in order, E, D, G, F, H, I, J, K
∴ Simple truss W
Truss of 6.21:
Start with ∆ABC , and add, in order, E, D, F, H, J, K, I, G, L, N, O, R, Q,
M, P, S, T, etc.
∴ Simple truss W
Truss of 6.23:
Start with ∆BDF , and add, in order, A, E, C, G.
∴ Simple truss W
PROBLEM 6.29
For the given loading, determine the zero-force members in the
truss shown.
SOLUTION
By inspection of joint B :
FBC = 0 W
Then by inspection of joint C:
FCD = 0 W
By inspection of joint J :
FIJ = 0 W
Then by inspection of joint I :
FIL = 0 W
By inspection of joint N :
FMN = 0 W
Then by inspection of joint M :
FLM = 0 W
PROBLEM 6.30
For the given loading, determine the zero-force members in the truss
shown.
SOLUTION
By inspection of joint C:
FBC = 0 W
Then by inspection of joint B:
FBE = 0 W
By inspection of joint G:
FFG = 0 W
Then by inspection of joint F:
FEF = 0 W
Then by inspection of joint E:
FDE = 0 W
By inspection of joint M :
FMN = 0 W
Then by inspection of joint N :
FKN = 0 W
By inspection of joint I :
FIJ = 0 W
PROBLEM 6.31
For the given loading, determine the zero-force members in
the truss shown.
SOLUTION
By inspection of joint D:
FDI = 0 W
By inspection of joint E:
FEI = 0 W
Then by inspection of joint I :
FAI = 0 W
By inspection of joint B:
FBJ = 0 W
By inspection of joint F:
FFK = 0 W
By inspection of joint G:
FGK = 0 W
Then by inspection of joint K :
FCK = 0 W
PROBLEM 6.32
For the given loading, determine the zero-force members
in the truss shown.
SOLUTION
By inspection of joint C:
FBC = 0 W
By inspection of joint M :
FLM = 0 W
PROBLEM 6.33
For the given loading, determine the zero-force members in the truss
shown.
SOLUTION
By inspection of joint F:
FBF = 0 W
Then by inspection of joint B:
FBG = 0 W
Then by inspection of joint G:
FGJ = 0 W
Then by inspection of joint J :
FHJ = 0 W
By inspection of joint D:
FDH = 0 W
Then by inspection of joint H :
FHE = 0 W
PROBLEM 6.34
For the given loading, determine the zero-force members in the truss
shown.
SOLUTION
By inspection of joint C:
FBC = 0 W
Then by inspection of joint B:
FBE = 0 W
Then by inspection of joint E:
FDE = 0 W
By inspection of joint H :
FFH = 0 W
and FHI = 0 W
By inspection of joint Q :
FOQ = 0 W
and FQR = 0 W
By inspection of joint J :
FIJ = 0 W
PROBLEM 6.35
Determine the zero-force members in the truss of (a) Prob. 6.9, (b)
Prob. 6.19.
SOLUTION
Truss of 6.9:
By inspection of joint C:
By inspection of joint G:
Truss of 6.19:
FBC = 0 W
FFG = 0 W
By inspection of joint C:
FBC = 0 W
By inspection of joint K :
FJK = 0 W
Then by inspection of joint J :
FIJ = 0 W
Then by inspection of joint I :
FHI = 0 W
PROBLEM 6.36
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = ( −5670 lb ) j and Q = 0.
SOLUTION
ΣFz = 0: B z = 0
FBD Truss:
ΣM BD = 0: ( 2.4 ft )( 5670 lb ) − ( 8.4 ft ) C y = 0
C y = (1620 lb ) j
ΣM x = 0: ( 6.3 ft ) D y − ( 6.3 ft ) By = 0
ΣFy = 0: By + D y − 5670 lb + 1620 lb = 0
By = Dy
B y = D y = ( 2025 lb ) j
ΣM y = 0: ( 6.3 ft ) Bx − ( 6.3 ft ) Dx = 0
B x = Dx = 0
ΣFx = 0: Bx + Dx = 0
Joint B:
Where
2.4i + 14.4 j − 6.3k
15.9
= FBA ( 0.1509i + 0.9057 j − 0.3962k )
FBA = FBA
FBC = FBC
8.4i − 6.3k
= FBC ( 0.8i − 0.6k )
10.5
ΣFy = 0: 0.9057 FBA + 2025 lb = 0
FBA = −2236 lb
FBA = 2.24 kips C W
By symmetry FAD = 2.24 kips C W
ΣFx = 0: 0.1509 ( −2236 lb ) + 0.8FBC = 0
FBC = 422 lb T W
By symmetry FDC = 422 lb T W
PROBLEM 6.36 CONTINUED
Joint C:
ΣFz = 0: −0.3962 ( −2236 lb ) − FBD − 0.6 ( 422 lb ) = 0
FBD = 633 lb T W
FAC = FAC
6i − 14.4 j
= FAC ( 0.3846i − 0.9231j)
15.6
ΣFy = 0: 1620 lb − ( 0.9231) FAC = 0
FAC = 1755 lb C W
PROBLEM 6.37
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = 0 and Q = ( 7420 lb ) i .
SOLUTION
ΣFz = 0: B z = 0
FBD Truss:
ΣM BD = 0: ( 8.4 ft ) C y − (14.4 ft )( 7420 lb ) = 0
C y = (12720 lb ) j
(
)
ΣM x = 0: ( 6.3 ft ) D y − By = 0
ΣFy = 0: By + Dy + C y = 0;
Dy = By
2 Dy + 12720 lb = 0
B y = D y = − ( 6360 lb ) j
so
ΣM y = 0: ( 6.3 ft )( Bx − Dx ) = 0;
Joint FBDs:
B x = D x = − ( 3710 lb ) i
ΣFx = 0: Bx + Dx + 7420 lb = 0;
FBA = FBA
Bx = Dx
( 2.4 ft i + 14.4 ft j − 6.3 ft k )
15.9 ft
= FBA ( 0.1509i + 0.9057 j − 0.3962k )
FBC = FBC
(8.4 ft i − 6.3 ft j)
10.5 ft
FBD = FBDk
= FBC ( 0.8i − 0.6 j)
ΣFy = 0: 0.9057 FBA − 6360 lb = 0
FBA = 7022 lb
FBA = 7.02 kips T W
FDA = 7.02 kips T W
By symmetry
ΣFx = 0: 0.1509 ( 7022 lb ) + 0.8FBC − 3710 lb = 0
FBC = 3313 lb
so FBC = 3.31 kips T W
FDC = 3.31 kips T W
By symmetry
ΣFz = 0: −0.3962 ( 7022 lb ) + 0.6 ( 3313 lb ) − FBD = 0
FBD = −4770 lb
FBD = 4.77 kips C W
PROBLEM 6.37 CONTINUED
FAC = FAC
( 6 ft i − 14.4 ft j)
15.6 ft
= FAC ( 0.3846i − 0.9231j)
ΣFy = 12720 lb − 0.9231 FAC = 0;
FAC = 13780 lb
FAC = 1.378 kips C W
PROBLEM 6.38
The truss shown consists of six members and is supported by a short
link at A, two short links at B, and a ball and socket at D. Determine
the force in each of the members for the given loading.
SOLUTION
ΣFz = 0: D z = 0
FBD Truss:
ΣM z = 0: ( 5 ft ) Ax − (12 ft )( 800 lb ) = 0
A x = (1920 lb ) i
ΣM y = 0: ( 3.5 ft ) ( Bx − Dx ) = 0; Bx = Dx
ΣFx = 0: Bx + Dx − 1920 lb = 0
so
(
B x = D x = ( 960 lb ) i
)
ΣM x = 0: ( 3.5 ft ) Dy − By = 0; Dy = By
ΣFy = 0: By + D y − 800 lb = 0
Joint FBDs:
FCA = FAC
FCD = FCD
so
( −12 ft i + 5 ft j)
13 ft
( −12 ft i − 3.5 ft k )
12.5 ft
FCB =
Similarly
ΣFz = 0:
B y = D y = ( 400 lb ) j
=
FAC
( −12i + 5j)
13
=
FCD
( −12i − 3.5k )
12.5
FCB
( −12i + 3.5k )
12.5
3.5
( FCB − FCD ) = 0;
12.5
5
ΣFy = 0: FAC   − 800 = 0
 13 
FCB = FCD
FAC = 2080 lb
FAC = 2.08 kips T W
FBA = FBA
( 5 ft j − 3.5 ft k )
6.1033 ft
=
FBA
( 5j − 3.5k )
6.1033
FBD = − FBD k
FBC = −FCB =
FCB
( +12i − 3.5k )
12.5
PROBLEM 6.38 CONTINUED
ΣFy = 0:
5FBA
+ 400 lb = 0
6.1033
FBA = −488 lb
so FBA = 488 lb C W
FAD = 488 lb C W
By symmetry:
 12 
ΣFx = 0: FBC 
 + 960 lb = 0
 12.5 
FBC = −1000 lb
FBC = 1.000 kip C W
FCD = 1.000 kip C W
By symmetry:
3.5
 3.5 
ΣFz = 0: − FBD − 488 lb 
=0
 + (1000 lb )
12.5
 6.1033 
FBD = −559.9 lb
FBD = 560 lb C W
PROBLEM 6.39
The portion of a power line transmission tower shown consists of nine
members and is supported by a ball and socket at B and short links at
C, D, and E. Determine the force in each of the members for the given
loading.
SOLUTION
ΣM BD = 0: (1 m )( 200 N − Ez ) = 0
FBD Truss:
E z = ( 200 N ) k
ΣM x = 0: (1 m )(1200 N − Ez − Dz ) = 0
Dz = 1200 N − 200 N = 1000 N DZ = (1000 N ) k
B z = (1200 N ) k
ΣFz = 0: Bz − 1000 N − 200 N = 0
ΣM Bz = 0: (.5 m )(1200 N ) − (1 m ) C y = 0
ΣFx = 0: Bx − 200 N = 0
C y = ( 600 N ) j
B x = ( 200 N ) i
ΣFy = 0: By + C y − 1200 N = 0
By = 1200 N − 600 N
B y = ( 600 N ) j
Joint FBDs:
ΣFz = 0:
FEA
− 200 N = 0
1.5
ΣFx = 0: FED +
.5
FEA = 0
1.5
FEA = 300 N T W
FED = −100 N
FED = 100 N C W
ΣFy = 0: −
FEA
− FEC = 0
1.5
FEC = −200 N
FEC = 200 N C W
ΣFz = 0:
FAD
− 1000 N = 0
1.5
ΣFx = 0: 100 N −
FCD =
0.5
1
FAD −
FCD = 0
1.5
2
1500 N 

2 100 N −
 = −400 2
3 

ΣFy = 0:
−
FAD = 1500 N T W
FCD = 566 N C W
FCD FAD
−
− FBD = 0
1.5
2
FBD = +400 − 1000 = −600 N
FBD = 600 N C W
PROBLEM 6.39 CONTINUED
ΣFy = 0: 600 N − 200 N − 400 N = 0
ΣFz = 0:
FAC
=0
1.25
ΣFx = 0: FBC − 400 N +
FAC = 0 W
.5
FAC0 = 0
1.25
FBC = 400 N T W
ΣFx = 0: −200 N − 400 N − FAB
FAB = 1200 1.25 N
.5
=0
1.25
FAB = 1342 N C W
PROBLEM 6.40
The truss shown consists of 18 members and is supported by a ball and
socket at A, two short links at B, and one short link at G. (a) Check that
this truss is a simple truss, that it is completely constrained, and that the
reactions at its supports are statically determinate. (b) For the given
loading, determine the force in each of the six members joined at E.
SOLUTION
(a) To check for simple truss, start with ABDE and add three members at a time which meet
at a single joint, thus successively adding joints F, G, H, and C, to complete the truss.
This is, therefore, a simple truss.W
There are six reaction force components, none of which are in-line, so they are determined
by the six equilibrium equations. Constraints prevent motion. Truss is completely constrained and
W
statically determinate W
(b) FBD Truss:
( )
ΣM BC = 0: ( 5.04 m )( 550 N ) + ( 5.5 m ) Ay = 0
A y = − ( 504 N ) j
ΣM BF = 0: ( 5.5 m )( Az + 480 N ) − ( 4.8 m )( 550 N ) = 0
Az = 0
By inspection of joint C:
FDC = FBC = FGC = 0
By inspection of joint A:
FAE = − Ay = 504 N T W
FAD = Az = 0
By inspection of joint H :
FDH = 0
FEH = 480 N C W
By inspection of joint F :
FEF = 0 W
PROBLEM 6.40 CONTINUED
FDE = 0 W
Then, since ED is the only non-zero member at D, not in the plane BDG,
Joint E:
ΣFz = 0: 480 N −
ΣFy = 0: −504 N −
4.8
FEG = 0
7.3
5.04
FBE = 0
7.46
FEG = 730 N T W
FBE = −746 N FBE = 746 N C W
PROBLEM 6.41
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at G.
SOLUTION
(a) See part (a) solution 6.40 above
(b) FBD Truss:
ΣM AB = 0:
( 4.8 m ) Gy + ( 5.04 m )( 480 N ) = 0
G y = −504 N
By inspection of joint C:
FCG = 0 W
By inspection of joint H :
FHG = 550 N C W
By inspection of joint F:
FFG = 0 W
Joint G:
ΣFx = 0: 550 N −
5.5
5.5
FDG −
FEG = 0
7.46
7.30
PROBLEM 6.41 CONTINUED
ΣFy = 0: −504 N −
ΣFz = 0:
Solving:
5.04
5.04
FDG −
FBG = 0
7.46
6.96
4.8
4.8
FEG +
FBG = 0
7.30
6.96
FBG = −696 N
FDG = 0
FEG = 730 N
FBG = 696 N C W
FDG = 0 W
FEG = 730 N T W
PROBLEM 6.42
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
FBD Truss:
ΣM K = 0: (15 m )( 3 kN ) + ( 20 m )( 3 kN ) − ( 25 m ) Ay = 0
A y = 4.2 kN
ΣM D = 0: ( 6 m ) FCE + ( 2.5 m )( 3 kN ) − ( 7.5 m )( 4.2 kN ) = 0
Section ABDC:
FCE = 4 kN
FCE = 4.00 kN T W
ΣFy = 0: 4.2 kN − 3 kN −
FDE = 1.3 kN
ΣFx = 0: FDF +
FDF = −
12
FDE = 0
13
FDE = 1.300 kN T W
5
FDE + FCE = 0
13
5
(1.3 kN ) − ( 4 kN ) = −4.5 kN
13
FDF = 4.50 kN C W
PROBLEM 6.43
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
FBD Truss:
ΣFx = 0: K x = 0
ΣM A = 0: ( 25 m ) K y − (10 m )( 3 kN ) − ( 5 m )( 3 kN ) = 0
K y = 1.8 kN
ΣM G = 0: (10 m )(1.8 kN ) + ( 6 m ) FFH = 0
Section FBD:
FFH = −3 kN
FFH = 3.00 kN C W
ΣM F = 0: (12.5 m )(1.8 kN ) − ( 6 m ) ( FEG ) = 0
FEG = 3.75 kN
ΣFy = 0:
FEG = 3.75 kN T W
12
FFG + 1.8 kN = 0
13
FFG = −1.95 kN
FFG = 1.950 kN C W
PROBLEM 6.44
A parallel chord Howe truss is loaded as shown. Determine the force
in members CE, DE, and DF.
SOLUTION
ΣFx = 0: A x = 0
FBD Truss:
ΣM M = 0: ( 4 ft ) 1( 0.3 kip ) + 2 ( 0.6 kip ) + 3 (1 kip )
+ 4 ( 0.6 kip ) + 5 ( 0.3 kip ) + 6 ( 0.3 kip ) − 6 Ay  = 0
A y = 1.7 kips
 4

ΣM D = 0: ( 4 ft )( 3 kips − 1.7 kips ) + ( 2 ft ) 
FCE = 0 
 4.1

FCE = 2.87 kips
FBD Section:
FCE = 2.87 kips T W
ΣM E = 0: ( 8 ft )( 3 kips − 1.7 kips ) + ( 4 ft )( 0.3 kip )
 4

− ( 2 ft ) 
FDF  = 0
 4.1

FDF = −5.125 kips
ΣFx = 0:
FDE = −
FDF = 5.13 kips C W
4
4
FDE = 0
( FDF + FCE ) +
4.1
17.21
17.21
( −5.125 + 2.87 ) kips
4.1
FDE = 2.28 kips
FDE = 2.28 kips T W
PROBLEM 6.45
A parallel chord Howe truss is loaded as shown. Determine the force
in members GI, GJ, and HJ.
SOLUTION
FBD Truss:
ΣM A = ( 24 ft ) M y − 4 ft ( 0.3 kip ) − 8 ft ( 0.6 kip ) − 12 ft (1 kip )
− 16 ft ( 0.6 kip ) − 20 ft ( 0.3 kip ) − 24 ft ( 0.3 kip ) = 0
M y = 1.7 kips
FBD Section:
 4

ΣM J = 8 ft (1.7 − 0.3) kips − 4 ft ( 0.3 kip ) − 2 ft 
FGI  = 0
 4.1

FGI = 5.125 kips
FGI = 5.3 kips T W
ΣM G = 12 ft (1.7 − 0.3) kips − 8 ft ( 0.3 kip )
 4

− 4 ft ( 0.6 kip ) + 2 ft 
FHJ  = 0
 4.1

FHJ = − 6.15 kips = 6.15 kips C W
ΣFx =
4
4
FGJ
( 6.15 − 5.125 ) kips −
4.1
4.94
FGJ = 1.235 kips T W
PROBLEM 6.46
A floor truss is loaded as shown. Determine the force in members CF,
EF, and EG.
SOLUTION
FBD Truss:
ΣM K = 0: (1 m ) 1 kN + 2 (1 kN ) + 3 (1.5 kN )
+ 4 ( 2 kN ) + 5 ( 2 kN ) + 6 (1 kN ) − 6 Ay  = 0
A y = 5.25 kN
ΣM E = 0: (1 m ) 1( 2 kN ) + 2 (1 kN − 5.25 kN )  + ( 0.5 m ) FCE = 0
FCF = 13.0 kN
FBD Section:
FCF = 13.00 kN T W
ΣM F = 0: (1 m ) 1( 2 kN ) + 2 ( 2 kN ) + 3 (1 kN − 5.25 kN ) 
− ( 0.5 m ) FEG = 0
FEG = −13.5 kN
FEG = 13.50 kN C W
ΣFy = 0: 5.25 kN − 1 kN − 2 kN − 2 kN −
FEF =
5
= 0.5590 kN
4
1
FEF = 0
5
FEF = 559 N T W
PROBLEM 6.47
A floor truss is loaded as shown. Determine the force in members FI,
HI, and HJ.
SOLUTION
FBD Truss:
ΣFx = 0: K x = 0
(
)
ΣM A = 0: (1m ) 6 K y − 0.5 kN − 5 (1 kN ) − 4 (1 kN ) − 3 (1.5 kN ) − 2 ( 2 kN ) − 1( 2 kN )  = 0


K y = 3.75 kN
FBD Section:
ΣM I = 0:
(1 m )( 3.75 kN − .5 kN ) − (.5 m ) FHJ
=0
FHJ = 6.5 kN
ΣFy = 0:
FHJ = 6.50 kN C W
1
FHI − 1 kN − 0.5 kN + 3.75 kN = 0
5
FHI = −2.25 5 kN
ΣFy = 0: −
FHI = 5.03 kN C W
2
FHI − FFI + FHJ = 0
5
FFI = 2 ( 2.25 kN ) + 6.50 kN
FFI = 11.00 kN T W
PROBLEM 6.48
A pitched flat roof truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
ΣFx = 0: A x = 0
FBD Truss:
By load symmetry: A y = I y = 8 kips
FBD Section:
ΣM D = 0: ( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0
FCE = 14 kips
FCE = 14.00 kips T W
ΣM E = 0: ( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips ) 
( 4.5 ft )
FDF =
7
FDF = 0
51.25
8 51.25
kips
4.5
ΣFy = 0: 8 kips − 2 kips − 4 kips +
FDE = −1.692 kips
FDF = 12.73 kips C W
1.5 8 51.25
kips −
51.25 4.5
3
FDE = 0
58
FDE = 1.692 kips C W
PROBLEM 6.49
A pitched flat roof truss is loaded as shown. Determine the force in
members EG, GH, and HJ.
SOLUTION
FBD Truss:
By load symmetry: A y = I y = 8 kips
FBD Section:
ΣM = 0: ( 7 ft )( 8 kips − 2 kips ) − ( 6 ft ) FEG = 0
FEG = 7.00 kips T W
ΣFx = 0:
FHJ =
ΣFy = 0:
7
FHJ − 7.00 kips = 0
51.25
FHJ = 7.16 kips C W
51.25 kips
1.5
51.25
(
)
51.25 kips − FHG + ( 8 − 2 ) kips = 0
FHG = 7.50 kips C W
PROBLEM 6.50
A Howe scissors roof truss is loaded as shown. Determine the force in
members DF, DG, and EG.
SOLUTION
FBD Truss:
Σ FX = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
15
m
6
2 5 5
yD = ⋅ = m
3 2 3
2
2
yE = ⋅ 1 = m
3
3
5
yF − yD = m
6
yG = 1 m
yF =
yD − yG =
2
m
3
PROBLEM 6.50 CONTINUED
ΣM D = 0: (1 m )
6
FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0
37
FEG =
8
37 kN
3
FEG = 16.22 kN T W
 1

 3

ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m ) 
FDG  − (1 m ) 
FDG  = 0
 10

 10

FDG =
ΣFx = 0:
4
10 kN
3
6
3
12
FEG −
FDG −
FDF = 0
13
37
10
FDF = 13 kN
FDG = 4.22 kN C W
16 − 4 −
12
FDF = 0
13
FDF = 13.00 kN C W
PROBLEM 6.51
A Howe scissors roof truss is loaded as shown. Determine the force in
members GI, HI, and HJ.
SOLUTION
FBD Truss:
Σ Fx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
so
2
3
2
=
3
yI =
m
yH
⋅
5 5
= m
2 3
yH − yI = 1 m
PROBLEM 6.51 CONTINUED
 12

ΣM I = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m )  FHJ  = 0
13


FHJ =
52
kN
3
FHJ = 17.33 kN C W
 6

ΣM H = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
FGI  = 0
 37

FGI =
8
37 kN
3
ΣM L = 0: ( 2 m )( 2 kN ) − ( 4 m ) FHI = 0
60
FGI = 16.22 kN T W
FHI = 1.000 kN T W
PROBLEM 6.52
A Fink roof truss is loaded as shown. Determine the force in members
BD, CD, and CE.
SOLUTION
FBD Truss:
Σ Fx = 0: A x = 0
By symmetry: A y = K y = 3.6 kips
FBD Section:
 16 
ft  FCE + ( 5 ft )(1.2 kips ) + (10 ft )( 0.6 kips ) − (10 ft )( 3.6 kips ) = 0
ΣM D = 0: 
 3 
FCE = 4.50 kips T W
4

ΣM A = 0: ( 6 ft )  FCD  − ( 5 ft )(1.2 kips ) = 0
5

FCD = 1.250 kips T W
ΣFy = 0: ( 3.6 − 0.6 ) kips − 1.2 kips +
FBD = 5.95 kips
4
8
(1.25 kips ) − FBD = 0
5
17
FBD = 5.95 kips C W
PROBLEM 6.53
A Fink roof truss is loaded as shown. Determine the force in members
FH, FG, and EG.
SOLUTION
FBD Truss:
Σ Fx = 0: A x = 0
By symmetry: A y = K y = 3.6 kips
FBD Section:
ΣM F = 0: (15 ft )( 3.6 − .6 ) kips − (10 ft )(1.2 kips ) − ( 5 ft ) (1.2 kips ) − ( 8ft ) FEG = 0
FEG = 3.375 kips
FEG = 3.38 kips T W
 8

ΣM K = 0: ( 5 ft )(1.2 kips ) + (10 ft )(1.2 kips ) − (12 ft ) 
FFG  = 0
 73

FFG =
ΣFy = 0
3
73 kips
16
8  3
 8
73 kips  −
FFH − 1.2 kips − 1.2 kips − 0.6 kip + 3.6 kips = 0

16
73 
 17
FFH = 4.4625 kips
62
FFG = 1.602 kips T W
FFH = 4.46 kips C W
PROBLEM 6.54
A Fink roof truss is loaded as shown. Determine the force in members
DF, DG, and EG. (Hint: First determine the force in member EK.)
SOLUTION
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = O y = 2000 lb
FBD Sections:
Forces in 1b
ΣM H = 0: ( a + 2a + 3a )( 500 lb ) + 4a ( 250 lb )
− 4a ( 2000 lb ) + 2aFEK = 0
FEK = 2000 lb T
ΣM D = 0: a ( 500 lb ) + 2a ( 250 lb ) − 2a ( 2000 lb ) + a ( 2000 lb )
3

4

+ a  FEG  + (12 ft − 2a )  FEG  = 0
5
5




FEG = 1000 lb
ΣFy = 0:
FEG = 1.000 kip T W
4
1
(1000 lb ) − FDF + ( 2000 − 250 − 500 − 500 ) lb = 0
5
5
FDF = 1550 5 lb
ΣFx = 0: 2000 lb +
FDF = 3.47 kips C W
(
)
3
2
1550 5 lb + FDG = 0
(1000 lb ) −
5
5
FDG = 500 lb T W
PROBLEM 6.55
A roof truss is loaded as shown. Determine the force in members FH,
GJ, and GI.
SOLUTION
By symmetry: A y = M y = 2.45 kips
ΣFx = 0: A x = 0
 4

ΣM J = 6 ft ( 0.3 kip ) − 12 ft ( 2.35 kips ) + 7 ft 
FGI  = 0
 17

FGI = 3.887 kips
FGI = 3.89 kips T W
ΣM G = ( 6 ft ) FHJ − ( 4 ft )(1.0 kips ) − (10 ft )( 0.3 kips ) − (16 ft )( 0.1 kips )
+
(16 ft )( 2.45 kips ) = 0
By inspection:
ΣFy = 0:
FHJ = −5.10 = 5.10 kips C
FFH = FHJ = 5.10 kips C W
1
3
FGJ = 0
( 3.887 kips ) + 2.45 kips − 1.4 kips −
17
13
FGJ = 2.40 kips T W
PROBLEM 6.56
A stadium roof truss is loaded as shown. Determine the force in
members AB, AG, and FG.
SOLUTION
FBD Section:
p = 8.4 ( 2.7 m ) = 1.89 m
Note: BG
12.0
ΣM A = 0: ( 2.7 m ) FFG − ( 3.6 m )( 8 kN ) − ( 7.8 m )( 8 kN ) − (12 m )( 4 kN ) = 0
FFG = 51.56 kN
FFG = 51.6 kN C W
 12

ΣM G = 0: (1.89 m ) 
FAB  − ( 4.2 m )( 8 kN ) − ( 8.4 m )( 4 kN ) = 0
 12.3

FAB = 36.44 kN
3

ΣM D = 0: ( 4.2 m )( 8 kN ) + ( 8.4 m )( 8 kN ) − ( 8.4 m )  FAG  = 0
5

FAB = 36.4 kN T W
FAG = 20.0 kN T W
PROBLEM 6.57
A stadium roof truss is loaded as shown. Determine the force in members
AE, EF, and FJ.
SOLUTION
FBD Section:
 8

ΣM F = 0: ( 2.7 m ) 
FAE  − ( 3.6 m )( 8 kN ) − ( 7.8 m )( 8 kN ) − (12 m )( 4 kN ) = 0
 145

FAE =
17.4
145 kN
2.7
ΣFx = 0: FEF −
8  17.4

145 kN  = 0

145  2.7

FEF = 51.555 kN
ΣFy = 0: FFJ −
FAE = 77.6 kN T W
9  17.4

145 kN  − ( 4 + 8 + 8 + 4 ) kN = 0

145  2.7

FEF = 51.6 kN C W
FFJ = 82.0 kN C W
PROBLEM 6.58
A vaulted roof truss is loaded as shown. Determine the force in members
BE, CE, and DF.
SOLUTION
FBD Truss:
ΣM N = 0: (1.2 m )( 0.2 kN ) + (1.8 m )( 0.5 kN ) + ( 2.88 m )( 0.5 kN ) + ( 4.2 m )(1.3 kN )
+ ( 4.8 m )( 0.6 kN ) + ( 6.6 m )( 0.7 kN ) + ( 8.4 m )( 0.1 kN ) − ( 8.4 m ) Ay = 0
A y = 1.95 kN
ΣFx = 0: A x = 0
FBD Section:
ΣM B = 0: ( 0.9 m ) FCE − (1.8 m )(1.95 kN − 0.1 kN ) = 0
FCE = 3.70 kN T W
PROBLEM 6.58 CONTINUED
 1


 2

ΣM A = 0: (1.8 m ) 
FBE  − .7 kN  + (.9 m )
FBE  = 0

 5

 5

FBE = 0.35 5 kN
ΣFx = 0: 3.70 kN −
(
FBE = 783 N C W
)
2
2
FBD = 0
0.35 5 kN −
5
5
FBD = 1.5 5 kN = 3.35 kN C
Then by inspection of joint D: FDF = FBD
so FDF = 3.35 kN C W
PROBLEM 6.59
A vaulted roof truss is loaded as shown. Determine the force in
members HJ, IJ, and GI.
SOLUTION
FBD Truss:
ΣM A = 0: (1.8 m )( 0.7 kN ) + ( 3.6 m )( 0.6 kN ) + ( 4.2 m )(1.3 kN ) + ( 5.52 m )( 0.5 kN )
( )
+ ( 6.6 m )( 0.5 kN ) + ( 7.2 m )( 0.2 kN ) + (8.4 m )( 0.1 kN ) − ( 8.4 m ) N y = 0
N y = 2.05 kN
FBD Section:
ΣM J = 0: (1.8 m )( 2.05 − 0.1) kN − ( 0.6 m )( 0.2 kN ) + ( 0.45 m ) FIK = 0
FIK = 7.533 kN
FIK = 7.53 kN T
PROBLEM 6.59 CONTINUED
 2

 1

ΣM I = 0: ( 2.88 m )( 2.05 − 0.1) kN − (1.68 m )( 0.2 kN ) − ( 0.45 m ) 
FHJ  − (1.08 m ) 
FHJ  = 0
 5

 5

FHJ = 2.3939 5 kN
ΣFy = 0: −
(
FHJ = 5.35 kN C W
)
1
5
FIJ + 2.05 kN − 0.5 kN − 0.2 kN − 0.1 kN = 0
2.3939 5 kN +
13
5
FIJ = 2.974 kN
FIJ = 2.97 kN C W
FBD Joint:
ΣFx = 0: − FGI −
12
( 2.974 kN ) + 7.533 kN = 0
13
FGI = 4.788 kN
FGI = 4.79 kN T W
PROBLEM 6.60
Determine the force in members AF and EJ of the truss shown when
P = Q = 2 kips. (Hint: Use section aa.)
SOLUTION
FBD Truss:
ΣM A = 0: ( 24 ft ) K x − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) = 0
K x = 4.5 kips
FBD Section:
ΣM F = 0: (12 ft )( 4.5 kips ) − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) + ( 36 ft ) FEJ = 0
FEJ = 1.500 kips T W
ΣM J = 0: (18 ft )( 2 kips ) + (12 ft )( 4.5 kips ) − ( 36 ft ) FAF = 0
FAF = 2.50 kips T W
PROBLEM 6.61
Determine the force in members AF and EJ of the truss shown when
P = 2 kips and Q = 0. (Hint: Use section aa.)
SOLUTION
FBD Truss:
ΣM A = 0: ( 24 ft ) K x − (18 ft )( 2 kips ) = 0
K x = 1.5 kips
FBD Section:
ΣM F = 0: (12 ft )(1.5 kips ) − (18 ft )( 2 kips ) + ( 36 ft ) FEJ = 0
FEJ = 0.500 kip T W
ΣM J = 0: (18 ft )( 2 kips ) + (12 ft )(1.5 kips ) − ( 36 ft ) FAF = 0
FAF = 1.500 kips T W
PROBLEM 6.62
Determine the force in members DG and FH of the truss shown.
(Hint: Use section aa.)
SOLUTION
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = N y = 3.2 kN
FBD Section:
ΣM F = 0: ( 2.4 m ) FDG − ( 3.2 m )( 3.2 kN ) + (1.6 m )(1 kN ) = 0
FDG = 3.60 kN C W
ΣM D = 0: ( 2.4 m ) FFH + (1.6 m )(1 kN ) − ( 3.2 m )( 3.2 kN ) = 0
FFH = 3.60 kN T W
PROBLEM 6.63
Determine the force in members IL, GJ, and HK of the truss shown.
(Hint: Begin with pins I and J and then use section bb.)
SOLUTION
FBD Truss:
ΣFx = 0:
Ax = 0
By symmetry: Ax = 0; Ay = N y = 3.2 kN
Joints: By inspection of joint I: FGI = FIL
By inspection of joint J:
and FIJ = 1.2 kN C
FGJ ( C ) = FHJ ( T )
PROBLEM 6.63 CONTINUED
FBD Section:
ΣFx = 0; FIL − FHK +
0

4
F
−
F
HJ
 GJ
=0
5

so
FIL − FHK = 0
ΣM J = 0: ( 3.2 m )( 3.2 kN ) − (1.2 m )( FIL + FHK ) − (1.6 m )(1 kN ) = 0
so
Solving: FIL = FHK = 3.6 kN
FIL + FHK = 7.2 kN
FIL = 3.60 kN C W
FHK = 3.60 kN T W
ΣFy = 0: 3.2 kN − 1 kN − 1.2 kN −
FGJ + FHJ =
5
kN
3
but
FGJ = FHJ =
5
kN
6
3
( FGJ + FHJ ) = 0
5
FGJ = .833 kN C W
PROBLEM 6.64
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known as
counters. Determine the forces in the counters which are acting under
the given loading.
SOLUTION
ΣFx = 0: Fx = 0
FBD Truss:
ΣM F = 0: ( 5.5 m ) H y + ( 2.75 m )( 24 kN ) − ( 2.75 m )( 24 kN )
− ( 5.5 m )(12 kN ) − ( 8.25 m )(12 kN ) = 0
H y = 30 kN
ΣFy = 0: Fy + 30 kN − 3 ( 24 kN ) − 2 (12 kN ) = 0
Fy = 66 kN
FBD Section:
Assume there is no pretension in any counter so that, as the truss is
loaded, one of each pair becomes taut while the other becomes slack.
Here tension is required in BG to provide downward force, so CF is
slack.
ΣFy = 0: 66 kN − 24 kN − 24 kN −
FBG = 27.375 kN
2.4
FBG = 0
3.65
FBG = 27.4 kN T W
FCF = 0 W
Here tension is required in GD to provide downward force, so CH is
slack.
FCH = 0 W
ΣFy = 0: 30 kN − 2 (12 kN ) −
FGD = 9.125 kN
2.4
FGD = 0
3.65
FGD = 9.13 kN T W
PROBLEM 6.65
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known as
counters. Determine the forces in the counters which are acting under
the given loading.
SOLUTION
Assume that there is no pretension in any counter. So that, as the truss is
loaded, one of each crossing pair becomes taut while the other becomes
slack.
FBD Section:
Note: Tension is required in CH to provide upward force; thus GD is
slack
2.4
ΣFy = 0:
FCH − 24 kN = 0
FCH = 36.5 kN T W
3.65
FBD Truss:
( )
ΣM G = 0: ( 5.5 m )( 24 kN ) + ( 2.75 m )( 24 kN ) − ( 2.75 m ) Fy
− ( 2.75 m )(12 kN ) − ( 5.5 m )(12 kN ) = 0
Fy = 36 kN
ΣFx = 0: Fx = 0
FBD Section:
Note: Tension is required in FC to provide upward force; so BG is
slack.
ΣFy = 0: ( 36 − 24 − 24 ) kN +
2.4
FFC = 0
3.65
FFC = 18.25 kN T W
PROBLEM 6.66
The diagonal members in the center panel of the truss shown are very
slender and can act only in tension; such members are known as
counters. Determine the force in members BD and CE and in the
counter which is acting when P = 3 kips.
SOLUTION
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0: ( 25 ft ) Fy − (17 ft )( 2 kips ) − ( 8 ft )( 3 kips ) = 0
Fy = 2.32 kips
Assume there is no pretension in either counter so that, as the truss is
loaded, one becomes taut while the other becomes slack.
FBD Section:
Here tension is required in CD to provide downward force, so
FBE = 0
ΣFy = 0: 2.32 kips − 2 kips −
FCD = 0.6057 kip
5.6
FCD = 0
10.6
FCD = 606 lb T W
ΣM D = 0: ( 8 ft )( 2.32 kips ) − (5.6 ft)FCE = 0
FCE = 3.314 kips
ΣFx = 0: FBD −
FCE = 3.31 kips T W
9
( 0.6057 kip ) − 3.314 kips = 0
10.6
FBD = 3.829 kips
FBD = 3.83 kips C W
PROBLEM 6.67
Solve Prob. 6.66 when P = 1.5 kips.
SOLUTION
FBD Truss:
ΣM A = 0: ( 25 ft ) Fy − (17 ft )( 2 kips ) − ( 8 ft )(1.5 kips ) = 0
Fy = 1.84 kips
Here tension is needed in BE to provide upward force,
FBD Section:
ΣFy = 0:
so FCD = 0
5.6
FBE + 1.84 kips − 2 kips = 0
10.6
FBE = 0.03029 kip
FBE = 303 lb T W
ΣM E = 0: ( 8 ft )(1.84 kips ) − ( 5.6 ft ) FBD = 0
FBD = 2.629 kips
FBD = 2.63 kips C W
ΣFx = 0: 2.629 kips −
9
( 0.3029 kip ) − FCE = 0
10.6
FCE = 2.371 kips
FCE = 2.37 kips T W
PROBLEM 6.68
The diagonal members CF and DE of the truss shown are very slender
and can act only in tension; such members are known as counters.
Determine the force in members CE and DF and in the counter which
is acting.
SOLUTION
FBD Section:
DE must be in tension to provide leftward force, so CF is slack.
ΣFx = 0: 4 kips −
4
FDE = 0
5
FDE = 5.00 kips T W
ΣM D = 0: ( 8 ft ) FCE − ( 6 ft )( 4 kips ) = 0
FCE = 3.00 kips T W
ΣM E = 0: ( 8 ft ) FDF − (12 ft )( 4 kips ) = 0
FDF = 6.00 kips C W
PROBLEM 6.69
The diagonal members EH and FG of the truss shown are very slender
and can act only in tension; such members are known as counters.
Determine the force in members EG and FH and in the counter which
is acting.
SOLUTION
FBD Section:
It is not obvious which counter is active, so assume FG is (and thus
EH is slack)
 6

ΣM F = 0: ( 8 ft ) 
FGE  − (12 ft )( 4 kips ) = 0
6.824


FGE = 6.824 kips T
 6

ΣM G = 0: (14.5 ft ) 
FFH  − (18 ft )( 4 kips ) = 0
 6.824

FFH = 5.647 kips C
6
 6 
ΣFy = 0: 
FFG = 0
 ( 5.647 kips − 6.824 kips ) −
12.75
 6.824 
This gives FFG < 0 which is impossible, so the assumption is wrong,
FG is slack, and EH is in tension.
Then
 6

ΣM E = 0: ( 8 ft ) 
FFH  − (12 ft )( 4 kips ) = 0
 6.824

FFH = 6.824 kips
FFH = 6.83 kips C W
 6

ΣM H = 0: (14.5 ft ) 
FGE  − (18 ft )( 4 kips ) = 0
 6.824

FGE = 5.647 kips
FGE = 5.65 kips T W
ΣFy = 0:
6
6
FEH = 0
( 6.824 kips − 5.647 kips ) −
6.824
12.75
FEH = 2.198 kips
FEH = 2.20 kips T W
PROBLEM 6.70
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension
and in compression.)
SOLUTION
Structure (a):
Simple truss with r = 4, m = 16, n = 10
So m + r = 20 = 2n so completely constrained and determinate W
Structure (b):
Compound truss with r = 3, m = 16, n = 10
So m + r = 19 < 2n = 20
so
Structure (c):
partially constrained W
Non-simple truss with r = 4, m = 12, n = 8
So m + r = 16 = 2n but must examine further, note that reaction
forces
A x and H x are aligned, so no equilibrium equation will resolve them.
∴ Statically indeterminate W
Also
consider
For ΣFy = 0: H y = 0, but then
ΣM A ≠ 0 in FBD Truss,
∴ Improperly constrained W
PROBLEM 6.71
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension and
in compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 16, n = 10
so m + r = 20 = 2n, but must examine further.
FBD Sections:
ΣM A = 0 ⇒
T1
II:
ΣFx = 0
⇒
T2
I:
ΣFx = 0
⇒
Ax
I:
ΣFy = 0
⇒
Ay
II:
ΣM E = 0 ⇒
Cy
II:
ΣFy = 0
Ey
FBD I:
⇒
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate. W
Non-simple truss with r = 3, m = 16, n = 10
Structure (b):
so
m + r = 19 < 2n = 20
∴ structure is partially constrained W
Structure (c):
Simple truss with r = 3, m = 17, n = 10
m + r = 20 = 2n, but the horizontal reaction forces Ax and Ex are
collinear and no equilibrium equation will resolve them, so the
W
structure is improperly constrained and indeterminate
PROBLEM 6.72
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension and
in compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n, check
for determinacy:
One can solve joint F for forces in EF, FG and then solve joint E for
E y and force in DE.
This leaves a simple truss ABCDGH with
r = 3, m = 9, n = 6
so
r + m = 12 = 2n
Structure is completely constrained and determinate W
Structure (b):
Simple truss (start with ABC and add joints alphabetically to complete
truss) with r = 4, m = 13, n = 8
so
Structure (c):
r + m = 17 > 2n = 16
Constrained but indeterminate W
Non-simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n . To
further examine, follow procedure in part (a) above to get truss at left.
Since F1 ≠ 0 (from solution of joint F),
ΣM A = aF1 ≠ 0 and there is no equilibrium.
Structure is improperly constrained W
PROBLEM 6.73
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension and
in compression.)
SOLUTION
Structure (a):
Simple truss (start with ABC and add joints alphabetical to complete
truss), with
r = 4, m = 13, n = 8
so
r + m = 17 > 2n = 16
Structure is completely constrained but indeterminate. W
Structure (b):
From FBD II: ΣM G = 0 ⇒
ΣFx = 0
FBD I:
FBD II:
Jy
⇒
Fx
ΣM A = 0 ⇒
Fy
ΣFy = 0 ⇒
Ay
ΣFx = 0
⇒
Ax
ΣFy = 0
⇒
Gy
Thus have two simple trusses with all reactions known,
so structure is completely constrained and determinate. W
Structure (c):
Structure has r = 4, m = 13, n = 9
so
r + m = 17 < 2n = 18,
structure is partially constrained W
PROBLEM 6.74
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension and
in compression.)
SOLUTION
Structure (a):
Rigid truss with r = 3, m = 14, n = 8
so
r + m = 17 > 2n = 16
so completely constrained but indeterminate W
Structure (b):
Simple truss (start with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8
so
r + m = 16 = 2n
so completely constrained and determinate W
Structure (c):
Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but
horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved
by any equilibrium equation.
∴ structure is improperly constrained W
PROBLEM 6.75
Classify each of the structures shown as completely, partially, or
improperly constrained; if completely constrained, further classify it as
determinate or indeterminate. (All members can act both in tension and
in compression.)
SOLUTION
Structure (a):
No. of members
m = 12
No. of joints
n=8
No. of react. comps. r = 4
m + r = 16 = 2n
unks = eqns
FBD of EH:
ΣM H = 0 → FDE ; ΣFx = 0 → FGH ; ΣFy = 0 → H y
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. W
Structure (b):
No. of members
m = 12
No. of joints
n=8
No. of react. comps. r = 3
m + r = 15 < 2n = 16
unks < eqns
partially constrained W
Note: Quadrilateral DEHG can collapse with joint D moving downward:
in (a) the roller at F prevents this action.
PROBLEM 6.75 CONTINUED
Structure (c):
No. of members
m = 13
No. of joints
n=8
No. of react. comps. r = 4
m + r = 17 > 2n = 16
unks > eqns
completely constrained but indeterminate W
PROBLEM 6.76
For the frame and loading shown, determine the force acting on
member ABC (a) at B, (b) at C.
SOLUTION
FBD member ABC:
Note: BD is a two-force member so FBD is through D.
(a)
ΣM C = 0:
(8 in.) 
4

FBD  − (12 in.)( 60 lb ) = 0
5

FBD = 112.5 lb
(b)
ΣFx = 0: 60 lb + Cx −
36.9° W
4
(112.5 lb ) = 0
5
Cx = 30 lb
ΣFy = 0: C y −
3
(112.5 lb ) = 0
5
C y = 67.5 lb
so C = 73.9 lb
66.0° W
PROBLEM 6.77
Determine the force in member AC and the reaction at B when (a)
θ = 30o , (b) θ = 60o.
SOLUTION
FBD member BCD:
Note: AC is two-force member so FAC is through A.
p = ( 8 in.) cosθ
BC
ΣM B = 0: ( 8 in.) cosθ ( FAC sin θ ) − (10 in.)( 20 lb )
− (14 in.)( 20 lb ) = 0
FAC =
60 lb
sin θ cosθ
ΣFx = 0: Bx − FAC cosθ = 0
Bx =
60 lb
sin θ
ΣFy = 0: By + FAC sin θ − 20 lb − 20 lb = 0
By = 40 lb −
(a) θ = 30°
60 lb
cos θ
FAC = 138.56 lb
FAC = 138.6 lb T W
Bx = 120.0 lb
By = −29.28 lb
B = 123.5 lb
(b) θ = 60°
13.71° W
FAC = 138.56 lb
FAC = 138.6 lb T W
Bx = 69.28 lb
By = −80 lb
B = 105.8 lb
49.1° W
PROBLEM 6.78
A circular ring of radius 200 mm is pinned at A and is supported by rod
BC, which is fitted with a collar at C that can be moved along the ring.
For the position when θ = 35°, determine (a) the force in rod BC,
(b) the reaction at A.
SOLUTION
FBD ring:
(a)
θ = 35°
ΣM A = 0: ( 0.2 m ) FBC cos 35° − ( 0.2 m )( 24 N ) = 0
FBC =
24 N
= 29.298 N
cos 35°
FBC = 29.3 N C W
(b)
ΣFx = 0: Ax −
24 N
cos 35° = 0
cos 35°
Ax = 24 N
ΣFy = 0: Ay +
24 N
sin 35° − 24 N = 0
cos35°
Ay = 7.195 N
so A = 25.1 N
16.69° W
PROBLEM 6.79
Solve Prob. 6.78 when θ = −20°.
SOLUTION
FBD ring:
(a)
θ = 20°
ΣM A = 0: ( 0.2 m )( FBC cos 20° ) − ( 0.2 m )( 24 N ) = 0
FBC =
24 N
= 25.54 N
cos 20°
FBC = 25.5 N C W
(b)
ΣFy = 0: Ay −
24 N
sin 20° − 24 N = 0
cos 20°
Ay = 32.735 lb
ΣM B = 0: ( 0.2 m ) Ax − ( 0.2 m )( 24 N ) = 0
Ax = 24 N
so A = 40.6 N
53.8° W
PROBLEM 6.80
For the frame and loading shown, determine the components of all
forces acting on member ABC.
SOLUTION
FBD Frame:
ΣM E = 0: (1.2 m ) Ax − (1.5 m )( 80 kN ) = 0
A x = 100.0 kN
ΣFy = 0: Ay − 80 kN = 0
FBD member ABC:
A y = 80.0 kN
W
W
Note: BE is two-force member so
By =
2
Bx = 0.4 Bx
5
ΣM C = 0: (1.2 m )(100 kN ) − ( 3.0 m )( 80 kN )
+ ( 0.6 m )( Bx ) + (1.5 m )( 0.4 Bx ) = 0
B x = 100.0 kN
W
so B y = 40.0 kN
W
ΣFx = 0: −100 kN − 100 kN + C x = 0
C x = 200 kN
W
ΣFy = 0: 80 kN − 40 kN − C y = 0
C y = 40.0 kN
W
PROBLEM 6.81
Solve Prob. 6.80 assuming that the 80-kN load is replaced by a
clockwise couple of magnitude 120 kN ⋅ m applied to member EDC
at point D.
SOLUTION
FBD Frame:
ΣFy = 0:
Ay = 0 W
ΣM E = 0: (1.2 m ) Ax − 120 kN ⋅ m = 0
A x = 100.0 kN
FBD member ABC:
W
Note: BE is two-force member, so
By =
2
Bx = 0.4 Bx
5
ΣM C = 0: (1.2 m )100 kN − ( 0.6 m ) Bx − (1.5 m )( 0.4 Bx ) = 0
B x = 100.0 kN
W
so B y = 40.0 kN
W
ΣFx = 0: −100 kN + 100 kN − C x = 0
Cx = 0 W
ΣFy = 0: 40 kN − C y = 0
C y = 40.0 kN
W
PROBLEM 6.82
For the frame and loading shown, determine the components of all
forces acting on member ABC.
SOLUTION
FBD Frame:
ΣM F = 0: (10.8 ft ) Ay − (12 ft )( 4.5 kips ) = 0
A y = 5.00 kips
W
ΣFx = 0: − Ax + 4.5 kips = 0
A x = 4.50 kips
W
FBD member ABC:
Note: BE is a two-force member
ΣM C = 0: (12 ft ) FBE + (10.8 ft )( 5 kips ) − (18 ft )( 4.5 kips ) = 0
FBE = 2.25 kips
W
ΣFx = 0: C x + 2.25 kips − 4.5 kips = 0
C x = 2.25 kips
W
ΣFy = 0: C y − 5 kips = 0
C y = 5.00 kips
W
PROBLEM 6.83
Solve Prob. 6.82 assuming that the 4.5-kip load is replaced by a
clockwise couple of magnitude 54 kip ⋅ ft applied to member CDEF
at point D.
SOLUTION
FBD Frame:
ΣFx = 0:
Ax = 0 W
ΣM E = 0: (10.8 ft ) Ay − 54 kip ⋅ ft = 0
A y = 5.00 kips
W
FBD member ABC:
ΣM C = 0: ( −12 ft ) FBE + (10.8 ft )( 5 kips ) = 0
FBE = 4.50 kips
W
ΣFx = 0: C x − 4.5 kips = 0
C x = 4.50 kips
W
ΣFy = 0: C y − 5 kips = 0
C y = 5.00 kips
W
PROBLEM 6.84
Determine the components of the reactions at A and E when a 160-lb force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
FBD Frame (part a):
Note: EC is a two-force member, so
Ey =
1
Ex
2
ΣM A = 0: (10 in.) Ex − ( 6 in.)(160 lb ) = 0
E x = 96.0 lb
W
so E y = 48.0 lb
W
A x = 96.0 lb
W
ΣFx = 0: − Ax + 96 lb = 0
ΣFy = 0: Ay − 160 lb + 48 lb = 0
A y = 112.0 lb
W
FBD member (part b):
Note: AC is a two-force member, so
Ax = 3 Ay
ΣM A = 0: same as part (a)
E x = 96.0 lb
W
A x = 96.0 lb
W
ΣFx = 0: same as part (a)
Here
Ay =
1
Ax
3
so A y = 32.0 lb
W
ΣFy = 0: 32 lb + E y − 160 lb = 0
E y = 128.0 lb
W
PROBLEM 6.85
Determine the components of the reactions at A and E when a 120-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
FBD ABC:
(a) CE is a two-force member
ΣM A = 0: ( 200 mm )
1
1
FCE + ( 50 mm )
FCE
2
2
− 150 mm (120 N ) = 0
FCE = 72 2 N
so E x = 72.0 N
E y = 72.0 N
ΣFx = 0: Ax +
1
FCE = 0
2
FBD CE:
W
Ax = −72 N W
A x = 72.0 N
1
FCE = 0
2
ΣFy = 0: Ay − 120 N +
W
A y = 48.0 N
W
W
(b) AC is a two-force member
 4

 1

ΣM E = 0: ( 75 mm ) 
FAC  + ( 75 mm ) 
FAC 
 17

 17

− ( 25 mm )(120 N ) = 0
ΣFx = 0: Ex −
ΣFy = 0: E y +
4
FAC = 0
17
1
FAC − 120 = 0
17
FAC = 8 17 N
E x = 32.0 N
E y = 112.0 N
and A x = 32.0 N
A y = 8.00 N
W
W
W
W
PROBLEM 6.86
Determine the components of the reactions at A and E when the frame
is loaded by a clockwise couple of magnitude 360 lb ⋅ in. applied
(a) at B, (b) at D.
SOLUTION
FBD Frame:
Note for analysis of the frame FBD, the location of the applied couple is
immaterial.
ΣM A = 0: (10 in.) Ex − 360 in ⋅ lb = 0
E x = 36.0 lb
W
ΣM E = 0: (10 in.) Ax − 360 in ⋅ lb = 0
A x = 36.0 lb
W
Part (a): If couple acts at B, EC is a two-force member, so
Ey =
1
Ex
2
E y = 18.0 lb
W
ΣFy = 0: Ay + 18 lb = 0
and then
A y = 18.00 lb
W
Part (b): If couple acts at D, AC is a two-force member, so
Ay =
Then
1
Ax
3
ΣFy = 0: 12 lb + E y = 0
A y = 12.00 lb
W
E y = −12 lb
E y = 12.00 lb
W
PROBLEM 6.87
Determine the components of the reactions at A and E when the frame is
loaded by a counterclockwise couple of magnitude 24 N · m applied
(a) at B, (b) at D.
SOLUTION
(a) FBD Frame:
Note: CE is a two-force member, so Ex = E y
ΣM A = 0: 24 N ⋅ m − ( 0.125 m ) Ex − ( 0.125 m ) E y = 0
Ex = E y = 96 N
E x = 96.0 N
E y = 96.0 N
ΣFx = 0: Ax − 96 N = 0
ΣFy = 0: Ay − 96 N = 0
(b) FBD Frame:
A x = 96.0 N
A y = 96.0 N
W
W
W
W
Note: AC is a two-force member, so Ax = 4 Ay
(
)
ΣM E = 0: 24 N ⋅ m + ( 0.125 m ) Ay − ( 0.125 m ) 4 Ay = 0
Ay = 64 N
A y = 64.0 N
A x = 256 N
ΣFy = 0: E y − 64 N = 0
ΣFx = 0: − Ex + 256 N = 0
E y = 64.0 N
E x = 256 N
W
W
W
W
PROBLEM 6.88
Determine the components of the reactions at A and B if (a) the 240-N
load is applied as shown, (b) the 240-N load is moved along its line of
action and is applied at E.
SOLUTION
FBD Frame:
Regardless of the point of application of the 240 N load;
ΣM A = 0: (1.25 m ) By − ( 2.25 m )( 240 N ) = 0
B y = 432 N
W
ΣM B = 0: (1.25 m ) Ay − (1.0 m )( 240 N ) = 0
A y = 192.0 N
so B x = 0 W
Part (a): If load at D, BCE is a two-force member,
Then
W
ΣFx = 0: Ax − Bx = 0
Ax = Bx = 0
Ax = 0 W
Part (b): If load at E , ACD is a two-force member, so Ax =
5
Ay
3
then A x = 320 N
and
W
ΣFx = 0: Ax − Bx = 0
B x = 320 N
W
PROBLEM 6.89
The 192-N load can be moved along the line of action shown and
applied at A, D, or E. Determine the components of the reactions at B
and F when the 192-N load is applied (a) at A, (b) at D, (c) at E.
SOLUTION
FBD Frame:
Note, regardless of the point of application of the 192 N load,
ΣM B = 0: ( 2 m )(192 N ) − ( 3 m ) Fx = 0
Fx = 128.0 N
W
ΣFx = 0: Bx − 128 N = 0
B x = 128.0 N
W
ΣFy = 0: By + Fy − 192 N = 0
(a) and (c): If load applied at either A or E, BC is a two-force member
5
Bx
16
so
By =
B y = 40.0 N
Then
ΣFy = 0: 40 N + Fy − 192 N = 0
Fy = 152.0 N
W
W
(b): If load applied at D, ACEF is a two-force member, so
Fy =
Then
7
Fy
16
Fy = 56.0 N
W
ΣFy = 0: By + 56 N − 192 N = 0
B y = 136.0 N
W
PROBLEM 6.90
The 192-N load is removed and a 288 N ⋅ m clockwise couple is
applied successively at A, D, and E. Determine the components of the
reactions at B and F when the couple is applied (a) at A, (b) at B, (c) at
E.
SOLUTION
FBD Frame:
Regardless of the location of applied couple,
ΣM B = 0: ( 3 m ) Fx − 288 N ⋅ m = 0
Fx = 96.0 N
W
ΣFx = 0: − Bx + 96 N = 0
B x = 96.0 N
W
ΣFy = 0: By + Fy = 0
(a) and (c): If couple applied anywhere on ACEF , BC is a two-force
member,
5
so
By =
Bx
B y = 30.0 N
16
Then
ΣFy = 0: 30 N + Fy = 0
W
Fy = −30 N
Fy = 30.0 N
W
(b): If couple is applied anywhere on BC , ACEF is a two-force
member,
so
Fy =
7
Fx
16
ΣFy = 0: By + 42 N = 0
Fy = 42.0 N
W
By = −42 N
B y = 42.0 N
W
PROBLEM 6.91
(a) Show that when a frame supports a pulley at A, an equivalent
loading of the frame and of each of its component parts can be
obtained by removing the pulley and applying at A two forces equal
and parallel to the forces that the cable exerted on the pulley.
(b) Show that if one end of the cable is attached to the frame at
point B, a force of magnitude equal to the tension in the cable should
also be applied at B.
SOLUTION
First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is
unchanged.
ΣM C = 0: rT1 − rT2 = 0
T1 = T2
(a) Replace each force with an equivalent force-couple.
(b) Cut cable and replace forces on pulley with equivalent pair of forces at A as above.
PROBLEM 6.92
Knowing that the pulley has a radius of 1.5 ft, determine the
components of the reactions at A and E.
SOLUTION
FBD Frame:
ΣM A = 0: ( 21 ft ) E y − (13.5 ft )( 210 lb ) = 0
E y = 135.0 lb
W
A y = 75.0 lb
W
ΣFy = 0: Ay − 210 lb + 135 lb = 0
FBD member ABC:
ΣFx = 0: Ax − Ex = 0
Ax = Ex
ΣM C = 0: ( 3 ft )( 210 lb ) − ( 3 ft )( 75 lb ) − ( 9 ft ) Ax = 0
A x = 45.0 lb
W
so E x = 45.0 lb
W
PROBLEM 6.93
Knowing that the pulley has a radius of 1.25 in., determine the
components of the reactions at B and E.
SOLUTION
ΣM E = 0: ( 3.75 in.) Bx + ( 8.75 in.)( 75 lb ) = 0
FBD Frame:
Bx = 175 lb
ΣFx = 0: Ex − Bx = 0
B x = 175.0 lb
W
E x = 175.0 lb
W
ΣFy = 0: E y + By − 75 lb = 0
By = 75 lb − E y
FBD member ACE:
ΣM C = 0: − (1.25 in.)( 75 lb ) + ( 3.75 in.)(175 lb ) − ( 4.5 in.) E y = 0
Thus
E y = 125.0 lb
W
B y = 50.0 lb
W
By = 75 lb − 125 lb = −50 lb
PROBLEM 6.94
Two 200-mm-diameter pipes (pipe 1 and pipe 2) are supported every
3 m by a small frame like the one shown. Knowing that the combined
mass per unit length of each pipe and its contents is 32 kg/m and
assuming frictionless surfaces, determine the components of the
reactions at A and E when a = 0.
SOLUTION
FBDs:
pipe 2:
3
ΣFx′ = 0: NG − W = 0
5
N G = 0.6W
4
W =0
5
N D = 0.8W
ΣFy′ = 0: N D −
pipe 1:
3
ΣFx′ = 0: N F cos16.26° − W − 0.6W = 0
5
N F = 1.25W
4
W =0
5
NC = 0.45W
ΣFy′ = 0: NC + 1.25W sin16.26° −
Frame:

4
 
ΣM A = 0: ( 0.96 m ) E y − ( 0.48 m )W −   0.48 + ( 0.2 )  m W = 0
5
 

E y = 1.16667W
E y = 1099 N
W
ΣFy = 0: Ay + 1.16667W − W − W = 0
Ay = 0.83333W
ΣFx = 0: Ax − Ex = 0
A y = 785 N
so
Ax = Ex
W
PROBLEM 6.94 CONTINUED
member BE:
ΣM B = 0: ( 0.075 m )(1.25W ) + ( 0.48 m )(1.16667W )
− ( 0.36 m ) Ex = 0
Ex = 1.816W
E x = 1710 N
W
thus A x = 1710 N
W
PROBLEM 6.95
Solve Prob. 6.94 when a = 280 mm.
SOLUTION
FBDs
pipe 2
W = 941.76 N
4
W =0
5
N D = 0.8W
3
ΣFx′ = 0: NG − W = 0
5
NG = 0.6W
4
W =0
5
NC = 0.8W
ΣFy′ = 0: N D −
pipe 1:
ΣFy′ = 0: NC −
ΣFx′ = 0: N F −
member BE:
3
96 g − NG = 0
5
N F = 1.2W
ΣM B = 0: ( 640 mm ) Ex + ( 480 mm ) E y + (100 mm ) N F = 0
ΣM A = 0: ( 280 mm ) Ex + ( 960 mm ) E y + (100 mm ) N F
− ( 700 mm ) NC − ( 900 mm ) N D = 0
Ex = −1.400W
E y = 1.617W
E x = 1318 N
E y = 1523 N
W
W
PROBLEM 6.95 CONTINUED
FBD Frame:
ΣFx = 0: Ax + Ex +
3
4
( NC + N D ) − N F = 0
5
5
Ax = 1.400W
ΣFy = 0: Ay + E y −
A x = 1318 N
W
4
3
( NC + N D ) − N F = 0
5
5
Ay = 0.3833W
A y = 361 N
W
PROBLEM 6.96
The tractor and scraper units shown are connected by a vertical pin
located 1.8 ft behind the tractor wheels. The distance from C to D
is 2.25 ft. The center of gravity of the 20-kip tractor unit is located at
Gt , while the centers of gravity of the 16-kip scraper unit and the 90kip load are located at Gs and Gl , respectively. Knowing that the
tractor is at rest with its brakes released, determine (a) the reactions
at each of the four wheels, (b) the forces exerted on the tractor unit at
C and D.
SOLUTION
FBD Entire machine:
(a)
ΣM A = 0: ( 21.6 ft ) 2 B − ( 7.2 ft )( 90 kips ) − (10.8 ft )(16 kips )
− ( 26.4 ft )( 20 kips ) = 0
B = 31.22 kips
B = 31.2 kips
W
ΣFy = 0: 2A + 2 ( 31.22 kips ) − ( 90 + 16 + 20 ) kips = 0
A = 31.78 kips
FBD Tractor:
A = 31.8 kips
W
(b )
ΣM D = 0: ( 2.25 ft ) C + (1.8 ft )( 62.44 kips ) − ( 6.6 ft )( 20 kips ) = 0
C = 8.7146 kips
ΣFx = 0: Dx − C = 0
C = 8.71 kips
W
Dx = 8.715 kips
ΣFy = 0: 62.44 kips − D y − 20 kips = 0
Dy = 42.44 kips
so D = 43.3 kips
78.4° W
PROBLEM 6.97
Solve Prob. 6.96 assuming that the 90-kip load has been removed.
SOLUTION
FBD Entire machine:
(a)
ΣM A = 0: ( 21.6 ft )( 2B ) − (10.8 ft )(16 kips ) − ( 26.4 ft )( 20 kips ) = 0
B = 16.222 kips
B = 16.22 kips
W
ΣFy = 0: 2A + 2 (16.222 kips ) − (16 + 20 ) kips = 0
A = 1.778 kips
FBD Tractor:
A = 1.778 kips
W
(b )
ΣM D = 0: ( 2.25 ft ) C + (1.8 ft )( 32.44 kips ) − ( 6.6 ft )( 20 kips ) = 0
C = 32.71 kips
ΣFx = 0: Dx − C = 0
C = 32.7 kips
W
Dx = 32.71 kips
ΣFy = 0: − Dy + 2 ( 32.44 kips ) − 20 kips = 0
Dy = 12.44 kips
so D = 35.0 kips
20.8° W
PROBLEM 6.98
For the frame and loading shown, determine the components of all
forces acting on member ABE.
SOLUTION
FBD Frame:
ΣM F = 0: (1.2 m )( 3.6 kN ) − ( 4.8 m ) E y = 0
E y = 0.9 kN
FBD member BC:
Cy =
4.8
8
Cx = Cx
5.4
9
ΣM C = 0: ( 2.4 m ) By − (1.2 m )( 3.6 kN ) = 0
By = 1.8kN
B y = 1.800 kN
on ABE:
ΣFy = 0: −1.8 kN + C y − 3.6 kN = 0
so
Cx =
9
Cy
8
ΣFx = 0: − Bx + C x = 0
W
W
C y = 5.40 kN
Cx = 6.075 kN
Bx = 6.075 kN
on ABE:
on BC
B x = 6.08 kN
W
PROBLEM 6.98 CONTINUED
FBD member AB0E:
ΣM A = 0: a ( 6.075 kN ) − 2aEx = 0
Ex = 3.038 kN
E x = 3.04 kN
W
A x = 3.04 kN
W
ΣFx = 0: − Ax + ( 6.075 − 3.038 ) kN = 0
ΣFy = 0: 0.9 kN + 1.8 kN − Ay = 0
A y = 2.70 kN
W
PROBLEM 6.99
For the frame and loading shown, determine the components of all
forces acting on member ABE.
SOLUTION
FBD Frame:
ΣM F = 0: ( 7.2 ft ) Fy − (1.2 ft )( 6 kips ) = 0
E y = 1.000 kip
ΣM B = 0: ( 4.8 ft ) C y − ( 7.2 ft )( 6 kips ) = 0
W
C y = 9 kips
FBD member BCD:
But C is ⊥ ACF , so Cx = 2C y ; Cx = 18 kips
ΣFx = 0: − Bx + C x = 0
Bx = 18.00 kips
Bx = C x = 18 kips
on BCD
ΣFy = 0: − By + 9 kips − 6 kips = 0
On ABE:
By = 3 kips
on BCD
B x = 18.00 kips
B y = 3.00 kips
W
W
ΣM A = 0: ( 4.8 ft )(18 kips ) − ( 2.4 ft )( 3 kips )
+ ( 3.6 ft )(1 kip ) − ( 7.2 ft ) ( Ex ) = 0
E x = 11.50 kips
W
ΣFx = 0: −11.50 kips + 18 kips − Ax = 0
A x = 6.50 kips
W
PROBLEM 6.99 CONTINUED
FBD member ABE:
ΣFy = 0: −1.00 kip + 3.00 kips − Ay = 0
A y = 2.00 kips
W
PROBLEM 6.100
For the frame and loading shown, determine the components of
the forces acting on member ABC at B and C.
SOLUTION
FBD Frame:
ΣM A = 0: ( 0.12 m ) D − ( 0.27 m )(192 N ) − ( 0.33 m )( 80 N ) = 0
D = 652 N
ΣFy = 0: A y = 0
FBD members:
ΣM E = 0: ( 0.39 m )( 652 N ) − ( 0.12 m ) Bx = 0
ΣFx = 0: Ex − 2119 N + 652 N = 0
Bx = 2119 N
Ex = 1467 N
PROBLEM 6.100 CONTINUED
ΣFx = 0: 80 N + 192 N − 1467 N + Cx = 0
Cx = 1195 N
ΣM E = 0: − ( 0.9 m ) C y + ( 0.12 m )(1195 N ) − ( 0.6 m )( 80 N ) = 0
C y = 1540 N
From above, on ABC
C x = 1.195 kN
C y = 1.540 kN
B x = 2.12 kN
ΣFy = 0: − By + 1540 N = 0
W
W
W
By = 1540 N
B y = 1.540 kN
W
PROBLEM 6.101
For the frame and loading shown, determine the components of the
forces acting on member CDE at C and D.
SOLUTION
FBD Frame:
Note: p
AF = 2 ( 0.32 m ) cos 30° = 0.5543 m
ΣM F = 0: ( 0.5543 m ) Ax − ( 0.48 m )(100 N ) = 0
A x = 86.603 N
ΣFy = 0: Ay − 100 N = 0
FBD members:
A y = 100 N
ΣM B = 0: ( 0.32 m )( cos 30° )( 86.603 N ) + ( 0.16 m )( cos 30° ) Dx
− ( 0.32 m )( sin 30° )(100 N ) − ( 0.16 m )( sin 30° ) Dy = 0
Dx = Dy tan 30° − 57.736 N
ΣM C = 0: ( 0.16 m ) Dy − ( 0.40 m )(100 N ) = 0
Dy = 250 N
D y = 250 N
Then, from above
Dx = 86.6 N
ΣFx = 0: Cx − 86.6 N = 0
ΣFy = 0: −C y + 250 N − 100 N = 0
W
D x = 86.6 N
W
C x = 86.6 N
W
C y = 150.0 N
W
PROBLEM 6.102
Knowing that P = 15 N and Q = 65 N, determine the components
of the forces exerted (a) on member BCDF at C and D, (b) on
member ACEG at E.
SOLUTION
FBD Frame:
P = 15 N
Q = 65 N
ΣM D = 0: ( 0.25 m )( P + Q ) − (.15 m )( P + Q ) − ( 0.08 m ) Ex = 0
Ex = 1.2 ( P + Q ) = 100 N
ΣFx = 0: Dx − Ex = 0 = Dx − 100 N
E x = 100.0 N
W
Dx = 100 N
D x = 100.0 N
W
ΣFy = 0: E y − Dy − 2P − 2Q = 0
E y = D y + 2 ( P + Q ) = Dy + 160 N
FBD member BF:
ΣM C = 0: ( 0.15 m )( 65 N ) − ( 0.1 m ) Dy − ( 0.04 m )(100 N )
− ( 0.25 m )(15 N ) = 0
Dy = 20 N
From above
E y = 20 N + 160 N = 180 N
D y = 20.0 N
W
E y = 180.0 N
W
ΣFx = 0: −Cx + 100 N = 0
C x = 100.0 N
W
ΣFy = 0: − 65 N + C y − 20 N − 15 N = 0
C y = 100.0 N
W
PROBLEM 6.103
Knowing that P = 25 N and Q = 55 N, determine the components
of the forces exerted (a) on member BCDF at C and D, (b) on
member ACEG at E.
SOLUTION
P = 25 N
FBD Frame:
Q = 55 N
ΣM D = 0: ( 0.25 m )( P + Q ) − ( 0.15 m )( P + Q ) − ( 0.08m ) Ex = 0
Ex = 1.20 ( P + Q ) = 100 N
E x = 100.0 N
W
D x = 100.0 N
W
ΣFx = Dx − 100 N = 0
ΣFy = E y − Dy − 2P − 2Q = 0
E y = D y + 2 ( P + Q ) = Dy + 160 N
FBD member BF:
ΣM C = 0: ( 0.15 m )( 55 N ) − ( 0.1 m ) Dy − ( 0.04 )(100 N )
− ( 0.25 m )( 25 N ) = 0
Dy = −20 N
From above
E y = −20 N + 160 N = 140 N
ΣFx = 0: −Cx + 100 N = 0
D y = 20.0 N
W
E y = 140.0 N
W
C x = 100.0 N
W
ΣFy = 0: −55 N + C y − ( −20 N ) − 25 N = 0
C y = 60.0 N
W
PROBLEM 6.104
Knowing that P = 822 N and Q = 0, determine for the frame and
loading shown (a) the reaction at D, (b) the force in member BF.
SOLUTION
FBD Frame:
ΣM = 0: ( 0.24 m ) Dx − ( 0.56 m )( 822 N ) = 0
Dx = 1918 N
D x = 1.918 kN
ΣM D = 0: ( 0.69 m )
FBD member DF:
8
3
FBF − ( 0.24 m ) FEC = 0
17
5
0.3247 FBF − 0.144 FEC = 0
ΣFx = 0: 1918 N −
15
4
FBF − FEC = 0
17
5
0.8824 FBF + 0.800 FEC = 1918 N
FBF = 714 N T W
Solving:
ΣM E = 0: ( 0.45 m )
8
( 714 N ) − ( 0.24 m ) Dy = 0
17
Dy = 630 N
so D = 2.02 kN
18.18° W
PROBLEM 6.105
Knowing that P = 0 and Q = 1096 N, determine for the frame and
loading shown (a) the reaction at D, (b) the force in member BF.
SOLUTION
FBD Frame:
ΣM A = 0: ( 0.24 m ) Dx − ( 0.69 m )(1096 N ) = 0
ΣM D = 0: ( 0.69 m )
FBD member DF:
D x = 3151 N
8
3
FBF − ( 0.69 m )(1096 N ) − ( 0.24 m ) FEC = 0
17
5
0.3247 FBF − 0.144 FEC = 756.24 N
ΣFx = 0: 3151 N −
4
15
FEC −
FBF = 0
5
17
0.8824 FBF + 0.800 FEC = 3151 N
Solving:
FBF = 2737 N
FBF = 2.74 kN T W
8

ΣM E = 0: ( 0.45 m )  ( 2737 N ) − 1096 N  − ( 0.24 m ) Dy = 0
17

D y = 360.06 N
D = 3.17 kN
6.52° W
PROBLEM 6.106
The axis of the three-hinge arch ABC is a parabola with vertex at B.
Knowing that P = 28 kips and Q = 35 kips, determine (a) the
components of the reaction at A, (b) the components of the force exerted
at B on segment AB.
SOLUTION
FBDs members:
From FBD I:
ΣM A = 0: ( 9.6 ft ) Bx − ( 24 ft ) By − (15 ft )( 28 kips ) = 0
1.2Bx − 3.0 By = 52.5 kips
FBD II:
ΣM C = 0: ( 5.4 ft ) Bx + (18 ft ) By − ( 9 ft )( 35 kips ) = 0
0.6 Bx − 2 By = 35 kips
Solving: Bx = 50 kips; By = 2.5 kips as drawn, so
on AB: B x = 50.0 kips
B y = 2.50 kips
FBD I:
ΣFx = 0: Ax − 50 kips = 0
ΣFx = 0: Ay − 28 kips − 2.5 kips = 0
A x = 50.0 kips
A y = 30.5 kips
W
W
W
W
PROBLEM 6.107
The axis of the three-hinge arch ABC is a parabola with vertex at B.
Knowing that P = 35 kips and Q = 28 kips , determine (a) the
components of the reaction at A, (b) the components of the force exerted
at B on segment AB.
SOLUTION
member FBDs:
From FBD I:
ΣM A = 0: ( 9.6 ft ) Bx + ( 24 ft ) By − (15 ft )( 35 kips ) = 0
3.2Bx + 8By = 175 kips
FBD I:
ΣM C = 0: ( 5.4 ft ) Bx − (18 ft ) By − ( 9 ft )( 28 kips ) = 0
0.6 Bx − 2 By = 28 kips
Solving: Bx = 51.25 kips; Bx = 1.375 kips as drawn, so
on AB: B x = 51.3 kips
B y = 1.375 kips
FBD I:
ΣFx = 0: Ax − 51.25 kips
ΣFy = 0: Ay − 35 kips + 1.375 kips
A x = 51.3 kips
A y = 33.6 kips
W
W
W
W
PROBLEM 6.108
For the frame and loading shown, determine the reactions at A, B, D,
and E. Assume that the surface at each support is frictionless.
SOLUTION
FBD Frame:
ΣM A = 0: ( 0.16 m ) E − ( 0.08 m )(1 kN ) cos 30°
− ( 0.06 m )(1 kN ) sin 30° = 0
E = 0.2455 kN
E = 246 N
ΣFy = 0: D + 0.2455 kN − (1 kN ) cos 30° = 0
D = 0.6205 kN
D = 621 N
ΣFx = 0: A − B + (1 kN ) sin 30° = 0
W
W
B = A + 0.5 kN
FBD member ACE:
ΣM C = 0: ( 0.8 m )( 0.2455 kN ) − ( 0.6 m )( A ) = 0
From above
A = 0.3274 kN
A = 327 N
W
B = 827 N
W
B = A + 0.05 kN
B = ( 0.327 + 0.50 ) kN = 0.827 kN
PROBLEM 6.109
Members ABC and CDE are pin-connected at C and are supported by
four links. For the loading shown, determine the force in each link.
SOLUTION
member FBDs:
FBD I:
FBD II:
ΣM I = 0: 2aC y + aC x − aP = 0
ΣM J = 0: 2aC y − aC x = 0
Solving: Cx =
FBD I:
2C y − Cx = 0
P
P
; Cy =
as shown
2
4
ΣFx = 0: −
1
FBG + C x = 0
2
ΣFy = 0: FAF − P +
FBD II:
2C y + Cx = P
FBG =
FBG = C x 2
1  2  P
P +
=0

2  2  4
ΣFx = 0: −C x +
1
FDG = 0
2
ΣFy = 0: −C y +
1  2 
P  + FEH = 0

2  2 
2
P CW
2
FAF =
FDG =
FDG = C x 2
FEH =
P P
P
−
=−
4
2
4
P
CW
4
2
P CW
2
FEH =
P
TW
4
PROBLEM 6.110
Members ABC and CDE are pin-connected at C and are supported by
four links. For the loading shown, determine the force in each link.
SOLUTION
member FBDs:
From FBD I:
FBD II:
Solving : Cx =
FBD I:
ΣM J = 0:
a
3a
a
Cx +
Cy − P = 0
2
2
2
ΣM K = 0:
a
3a
Cx −
Cy = 0
2
2
Cx + 3C y = P
Cx − 3C y = 0
P
P
; Cy =
as drawn
2
6
ΣM B = 0: aC y − a
ΣFx = 0: −
1
FAG = 0
2
FAG =
1
1
FAG +
FBF − Cx = 0
2
2
2C y =
2
P
6
FBF = FAG + Cx 2 =
FAG =
2
2
P+
P
6
2
FBF =
FBD II:
ΣM D = 0: a
1
FEH + aC y = 0
2
ΣFx = 0: C x −
FEH = − 2C y = −
1
1
FDI +
FEH = 0
2
2
2
P
6
FDI = FEH + Cx 2 = −
2
P CW
6
2 2
P CW
3
FEH =
2
P TW
6
2
2
P+
P
6
2
FDI =
2
P CW
3
PROBLEM 6.111
Members ABC and CDE are pin-connected at C and are supported
by four links. For the loading shown, determine the force in each link.
SOLUTION
member FBDs:
FBD I:
ΣM B = 0: aC y − a
1
FAF = 0
2
FAF =
2C y
FBD II:
M D = 0: aC y − a
1
FEH = 0
2
FEH =
2C y
FBDs combined:
ΣM G = 0: aP − a
1
1
FAF − a
FEH = 0
2
2
Cy =
FBD I:
ΣFy = 0:
1
1
2C y +
2C y
2
2
P
2
1
1
FDG −
FEH = 0
2
2
so FAF =
2
P C
2
FEH =
2
P T
2
P
1
P
FBG − P +
+
=0
2
2
2
1
1
FAF +
FBG − P + C y = 0
2
2
FBD II: ΣFy = 0: − C y +
P=
−
P
1
P
FDG −
+
=0
2
2
2
FBG = 0
FDG =
2P C
PROBLEM 6.112
Solve Prob. 6.109 assuming that the force P is replaced by a
clockwise couple of moment M 0 applied to member CDE at D.
SOLUTION
FBDs members:
1
FBG = 0
2
FBD I:
ΣM A = 0: 2aC y − a
FBD II:
ΣM E = 0: 2aC y − M 0 + a
FBDs combined:
ΣFx = 0:
FBG = 2 2C y
1
FDG = 0
2
FBG = 2 2
FBD I: ΣFy = 0: FAF −
FBD II:
2
M0
a
1
1
FBG + C x − C x −
FDG = 0
2
2
FBG = FDG : 2 2C y = −2 2C y +
FDG =
FDG = −2 2C y +
2
M0
a
Cy =
M0
4a
2
M
M0 − 2 2 0
a
4a
1
FBG + C y = 0
2
ΣH D = 0: aC y − M 0 + aFEH = 0
FAF =
1 2 M0 M0
−
4a
2 2 a
aFEH = M 0 − a
M0
4a
M0
4a
FBG =
2 M0
T
2 a
FDG =
2 M0
2 a
T
M0
4a
C
3 M0
4 a
C
FAF =
FEH =
PROBLEM 6.113
Four wooden beams, each of length 2a, are nailed together at their
midpoints to form the support system shown. Assuming that only
vertical forces are exerted at the connections, determine the vertical
reactions at A, D, E, and H.
SOLUTION
Note that, if we assume P is applied to EG, each individual member FBD
looks like
so
2Fleft = 2 Fright = Fmiddle
Labeling each interaction force with the letter corresponding to the joint
of its application, we see that
B = 2 A = 2F
C = 2B = 2D
G = 2C = 2H
P + F = 2G ( = 4C = 8B = 16 F ) = 2E
From
P + F = 16 F , F =
P
15
so A =
P
15
D=
2P
15
H=
4P
15
E=
8P
15
PROBLEM 6.114
Three wooden beams, each of length of 3a, are nailed together to
form the support system shown. Assuming that only vertical forces
are exerted at the connections, determine the vertical reactions at A,
D, and F.
SOLUTION
Note that, if we assume P is applied to BF, each individual member FBD
looks like:
Fshort = 2Flong =
so
2
Fload
3
(by moment equations about S and L).
Labeling each interaction force with the letter corresponding to the joint
of application, we have:
2( P + E )
= 2B
3
C
D
=
E =
3
2
B
A
C =
=
3
2
F =
so
2(P + E)
= 2B = 6C = 18E
3
A = 2C = 3E
F =
2
P
P +

3
26 
P + E = 27 E
E=
P
26
so D = 2E =
P
13
A=
3P
13
F=
9P
13
PROBLEM 6.115
Each of the frames shown consists of two L-shaped members
connected by two rigid links. For each frame, determine the reactions
at the supports and indicate whether the frame is rigid.
SOLUTION
(a) member FBDs:
FBD I:
ΣM A = 0: aF1 − 2aP = 0
FBD II:
ΣM B = 0: − aF2 = 0
F1 = 2 P;
ΣFy = 0: Ay − P = 0
F2 = 0
ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2P
B x = 2P
ΣFy = 0: By = 0
FBD I:
Ay = P
ΣFx = 0: − Ax − F1 + F2 = 0
Ax = F2 − F1 = 0 − 2P
so B = 2P
A x = 2P
so A = 2.24 P
26.6°
frame is rigid
(b) FBD left:
FBD I:
FBD II:
FBD whole:
a
a
5a
P + Ax −
Ay = 0
2
2
2
Ax − 5 Ay = − P
ΣM B = 0: 3aP + aAx − 5aAy = 0
Ax − 5 Ay = −3P
ΣM E = 0:
This is impossible unless P = 0 ∴ not rigid
PROBLEM 6.115 CONTINUED
(c) member FBDs:
FBD I: ΣFy = 0: A − P = 0
A= P
ΣM D = 0: aF1 − 2aA = 0
ΣFx = 0: F2 − F1 = 0
FBD II:
ΣM B = 0: 2aC − aF1 = 0
C =
F2 = 2 P
F1
= P
2
ΣFx = 0: F1 − F2 + Bx = 0
ΣFx = 0: By + C = 0
F1 = 2 P
C= P
Bx = P − P = 0
By = −C = − P
B= P
Frame is rigid
PROBLEM 6.116
Each of the frames shown consists of two L-shaped members connected by
two rigid links. For each frame, determine the reactions at the supports and
indicate whether the frame is rigid.
SOLUTION
(a) member FBDs:
FBD II:
ΣFy = 0: By = 0
ΣM B = 0: aF2 = 0
FBD I:
ΣM A = 0: aF2 − 2aP = 0
F2 = 0
but F2 = 0
so P = 0
not rigid for P ≠ 0
(b) member FBDs:
(
)
Note: 7 unknowns Ax , Ay , Bx , By , F1, F2 , C but only 6 independent equations.
System is statically indeterminate
System is, however, rigid
(c) FBD whole:
FBD right:
PROBLEM 6.116 CONTINUED
FBD I:
ΣM A = 0: 5aBy − 2aP = 0
ΣFy = 0: Ay − P +
FBD II:
FBD I:
ΣM c = 0:
2
P=0
5
By =
Ay =
a
5a
Bx −
By = 0
2
2
ΣFx = 0: Ax + Bx = 0
2
P
5
3
P
5
Bx = 5By
Ax = − Bx
B x = 2P
A x = 2P
A = 2.09 P
16.70°
B = 2.04P
11.31°
System is rigid
PROBLEM 6.117
Each of the frames shown consists of two L-shaped members
connected by two rigid links. For each frame, determine the reactions
at the supports and indicate whether the frame is rigid.
SOLUTION
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the
third force, at B, must be parallel to the link forces.
(a) FBD whole:
ΣM A = 0: − 2aP −
ΣFx = 0: Ax −
a 4
1
B + 5a
B=0
4 17
17
4
B=0
17
ΣFy = 0: Ay − P +
1
B=0
17
B = 2.06 P
B = 2.06P
14.04° W
A = 2.06P
14.04° W
A x = 2P
Ay =
P
2
rigid W
(b) FBD whole:
Since B passes through A,
∴ no equilibrium if P ≠ 0
ΣM A = 2aP = 0 only if P = 0
not rigid W
PROBLEM 6.117 CONTINUED
(c) FBD whole:
ΣM A = 0: 5a
ΣFx = 0: Ax +
1
3a 4
B+
B − 2aP = 0
4 17
17
4
B=0
17
ΣFy = 0: Ay − P +
1
B=0
17
B=
17
P
4
B = 1.031P
14.04° W
Ax = − P
Ay = P −
P 3P
=
4
4
A = 1.250P
36.9° W
System is rigid W
PROBLEM 6.118
A 50-N force directed vertically downward is applied to the toggle vise
at C. Knowing that link BD is 150 mm long and that a = 100 mm,
determine the horizontal force exerted on block E.
SOLUTION
FBD machine:
Note: ( 0.1 m ) sin 15° = ( 0.15 m ) sin θ
θ = sin −1 ( 0.17255 ) = 9.9359°
p
AD = ( 0.1 m ) cos 15° + ( 0.15 m ) cos θ
= 0.24434 m
ΣM A = 0: ( 0.24434 m ) D sin θ − ( 0.25m )( cos 15° )( 50 N ) = 0
D=
FBD part D :
0.25 cos 15°
50 N = 286.38 N
( 0.24434 )( 0.17255)
ΣFx = 0: D cosθ − E = 0
E = D cosθ = 282.1 N
Eblock = 282 N
W
PROBLEM 6.119
A 50-N force directed vertically downward is applied to the toggle
vise at C. Knowing that link BD is 150 mm long and that a = 200 mm,
determine the horizontal force exerted on block E.
SOLUTION
FBD machine:
Note:
( 0.2 m ) sin15° = ( 0.15 m ) sin θ
θ = sin −1 ( 0.3451) = 20.187°
p
AD = ( 0.2 m ) cos15° + ( 0.15m ) cos θ
= 0.33397 m
ΣM A = 0: ( 0.33397 m ) D sin 20.187° − ( 0.35m )( cos15° )( 50 N ) = 0
FBD part D:
D = 146.67 N
ΣFx = 0: D cosθ − E = 0
E = (146.67 N ) cos 20.187°
= 137.66 N
Eon block = 137.7 N
PROBLEM 6.120
The press shown is used to emboss a small seal at E. Knowing that
P = 60 lb, determine (a) the vertical component of the force
exerted on the seal, (b) the reaction at A.
SOLUTION
FBD machine:
ΣM A = 0: ( 5 in.) cos 60°  D cos 20° + ( 5 in.) sin 60° D sin 20°
− ( 5 in.) cos 60° + (10 in.) cos15° ( 60 lb ) = 0
D = 190.473 lb
ΣFx = 0: Ax − D sin 20° = 0
Ax = 65.146 lb
ΣFy = 0: Ay + D cos 20° − 60 lb = 0
Ax = −118.99 lb
so A = 135.7 lb
61.3°
FBD part D:
ΣFy = 0: E − D cos 20° = 0
E = (190.47 lb ) cos 20°
= 178.98 lb
Eon seal = 179.0 lb
PROBLEM 6.121
The press shown is used to emboss a small seal at E. Knowing that
the vertical component of the force exerted on the seal must be 240 lb,
determine (a) the required vertical force P, (b) the corresponding
reaction at A.
SOLUTION
FBD part D:
(a)
ΣFy = 0: E − D cos 20° = 0
D=
240 lb
= 255.40 lb
cos 20°
ΣM A = 0: ( 5in.) cos 60°  D cos 20° + ( 5 in.) cos 60° D sin 20°
FBD machine:
− ( 5in.) cos 60° + 10in.) cos15° P = 0
P = 80.453 lb
(b)
P = 80.5 lb
ΣFx = 0: Ax − D sin 20° = 0
Ax = 87.35 lb
ΣFy = 0: Ay + 240 lb − 80.5 lb = 0
Ay = 159.5 lb
A = 181.9 lb
61.3°
PROBLEM 6.122
The double toggle latching mechanism shown is used to hold member
G against the support. Knowing that α = 60°, determine the force
exerted on G.
SOLUTION
member FBDs:
tan θ =
Note:
( 0.05 m + 0.02 m ) sin 60°
( 0.09 m ) + ( 0.03 + 0.05 − 0.02 ) m  cos 60°
= 0.50518
θ = 26.802°
FBD I:
ΣM C = 0: ( 0.15 m ) 80 N − ( 0.02 m ) FDF cos ( 30° − 26.802° ) = 0
FDF = 600.94 N
ΣFx = 0: ( 600.94 N ) cos 26.802° − ( 80 N ) sin 60° − Cx = 0
Cx = 467.10 N
ΣFy = 0: − C y + ( 600.94 N ) sin 26.802° − ( 80 N ) cos 60° = 0
C y = 230.97 N
FBD II:
ΣM G = 0: ( 0.03 m )( cos 30° )  Bx
+ 0.12 m + ( 0.03m ) cos 60°  By
+ ( 0.03 m ) ( 600.94 N ) sin 26.802°
− ( .026 m ) ( 600.94 N ) cos 26.802° = 0
0.015 3 Bx + 0.135By = 5.8065 N
(1)
PROBLEM 6.122 CONTINUED
FBD III:
ΣM A = 0: ( 0.03m )( sin 60° )  Bx + ( 0.03 m )( cos 60° )  By
− ( 0.08 m )( sin 60° )  467.10 N
+ ( 0.08 m )( cos 60° )  230.97 N = 0
0.015 3 Bx + 0.015By = 23.123 N
Solving (1) and (2)
(2)
Bx = 973.31 N
By = −144.303 N
FBD II:
ΣFx = 0: − Gx − ( 600.94 N ) cos 26.802° + 973.31 N = 0
G x = 436.93 N
ΣFy = 0: − ( −144.303 N ) + G y − ( 600.94 N ) sin 26.802° = 0
G y = 126.67 N
Therefore, the force acting on member G is
G = 455 N
16.17°
PROBLEM 6.123
The double toggle latching mechanism shown is used to hold member
G against the support. Knowing that α = 75°, determine the force
exerted on G.
SOLUTION
FBDs:
tan θ =
Note:
( 0.07 m ) cos 30°
0.09 m + ( 0.03 m ) cos 75° + ( 0.03 m ) sin 30°
θ = 28.262°
FBD I:
ΣM C = 0: ( 0.15 m ) ( 80 N ) − ( 0.02 m ) FDF cos ( 30° − 28.262° ) = 0
FDF = 600.28 N
ΣFx = 0: − Cx − ( 80 N ) cos 30° + ( 600.28 N ) cos 28.262° = 0
Cx = 459.44 N
ΣFy = 0: − C y − ( 80 N ) sin 30° − ( 600.28 N ) sin 28.262° = 0
C y = 244.24 N
FBD II:
ΣM G = 0: − ( 0.03m ) sin 75° Bx +  0.120 m + ( 0.03m ) cos 75°  By
+  ( 0.03 m ) sin 75° ( 600.28 N ) cos 28.262°
− ( 0.03 m ) ( 600.28 N ) sin 28.262° = 0
0.9659 Bx − 4.2588By = 226.47 N
(1)
PROBLEM 6.123 CONTINUED
FBD III:
ΣM A = 0: ( 0.03 m ) sin 75° Bx − ( 0.03 m ) cos 75° By
− ( 0.03 m ) sin 75° + ( 0.05 m ) sin 60° ( 459.44 N )
+ ( 0.03 m ) cos 75° + ( 0.05 m ) cos 60°  ( 244.24 N ) = 0
0.9659 Bx − 0.2588By = 840.18 N
Solving (1) and (2):
(2)
Bx = 910.93 N
By = 153.428 N
FBD II:
ΣFx = 0: − Gx − ( 600.28 N ) cos 28.262° + 910.93 N = 0
G x = 382.21 N
ΣFy = 0: G y − ( 600.28 N ) sin 28.262° + 153.428 N = 0
G y = 130.81 N
Therefore, the force acting on member G is:
G = 404 N
18.89°
PROBLEM 6.124
For the system and loading shown, determine (a) the force P required
for equilibrium, (b) the corresponding force in member BD, (c) the
corresponding reaction at C.
SOLUTION
FBD I :
member FBDs:
ΣM C = 0: R ( FBD sin 30° )
−  R (1 − cos 30° )  ( 200 N ) − R ( 100 N ) = 0
FBD = 253.6 N
FBD = 254 N T
ΣFx = 0: − Cx + ( 253.6 N ) cos 30° = 0
C x = 219.6 N
ΣFy = 0: C y + ( 253.6 N ) sin 30° − 200 N − 100 N = 0
C y = 173.2 N
FBD II :
so C = 280 N
ΣM A = 0: aP − a ( 253.6 N ) cos 30°  = 0
P = 220 N
38.3°
PROBLEM 6.125
A couple M of magnitude 15 kip ⋅ in. is applied to the crank of
the engine system shown. For each of the two positions shown,
determine the force P required to hold the system in equilibrium.
SOLUTION
(a) FBDs:
Note:
tan θ =
=
1.5 in.
5.25 in.
2
7
FBD whole:
ΣM A = 0: ( 7.50 in.) C y − 15 kip ⋅ in. = 0
FBD piston:
ΣFy = 0: C y − FBC sin θ = 0
FBC =
C y = 2.00 kips
Cy
sin θ
=
2 kips
sinθ
ΣFx = 0: FBC cosθ − P = 0
P = FBC cosθ =
2 kips
= 7 kips
tan θ
P = 7.00 kips
PROBLEM 6.125 CONTINUED
(b) FBDs:
Note:
FBD whole:
tan θ =
2
as above
7
ΣM A = 0: ( 3 in.) C y − 15 kip ⋅ in. = 0
C y = 5 kips
ΣFy = 0: C y − FBC sin θ = 0
FBC =
Cy
sin θ
ΣFx = 0: FBC cosθ − P = 0
P = FBC cosθ =
Cy
tan θ
=
5 kips
2/7
P = 17.50 kips
PROBLEM 6.126
A force P of magnitude 4 kips is applied to the piston of the engine
system shown. For each of the two positions shown, determine the
couple M required to hold the system in equilibrium.
SOLUTION
(a) FBDs:
Note:
tan θ =
=
FBD piston:
FBD whole:
1.5 in.
5.25 in.
2
7
P
cosθ
ΣFx = 0: FBC cosθ − P = 0
FBC =
ΣFy = 0: C y − FBC sin θ = 0
C y = FBC sin θ = P tan θ =
ΣM A = 0: ( 7.50 in.) C y − M = 0
M = 7.5 in.
Cy =
2
P
7
15 in.
P
7
M = 8.57 kip ⋅ in.
PROBLEM 6.126 CONTINUED
(b) FBDs:
Note:
tan θ =
2
as above
7
FBD piston: as above C y = P tan θ =
FBD whole:
2
P
7
ΣM A = 0: ( 3.0 in.) C y − M = 0
M =
2
M = ( 3.0 in.) P
7
24
kip ⋅ in.
7
M = 3.43 kip ⋅ in.
PROBLEM 6.127
Arm BCD is connected by pins to crank AB at B and to a collar at C.
Neglecting the effect of friction, determine the couple M required to
hold the system in equilibrium when θ = 0.
SOLUTION
member FBDs:
FBD II:
ΣFy = 0: By = 0
ΣM C = 0:
FBD I:
( 6.4 in.) Bx − ( 5 in.)16 lb = 0
ΣM A = 0: (7.5 in.) Bx − M = 0
Bx = 12.5 lb
M = ( 7.5 in.)(12.5 lb ) = 93.8 lb ⋅ in.
M = 93.8 lb ⋅ in.
PROBLEM 6.137
The drum lifter shown is used to lift a steel drum. Knowing that the mass of
the drum and its contents is 240 kg, determine the forces exerted at F and H
on member DFH.
SOLUTION
ΣFy = 0;
FBD System:
P −W = 0
P =W
(
= ( 240 kg ) 9.81 m/s 2
P = 2354.4 N
FBD machine:
Symmetry: H x = I x
Hy = Iy
ΣFy = 0: P − 2 H y = 0
Hy =
FBD ABC:
P
= 1177.2 N
2
Symmetry: C = B
4
ΣFy = 0: P − 2 B = 0
5
B=
5
P = 1471.5 N
8
)
PROBLEM 6.137 CONTINUED
FBD DFH:
3

ΣM H = 0: − ( 0.1 m ) F + ( 0.39 m )  1471.5 N 
5

4

− ( 0.055 m )  1471.5 N  = 0
5

F = 3973.05 N
F = 3.97 kN
W
3
ΣFx = 0: 3973.05 N − 1471.5 N − H x = 0
5
H x = 3090.15 N
H = 3.31 kN
20.9° W
PROBLEM 6.138
A small barrel having a mass of 72 kg is lifted by a pair of tongs as
shown. Knowing that a = 100 mm, determine the forces exerted at
B and D on tong ABD.
SOLUTION
Notes: From FBD whole, by inspection,
(
P = W = mg = ( 72 kg ) 9.81 m/s2
FBB ABD:
)
P = 706.32 N
BC is a two-force member: Bx = 3By
ΣM D = 0: ( 0.1 m ) Bx + ( 0.06 m ) By − ( 0.18 m ) P = 0
3By + 0.6 By = 1.8P
By =
so
P
= 353.16 N
2
Bx =
3P
= 1059.48 N
2
and B = 1.117 kN
18.43° W
ΣFx = 0: − Bx + Dx = 0
Dx =
3P
= 1059.48 N
2
ΣFy = 0: P − By − Dy = 0
Dy = P −
P
= 353.16 N
2
so D = 1.117 kN
18.43° W
PROBLEM 6.139
Determine the magnitude of the ripping forces exerted along
line aa on the nut when two 240-N forces are applied to the handles
as shown. Assume that pins A and D slide freely in slots cut in the
jaws.
SOLUTION
FBD jaw AB:
ΣFx = 0: Bx = 0
ΣM B = 0: ( 0.01 m ) G − ( 0.03 m ) A = 0
A=
G
3
ΣFy = 0: A + G − By = 0
By = A + G =
4G
3
FBD handle ACE:
By symmetry and FBD jaw DE: D = A =
E y = By =
G
, Ex = Bx = 0,
3
4G
3
ΣM C = 0: ( 0.105 m )( 240 N ) + ( 0.015 m )
G
4G
− ( 0.015 m )
=0
3
3
G = 1680 N W
PROBLEM 6.140
In using the bolt cutter shown, a worker applies two 75-lb forces to
the handles. Determine the magnitude of the forces exerted by the
cutter on the bolt.
SOLUTION
FBD Cutter AB:
FBD I:
FBD handle BC:
ΣFx = 0: Bx = 0
FBD II:
ΣM C = 0: ( 0.5 in.) By − (19.5 in.) 75 lb = 0
By = 2925 lb
Then
FBD I:
ΣM A = 0: ( 4 in.) By − (1 in.) F = 0
F = 4 By
F = 11700 lb = 11.70 kips W
PROBLEM 6.141
Determine the magnitude of the gripping forces produced when two 50-lb
forces are applied as shown.
SOLUTION
FBD handle CD:
ΣM D = 0: − ( 4.2 in.)( 50 lb ) − ( 0.2 in.)
2.8
A
8.84
 1

+ (1 in.) 
A = 0
 8.84 
A = 477.27 8.84 lb
FBD handle AD:
ΣM D = 0: ( 4.4 in.)( 50 lb ) − ( 4 in.)
(
1
477.27 8.84 lb
8.84
)
+ (1.2 in.) F = 0
F = 1.408 kips W
PROBLEM 6.142
The compound-lever pruning shears shown can be adjusted by
placing pin A at various ratchet positions on blade ACE. Knowing
that 1.5-kN vertical forces are required to complete the pruning of a small
branch, determine the magnitude P of the forces that must be applied to the
handles when the shears are adjusted as shown.
SOLUTION
FBD cutter AC:
ΣM C = 0: ( 32 mm )1.5 KN − ( 28 mm ) Ay − (10 mm ) Ax = 0
 11 
10 Ax + 28  Ax  = 48 kN
 13 
Ax = 1.42466 kN
Ay = 1.20548 kN
FBD handle AD:
ΣM D = 0: (15 mm )(1.20548 kN ) − ( 5 mm )(1.42466 kN )
− (70 mm) P = 0
P = 0.1566 kN = 156.6 N W
PROBLEM 6.143
Determine the force P which must be applied to the toggle BCD
To maintain equilibrium in the position shown.
SOLUTION
θ = 30° + α
Note:
FBD joint B:
= 30° + tan −1
20
200
= 30° + 5.711°
= 35.711°
ΣFx′ = 0: FBC cos 35.711° − ( 240 N ) cos 30° = 0
FBC = 255.98 N T
FBD joint C:
By symmetry: FCD = 255.98 N
ΣFx′′ = 0: P − 2 ( 255.98 N ) sin 5.711° = 0
P = 50.9 N
30.0° W
PROBLEM 6.144
In the locked position shown, the toggle clamp exerts at A a vertical
1.2-kN force on the wooden block, and handle CF rests against the
stop at G. Determine the force P required to release the clamp.
(Hint: To release the clamp, the forces of contact at G must be
zero.)
SOLUTION
FBD BC:
ΣM B = 0: ( 24 mm ) C x − ( 66 mm )(1.2 kN ) = 0
Cx = 3.3 kN
ΣFx = 0: Dx − C x = 0
Dx = 3.3 kN
ΣM C = 0: ( 86 mm ) P
FBD CDF:
− 18.5 mm − (16 mm ) tan 40°  ( FDE cos 40° ) = 0
FDE = 22.124 P
ΣFx = 0: Cx − FDE cos 40° + P sin 30° = 0
3.3 kN − ( 22.124 P ) cos 40° + P sin 30° = 0
P = 201 N
60° W
PROBLEM 6.145
The garden shears shown consist of two blades and two handles.
The two handles are connected by pin C and the two blades are
connected by pin D. The left blade and the right handle are connected
by pin A; the right blade and the left handle are connected by pin B.
Determine the magnitude of the forces exerted on the small branch
E when two 20-lb forces are applied to the handles as shown.
SOLUTION
Note: By symmetry the vertical components of pin forces C and D
are zero.
FBD handle ACF: (not to scale)
ΣFy = 0: Ay = 0
ΣM C = 0: (13.5 in.)( 20 lb ) − (1.5 in.) Ax = 0
ΣFx = 0: C − Ax − 20 lb = 0
Ax = 180 lb
C = (180 + 20 ) lb = 200 lb
FBD blade DE:
ΣM D = 0: ( 9 in.) E − ( 3 in.)(180 lb ) = 0
E = 60.0 lb W
PROBLEM 6.146
The bone rongeur shown is used in surgical procedures to cut small
bones. Determine the magnitude of the forces exerted on the bone
at E when two 25-lb forces are applied as shown.
SOLUTION
Note: By symmetry the horizontal components of pin forces at A and D
are zero.
FBD handle AB:
ΣFx = 0: Bx = 0
ΣM A = 0: (1.1 in.) By − ( 4.4 in.)( 25 lb )
By = 100 lb
FBD Blade BD:
ΣM A = 0: (1.6 in.)(100 lb ) − (1.2 in.)( E ) = 0
E = 133.3 lb W
PROBLEM 6.147
The telescoping arm ABC is used to provide an elevated platform for
construction workers. The workers and the platform together have a
mass of 240 kg and have a combined center of gravity located directly
above C. For the position when θ = 24o , determine (a) the force
exerted at B by the single hydraulic cylinder BD, (b) the force exerted
on the supporting carriage at A.
SOLUTION
θ = tan −1
Note:
FBD boom:
( 3.2sin 24° − 1) m
( 3.2cos 24° − 0.6 ) m
θ = 44.73°
(a )
ΣM A = 0: ( 6.4 m ) cos 24° ( 2.3544 kN )
− ( 3.2 m ) cos 24° B sin 44.73°
+ ( 3.2 m ) sin 24° B cos 44.73° = 0
B = 12.153 kN
B = 12.15 kN
44.7° W
ΣFx = 0: Ax − (12.153 kN ) cos 44.73° = 0
A x = 8.633 kN
(b )
ΣFy = 0: − 2.3544 kN + (12.153 kN ) sin 44.73° − Ay = 0
A y = 6.198 kN
On boom:
On carriage:
A = 10.63 kN
35.7°
A = 10.63 kN
35.7° W
PROBLEM 6.148
The telescoping arm ABC can be lowered until end C is close to the ground,
so that workers can easily board the platform. For the position when
θ = −18o , determine (a) the force exerted at B by the single hydraulic
cylinder BD, (b) the force exerted on the supporting carriage at A.
SOLUTION
FBD boom:
1 m − 3.2 m sin18°
3.2 m cos18° − 0.6 m
θ = 0.2614°
θ = tan −1
W = ( 240 kg )( 9.81 N kg ) = 2354.4 N
(a)
ΣM A = 0: ( 6.4 m ) cos18°  2.3544 kN − ( 3.2 m ) cos18° B sin ( 0.2614° )
− ( 3.2 m ) sin18°  B cos ( 0.2614° ) = 0
B = 14.292 kN
(b)
B = 14.29 kN
.261° W
A = 14.47 kN
9.10°
ΣFx = 0: Ax − B cos ( 0.2614° ) = 0
Ax = (14.292 kN ) cos ( 0.2614° ) = 14.292 kN
ΣFy = 0: Ay + B sin ( 0.2614° ) − 2.3544 kN = 0
Ay = 2.3544 kN − (14.292 kN ) sin ( 0.2614° )
Ay = 2.2892 kN
On boom:
On carriage:
A = 14.47 kN
9.10° W
PROBLEM 6.149
The bucket of the front-end loader shown carries a 12-kN load.
The motion of the bucket is controlled by two identical
mechanisms, only one of which is shown. Knowing that the
mechanism shown supports one-half of the 12-kN load,
determine the force exerted (a) by cylinder CD, (b) by
cylinder FH.
SOLUTION
FBD bucket:
(a )
ΣM D = 0: ( 0.3 m )(12 kN ) − ( 0.32 m )( 2FAB ) = 0 FAB = 5.625 kN
ΣFx = 0: 2FAB − 2 Dx = 0
Dx = FAB = 5.625 kN
ΣFy = 0: Dy − 12 kN = 0
Dy = 12 kN
FBD link BCE:
θ = tan −1
160
= 21.801°
400
ΣM E = 0: ( 0.46 m )( 5.625 kN ) + ( 0.1 m )( FCD sin 21.801° )
− ( 0.3 m )( FCD cos 21.801° ) = 0
FCD = 10.7185 kN
On BCE:
(C)
FCD = 10.72 kN
21.8° W
FBD boom & bucket mechanism: (b)
ΣM G = 0: (1.5 m )(12 kN ) + ( 0.12 m )( 2 FFH cos 45° )
− ( 0.48 m )( 2FFH sin 45° ) = 0
FFH = 35.4 kN C
On DFG:
FFH = 35.4 kN
45° W
PROBLEM 6.150
The motion of the bucket of the front-end loader shown is controlled
by two arms and a linkage which are pin-connected at D. The arms
are located symmetrically with respect to the central, vertical, and
longitudinal plane of the loader; one arm AFJ and its control cylinder
EF are shown. The single linkage GHBD and its control cylinder
BC are located in the plane of symmetry. For the position shown,
determine the force exerted (a) by cylinder BC, (b) by cylinder EF.
SOLUTION
FBD bucket:
(a )
ΣM J = 0: ( 0.5 m )( 20 kN ) − ( 0.55 m ) FGH = 0
FGH = 18.1818 kN
ΣM D = 0: ( 0.5 m ) FBC − ( 0.6 m ) FGH = 0
FBD link BH:
FBC =
6
6
FGH = 18.1818 kN = 21.818 kN
5
5
On BH:
FBC = 21.8 kN
W
FBD mechanism with bucket:
(b)
θ = tan −1
1.625 m
= 74.521°
0.45 m
ΣM A = ( 3.075 m )( 20 kN ) − ( 0.3 m )( 28.818 kN )
− ( 0.6 m )( 2FEF sin 74.521° ) = 0
On AF:
FEF = 47.5 kN
15.48° W
PROBLEM 6.151
In the planetary gear system shown, the radius of the central gear
A is a = 20 mm, the radius of the planetary gear is b, and the radius
of the outer gear E is ( a + 2b ) . A clockwise couple of magnitude
M A = 20 N ⋅ m is applied to the central gear A, and a counterclockwise couple of magnitude M S = 100 N ⋅ m is applied to the
spider BCD. If the system is to be in equilibrium, determine (a) the
required radius b of the planetary gears, (b) the couple ME that must
be applied to the outer gear E.
SOLUTION
FBD Gear A:
(a)
By symmetry F1 = F2 = F3 = F
ΣM A = 0: 3aF − 20 N ⋅ m = 0
F =
FBD Gear C:
20
N⋅m
3a
ΣFx′ = 0: Cx′ = 0
ΣM C = 0: bF − bF4 = 0
ΣFy′ = 0: C y′ − F − F4 = 0
F4 = F =
20 N ⋅ m
3a
C y′ = 2 F =
40 N ⋅ m
3a
By symmetry central forces on gears B and D are the same
FBD Spider:
Smaller scale
ΣM A = 0: M S − ( a + b ) 2 F = 0
100 N ⋅ m = 6 ( a + b ) F = ( a + b )
100
b
=1+
40
a
40
N⋅m
a
b
3
=
a
2
a = 20 mm so that b = 30.0 mm W
PROBLEM 6.151 CONTINUED
FBD Outer gear:
(b)
ΣM A = 0: 3 ( a + 2b ) F − M E = 0
M E = 3 ( 20 mm + 60 mm )
20 N ⋅ m
= 80.0 N ⋅ m W
3 ( 20 mm )
PROBLEM 6.152
Gears A and D are rigidly attached to horizontal shafts that are held
by frictionless bearings. Determine (a) the couple M 0 that must be
applied to shaft DEF to maintain equilibrium, (b) the reactions at
G and H.
SOLUTION
FBD Gear A: looking from C
(a)
M A = 15 lb ⋅ ft
rA = 4 in.
ΣM A = 0: M A − P rA = 0
P=
M A 180 lb ⋅ in.
=
rA
4 in.
P = 45 lb
FBD Gear B: looking from F
ΣM B = 0: M 0 − rB P = 0
M 0 = rB P = ( 2.5 in.)( 45 lb ) = 112.5 lb ⋅ in.
M 0 = 112.5 lb ⋅ in. i W
FBD ABC: looking down
(b)
ΣM B = 0: ( 2 in.)( 45 lb ) − ( 5 in.) C = 0
ΣFz = 0: 45 lb − B + 18 lb = 0
C = 18 lb k
B = −63 lb k
PROBLEM 6.152 CONTINUED
FBD BEG:
By analogy, using FBD DEF
E = 63 lb k
F = 18 lb k
ΣFz = 0: Gz + 63 lb − 63 lb = 0
Gz = 0
ΣFy = 0
ΣM G = 0
Gy = 0
M G − ( 6.5 in.)( 63 lb ) = 0
M G = ( 410 lb ⋅ in.) i W
FBD CFH:
ΣF = 0: H z = H y = 0
ΣM H = 0
M H = − ( 6.5 in.)(18 lb )
= −117 lb ⋅ in.
M G = − (117.0 lb ⋅ in.) i W
PROBLEM 6.153
Two shafts AC and CF, which lie in the vertical xy plane, are
connected by a universal joint at C. The bearings at B and D do
ot exert any axial force. A couple of magnitude 50 N ⋅ m (clockwise
when viewed from the positive x axis) is applied to shaft CF at F.
At a time when the arm of the crosspiece attached to shaft CF is
horizontal, determine (a) the magnitude of the couple which must
be applied to shaft AC at A to maintain equilibrium, (b) the reactions
at B, D, and E. (Hint: The sum of the couples exerted on the
crosspiece must be zero).
SOLUTION
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither
yoke can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples
must be normal to the plane of the crosspiece.
If the crosspiece arm attached to shaft CF is horizontal, the plane of the crosspiece is normal to shaft AC,
so couple M C is along AC.
FBDs shafts with yokes:
(a)
FBD CDE:
ΣM x = 0:
M C cos30° − 50 N ⋅ m = 0
M C = 57.735 N ⋅ m
FBD BC: ΣM x′ = 0: M A − M C = 0
(b)
M A = 57.7 N ⋅ m W
ΣM C = 0: M A i′ + ( 0.5 m ) Bz j′ − ( 0.5 m ) By′ k = 0
ΣF = 0: B + C = 0
so
B = 0W
C=0
FBD CDF : ΣM Dy = 0: − ( 0.6 m ) Ez + ( 57.735 N ⋅ m ) sin 30° = 0
E z = 48.1 N k
ΣFx = 0: Ex = 0
ΣM Dz = 0: ( 0.6 m ) E y = 0
0
ΣF = 0: C + D + E = 0
E y = 0 so E = ( 48.1 N ) k W
D = −E = − ( 48.1 N ) k W
PROBLEM 6.154
Solve Prob. 6.153 assuming that the arm of the crosspiece attached to
shaft CF is vertical.
SOLUTION
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither
yoke can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples
must be normal to the plane of the crosspiece.
If the crosspiece arm attached to CF is vertical, the plane of the crosspiece is normal to CF, so the couple
M C is along CF.
(a) FBD CDE:
FBD BC:
ΣM x = 0: M C − 50 N ⋅ m = 0
M C = 50 N ⋅ m
ΣM x′ = 0: M A − M C cos 30° = 0
M A = ( 50 N ⋅ m ) cos 30°
M A = 43.3 N ⋅ m W
(b)
ΣM Cy′ = 0: M C sin 30° + ( 0.5 m ) Bz = 0
ΣM Cz = 0: − ( 0.5 m ) By = 0
ΣF = 0: B + C = 0
FBD CDE:
C = −B
Bz = −
( 50 N ⋅ m )( 0.5)
By = 0
so
ΣM Dy = 0: − ( 0.4 m ) Cz − ( 0.6 m ) Ez = 0
0.5 m
= −50 N
so B = − ( 50.0 N ) k W
C = ( 50 N ) k on BC
4
Ez = − ( 50 N )   = −33.3 N
6
ΣM Dz = 0: E y = 0
ΣFx = 0: Ex = 0
ΣF = 0: C + D + E = 0
so E = − ( 33.3 N ) k W
− ( 50 N ) k + D − ( 33.3 N ) k = 0
D = ( 83.3 N ) k W
PROBLEM 6.155
The large mechanical tongs shown are used to grab and lift a thick
1800-lb steel slab HJ. Knowing that slipping does not occur between
the tong grips and the slab at H and J, determine the components of
all forces acting on member EFH. (Hint: Consider the symmetry of
the tongs to establish relationships between the components of the
force acting at E on EFH and the components of the force acting at
D on CDF.)
SOLUTION
FBD A:
By inspection of FBD whole: FA = W = 1800 lb
By symmetry: FAB = FAC = T (say)
ΣFy = 0: 1800 lb − 2
1
T =0
5
T = 900 5 lb = FAC = FAB
ΣFx = 0: Dx − Fx −
FBD CDF:
(
)
2
900 5 lb = 0
5
Dx − Fx = 1800 lb
ΣFy = 0: − Dy + Fy +
(
(1)
)
1
900 5 lb = 0
5
Dy − Fy = 900 lb
(
(2)
)
(
)
 1

 2

ΣM D = 0: (6.9 ft) 
900 5 lb  + ( 0.9 ft ) 
900 5 lb 
 5

 5

FBD EFH:
− (1.5 ft ) Fy − ( 5.4 ft ) Fx = 0
5.4 Fx + 1.5Fy = 7830 lb
(3)
Note: By symmetry Ex = Dx ; E y = Dy
ΣM F = 0: ( 4.5 ft ) Dy − ( 5.4 ft ) Dy − (1.5 ft ) H x
+ ( 0.9 ft ) 900 lb = 0
5.4 Dx − 4.5D y + 1.5H x = 810 lb
(4)
ΣFx = 0: Dx + Fx − H x = 0
Fx = 648 lb
(5)
W
PROBLEM 6.155 CONTINUED
Solving equations (1) through (5):
Fx = 648 lb
W
Fy = 2.89 kips
W
E x = D x = 2.45 kips
E y = D y = 3.79 kips
H x = 3.10 kip
and, as noted H y = 900 lb
W
W
W
W
PROBLEM 6.156
For the frame and loading shown, determine the force acting on member
ABC (a) at B, (b) at C.
SOLUTION
FBD ABC:
Note: BD is two-force member
(a)
3

ΣM C = 0: ( 0.09 m )( 200 N ) − ( 2.4 m )  FBD  = 0
5

FBD = 125.0 N
(b)
ΣFx = 0: 200 N −
ΣFy = 0:
4
(125 N ) − Cx = 0
5
3
FBD − C y = 0
5
Cy =
36.9° W
C x = 100 N
3
(125 N ) = 75 N
5
C = 125.0 N
36.9° W
PROBLEM 6.157
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
ΣFx = 0: H x = 0
FBD Truss:
ΣM H = 0: ( 32 ft )(12 kips ) − ( 8 ft ) G = 0
G = 48 kips
ΣFy = 0: −12 kips + G − H y = 0
H y = 48 kips − 12 kips = 36 kips
H y = 36 kips
12 kips
F
F
= AB = AC
3
8
73
FBD joint A:
so FAB = 32.0 kips T W
FAC = 4 73 kips;
FAC = 34.2 kips C W
FBC = 0 W
By inspection of joint C :
FCE = 34.2 kips C W
FBE = 0 W
Then by inspection of joint B :
FBD = 32.0 kips T W
FDE = 0 W
Then by inspection of joint E :
FEG = 34.2 kips C W
FDG = 0 W
Then by inspection of joint D :
FBD joint G:
FDF = 32.0 kips T W
FFG = 0 W
By inspection of joint F :
FFH = 32.0 kips T W
ΣFx = 0:
(
)
8
8
4 73 kips −
FGH = 0
73
145
FGH = 4 145 kips
FGH = 48.2 kips C W
PROBLEM 6.158
The tongs shown are used to apply a total upward force of 45 kN on a
pipe cap. Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
By symmetry A = B = 22.5 kN
ΣM F = 0: ( 75 mm ) CD − (100 mm )( 22.5 kN ) = 0
CD = 30 kN
FBD ADF:
ΣFx = 0: Fx − CD = 0
ΣFy = 0: 22.5 kN − Fy = 0
W
Fx = CD = 30 kN
Fy = 22.5 kN
so F = 37.5 kN
36.9° W
PROBLEM 6.159
A stadium roof truss is loaded as shown. Determine the force in
members BC, BH, and GH.
SOLUTION
ΣM B = 0: ( 6.3 ft ) FGH − (14 ft )(1.8 kips ) − ( 28 ft )( 0.9 kip ) = 0
FBD Section:
FGH = 8.00 kips C W
 40

ΣM H = 0: ( 3.15 ft )  FBC  − (14 ft )( 0.9 kip ) = 0
 41

FBC = 4.10 kips T W
ΣFY = 0:
FBH =
9
9
FBC − 1.8 kips − 0.9 kip +
FBH = 0
41
21.93
21.93 
9
2.7 kips −
( 4.10 kips ) = 4.386 kips
9 
41

FBH = 4.39 kips T W
PROBLEM 6.160
A stadium roof truss is loaded as shown. Determine the force in
members EJ, FJ, and EI.
SOLUTION
FBD Truss:
(
ΣM K = 0: (16 ft ) Ly − 0.9 kip
)
− ( 28 ft )(1.8 kips )
− ( 42ft )(1.8 kips )
− ( 56 ft )( 0.9 kip ) = 0
Ly = 11.925 kips
ΣM E = 0: ( 8 ft )(11.925 kips − 0.9 kip )
− ( 20 ft ) (1.8 kips ) − ( 34 ft ) (1.8 kips )
− ( 48ft ) ( 0.9 kip ) + ( 31.5ft ) Lx = 0
Lx = 1.65714 kips
Joint L:
ΣFx = 0: −
8
FIL + 1.65714 kips = 0
32.5
FIL = 6.7321 kips
ΣFy = 0:
31.5
( 6.7321 kips ) + 11.925 kips
32.5
− FJL = 0
FJL = 18.4500 kips
By inspection of joint I,
FIJ = 0 and FEI = FIL = 6.73 kips T W
Then by inspection of joint J,
FEJ = 0 W
and FFJ = FJL = 18.45 kips C W
PROBLEM 6.161
For the frame and loading shown, determine the components of the
forces acting on member DABC at B and at D.
SOLUTION
FBD Frame:
ΣM G = 0: ( 0.6 m ) H y − ( 0.5 m ) 6 kN − (1.0 m )(12 kN ) = 0
H y = 25 kN
FBD BEH:
ΣM E = 0: ( 0.5 m ) Bx − ( 0.2 m )( 25 kN ) = 0
Bx = 10 kN
on DABC B x = 10.00 kN
W
ΣFx = 0: − Dx + Bx + 12 kN = 0
FBD DABC:
Dx = (10 kN + 12 kN ) = 22 kN
D x = 22.0 kN
W
ΣM B = 0: ( 0.8 m ) Dy − ( 0.5 m ) Dx = 0
Dy = 13.75 kN
D y = 13.75 kN
W
B y = 13.75 kN
W
ΣFy = 0: By − Dy = 0
By = 13.75 kN
PROBLEM 6.162
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Joint FBDs:
8.4 kN
F
F
= AC = AB
2.8
4.5
5.3
A:
FAB = 15.90 kN C W
FAC = 13.50 kN T W
C:
ΣFy = 0: 13.5 kN −
4.5
FCD = 0
5.3
FCD = 15.90 kN T W
ΣFy = 0: FBC −
2.8
(15.9 kN ) − 8.4 kN = 0
5.3
FBC = 16.80 kN C W
D:
ΣFy = 0:
4.5
(15.9 kN ) − FBD = 0
5.3
FBD = 13.50 kN C W
PROBLEM 6.163
For the frame and loading shown, determine the components of
the forces acting on member CFE at C and at F.
SOLUTION
FBD Frame:
ΣM D = 0: (13 in.)( 40 lb ) − (10 in.) Ax = 0
FBD ABF:
A x = 52 lb
ΣM B = 0: ( 4 in.) Fx − ( 6 in.)( 52 lb ) = 0
Fx = 78 lb
on ABF
Fx = 78.0 lb
on CFE from above
W
ΣM c = 0: ( 9 in.)( 40 lb ) − ( 4 in.) Fy − ( 4 in.)( 78 lb ) = 0
FBD CFE:
Fy = 12.00 lb
ΣFx = 0: C x − Fx = 0
W
C x = 78 lb
C x = 78.0 lb
W
ΣFy = 0: − 40 lb + Fy + C y = 0
C y = 40 lb − 12 lb = 28 lb
C y = 28.0 lb
W
PROBLEM 6.164
A Mansard roof truss is loaded as shown. Determine the force in
members DF, DG, and EG.
SOLUTION
FBD Truss:
ΣFx = 0: Ax = 0
By symmetry: Ay = Ly =
FBD Section:
5P
2
or
Ay = Ly = 3 kN
ΣM D = 0: ( 3 m ) FEG + ( 4 m )(1.2 kN ) − ( 6.25 m )( 3 kN ) = 0
FEG = 4.65 kN T W
ΣFy = 0: 3 kN − 2 (1.2 kN ) −
3
FDG = 0
5
FDG = 1.000 kN T W
ΣFx = 0: FEG +
FDF = 4.65 kN +
4
FDG − FDF = 0
5
4
(1 kN ) = 5.45 kN
5
FDF = 5.45 kN C W
PROBLEM 6.165
A Mansard roof truss is loaded as shown. Determine the force in
members GI, HI, and HJ.
SOLUTION
FBD Truss:
ΣFx = 0: Ax = 0
By symmetry: Ay = Ly =
FBD Section:
5P
2
or
Ay = Ly = 3 kN
ΣM I = 0: ( 6.25 m )( 3 kN ) − ( 4 m )(1.2 kN ) − ( 3 m ) FHJ = 0
FHJ = 4.65 kN C W
ΣFx = 0: FHJ − FGI = 0
FGI = FHJ
FGI = 4.65 kN T W
ΣFy = 0: − FHI − 1.2 kN + 3 kN = 0
FHI = 1.800 kN C W
PROBLEM 6.166
Rod CD is fitted with a collar at D that can be moved along rod AB,
which is bent in the shape of a circular arc. For the position when
θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.
SOLUTION
FBD:
(a)
(
)
ΣM C = 0: (15 in.) 20 lb − By = 0
B y = 20 lb
ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0
FCD = 80.0 lb T W
(b)
ΣFx = 0: ( 80 lb ) cos 30° − Bx = 0
B x = 69.282 lb
so B = 72.1 lb
16.10° W
PROBLEM 6.167
A log weighing 800 lb is lifted by a pair of tongs as shown. Determine
the forces exerted at E and at F on tong DEF.
SOLUTION
FBD AB:
By symmetry: Ay = By = 400 lb
and
FBD DEF:
Note:
Ax = Bx =
D = −B
6
( 400 lb ) = 480 lb
5
so
Dx = 480 lb
Dy = 400 lb
ΣM F = (10.5 in.)( 400 lb ) + (15.5 in.)( 480 lb ) − (12 in.) Ex = 0
Ex = 970 lb
E = 970 lb
ΣFx = 0: − 480 lb + 970 lb − Fx = 0
ΣFy = 0: 400 lb − Fy = 0
W
Fx = 490 lb
Fy = 400 lb
F = 633 lb
39.2° W
PROBLEM 7.1
Determine the internal forces (axial force, shearing force, and bending
moment) at point J of the structure indicated:
Frame and loading of Prob. 6.77.
SOLUTION
ΣFx = 0: − F = 0
FBD JD:
F=0
ΣFy = 0: V − 20 lb − 20 lb = 0
V = 40.0 lb
ΣM J = 0: M − ( 2 in.)( 20 lb ) − ( 6 in.)( 20 lb ) = 0
M = 160.0 lb ⋅ in.
PROBLEM 7.2
Determine the internal forces (axial force, shearing force, and bending
moment) at point J of the structure indicated:
Frame and loading of Prob. 6.76.
SOLUTION
FBD AJ:
ΣFx = 0: 60 lb − V = 0
V = 60.0 lb
ΣFy = 0: − F = 0
F=0
ΣM J = 0: M − (1 in.)( 60 lb ) = 0
M = 60.0 lb ⋅ in.
PROBLEM 7.3
For the frame and loading of Prob. 6.80, determine the internal forces
at a point J located halfway between points A and B.
SOLUTION
FBD Frame:
ΣFy = 0: Ay − 80 kN = 0
A y = 80 kN
ΣM E = 0: (1.2 m ) Ax − (1.5 m )( 80 kN ) = 0
A x = 100 kN
 0.3 m 
θ = tan −1 
 = 21.801°
 0.75 m 
FBD AJ:
ΣFx′ = 0: F − ( 80 kN ) sin 21.801° − (100 kN ) cos 21.801° = 0
F = 122.6 kN
ΣFy′ = 0: V + ( 80 kN ) cos 21.801° − (100 kN ) sin 21.801° = 0
V = 37.1 kN
ΣM J = 0: M + (.3 m )(100 kN ) − (.75 m )( 80 kN ) = 0
M = 30.0 kN ⋅ m
PROBLEM 7.4
For the frame and loading of Prob. 6.101, determine the internal forces
at a point J located halfway between points A and B.
SOLUTION
FBD Frame:
ΣFy = 0: Ay − 100 N = 0
A y = 100 N
ΣM F = 0:  2 ( 0.32 m ) cos 30°  Ax − ( 0.48 m )(100 N ) = 0
A x = 86.603 N
FBD AJ:
ΣFx′ = 0: F − (100 N ) cos 30° − ( 86.603 N ) sin 30° = 0
F = 129.9 N
ΣFy′ = 0: V + (100 N ) sin 30° − ( 86.603 N ) cos 30° = 0
V = 25.0 N
ΣM J = 0: ( 0.16 m ) cos 30° ( 86.603 N )
− ( 0.16 m ) sin 30° (100 N ) − M = 0
M = 4.00 N ⋅ m
PROBLEM 7.5
Determine the internal forces at point J of the structure shown.
SOLUTION
FBD Frame:
AB is two-force member, so
Ay
Ax
=
0.36 m 0.15 m
Ay =
5
Ax
12
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N
Ay =
5
Ax = 260 N or A y = 260 N
12
ΣFx = 0: F − 624 N = 0
F = 624 N
FBD AJ:
ΣFy = 0: 260 N − V = 0
V = 260 N
ΣM J = 0: M − ( 0.2 m )( 260 N ) = 0
M = 52.0 N ⋅ m
PROBLEM 7.6
Determine the internal forces at point K of the structure shown.
SOLUTION
FBD Frame:
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N
AB is two-force member, so
Ay
Ax
5
=
→ Ay =
Ax
0.36 m 0.15 m
12
ΣFx = 0: − Ax + C x = 0
A y = 260 N
C x = A x = 624 N
ΣFy = 0: Ay + C y − 390 N = 0
C y = 390 N − 260 N = 130 N or C y = 130 N
FBD CK:
ΣFx′ = 0: F +
12
5
( 624 N ) + (130 N ) = 0
13
13
F = −626 N
ΣFy′ = 0:
F = 626 N
12
5
(130 N ) − ( 624 N ) − V = 0
13
13
V = −120 N
V = 120.0 N
ΣM K = 0: ( 0.1 m )( 624 N ) − ( 0.24 m )(130 N ) − M = 0
M = 31.2 N ⋅ m
PROBLEM 7.7
A semicircular rod is loaded as shown. Determine the internal forces
at point J.
SOLUTION
FBD Rod:
ΣM B = 0: Ax ( 2r ) = 0
Ax = 0
ΣFx′ = 0: V − ( 30 lb ) cos 60° = 0
V = 15.00 lb
FBD AJ:
ΣFy′ = 0: F + ( 30 lb ) sin 60° = 0
F = −25.98 lb
F = 26.0 lb
ΣM J = 0: M − [ (9 in.) sin 60°] ( 30 lb ) = 0
M = −233.8 lb ⋅ in.
M = 234 lb ⋅ in.
PROBLEM 7.8
A semicircular rod is loaded as shown. Determine the internal forces
at point K.
SOLUTION
FBD Rod:
ΣFy = 0: By − 30 lb = 0
ΣM A = 0: 2rBx = 0
B y = 30 lb
Bx = 0
ΣFx′ = 0: V − ( 30 lb ) cos 30° = 0
FBD BK:
V = 25.98 lb
V = 26.0 lb
ΣFy′ = 0: F + ( 30 lb ) sin 30° = 0
F = −15 lb
F = 15.00 lb
ΣM K = 0: M − ( 9 in.) sin 30°  ( 30 lb ) = 0
M = 135.0 lb ⋅ in.
PROBLEM 7.9
An archer aiming at a target is pulling with a 210-N force on the
bowstring. Assuming that the shape of the bow can be approximated
by a parabola, determine the internal forces at point J.
SOLUTION
FBD Point A:
By symmetry T1 = T2
3 
ΣFx = 0: 2  T1  − 210 N = 0
5 
T1 = T2 = 175 N
Curve CJB is parabolic: y = ax 2
FBD BJ:
At B :
x = 0.64 m,
y = 0.16 m
a=
0.16 m
( 0.64 m )
2
=
1
2.56 m
1
( 0.32 m )2 = 0.04 m
2.56 m
So, at J : yJ =
Slope of parabola = tan θ =
dy
= 2ax
dx
 2
At J : θ J = tan −1 
( 0.32 m ) = 14.036°
 2.56 m

So
α = tan −1
4
− 14.036° = 39.094°
3
ΣFx′ = 0: V − (175 N ) cos ( 39.094° ) = 0
V = 135.8 N
ΣFy′ = 0: F + (175 N ) sin ( 39.094° ) = 0
F = −110.35 N
F = 110.4 N
PROBLEM 7.9 CONTINUED
3

Σ M J = 0 : M + ( 0.32 m )  (175 N ) 
5

4

+ ( 0.16 − 0.04 ) m   (175 N )  = 0
5


M = 50.4 N ⋅ m
PROBLEM 7.10
For the bow of Prob. 7.9, determine the magnitude and location of
the maximum (a) axial force, (b) shearing force, (c) bending moment.
SOLUTION
By symmetry
T1 = T2 = T
3
ΣFx = 0: 2T1   − 210 N = 0
5
FBD Point A:
4
(175 N ) = 0
5
FC = 140 N
3
(175 N ) − VC = 0
5
VC = 105 N
ΣFy = 0: FC −
ΣFx = 0:
FBD BC:
T1 = 175 N
3

4

ΣM C = 0: M C − ( 0.64 m )  (175 N )  − ( 0.16 m )  (175 N )  = 0
5

5

M C = 89.6 N ⋅ m
Also: if y = ax 2 and, at B, y = 0.16 m, x = 0.64 m
0.16 m
Then
a=
And
θ = tan −1
( 0.64 m )
2
=
1
;
2.56 m
FBD CK:
dy
= tan −1 2ax
dx
ΣFx′ = 0: (140 N ) cos θ − (105 N ) sin θ + F = 0
So
F = (105 N ) sin θ − (140 N ) cos θ
dF
= (105 N ) cos θ + (140 N ) sin θ
dθ
ΣFy′ = 0: V − (105 N ) cos θ − (140 N ) sin θ = 0
So
V = (105 N ) cos θ + (140 N ) sin θ
PROBLEM 7.10 CONTINUED
And
dV
= − (105 N ) sin θ + (140 N ) cos θ
dθ
ΣM K = 0: M + x (105 N ) + y (140 N ) − 89.6 N ⋅ m = 0
M = − (105 N ) x −
(140 N ) x 2
( 2.56 m )
+ 89.6 N ⋅ m
dM
= − (105 N ) − (109.4 N/m ) x + 89.6 N ⋅ m
dx
Since none of the functions, F, V, or M has a vanishing derivative in
the valid range of 0 ≤ x ≤ 0.64 m ( 0 ≤ θ ≤ 26.6° ) , the maxima are at
the limits ( x = 0, or x = 0.64 m ) .
Therefore,
(a)
Fmax = 140.0 N
at C
(b)
Vmax = 156.5 N
at B
(c)
M max = 89.6 N ⋅ m
at C
PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal forces at
point J knowing that θ = 30o.
SOLUTION
FBD AB:
4 
3 
ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0
5 
5 
C = 100 lb
ΣFx = 0: − Ax +
4
(100 lb ) = 0
5
A x = 80 lb
ΣFy = 0: Ay +
3
(100 lb ) − 70 lb = 0
5
A y = 10 lb
FBD AJ:
ΣFx′ = 0: F − ( 80 lb ) sin 30° − (10 lb ) cos 30° = 0
F = 48.66 lb
F = 48.7 lb
60°
ΣFy′ = 0: V − ( 80 lb ) cos 30° + (10 lb ) sin 30° = 0
V = 64.28 lb
V = 64.3 lb
30°
ΣM 0 = 0: ( 8 in.)( 48.66 lb ) − ( 8 in.)(10 lb ) − M = 0
M = 309.28 lb ⋅ in.
M = 309 lb ⋅ in.
PROBLEM 7.12
A semicircular rod is loaded as shown. Determine the magnitude and
location of the maximum bending moment in the rod.
SOLUTION
4 
3 
ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0
5 
5 
FBD AB:
C = 100 lb
ΣFx = 0: − Ax +
4
(100 lb ) = 0
5
A x = 80 lb
ΣFy = 0: Ay +
3
(100 lb ) − 70 lb = 0
5
A y = 10 lb
ΣM J = 0: M − ( 8 in.)(1 − cosθ )(10 lb ) − ( 8 in.)( sin θ )( 80 lb ) = 0
FBD AJ:
M = ( 640 lb ⋅ in.) sin θ + ( 80 lb ⋅ in.) ( cos θ − 1)
dM
= ( 640 lb ⋅ in.) cos θ − ( 80 lb ⋅ in.) sin θ = 0
dθ
θ = tan −1 8 = 82.87° ,
for
where
d 2M
= − ( 640 lb ⋅ in.) sin θ − ( 80 lb ⋅ in.) cos θ < 0
dθ 2
M = 565 lb ⋅ in. at θ = 82.9° is a max for AC
So
FBD BK:
ΣM K = 0: M − ( 8 in.)(1 − cos β )( 70 lb ) = 0
M = ( 560 lb ⋅ in.)(1 − cos β )
dM
= ( 560 lb ⋅ in.) sin β = 0
dβ
So, for β =
π
2
for β = 0, where M = 0
, M = 560 lb ⋅ in. is max for BC
∴ M max = 565 lb ⋅ in. at θ = 82.9°
PROBLEM 7.13
Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D.
Determine the internal forces at point J.
SOLUTION
FBD Frame:
 24 
ΣM A = 0: ( 0.336 m )  C  − ( 0.252 m )( 480 N ) = 0
 25 
C = 375 N
ΣFy = 0: Ax −
24
C =0
25
Ax =
24
( 375 N ) = 360 N
25
A x = 360 N
ΣFy = 0: Ay − 480 N +
FBD CD:
7
( 375 N ) = 0
24
A y = 375 N
ΣM C = 0: ( 0.324 m )( 480 N ) − ( 0.27 m ) B = 0
B = 576 N
ΣFx = 0: C x −
24
( 375 N ) = 0
25
C x = 360 N
ΣFy = 0: −480 N +
7
( 375 N ) + ( 576 N ) − C y = 0
25
C y = 201 N
FBD CJ:
ΣFx′ = 0: V − ( 360 N ) cos 30° − ( 201 N ) sin 30° = 0
V = 412 N
ΣFy′ = 0: F + ( 360 N ) sin 30° − ( 201 N ) cos 30° = 0
F = −5.93 N
F = 5.93 N
ΣM 0 = 0: ( 0.168 m )( 201 N + 5.93 N ) − M = 0
M = 34.76 N ⋅ m
M = 34.8 N ⋅ m
PROBLEM 7.14
Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D.
Determine the internal forces at point K.
SOLUTION
FBD CD:
ΣFx = 0: C x = 0
ΣM B = 0: ( 0.054 m )( 480 N ) − ( 0.27 m ) C y = 0
C y = 96 N
ΣFy = 0: B − C y = 0
FBD CK:
B = 96 N
ΣFy′ = 0: V − ( 96 N ) cos 30° = 0
V = 83.1 N
ΣFx′ = 0: F − ( 96 N ) sin 30° = 0
F = 48.0 N
ΣM K = 0: M − ( 0.186 m )( 96 N ) = 0
M = 17.86 N ⋅ m
PROBLEM 7.15
Knowing that the radius of each pulley is 7.2 in. and neglecting friction,
determine the internal forces at point J of the frame shown.
SOLUTION
FBD Frame:
Note: Tension T in cord is 90 lb at any cut. All radii = 0.6 ft
ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B x = 140 lb
ΣM E = 0: ( 5.4 ft )(140 lb ) − ( 7.2 ft ) By
+ ( 4.8 ft ) 90 lb − ( 0.6 ft ) 90 lb = 0
FBD BCE with pulleys and
cord:
B y = 157.5 lb
ΣFx = 0: Ex − 140 lb = 0
E x = 140 lb
ΣFy = 0: 157.5 lb − 90 lb − 90 lb + E y = 0
E y = 22.5 lb
FBD EJ:
ΣFx′ = 0: −V +
3
4
(140 lb ) + ( 22.5 lb − 90 lb ) = 0
5
5
V = 30 lb
ΣFy′ = 0: F + 90 lb −
V = 30.0 lb
4
3
(140 lb ) − ( 90 lb − 22.5 lb ) = 0
5
5
F = 62.5lb
F = 62.5 lb
ΣM J = 0: M + (1.8 ft )(140 lb ) + ( 0.6 ft )( 90 lb )
+ ( 2.4 ft )( 22.5 lb ) − ( 3.0 ft )( 90 lb ) = 0
M = − 90 lb ⋅ ft
M = 90.0 lb ⋅ ft
PROBLEM 7.16
Knowing that the radius of each pulley is 7.2 in. and neglecting
friction, determine the internal forces at point K of the frame shown.
SOLUTION
FBD Whole:
Note: T = 90 lb
ΣM B = 0: ( 5.4 ft ) Ax − ( 6 ft )( 90 lb ) − ( 7.8 ft )( 90 lb ) = 0
A x = 2.30 lb
FBD AE:
Note: Cord tensions moved to point D as per Problem 6.91
ΣFx = 0: 230 lb − 90 lb − Ex = 0
E x = 140 lb
ΣM A = 0: (1.8 ft )( 90 lb ) − ( 7.2 ft ) E y = 0
E y = 22.5 lb
FBD KE:
ΣFx = 0: F − 140 lb = 0
F = 140.0 lb
ΣFy = 0: V − 22.5 lb = 0
V = 22.5 lb
ΣM K = 0: M − ( 2.4 ft )( 22.5 lb ) = 0
M = 54.0 lb ⋅ ft
PROBLEM 7.17
Knowing that the radius of each pulley is 7.2 in. and neglecting friction,
determine the internal forces at point J of the frame shown.
SOLUTION
FBD Whole:
ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) = 0
B x = 130 lb
FBD BE with pulleys and cord:
ΣM E = 0: ( 5.4 ft )(130 lb ) − ( 7.2 ft ) By
+ ( 4.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B y = 150 lb
ΣFx = 0: Ex − 130 lb = 0
E x = 130 lb
ΣFy = 0: E y + 150 lb − 90 lb − 90 lb = 0
E y = 30 lb
FBD JE and pulley:
ΣFx′ = 0: − F − 90 lb +
4
3
(130 lb ) + ( 90 lb − 30 lb ) = 0
5
5
F = 50.0 lb
ΣFy′ = 0: V +
3
4
(130 lb ) + ( 30 lb − 90 lb ) = 0
5
5
V = −30 lb
V = 30.0 lb
ΣM J = 0: − M + (1.8 ft )(130 lb ) + ( 2.4 ft )( 30 lb ) + ( 0.6 ft )( 90 lb )
− ( 3.0 ft )( 90 lb ) = 0
M = 90.0 lb ⋅ ft
PROBLEM 7.18
Knowing that the radius of each pulley is 7.2 in. and neglecting
friction, determine the internal forces at point K of the frame shown.
SOLUTION
FBD Whole:
ΣM B = 0: ( 5.4 ft ) Ax − ( 7.8 ft )( 90 lb ) = 0
A x = 130 lb
FBD AE:
ΣM E = 0: − ( 7.2 ft ) Ay − ( 4.8 ft )( 90 lb ) = 0
Ay = −60 lb
A y = 60 lb
ΣFx = 0:
FBD AK:
F=0
ΣFy = 0: −60 lb + 90 lb − V = 0
V = 30.0 lb
ΣM K = 0: ( 4.8 ft )( 60 lb ) − ( 2.4 ft )( 90 lb ) − M = 0
M = 72.0 lb ⋅ ft
PROBLEM 7.19
A 140-mm-diameter pipe is supported every 3 m by a small frame
consisting of two members as shown. Knowing that the combined
mass per unit length of the pipe and its contents is 28 kg/m and
neglecting the effect of friction, determine the internal forces at
point J.
SOLUTION
FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N
ΣM A = ( 0.6 m ) Cx − ( 0.315 m )( 824.04 N ) = 0
C x = 432.62 N
FBD pipe:
By symmetry: N1 = N 2
ΣFy = 0: 2
N1 =
21
N1 − W = 0
29
29
(824.04 N )
42
= 568.98 N
Also note:
 20 
a = r tan θ = 70 mm  
 21 
a = 66.67 mm
FBD BC:
ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) C y
+ ( 0.06667 m )( 568.98 N ) = 0
C y = 532.42 N
PROBLEM 7.19 CONTINUED
FBD CJ:
ΣFx′ = 0: F −
21
20
( 432.62 N ) − ( 532.42 N ) = 0
29
29
F = 680 N
ΣFy′ = 0:
21
20
( 532.42 N ) − ( 432.62 N ) − V = 0
29
29
V = 87.2 N
ΣM J = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 532.42 N ) + M = 0
M = 18.96 N ⋅ m
PROBLEM 7.20
A 140-mm-diameter pipe is supported every 3 m by a small frame
consisting of two members as shown. Knowing that the combined mass
per unit length of the pipe and its contents is 28 kg/m and neglecting the
effect of friction, determine the internal forces at point K.
SOLUTION
FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N
ΣM C = 0: (.6 m ) Ax − (.315 m )( 824.04 N ) = 0
A x = 432.62 N
FBD pipe
By symmetry: N1 = N 2
ΣFy = 0: 2
N2 =
21
N1 − W = 0
29
29
824.04 N
42
= 568.98 N
Also note:
FBD AD:
a = r tan θ = ( 70 mm )
20
21
a = 66.67 mm
ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) Ay
− ( 0.06667 m )( 568.98 N ) = 0
A y = 291.6 N
PROBLEM 7.20 CONTINUED
FBD AK:
ΣFx′ = 0:
21
20
( 432.62 N ) + ( 291.6 N ) − F = 0
29
29
F = 514 N
ΣFy′ = 0:
21
20
( 291.6 N ) − ( 432.62 N ) + V = 0
29
29
V = 87.2 N
ΣM K = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 291.6 N ) − M = 0
M = 18.97 N ⋅ m
PROBLEM 7.21
A force P is applied to a bent rod which is supported by a roller and
a pin and bracket. For each of the three cases shown, determine the
internal forces at point J.
SOLUTION
(a) FBD Rod:
ΣM D = 0: aP − 2aA = 0
A=
P
2
ΣFx = 0: V −
P
=0
2
V =
FBD AJ:
ΣFy = 0:
F=0
ΣM J = 0: M − a
P
=0
2
M =
(b) FBD Rod:
ΣM D = 0: aP −
A=
5P
2
P
2
a4 
 A = 0
25 
aP
2
PROBLEM 7.21 CONTINUED
FBD AJ:
ΣFx = 0:
3 5P
−V = 0
5 2
V =
ΣFy = 0:
3P
2
4 5P
−F =0
5 2
F = 2P
M =
(c) FBD Rod:
3
aP
2
3 
4 
ΣM D = 0: aP − 2a  A  − 2a  A  = 0
5 
5 
A=
5P
14
 3 5P 
ΣFx = 0: V − 
=0
 5 14 
V =
ΣFy = 0:
3P
14
4 5P
−F =0
5 14
F=
2P
7
 3 5P 
ΣM J = 0: M − a 
=0
 5 14 
M =
3
aP
14
PROBLEM 7.22
A force P is applied to a bent rod which is supported by a roller and
a pin and bracket. For each of the three cases shown, determine the
internal forces at point J.
SOLUTION
(a) FBD Rod:
ΣFx = 0: Ax = 0
ΣM D = 0: aP − 2aAy = 0
Ay =
P
2
ΣFx = 0: V = 0
FBD AJ:
ΣFy = 0:
P
−F =0
2
F=
P
2
ΣM J = 0: M = 0
(b) FBD Rod:
ΣM A = 0
4 
3 
2a  D  + 2a  D  − aP = 0
5 
5 
D=
5P
14
ΣFx = 0: Ax −
4 5
P=0
5 14
Ax =
2P
7
ΣFy = 0: Ay − P +
3 5
P=0
5 14
Ay =
11P
14
PROBLEM 7.22 CONTINUED
FBD AJ:
ΣFx = 0:
2
P −V = 0
7
V =
2P
7
11P
−F =0
14
ΣFy = 0:
F=
ΣM J = 0: a
2P
−M =0
7
M =
(c) FBD Rod:
ΣM A = 0:
a  4D 

 − aP = 0
2 5 
4 5P
=0
5 2
ΣFx = 0: Ax −
ΣFy = 0: Ay − P −
FBD AJ:
11P
14
3 5P
=0
5 2
D=
2
aP
7
5P
2
Ax = 2P
Ay =
5P
2
ΣFx = 0: 2 P − V = 0
V = 2P
ΣFy = 0:
5P
−F =0
2
F=
5P
2
ΣM J = 0: a ( 2 P ) − M = 0
M = 2aP
PROBLEM 7.23
A rod of weight W and uniform cross section is bent into the circular
arc of radius r shown. Determine the bending moment at point J
when θ = 30°.
SOLUTION
FBD CJ:
Note α =
180° − 60°
π
= 60° =
2
3
r =
r
α
sin α =
3r
3
3 3
r
=
π 2
2π
Weight of section = W
ΣFy′ = 0: F −
120
4
= W
270 9
4
W cos 30° = 0
9
ΣM 0 = 0: rF − ( r sin 60° )
2 3 3 3 3
−
M = r
2π 2
 9
F =
2 3
W
9
4W
−M =0
9
2 3 1 
4
−  Wr
W = 
π 
9 
 9
M = 0.0666Wr
PROBLEM 7.24
A rod of weight W and uniform cross section is bent into the circular
arc of radius r shown. Determine the bending moment at point J
when θ = 120°.
SOLUTION
ΣFx = 0: Ax = 0
(a) FBD Rod:
ΣM B = 0: rAy +
2r W 2r 2W
−
=0
π 3
π 3
Ay =
2W
3π
FBD AJ:
α =
Note:
60°
π
= 30° =
2
6
Weight of segment = W
F =
r
α
sin α =
ΣM J = 0: ( r cos α − r sin 30° )
M =
2W
9
60
2W
=
270
9
r
3r
sin 30° =
π /6
π
2W
2W
+ ( r − r sin 30° )
−M =0
9
3π
 3r 3 r
 3 1
3r 
1 
− +
− +

 = Wr 

2 2π 
9 3π 
π 2
 3π
M = 0.1788Wr
PROBLEM 7.25
A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J when
θ = 30o.
SOLUTION
FBD Rod:
ΣFx = 0: A x = 0
ΣM B = 0:
2r
π
W − rAy = 0
α = 15°, weight of segment = W
FBD AJ:
r =
r
α
sin α =
ΣFy′ = 0:
2W
π
F=
Ay =
2W
π
30° W
=
90°
3
r
sin15° = 0.9886r
π /12
cos 30° −
W
cos 30° − F = 0
3
W 3  2 1
 − 
2 π 3
2W 
W

ΣM 0 = M + r  F −
=0
 + r cos15°
π
3


M = 0.0557 Wr
PROBLEM 7.26
A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J when
θ = 30o.
SOLUTION
FBD Rod:
ΣM A = 0: rB −
B=
r =
r
π /12
π
π
π
12
sin15° = 0.98862r
Weight of segment = W
ΣFy′ = 0: F −
W =0
2W
α = 15° =
FBD BJ:
2r
30° W
=
90°
3
W
2W
cos 30° −
sin 30° = 0
π
3
 3 1
+ W
F=
 6
π 

ΣM 0 = 0: rF − ( r cos15° )
W
−M =0
3
 3 1 
cos15° 
M = rW 
+  −  0.98862
 Wr
 6

π 
3 

M = 0.289Wr
PROBLEM 7.27
For the rod of Prob.7.26, determine the magnitude and location of the
maximum bending moment.
SOLUTION
FBD Bar:
ΣM A = 0: rB −
α =
2r
π
θ
W =0
r =
r
α
=
F =
2W
=
2W
π
π
4α
π
( sin 2α
M =
or
W
2W
π
sin 2α = 0
+ θ cos θ )
4α
π
W −M =0
r
 4α
Wr ( sin θ + θ cos θ ) −  sin α cos α 
W
π
α
 π
2
But,
so
π
+ 2α cos 2α )
ΣM 0 = 0: rF − ( r cos α )
FBD BJ:
4
2α
π /2
4α
W cos 2α −
( sin θ
π
sin α ,
Weight of segment = W
ΣFx′ = 0: F −
2W
π
0≤α ≤
so
2
B=
sin α cos α =
M =
2Wr
π
1
1
sin 2α = sin θ
2
2
( sin θ
M =
+ θ cosθ − sin θ )
2
π
Wrθ cosθ
dM
2
= Wr ( cos θ − θ sin θ ) = 0 at θ tan θ = 1
dθ
π
PROBLEM 7.27 CONTINUED
Solving numerically
θ = 0.8603 rad
and
M = 0.357Wr
at θ = 49.3°
(Since M = 0 at both limits, this is the maximum)
PROBLEM 7.28
For the rod of Prob.7.25, determine the magnitude and location of the
maximum bending moment.
SOLUTION
ΣFx = 0: Ax = 0
FBD Rod:
2r
ΣM B = 0:
W − rAy = 0
π
α =
θ
2
r =
,
r
α
F =
2W
π
4α
π
2W

ΣM 0 = 0: M +  F −
π

FBD AJ:
M =
2W
π
(1 + θ cosθ
M =
so
2r
π
2W
π
2W
π
cos 2α = 0
(1 − θ ) cosθ
4α

W =0
 r + ( r cos α )
π

− cosθ ) r −
sin α cos α =
But,
=
π
2α
4α
W
=
π /2
π
W cos 2α +
(1 − 2α ) cos 2α
2W
sin α
Weight of segment = W
ΣFx′ = 0: − F −
Ay =
4αW r
π
α
sin α cos α
1
1
sin 2α = sin θ
2
2
W (1 − cosθ + θ cosθ − sin θ )
dM
2rW
=
( sin θ − θ sin θ + cosθ − cosθ ) = 0
dθ
π
(1 − θ ) sin θ
for
dM
=0
dθ
for
=0
θ = 0, 1, nπ ( n = 1, 2,")
Only 0 and 1 in valid range
At
θ = 0 M = 0,
at θ = 57.3°
at θ = 1 rad
M = M max = 0.1009 Wr
PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
FBD beam:
(a) By symmetry: Ay = D =
1
L
( w)
2
2
Ay = D =
wL
4
Along AB:
ΣFy = 0:
ΣM J = 0: M − x
wL
−V = 0
4
wL
=0
4
M =
V =
wL
x (straight)
4
Along BC:
ΣFy = 0:
wL
− wx1 − V = 0
4
V =
ΣM k = 0: M +
M =
wL
− wx1
4
V =0
straight with
wL
4
at
x1 =
L
4
x1
L
 wL
wx1 −  + x1 
=0
2
4
 4

w  L2 L
+ x1 − x12 

2 8
2

PROBLEM 7.29 CONTINUED
Parabola with
M =
3
L
wL2 at x1 =
32
4
Section CD by symmetry
(b) From diagrams:
V
max
M
=
max
wL
on AB and CD
4
=
3wL2
at center
32
PROBLEM 7.30
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) Along AB:
ΣFy = 0: −wx − V = 0
straight with
V =−
ΣM J = 0: M +
wL
2
L
2
x=
at
x
wx = 0
2
M =−
parabola with
V = − wx
1
M = − wx 2
2
wL2
L
at x =
8
2
Along BC:
ΣFy = 0: − w
L
−V = 0
2
1
V = − wL
2
L L

ΣM k = 0: M +  x1 +  w = 0
4 2

M =−
straight with
(b) From diagrams:
wL  L

 + x1 
2 4

3
L
M = − wL2 at x1 =
8
2
V
M
max
max
=
wL
on BC
2
=
3wL2
at C
8
PROBLEM 7.31
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) Along AB:
ΣFy = 0: P − V = 0
V = P
ΣM J = 0: M − Px = 0
straight with M =
M = Px
PL
at B
2
Along BC:
ΣFy = 0: P − P − V = 0
V =0
L

ΣM K = 0: M + Px1 − P  + x1  = 0
2

M =
PL
2
(constant)
(b) From diagrams:
V
M
max
max
=
= P along AB
PL
along BC
2
PROBLEM 7.32
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM C = 0: LAy − M 0 = 0
(a) FBD Beam:
Ay =
M0
L
ΣFy = 0: − Ay + C = 0
M0
L
C=
Along AB:
ΣFy = 0: −
M0
−V = 0
L
ΣM J = 0: x
V =−
M0
+M =0
L
M0
L
M =−
straight with M = −
M0
x
L
M0
at B
2
Along BC:
ΣFy = 0: −
ΣM K = 0: M + x
M0
−V = 0
L
M0
− M0 = 0
L
straight with M =
(b) From diagrams:
V =−
M0
L
x

M = M 0 1 − 
L


M0
at B
2
V
max
M = 0 at C
= P everywhere
M
max
=
M0
at B
2
PROBLEM 7.33
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM B = 0:
(.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay
A y = 12.625 kips
ΣFy = 0: 12.625 kips − 10 kips − 8 kips − 4 kips + B = 0
B = 9.375 kips
Along AC:
ΣFy = 0: 12.625 kips − V = 0
V = 12.625 kips
ΣM J = 0: M − x (12.625 kips ) = 0
M = (12.625 kips ) x
M = 22.725 kip ⋅ ft at C
Along CD:
ΣFy = 0: 12.625 kips − 10 kips − V = 0
V = 2.625 kips
ΣM K = 0: M + ( x − 1.8 ft )(10 kips ) − x (12.625 kips ) = 0
M = 18 kip ⋅ ft + ( 2.625 kips ) x
M = 29.8125 kip ⋅ ft at D ( x = 4.5 ft )
=0
PROBLEM 7.33 CONTINUED
Along DE:
Along DE:
ΣFy = 0: (12.625 − 10 − 8 ) kips − V = 0
V = −5.375 kips
ΣM L = 0: M + x1 ( 8 kips ) + ( 2.7 ft + x1 )(10 kips )
− ( 4.5 ft + x1 )(12.625 kips ) = 0
M = 29.8125 kip ⋅ ft − ( 5.375 kips ) x1
M = 5.625 kip ⋅ ft at E
Along EB:
( x1 = 4.5 ft )
Along EB:
ΣFy = 0: V + 9.375 kips = 0
V = 9.375 kips
ΣM N = 0: x2 ( 9.375 kip ) − M = 0
M = ( 9.375 kips ) x2
M = 5.625 kip ⋅ ft at E
(b) From diagrams:
V
max
M
= 12.63 kips on AC
max
= 29.8 kip ⋅ ft at D
PROBLEM 7.34
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM C = 0:
(1.2 m )( 4 kN ) − (1 m )(16 kN ) + ( 2 m )(8 kN ) + ( 3.2 m ) B = 0
B = −1.5 kN
ΣFy = 0: −4 kN + C y − 16 kN + 8 kN − 1.5 kN = 0
C y = 13.5 kN
Along AC:
ΣFy = 0: −4 kN − V = 0
V = −4 kN
ΣM J = 0: M + x ( 4 kN ) = 0
M = −4 kN x
M = −4.8 kN ⋅ m at C
Along CD:
ΣFy = 0: −4 kN + 13.5 kN − V = 0
V = 9.5 kN
ΣM K = 0: M + (1.2 m + x1 )( 4 kN ) − x1 (13.5 kN ) = 0
M = −4.8 kN ⋅ m + ( 9.5 kN ) x1
M = 4.7 kN ⋅ m at D ( x1 = 1 m )
PROBLEM 7.34 CONTINUED
Along DE:
ΣFy = 0: V + 8 kN − 1.5 kN = 0
V = −6.5 kN
ΣM L = 0: M − x3 ( 8 kN ) + ( x3 + 1.2 m ) (1.5 kN ) = 0
M = −1.8 kN ⋅ m + ( 6.5 kN ) x3
M = 4.7 kN ⋅ m at D ( x3 = 1 m )
Along EB:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM N = 0: x2 (1.5 kN ) + M = 0
M = − (1.5 kN ) x2
(b) From diagrams:
M = −1.8 kN ⋅ m at E
V
M
max
max
= 9.50 kN ⋅ on CD
= 4.80 kN ⋅ m at C
PROBLEM 7.35
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) Along AC:
ΣFy = 0: −3 kip − V = 0
V = −3 kips
ΣM J = 0: M + x ( 3 kips ) = 0
M = ( 3 kips ) x
M = −9 kip ⋅ ft at C
Along CD:
ΣFy = 0: −3 kips − 5 kips − V = 0
V = −8 kips
ΣM K = 0: M + ( x − 3 ft )( 5 kips ) + x ( 3 kips ) = 0
M = +15 kip ⋅ ft − ( 8 kips ) x
M = −16.2 kip ⋅ ft at D ( x = 3.9 ft )
Along DE:
ΣFy = 0: −3 kips − 5 kips + 6 kips − V = 0
V = −2 kips
ΣM L = 0: M − x1 ( 6 kips ) + (.9 ft + x1 )( 5 kips )
+ ( 3.9 ft + x1 )( 3 kips ) = 0
M = −16.2 kip ⋅ ft − ( 2 kips ) x1
M = −18.6 kip ⋅ ft at E
( x1 = 1.2 ft )
PROBLEM 7.35 CONTINUED
Along EB:
ΣFy = 0: −3 kips − 5 kips + 6 kips − 4 kips − V = 0
V = −6 kips
ΣM N = 0: M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips )
+ ( 5.1 ft + x2 )( 3 kips ) − (1.2 ft + x2 )( 6 kips ) = 0
M = −18.6 kip ⋅ ft − ( 6 kips ) x2
M = −33 kip ⋅ ft at B
(b) From diagrams:
( x2
= 2.4 ft )
V
max
M
max
= 8.00 kips on CD
= 33.0 kip ⋅ ft at B
PROBLEM 7.36
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM E = 0:
(a) FBD Beam:
(1.1 m )( 540 N ) − ( 0.9 m ) C y + ( 0.4 m )(1350 N ) − ( 0.3 m )( 540 N ) = 0
C y = 1080 N
ΣFy = 0: −540 N + 1080 N − 1350 N
−540 N + E y = 0
E y = 1350 N
Along AC:
ΣFy = 0: −540 N − V = 0
V = −540 N
ΣM J = 0: x ( 540 N ) + M = 0
M = − ( 540 N ) x
Along CD:
ΣFy = 0: −540 N + 1080 N − V = 0
V = 540 N
ΣM K = 0: M + ( 0.2 m + x1 )( 540 N ) − x1 (1080 N ) = 0
M = −108 N ⋅ m + ( 540 N ) x1
M = 162 N ⋅ m at D ( x1 = 0.5 m )
PROBLEM 7.36 CONTINUED
Along DE:
ΣFy = 0: V + 1350 N − 540 N = 0
V = −810 N
ΣM N = 0: M + ( x3 + 0.3 m )( 540 N ) − x3 (1350 N ) = 0
M = −162 N ⋅ m + ( 810 N ) x3
M = 162 N ⋅ m at D ( x3 = 0.4 )
Along EB:
ΣFy = 0: V − 540 N = 0
V = 540 N
ΣM L = 0: M + x2 ( 540 N ) = 0
( x2
M = −162 N ⋅ m at E
(b) From diagrams
M = −540 N x2
= 0.3 m )
V
M
max
max
= 810 N on DE W
= 162.0 N ⋅ m at D and E W
PROBLEM 7.37
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0
A y = 2 kips
ΣM A = 0: M A + ( 3 ft )( 6 ft )( 2 kips/ft )
− (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0
M A = 114 kip ⋅ ft
Along AC:
ΣFy = 0: 2 kips + x ( 2 kips/ft ) − V = 0
V = 2 kips + ( 2 kips/ft ) x
V = 14 kips at C ( x = 6 ft )
ΣM J = 0: 114 kip ⋅ ft − x ( 2 kips )
−
x
x ( 2 kips/ft ) + M = 0
2
M = (1 kip/ft ) x 2 + ( 2 kips ) x − 114 kip ⋅ ft
M = −66 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: V − 12 kips − 2 kips = 0
V = 14 kips
ΣM k = 0: −M − x1 (12 kips ) − (1.5 ft + x1 )( 2 kips ) = 0
PROBLEM 7.37 CONTINUED
M = −3 kip ⋅ ft − (14 kips ) x1
M = −66 kip ⋅ ft at C
( x1 = 4.5 ft )
Along DB:
ΣFy = 0:
V − 2 kips = 0
V = + 2 kips
ΣM L = 0: −M − 2 kip x3 = 0
M = − ( 2 kips ) x3
M = −3 kip ⋅ ft at D ( x = 1.5 ft )
(b) From diagrams:
V
max
M
max
= 14.00 kips on CD W
= 114.0 kip ⋅ ft at A W
PROBLEM 7.38
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM A = (15 ft ) B − (12 ft )( 2 kips/ft )( 6 ft ) − ( 6 ft )(12 kips ) = 0
B = 14.4 kips
ΣFy = 0: Ay − 12 kips − ( 2 kips/ft )( 6 ft ) + 14.4 kips
A y = 9.6 kips
Along AC:
ΣFy = 0: 9.6 kips − V = 0
V = 9.6 kips
ΣM J = 0: M − x ( 9.6 kips ) = 0
M = ( 9.6 kips ) x
M = 57.6 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: 9.6 kips − 12 kips − V = 0
V = −2.4 kips
ΣM K = 0: M + x1 (12 kips ) − ( 6 ft + x1 )( 9.6 kips ) = 0
M = 57.6 kip ⋅ ft − ( 2.4 kips ) x1
M = 50.4 kip ⋅ ft at D
PROBLEM 7.38 CONTINUED
Along DB:
ΣFy = 0: V − x3 ( 2 kips/ft ) + 14.4 kips = 0
V = −14.4 kips + ( 2 kips/ft ) x3
V = −2.4 kips at D
ΣM L = 0: M +
x3
( 2 kips/ft )( x3 ) − x3 (14.4 kips ) = 0
2
M = (14.4 kips ) x3 − (1 kip/ft ) x32
M = 50.4 kip ⋅ ft at D ( x3 = 6 ft )
(b) From diagrams:
V
M
max
max
= 14.40 kips at B W
= 57.6 kip ⋅ ft at C W
PROBLEM 7.39
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) By symmetry:
Ay = B = 8 kN +
1
( 4 kN/m )( 5 m )
2
A y = B = 18 kN
Along AC:
ΣFy = 0: 18 kN − V = 0
V = 18 kN
ΣM J = 0: M − x (18 kN )
M = (18 kN ) x
M = 36 kN ⋅ m at C ( x = 2 m )
Along CD:
ΣFy = 0: 18 kN − 8 kN − ( 4 kN/m ) x1 − V = 0
V = 10 kN − ( 4 kN/m ) x1
V = 0 at x1 = 2.5 m ( at center )
ΣM K = 0: M +
x1
( 4 kN/m ) x1 + (8 kN ) x1 − ( 2 m + x1 )(18 kN ) = 0
2
M = 36 kN ⋅ m + (10 kN/m ) x1 − ( 2 kN/m ) x12
M = 48.5 kN ⋅ m at x1 = 2.5 m
Complete diagram by symmetry
(b) From diagrams:
V
max
M
= 18.00 kN on AC and DB W
max
= 48.5 kN ⋅ m at center W
PROBLEM 7.40
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
ΣM D = 0: ( 2 m )( 2 kN/m )( 2 m ) − ( 2.5 m )( 4 kN/m )( 3 m )
− ( 4 m ) F − ( 5 m )( 22 kN ) = 0
F = 22 kN
ΣFy = 0: − ( 2 m )( 2 kN/m ) + D y
− ( 3 m )( 4 kN/m ) − 22 kN + 22 kN = 0
D y = 16 kN
Along AC:
ΣFy = 0: − x ( 2 kN/m ) − V = 0
V = − ( 2 kN/m ) x
V = −4 kN at C
ΣM J = 0: M +
x
( 2 kN/m )( x ) ≠ 0
2
M = − (1 kN/m ) x 2
M = −4 kN ⋅ m at C
Along CD:
ΣFy = 0: − ( 2 m )( 2 kN/m ) − V = 0
V = −4 kN
ΣM K = 0: (1 m + x1 )( 2 kN/m )( 2 m ) = 0
M = −4 kN ⋅ m − ( 4 kN/m ) x1
M = −8 kN ⋅ m at D
PROBLEM 7.40 CONTINUED
Along DE:
ΣFy = 0: − ( 2 kN/m )( 2 m ) + 16 kN − V = 0
V = 12 kN
ΣM L = 0: M − x2 (16 kN ) + ( x2 + 2 m )( 2 kN/m )( 2 m ) = 0
M = −8 kN ⋅ m + (12 kN ) x2
M = 4 kN ⋅ m at E
Along EF:
ΣFy = 0: V − x4 ( 4 kN/m ) − 22 kN + 22 kN = 0
V = ( 4 kN/m ) x4
ΣM 0 = 0: M +
V = 12 kN at E
x4
( 4 kN/m ) x4 − (1 m )( 22 kN ) = 0
2
M = 22 kN ⋅ m − ( 2 kN/m ) x42
M = 4 kN ⋅ m at E
Along FB:
ΣFy = 0: V + 22 kN = 0
V = 22 kN
ΣM N = 0: M − x3 ( 22 kN ) = 0
M = ( 22 kN ) x3
M = 22 kN ⋅ m at F
(b) From diagrams:
V
M
max
max
= 22.0 kN on FB W
= 22.0 kN ⋅ m at F W
PROBLEM 7.41
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
SOLUTION
ΣFy = 0: (12 m ) w − ( 6 m )( 3 kN/m ) = 0
(a)
w = 1.5 kN/m
Along AC:
ΣFy = 0: x (1.5 kN/m ) − V = 0
V = (1.5 kN/m ) x
V = 4.5 kN at C
ΣM J = 0: M −
x
(1.5 kN/m )( x ) = 0
2
M = ( 0.75 kN/m ) x 2
M = 6.75 N ⋅ m at C
Along CD:
ΣFy = 0: x (1.5 kN/m ) − ( x − 3 m )( 3 kN/m ) − V = 0
V = 9 kN − (1.5 kN/m ) x
V = 0 at x = 6 m
x
 x − 3m
ΣM K = 0: M + 
 ( 3 kN/m )( x − 3 m ) − (1.5 kN/m ) x = 0
2
2


M = −13.5 kN ⋅ m + ( 9 kN ) x − ( 0.75 kN/m ) x 2
( x = 6 m)
M = 13.5 kN ⋅ m at center
Finish by symmetry
(b) From diagrams:
V
M
max
max
= 4.50 kN at C and D
= 13.50 kN ⋅ m at center
PROBLEM 7.42
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
SOLUTION
(a) FBD Beam:
ΣFy = 0: ( 4 m )( w ) − ( 2 m )(12 kN/m ) = 0
w = 6 kN/m
Along AC:
ΣFy = 0: − x ( 6 kN/m ) − V = 0
V = − ( 6 kN/m ) x
V = −6 kN at C ( x = 1 m )
ΣM J = 0: M +
x
( 6 kN/m )( x ) = 0
2
M = − ( 3 kN/m ) x 2
M = −3 kN ⋅ m at C
Along CD:
ΣFy = 0: − (1 m )( 6 kN/m ) + x1 ( 6 kN/m ) − v = 0
V = ( 6 kN/m )(1 m − x1 )
V = 0 at x1 = 1 m
ΣM K = 0: M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) −
x1
( 6 kN/m ) x1 = 0
2
M = −3 kN ⋅ m − ( 6 kN ) x1 + ( 3 kN/m ) x12
M = −6 kN ⋅ m at center
( x1 = 1 m )
Finish by symmetry
(b) From diagrams:
V
max
M
= 6.00 kN at C and D
max
= 6.00 kN at center
PROBLEM 7.43
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed and knowing that a = 0.9 ft, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0
w = 400 lb/ft
Along AC:
ΣFy = 0: x ( 400 lb ) − V = 0
V = 360 lb at C
ΣM J = 0: M −
V = ( 400 lb ) x
( x = 0.9 ft )
x
( 400 lb/ft ) x = 0
2
M = ( 200 lb/ft ) x 2
M = 162 lb ⋅ ft at C
Along CD:
ΣFy = 0: ( 0.9 ft + x1 )( 400 lb/ft ) − 900 lb − V = 0
V = −540 lb + ( 400 lb/ft ) x1
ΣM K = 0: M + x1 ( 900 lb ) −
V = 0 at x1 = 1.35 ft
0.9 ft + x1
( 400 lb/ft )( 0.9 ft + x1 ) = 0
2
M = 162 lb ⋅ ft − ( 540 lb ) x1 + ( 200 lb/ft ) x12
M = −202.5 lb ⋅ ft at center
( x1 = 1.35 ft )
Finish by symmetry
(b) From diagrams:
V
max
M
= 540 lb at C + and D −
max
= 203 lb ⋅ ft at center
PROBLEM 7.44
Solve Prob. 7.43 assuming that a = 1.5 ft.
SOLUTION
(a) FBD Beam:
ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0
w = 400 lb/ft
Along AC:
x ( 400 lb/ft ) − V = 0
ΣFy = 0:
V = ( 400 lb/ft ) x
V = 600 lb at C
ΣM J = 0: M −
( x = 1.5 ft )
x
( 400 lb/ft ) x = 0
2
M = ( 200 lb/ft ) x 2
M = 450 lb ⋅ ft at C
Along CD:
ΣFy = 0: x ( 400 lb/ft ) − 900 lb − V = 0
V = −900 lb + ( 400 lb/ft ) x
V = −300 at x = 1.5 ft
V = 0 at x = 2.25 ft
ΣM K = 0: M + ( x − 1.5 ft )( 900 lb ) −
x
( 400 lb/ft ) x = 0
2
M = 1350 lb ⋅ ft − ( 900 lb ) x + ( 200 lb/ft ) x 2
M = 450 lb ⋅ ft at x = 1.5 ft
M = 337.5 lb ⋅ ft at x = 2.25 ft ( center )
Finish by symmetry
(b) From diagrams:
V
M
max
max
= 600 lb at C − and D +
= 450 lb ⋅ ft at C and D
PROBLEM 7.45
Two short angle sections CE and DF are bolted to the uniform
beam AB of weight 3.33 kN, and the assembly is temporarily
supported by the vertical cables EG and FH as shown. A second
beam resting on beam AB at I exerts a downward force of 3 kN
on AB. Knowing that a = 0.3 m and neglecting the weight of the
ngle sections, (a) draw the shear and bending-moment diagrams for
beam AB, (b) determine the maximum absolute values of the shear
and bending moment in the beam.
SOLUTION
FBD angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
M C = 0.3165 kN ⋅ m
PD = 3.165 kN; M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − x (1.11 kN/m ) − V = 0
V = − (1.11 kN/m ) x
V = −1.332 kN at C
ΣM J = 0: M +
M = ( 0.555 kN/m ) x 2
( x = 1.2 m )
x
(1.11 kN/m ) x = 0
2
M = − 0.7992 kN ⋅ m at C
Along CI:
ΣFy = 0: − (1.11 kN/m ) x + 3.165 kN − V = 0
V = 3.165 kN − (1.11 kN/m ) x
V = 1.5 kN at I
( x = 1.5 m )
ΣM k = 0:
M + (1.11 kN/m ) x − ( x − 1.2 m )( 3.165 kN ) − ( 0.3165 kN ⋅ m ) = 0
PROBLEM 7.45 CONTINUED
M = 3.4815 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2
M = − 0.4827 kN ⋅ m at C
M = 0.01725 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3
kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B,
the diagram can be completed by symmetry.
(b) From diagrams:
V
M
max
max
= 1.833 kN at C and D
= 799 N ⋅ m at C and D
PROBLEM 7.46
Solve Prob. 7.45 when a = 0.6 m.
SOLUTION
FBD angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
PD = 3.165 kN
M C = 0.3165 kN ⋅ m
M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − (1.11 kN/m ) x − V = 0
V = − (1.11 kN/m ) x
V = − 0.999 kN at C
ΣM J = 0: M +
( x = 0.9 m )
x
(1.11 kN/m ) x = 0
2
M = − ( 0.555 kN/m ) x 2
M = − 0.44955 kN ⋅ m at C
Along CI:
ΣFy = 0: − x (1.11 kN/m ) + 3.165 kN − V = 0
V = 3.165 kN − (1.11 kN/m ) x
V = 1.5 kN at I
V = 2.166 kN at C
( x = 1.5 m )
ΣM K = 0:
M − 0.3165 kN ⋅ m + ( x − 0.9 m )( 3.165 kN ) +
x
(1.11 kN/m ) x = 0
2
PROBLEM 7.46 CONTINUED
M = −2.532 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2
M = − 0.13305 kN ⋅ m at C
M = 0.96675 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3
kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B,
the diagram can be completed by symmetry.
(b) From diagrams:
V
max
= 2.17 kN at C and D
M
max
= 967 N ⋅ m at I
PROBLEM 7.47
Draw the shear and bending-moment diagrams for the beam AB,
and determine the shear and bending moment (a) just to the left
of C, (b) just to the right of C.
SOLUTION
FBD CD:
ΣFy = 0: −1.2 kN + C y = 0
ΣM C = 0: ( 0.4 m )(1.2 kN ) − M C = 0
C y = 1.2 kN
M C = 0.48 kN ⋅ m
FBD Beam:
ΣM A = 0: (1.2 m ) B + 0.48 kN ⋅ m − ( 0.8 m )(1.2 kN ) = 0
B = 0.4 kN
ΣFy = 0: Ay − 1.2 kN + 0.4 kN = 0
A y = 0.8 kN
Along AC:
ΣFy = 0: 0.8 kN − V = 0
ΣM J = 0: M − x ( 0.8 kN ) = 0
V = 0.8 kN
M = ( 0.8 kN ) x
M = 0.64 kN ⋅ m at x = 0.8 m
Along CB:
ΣFy = 0: V + 0.4 kN = 0
ΣM K = 0: x1 ( 0.4 kN ) − M = 0
V = −0.4 kN
M = ( 0.4 kN ) x1
M = 0.16 kN ⋅ m at x1 = 0.4 m
(a)
Just left of C:
V = 800 N
M = 640 N ⋅ m
(b)
Just right of C:
V = −400 N
M = 160.0 N ⋅ m
PROBLEM 7.48
Draw the shear and bending-moment diagrams for the beam AB,
and determine the maximum absolute values of the shear and bending
moment.
SOLUTION
FBD angle:
ΣFy = 0: Fy − 600 N = 0
Fy = 600 N
ΣM Base = 0: M − ( 0.3 m )( 600 N ) = 0
M = 180 N ⋅ m
All three angles are the same.
FBD Beam:
ΣM A = 0: (1.8 m ) B − 3 (180 N ⋅ m )
− ( 0.3 m + 0.9 m + 1.5 m )( 600 N ) = 0
B = 1200 N
ΣFy = 0: Ay − 3 ( 600 N ) + 1200 N = 0
A y = 600 N
Along AC:
ΣFy = 0: 600 N − V = 0
V = 600 N
ΣM J = 0: M − x ( 600 N ) = 0
M = ( 600 N ) x = 180 N ⋅ m at x = .3 m
Along CD:
ΣFy = 0: 600 N − 600 N − V = 0
V =0
ΣM K = 0: M + ( x − 0.3 m )( 600 N ) − 180 N ⋅ m − x ( 600 N ) = 0
M = 360 N ⋅ m
PROBLEM 7.48 CONTINUED
Along DE:
ΣFy = 0: V − 600 N + 1200 N = 0
V = −600 N
ΣM N = 0: −M − 180 N ⋅ m − x2 ( 600 N ) + ( x2 + 0.3 m )(1200 N ) = 0
M = 180 N ⋅ m + ( 600 N ) x2 = 540 N ⋅ m at D, x2 = 0.6 m
M = 180 N ⋅ m at E (x2 = 0)
Along EB:
ΣFy = 0: V + 1200 N = 0
ΣM L = 0: x1 (1200 N ) − M = 0
V = −1200 N
M = (1200 N ) x1
M = 360 N ⋅ m at x1 = 0.3 m
From diagrams:
V
M
max
max
= 1200 N on EB
= 540 N ⋅ m at D +
PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB,
and determine the maximum absolute values of the shear and
bending moment.
SOLUTION
FBD Whole:
ΣM D = 0: (1.2 m )(1.5 kN ) − (1.2 m )( 6 kN )
− ( 3.6 m )(1.5 kN ) + (1.6 m ) G = 0
G = 6.75 kN
ΣFx = 0: − Dx + G = 0
Beam AB, with forces D and G
replaced by equivalent
force/couples at C and F
D x = 6.75 kN
ΣFy = 0: Dy − 1.5 kN − 6 kN − 1.5 kN = 0
D y = 9 kN
Along AD:
ΣFy = 0: −1.5 kN − V = 0
ΣM J = 0: x (1.5 kN ) + M = 0
V = −1.5 kN
M = − (1.5 kN ) x
M = −1.8 kN at x = 1.2 m
Along DE:
ΣFy = 0: −1.5 kN + 9 kN − V = 0
V = 7.5 kN
ΣM K = 0: M + 5.4 kN ⋅ m − x1 ( 9 kN ) + (1.2 m + x1 )(1.5 kN ) = 0
M = 7.2 kN ⋅ m + ( 7.5 kN ) x1
M = 1.8 kN ⋅ m at x1 = 1.2 m
PROBLEM 7.49 CONTINUED
Along EF:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM N = 0: − M + 5.4 kN ⋅ m − ( x4 + 1.2 m )(1.5 kN )
M = 3.6 kN ⋅ m − (1.5 kN ) x4
M = 1.8 kN ⋅ m at x4 = 1.2 m;
M = 3.6 kN ⋅ m at x4 = 0
Along FB:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM L = 0: − M − x3 (1.5 kN ) = 0
M = ( −1.5 kN ) x3
M = −1.8 kN ⋅ m at x3 = 1.2 m
From diagrams:
V
M
max
max
= 7.50 kN on DE
= 7.20 kN ⋅ m at D +
PROBLEM 7.50
Neglecting the size of the pulley at G, (a) draw the shear and
bending-moment diagrams for the beam AB, (b) determine the
maximum absolute values of the shear and bending moment.
SOLUTION
FBD Whole:
(a)
ΣM A = 0: ( 0.5 m )
12
5
D + (1.2 m ) D − ( 2.5 m )( 480 N ) = 0
13
13
D = 1300 N
ΣFy = 0: Ay +
5
(1300 N ) − 480 N = 0
13
Ay = −20 N
A y = 20 N
Beam AB with pulley forces and Along AE:
force at D replaced by equivalent
force-couples at B, F, E
ΣFy = 0: −20 N − V = 0
ΣM J = 0: M + x ( 20 N )
V = −20 N
M = − ( 20 N ) x
M = −24 N ⋅ m at x = 1.2 m
Along EF:
ΣFy = 0: V − 288 N − 480 N = 0
V = 768 N
ΣM L = 0: −M − x2 ( 288 N ) − ( 28.8 N ⋅ m ) − ( x2 + 0.6 m )( 480 N ) = 0
M = −316.8 N ⋅ m − ( 768 N ) x2
M = −316.8 N ⋅ m at x2 = 0 ; M = −624 N ⋅ m at x2 = 0.4 m
Along FB:
ΣFy = 0: V − 480 N = 0
ΣM K = 0: − M − x1 ( 480 N ) = 0
V = 480 N
M = − ( 480 N ) x1
M = −288 N ⋅ m at x1 = 0.6 m
PROBLEM 7.50 CONTINUED
(b) From diagrams:
V
max
M
= 768 N along EF
max
= 624 N ⋅ m at E +
PROBLEM 7.51
For the beam of Prob. 7.43, determine (a) the distance a for which the
maximum absolute value of the bending moment in the beam is as small
as possible, (b) the corresponding value of M max .
(Hint: Draw the bending-moment diagram and then equate the absolute
values of the largest positive and negative bending moments obtained.)
SOLUTION
FBD Beam:
ΣFy = 0: Lw − 2 P = 0
w=2
ΣM J = 0: M −
M =
Along AC:
P
L
x  2P 
x = 0

2 L 
P 2
x
L
P 2
a at x = a
L
ΣM K = 0: M + ( x − a ) P −
M = P (a − x) +
M =
x  2P 
x = 0

2 L 
P 2 Pa 2
x =
L
L
at
x=a
L
L

M = P  a −  at x =
2
4


Along CD:
This is M min by symmetry–see moment diagram completed by
symmetry.
For minimum M
max
, set M max = −M min :
P
a2
L

= −P  a − 
L
4

a 2 + La −
or
a=
Solving:
Positive answer
L2
=0
4
−1 ± 2
L
2
a = 0.20711L = 0.932 ft
(a)
(b)
M
max
= 0.04289PL = 173.7 lb ⋅ ft
PROBLEM 7.52
For the assembly of Prob. 7.45, determine (a) the distance a for which
the maximum absolute value of the bending moment in beam AB is as
small as possible, (b) the corresponding value of M max . (See hint for
Prob. 7.51.)
SOLUTION
By symmetry of whole arrangement:
FBD Angle:
T =
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − F = 0
F = 3.165 kN
ΣM 0 = 0: M − ( 0.1 m )( 3.165 kN ) = 0
ΣM J = 0: M +
M = 0.3165 kN ⋅ m
x
(1.11 kN/m ) x = 0
2
M = − ( 0.555 kN/m ) x 2 = − ( 0.555 kN/m )(1.5 m − a )
Along AC:
at C
2
( this is M min )
x
(1.11 kN/m ) x
2
−  x − (1.5 m − a )  ( 3.165 kN ) = 0
ΣM K = 0: M − 0.3165 kN ⋅ m +
Along CI:
M = −4.431 kN ⋅ m + ( 3.165 kN )( x + a ) − ( 0.555 kN/m ) x 2
M max ( at x = 1.5 m ) = − 0.93225 kN ⋅ m + ( 3.165 kN ) a
For minimum M
max
, set M max = −M min :
− 0.93225 kN ⋅ m + ( 3.165 kN ) a = ( 0.555 kN/m )(1.5 m − a )
Yielding:
a 2 − ( 8.7027 m ) a + 3.92973 m 2 = 0
Solving:
a = 4.3514 ± 13.864 = 0.4778 m, 8.075 m
Second solution out of range, so
(a)
2
a = 0.478 m
M max = 0.5801 kN ⋅ m
(b)
M max = 580 N ⋅ m
PROBLEM 7.53
For the beam shown, determine (a) the magnitude P of the two
upward forces for which the maximum value of the bending
moment is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
Ay = B = 60 kN − P
By symmetry:
Along AC:
ΣM J = 0: M − x ( 60 kN − P ) = 0
M = ( 60 kN − P ) x
M = 120 kN ⋅ m − ( 2 m ) P at x = 2 m
Along CD:
ΣM K = 0: M + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − Px
M = 120 kN ⋅ m − ( 4 m ) P at x = 4 m
Along DE:
ΣM L = 0: M − ( x − 4 m ) P + ( x − 2 m )( 60 kN )
− x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − ( 4 m ) P
(const)
Complete diagram by symmetry
For minimum M
max
, set M max = −M min
120 kN ⋅ m − ( 2 m ) P = − 120 kN ⋅ m − ( 4 m ) P 
P = 40.0 kN
(a)
M min = 120 kN ⋅ m − ( 4 m) P
(b)
M
max
= 40.0 kN ⋅ m
PROBLEM 7.54
For the beam and loading shown, determine (a) the distance a for
which the maximum absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
ΣM A = 0: M A − (1.5 ft )(1 kip ) − ( 3.5 ft )( 4 kips )
FBD Beam:
+ ( 3.5 ft + a )( 2 kips ) = 0
M A = 8.5 kip ⋅ ft − ( 2 kips ) a 
ΣFy = 0: Ay − 1 kip − 4 kips + 2 kips = 0
A y = 3 kips
Along AC:
ΣM J = 0: M − x ( 3 kips ) + 8.5 kip ⋅ ft − ( 2 kips ) a = 0
M = ( 3 kips ) x + ( 2 kips ) a − 8.5 kip ⋅ ft
M = ( 2 kips ) a − 4 kip ⋅ ft at C ( x = 1.5 ft )
M = ( 2 kips ) a − 8.5 kip ⋅ ft at A ( M min )
Along DB:
ΣM K = 0: − M + x1 ( 2 kips ) = 0
M = ( 2 kips ) x1
M = ( 2 kips ) a at D
ΣM L = 0: ( x2 + a )( 2 kips ) − x2 ( 4 kips ) − M = 0
M = ( 2 kips ) a − ( 2 kips ) x2
Along CD:
M = ( 2 kips ) a − 4 kip ⋅ ft at C
For minimum M
max
( see above )
, set M max ( at D ) = −M min ( at A)
( 2 kips ) a = − ( 2 kips ) a − 8.5 kip ⋅ ft 
4a = 8.5 ft
a = 2.125 ft
a = 2.13 ft
(a)
So
M max = ( 2 kips ) a = 4.25 kip ⋅ ft
(b)
M
max
= 4.25 kip ⋅ ft
PROBLEM 7.55
Knowing that P = Q = 375 lb, determine (a) the distance a for
which the maximum absolute value of the bending moment in
beam AB is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
ΣM A = 0: ( a ft ) D − ( 4 ft )( 375 lb ) − ( 8 ft )( 375 lb ) = 0
D=
FBD Beam:
4500
lb
a
ΣFy = 0: Ay − 2 ( 375 lb ) +
4500
lb = 0
a
4500 

A y =  750 −
 lb
a 

It is apparent that M = 0 at A and B, and that all segments of the M
diagram are straight, so the max and min values of M must occur
at C and D
4500 

ΣM C = 0: M − ( 4 ft )  750 −
 lb = 0
a 

Segment AC:
18000 

M =  3000 −
 lb ⋅ ft
a 

ΣM D = 0: − ( 8 − a ) ft  ( 375 lb ) − M = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M
So
max
, set M max = −M min
3000 −
a 2 = 48
18000
= 375 ( 8 − a )
a
a = 6.9282 ft
a = 6.93 ft
(a)
M max = 375 ( 8 − a ) = 401.92 lb ⋅ ft
(b)
M
max
= 402 lb ⋅ ft
PROBLEM 7.56
Solve Prob. 7.55 assuming that P = 750 lb and Q = 375 lb.
SOLUTION
ΣM D = 0: − ( a ft ) Ay + ( a − 4 ) ft  ( 750 lb )
FBD Beam:
− ( 8 − a ) ft  ( 375 lb ) = 0
6000 

A y = 1125 −
 lb
a 

It is apparent that M = 0 at A and B, and that all segments of the
M-diagram are straight, so M max and M min occur at C and D.
6000 

ΣM C = 0: M − ( 4 ft ) 1125 −
 lb = 0
a 

Segment AC:
24000 

M =  4500 −
 lb ⋅ ft
a 

ΣM D = 0: − M − ( 8 − a ) ft  ( 375 lb ) = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M max , set M max = −M min
4500 −
24000
= 375 ( 8 − a )
a
a 2 + 4a − 64 = 0
a = 6.2462 ft
Then
a = −2 ± 68 ( need + )
a = 6.25 ft
(a)
M max = 375 ( 8 − a ) = 657.7 lb ⋅ ft
(b)
M
max
= 658 lb ⋅ ft
PROBLEM 7.57
In order to reduce the bending moment in the cantilever beam
AB, a cable and counterweight are permanently attached at end B.
Determine the magnitude of the counterweight for which the
maximum absolute value of the bending moment in the beam is
as small as possible and the corresponding value of M max .
Consider (a) the case when the distributed load is permanently
applied to the beam, (b) the more general case when the distributed
load may either be applied or removed.
SOLUTION
x
wx = 0
2
ΣM J = 0: −M −
M due to distributed load:
1
M = − wx 2
2
ΣM J = 0: −M + xw = 0
M due to counter weight:
(a) Both applied:
M = wx
M = Wx −
w 2
x
2
And here M =
dM
W
= W − wx = 0 at x =
dx
w
W2
> 0 so M max ; M min must be at x = L
2w
So M min = WL −
1 2
wL . For minimum M
2
max
set M max = −M min , so
W2
1
= −WL + wL2 or W 2 + 2wLW − w2 L2 = 0
2w
2
W =
W = −wL ± 2w2 L2 (need +)
M max
(b) w may be removed:
W2
=
=
2w
Without w,
With w (see part a)
(
)
2 −1
2
(
)
2 − 1 wL = 0.414wL
2
M max = 0.858wL2
wL2
M = Wx, M max = WL at A
M = Wx −
M min = WL −
w 2
x ,
2
M max =
1 2
wL at x = L
2
W2
W
at x =
2w
w
PROBLEM 7.57 CONTINUED
For minimum M max , set M max ( no w ) = −M min ( with w )
WL = −WL +
With
1 2
1
wL → W = wL →
2
4
M max =
1 2
wL
4
W =
1
wL
4
PROBLEM 7.58
Using the method of Sec. 7.6, solve Prob. 7.29.
SOLUTION
(a) and (b)
By symmetry: Ay = D =
Shear Diag:
1  L  wL
w  =
2 2 
4
V jumps to Ay =
or A y = D =
wL
4
wL
at A,
4
and stays constant (no load) to B. From B to C, V is linear
wL
L
wL
 dV

−w =−
= −w  , and it becomes
at C.

4
2
4
 dx

(Note: V = 0 at center of beam. From C to D, V is again constant.)
Moment Diag: M starts at zero at A
wL 
 dM
=
and increases linearly 

4  to B.
 dV
MB = 0 +
L  wL  wL2
.

=
4 4 
16
From B to C M is parabolic
wL
 dM

= V , which decreases to zero at center and −
at C  ,

4
 dx

M is maximum at center.
M max =
Then, M is linear with
wL2 1  L  wL 
+  

16
2  4  4 
dM
wL
=−
to D
dy
4
MD = 0
V
M
max
max
=
=
wL
4
3wL2
32
Notes: Symmetry could have been invoked to draw second half.
Smooth transitions in M at B and C, as no discontinuities in V.
PROBLEM 7.59
Using the method of Sec. 7.6, solve Prob. 7.30.
SOLUTION
(a) and (b)
Shear Diag:
V = 0 at A and is linear
wL
 dV

L
 dV

= −w  to − w   = −
at B. V is constant 
= 0  from

2
 dx

 dx

2
B to C.
V
max
=
wL
2
Moment Diag: M = 0 at A and is
 dM

decreasing with V  to B.
parabolic 
 dx

MB =
1  L  wL 
wL2
−
=
−
 

2  2  2 
8
wL 
 dM
=−
From B to C, M is linear 

dx
2 

MC = −
wL2  L  wL 
3wL2
−  
=
−

8
8
 2  2 
M
max
=
Notes: Smooth transition in M at B, as no discontinuity in V.
It was not necessary to predetermine reactions at C.
In fact they are given by −VC and − M C .
3wL2
8
PROBLEM 7.60
Using the method of Sec. 7.6, solve Prob. 7.31.
SOLUTION
Shear Diag:
(a) and (b)
 dV

= 0  to B. V jumps down P
V jumps to P at A, then is constant 
 dx

to zero at B, and is constant (zero) to C.
V
max
= P
Moment Diag:
 dM

M is linear 
= V = P  to B.
dy


PL
L
M B = 0 +  ( P) =
.
2
2
PL
 dM

= 0  at
to C
M is constant 
2
 dx

M
max
=
PL
2
Note: It was not necessary to predetermine reactions at C. In fact they
are given by −VC and − M C .
PROBLEM 7.61
Using the method of Sec. 7.6, solve Prob. 7.32.
SOLUTION
(a) and (b)
ΣM C = 0: LAy − M 0 = 0
Ay =
M0
L
Shear Diag:
V jumps to −
M0
 dV

at A and is constant 
= 0  all the way to C
L
dx


V
max
=
M0
W
L
=
M0
W
2
Moment Diag:
M 
 dM
M is zero at A and linear 
= V = − 0  throughout.
L 
 dx
LM 
M
M B− = −  0  = − 0 ,
2 L 
2
M
but M jumps by + M 0 to + 0 at B.
2
M
LM 
MC = 0 −  0  = 0
2
2 L 
M
max
PROBLEM 7.62
Using the method of Sec. 7.6, solve Prob. 7.33.
SOLUTION
(a) and (b)
ΣM B = 0: ( 0.6 ft )( 4 kips ) + ( 5.1 ft )( 8 kips )
+ ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips
Shear Diag:
 dV

V is piecewise constant, 
= 0  with discontinuities at each
 dx

concentrated force. (equal to force)
V
= 12.63 kips W
max
Moment Diag:
 dM

= V  throughout.
M is zero at A, and piecewise linear 
 dx

M C = (1.8 ft )(12.625 kips ) = 22.725 kip ⋅ ft
M D = 22.725 kip ⋅ ft + ( 2.7 ft )( 2.625 kips )
= 29.8125 kip ⋅ ft
M E = 29.8125 kip ⋅ ft − ( 4.5 ft )( 5.375 kips )
= 5.625 kip ⋅ ft
M B = 5.625 kip ⋅ ft − ( 0.6 ft )( 9.375 kips ) = 0
M
max
= 29.8 kip ⋅ ft W
PROBLEM 7.63
Using the method of Sec. 7.6, solve Prob. 7.36.
SOLUTION
(a) and (b)
FBD Beam:
ΣM E = 0: (1.1 m )( 0.54 kN ) − ( 0.9 m ) C y
+ ( 0.4 m )(1.35 kN ) − ( 0.3 m )( 0.54 kN ) = 0
C y = 1.08 kN
ΣFy = 0: − 0.54 kN + 1.08 kN − 1.35 kN + E − 0.54 kN = 0
E = 1.35 kN
Shear Diag:
 dV

V is piecewise constant, 
= 0 everywhere  with discontinuities at
dx


each concentrated force. (equal to the force)
V
max
= 810 N W
Moment Diag:
M is piecewise linear starting with M A = 0
M C = 0 − 0.2 m ( 0.54 kN ) = 0.108 kN ⋅ m
M D = 0.108 kN ⋅ m + ( 0.5 m )( 0.54 kN ) = 0.162 kN ⋅ m
M E = 0.162 kN ⋅ m − ( 0.4 m )( 0.81 kN ) = − 0.162 kN ⋅ m
M B = 0.162 kN ⋅ m + ( 0.3 m )( 0.54 kN ) = 0
M
max
= 0.162 kN ⋅ m = 162.0 N ⋅ m W
PROBLEM 7.64
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) and (b)
Shear Diag:
 dV

V = 0 at A and linear 
= −2 kN/m  to C
 dx

VC = −1.2 m ( 2 kN/m ) = −2.4 kN.
At C, V jumps 6 kN to 3.6 kN, and is constant to D. From there, V is
 dV

= +3 kN/m  to B
linear 
 dx

VB = 3.6 kN + (1 m )( 3 kN/m ) = 6.6 kN
V
max
= 6.60 kN W
M A = 0.
Moment Diag:
 dM

decreasing with V  .
From A to C, M is parabolic, 
 dx

MC = −
1
(1.2 m )( 2.4 kN ) = −1.44 kN ⋅ m
2
 dM
From C to D, M is linear 
 dx

= 3.6 kN 

M D = −1.44 kN ⋅ m + ( 0.6 m )( 3.6 kN )
= 0.72 kN ⋅ m.
 dM

From D to B, M is parabolic 
increasing with V 
 dx

M B = 0.72 kN ⋅ m +
1
(1 m )( 3.6 + 6.6 ) kN
2
= 5.82 kN ⋅ m
M
max
= 5.82 kN ⋅ m W
Notes: Smooth transition in M at D. It was unnecessary to predetermine
reactions at B, but they are given by −VB and −M B
PROBLEM 7.65
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
ΣM B = 0: ( 3 ft )(1 kip/ft )( 6 ft ) + ( 8 ft )( 6 kips )
(a) and (b)
+ (10 ft )( 6 kips ) − (12 ft ) Ay = 0
A y = 10.5 kips
Shear Diag:
V is piecewise constant from A to E, with discontinuities at A, C, and E
equal to the forces. VE = −1.5 kips. From E to B, V is linear
 dV

= −1 kip/ft  ,

dx


so
VB = −1.5 kips − ( 6 ft )(1 kip/ft ) = −7.5 kips
V
max
= 10.50 kips W
Moment Diag: M A = 0, then M is piecewise linear to E
M C = 0 + 2 ft (10.5 kips ) = 21 kip ⋅ ft
M D = 21 kip ⋅ ft + ( 2 ft )( 4.5 kips ) = 30 kip ⋅ ft
M E = 30 kip ⋅ ft − ( 2 ft )(1.5 kips ) = 27 kip ⋅ ft
 dM

decreasing with V  , and
From E to B, M is parabolic 
 dx

M B = 27 kip ⋅ ft −
1
( 6 ft )(1.5 kips + 7.5 kips ) = 0
2
M
max
= 30.0 kip ⋅ ft W
PROBLEM 7.66
Using the method of Sec. 7.6, solve Prob. 7.37.
SOLUTION
(a) and (b)
FBD Beam:
ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0
A y = 2 kips
ΣM A = 0: M A + ( 3 ft )( 2 kips/ft )( 6 ft )
− (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0
M A = 114 kip ⋅ ft
Shear Diag:
 dV

VA = Ay = 2 kips. Then V is linear 
= 2 kips/ft  to C, where
dx


VC = 2 kips + ( 6 ft )( 2 kips/ft ) = 14 kips.
V is constant at 14 kips to D, then jumps down 12 kips to 2 kips and is
constant to B
V
Moment Diag:
max
= 14.00 kips W
M A = −114 kip ⋅ ft.
 dM

increasing with V  and
From A to C, M is parabolic 
dx


M C = −114 kip ⋅ ft +
1
( 2 kips + 14 kips )( 6 ft )
2
M C = −66 kip ⋅ ft.
Then M is piecewise linear.
M D = −66 kip ⋅ ft + (14 kips )( 4.5 ft ) = −3 kip ⋅ ft
M B = −3 kip ⋅ ft + ( 2 kips )(1.5 ft ) = 0
M
max
= 114.0 kip ⋅ ft W
PROBLEM 7.67
Using the method of Sec. 7.6, solve Prob. 7.38.
SOLUTION
(a) and (b)
FBD Beam:
 kips 
ΣM B = 0: ( 3 ft )  2
 ( 6 ft ) + ( 9 ft )(12 kips ) − (15 ft ) Ay = 0
 ft 
A y = 9.6 kips
Shear Diag:
V jumps to Ay = 9.6 kips at A, is constant to C, jumps down 12 kips
to −2.4 kips at C, is constant to D, and then is linear
 dV

= −2 kips/ft  to B

 dx

VB = −2.4 kips − ( 2 kips/ft )( 6 ft )
= −14.4 kips
V
max
= 14.40 kips W
Moment Diag:
 dM

= 9.6 kips/ft 

dx


M is linear from A to C
M C = 9.6 kips ( 6 ft ) = 57.6 kip ⋅ ft,
M is linear from C to D
 dM

 dx

= −2.4 kips/ft 

M D = 57.6 kip ⋅ ft − 2.4 kips ( 3 ft )
M D = 50.4 kip ⋅ ft.
 dM

M is parabolic 
decreasing with V  to B.
 dx

M B = 50.4 kip ⋅ ft −
1
( 2.4 kips + 14.4 kips )( 6 ft ) = 0
2
=0
M
max
= 57.6 kip ⋅ ft W
PROBLEM 7.68
Using the method of Sec. 7.6, solve Prob. 7.39.
SOLUTION
(a) and (b)
FBD Beam:
By symmetry:
1
( 5 m )( 4 kN/m ) + 8 kN
2
or A y = B = 18 kN
Ay = B =
Shear Diag:
V jumps to 18 kN at A, and is constant to C, then drops 8 kN to 10 kN.
 dV

= −4 kN/m  , reaching −10 kN at
After C, V is linear 
 dx

D VD = 10 kN − ( 4 kN/m )( 5 m )  passing through zero at the beam
center. At D, V drops 8 kN to −18 kN and is then constant to B
V
max
= 18.00 kN W
Moment Diag:
 dM

M A = 0. Then M is linear 
= 18 kN/m  to C
 dx

M C = (18 kN )( 2 m ) = 36 kN ⋅ m, M is parabolic to D
 dM

decreases with V to zero at center 

dx


M center = 36 kN ⋅ m +
1
(10 kN )( 2.5 m ) = 48.5 kN ⋅ m = M max
2
M
Complete by invoking symmetry.
max
= 48.5 kN ⋅ m W
PROBLEM 7.69
Using the method of Sec. 7.6, solve Prob. 7.40.
SOLUTION
(a) and (b)
FBD Beam:
ΣM F = 0: (1 m )( 22 kN ) + (1.5 m )( 4 kN/m )( 3 m )
− ( 4 m ) Dy + ( 6 m )( 2 kN/m )( 2 m ) = 0
D y = 16 kN
ΣFy = 0: 16 kN + 22 kN − Fy − ( 2 kN/m )( 2 m )
− ( 4 kN/m )( 3 m ) = 0
Fy = 22 kN
Shear Diag:
 dV

VA = 0, then V is linear 
= −2 kN/m  to C;
 dx

VC = −2 kN/m ( 4 m ) = −4 kN
V is constant to D, jumps 16 kN to 12 kN and is constant to E.
 dV

= −4 kN/m  to F.
Then V is linear 
 dx

VF = 12 kN − ( 4 kN/m )( 3 m ) = 0.
V jumps to −22 kN at F, is constant to B, and returns to zero.
V
Moment Diag:
 dM

M A = 0, M is parabolic 
decreases with V  to C.
 dx

1
M C = − ( 4 kN )( 2 m ) = −4 kN ⋅ m.
2
max
= 22.0 kN W
PROBLEM 7.69 CONTINUED
 dM

Then M is linear 
= −4 kN  to D.
dx


M D = −4 kN ⋅ m − ( 4 kN )(1 m ) = −8 kN ⋅ m
 dM

= 12 kN  , and
From D to E, M is linear 
 dx

M E = −8 kN ⋅ m + (12 kN )(1m )
M E = 4 kN ⋅ m
 dM

M is parabolic 
decreasing with V  to F.
dx


M F = 4 kN ⋅ m +
1
(12 kN )( 3 m ) = 22 kN ⋅ m.
2
 dM

Finally, M is linear 
= −22 kN  , back to zero at B.
dx


M
max
= 22.0 kN ⋅ m W
PROBLEM 7.70
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) and (b)
FBD Beam:
ΣM B = 0: (1.5 m )(16 kN )
+ ( 3 m )( 8 kN ) + 6 kN ⋅ m − ( 4.5 m ) Ay = 0
A y = 12 kN
Shear Diag:
V is piecewise constant with discontinuities equal to the concentrated
forces at A, C, D, B
V max = 12.00 kN W
Moment Diag:
 dM

=V
After a jump of −6 kN ⋅ m at A, M is piecewise linear 
 dx

So
M C = −6 kN ⋅ m + (12 kN )(1.5 m ) = 12 kN ⋅ m
M D = 12 kN ⋅ m + ( 4 kN )(1.5 m ) = 18 kN ⋅ m
M B = 18 kN ⋅ m − (12 kN )(1.5 m ) = 0
M
max
= 18.00 kN ⋅ m W
PROBLEM 7.71
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
FBD Beam:
ΣM A = 0: ( 8 m ) F + (11 m )( 2 kN ) + 10 kN ⋅ m − ( 6 m )( 8 kN )
− 12 kN ⋅ m − ( 2 m )( 6 kN ) = 0 F = 5 kN
ΣFy = 0: Ay − 6 kN − 8 kN + 5 kN + 2 kN = 0
A y = 7 kN
Shear Diag:
V is piecewise constant with discontinuities equal to the concentrated
forces at A, C, E, F, G
Moment Diag:
M is piecewise linear with a discontinuity equal to the couple at D.
M C = ( 7 kN )( 2 m ) = 14 kN ⋅ m
M D− = 14 kN ⋅ m + (1 kN )( 2 m ) = 16 kN ⋅ m
M D+ = 16 kN ⋅ m + 12 kN ⋅ m = 28 kN ⋅ m
M E = 28 kN ⋅ m + (1 kN )( 2 m ) = 30 kN ⋅ m
M F = 30 kN ⋅ m − ( 7 kN )( 2 m ) = 16 kN ⋅ m
M G = 16 kN ⋅ m − ( 2 kN )( 3 m ) = 10 kN ⋅ m
(b)
V
M
max
max
= 7.00 kN
= 30.0 kN
PROBLEM 7.72
For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the magnitude and location
of the maximum bending moment.
SOLUTION
(a)
FBD Beam:
ΣM B = 0: ( 3 ft )(1.2 kips/ft )( 6 ft ) − ( 8 ft ) Ay = 0
A y = 2.7 kips
Shear Diag:
V = Ay = 2.7 kips at A, is constant to C, then linear
 dV

= −1.2 kips/ft  to B.

dx


VB = 2.7 kips − (1.2 kips/ft )( 6 ft )
VB = −4.5 kips
V = 0 = 2.7 kips − (1.2 kips/ft ) x1 at
x1 = 2.25 ft
Moment Diag:
 dM

M A = 0, M is linear 
= 2.7 kips  to C.
 dx

M C = ( 2.7 kips )( 2 ft ) = 5.4 kip ⋅ ft
 dM

decreasing with V 
Then M is parabolic 
 dx

(b)
M max occurs where
dM
=V =0
dx
M max = 5.4 kip ⋅ ft +
1
( 2.7 kips ) x1;
2
x1 = 2.25 m
M max = 8.4375 kip ⋅ ft
M max = 8.44 kip ⋅ ft, 2.25 m right of C
Check:
M B = 8.4375 kip ⋅ ft −
1
( 4.5 kips )( 3.75 ft ) = 0
2
PROBLEM 7.73
For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the magnitude and location
of the maximum bending moment.
SOLUTION
FBD Beam:
(a)
ΣM B = 0: ( 6 ft )( 2 kips/ft )( 8 ft ) − (15 ft ) Ay = 0
A y = 6.4 kips
Shear Diag:
V = Ay = 6.4 kips at A, and is constant to C, then linear
 dV

= −2 kips/ft  to D,

 dx

VD = 6.4 kips − ( 2 kips/ft )( 8 ft ) = −9.6 kips
V = 0 = 6.4 kips − ( 2 kips/ft ) x1 at x1 = 3.2 ft
Moment Diag:
 dM

M A = 0, then M is linear 
= 6.4 kips  to M C = ( 6.4 kips )( 5 ft ) .
dx


dM


M C = 32 kip ⋅ ft. M is then parabolic 
decreasing with V  .
dx


(b)
M max occurs where
M max = 32 kip ⋅ ft +
dM
= V = 0.
dx
1
( 6.4 kips ) x1;
2
x1 = 3.2 ft
M max = 42.24 kip ⋅ ft
M max = 42.2 kip ⋅ ft, 3.2 ft right of C
M D = 42.24 kip ⋅ ft −
M is linear from D to zero at B.
1
( 9.6 kips )( 4.8 ft ) = 19.2 kip ⋅ ft
2
PROBLEM 7.74
For the beam shown, draw the shear and bending-moment diagrams and
determine the maximum absolute value of the bending moment knowing
that (a) P = 14 kN, (b) P = 20 kN.
SOLUTION
(a)
FBD Beam:
ΣFy = 0: Ay − (16 kN/m )( 2 m ) − 6 kN + P = 0
Ay = 38 kN − P
(a)
A y = 24 kN
(b)
A y = 18 kN
ΣM A = 0: ( 5 m ) P − ( 3.5 m )( 6 kN ) − 1 m (16 kN/m )( 2 m ) − M A = 0
M A = ( 5 m ) P − 53 kN ⋅ m
(b)
(a)
M A = 17 kN ⋅ m
(b)
M A = 47 kN ⋅ m
Shear Diags:
 dV

VA = Ay . Then V is linear 
= −16 kN/m  to C.
 dx

VC = VA − (16 kN/m )( 2 m ) = VA − 32 kN
(a)
VC = −8 kN
(b)
VC = −14 kN
V = 0 = VA − (16 kN/m ) x1
(a)
x1 = 1.5 m
(b)
x1 = 1.125 m
V is constant from C to D, decreases by 6 kN at D and is constant to
B (at − P)
PROBLEM 7.74 CONTINUED
Moment Diags:
 dM

M A = M A reaction. Then M is parabolic 
decreasing with V  .
 dx

The maximum occurs where V = 0. M max = M A +
(a)
M max = 17 kN ⋅ m +
1
VA x1.
2
1
( 24 kN )(1.5 m ) = 35.0 kN ⋅ m
2
1.5 ft from A
(b)
M max = 47 kN ⋅ m +
1
(18 kN )(1.125 m ) = 57.125 kN ⋅ m
2
M max = 57.1 kN ⋅ m 1.125 ft from A
M C = M max +
(a)
(b)
M C = 35 kN ⋅ m −
M C = 57.125 kN ⋅ m −
1
VC ( 2 m − x1 )
2
1
(8 kN )( 0.5 m ) = 33 kN ⋅ m
2
1
(14 kN )( 0.875 m ) = 51 kN ⋅ m
2
M is piecewise linear along C, D, B, with M B = 0 and
M D = (1.5 m ) P
(a)
M D = 21 kN ⋅ m
(b)
M D = 30 kN ⋅ m
PROBLEM 7.75
For the beam shown, draw the shear and bending-moment diagrams,
and determine the magnitude and location of the maximum absolute
value of the bending moment knowing that (a) M = 0 ,
(b) M = 12 kN ⋅ m.
SOLUTION
FBD Beam:
ΣM A = 0: ( 4 m ) B − (1 m )( 20 kN/m )( 2 m ) − M = 0
B = 10 kN +
(a)
M
4m
(a)
B = 10 kN
(b)
B = 13 kN
ΣFy = 0: Ay − ( 20 kN/m )( 2 m ) + B = 0
Ay = 40 kN − B
(a)
A y = 30 kN
(b)
A y = 27 kN
Shear Diags:
(b)
 dV

= −20 kN/m  to C.
VA = Ay , then V is linear 
 dx

VC = Ay − ( 20 kN/m )( 2 m ) = Ay − 40 kN
(a)
VC = −10 kN
(b)
VC = −13 kN
V = 0 = Ay − ( 20 kN/m ) x1 at x1 =
(a)
x1 = 1.5 m
(b)
x1 = 1.35 m
V is constant from C to B.
Ay m
20 kN
PROBLEM 7.75 CONTINUED
Moment Diags:
 dM

M A = applied M . Then M is parabolic 
decreases with V 
 dx

M is max where V = 0. M max = M +
(a)
M
=
max
1
Ay x1.
2
1
( 30 kN )(1.5 m ) = 22.5 kN ⋅ m
2
1.500 m from A
(b)
M max = 12 kN ⋅ m +
1
( 27 kN )(1.35 m ) = 30.225 kN ⋅ m
2
M
M C = M max −
max
= 30.2 kN 1.350 m from A
1
VC ( 2 m − x1 )
2
(a)
M C = 20 kN ⋅ m
(b)
M C = 26 kN ⋅ m
 dM

Finally, M is linear 
= VC  to zero at B.
 dx

PROBLEM 7.76
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum
absolute value of the bending moment.
SOLUTION
FBD Beam:
(a)
ΣM B = 0: ( 3 m )( 40 kN/m )( 6 m ) − ( 30 kN ⋅ m ) − ( 6 m ) Ay = 0
A y = 115 kN
Shear Diag:
 dM

= − 40 kN/m  to B.
VA = Ay = 115 kN, then V is linear 
 dx

VB = 115 kN − ( 40 kN/m )( 6 m ) = −125 kN.
V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 2.875 m
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0,
M max =
1
(115 kN/m )( 2.875 m ) = 165.3125 kN ⋅ m
2
1
(125 kN )( 6 m − x1 )
2
1
= 165.3125 kN ⋅ m − (125 kN )( 6 − 2.875 ) m
2
= −30 kN ⋅ m as expected.
M B = M max −
(b)
M
max
= 165.3 kN ⋅ m ( 2.88 m from A )
PROBLEM 7.77
Solve Prob. 7.76 assuming that the 30 kN ⋅ m couple applied at B is
counterclockwise
SOLUTION
(a)
FBD Beam:
ΣM B = 0: 30 kN ⋅ m + ( 3 m )( 40 kN/m )( 6 m ) − ( 6 m ) Ay = 0
A y = 125 kN
Shear Diag:
 dV

VA = Ay = 125 kN, V is linear 
= −40 kN/m  to B.
 dx

VB = 125 kN − ( 40 kN/m )( 6 m ) = −115 kN
V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 3.125 m
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreases with V  . Max M occurs
 dx

where V = 0,
M max =
1
(125 kN )( 3.125 m ) = 195.3125 kN ⋅ m
2
(b)
M
max
M B = 195.3125 kN ⋅ m −
= 195.3 kN ⋅ m ( 3.125 m from A )
1
(115 kN )( 6 − 3.125) m
2
M B = 30 kN ⋅ m as expected.
PROBLEM 7.78
For beam AB, (a) draw the shear and bending-moment diagrams, (b)
determine the magnitude and location of the maximum absolute value
of the bending moment.
SOLUTION
(a)
Replacing the load at E with
equivalent force-couple at C:
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) − ( 3 m )( 4 kN )
− (1.5 m )( 8 kN/m )( 3 m ) − 4 kN ⋅ m = 0
D = 10 kN
ΣFy = 0: Ay + 10 kN − 2 kN − 4 kN − ( 8 kN/m )( 3 m ) = 0
A y = 20 kN
Shear Diag:
 dV

VA = Ay = 20 kN, then V is linear 
= −8 kN/m  to C.
 dx

VC = 20 kN − ( 8 kN/m )( 3 m ) = −4 kN
V = 0 = 20 kN − ( 8 kN/m ) x1 at x1 = 2.5 m
At C, V decreases by 4 kN to −8 kN.
At D, V increases by 10 kN to 2 kN.
Moment Diag:
 dM

M A = 0, then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0.
M max =
1
( 20 kN )( 2.5 m ) = 25 kN ⋅ m
2
(b)
M max = 25.0 kN ⋅ m, 2.50 m from A
PROBLEM 7.78 CONTINUED
M C = 25 kN ⋅ m −
1
( 4 kN )( 0.5 m ) = 24 kN ⋅ m.
2
At C, M decreases by 4 kN ⋅ m to 20 kN ⋅ m. From C to B, M is piecewise
dM
dM
linear with
= +2 kN to B.
= −8 kN to D, then
dx
dx
M D = 20 kN ⋅ m − ( 8 kN )( 3 m ) = −4 kN ⋅ m.
MB = 0
PROBLEM 7.79
Solve Prob. 7.78 assuming that the 4-kN force applied at E is directed
upward.
SOLUTION
(a)
Replacing the load at E with
equivalent force-couple at C.
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) + ( 3 m )( 4 kN )
− 4 kN ⋅ m − (1.5 m )( 8 kN/m )( 3 m ) = 0
D=
ΣFy = 0: Ay +
22
kN
3
22
kN − ( 8 kN/m )( 3 m ) + 4 kN − 2 kN = 0
3
Ay =
44
kN
3
Shear Diag:
VA = Ay =
44
 dV

kN, then V is linear 
= −8 kN/m  to C.
3
 dx

VC =
44
28
kN − ( 8 kN/m )( 3 m ) = −
kN
3
3
V =0=
44
11
kN − ( 8 kN/m ) x1 at x1 =
m.
3
6
At C, V increases 4 kN to −
At D, V increases
16
kN.
3
22
kN to 2 kN.
3
PROBLEM 7.79 CONTINUED
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0.
M max =
1  44
 11  121
kN 
m =
kN ⋅ m

2 3
9
 6

(b)
MC =
M max = 13.44 kN ⋅ m at 1.833 m from A
121
1  28
 7 
kN ⋅ m − 
kN  m  = 8 kN ⋅ m.
9
2 3
 6 
At C, M increases by 4 kN ⋅ m to 12 kN ⋅ m. Then M is linear
16
 dM

=−
kN  to D.

3
 dx

 16

kN  ( 3 m ) = −4 kN ⋅ m. M is again linear
M D = 12 kN ⋅ m − 
3


 dM

= 2 kN  to zero at B.

dx


PROBLEM 7.80
For the beam and loading shown, (a) derive the equations of the shear and
bending-moment curves, (b) draw the shear and bending-moment diagrams,
(c) determine the magnitude and location of the maximum bending moment.
SOLUTION
x

Distributed load w = w0 1 − 
L

ΣM A = 0:
1


 total = w0 L 
2


L 1

 w0 L  − LB = 0
3 2

B=
w0 L
6
1
wL
w0 L + 0 = 0
2
6
Ay =
w0 L
3
ΣFy = 0: Ay −
Shear:
VA = Ay =
w0 L
,
3
x 
dV
x
= − w → V = VA − ∫ w0 1 −  dx
0 
L
dx
Then
 1 x 1  x 2 
1 w0 2
w L
V =  0  − w0 x +
x = w0 L  − +   
2 L
 3 
 3 L 2  L  
Note: At x = L, V = −
w0 L
;
6
2
x
x
x 2
V = 0 at   − 2   + = 0 →
= 1−
L
L
L 3
1
3
Moment:
M A = 0,
Then
x
x/ L  x   x 
 dM 

 = V → M = ∫0 Vdx = L ∫0 V   d  
 dx 
L L
x / L  1 x 1  x 2   x 
 − +
d
M = w0 L ∫
0  3 L 2  L    L 
2


 1  x  1  x  2 1  x 3 
M = w0 L2    −   +   
6  L  
 3  L  2  L 
PROBLEM 7.80 CONTINUED
 x
M max  at
=1−
 L

1
2
 = 0.06415w0 L
3
 1 x 1  x 2 
V = w0 L  − +   
 3 L 2  L  
(a)
 1  x  1  x  2 1  x 3 
M = w0 L    −   +   
6  L  
 3  L  2  L 
2
(c)
M max = 0.0642 w0 L2
at x = 0.423L
PROBLEM 7.81
For the beam and loading shown, (a) derive the equations of the shear
and bending-moment curves, (b) draw the shear and bending-moment
diagrams, (c) determine the magnitude and location of the maximum
bending moment.
SOLUTION
 x

w = w0  4   − 1
 L

Distributed load
dV
= −w, and V ( 0 ) = 0, so
dx
Shear:
V =
x
∫0
− wdx = −
x/L
∫0
x
Lwd  
L
2
 x
x/L

 x   x 
  x 

V =∫
wo L 1 − 4    d   = w L   − 2 
0  L   L  
0
 L   L 



Notes: At x = L, V = −w0 L
x
x
  = 2 
L
L
And V = 0 at
2
or
x
1
=
L
2
M ( 0 ) = 0,
dM
=V
dx
1
x
Also V is max where w = 0  = 
4
L
Vmax =
1
w0 L
8
Moment:
x x
V  d  
L L
2
x / L  x 
  − 2  x   d  x 
M = w0 L2 ∫
0
 L 
L  L


M =
x
x/ L
∫0 vdx = L∫0
(a)
2
 x 
x 
V = w0 L   − 2   
 L  
 L 
 1  x  2 2  x 3 
M = w0 L2    −   
3  L  
 2  L 
PROBLEM 7.81 CONTINUED
M max =
1
L
w0 L2 at x =
24
2
1
M min = − w0 L2 at x = L
6
M max =
w0 L2
L
at x =
24
2
(c)
M
max
= −M min =
w0 L2
at B
6
PROBLEM 7.82
For the beam shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the
maximum bending moment. (Hint: Derive the equations of the
shear and bending-moment curves for portion CD of the beam.)
SOLUTION
(a)
FBD Beam:
ΣM B = 0:
( 3a ) 
1

w0 ( 3a )  − 5aAy = 0
A y = 0.9w0a
2

1
ΣFy = 0: 0.9w0a − w0 ( 3a ) + B = 0
2
B = 0.6w0a
Shear Diag:
V = Ay = 0.9w0a from A to C and V = B = − 0.6w0a from B to D.
x1
. If x1 is measured right to left,
3a
dV
x w
dM
= + w and
= − V . So, from D, V = − 0.6w0a + ∫0 1 0 x1dx1,
3a
dx1
dx1
Then from D to C, w = w0
2

1  x1  
V = w0a  − 0.6 +   
6  a  

2
x 
Note: V = 0 at  1  = 3.6, x1 =
a
3.6 a
Moment Diag:
 dM

M = 0 at A, increasing linearly 
= 0.9w0a  to M C = 0.9w0a 2.
 dx1

 dM

= 0.6w0a  to
Similarly M = 0 at B increasing linearly 
 dx

M D = 0.6w0a 2. Between C and D
2
x1 
1  x1  

M = 0.6w0a + w0a ∫ 0.6 −   dx1,
0 
6 a  


2
3

x  1 x  
M = w0a 2 0.6 + 0.6  1  −  1  
 a  18  a  

PROBLEM 7.82 CONTINUED
(b)
At
x1
=
a
3.6, M = M max = 1.359w0a 2
x1 = 1.897a left of D
PROBLEM 7.83
Beam AB, which lies on the ground, supports the parabolic load
shown. Assuming the upward reaction of the ground to be uniformly
distributed, (a) write the equations of the shear and bending-moment
curves, (b) determine the maximum bending moment.
SOLUTION
ΣFy = 0: wg L −
(a)
wg L =
Define ξ =
or
L 4w0
∫0
L2
( Lx − x ) dx = 0
2
4w0  1 2 1 3  2
 LL − L  = w0 L
3  3
L2  2
2w0
3
wg =
 x  x 2  2
 −    − w0
 L  L   3
x
dx
so dξ =
→ net load w = 4w0
L
L
 1

w = 4w0  − + ξ − ξ 2 
6


V = V(0) −
ξ
 1
1

1
∫0 4w0 L  − 6 + ξ − ξ
2
 dξ =



1
0 + 4w0 L  ξ + ξ 2 − ξ 3 
6
2
3
(
2
w0 L ξ − 3ξ 2 + 2ξ 3
3
V =
M = M0 +
=
(b)
x
2
ξ
∫0 Vdx = 0 + 3 w0 L ∫0
2
(ξ − 3ξ
2
)
)
+ 2ξ 3 d ξ
(
2
1  1
1
w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 ξ 2 − 2ξ 3 + ξ 4
3
2  3
2
Max M occurs where V = 0 → 1 − 3ξ + 2ξ 2 = 0 → ξ =
)
1
2
1 1
1  w L2

1 2
M  ξ =  = w0 L2  − +  = 0
2 3
48

 4 8 16 
M max =
w0 L2
at center of beam
48
PROBLEM 7.84
The beam AB is subjected to the uniformly distributed load shown
and to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +325 lb ft
at D and +800 lb ft at E, (a) determine P and Q, (b) draw the shear and
bending-moment diagrams for the beam.
SOLUTION
FBD ACD:
(a)
ΣM D − = 0: 0.325 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0
C y = 3.325 kips
FBD EB:
ΣM E = 0: (1 ft ) B − 0.8 kip ⋅ ft = 0
B = 0.8 kip
FBD Beam:
ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft ) − (1 ft )( 3.325 kips )
− (1 ft ) Q + 2 ft ( 0.8 kips ) = 0
Q = 1.275 kips
ΣFy = 0: 3.325 kips + 0.8 kips − 1.275 kips
− ( 2 kips/ft )(1ft ) − P = 0
(a)
P = 0.85 kip
P = 850 lb
Q = 1.275 kips
(b) Shear Diag:
 dV

= −2 kips/ft  from 0 at A to
V is linear 
 dx

− ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise constant with
discontinuities equal to forces at C, D, E, B
Moment Diag:
 dM

decreasing with V  from 0 at A to
M is parabolic 
 dx

1
− ( 2 kips )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with
2
PROBLEM 7.84 CONTINUED
M D = −1 kip ⋅ ft + (1.325 kips )(1 ft ) = 0.325 kip ⋅ ft
M E = 0.325 kip ⋅ ft + ( 0.475 kips )(1 ft ) = 0.800 kip ⋅ ft
M B = 0.8 kip ⋅ ft − ( 0.8 kip )(1 ft ) = 0
PROBLEM 7.85
Solve Prob. 7.84 assuming that the bending moment was found to be
+260 lb ft at D and +860 lb ft at E.
SOLUTION
FBD ACD:
(a)
ΣM D = 0: 0.26 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0
C y = 3.26 kips
FBD DB:
ΣM E = 0: (1 ft ) B − 0.86 kip ⋅ ft
B = 0.86 kip
FBD Beam:
ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft )
− (1 ft )( 3.26 kips ) + (1 ft ) Q + ( 2 ft )( 0.86 kips ) = 0
Q = 1.460 kips
Q = 1.460 kips
ΣFy = 0: 3.26 kips + 0.86 kips − 1.460 kips
− P − ( 2 kips/ft )(1 ft ) = 0
P = 0.66 kips
P = 660 lb
(b) Shear Diag:
 dV

V is linear 
= −2 kips/ft  from 0 at A to
 dx

− ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise
constant with discontinuities equal to forces at C, D, E, B.
Moment Diag:
 dM

decreasing with V  from 0 at A to
M is parabolic 
 dx

1
− ( 2 kips/ft )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with
2
PROBLEM 7.85 CONTINUED
M 0 = 0.26 kip ⋅ ft
M E = 0.86 kip ⋅ ft, M B = 0
PROBLEM 7.86
The beam AB is subjected to the uniformly distributed load shown
and to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +7 kN · m
at D and +5 kN · m at E, (a) determine P and Q, (b) draw the shear
and bending-moment diagrams for the beam.
SOLUTION
FBD AD:
ΣM D = 0: 7 kN ⋅ m + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0
2 Ay − P = 8.2 kN
(1)
ΣM E = 0: ( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m )
− 5 kN ⋅ m = 0
FBD EB:
2B − Q = 6.2 kN
(2)
( 6 m ) B − (1 m ) P − ( 5 m ) Q
− ( 3 m )( 0.6 kN/m )( 6 m ) = 0
ΣM A = 0:
6 B − P − 5Q = 10.8 kN
(3)
ΣM B = 0: (1 m ) Q + ( 5 m ) P + ( 3 m )( 0.6 kN/m )( 6 m )
− (6 m) A = 0
6 A − Q − 5P = 10.8 kN
P = 6.60 kN , Q = 600 N
Solving (1)–(4):
A y = 7.4 kN ,
(4)
B = 3.4 kN
(b) Shear Diag:
dV
= − 0.6 kN/m throughout, and
dx
discontinuities equal to forces at A, C, F, B.
V is piecewise linear with
Note V = 0 = 0.2 kN − ( 0.6 kN/m ) x at x =
1
m
3
PROBLEM 7.86 CONTINUED
Moment Diag:
 dM

decreasing with V  with “breaks” in
M is piecewise parabolic 
 dx

slope at C and F.
MC =
M max = 7.1 kN ⋅ m +
1
( 7.4 + 6.8) kN (1 m ) = 7.1 kN ⋅ m
2
1
1
( 0.2 kN )  m  = 7.133 kN ⋅ m
2
3 
1
 2 
M F = 7.133 kN ⋅ m − ( 2.2 kN )  3 m  = 3.1 kN ⋅ m
2
 3 
PROBLEM 7.87
Solve Prob. 7.86 assuming that the bending moment was found to be
+3.6 kN · m at D and +4.14 kN · m at E.
SOLUTION
FBD AD:
ΣM D = 0: 3.6 kN ⋅ m + (1 m ) P + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0
2 Ay − P = 4.8 kN
FBD EB:
ΣM E = 0:
(1)
( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m )
− 4.14 kN ⋅ m = 0
2B − Q = 5.34 kN
(2)
ΣM A = 0: ( 6 m ) B − ( 5 m ) Q − (1 m ) P − ( 3 m )( 0.6 kN/m )( 6 m ) = 0
6 B − P − 5Q = 10.8 kN
(3)
By symmetry:
6 A − Q − 5P = 10.8 kN
Solving (1)–(4)
(4)
P = 660 N , Q = 2.28 kN
Ay = 2.73 kN , B = 3.81 kN
(b) Shear Diag:
 dV

= − 0.6 kN/m  throughout, and
V is piecewise linear with 
 dx

discontinuities equal to forces at A, C, F, B.
Note that V = 0 = 1.47 kN − ( 0.6 kN/m ) x at x = 2.45 m
Moment Diag:
 dM

decreasing with V  , with “breaks” in
M is piecewise parabolic 
 dx

slope at C and F.
PROBLEM 7.87 CONTINUED
MC =
1
( 2.73 + 2.13) kN (1 m ) = 2.43 kN ⋅ m
2
M max = 2.43 kN ⋅ m +
M F = 4.231 kN ⋅ m −
1
(1.47 kN )( 2.45 m ) = 4.231 kN ⋅ m
2
1
( 0.93 kN )(1.55 m ) = 3.51 kN ⋅ m
2
PROBLEM 7.88
Two loads are suspended as shown from cable ABCD. Knowing
that dC = 1.5 ft, determine (a) the distance dB, (b) the components
of the reaction at A, (c) the maximum tension in the cable.
SOLUTION
ΣM A = 0: (10 ft ) Dy − 8 ft ( 450 lb ) − 4 ft ( 600 lb ) = 0
FBD cable:
Dy = 600 1b
ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0
Ay = 450 lb
ΣFx = 0: Ax − Dx = 0
FBD pt D:
(1)
600 lb
D
T
= x = CD : Dx = 800 lb
3
4
5
= Ax
So
TCD = 1000 lb
And
800 lb 450 lb
=
dB
4 ft
FBD pt A:
d B = 2.25 ft
(a)
(b)
A x = 800 lb
A y = 450 lb
TAB =
So
(800 lb )2 + ( 450 lb )2
(c)
= 918 lb
Tmax = TCD = 1000 lb
Note: TCD is Tmax as cable slope is largest in section CD.
PROBLEM 7.89
Two loads are suspended as shown from cable ABCD. Knowing that the
maximum tension in the cable is 720 lb, determine (a) the distance dB,
(b) the distance dC.
SOLUTION
ΣM A = 0: (10 ft ) Dy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
FBD cable:
D y = 600 1b
ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0
A y = 450 lb
FBD pt D:
ΣFx = 0: Ax − Bx = 0
Since Ax = Bx ; And Dy > Ay , Tension TCD > TAB
So
TCD = Tmax = 720 lb
Dx =
( 720 lb )2 − ( 600 lb )2
dC
2ft
=
600 lb 398lb
FBD pt. A:
dB
4ft
=
450 lb 398lb
= 398 lb = Ax
dC = 3.015 ft
(a)
d B = 4.52 ft
(b)
dC = 3.02 ft
PROBLEM 7.90
Knowing that dC = 4 m, determine (a) the reaction at A, (b) the reaction
at E.
SOLUTION
(a) FBD cable:
ΣM E = 0:
( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN )
− ( 3 m ) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
FBD ABC:
ΣM C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax − ( 8 m ) Ay = 0
Ax − 8 Ay = −4.8 kN
(2)
Ax = 3.2 kN
Solving (1) and (2)
Ay = 1 kN
So A = 3.35 kN
(b)
cable:
17.35°
ΣFx = 0: − Ax + Ex = 0
Ex = Ax = 3.2 kN
ΣFy = 0: Ay − (1.2 + 0.8 + 1.2 ) kN + E y = 0
E y = 3.2 kN − Ay = ( 3.2 − 1) kN = 2.2 kN
So E = 3.88 kN
34.5°
PROBLEM 7.91
Knowing that dC = 2.25 m, determine (a) the reaction at A, (b) the
reaction at E.
SOLUTION
FBD Cable:
(a)
ΣM E = 0: ( 4 m )(1.2 kN ) + ( 8 m )( 0.8 kN )
+ (12 m )(1.2 kN ) − (3 m) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
ΣM C = 0: ( 4 m )(1.2 kN ) − ( 0.75 m ) Ax − ( 8 m ) Ay = 0
0.75 Ax + 8 Ay = 4.8 kN
FBD ABC:
Solving (1) and (2)
Ax =
(2)
32
kN,
3
Ay = − 0.4 kN
So
A = 10.67 kN
2.15°
Note: this implies d B < 3 m (in fact d B = 2.85 m)
(b) FBD cable:
ΣFx = 0: −
32
kN + Ex = 0
3
Ex =
32
kN
3
ΣFy = 0: − 0.4 kN − 1.2 kN − 0.8 kN − 1.2 kN + E y = 0
E y = 3.6 kN
E = 11.26 kN
18.65°
PROBLEM 7.92
Cable ABCDE supports three loads as shown. Knowing that
dC = 3.6 ft, determine (a) the reaction at E, (b) the distances
dB and dD.
SOLUTION
FBD Cable:
ΣM A = 0: ( 2.4 ft ) Ex + ( 8 ft ) E y − ( 2 ft )( 360 )
(a)
− ( 4 ft )( 720 lb ) − ( 6 ft )( 240 lb ) = 0
0.3Ex + E y = 630 lb
(1)
ΣM C = 0: − (1.2 ft ) Ex + ( 4 ft ) E y − ( 2 ft )( 240 lb ) = 0
FBD CDE:
− 0.3Ex + E y = +120 lb
Solving (1) and (2)
Ex = 850 lb
(a)
(b) cable:
ΣFx = 0: − Ax + Ex = 0
(2)
E y = 375 lb
E = 929 lb
23.8°
Ax = Ex = 850 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + 375 lb = 0
Point A:
Ay = 945 lb
dB
945 lb
=
2 ft 850 lb
d B = 2.22 ft
PROBLEM 7.92 CONTINUED
ΣM D = 0: ( 2 ft )( 375 lb ) − ( d D − 2.4 ft )( 850 lb ) = 0
Segment DE:
d D = 3.28 ft
PROBLEM 7.93
Cable ABCDE supports three loads as shown. Determine (a) the
distance dC for which portion CD of the cable is horizontal, (b) the
corresponding reactions at the supports.
SOLUTION
Segment DE:
ΣFy = 0: E y − 240 lb = 0
E y = 240 lb
ΣM A = ( 2.4 ft ) Ex + ( 8 ft )( 240 lb ) − ( 6 ft )( 240 lb )
FBD Cable:
− ( 4 ft )( 720 lb ) − ( 2 ft )( 360 lb ) = 0
E x = 1300 lb
So
From Segment DE:
ΣM D = 0: ( 2 ft ) E y − ( dC − 2.4 ft ) Ex = 0
dC = 2.4 ft +
Ey
Ex
( 2 ft ) = ( 2.4 ft ) +
240 lb
( 2 ft ) = 2.7692 ft
1300 lb
(a)
dC = 2.77 ft
From FBD Cable:
ΣFx = 0: − Ax + Ex = 0
A x = 1300 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + E y = 0
A y = 1080 lb
(b)
A = 1.690 kips
39.7°
E = 1.322 kips
10.46°
PROBLEM 7.94
An oil pipeline is supported at 6-m intervals by vertical hangers
attached to the cable shown. Due to the combined weight of the
pipe and its contents, the tension in each hanger is 4 kN. Knowing
that dC = 12 m, determine (a) the maximum tension in the cable,
(b) the distance d D .
SOLUTION
FBD Cable:
Note: A y and Fy shown are forces on cable, assuming the 4 kN loads
at A and F act on supports.
ΣM F = 0: ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN ) 
− ( 30 m ) Ay − ( 5 m ) Ax = 0
Ax + 6 Ay = 48 kN
FBD ABC:
(1)
ΣM C = 0: ( 6 m )( 4 kN ) + ( 7 m ) Ax − (12 m ) Ay = 0
7 Ax − 12 Ay = −24 kN
Solving (1) and (2)
A x = 8 kN
(2)
20
kN
3
Ay =
From FBD Cable:
ΣFx = 0: − Ax + Fx = 0
Fx = Ax = 8 kN
ΣFy = 0: Ay − 4 ( 4 kN ) + Fy = 0
20 
28

Fy = 16 kN − Ay = 16 −
kN > Ay
 kN =
3 
3

So
TEF > TAB
Tmax = TEF =
Fx2 + Fy2
FBD DEF:
(a)
Tmax =
(18 kN )
2
2
 28

+
kN  = 12.29 kN
 3

 28

ΣM D = 0: (12 m ) 
kN  − d D ( 8 kN ) − ( 6 m )( 4 kN ) = 0
 3

(b)
d D = 11.00 m
PROBLEM 7.95
Solve Prob. 7.94 assuming that dC = 9 m.
SOLUTION
Note: 4 kN loads at A and F act directly on supports, not on cable.
FBD Cable:
ΣM A = 0: ( 30 m ) Fy − ( 5 m ) Fx
− ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN )  = 0
Fx − 6 Fy = −48 kN
(1)
ΣM C = 0: (18 ) Fy − ( 9 m ) Fx − (12 m )( 4 kN ) − ( 6 m )( 4 kN ) = 0
FBD CDEF:
Fx − 2Fy = −8 kN
(2)
Fx = 12 kN
Solving (1) and (2)
TEF =
Fy = 10 kN
(10 kN )2 + (12 kN )2
= 15.62 kN
Since slope EF > slope AB this is Tmax
Tmax = 15.62 kN
(a)
Also could note from FBD cable
ΣFy = 0: Ay + Fy − 4 ( 4 kN ) = 0
Ay = 16 kN − 12 kN = 4 kN
Thus
FBD DEF:
(b)
Ay < Fy
and
TAB < TEF
ΣM D = 0: (12 m )(10 kN ) − d D (12 kN ) − ( 6 m )( 4 kN ) = 0
d D = 8.00 m
PROBLEM 7.96
Cable ABC supports two boxes as shown. Knowing that b = 3.6 m,
determine (a) the required magnitude of the horizontal force P,
(b) the corresponding distance a.
SOLUTION
(
)
W = ( 8 kg ) 9.81 m/s 2 = 78.48 N
FBD BC:
ΣM A = 0: ( 3.6 m ) P − ( 2.4 m )
3W
− aW = 0
2
a 

P = W 1 +

3.6 m 

ΣFx = 0: −T1x + P = 0
ΣFy = 0: T1y − W −
T1 y
But
T1x
=
3
W =0
2
2.8 m
a
P=
So
T1x = P
5W
2
T1 y =
5W
2.8 m
=
a
2P
so
5Wa
5.6 m
(2)
a = 1.6258 m,
Solving (1) and (2):
(1)
P = 1.4516W
(a)
P = 1.4516 ( 78.48 ) = 113.9 N
(b)
a = 1.626 m
PROBLEM 7.97
Cable ABC supports two boxes as shown. Determine the distances
a and b when a horizontal force P of magnitude 100 N is applied at C.
SOLUTION
FBD pt C:
Segment BC:
2.4 m − a
0.8 m
=
100 N
117.72 N
a = 1.7204 m
a = 1.720 m
ΣM A = 0: b (100 N ) − ( 2.4 m )(117.72 N )
2

− (1.7204 m )  117.72 N  = 0
3

b = 4.1754 m
b = 4.18 m
PROBLEM 7.98
Knowing that WB = 150 lb and WC = 50 lb, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0
3Dx = 4 Dy
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 50 lb ) = 0
FBD BCD:
2D y − Dx = 60 lb
Solving (1) and (2)
FBD Cable:
D x = 120 lb
(2)
D y = 90 lb
ΣM A = 0: ( 42 ft )( 90 lb ) − ( 30 ft )( 50 lb )
− (12 ft )(150 lb ) − (15 ft ) P = 0
P = 32.0 lb
PROBLEM 7.99
Knowing that WB = 40 lb and WC = 22 lb, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0
4D y = 3Dx
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 22 lb ) = 0
FBD BCD:
10 Dy − 5Dx = 132 lb
Solving (1) and (2)
D x = 52.8 lb
(2)
D y = 39.6 lb
FBD Whole:
ΣM A = 0: ( 42 ft )( 39.6 lb ) − ( 30 ft )( 22 lb )
− (12 ft )( 40 lb ) − (15 ft ) P = 0
P = 34.9 lb
PROBLEM 7.100
If a = 4.5 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
By symmetry:
FBD pt C:
TBC = TCD = T
 1 
ΣFy = 0: 2 
T  − 180 kN = 0
 5 
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
Segment DE:
ΣM E = 0: ( 7.5 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 108.0 kN
Segment AB:
ΣM A = 0: ( 4.5 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0
Q = 45.0 kN
PROBLEM 7.101
If a = 6 m, determine the magnitudes of P and Q required to maintain the
cable in the shape shown.
SOLUTION
FBD pt C:
TBC = TCD = T
By symmetry:
 1 
ΣFy = 0: 2 
T  − 180 kN = 0
 5 
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
FBD DE:
ΣM E = 0: ( 9 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 120.0 kN
FBD AB:
ΣM A = 0: ( 6 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0
Q = 90.0 kN
PROBLEM 7.102
A transmission cable having a mass per unit length of 1 kg/m is strung
between two insulators at the same elevation that are 60 m apart. Knowing
that the sag of the cable is 1.2 m, determine (a) the maximum tension in
the cable, (b) the length of the cable.
SOLUTION
(a) Since h = 1.2 m L = 30 m we can approximate the load as evenly distributed in the horizontal
direction.
(
)
w = 1 kg/m 9.81 m/s 2 = 9.81 N/m.
w = ( 60 m )( 9.81 N/m )
w = 588.6 N
Also we can assume that the weight of half the cable acts at the
1
chord point.
4
FBD half-cable:
ΣM B = 0: (15 m )( 294.3 N ) − (1.2 m ) Tmin = 0
Tmin = 3678.75 N = Tmax
ΣFy = 0: Tmax
Tmax
y
y
x
− 294.3 N = 0
= 294.3 N
Tmax = 3690.5 N
Tmax = 3.69 kN
(b)
2
4


2 y 
2 y 
sB = xB 1 +  B  −  B  + "
3  xB 
5  xB 




2
4


2  1.2 
2  1.2 
= ( 30 m ) 1 + 
 − 
 + " = 30.048 m
3  30 
5  30 


so
s = 2sB = 60.096 m
s = 60.1 m
Note: The more accurate methods of section 7.10, which assume the load is evenly distributed along the length
instead of horizontally, yield Tmax = 3690.5 N and s = 60.06 m. Answers agree to 3 digits at least.
PROBLEM 7.103
Two cables of the same gauge are attached to a transmission tower at B.
Since the tower is slender, the horizontal component of the resultant of
the forces exerted by the cables at B is to be zero. Knowing that the mass
per unit length of the cables is 0.4 kg/m, determine (a) the required sag h,
(b) the maximum tension in each cable.
SOLUTION
Half-cable FBDs:
T1x = T2 x to create zero horizontal force on tower → thus T01 = T02
ΣM B = 0: (15 m )  w ( 30 m )  − h1T0 = 0
FBD I:
( 450 m ) w
=
2
h1
FBD II:
T0
ΣM B = 0: ( 2 m ) T0 − (10 m )  w ( 20 m )  = 0
T0 = (100 m ) w
( 450 m ) w = 4.50 m
=
2
h1
(a)
ΣFx = 0: T1x − T0 = 0
T1x = (100 m ) w
FBD I:
ΣFy = 0: T1y − ( 30 m ) w = 0
T1y = ( 30 m ) w
T1 =
(100 m )2 + ( 30 m )2 w
(
= (104.4 m )( 0.4 kg/m ) 9.81 m/s 2
= 409.7 N
)
(100 m ) w
PROBLEM 7.103 CONTINUED
ΣFy = 0: T2y − ( 20 m ) w = 0
FBD II:
T2y = ( 20 m ) w
T2 x = T1x = (100 m ) w
T2 =
(100 m )2 + ( 20 m )2 w = 400.17 N
(b)
T1 = 410 N
T2 = 400 N
*
Since h L it is reasonable to approximate the cable weight as being distributed uniformly along the
horizontal. The methods of section 7.10 are more accurate for cables sagging under their own weight.
PROBLEM 7.104
The center span of the George Washington Bridge, as originally
constructed, consisted of a uniform roadway suspended from four
cables. The uniform load supported by each cable was w = 9.75
kips/ft along the horizontal. Knowing that the span L is 3500 ft and
that the sag h is 316 ft, determine for the original configuration (a)
the maximum tension in each cable, (b) the length of each cable.
SOLUTION
W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips
FBD half-span:
ΣM B = 0: ( 875 ft )(17, 065 kips ) − ( 316 ft ) T0 = 0
T0 = 47, 246 kips
Tmax = T02 + W 2 =
( 47, 246 kips )2 + (17, 063 kips )2
(a)
Tmax = 50, 200 kips
2
4


2 y
2 y
s = x 1 +   −   + "
3 x 
5 x


2
4


2  316 ft 
2  316 ft 
−
sB = (1750 ft ) 1 + 


 + "
3  1750 ft 
5  1750 ft 


= 1787.3 ft
(b)
*
To get 3-digit accuracy, only two terms are needed.
l = 2sB = 3575 ft
PROBLEM 7.105
Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft
along the horizontal. Knowing that the span L is 4150 ft and that the
sag h is 464 ft, determine (a) the maximum tension in each cable,
(b) the length of each cable.
SOLUTION
FBD half-span:
(a)
 2075 ft 
ΣM B = 0: 
 ( 23032.5 kips ) − ( 464 ft ) T0 = 0
 2 
T0 = 47, 246 kips
Tmax = T02 + W 2 =
(b)
( 47, 246 kips )2 + ( 23,033 kips )2
2

2 y
2
s = x 1 +   − 
3
5
x
 


= 56, 400 kips
4

y
 + "
x

2
4


2  464 ft 
2  464 ft 
sB = ( 2075 ft ) 1 + 
 − 
 + "
3  2075 ft 
5  2075 ft 


sB = 2142 ft
l = 2s B
l = 4284 ft
PROBLEM 7.106
To mark the positions of the rails on the posts of a fence, a homeowner
ties a cord to the post at A, passes the cord over a short piece of pipe
attached to the post at B, and ties the free end of the cord to a bucket
filled with bricks having a total mass of 20 kg. Knowing that the mass
per unit length of the rope is 0.02 kg/m and assuming that A and B are
at the same elevation, determine (a) the sag h, (b) the slope of the cable
at B. Neglect the effect of friction.
SOLUTION
FBD pulley:
ΣM P = 0: (Tmax − WB ) r = 0
Tmax = WB = 196.2 N
FBD half-span:*
2
T0 = Tmax
−W2 =
(196.2 N )2 − ( 4.91 N )2
= 196.139 N
 25 m 
ΣM B = 0: 
 ( 4.905 N ) − h (196.139 N ) = 0
 2 
h = 0.3126 m = 313 mm
(a)
(b)
*See note Prob. 7.103
θ B = sin −1
W
 4.905 N 
= sin −1 
 = 1.433°
Tmax
 196.2 N 
PROBLEM 7.107
A small ship is tied to a pier with a 5-m length of rope as shown.
Knowing that the current exerts on the hull of the ship a 300-N force
directed from the bow to the stern and that the mass per unit length of the
rope is 2.2 kg/m, determine (a) the maximum tension in the rope,
(b) the sag h. [Hint: Use only the first two terms of Eq. (7.10).]
SOLUTION
(a) FBD ship:
ΣFx = 0: T0 − 300 N = 0
T0 = 300 N
FBD half-span:*
Tmax = T02 + W 2 =
(b)
ΣM A = 0: hTx −
L
W =0
4
2


2 4
s = x 1 +   + "
3 x


L
( 2.5 m ) =
2
but
= ( 54 N ) = 305 N
2
LW
4Tx
yA = h =
LW
4Tx
so
yA
W
=
2Tx
xA
2


2  53.955 N 
1 + 
 − " → L = 4.9732 m
3  600 N 


So h =
*See note Prob. 7.103
h=
( 300 N )2
LW
= 0.2236 m
4Tx
h = 224 mm
PROBLEM 7.108
The center span of the Verrazano-Narrows Bridge consists of two
uniform roadways suspended from four cables. The design of the bridge
allowed for the effect of extreme temperature changes which cause the
sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in
summer. Knowing that the span is L = 4260 ft, determine the change in
length of the cables due to extreme temperature changes.
SOLUTION
2

2 y
2
s = x 1 +   − 
3
x
5
 


Knowing
l = 2sTOT
4

y
 + "
x

2
2


2 h 
2 h 
= L 1 + 
 − 
 + "
3  L/2 
5  L/2 




Winter:
2
4


2  386 ft 
2  386 ft 
lw = ( 4260 ft ) 1 + 
"
−
+
 = 4351.43 ft



3  2130 ft 
5  2130 ft 


Summer:
2
4


2  394 ft 
2  394 ft 
ls = ( 4260 ft )  1 + 
 − 
 + " = 4355.18 ft
3  2130 ft 
5  2130 ft 


∆l = ls − lw = 3.75 ft
PROBLEM 7.109
A cable of length L + ∆ is suspended between two points which are at
the same elevation and a distance L apart. (a) Assuming that ∆ is small
compared to L and that the cable is parabolic, determine the approximate
sag in terms of L and ∆ . (b) If L = 30 m and ∆ = 1.2 m, determine the
approximate sag. [Hint: Use only the first two terms of Eq. (7.10).]
SOLUTION
(a)
2


2 y
s = x 1 +   − "
3 x


L + ∆ = 2sTOT
2


2 h 
= L 1 + 
 − "
3  L/2 




2
2
∆
2  2h 
8 h 
=   =   →h=
L
3 L 
3 L 
(b)
For
L = 30 m,
∆ = 1.2 m
3
L∆
8
h = 3.67 m
PROBLEM 7.110
Each cable of the side spans of the Golden Gate Bridge supports a load
w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the
maximum vertical distance h from each cable to the chord AB is 30 ft and
occurs at midspan, determine (a) the maximum tension in each cable,
(b) the slope at B.
SOLUTION
FBD AB:
ΣM A = 0: (1100 ft ) TBy − ( 496 ft ) TBx − ( 550 ft )W = 0
11TBy − 4.96TBx = 5.5W
(1)
FBD CB:
ΣM C = 0: ( 550 ft ) TBy − ( 278 ft ) TBx − ( 275 ft )
W
=0
2
11TBy − 5.56TBx = 2.75W
Solving (1) and (2)
TBy = 28,798 kips
Solving (1) and (2)
TBx = 51, 425 kips
Tmax = TB = TB2x + TB2y
So that
(2)
tan θ B =
TBy
TBx
(a)
Tmax = 58,900 kips
(b)
θ B = 29.2°
PROBLEM 7.111
A steam pipe weighting 50 lb/ft that passes between two buildings 60 ft
apart is supported by a system of cables as shown. Assuming that the
weight of the cable is equivalent to a uniformly distributed loading of 7.5
lb/ft, determine (a) the location of the lowest point C of the cable, (b) the
maximum tension in the cable.
SOLUTION
FBD AC:
FBD CB:
ΣM A = 0: (13.5 ft ) T0 −
a
( 57.5 lb/ft ) a = 0
2
(
)
T0 = 2.12963 lb/ft 2 a 2
(1)
60 ft − a
( 57.5 lb/ft )( 60 ft − a ) − ( 6 ft ) T0 = 0
2
ΣM B = 0:
(
)
6T0 = 28.75 lb/ft 2 3600 ft 2 − (120 ft ) a + a 2 
Using (1) in (2)
(2)
0.55a 2 − (120 ft ) a + 3600 ft 2 = 0
Solving: a = (108 ± 72 ) ft
a = 36 ft (180 ft out of range)
So
(a)
C is 36 ft from A
C is 6 ft below and 24 ft left of B
T0 = 2.1296 lb/ft 2 ( 36 ft ) = 2760 lb
2
W1 = ( 57.5 lb/ft )( 36 ft ) = 2070 lb
(b)
Tmax = TA = T02 + W12 =
( 2760 lb )2 + ( 2070 lb )2
= 3450 lb
PROBLEM 7.112
Chain AB supports a horizontal, uniform steel beam having a mass
per unit length of 85 kg/m. If the maximum tension in the cable is
not to exceed 8 kN, determine (a) the horizontal distance a from A
to the lowest point C of the chain, (b) the approximate length of the
chain.
SOLUTION
ΣM A = 0: y AT0 −
yA =
a
wa = 0
2
ΣM B = 0: − yBT0 +
wa 2
2T0
yB =
d = ( y B − yB ) =
Using
L = 6 m,
yields
w 2
b − a2
2T0
)
2
2
2
− ( wb ) 
( 2d )2 Tmax

or
wb 2
2T0
2
T0 = TB2 − ( wb ) = Tmax
− ( wb )
But
∴
(
b
wb = 0
2
(
= w2 b 2 − a 2
)
2
(
= L2 w2 4b 2 − 4 Lb + L2

T2
4 L2 + d 2 b 2 − 4L3b +  L4 − 4d 2 max
w2

(
d = 0.9 m,
)
Tmax = 8 kN,
b = ( 2.934 ± 1.353) m
2
)

 = 0

(
)
w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m
b = 4.287 m
( since b > 3 m )
(a)
a = 6 m − b = 1.713 m
PROBLEM 7.112 CONTINUED
2
T0 = Tmax
− ( wb ) = 7156.9 N
2
yA
wa
=
= 0.09979
xA
2T0
yB
wb
=
= 0.24974
xB
2T0
2
2




2  yA 
2  yB 



l = s A + sB = a 1 + 
 +" + b 1+ 
 + "
3  xA 
3  xB 








2
2
2
2


= (1.713 m ) 1 + ( 0.09979 )  + ( 4.287 m ) 1 + ( 0.24974 )  = 6.19 m
3
3




(b)
l = 6.19 m
PROBLEM 7.113
Chain AB of length 6.4 m supports a horizontal, uniform steel beam
having a mass per unit length of 85 kg/m. Determine (a) the horizontal
distance a from A to the lowest point C of the chain, (b) the maximum
tension in the chain.
SOLUTION
ΣM P = 0:
Geometry:
y =
wx 2
2T0
x
wx − yT0 = 0
2
y
wx
=
x
2T0
so
and d = yB − y A =
FBD Segment:
(
w 2
b − a2
2T0
)
2
2


2 y  
2 y  
l = s A + sB = a 1 +  A   + b 1 +  B  
3  a  
3  b  


Also
2  y A 
 yB 

 +

3  a 
 b 
2
l−L=
1
4d 2
=
6 b2 − a 2
(
)
2
(a
3
2
(
w2 3
a + b3
 =
2
 6T0
+b
3
)
(
)
2
3
3
2d a +b
=
3 b2 − a 2 2
(
)
)
Using l = 6.4 m, L = 6 m, d = 0.9 m, b = 6 m − a, and solving for a,
knowing that a < 3 ft
a = 2.2196 m
(a)
T0 =
Then
(
w 2
b − a2
2d
(
a = 2.22 m
)
)
w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m
And with
b = 6 m − a = 3.7804 m
And
Tmax = TB = T02 + ( wb )
=
T0 = 4338 N
2
( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2
Tmax = 5362 N
(b)
Tmax = 5.36 kN
PROBLEM 7.114
A cable AB of span L and a simple beam A′B′ of the same span
are subjected to identical vertical loadings as shown. Show that the
magnitude of the bending moment at a point C ′ in the beam is
equal to the product , T0h where T0 is the magnitude of the horizontal
component of the tension force in the cable and h is the vertical
distance between point C and the chord joining the points of support
A and B.
SOLUTION
FBD Cable:
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
(Where ΣM B loads includes all applied loads)
x

ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0
L


FBD AC:
(Where ΣM C left includes all loads left of C)
x
(1) − ( 2 ):
L
FBD Beam:
(2)
hT0 −
x
ΣM B loads + ΣM C left = 0
L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
ΣM C = 0: xABy − ΣM C left − M C = 0
(5)
FBD AC:
x
( 4 ) − ( 5):
L
Comparing (3) and (6)
−
x
ΣM B loads + ΣM C left + M C = 0
L
M C = hT0
Q.E.D.
(6)
PROBLEM 7.115
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.89a.
SOLUTION
FBD Beam:
ΣM D = 0: ( 2 ft )( 450 lb ) + ( 6 ft )( 600 lb ) − (10 ft ) A By = 0
A By = 450 lb
ΣM B = 0: M B − ( 4 ft )( 450 lb ) = 0
Section AB:
M B = 1800 lb ⋅ ft
ΣM A = 0: (10 ft ) DCy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
Cable:
DCy = 600 lb
( Note: Dy > Ay so Tmax = TCD )
2
2
T0 = Tmax
− DCy
T0 =
( 720 lb )2 − ( 600 lb )2
T0 = 398 lb
dB =
M B 1800 lb ⋅ ft
=
= 4.523 ft
T0
398 lb
d B = 4.52 ft
PROBLEM 7.116
Making use of the property established in Prob. 7.114, solve the problem
indicated by first solving the corresponding beam problem. Prob. 7.92b.
SOLUTION
FBD Beam:
ΣM E = 0: ( 2 ft )( 240 lb ) + ( 4 ft )( 720 lb )
+ ( 6 ft )( 360 lb ) − ( 8 ft ) ABy = 0
A By = 690 lb
ΣM B = 0: M B − ( 2 ft )( 690 lb ) = 0
M B = 1380 lb ⋅ ft
ΣM B = 0: M C + ( 2 ft )( 360 lb ) − ( 4 ft )( 690 lb ) = 0
M C = 2040 lb ⋅ ft
ΣM D = 0: M D + ( 2 ft )( 720 lb ) + ( 4 ft )( 360 lb )
− ( 6 ft )( 690 lb ) = 0
M D = 1260 lb ⋅ ft
hC = dC − 1.2 ft = 3.6 ft − 1.2 ft = 2.4 ft
T0 =
Cable:
MC
2040 lb ⋅ ft
=
= 850 lb
hC
2.4 ft
M B 1380 lb ⋅ ft
=
= 1.6235 ft
T0
850 lb
hB =
d B = hB + 0.6 ft
h0 =
d B = 2.22 ft
M D 1260 lb ⋅ ft
=
= 1.482 ft
T0
850 lb
d B = h0 + 1.8 ft
d D = 3.28 ft
PROBLEM 7.117
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.94b.
SOLUTION
FBD Beam:
A By = F = 8 kN
By symmetry:
M B = M E;
MC = M D
AC:
ΣM C = 0: M C + ( 6 m )( 4 kN ) − (12 m )( 8 kN ) = 0
M C = 72 kN ⋅ m
so
M D = 72 kN ⋅ m
Cable:
Since
M D = MC
hD = hC = 12 m − 3 m = 9 m
d D = hD + 2 m = 11 m
d D = 11.00 m
PROBLEM 7.118
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.95b.
SOLUTION
FBD Beam:
By symmetry: M B = M E
and
MC = M D
Cable:
Since
M D = M C , hD = hC
hD = hC = dC − 3 m = 9 m − 3 m = 6 m
Then
d D = hD + 2 m = 6 m + 2 m = 8 m
d D = 8.00 m
PROBLEM 7.119
Show that the curve assumed by a cable that carries a distributed load
w( x) is defined by the differential equation d 2 y / dx 2 = w( x ) / T0 ,
where T0 is the tension at the lowest point.
SOLUTION
ΣFy = 0: Ty ( x + ∆x ) − Ty ( x ) − w ( x ) ∆ x = 0
FBD Elemental segment:
Ty ( x + ∆x )
So
T0
Ty ( x )
−
T0
Ty
But
T0
dy
dx
So
−
x + ∆x
∆x
In
lim :
∆x → 0
=
dy
dx
x
=
w( x)
∆x
T0
dy
dx
=
w( x)
T0
w( x)
d2y
=
2
T0
dx
Q.E.D.
PROBLEM 7.120
Using the property indicated in Prob. 7.119, determine the curve
assumed by a cable of span L and sag h carrying a distributed load
w = w0 cos(π x / L ) , where x is measured from mid-span. Also
determine the maximum and minimum values of the tension in
the cable.
SOLUTION
w ( x ) = w0 cos
πx
L
From Problem 7.119
w( x)
πx
d2y
w
=
= 0 cos
2
T0
T0
L
dx
So

dy
 using
dx

dy W0 L
πx
=
sin
dx
T0π
L
y =
But
And
0

= 0

w0 L2 
πx
1 − cos
  using y ( 0 ) = 0 
2 
L  
T0π 
w L2 
π
L
y   = h = 0 2 1 − cos 
2
T0π 
2
T0 =
so
T0 = Tmin
w0 L2
π 2h
Tmin =
so
Tmax = TA = TB :
TBy
TBy =
w0 L
T0
=
dy
dx
=
x = L2
w0 L2
π 2h
w0 L
T0π
π
2
TB = TBy
+ T02 =
w0 L
π
 L 
1+ 

πh 
2
PROBLEM 7.121
If the weight per unit length of the cable AB is w0 / cos 2 θ , prove
that the curve formed by the cable is a circular arc. (Hint: Use the
property indicated in Prob. 7.119.)
SOLUTION
Elemental Segment:
Load on segment*
But
dx = cosθ ds,
w0
ds
cos 2 θ
w0
w( x) =
cos3 θ
w ( x ) dx =
so
From Problem 7.119
d2y
w( x)
w0
=
=
T0
dx 2
T0 cos3 θ
In general
d2y
d  dy 
d
dθ
=
( tan θ ) = sec2 θ
 =
2
dx  dx  dx
dx
dx
So
dθ
w0
w0
=
=
3
2
dx
T0 cosθ
T0 cos θ sec θ
or
T0
cosθ dθ = dx = rdθ cosθ
w0
Giving r =
T0
= constant. So curve is circular arc
w0
*For large sag, it is not appropriate to approximate ds by dx.
Q.E.D.
PROBLEM 7.122
Two hikers are standing 30-ft apart and are holding the ends of a
35-ft length of rope as shown. Knowing that the weight per unit length
of the rope is 0.05 lb/ft, determine (a) the sag h, (b) the magnitude
of the force exerted on the hand of a hiker.
SOLUTION
Half-span:
w = 0.05 lb/ft,
L = 30 ft,
sB = c sinh
sB =
35
ft
2
yB
xB
 15 ft 
17.5 ft = c sinh 

 c 
c = 15.36 ft
Solving numerically,
Then
yB = c cosh
xB
15 ft
= (15.36 ft ) cosh
= 23.28 ft
c
15.36 ft
(a)
hB = yB − c = 23.28 ft − 15.36 ft = 7.92 ft
(b)
TB = wyB = ( 0.05 lb/ft )( 23.28 ft ) = 1.164 lb
PROBLEM 7.123
A 60-ft chain weighing 120 lb is suspended between two points at
the same elevation. Knowing that the sag is 24 ft, determine (a) the
distance between the supports, (b) the maximum tension in the chain.
SOLUTION
sB = 30 ft,
w=
hB = 24 ft,
120 lb
= 2 lb/ft
60 ft
xB =
yB2 = c 2 + sB2 = ( hB + c )
L
2
2
= hB2 + 2chB + c 2
( 30 ft ) − ( 24 ft )
sB2 − hB2
=
2hB
2 ( 24 ft )
2
c=
2
c = 6.75 ft
Then
sB = c sinh
xB
s
→ xB = c sinh −1 B
c
c
 30 ft 
xB = ( 6.75 ft ) sinh −1 
 = 14.83 ft
 6.75 ft 
(a)
L = 2 xB = 29.7 ft
Tmax = TB = wyB = w ( c + hB ) = ( 2 lb/ft )( 6.75 ft + 24 ft ) = 61.5 lb
(b)
Tmax = 61.5 lb
PROBLEM 7.124
A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched
between two points at the same elevation and pulled until the tension
at each end is 16 lb, determine the horizontal distance between the
ends of the tape. Neglect the elongation of the tape due to the tension.
SOLUTION
sB = 100 ft,
w=
4 lb
= 0.02 lb/ft
200 ft
Tmax = 16 lb
Tmax = TB = wyB
yB =
TB
16 lb
=
= 800 ft
w
0.02 lb/ft
c 2 = yB2 − sB2
c=
But
(800 ft )2 − (100 ft )2
yB = xB cosh
= 793.73 ft
xB
y
→ xB = c cosh −1 B
c
c
 800 ft 
= ( 793.73 ft ) cosh −1 
 = 99.74 ft
 793.73 ft 
L = 2 xB = 2 ( 99.74 ft ) = 199.5 ft
PROBLEM 7.125
An electric transmission cable of length 130 m and mass per unit
length of 3.4 kg/m is suspended between two points at the same
elevation. Knowing that the sag is 30 m, determine the horizontal
distance between the supports and the maximum tension.
SOLUTION
sB = 65 m,
(
hB = 30 m
)
w = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m
yB2 = c 2 + s 2B
( c + hB )2
= c 2 + sB2
( 65 m ) − ( 30 m )
sB2 − hB2
=
2hB
2 ( 30 m )
2
c=
2
= 55.417 m
Now
sB = c sinh
xB
s
 65 m 
→ xB = c sinh −1 B = ( 55.417 m ) sinh −1 

c
c
 55.417 m 
= 55.335 m
L = 2 xB = 2 ( 55.335 m ) = 110.7 m
Tmax = wyB = w ( c + hB ) = ( 33.35 N/m )( 55.417 m + 30 m ) = 2846 N
Tmax = 2.85 kN
PROBLEM 7.126
A 30-m length of wire having a mass per unit length of 0.3 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the force P for which h = 12 m,
(b) the corresponding span L.
SOLUTION
FBD Cable:
s = 30 m
30 m


= 15 m 
 so sB =
2


(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
hB = 12 m
yB2 = ( c + hB ) = c 2 + s 2B
2
c=
So
c=
Now
sB = c sinh
sB2 − hB2
2hB
(15 m )2 − (12 m )2
2 (12 m )
= 3.375 m
xB
s
 15 m 
→ xB = c sinh −1 B = ( 3.375 m ) sinh −1 

c
c
 3.375 m 
xB = 7.4156 m
P = T0 = wc = ( 2.943 N/m )( 3.375 m )
L = 2 xB = 2 ( 7.4156 m )
(a)
(b)
P = 9.93 N
L = 14.83 m
PROBLEM 7.127
A 30-m length of wire having a mass per unit length of 0.3 kg/m is
attached to a fixed support at A and to a collar at B. Knowing that the
magnitude of the horizontal force applied to the collar is P = 30 N,
determine (a) the sag h, (b) the corresponding span L.
SOLUTION
FBD Cable:
sT = 30 m,
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
P = T0 = wc
c=
c=
P
w
30 N
= 10.1937 m
2.943 N/m
yB2 = ( hB + c ) = c 2 + sB2
2
h 2 + 2ch − sB2 = 0
sB =
30 m
= 15 m
2
h 2 + 2 (10.1937 m ) h − 225 m 2 = 0
h = 7.9422 m
sB = c sinh
(a)
h = 7.94 m
xA
s
 15 m 
→ xB = c sinh −1 B = (10.1937 m ) sinh −1 

c
c
 10.1937 m 
= 12.017 m
L = 2 xB = 2 (12.017 m )
(b)
L = 24.0 m
PROBLEM 7.128
A 30-m length of wire having a mass per unit length of 0.3 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the sag h for which L = 22.5 m,
(b) the corresponding force P.
SOLUTION
FBD Cable:
sT = 30 m → sB =
30 m
= 15 m
2
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
L = 22.5 m
sB = c sinh
xB
L/2
= c sinh
c
c
15 m = c sinh
11.25 m
c
Solving numerically: c = 8.328 m
yB2 = c 2 + sB2 = ( 8.328 m ) + (15 m ) = 294.36 m 2
2
2
yB = 17.157 m
hB = yB − c = 17.157 m − 8.328 m
(a)
P = wc = ( 2.943 N/m )( 8.328 m )
(b)
hB = 8.83 m
P = 24.5 N
PROBLEM 7.129
A 30-ft wire is suspended from two points at the same elevation
that are 20 ft apart. Knowing that the maximum tension is 80 lb,
determine (a) the sag of the wire, (b) the total weight of the wire.
SOLUTION
L = 20 ft
sB =
xB =
30 ft
= 15 ft
2
sB = c sinh
xB
10 ft
= c sinh
c
c
Solving numerically:
yB = c cosh
20 ft
= 10 ft
2
c = 6.1647 ft
xB
 10 ft 
= ( 6.1647 ft ) cosh 

c
 1.1647 ft 
yB = 16.217 ft
hB = yB − c = 16.217 ft − 6.165 ft
Tmax = wyB
So
W =
and
(a)
hB = 10.05 ft
(b)
Wm = 148.0 lb
W = w ( 2 sB )
Tmax
80 lb
( 2sB ) =
( 30 ft )
yB
16.217 ft
PROBLEM 7.130
Determine the sag of a 45-ft chain which is attached to two points at
the same elevation that are 20 ft apart.
SOLUTION
sB =
45 ft
= 22.5 ft
2
xB =
L = 20 ft
L
= 10 ft
2
sB = c sinh
xB
c
22.5 ft = c sinh
Solving numerically:
yB = c cosh
10 ft
c
c = 4.2023 ft
xB
c
= ( 4.2023 ft ) cosh
10 ft
= 22.889 ft
4.2023 ft
hB = yB − c = 22.889 ft − 4.202 ft
hB = 18.69 ft
PROBLEM 7.131
A 10-m rope is attached to two supports A and B as shown.
Determine (a) the span of the rope for which the span is equal to
the sag, (b) the corresponding angle θB.
SOLUTION
y = c cosh
We know
yB = c + h = c cosh
At B,
1 = cosh
or
sB = c sinh
c=
h
2c
h
h
−
2c c
h
= 4.933
c
Solving numerically
So
x
c
xB
s
h
→ T = c sinh
c
2
2c
sT
10 m
=
= 0.8550 m
 h 
 4.933 
2sinh   2 sinh 

 2c 
 2 
h = 4.933c = 4.933 ( 0.8550 ) m = 4.218 m
h = 4.22 m
(a)
From
x
y = c cosh ,
c
At B,
tan θ =
dy
dx
θ = tan −1 5.848
dy
x
= sinh
dx
c
= sinh
B
L = h = 4.22 m W
L
4.933
= sinh
= 5.848
2c
2
(b)
θ = 80.3° W
PROBLEM 7.132
A cable having a mass per unit length of 3 kg/m is suspended
between two points at the same elevation that are 48 m apart.
Determine the smallest allowable sag of the cable if the maximum
tension is not to exceed 1800 N.
SOLUTION
(
)
w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m
L = 48 m,
Tmax ≤ 1800 N
Tmax = wyB → yB =
yB ≤
yB = c cosh
xB
c
Tmax
w
1800 N
= 61.162 m
29.43 N/m
61.162 m = c cosh
Solving numerically
48m / 2
*
c
c = 55.935 m
h = yB − c = 61.162 m − 55.935 m
h = 5.23 m W
*Note: There is another value of c which will satisfy this equation. It is much smaller, thus
corresponding to a much larger h.
PROBLEM 7.133
An 8-m length of chain having a mass per unit length of 3.72 kg/m
is attached to a beam at A and passes over a small pulley at B as
shown. Neglecting the effect of friction, determine the values of
distance a for which the chain is in equilibrium.
SOLUTION
TB = wa
Neglect pulley size and friction
But
TB = wyB
so
yB = c cosh
c cosh
But sB = c sinh
So
yB = a
xB
c
1m
=a
c
8m − a
1m
= c sinh
c
2
xB
c
4 m = c sinh
(
1m c
1m
+ cosh
c
c
2
16 m = c 3e1/c − e−1/c
Solving numerically
)
c = 0.3773 m, 5.906 m
1m

( 0.3773 m ) cosh 0.3773 m = 2.68 m W
1m 
=
a = c cosh
c
 ( 5.906 m ) cosh 1 m = 5.99 m W

5.906 m
PROBLEM 7.134
A motor M is used to slowly reel in the cable shown. Knowing that
the weight per unit length of the cable is 0.5 lb/ft, determine the
maximum tension in the cable when h = 15 ft.
SOLUTION
w = 0.5 lb/ft
L = 30 ft
yB = c cosh
hB = 15 ft
xB
c
hB + c = c cosh
L
2c
15 ft


− 1
15 ft = c  cosh
c


Solving numerically c = 9.281 ft
yB = ( 9.281 ft ) cosh
15 ft
= 24.281 ft
9.281 ft
Tmax = TB = wyB = ( 0.5 lb/ft )( 24.281 ft )
Tmax = 12.14 lb W
PROBLEM 7.135
A motor M is used to slowly reel in the cable shown. Knowing that
the weight per unit length of the cable is 0.5 lb/ft, determine the
maximum tension in the cable when h = 9 ft.
SOLUTION
w = 0.5 lb/ft,
L = 30 ft,
yB = hB + c = c cosh
hB = 9 ft
xB
L
= c cosh
2c
c
15 ft


9 ft = c  cosh
− 1
c


Solving numerically c = 13.783 ft
yB = hB + c = 9 ft + 13.783 ft = 21.783 ft
Tmax = TB = wyB = ( 0.5 lb/ft )( 21.78 ft )
Tmax = 11.39 lb W
PROBLEM 7.136
To the left of point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the weight per unit length
of the cable is 1.5 lb/ft, determine the force F when a = 10.8 ft.
SOLUTION
yD = c cosh
xD
c
h + c = c cosh
a
c
10.8 m


− 1
12 m = c  cosh
c


Solving numerically
Then
c = 6.2136 m
yB = ( 6.2136 m ) cosh
10.8 m
= 18.2136 m
6.2136 m
F = Tmax = wyB = (1.5 lb/ft )(18.2136 m )
F = 27.3 lb
W
PROBLEM 7.137
To the left of point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the weight per unit length
of the cable is 1.5 lb/ft, determine the force F when a = 18 ft.
SOLUTION
yD = c cosh
xD
c
c + h = c cosh
a
c
a


h = c  cosh − 1
c


18 ft


12 ft = c  cosh
− 1
c


Solving numerically
c = 15.162 ft
yB = h + c = 12 ft + 15.162 ft = 27.162 ft
F = TD = wyD = (1.5 lb/ft )( 27.162 ft ) = 40.74 lb
F = 40.7 lb
W
PROBLEM 7.138
A uniform cable has a mass per unit length of 4 kg/m and is held in
the position shown by a horizontal force P applied at B. Knowing that
P = 800 N and θA = 60°, determine (a) the location of point B, (b) the
length of the cable.
SOLUTION
(
)
w = 4 kg/m 9.81 m/s 2 = 39.24 N/m
P = T0 = wc
c=
P
800 N
=
w 39.24 N/m
c = 20.387 m
y = c cosh
x
c
dy
x
= sinh
dx
c
tan θ = −
dy
dx
= − sinh
−a
−a
a
= sinh
c
c
a = c sinh −1 ( tan θ ) = ( 20.387 m ) sinh −1 ( tan 60° )
a = 26.849 m
y A = c cosh
a
26.849 m
= ( 20.387 m ) cosh
= 40.774 m
c
20.387 m
b = y A − c = 40.774 m − 20.387 m = 20.387 m
So
s = c sinh
(a)
a
26.849 m
= ( 20.387 m ) sinh
= 35.31 m
c
20.387 m
B is 26.8 m right and 20.4 m down from A W
(b)
s = 35.3 m W
PROBLEM 7.139
A uniform cable having a mass per unit length of 4 kg/m is held in
the position shown by a horizontal force P applied at B. Knowing
that P = 600 N and θA = 60°, determine (a) the location of point B,
(b) the length of the cable.
SOLUTION
(
)
w = ( 4 kg/m ) 9.81 m/s 2 = 39.24 N/m
P = T0 = wc
c=
P
600 N
=
w 39.24 N/m
c = 15.2905 m
y = c cosh
At A:
So
tan θ = −
dy
dx
x
c
dy
x
= sinh
dx
c
= − sinh
−a
−a
a
= sinh
c
c
a = c sinh −1 ( tan θ ) = (15.2905 m ) sinh −1 ( tan 60° ) = 20.137 m
a
c
yB = h + c = c cosh
a


h = c  cosh − 1
c




20.137 m
= (15.2905 m )  cosh
− 1
15.2905 m


= 15.291 m
So
s = c sinh
(a)
B is 20.1 m right and 15.29 m down from A
a
20.137 m
= (15.291 m ) sinh
= 26.49 m
c
15.291 m
(b)
s = 26.5 m
PROBLEM 7.140
The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of the
cable is located at a distance a = 1.8 ft below the support A, determine
(a) the location of the lowest point C, (b) the maximum tension in the
cable.
SOLUTION
y A = c cosh
−a
= c + 1.8 ft
c
1.8 ft 

a = c cosh −1 1 +

c 

yB = c cosh
b
= c + 7.2 ft
c
7.2 ft 

b = c cosh −1 1 +

c 

But

1.8 ft 
7.2 ft  

−1 
a + b = 36 ft = c cosh −1 1 +
 + cosh  1 +

c 
c  



Solving numerically
Then
c = 40.864 ft

7.2 ft 
b = ( 40.864 ft ) cosh −1  1 +
 = 23.92 ft
40.864 ft 

(a)
Tmax = wyB = ( 0.3 lb/ft )( 40.864 ft + 7.2 ft )
C is 23.9 ft left of and 7.20 ft below B W
(b)
Tmax = 14.42 lb W
PROBLEM 7.141
The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of
the cable is located at a distance a = 6 ft below the support A,
determine (a) the location of the lowest point C, (b) the maximum
tension in the cable.
SOLUTION
y A = c cosh
−a
= c + 6 ft
c
6 ft 

a = c cosh −1 1 +

c 

yB = c cosh
b
= c + 11.4 ft
c
11.4 ft 

b = c cosh −1 1 +

c 

So
Solving numerically

6 ft 
11.4 ft  

−1 
a + b = c cosh −1  1 +
 + cosh  1 +
 = 36 ft
c 
c  



c = 20.446 ft
11.4 ft 

b = ( 20.446 ft ) cosh −1 1 +
 = 20.696 ft
20.446 ft 

(a) C is 20.7 ft left of and 11.4 ft below B
 20.696 ft 
Tmax = wyB = ( 0.3 lb/ft )( 20.446 ft ) cosh 
 = 9.554 lb
 20.446 ft 
(b)
Tmax = 9.55 lb
PROBLEM 7.142
Denoting by θ the angle formed by a uniform cable and the horizontal,
show that at any point (a) s = c tan θ , (b) y = c sec θ .
SOLUTION
tan θ =
(a)
s = c sinh
(b)
Also
dy
x
= sinh
dx
c
x
= c tan θ Q.E.D.
c
(
)
y 2 = s 2 + c 2 cosh 2 x = sinh 2 x + 1
(
)
So
y 2 = c 2 tan 2 θ + 1 = c 2 sec2 θ
And
y = c sec θ Q.E.D.
PROBLEM 7.143
(a) Determine the maximum allowable horizontal span for a uniform
cable of mass per unit length m′ if the tension in the cable is not to
exceed a given value Tm . (b) Using the result of part a, determine the
maximum span of a steel wire for which m′ = 0.34 kg/m and
Tm = 32 kN.
SOLUTION
TB = Tmax = wyB
= wc cosh
Let
ξ =
L
2c
so
xB
L  2c 
L
= w   cosh
c
2 L 
2c
Tmax =
wL
cosh ξ
2ξ

dTmax
wL 
1
=
 sinh ξ − cosh ξ 
dξ
ξ
2ξ 

For
min Tmax ,
tanh ξ −
1
ξ
=0
Solving numerically ξ = 1.1997
(Tmax )min
=
wL
cosh (1.1997 ) = 0.75444wL
2 (1.9997 )
(a)
If
Lmax =
(
Tmax
T
= 1.3255 max
0.75444w
w
)
Tmax = 32 kN and w = ( 0.34 kg/m ) 9.81 m/s 2 = 3.3354 N/m
Lmax = 1.3255
32.000 N
= 12 717 m
3.3354 N/m
(b)
Lmax = 12.72 km
PROBLEM 7.144
A cable has a weight per unit length of 2 lb/ft and is supported as shown.
Knowing that the span L is 18 ft, determine the two values of the sag h
for which the maximum tension is 80 lb.
SOLUTION
ymax = c cosh
Tmax = wymax
ymax =
ymax =
Tmax
w
80 lb
= 40 ft
2 lb/ft
c cosh
Solving numerically
L
=h+c
2c
9 ft
= 40 ft
c
c1 = 2.6388 ft
c2 = 38.958 ft
h = ymax − c
h1 = 40 ft − 2.6388 ft
h1 = 37.4 ft
h2 = 40 ft − 38.958 ft
h2 = 1.042 ft
PROBLEM 7.145
Determine the sag-to-span ratio for which the maximum tension in
the cable is equal to the total weight of the entire cable AB.
SOLUTION
Tmax = wyB = 2wsB
y B = 2sB
c cosh
L
L
= 2c sinh
2c
2c
tanh
L
1
=
2c
2
L
1
= tanh −1 = 0.549306
2c
2
hB
y −c
L
= B
= cosh
−1
c
c
2c
= 0.154701
hB
hB / c
=
L
2( L / 2c)
=
0.5 ( 0.154701)
= 0.14081
0.549306
hB
= 0.1408
L
PROBLEM 7.146
A cable of weight w per unit length is suspended between two points
at the same elevation that are a distance L apart. Determine (a) the sagto-span ratio for which the maximum tension is as small as possible,
(b) the corresponding values of θ B and Tm .
SOLUTION
Tmax = wyB = wc cosh
(a)
L
2c
dTmax
L
L
L

sinh 
= w  cosh
−
dc
2c 2c
2c 

min Tmax ,
For
tanh
dTmax
=0
dc
L
L
2c
= 1.1997
=
→
2c
2c
L
yB
L
= cosh
= 1.8102
c
2c
h
y
= B − 1 = 0.8102
c
c
h  1 h  2c  
0.8102
=
= 0.3375
  =
L  2 c  L   2 (1.1997 )
T0 = wc
(b)
But
So
Tmax = wc cosh
T0 = Tmax cosθ B
L
2c
h
= 0.338
L
Tmax
L
y
= cosh
= B
2c
T0
c
Tmax
= secθ B
T0
y 
θ B = sec−1  B  = sec−1 (1.8102 )
 c 
= 56.46°
Tmax = wyB = w
θ B = 56.5°
yB  2c  L 
L
   = w (1.8102 )
c  L  2 
2 (1.1997 )
Tmax = 0.755wL
PROBLEM 7.147
For the beam and loading shown, (a) draw the shear and bending
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
FBD Beam:
(a)
ΣM A = 0: ( 6 ft ) E − ( 8 ft )( 4.5 kips )
− ( 4 ft )(12 kips ) − ( 2 ft )( 6 kips ) = 0
E = 16 kips
ΣM E = 0: − ( 6 ft ) Ay + ( 4 ft )( 6 kips )
+ ( 2 ft )(12 kips ) − ( 2 ft )( 4.5 kips ) = 0
A y = 6.5 kips
Shear Diag: V is piece wise constant with discontinuities equal to the
forces at A, C, D, E, B
Moment Diag: M is piecewise linear with slope changes at C, D, E
MA = 0
M C = ( 6.5 kips )( 2 ft ) = 13 kip ⋅ ft
M C = 13 kip ⋅ ft + ( 0.5 kips )( 2 ft ) = 14 kip ⋅ ft
M D = 14 kip ⋅ ft − (11.5 kips )( 2 ft ) = −9 kip ⋅ ft
M B = − 9 kip ⋅ ft + ( 4.5 kips )( 2 ft ) = 0
(b)
V
M
max
max
= 11.50 kips on DE
= 14.00 kip ⋅ ft at D
PROBLEM 7.148
For the beam and loading shown, (a) draw the shear and bending
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
FBD Beam:
(a)
ΣM B = 0: ( 4 ft )(12 kips ) + ( 7 ft )( 2.5 kips/ft )( 6 ft )
− (10 ft ) Ay = 0
A y = 15.3 kips
Shear Diag: VA = Ay = 15.3 kips, then V is linear
 dV

= −2.5 kips/ft  to C.

dx


VC = 15.3 kips − ( 2.5 kips/ft )( 6 ft ) = 0.3 kips
At C , V decreases by 12 kips to − 11.7 kips and is constant to B.
Moment Diag: M A = 0 and M is parabolic
 dM

decreasing with V  to C

dx


MC =
1
(15.3 kips + 0.3 kip )( 6 ft ) = 46.8 kip ⋅ ft
2
M B = 46.8 kip ⋅ ft − (11.7 kips )( 4 ft ) = 0
(b)
V max = 15.3 kips
M
max
= 46.8 kip ⋅ ft
PROBLEM 7.149
Two loads are suspended as shown from the cable ABCD. Knowing
that hB = 1.8 m, determine (a) the distance hC, (b) the components
of the reaction at D, (c) the maximum tension in the cable.
SOLUTION
ΣFx = 0: − Ax + Dx = 0
FBD Cable:
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0
Dy = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
FBD AB:
ΣM B = 0: (1.8 m ) Ax − ( 3 m )( 8.2 kN ) = 0
Ax =
From above
FBD CD:
41
kN
3
Dx = Ax =
41
kN
3
 41

kN  = 0
ΣM C = 0: ( 4 m )( 7.8 kN ) − hC 
 3

hC = 2.283 m
hC = 2.28 m
(a)
Dx = 13.67 kN
(b)
Dy = 7.80 kN
Since Ax = Bx and Ay > By , max T is TAB
2
TAB =
Ax2 + Ay2 =
2
 41

kN  + ( 8.2 kN )

 3

(c)
Tmax = 15.94 kN
PROBLEM 7.150
Knowing that the maximum tension in cable ABCD is 15 kN,
determine (a) the distance hB, (b) the distance hC.
SOLUTION
ΣFx = 0: − Ax + Dx = 0
FBD Cable:
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0
D y = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
Since
Ax = Dx
and
Ay > Dy ,
Tmax = TAB
ΣFy = 0: 8.2 kN − (15 kN ) sin θ A = 0
FBD pt A:
θ A = sin −1
8.2 kN
= 33.139°
15 kN
ΣFx = 0: − Ax + (15 kN ) cos θ A = 0
Ax = (15 kN ) cos ( 33.139° ) = 12.56 kN
FBD CD:
From FBD cable:
hB = ( 3 m ) tan θ A = ( 3 m ) tan ( 33.139° )
(a)
hB = 1.959 m
ΣM C = 0: ( 4 m )( 7.8 kN ) − hC (12.56 kN ) = 0
(b)
hC = 2.48 m
PROBLEM 7.151
A semicircular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J
when θ = 60°.
SOLUTION
FBD Rod:
ΣM A = 0:
2r
π
B=
ΣFy′ = 0: F +
FBD BJ:
W − 2rB = 0
W
π
W
W
sin 60° − cos 60° = 0
π
3
F = − 0.12952W
W  3r  W 

ΣM 0 = 0: r  F −  +
+M =0
π  2π  3 

1
1 

M = Wr  0.12952 + −
= 0.28868Wr
π 2π 

On BJ
M J = 0.289Wr
PROBLEM 7.152
A semicircular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J
when θ = 150°.
SOLUTION
FBD rod:
ΣFy = 0: Ay − W = 0
Ay = W
2r
ΣM B = 0:
W − 2rAx = 0
W π
Ax =
π
FBD AJ:
ΣFx′ = 0:
W
π
cos 30° +
5W
sin 30° − F = 0 F = 0.69233W
6
W
W 

ΣM 0 = 0: 0.25587r   + r  F −  − M = 0
π 
 6

1
 0.25587
M = Wr 
+ 0.69233 − 
π
 6
M = Wr ( 0.4166 )
On AJ
M = 0.417Wr
PROBLEM 7.153
Determine the internal forces at point J of the structure shown.
SOLUTION
FBD ABC:
ΣM D = 0: ( 0.375 m )( 400 N ) − ( 0.24 m ) C y = 0
C y = 625 N
ΣM B = 0: − ( 0.45 m ) C x + ( 0.135 m )( 400 N ) = 0
C x = 120 N
FBD CJ:
ΣFy = 0: 625 N − F = 0
F = 625 N
ΣFx = 0: 120 N − V = 0
V = 120.0 N
ΣM J = 0: M − ( 0.225 m )(120 N ) = 0
M = 27.0 N ⋅ m
PROBLEM 7.154
Determine the internal forces at point K of the structure shown.
SOLUTION
FBD AK:
ΣFx = 0: V = 0
V =0
ΣFy = 0: F − 400 N = 0
F = 400 N
ΣM K = 0: ( 0.135 m )( 400 N ) − M = 0
M = 54.0 N ⋅ m
PROBLEM 7.155
Two small channel sections DF and EH have been welded to the uniform
beam AB of weight W = 3 kN to form the rigid structural member shown.
This member is being lifted by two cables attached at D and E. Knowing the
θ = 30° and neglecting the weight of the channel sections, (a) draw the
shear and bending-moment diagrams for beam AB, (b) determine the
maximum absolute values of the shear and bending moment in the beam.
SOLUTION
FBD Beam + channels:
(a)
T1 = T2 = T
By symmetry:
ΣFy = 0: 2T sin 60° − 3 kN = 0
T =
FBD Beam:
With cable force replaced by
equivalent force-couple system at
F and G
3
kN
3
M = ( 0.5 m )
T1x =
3
2 3
3
T1y =
2 3
3
kN
2
kN = 0.433 kN ⋅ m
Shear Diagram: V is piecewise linear
 dV

= − 0.6 kN/m  with 1.5 kN

dx


discontinuities at F and H.
V F − = − ( 0.6 kN/m )(1.5 m ) = 0.9 kN
V increases by 1.5 kN to + 0.6 kN at F +
VG = 0.6 kN − ( 0.6 kN/m )(1 m ) = 0
Finish by invoking symmetry
Moment Diagram: M is piecewise parabolic
 dM

decreasing with V  with discontinuities of .433 kN at F and H.

dx


1
M F − = − ( 0.9 kN )(1.5 m ) = − 0.675 kN ⋅ m
2
M increases by 0 .433 kN ⋅ m to − 0.242 kN ⋅ m at F +
M G = − 0.242 kN ⋅ m +
1
( 0.6 kN )(1 m ) = 0.058 kN ⋅ m
2
Finish by invoking symmetry
V
(b)
max
= 900 N
at F − and G +
M
max
= 675 N ⋅ m
at F and G
PROBLEM 7.156
(a) Draw the shear and bending moment diagrams for beam AB,
(b) determine the magnitude and location of the maximum absolute value of
the bending moment.
PROBLEM 7.157
Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine
(a) the required magnitude of the horizontal force P, (b) the corresponding
distance a.
SOLUTION
FBD ABC:
ΣFy = 0: −40 lb − 80 lb + C y = 0
C y = 120 lb
FBD BC:
ΣM B = 0: ( 4 ft )(120 lb ) − (10 ft ) Cx = 0
C x = 48 lb
From ABC:
ΣFx = 0: − P + Cx = 0
P = C x = 48 lb
(a)
P = 48.0 lb
ΣM C = 0: ( 4 ft )( 80 lb ) + a ( 40 lb ) − (15 ft )( 48 lb ) = 0
(b)
a = 10.00 ft
PROBLEM 7.158
Cable ABC supports two loads as shown. Determine the distances a and b
when a horizontal force P of magnitude 60 lb is applied at A.
SOLUTION
FBD ABC:
ΣFx = 0: C x − P = 0
Cx = 60 lb
ΣFy = 0: C y − 40 lb − 80 lb = 0
C y = 120 lb
FBD BC:
ΣM B = 0: b (120 lb ) − (10 ft )( 60 lb ) = 0
b = 5.00 ft
FBD AB:
ΣM B = 0: ( a − b )( 40 lb ) − ( 5 ft ) 60 lb = 0
a − b = 7.5 ft
a = b + 7.5 ft
= 5 ft + 7.5 ft
a = 12.50 ft
PROBLEM 8.1
Determine whether the block shown is in equilibrium, and find the
magnitude and direction of the friction force when θ = 30o and
P = 200 N.
SOLUTION
FBD block:
ΣFn = 0: N − (1000 N ) cos 30° − ( 200 N ) sin 30° = 0
N = 966.03 N
Assume equilibrium:
ΣFt = 0: F + ( 200 N ) cos 30° − (1000 N ) sin 30° = 0
F = 326.8 N = Feq.
But
Fmax = µ s N = ( 0.3) 966 N = 290 N
Feq. > Fmax
and
impossible ⇒ Block moves
F = µk N
= ( 0.2 )( 966.03 N )
Block slides down
F = 193.2 N
PROBLEM 8.2
Determine whether the block shown is in equilibrium, and find the
magnitude and direction of the friction force when θ = 35o and
P = 400 N.
SOLUTION
FBD block:
ΣFn = 0: N − (1000 N ) cos35° − ( 400 N ) sin 35° = 0
N = 1048.6 N
Assume equilibrium:
ΣFt = 0: F − (1000 N ) sin 35° + ( 400 N ) cos 35° = 0
F = 246 N = Feq.
Fmax = µ s N = ( 0.3)(1048.6 N ) = 314 N
Feq. < Fmax
OK
equilibrium
∴ F = 246 N
PROBLEM 8.3
Determine whether the 20-lb block shown is in equilibrium, and find the
magnitude and direction of the friction force when P = 8 lb and θ = 20°.
SOLUTION
FBD block:
ΣFn = 0: N − ( 20 lb ) cos 20° + ( 8 lb ) sin 20° = 0
N = 16.0577 lb
Fmax = µ s N = ( 0.3)(16.0577 lb ) = 4.817 lb
Assume equilibrium:
ΣFt = 0:
(8 lb ) cos 20° − ( 20 lb ) sin 20° − F
=0
F = 0.6771 lb = Feq.
Feq. < Fmax
and
OK
equilibrium
F = 0.677 lb
PROBLEM 8.4
Determine whether the 20-lb block shown is in equilibrium, and find the
magnitude and direction of the friction force when P = 12.5 lb and
θ = 15°.
SOLUTION
FBD block:
ΣFn = 0: N − ( 20 lb ) cos 20° + (12.5 lb ) sin15° = 0
N = 15.559 lb
Fmax = µ s N = ( 0.3)(15.559 lb ) = 4.668 lb
Assume equilibrium:
ΣFt = 0:
(12.5 lb ) cos15° − ( 20 lb ) sin 20° − F
=0
F = 5.23 lb = Feq.
but Feq. > Fmax impossible, so block slides up
and
F = µk N = ( 0.25 )(15.559 lb )
F = 3.89 lb
PROBLEM 8.5
Knowing that θ = 25°, determine the range of values of P for which
equilibrium is maintained.
SOLUTION
FBD block:
Block is in equilibrium:
ΣFn = 0: N − ( 20 lb ) cos 20° + P sin 25° = 0
N = 18.794 lb − P sin 25°
ΣFt = 0: F − ( 20 lb ) sin 20° + P cos 25° = 0
F = 6.840 lb − P cos 25°
or
Impending motion up:
Therefore,
F = µs N ;
Impending motion down: F = − µ s N
6.840 lb − P cos 25° = ± ( 0.3)(18.794 lb − P sin 25° )
Pup = 12.08 lb
Pdown = 1.542 lb
1.542 lb ≤ Peq. ≤ 12.08 lb
PROBLEM 8.6
Knowing that the coefficient of friction between the 60-lb block and the
incline is µ s = 0.25, determine (a) the smallest value of P for which
motion of the block up the incline is impending, (b) the corresponding
value of β.
SOLUTION
FBD block (impending motion up)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.04°
(a) Note: For minimum P, P ⊥ R so β = φ s
Then
P = W sin ( 30° + φ s )
= ( 60 lb ) sin 44.04° = 41.71 lb
Pmin = 41.7 lb
(b) Have β = φ s
β = 14.04°
PROBLEM 8.7
Considering only values of θ less than 90° , determine the smallest value
of θ for which motion of the block to the right is impending when
(a) m = 30 kg, (b) m = 40 kg.
SOLUTION
FBD block (impending motion to
the right)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
P
W
=
sin φs
sin (θ − φ s )
sin (θ − φs ) =
(a)
W
sin φ s
P
m = 30 kg: θ − φ s = sin
W = mg
(
 ( 30 kg ) 9.81 m/s 2
−1 

120 N
) sin14.036°

= 36.499°
∴ θ = 36.499° + 14.036°
(b)
or θ = 50.5°
(
)
 ( 40 kg ) 9.81 m/s 2

sin14.036° 
m = 40 kg: θ − φs = sin −1 


120 N
= 52.474°
∴ θ = 52.474° + 14.036°
or θ = 66.5°
PROBLEM 8.8
Knowing that the coefficient of friction between the 30-lb block and the
incline is µ s = 0.25 , determine (a) the smallest value of P required to
maintain the block in equilibrium, (b) the corresponding value of β .
SOLUTION
FBD block (impending motion
downward)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
(a) Note: For minimum P,
So
and
P⊥R
β = α = 90° − ( 30° + 14.036° ) = 45.964°
P = ( 30 lb ) sin α = ( 30 lb ) sin ( 45.964° ) = 21.567 lb
P = 21.6 lb
(b)
β = 46.0°
PROBLEM 8.9
A 6-kg block is at rest as shown. Determine the positive range of values
of θ for which the block is in equilibrium if (a) θ is less than 90°,
(b) θ is between 90° and 180°.
SOLUTION
FBD block (impending motion)
φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801°
(a) 0° ≤ θ ≤ 90° :
58.86 N
40 N
=
sin (θ − φs ) sinφ s
θ − φ s = sin −1
58.86 N
sin ( 21.801° )
40 N
= 33.127°, 146.873°
θ = 54.9°
∴
(b) 90° ≤ θ ≤ 180° :
and
θ = 168.674°
(a)
Equilibrium for 0 ≤ θ ≤ 54.9°
(b)
and for 168.7° ≤ θ ≤ 180.0°
PROBLEM 8.10
Knowing that P = 25 lb, determine the range of values of θ for which
equilibrium of the 18-lb block is maintained.
SOLUTION
FBD block (impending motion
down)
φ s = tan −1 µ s = tan −1 ( 0.45 ) = 24.228°
25 lb
18 lb
=
sin ( 90° − φ s ) sin (θ + φs )
 18 lb

θ + φs = sin −1 
sin ( 90° − 24.228° )  = 41.04°
 25 lb

θ = 16.81°
Impending motion up:
25 lb
18 lb
=
sin ( 90° + φs ) sin (θ − φs )
 18 lb

θ − φ s = sin −1 
sin ( 90° + 24.228° )  = 41.04°
 25 lb

θ = 65.27°
Equilibrium for 16.81° ≤ θ ≤ 65.3°
PROBLEM 8.1
Determine whether the block shown is in equilibrium, and find the
magnitude and direction of the friction force when θ = 30o and
P = 200 N.
SOLUTION
FBD block:
ΣFn = 0: N − (1000 N ) cos 30° − ( 200 N ) sin 30° = 0
N = 966.03 N
Assume equilibrium:
ΣFt = 0: F + ( 200 N ) cos 30° − (1000 N ) sin 30° = 0
F = 326.8 N = Feq.
But
Fmax = µ s N = ( 0.3) 966 N = 290 N
Feq. > Fmax
and
impossible ⇒ Block moves
F = µk N
= ( 0.2 )( 966.03 N )
Block slides down
F = 193.2 N
PROBLEM 8.2
Determine whether the block shown is in equilibrium, and find the
magnitude and direction of the friction force when θ = 35o and
P = 400 N.
SOLUTION
FBD block:
ΣFn = 0: N − (1000 N ) cos35° − ( 400 N ) sin 35° = 0
N = 1048.6 N
Assume equilibrium:
ΣFt = 0: F − (1000 N ) sin 35° + ( 400 N ) cos 35° = 0
F = 246 N = Feq.
Fmax = µ s N = ( 0.3)(1048.6 N ) = 314 N
Feq. < Fmax
OK
equilibrium
∴ F = 246 N
PROBLEM 8.3
Determine whether the 20-lb block shown is in equilibrium, and find the
magnitude and direction of the friction force when P = 8 lb and θ = 20°.
SOLUTION
FBD block:
ΣFn = 0: N − ( 20 lb ) cos 20° + ( 8 lb ) sin 20° = 0
N = 16.0577 lb
Fmax = µ s N = ( 0.3)(16.0577 lb ) = 4.817 lb
Assume equilibrium:
ΣFt = 0:
(8 lb ) cos 20° − ( 20 lb ) sin 20° − F
=0
F = 0.6771 lb = Feq.
Feq. < Fmax
and
OK
equilibrium
F = 0.677 lb
PROBLEM 8.4
Determine whether the 20-lb block shown is in equilibrium, and find the
magnitude and direction of the friction force when P = 12.5 lb and
θ = 15°.
SOLUTION
FBD block:
ΣFn = 0: N − ( 20 lb ) cos 20° + (12.5 lb ) sin15° = 0
N = 15.559 lb
Fmax = µ s N = ( 0.3)(15.559 lb ) = 4.668 lb
Assume equilibrium:
ΣFt = 0:
(12.5 lb ) cos15° − ( 20 lb ) sin 20° − F
=0
F = 5.23 lb = Feq.
but Feq. > Fmax impossible, so block slides up
and
F = µk N = ( 0.25 )(15.559 lb )
F = 3.89 lb
PROBLEM 8.5
Knowing that θ = 25°, determine the range of values of P for which
equilibrium is maintained.
SOLUTION
FBD block:
Block is in equilibrium:
ΣFn = 0: N − ( 20 lb ) cos 20° + P sin 25° = 0
N = 18.794 lb − P sin 25°
ΣFt = 0: F − ( 20 lb ) sin 20° + P cos 25° = 0
F = 6.840 lb − P cos 25°
or
Impending motion up:
Therefore,
F = µs N ;
Impending motion down: F = − µ s N
6.840 lb − P cos 25° = ± ( 0.3)(18.794 lb − P sin 25° )
Pup = 12.08 lb
Pdown = 1.542 lb
1.542 lb ≤ Peq. ≤ 12.08 lb
PROBLEM 8.6
Knowing that the coefficient of friction between the 60-lb block and the
incline is µ s = 0.25, determine (a) the smallest value of P for which
motion of the block up the incline is impending, (b) the corresponding
value of β.
SOLUTION
FBD block (impending motion up)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.04°
(a) Note: For minimum P, P ⊥ R so β = φ s
Then
P = W sin ( 30° + φ s )
= ( 60 lb ) sin 44.04° = 41.71 lb
Pmin = 41.7 lb
(b) Have β = φ s
β = 14.04°
PROBLEM 8.7
Considering only values of θ less than 90° , determine the smallest value
of θ for which motion of the block to the right is impending when
(a) m = 30 kg, (b) m = 40 kg.
SOLUTION
FBD block (impending motion to
the right)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
P
W
=
sin φs
sin (θ − φ s )
sin (θ − φs ) =
(a)
W
sin φ s
P
m = 30 kg: θ − φ s = sin
W = mg
(
 ( 30 kg ) 9.81 m/s 2
−1 

120 N
) sin14.036°

= 36.499°
∴ θ = 36.499° + 14.036°
(b)
or θ = 50.5°
(
)
 ( 40 kg ) 9.81 m/s 2

sin14.036° 
m = 40 kg: θ − φs = sin −1 


120 N
= 52.474°
∴ θ = 52.474° + 14.036°
or θ = 66.5°
PROBLEM 8.8
Knowing that the coefficient of friction between the 30-lb block and the
incline is µ s = 0.25 , determine (a) the smallest value of P required to
maintain the block in equilibrium, (b) the corresponding value of β .
SOLUTION
FBD block (impending motion
downward)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
(a) Note: For minimum P,
So
and
P⊥R
β = α = 90° − ( 30° + 14.036° ) = 45.964°
P = ( 30 lb ) sin α = ( 30 lb ) sin ( 45.964° ) = 21.567 lb
P = 21.6 lb
(b)
β = 46.0°
PROBLEM 8.9
A 6-kg block is at rest as shown. Determine the positive range of values
of θ for which the block is in equilibrium if (a) θ is less than 90°,
(b) θ is between 90° and 180°.
SOLUTION
FBD block (impending motion)
φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801°
(a) 0° ≤ θ ≤ 90° :
58.86 N
40 N
=
sin (θ − φs ) sinφ s
θ − φ s = sin −1
58.86 N
sin ( 21.801° )
40 N
= 33.127°, 146.873°
θ = 54.9°
∴
(b) 90° ≤ θ ≤ 180° :
and
θ = 168.674°
(a)
Equilibrium for 0 ≤ θ ≤ 54.9°
(b)
and for 168.7° ≤ θ ≤ 180.0°
PROBLEM 8.10
Knowing that P = 25 lb, determine the range of values of θ for which
equilibrium of the 18-lb block is maintained.
SOLUTION
FBD block (impending motion
down)
φ s = tan −1 µ s = tan −1 ( 0.45 ) = 24.228°
25 lb
18 lb
=
sin ( 90° − φ s ) sin (θ + φs )
 18 lb

θ + φs = sin −1 
sin ( 90° − 24.228° )  = 41.04°
 25 lb

θ = 16.81°
Impending motion up:
25 lb
18 lb
=
sin ( 90° + φs ) sin (θ − φs )
 18 lb

θ − φ s = sin −1 
sin ( 90° + 24.228° )  = 41.04°
 25 lb

θ = 65.27°
Equilibrium for 16.81° ≤ θ ≤ 65.3°
PROBLEM 8.11
The coefficients of friction are µ s = 0.40 and µ k = 0.30 between all
surfaces of contact. Determine the force P for which motion of the 60-lb
block is impending if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) Note: With the cable, motion must impend at both contact surfaces.
FBDs
Top block:
ΣFy = 0: N1 − 40 lb = 0
F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb
Impending slip:
ΣFx = 0: T − F1 = 0
T − 16 lb = 0
ΣFy = 0: N 2 − 40 lb − 60 lb = 0
Bottom block:
Impending slip:
N1 = 40 lb
T = 16 lb
N 2 = 100 lb
F2 = µ s N 2 = 0.4 (100 lb ) = 40 lb
ΣFx = 0:
− P + 16 lb + 16 lb + 40 lb = 0
P = 72.0 lb
FBD blocks:
(b) Without the cable, both blocks will stay together and motion will
impend only at the floor.
ΣFy = 0: N − 40 lb − 60 lb = 0
N = 100 lb
Impending slip:
F = µ s N = 0.4 (100 lb ) = 40 lb
ΣFx = 0: 40 lb − P = 0
P = 40.0 lb
PROBLEM 8.12
The coefficients of friction are µ s = 0.40 and µ k = 0.30 between all
surfaces of contact. Determine the force P for which motion of the 60-lb
block is impending if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) With the cable, motion must impend at both surfaces.
FBDs
Top block:
ΣFy = 0: N1 − 40 lb = 0
Impending slip:
N1 = 40 lb
F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb
ΣFy = 0: N 2 − 40 lb − 60 lb = 0
Impending slip:
Bottom block:
N 2 = 100 lb
F2 = µ N 2 = 0.4 (100 lb ) = 40 lb
ΣFx = 0: 16 lb + 40 lb − P = 0
P = 56 lb
P = 56.0 lb
FBD blocks:
(b) Without the cable, both blocks stay together and motion will impend
at the floor surface only.
ΣFy = 0: N − 40 lb − 60 lb = 0
N = 100 lb
Impending slip:
F = µ s N = 0.4 (100 lb ) = 40 lb
ΣFx = 0: − P + 40 lb = 0
P = 40 lb
P = 40.0 lb
PROBLEM 8.13
The 8-kg block A is attached to link AC and rests on the 12-kg block B.
Knowing that the coefficient of static friction is 0.20 between all surfaces
of contact and neglecting the mass of the link, determine the value of θ
for which motion of block B is impending.
SOLUTION
FBDs:
Motion must impend at both contact surfaces
Block A:
ΣFy = 0: N1 − WA = 0
N1 = WA
Block B:
ΣFy = 0: N 2 − N1 − WB = 0
N 2 = N1 + WB = WA + WB
Impending motion:
F1 = µ s N1 = µ sWA
F2 = µ s N 2 = µ s ( N1 + WB )
Block B:
ΣFx = 0: 50 N − F1 − F2 = 0
50 N = µ s ( N1 + N1 + WB ) = 0.2 ( 2 N1 + 117.72 N )
or
N1 = 66.14 N
Block A:
F1 = 0.2 ( 66.14 N ) = 13.228 N
ΣFx = 0: 13.228 N − FAC cosθ = 0
FAC cosθ = 13.228 N
or
(1)
ΣFy = 0: 66.14 N − 78.48 N + FAC sinθ = 0
FAC sin θ = 78.48 N − 66.14 N
or
Then,
Eq. (2)
Eq. (1)
tan θ =
(2)
78.48 N − 66.14 N
13.228 N
θ = 43.0°
PROBLEM 8.14
The 8-kg block A and the 16-kg block B are at rest on an incline as
shown. Knowing that the coefficient of static friction is 0.25 between all
surfaces of contact, determine the value of θ for which motion is
impending.
SOLUTION
FBDs:
Block A:
Impending motion:
ΣFy = 0: N1 − WA = 0
F1 = µ s N1 = µ sWA
ΣFx = 0: F1 − T = 0
Block B:
N1 = WA
T = F1 = µ sWA
ΣFy′ = 0: N 2 − ( N1 + WB ) cosθ − F1 sin θ = 0
N 2 = 3WA cosθ + µ sWA sin θ
= WA ( 3cosθ + 0.25sin θ )
Impending motion:
F2 = µ s N 2 = 0.25WA ( 3cosθ + 0.25sin θ )
ΣFx′ = 0: − T − F2 − F1 cosθ + ( N1 + WB ) sin θ = 0
 −0.25 − 0.25 ( 3cosθ + 0.25sin θ ) − 0.25cosθ + 3sinθ  WA = 0
or
Solving numerically
47sinθ − 16cosθ − 4 = 0
θ = 23.4°
PROBLEM 8.15
A 48-kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and each
caster is 0.30. Knowing that h = 640 mm, determine the magnitude of
the force P required for impending motion of the cabinet to the right (a) if
all casters are locked, (b) if the casters at B are locked and the casters at A
are free to rotate, (c) if the casters at A are locked and the casters at B are
free to rotate.
SOLUTION
N A = FA = 0
Note: For tipping,
FBD cabinet:
ΣM B = 0:
( 0.24 m )W − ( 0.64 m ) Ptip
ΣFy = 0: N A + N B − W = 0
N A + NB = W
FA + FB = µ sW
P = FA + FB = µ sW
ΣFx = 0: P − FA − FB = 0
∴ P = 0.3 ( 470.88 N )
( P = 0.3W
(
W = 48 kg 9.81 m/s 2
= 470.88 N
µ s = 0.3
)
Ptip = 0.375W
FA = µ s N A , FB = µ s N B
(a) All casters locked: Impending slip:
So
=0
< Ptip
P = 141.3 N
or
OK
)
FA = 0
(b) Casters at A free, so
FB = µ s N B
Impending slip:
ΣFx = 0: P − FB = 0
P = FB = µ s N B
ΣM A = 0:
NB =
P
µs
( 0.64 m ) P + ( 0.24 m )W − ( 0.48 m ) N B
8P + 3W − 6
P
=0
0.3
( P = 0.25W
< Ptip
∴ P = 0.25 ( 470.88 N )
=0
P = 0.25W
OK
)
P = 117.7 N
PROBLEM 8.15 CONTINUED
FB = 0
(c) Casters at B free, so
FA = µ s N A
Impending slip:
P = FA = µ s N A
ΣFx = 0: P − FA = 0
NA =
ΣM B = 0:
P
µs
=
P
0.3
( 0.24 m )W − ( 0.64 m ) P − ( 0.48 m ) N A
3W − 8P − 6
P
=0
0.3
( P < Ptip
=0
P = 0.10714W = 50.45 N
OK
)
P = 50.5 N
PROBLEM 8.16
A 48-kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and
each caster is 0.30. Assuming that the casters at A and B are locked,
determine (a) the force P required for impending motion of the cabinet
to the right, (b) the largest allowable height h if the cabinet in not to tip
over.
SOLUTION
ΣFy = 0: N A + N B − W = 0;
(a)
FBD cabinet:
Impending slip:
N A + NB = W
FA = µ s N A , FB = µ s N B
FA + FB = µ sW
So
P = FA + FB = µ sW
ΣFx = 0: P − FA − FB = 0
P = 0.3 ( 470.88 N ) = 141.26 N
P = 141.3 N
(b) For tipping,
N A = FA = 0
ΣM B = 0: hP − ( 0.24 m )W = 0
(
W = 48 kg 9.81 m/s 2
= 470.88 N
)
hmax = ( 0.24 m )
W
1
0.24 m
= ( 0.24 m )
=
µs
0.3
P
hmax = 0.800 m
PROBLEM 8.17
The cylinder shown is of weight W and radius r, and the coefficient of
static friction µ s is the same at A and B. Determine the magnitude of the
largest couple M which can be applied to the cylinder if it is not to rotate.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ s N A, FB = µ s N B
ΣFx = 0: N A − FB = 0
N A = FB = µ s N B
FA = µ s N A = µ s2 N B
N B + µ s2 N B = W
ΣFy = 0: N B + FA − W = 0
or
NB =
W
1 + µ s2
and
FB =
µ sW
1 + µ s2
FA =
µ s2W
1 + µ2
ΣM C = 0: M − r ( FA + FB ) = 0
(
M = r µ s + µ s2
) 1 +Wµ
2
s
M max = Wr µ s
1 + µs
1 + µ s2
PROBLEM 8.18
The cylinder shown is of weight W and radius r. Express in terms of W
and r the magnitude of the largest couple M which can be applied to the
cylinder if it is not to rotate assuming that the coefficient of static friction
is (a) zero at A and 0.36 at B, (b) 0.30 at A and 0.36 at B.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ A N A;
FB = µ B N B
N A = FB = µ B N B
ΣFx = 0: N A − FB = 0
FA = µ A N A = µ Aµ B N B
N B (1 + µ Aµ B ) = W
ΣFy = 0: N B + FA − W = 0
1
W
1 + µ Aµ B
or
NB =
and
FB = µ B N B =
µB
1 + µ Aµ B
FA = µ Aµ B N B =
µ Aµ B
1 + µ Aµ B
ΣM C = 0: M − r ( FA + FB ) = 0
(a) For
µA = 0
and
W
W
M = Wr µ B
1 + µA
1 + µ Aµ B
µ B = 0.36
M = 0.360Wr
(b) For
µ A = 0.30
and
µ B = 0.36
M = 0.422Wr
PROBLEM 8.19
The hydraulic cylinder shown exerts a force of 680 lb directed to the right
on point B and to the left on point E. Determine the magnitude of the
couple M required to rotate the drum clockwise at a constant speed.
SOLUTION
FBDs
Drum:
Rotating drum ⇒ slip at both sides; constant speed ⇒ equilibrium
∴ F1 = µ k N1 = 0.3N1;
AB:
ΣM A = 0:
F2 = µk N 2 = 0.3N 2
( 6 in.)( 680 lb ) + ( 6 in.)( F1 ) − (18 in.) N1 = 0
 18 in.

− 6 in.  = ( 6 in.)( 680 lb )
F1 
 0.3

DE:
ΣM D = 0:
F1 = 75.555 lb
( 6 in.) F2 + (18 in.) N 2 − ( 6 in.)( 680 lb ) = 0
18 in. 

F2  6 in. +
 = ( 6 in.)( 680 lb )
0.3 

Drum:
or
or
F2 = 61.818 lb
ΣM C = 0: r ( F1 + F2 ) − M = 0
M = (10 in.)( 75.555 + 61.818 ) lb
M = 1374 lb ⋅ in.
PROBLEM 8.20
A couple M of magnitude 70 lb ⋅ ft is applied to the drum as shown.
Determine the smallest force which must be exerted by the hydraulic
cylinder on joints B and E if the drum is not to rotate.
SOLUTION
FBDs
DE:
Drum:
For minimum T, slip impends at both sides, so
F1 = µ s N1 = 0.4 N1
AB:
ΣM A = 0:
F2 = µ s N 2 = 0.4 N 2
( 6 in.)T + ( 6 in.) F1 − (18 in.) N1 = 0
 18 in.

− 6 in.  = ( 6 in.) T
F1 
 0.4

DE:
ΣM D = 0:
( 6 in.) F2 + (18 in.) N 2 − ( 6 in.) T
18 in. 

F2  6 in. +
 = ( 6 in.) T
0.4 

Drum:
ΣM C = 0:
or
F1 =
T
6.5
F2 =
T
8.5
=0
or
(10 in.) (F1 + F2 ) − 840 lb ⋅ in. = 0
1 
 1
+
T
 = 84 lb
6.5
8.5


T = 309 lb
PROBLEM 8.21
A 19.5-ft ladder AB leans against a wall as shown. Assuming that the
coefficient of static friction µ s is the same at A and B, determine the
smallest value of µ s for which equilibrium is maintained.
SOLUTION
Motion impends at both A and B.
FBD ladder:
FA = µ s N A
FB = µ s N B
ΣFx = 0: FA − N B = 0
Then
FB = µ s N B = µ s2 N A
ΣFy = 0: N A − W + FB = 0
ΣM O = 0: bN B +
or
a = 7.5 ft
N B = FA = µ s N A
or
b = 18 ft
(
)
(
)
N A 1 + µ s2 = W
a
W − aN A = 0
2
aN A − bµ s N A =
µ s2 +
or
a
a
W = N A 1 + µ s2
2
2
2b
µs − 1 = 0
a
2
µs = −
b
b
±   + 1 = −2.4 ± 2.6
a
a
The positive root is physically possible. Therefore,
µ s = 0.200
PROBLEM 8.22
A 19.5-ft ladder AB leans against a wall as shown. Assuming that the
coefficient of static friction µ s is the same at A and B, determine the
smallest value of µ s for which equilibrium is maintained.
SOLUTION
Motion impends at both A and B, so
FBD ladder:
FA = µ s N A
a
W =0
2
ΣM A = 0: lN B −
or
FB = µ s N B = µ s
ΣFx = 0: FA +
a
5
=
l
13
b 12
=
l
13
12.5
(13)
2.5W
13
(13)
2
µ sW −
30
(13)2
W =0
( 30 − 12.5µ s )
W
ΣFy = 0: N A − W +
l = 19.5 ft
a
7.5 ft
W =
W
2l
39 ft
5
12
FB −
NB = 0
13
13
µs N A +
NA −
NB =
or
2.5
W
13
NB =
Then
a = 7.5 ft
FB = µ s N B
and
2
µs
12
5
FB +
NB = 0
13
13
 30 − 12.5µ s
 W
+ 30µ s + 12.5 
=W

2
µ
s

 (13)
or
µ s2 − 5.6333µ s + 1 = 0
µ s = 2.8167 ± 2.6332
or
µ s = 0.1835
and
µ s = 5.45
The larger value is very unlikely unless the surface is treated with
some “non-skid” material.
In any event, the smallest value for equilibrium is µ s = 0.1835
PROBLEM 8.23
End A of a slender, uniform rod of weight W and length L bears on a
horizontal surface as shown, while end B is supported by a cord BC of
length L. Knowing that the coefficient of static friction is 0.40, determine
(a) the value of θ for which motion is impending, (b) the corresponding
value of the tension in the cord.
SOLUTION
FBD rod:
(a) Geometry:
BE =
L
cosθ
2
EF = L sin θ
So
or
Also,
or
L

DE =  cosθ  tan β
2

DF =
L cosθ
2 tan φ s
1
 L cosθ
L  cosθ tan β + sin θ  =
2
 2 tan φs
tan β + 2 tan θ =
1
1
1
=
=
= 2.5
tan φ s
µ s 0.4
L sin θ + L sin β = L
sin θ + sin β = 1
Solving Eqs. (1) and (2) numerically
θ1 = 4.62°
(2)
β1 = 66.85°
θ 2 = 48.20° β 2 = 14.75°
θ = 4.62° and θ = 48.2°
Therefore,
(b) Now
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
and
T
W
=
sin φs
sin ( 90 + β − φ s )
or
T =W
For
(1)
sin φs
sin ( 90 + β − φ s )
θ = 4.62°
T = 0.526W
θ = 48.2°
T = 0.374W
PROBLEM 8.24
A slender rod of length L is lodged between peg C and the vertical wall
and supports a load P at end A. Knowing that the coefficient of static
friction between the peg and the rod is 0.25 and neglecting friction at the
roller, determine the range of values of the ratio L/a for which
equilibrium is maintained.
SOLUTION
ΣM B = 0:
FBD rod:
a
N − L sin 30°P = 0
sin 30°
N =
L 2
LP
sin 30°P =
a
a 4
Impending motion at C : down → F = µ s N 
N
F = ±
4
up → F = − µ s N 
ΣFy = 0: F cos 30° + N sin 30° − P = 0
±
L P 3 LP1
+
= P
a 16 2
a 42
L 1
3
 ±
 =1
a  8 32 
L
32
=
a
4± 3
or
For equilibrium:
L
= 5.583
a
and
L
= 14.110
a
5.58 ≤
L
≤ 14.11
a
PROBLEM 8.25
The basic components of a clamping device are bar AB, locking plate CD,
and lever EFG; the dimensions of the slot in CD are slightly larger than
those of the cross section of AB. To engage the clamp, AB is pushed
against the workpiece, and then force P is applied. Knowing that
P = 160 N and neglecting the friction force between the lever and the
plate, determine the smallest allowable value of the static coefficient of
friction between the bar and the plate.
SOLUTION
FBD Plate:
DC is three-force member and motion impends at C and D (for minimum
µ s ).
OCG = 20° + φs
ODG = 20° − φ s
 24 mm

OG = (10 mm ) tan ( 20° + φs ) = 
+ 10 mm  tan ( 20° − φs )
 sin70°

or
tan ( 20° + φ s ) = 3.5540 tan ( 20° − φs )
Solving numerically
Now
so that
φ s = 10.565°
µ s = tan φs
µ s = 0.1865
PROBLEM 8.26
A window sash having a mass of 4 kg is normally supported by two 2-kg
sash weights. Knowing that the window remains open after one sash cord
has broken, determine the smallest possible value of the coefficient of
static friction. (Assume that the sash is slightly smaller that the frame and
will bind only at points A and D.)
SOLUTION
(
ΣFx = 0:
N A − ND = 0
FA = µ s N A
Impending motion:
ΣM D = 0:
NA =
)
W = ( 4 kg ) 9.81 m/s 2 = 39.24 N
W
2
N A = ND
FD = µ s N D
( 0.36 m )W − ( 0.54 m ) N A − ( 0.72 m ) FA
W =
(
)
T = ( 2 kg ) 9.81 m/s 2 = 19.62 N =
FBD window:
=0
3
N A + 2µ s N A
2
2W
3 + 4µ s
ΣFy = 0: FA − W + T + FD = 0
FA + FD = W − T
=
Now
Then
or
W
2
FA + FD = µ s ( N A + N D ) = 2µ s N A
W
2W
= 2µ s
2
3 + 4µ s
µ s = 0.750
PROBLEM 8.27
The steel-plate clamp shown is used to lift a steel plate H of mass 250 kg.
Knowing that the normal force exerted on steel cam EG by pin D forms
an angle of 40° with the horizontal and neglecting the friction force
between the cam and the pin, determine the smallest allowable value of
the coefficient of static friction.
SOLUTION
FBDs:
(Note: P is vertical as AB is two force member; also P = W since
clamp + plate is a two force FBD)
BCD:
ΣM C = 0:
( 0.37 m ) P − ( 0.46 m ) N D cos 40°
− ( 0.06 m ) N D sin 40° = 0
N D = 0.94642P = 0.94642W
or
EG:
ΣM E = 0:
( 0.18 m ) NG − ( 0.26 m ) FG − ( 0.26 m ) N D cos 40° = 0
Impending motion:
Combining
FG = µ s NG
(18 + 26µs ) NG
= 19.9172 N D
= 18.850W
PROBLEM 8.27 CONTINUED
Plate:
From plate:
Then
FG =
W
2
(18 + 26µs )
so that NG =
W
2µ s
W
= 18.85W
2µ s
µ s = 0.283
PROBLEM 8.28
The 5-in.-radius cam shown is used to control the motion of the plate CD.
Knowing that the coefficient of static friction between the cam and the
plate is 0.45 and neglecting friction at the roller supports, determine
(a) the force P for which motion of the plate is impending knowing that
the plate is 1 in. thick, (b) the largest thickness of the plate for which the
mechanism is self-locking, (that is, for which the plate cannot be moved
however large the force P may be).
SOLUTION
FBDs:
ΣFx = 0: F − P = 0
From plate:
cosθ =
From cam geometry:
F = P
5 in. − t
5 in.
ΣM A = 0: ( 5 in.) sin θ  N − ( 5 in.) cosθ  F − ( 5 in.) Q = 0
F = µs N
Impending motion:
N sin θ − µ s N cosθ = Q = 15 lb
So
N =
Q
sin θ − µ s cosθ
µ sQ
sin θ − µ s cosθ
P = F = µs N =
So
t = 1 in. ⇒ cosθ =
(a)
P=
4 in.
= 0.8; sin θ = 0.6
5 in.
( 0.45)(15 lb )
0.6 − ( 0.45 )( 0.8 )
= 28.125 lb; P = 28.1 lb
P → ∞ : sin θ − µ s cosθ =
(b)
Thus
But
tan θ → µ s = 0.45
( 5 in.) cosθ
= 5 in. − t
so that
or
µ sQ
P
0
θ = 24.228°
t = ( 5 in.) (1 − cosθ )
t = 0.440 in.
PROBLEM 8.11
The coefficients of friction are µ s = 0.40 and µ k = 0.30 between all
surfaces of contact. Determine the force P for which motion of the 60-lb
block is impending if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) Note: With the cable, motion must impend at both contact surfaces.
FBDs
Top block:
ΣFy = 0: N1 − 40 lb = 0
F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb
Impending slip:
ΣFx = 0: T − F1 = 0
T − 16 lb = 0
ΣFy = 0: N 2 − 40 lb − 60 lb = 0
Bottom block:
Impending slip:
N1 = 40 lb
T = 16 lb
N 2 = 100 lb
F2 = µ s N 2 = 0.4 (100 lb ) = 40 lb
ΣFx = 0:
− P + 16 lb + 16 lb + 40 lb = 0
P = 72.0 lb
FBD blocks:
(b) Without the cable, both blocks will stay together and motion will
impend only at the floor.
ΣFy = 0: N − 40 lb − 60 lb = 0
N = 100 lb
Impending slip:
F = µ s N = 0.4 (100 lb ) = 40 lb
ΣFx = 0: 40 lb − P = 0
P = 40.0 lb
PROBLEM 8.12
The coefficients of friction are µ s = 0.40 and µ k = 0.30 between all
surfaces of contact. Determine the force P for which motion of the 60-lb
block is impending if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) With the cable, motion must impend at both surfaces.
FBDs
Top block:
ΣFy = 0: N1 − 40 lb = 0
Impending slip:
N1 = 40 lb
F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb
ΣFy = 0: N 2 − 40 lb − 60 lb = 0
Impending slip:
Bottom block:
N 2 = 100 lb
F2 = µ N 2 = 0.4 (100 lb ) = 40 lb
ΣFx = 0: 16 lb + 40 lb − P = 0
P = 56 lb
P = 56.0 lb
FBD blocks:
(b) Without the cable, both blocks stay together and motion will impend
at the floor surface only.
ΣFy = 0: N − 40 lb − 60 lb = 0
N = 100 lb
Impending slip:
F = µ s N = 0.4 (100 lb ) = 40 lb
ΣFx = 0: − P + 40 lb = 0
P = 40 lb
P = 40.0 lb
PROBLEM 8.13
The 8-kg block A is attached to link AC and rests on the 12-kg block B.
Knowing that the coefficient of static friction is 0.20 between all surfaces
of contact and neglecting the mass of the link, determine the value of θ
for which motion of block B is impending.
SOLUTION
FBDs:
Motion must impend at both contact surfaces
Block A:
ΣFy = 0: N1 − WA = 0
N1 = WA
Block B:
ΣFy = 0: N 2 − N1 − WB = 0
N 2 = N1 + WB = WA + WB
Impending motion:
F1 = µ s N1 = µ sWA
F2 = µ s N 2 = µ s ( N1 + WB )
Block B:
ΣFx = 0: 50 N − F1 − F2 = 0
50 N = µ s ( N1 + N1 + WB ) = 0.2 ( 2 N1 + 117.72 N )
or
N1 = 66.14 N
Block A:
F1 = 0.2 ( 66.14 N ) = 13.228 N
ΣFx = 0: 13.228 N − FAC cosθ = 0
FAC cosθ = 13.228 N
or
(1)
ΣFy = 0: 66.14 N − 78.48 N + FAC sinθ = 0
FAC sin θ = 78.48 N − 66.14 N
or
Then,
Eq. (2)
Eq. (1)
tan θ =
(2)
78.48 N − 66.14 N
13.228 N
θ = 43.0°
PROBLEM 8.14
The 8-kg block A and the 16-kg block B are at rest on an incline as
shown. Knowing that the coefficient of static friction is 0.25 between all
surfaces of contact, determine the value of θ for which motion is
impending.
SOLUTION
FBDs:
Block A:
Impending motion:
ΣFy = 0: N1 − WA = 0
F1 = µ s N1 = µ sWA
ΣFx = 0: F1 − T = 0
Block B:
N1 = WA
T = F1 = µ sWA
ΣFy′ = 0: N 2 − ( N1 + WB ) cosθ − F1 sin θ = 0
N 2 = 3WA cosθ + µ sWA sin θ
= WA ( 3cosθ + 0.25sin θ )
Impending motion:
F2 = µ s N 2 = 0.25WA ( 3cosθ + 0.25sin θ )
ΣFx′ = 0: − T − F2 − F1 cosθ + ( N1 + WB ) sin θ = 0
 −0.25 − 0.25 ( 3cosθ + 0.25sin θ ) − 0.25cosθ + 3sinθ  WA = 0
or
Solving numerically
47sinθ − 16cosθ − 4 = 0
θ = 23.4°
PROBLEM 8.15
A 48-kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and each
caster is 0.30. Knowing that h = 640 mm, determine the magnitude of
the force P required for impending motion of the cabinet to the right (a) if
all casters are locked, (b) if the casters at B are locked and the casters at A
are free to rotate, (c) if the casters at A are locked and the casters at B are
free to rotate.
SOLUTION
N A = FA = 0
Note: For tipping,
FBD cabinet:
ΣM B = 0:
( 0.24 m )W − ( 0.64 m ) Ptip
ΣFy = 0: N A + N B − W = 0
N A + NB = W
FA + FB = µ sW
P = FA + FB = µ sW
ΣFx = 0: P − FA − FB = 0
∴ P = 0.3 ( 470.88 N )
( P = 0.3W
(
W = 48 kg 9.81 m/s 2
= 470.88 N
µ s = 0.3
)
Ptip = 0.375W
FA = µ s N A , FB = µ s N B
(a) All casters locked: Impending slip:
So
=0
< Ptip
P = 141.3 N
or
OK
)
FA = 0
(b) Casters at A free, so
FB = µ s N B
Impending slip:
ΣFx = 0: P − FB = 0
P = FB = µ s N B
ΣM A = 0:
NB =
P
µs
( 0.64 m ) P + ( 0.24 m )W − ( 0.48 m ) N B
8P + 3W − 6
P
=0
0.3
( P = 0.25W
< Ptip
∴ P = 0.25 ( 470.88 N )
=0
P = 0.25W
OK
)
P = 117.7 N
PROBLEM 8.15 CONTINUED
FB = 0
(c) Casters at B free, so
FA = µ s N A
Impending slip:
P = FA = µ s N A
ΣFx = 0: P − FA = 0
NA =
ΣM B = 0:
P
µs
=
P
0.3
( 0.24 m )W − ( 0.64 m ) P − ( 0.48 m ) N A
3W − 8P − 6
P
=0
0.3
( P < Ptip
=0
P = 0.10714W = 50.45 N
OK
)
P = 50.5 N
PROBLEM 8.16
A 48-kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and
each caster is 0.30. Assuming that the casters at A and B are locked,
determine (a) the force P required for impending motion of the cabinet
to the right, (b) the largest allowable height h if the cabinet in not to tip
over.
SOLUTION
ΣFy = 0: N A + N B − W = 0;
(a)
FBD cabinet:
Impending slip:
N A + NB = W
FA = µ s N A , FB = µ s N B
FA + FB = µ sW
So
P = FA + FB = µ sW
ΣFx = 0: P − FA − FB = 0
P = 0.3 ( 470.88 N ) = 141.26 N
P = 141.3 N
(b) For tipping,
N A = FA = 0
ΣM B = 0: hP − ( 0.24 m )W = 0
(
W = 48 kg 9.81 m/s 2
= 470.88 N
)
hmax = ( 0.24 m )
W
1
0.24 m
= ( 0.24 m )
=
µs
0.3
P
hmax = 0.800 m
PROBLEM 8.17
The cylinder shown is of weight W and radius r, and the coefficient of
static friction µ s is the same at A and B. Determine the magnitude of the
largest couple M which can be applied to the cylinder if it is not to rotate.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ s N A, FB = µ s N B
ΣFx = 0: N A − FB = 0
N A = FB = µ s N B
FA = µ s N A = µ s2 N B
N B + µ s2 N B = W
ΣFy = 0: N B + FA − W = 0
or
NB =
W
1 + µ s2
and
FB =
µ sW
1 + µ s2
FA =
µ s2W
1 + µ2
ΣM C = 0: M − r ( FA + FB ) = 0
(
M = r µ s + µ s2
) 1 +Wµ
2
s
M max = Wr µ s
1 + µs
1 + µ s2
PROBLEM 8.18
The cylinder shown is of weight W and radius r. Express in terms of W
and r the magnitude of the largest couple M which can be applied to the
cylinder if it is not to rotate assuming that the coefficient of static friction
is (a) zero at A and 0.36 at B, (b) 0.30 at A and 0.36 at B.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ A N A;
FB = µ B N B
N A = FB = µ B N B
ΣFx = 0: N A − FB = 0
FA = µ A N A = µ Aµ B N B
N B (1 + µ Aµ B ) = W
ΣFy = 0: N B + FA − W = 0
1
W
1 + µ Aµ B
or
NB =
and
FB = µ B N B =
µB
1 + µ Aµ B
FA = µ Aµ B N B =
µ Aµ B
1 + µ Aµ B
ΣM C = 0: M − r ( FA + FB ) = 0
(a) For
µA = 0
and
W
W
M = Wr µ B
1 + µA
1 + µ Aµ B
µ B = 0.36
M = 0.360Wr
(b) For
µ A = 0.30
and
µ B = 0.36
M = 0.422Wr
PROBLEM 8.19
The hydraulic cylinder shown exerts a force of 680 lb directed to the right
on point B and to the left on point E. Determine the magnitude of the
couple M required to rotate the drum clockwise at a constant speed.
SOLUTION
FBDs
Drum:
Rotating drum ⇒ slip at both sides; constant speed ⇒ equilibrium
∴ F1 = µ k N1 = 0.3N1;
AB:
ΣM A = 0:
F2 = µk N 2 = 0.3N 2
( 6 in.)( 680 lb ) + ( 6 in.)( F1 ) − (18 in.) N1 = 0
 18 in.

− 6 in.  = ( 6 in.)( 680 lb )
F1 
 0.3

DE:
ΣM D = 0:
F1 = 75.555 lb
( 6 in.) F2 + (18 in.) N 2 − ( 6 in.)( 680 lb ) = 0
18 in. 

F2  6 in. +
 = ( 6 in.)( 680 lb )
0.3 

Drum:
or
or
F2 = 61.818 lb
ΣM C = 0: r ( F1 + F2 ) − M = 0
M = (10 in.)( 75.555 + 61.818 ) lb
M = 1374 lb ⋅ in.
PROBLEM 8.20
A couple M of magnitude 70 lb ⋅ ft is applied to the drum as shown.
Determine the smallest force which must be exerted by the hydraulic
cylinder on joints B and E if the drum is not to rotate.
SOLUTION
FBDs
DE:
Drum:
For minimum T, slip impends at both sides, so
F1 = µ s N1 = 0.4 N1
AB:
ΣM A = 0:
F2 = µ s N 2 = 0.4 N 2
( 6 in.)T + ( 6 in.) F1 − (18 in.) N1 = 0
 18 in.

− 6 in.  = ( 6 in.) T
F1 
 0.4

DE:
ΣM D = 0:
( 6 in.) F2 + (18 in.) N 2 − ( 6 in.) T
18 in. 

F2  6 in. +
 = ( 6 in.) T
0.4 

Drum:
ΣM C = 0:
or
F1 =
T
6.5
F2 =
T
8.5
=0
or
(10 in.) (F1 + F2 ) − 840 lb ⋅ in. = 0
1 
 1
+
T
 = 84 lb
6.5
8.5


T = 309 lb
PROBLEM 8.21
A 19.5-ft ladder AB leans against a wall as shown. Assuming that the
coefficient of static friction µ s is the same at A and B, determine the
smallest value of µ s for which equilibrium is maintained.
SOLUTION
Motion impends at both A and B.
FBD ladder:
FA = µ s N A
FB = µ s N B
ΣFx = 0: FA − N B = 0
Then
FB = µ s N B = µ s2 N A
ΣFy = 0: N A − W + FB = 0
ΣM O = 0: bN B +
or
a = 7.5 ft
N B = FA = µ s N A
or
b = 18 ft
(
)
(
)
N A 1 + µ s2 = W
a
W − aN A = 0
2
aN A − bµ s N A =
µ s2 +
or
a
a
W = N A 1 + µ s2
2
2
2b
µs − 1 = 0
a
2
µs = −
b
b
±   + 1 = −2.4 ± 2.6
a
a
The positive root is physically possible. Therefore,
µ s = 0.200
PROBLEM 8.22
A 19.5-ft ladder AB leans against a wall as shown. Assuming that the
coefficient of static friction µ s is the same at A and B, determine the
smallest value of µ s for which equilibrium is maintained.
SOLUTION
Motion impends at both A and B, so
FBD ladder:
FA = µ s N A
a
W =0
2
ΣM A = 0: lN B −
or
FB = µ s N B = µ s
ΣFx = 0: FA +
a
5
=
l
13
b 12
=
l
13
12.5
(13)
2.5W
13
(13)
2
µ sW −
30
(13)2
W =0
( 30 − 12.5µ s )
W
ΣFy = 0: N A − W +
l = 19.5 ft
a
7.5 ft
W =
W
2l
39 ft
5
12
FB −
NB = 0
13
13
µs N A +
NA −
NB =
or
2.5
W
13
NB =
Then
a = 7.5 ft
FB = µ s N B
and
2
µs
12
5
FB +
NB = 0
13
13
 30 − 12.5µ s
 W
+ 30µ s + 12.5 
=W

2
µ
s

 (13)
or
µ s2 − 5.6333µ s + 1 = 0
µ s = 2.8167 ± 2.6332
or
µ s = 0.1835
and
µ s = 5.45
The larger value is very unlikely unless the surface is treated with
some “non-skid” material.
In any event, the smallest value for equilibrium is µ s = 0.1835
PROBLEM 8.23
End A of a slender, uniform rod of weight W and length L bears on a
horizontal surface as shown, while end B is supported by a cord BC of
length L. Knowing that the coefficient of static friction is 0.40, determine
(a) the value of θ for which motion is impending, (b) the corresponding
value of the tension in the cord.
SOLUTION
FBD rod:
(a) Geometry:
BE =
L
cosθ
2
EF = L sin θ
So
or
Also,
or
L

DE =  cosθ  tan β
2

DF =
L cosθ
2 tan φ s
1
 L cosθ
L  cosθ tan β + sin θ  =
2
 2 tan φs
tan β + 2 tan θ =
1
1
1
=
=
= 2.5
tan φ s
µ s 0.4
L sin θ + L sin β = L
sin θ + sin β = 1
Solving Eqs. (1) and (2) numerically
θ1 = 4.62°
(2)
β1 = 66.85°
θ 2 = 48.20° β 2 = 14.75°
θ = 4.62° and θ = 48.2°
Therefore,
(b) Now
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
and
T
W
=
sin φs
sin ( 90 + β − φ s )
or
T =W
For
(1)
sin φs
sin ( 90 + β − φ s )
θ = 4.62°
T = 0.526W
θ = 48.2°
T = 0.374W
PROBLEM 8.24
A slender rod of length L is lodged between peg C and the vertical wall
and supports a load P at end A. Knowing that the coefficient of static
friction between the peg and the rod is 0.25 and neglecting friction at the
roller, determine the range of values of the ratio L/a for which
equilibrium is maintained.
SOLUTION
ΣM B = 0:
FBD rod:
a
N − L sin 30°P = 0
sin 30°
N =
L 2
LP
sin 30°P =
a
a 4
Impending motion at C : down → F = µ s N 
N
F = ±
4
up → F = − µ s N 
ΣFy = 0: F cos 30° + N sin 30° − P = 0
±
L P 3 LP1
+
= P
a 16 2
a 42
L 1
3
 ±
 =1
a  8 32 
L
32
=
a
4± 3
or
For equilibrium:
L
= 5.583
a
and
L
= 14.110
a
5.58 ≤
L
≤ 14.11
a
PROBLEM 8.25
The basic components of a clamping device are bar AB, locking plate CD,
and lever EFG; the dimensions of the slot in CD are slightly larger than
those of the cross section of AB. To engage the clamp, AB is pushed
against the workpiece, and then force P is applied. Knowing that
P = 160 N and neglecting the friction force between the lever and the
plate, determine the smallest allowable value of the static coefficient of
friction between the bar and the plate.
SOLUTION
FBD Plate:
DC is three-force member and motion impends at C and D (for minimum
µ s ).
OCG = 20° + φs
ODG = 20° − φ s
 24 mm

OG = (10 mm ) tan ( 20° + φs ) = 
+ 10 mm  tan ( 20° − φs )
 sin70°

or
tan ( 20° + φ s ) = 3.5540 tan ( 20° − φs )
Solving numerically
Now
so that
φ s = 10.565°
µ s = tan φs
µ s = 0.1865
PROBLEM 8.26
A window sash having a mass of 4 kg is normally supported by two 2-kg
sash weights. Knowing that the window remains open after one sash cord
has broken, determine the smallest possible value of the coefficient of
static friction. (Assume that the sash is slightly smaller that the frame and
will bind only at points A and D.)
SOLUTION
(
ΣFx = 0:
N A − ND = 0
FA = µ s N A
Impending motion:
ΣM D = 0:
NA =
)
W = ( 4 kg ) 9.81 m/s 2 = 39.24 N
W
2
N A = ND
FD = µ s N D
( 0.36 m )W − ( 0.54 m ) N A − ( 0.72 m ) FA
W =
(
)
T = ( 2 kg ) 9.81 m/s 2 = 19.62 N =
FBD window:
=0
3
N A + 2µ s N A
2
2W
3 + 4µ s
ΣFy = 0: FA − W + T + FD = 0
FA + FD = W − T
=
Now
Then
or
W
2
FA + FD = µ s ( N A + N D ) = 2µ s N A
W
2W
= 2µ s
2
3 + 4µ s
µ s = 0.750
PROBLEM 8.27
The steel-plate clamp shown is used to lift a steel plate H of mass 250 kg.
Knowing that the normal force exerted on steel cam EG by pin D forms
an angle of 40° with the horizontal and neglecting the friction force
between the cam and the pin, determine the smallest allowable value of
the coefficient of static friction.
SOLUTION
FBDs:
(Note: P is vertical as AB is two force member; also P = W since
clamp + plate is a two force FBD)
BCD:
ΣM C = 0:
( 0.37 m ) P − ( 0.46 m ) N D cos 40°
− ( 0.06 m ) N D sin 40° = 0
N D = 0.94642P = 0.94642W
or
EG:
ΣM E = 0:
( 0.18 m ) NG − ( 0.26 m ) FG − ( 0.26 m ) N D cos 40° = 0
Impending motion:
Combining
FG = µ s NG
(18 + 26µs ) NG
= 19.9172 N D
= 18.850W
PROBLEM 8.27 CONTINUED
Plate:
From plate:
Then
FG =
W
2
(18 + 26µs )
so that NG =
W
2µ s
W
= 18.85W
2µ s
µ s = 0.283
PROBLEM 8.28
The 5-in.-radius cam shown is used to control the motion of the plate CD.
Knowing that the coefficient of static friction between the cam and the
plate is 0.45 and neglecting friction at the roller supports, determine
(a) the force P for which motion of the plate is impending knowing that
the plate is 1 in. thick, (b) the largest thickness of the plate for which the
mechanism is self-locking, (that is, for which the plate cannot be moved
however large the force P may be).
SOLUTION
FBDs:
ΣFx = 0: F − P = 0
From plate:
cosθ =
From cam geometry:
F = P
5 in. − t
5 in.
ΣM A = 0: ( 5 in.) sin θ  N − ( 5 in.) cosθ  F − ( 5 in.) Q = 0
F = µs N
Impending motion:
N sin θ − µ s N cosθ = Q = 15 lb
So
N =
Q
sin θ − µ s cosθ
µ sQ
sin θ − µ s cosθ
P = F = µs N =
So
t = 1 in. ⇒ cosθ =
(a)
P=
4 in.
= 0.8; sin θ = 0.6
5 in.
( 0.45)(15 lb )
0.6 − ( 0.45 )( 0.8 )
= 28.125 lb; P = 28.1 lb
P → ∞ : sin θ − µ s cosθ =
(b)
Thus
But
tan θ → µ s = 0.45
( 5 in.) cosθ
= 5 in. − t
so that
or
µ sQ
P
0
θ = 24.228°
t = ( 5 in.) (1 − cosθ )
t = 0.440 in.
PROBLEM 8.29
A child having a mass of 18 kg is seated halfway between the ends of a
small, 16-kg table as shown. The coefficient of static friction is 0.20
between the ends of the table and the floor. If a second child pushes on
edge B of the table top at a point directly opposite to the first child with a
force P lying in a vertical plane parallel to the ends of the table and
having a magnitude of 66 N, determine the range of values of θ for
which the table will (a) tip, (b) slide.
SOLUTION
FBD table + child:
(
)
= 16 kg ( 9.81 m/s ) = 156.96 N
WC = 18 kg 9.81 m/s 2 = 176.58 N
WT
2
(a) Impending tipping about E, N F = FF = 0, and
ΣM E = 0:
( 0.05 m )(176.58 N ) − ( 0.4 m )(156.96 N ) + ( 0.5 m ) P cosθ − ( 0.7 m ) P sin θ
=0
33cosθ − 46.2sin θ = 53.955
Solving numerically
θ = −36.3°
θ = −72.6°
and
−72.6° ≤ θ ≤ −36.3°
Therefore
Impending tipping about F is not possible
(b) For impending slip:
FE = µ s N E = 0.2 N E
ΣFx = 0: FE + FF − P cosθ = 0
or
FF = µ s N F = 0.2 N F
0.2 ( N E + N F ) = ( 66 N ) cosθ
ΣFy = 0: N E + N F − 176.58 N − 156.96 N − P sin θ = 0
N E + N F = ( 66sin θ + 333.54 ) N
So
Solving numerically,
Therefore,
330cosθ = 66sin θ + 333.54
θ = −3.66°
and
θ = −18.96°
−18.96° ≤ θ ≤ −3.66°
PROBLEM 8.30
A pipe of diameter 3 in. is gripped by the stillson wrench shown. Portions
AB and DE of the wrench are rigidly attached to each other, and portion
CF is connected by a pin at D. If the wrench is to grip the pipe and be
self-locking, determine the required minimum coefficients of friction at A
and C.
SOLUTION
ΣM D = 0:
FBD ABD:
( 0.75 in.) N A − ( 5.5 in.) FA
=0
FA = µ A N A
Impending motion:
0.75 − 5.5µ A = 0
Then
µ A = 0.13636
or
µ A = 0.1364
ΣFx = 0: FA − Dx = 0
Dx = FA
Pipe:
ΣFy = 0: NC − N A = 0
NC = N A
FBD DF:
ΣM F = 0:
( 27.5 in.) FC − ( 0.75 in.) NC − ( 25 in.) Dx
Impending motion:
FC = µC NC
Then
27.5µC − 0.75 = 25
But
So
NC = N A
and
=0
FA
NC
FA
= µ A = 0.13636
NA
27.5µC = 0.75 + 25 ( 0.13636 )
µC = 0.1512
PROBLEM 8.31
Solve Problem 8.30 assuming that the diameter of the pipe is 1.5 in.
SOLUTION
ΣM D = 0:
FBD ABD:
Impending motion:
Then
( 0.75
in .) N A − (4 in.)FA = 0
FA = µ A N A
0.75 in. − (4 in.)µ A = 0
µ A = 0.1875
ΣFx = 0: FA − Dx = 0
Dx = FA = 0.1875 N A
so that
FBD Pipe:
ΣFy = 0: NC − N A = 0
NC = N A
FBD DF:
ΣM F = 0:
Impending motion:
( 27.5 in.) FC − ( 0.75 in.) NC − ( 25 in.) Dx
=0
FC = µC NC
27.5µC − 0.75 = 25(0.1875)
But N A = NC (from pipe FBD) so
NA
NC
NA
=1
NC
and µC = 0.1977
PROBLEM 8.32
The 25-kg plate ABCD is attached at A and D to collars which can slide
on the vertical rod. Knowing that the coefficient of static friction is 0.40
between both collars and the rod, determine whether the plate is in
equilibrium in the position shown when the magnitude of the vertical
force applied at E is (a) P = 0, (b) P = 80 N.
SOLUTION
(a) P = 0; assume equilibrium:
FBD plate:
ΣM A = 0:
( 0.7 m ) N D − (1 m )W
ΣFx = 0: N D − N A = 0
( FA )max
W = 25 kg ( 9.81 N/kg )
So
= 245.25 N
( FA + FD )max
=0
ND =
N A = ND =
10W
7
( FD )max
= µs N A
= µs ( N A + N D ) =
10W
7
= µs N D
20µ sW
= 1.143W
7
ΣFy = 0: FA + FD − W = 0
∴
FA + FD = W < ( FA + FD )max
OK.
Plate is in equilibrium
(b) P = 80 N; assume equilibrium:
ΣM A = 0:
or
(1.75 m ) P + ( 0.7 m ) N D − (1 m )W
ND =
W − 1.75P
0.7
ΣFx = 0: N D − N A = 0
( FA )max
So
=0
( FB )max
= µs N A
( FA + FB )max
W − 1.75P
0.7
ND = N A =
= 0.4
= µs N B
2W − 3.5P
= 120.29 N
0.7
ΣFy = 0: FA + FD − W + P = 0
FA + FD = W − P = 165.25 N
( FA + FD )equil
> ( FA + FD )max
Impossible, so plate slides downward
PROBLEM 8.33
In Problem 8.32, determine the range of values of the magnitude P of the
vertical force applied at E for which the plate will move downward.
SOLUTION
ΣM A = 0:
FBD plate:
( 0.7 m ) N D − (1 m ) W + (1.75 m ) P = 0
ND =
W − 1.75P
0.7
ΣFx = 0: N D − N A = 0
so that
Note: NA and ND will be > 0 if P <
(
W = ( 25 kg ) 9.81 m/s 2
)
N A = ND =
W − 1.75P
0.7
4
4
W and < 0 if P > W .
7
7
Impending motion downward: FA and FB are both > 0, so
= 245.25 N
FA = µ s N A =
0.4
4
W − 1.75P = W − P
0.7
7
FD = µ S N D =
4
W −P
7
ΣFy = 0: FA + FD − W + P = 0
4
2 W − P −W + P = 0
7
For P <
4
W;
7
P=
W
= 35.04 N
7
For P >
4
W;
7
P=
5W
= 175.2 N
7
Downward motion for 35.0 N < P < 175.2 N
Alternative Solution
We first observe that for smaller values of the magnitude of P that (Case 1)
the inner left-hand and right-hand surfaces of collars A and D, respectively,
will contact the rod, whereas for larger values of the magnitude of P that
(Case 2) the inner right-hand and left-hand surfaces of collars A and D,
respectively, will contact the rod.
First note:
W = ( 25 kg ) ( 9.81 m/s 2 )
= 245.25 N
PROBLEM 8.33 CONTINUED
Case 1
ΣM D = 0:
( 0.7 m ) N A − (1 m) ( 245.25 N ) + (1.75 m ) P = 0
NA =
or
10 
7 
 245.25 − P  N
7
4 
ΣFx = 0: −N A + N D = 0
ND = N A
or
ΣFy = 0: FA + FD + P − 245.25 N = 0
FA + FD = ( 245.25 − P ) N
or
Now
( FA )max
so that
( FA )max + ( FD )max
( FD )max
= µs N A
= µs N D
= µs ( N A + N D )
10 
7 
= 2 ( 0.4 )   245.25 − P  
4 
7
Case 2
For motion:
FA + FD > ( FA )max + ( FD )max
Substituting
245.25 − P >
or
P > 35.0 N
From Case 1:
ND = N A
8
7 
 245.25 − P 
7
4 
FA + FD = ( 245.25 − P ) N
( FA )max + ( FD )max
= 2µ s N A
ΣM D = 0: − ( 0.7 m ) N A − (1 m) ( 245.25 N ) + (1.75 m ) P = 0
10  7

 P − 245.25  N
7 4

or
NA =
For motion:
FA + FD > ( FA )max + ( FD )max
Substituting:
10  7

245.25 − P > 2 ( 0.4 )   P − 245.25  

 7 4
or
P < 175.2 N
Therefore, have downward motion for
35.0 N < P < 175.2 N
PROBLEM 8.34
A collar B of weight W is attached to the spring AB and can move along
the rod shown. The constant of the spring is 1.5 kN/m and the spring is
unstretched when θ = 0. Knowing that the coefficient of static friction
between the collar and the rod is 0.40, determine the range of values of W
for which equilibrium is maintained when (a) θ = 20o , (b) θ = 30o.
SOLUTION
FBD collar:
Stretch of spring x = AB − a =
Impending motion down:
a
−a
cosθ
 a

 1

Fs = kx = k 
− a  = (1.5 kN/m )( 0.5 m ) 
− 1
θ
θ
cos
cos




 1

= ( 0.75 kN ) 
− 1
 cos θ

ΣFx = 0: N − Fs cosθ = 0
N = Fs cosθ = ( 0.75 kN )(1 − cosθ )
Impending motion up:
Impending slip:
F = µ s N = ( 0.4 )( 0.75 kN )(1 − cosθ )
= ( 0.3 kN )(1 − cosθ )
+ down, – up
ΣFy = 0: Fs sin θ ± F − W = 0
( 0.75 kN )( tan θ
or
(a) θ = 20°:
− sin θ ) ± ( 0.3 kN )(1 − cosθ ) − W = 0
W = ( 0.3 kN ) [ 2.5 ( tan θ − sin θ ) ± (1 − cosθ )]
( impossible )
Wup = −0.00163 kN
Wdown = 0.03455 kN
( OK )
Equilibrium if 0 ≤ W ≤ 34.6 N
(b) θ = 30°:
Wup = 0.01782 kN
Wdown = 0.0982 kN
( OK )
( OK )
Equilibrium if 17.82 N ≤ W ≤ 98.2 N
PROBLEM 8.35
A collar B of weight W is attached to the spring AB and can move along
the rod shown. The constant of the spring is 1.5 kN/m and the spring is
unstretched when θ = 0. Knowing that the coefficient of static friction
between the collar and the rod is 0.40, determine the range of values of W
for which equilibrium is maintained when (a) θ = 20o , (b) θ = 30o.
SOLUTION
FBD collar:
Stretch of spring x = AB − a =
a
−a
cosθ
 a

 1

Fs = k 
− a  = (1.5 kN/m )( 0.5 m ) 
− 1
 cos θ

 cos θ

 1

= ( 0.75 kN ) 
− 1 = ( 750 N )( sec θ − 1)
θ
cos


ΣFy = 0: Fs cosθ − W + N = 0
W = N + ( 750 N ) (1 − cosθ )
or
Impending slip:
F = µ s N (F must be +, but N may be positive or negative)
ΣFx = 0: Fs sin θ − F = 0
or
F = Fs sin θ = ( 750 N )( tan θ − sin θ )
(a) θ = 20°:
F = ( 750 N )( tan 20° − sin 20° ) = 16.4626 N
Impending motion:
(Note: for
N =
F
µs
=
16.4626 N
= 41.156 N
0.4
N < 41.156 N, motion will occur, equilibrium for
N > 41.156)
W = N + ( 750 N )(1 − cos 20° ) = N + 45.231 N
But
So equilibrium for W ≤ 4.07 N and W ≥ 86.4 N
(b) θ = 30°:
F = ( 750 N )( tan 30° − sin 30° ) = 58.013 N
Impending motion:
N =
F
µs
=
58.013
= 145.032 N
0.4
W = N + ( 750 N )(1 − cos 30° ) = N ± 145.03 N
= −44.55 N ( impossible ) , 245.51 N
Equilibrium for W ≥ 246 N
PROBLEM 8.36
The slender rod AB of length l = 30 in. is attached to a collar at B and
rests on a small wheel located at a horizontal distance a = 4 in. from the
vertical rod on which the collar slides. Knowing that the coefficient of
static friction between the collar and the vertical rod is 0.25 and
neglecting the radius of the wheel, determine the range of values of P for
which equilibrium is maintained when Q = 25 lb and θ = 30o.
SOLUTION
FBD rod + collar:
Note: d =
a
4 in.
=
= 8 in., so AC = 22 in.
sin θ
sin 30°
Neglect weights of rod and collar.
ΣM B = 0:
( 30 in.)( sin 30° )( 25 lb ) − (8 in.) C
=0
C = 46.875 lb
ΣFx = 0: N − C cos 30° = 0
N = ( 46.875 lb ) cos 30° = 40.595 lb
F = µ s N = 0.25 ( 40.595 lb )
Impending motion up:
= 10.149 lb
ΣFy = 0: − 25 lb + ( 46.875 lb ) sin 30° − P − 10.149 lb = 0
or
P = −1.563 lb − 10.149 lb = −11.71 lb
Impending motion down: Direction of F is now upward, but still have
F = µ s N = 10.149 lb
ΣFy = 0: − 25 lb + ( 46.875 lb ) sin 30° − P + 10.149 lb = 0
or
P = −1.563 lb + 10.149 lb = 8.59 lb
∴
Equilibrium for −11.71 lb ≤ P ≤ 8.59 lb
PROBLEM 8.37
The 4.5-kg block A and the 3-kg block B are connected by a slender rod
of negligible mass. The coefficient of static friction is 0.40 between all
surfaces of contact. Knowing that for the position shown the rod is
horizontal, determine the range of values of P for which equilibrium is
maintained.
SOLUTION
Note: φ s = tan −1 µ s = tan −1 0.4 = 21.801°
FBDs:
(b) Block A impending slip
(a) Block A impending slip
FAB = WA ctn ( 45° − φ s )
FAB = WA tan ( 45° − φ s )
(
(
)
= ( 4.5 kg ) 9.81 m/s tan ( 23.199° )
2
= 103.005 N
= 18.9193 N
Block B:
(
WB = ( 3 kg ) 9.81 m/s 2
)
= 29.43 N
From Block B:
)
= ( 4.5 kg ) 9.81 m/s 2 ctn ( 23.199° )
ΣFy′ = 0: N − WB cos 30° − FAB sin 30° = 0
PROBLEM 8.37 CONTINUED
N = ( 29.43 N ) cos 30° + (18.9193 N ) sin 30° = 34.947 N
Case (a)
F = µ s N = 0.4 ( 34.947 N ) = 13.979 N
Impending motion:
ΣFx′ = 0: FAB cos 30° − WB sin 30° − 13.979 N − P = 0
P = (18.9193 N ) cos 30° − ( 29.43 N ) sin 30° − 13.979 N
= −12.31 N
N = ( 29.43 N ) cos 30° + (103.005 N ) sin 30° = 76.9896 N
Case (b)
F = 0.4 ( 76.9896 N ) = 30.7958 N
Impending motion:
ΣFx′ = 0:
(103.005 N ) cos 30° − ( 29.43 N ) sin 30° + 30.7958 N − P = 0
P = 105.3 N
For equilibrium −12.31 N ≤ P ≤ 105.3 N
PROBLEM 8.38
Bar AB is attached to collars which can slide on the inclined rods shown.
A force P is applied at point D located at a distance a from end A.
Knowing that the coefficient of static friction µ s between each collar and
the rod upon which it slides is 0.30 and neglecting the weights of the bar
and of the collars, determine the smallest value of the ratio a/L for which
equilibrium is maintained.
SOLUTION
FBD bar + collars:
Impending motion
φ s = tan −1 µ s = tan −1 0.3 = 16.6992°
Neglect weights: 3-force FBD and ( ACB = 90°
So
AC =
a
= l sin ( 45° − φs )
cos ( 45° + φs )
a
= sin ( 45° − 16.6992° ) cos ( 45° + 16.6992° )
l
a
= 0.225
l
PROBLEM 8.39
The 6-kg slender rod AB is pinned at A and rests on the 18-kg cylinder C.
Knowing that the diameter of the cylinder is 250 mm and that the
coefficient of static friction is 0.35 between all surfaces of contact,
determine the largest magnitude of the force P for which equilibrium is
maintained.
SOLUTION
FBD rod:
FBD cylinder:
ΣM A = 0:
( 0.4 m ) N1 − ( 0.25 m ) Wr
=0
N1 = 0.625Wr = 36.7875 N
Cylinder:
ΣFy = 0: N 2 − N1 − WC = 0
ΣM D = 0:
or
N 2 = 0.625Wr + 3Wr = 3.625Wr = 5.8N1
( 0.165 m ) F1 − ( 0.085 m ) F2
=0
or
F2 = 1.941F1
Since µ s1 = µ s 2 , motion will impend first at top of the cylinder
So
F1 = µ s N1 = 0.35 ( 36.7875 N ) = 12.8756 N
and
F2 = 1.941 (12.8756 N ) = 24.992 N
[Check
F2 = 25 N < µ S N 2 = 74.7 N
OK ]
ΣFx = 0: P − F1 − F2 = 0
or
P = 12.8756 N + 24.992 N
or P = 37.9 N
PROBLEM 8.40
Two rods are connected by a collar at B. A couple M A of magnitude
12 lb ⋅ ft is applied to rod AB. Knowing that µ s = 0.30 between the
collar and rod AB, determine the largest couple M C for which
equilibrium will be maintained.
SOLUTION
FBD AB:
ΣM A = 0:
8 in 2 + 4 in 2 ( N ) − M A = 0
N =
Impending motion:
(12 lb ⋅ ft )(12 in./ft )
8.9443 in.
= 16.100 lb
F = µ s N = 0.3 (16.100 lb ) = 4.83 lb
(Note: For max, MC, need F in direction shown; see FBD BC.)
FBD BC + collar:
ΣM C = 0: M C − (17 in.)
or
MC =
1
2
2
N − ( 8 in.)
N − (13 in.)
F =0
5
5
5
17 in.
16 in.
26 in.
(16.100 lb ) +
(16.100 lb ) +
( 4.830 lb ) = 293.77 lb ⋅ in.
5
5
5
( MC )max
= 24.5 lb ⋅ ft
PROBLEM 8.29
A child having a mass of 18 kg is seated halfway between the ends of a
small, 16-kg table as shown. The coefficient of static friction is 0.20
between the ends of the table and the floor. If a second child pushes on
edge B of the table top at a point directly opposite to the first child with a
force P lying in a vertical plane parallel to the ends of the table and
having a magnitude of 66 N, determine the range of values of θ for
which the table will (a) tip, (b) slide.
SOLUTION
FBD table + child:
(
)
= 16 kg ( 9.81 m/s ) = 156.96 N
WC = 18 kg 9.81 m/s 2 = 176.58 N
WT
2
(a) Impending tipping about E, N F = FF = 0, and
ΣM E = 0:
( 0.05 m )(176.58 N ) − ( 0.4 m )(156.96 N ) + ( 0.5 m ) P cosθ − ( 0.7 m ) P sin θ
=0
33cosθ − 46.2sin θ = 53.955
Solving numerically
θ = −36.3°
θ = −72.6°
and
−72.6° ≤ θ ≤ −36.3°
Therefore
Impending tipping about F is not possible
(b) For impending slip:
FE = µ s N E = 0.2 N E
ΣFx = 0: FE + FF − P cosθ = 0
or
FF = µ s N F = 0.2 N F
0.2 ( N E + N F ) = ( 66 N ) cosθ
ΣFy = 0: N E + N F − 176.58 N − 156.96 N − P sin θ = 0
N E + N F = ( 66sin θ + 333.54 ) N
So
Solving numerically,
Therefore,
330cosθ = 66sin θ + 333.54
θ = −3.66°
and
θ = −18.96°
−18.96° ≤ θ ≤ −3.66°
PROBLEM 8.30
A pipe of diameter 3 in. is gripped by the stillson wrench shown. Portions
AB and DE of the wrench are rigidly attached to each other, and portion
CF is connected by a pin at D. If the wrench is to grip the pipe and be
self-locking, determine the required minimum coefficients of friction at A
and C.
SOLUTION
ΣM D = 0:
FBD ABD:
( 0.75 in.) N A − ( 5.5 in.) FA
=0
FA = µ A N A
Impending motion:
0.75 − 5.5µ A = 0
Then
µ A = 0.13636
or
µ A = 0.1364
ΣFx = 0: FA − Dx = 0
Dx = FA
Pipe:
ΣFy = 0: NC − N A = 0
NC = N A
FBD DF:
ΣM F = 0:
( 27.5 in.) FC − ( 0.75 in.) NC − ( 25 in.) Dx
Impending motion:
FC = µC NC
Then
27.5µC − 0.75 = 25
But
So
NC = N A
and
=0
FA
NC
FA
= µ A = 0.13636
NA
27.5µC = 0.75 + 25 ( 0.13636 )
µC = 0.1512
PROBLEM 8.31
Solve Problem 8.30 assuming that the diameter of the pipe is 1.5 in.
SOLUTION
ΣM D = 0:
FBD ABD:
Impending motion:
Then
( 0.75
in .) N A − (4 in.)FA = 0
FA = µ A N A
0.75 in. − (4 in.)µ A = 0
µ A = 0.1875
ΣFx = 0: FA − Dx = 0
Dx = FA = 0.1875 N A
so that
FBD Pipe:
ΣFy = 0: NC − N A = 0
NC = N A
FBD DF:
ΣM F = 0:
Impending motion:
( 27.5 in.) FC − ( 0.75 in.) NC − ( 25 in.) Dx
=0
FC = µC NC
27.5µC − 0.75 = 25(0.1875)
But N A = NC (from pipe FBD) so
NA
NC
NA
=1
NC
and µC = 0.1977
PROBLEM 8.32
The 25-kg plate ABCD is attached at A and D to collars which can slide
on the vertical rod. Knowing that the coefficient of static friction is 0.40
between both collars and the rod, determine whether the plate is in
equilibrium in the position shown when the magnitude of the vertical
force applied at E is (a) P = 0, (b) P = 80 N.
SOLUTION
(a) P = 0; assume equilibrium:
FBD plate:
ΣM A = 0:
( 0.7 m ) N D − (1 m )W
ΣFx = 0: N D − N A = 0
( FA )max
W = 25 kg ( 9.81 N/kg )
So
= 245.25 N
( FA + FD )max
=0
ND =
N A = ND =
10W
7
( FD )max
= µs N A
= µs ( N A + N D ) =
10W
7
= µs N D
20µ sW
= 1.143W
7
ΣFy = 0: FA + FD − W = 0
∴
FA + FD = W < ( FA + FD )max
OK.
Plate is in equilibrium
(b) P = 80 N; assume equilibrium:
ΣM A = 0:
or
(1.75 m ) P + ( 0.7 m ) N D − (1 m )W
ND =
W − 1.75P
0.7
ΣFx = 0: N D − N A = 0
( FA )max
So
=0
( FB )max
= µs N A
( FA + FB )max
W − 1.75P
0.7
ND = N A =
= 0.4
= µs N B
2W − 3.5P
= 120.29 N
0.7
ΣFy = 0: FA + FD − W + P = 0
FA + FD = W − P = 165.25 N
( FA + FD )equil
> ( FA + FD )max
Impossible, so plate slides downward
PROBLEM 8.33
In Problem 8.32, determine the range of values of the magnitude P of the
vertical force applied at E for which the plate will move downward.
SOLUTION
ΣM A = 0:
FBD plate:
( 0.7 m ) N D − (1 m ) W + (1.75 m ) P = 0
ND =
W − 1.75P
0.7
ΣFx = 0: N D − N A = 0
so that
Note: NA and ND will be > 0 if P <
(
W = ( 25 kg ) 9.81 m/s 2
)
N A = ND =
W − 1.75P
0.7
4
4
W and < 0 if P > W .
7
7
Impending motion downward: FA and FB are both > 0, so
= 245.25 N
FA = µ s N A =
0.4
4
W − 1.75P = W − P
0.7
7
FD = µ S N D =
4
W −P
7
ΣFy = 0: FA + FD − W + P = 0
4
2 W − P −W + P = 0
7
For P <
4
W;
7
P=
W
= 35.04 N
7
For P >
4
W;
7
P=
5W
= 175.2 N
7
Downward motion for 35.0 N < P < 175.2 N
Alternative Solution
We first observe that for smaller values of the magnitude of P that (Case 1)
the inner left-hand and right-hand surfaces of collars A and D, respectively,
will contact the rod, whereas for larger values of the magnitude of P that
(Case 2) the inner right-hand and left-hand surfaces of collars A and D,
respectively, will contact the rod.
First note:
W = ( 25 kg ) ( 9.81 m/s 2 )
= 245.25 N
PROBLEM 8.33 CONTINUED
Case 1
ΣM D = 0:
( 0.7 m ) N A − (1 m) ( 245.25 N ) + (1.75 m ) P = 0
NA =
or
10 
7 
 245.25 − P  N
7
4 
ΣFx = 0: −N A + N D = 0
ND = N A
or
ΣFy = 0: FA + FD + P − 245.25 N = 0
FA + FD = ( 245.25 − P ) N
or
Now
( FA )max
so that
( FA )max + ( FD )max
( FD )max
= µs N A
= µs N D
= µs ( N A + N D )
10 
7 
= 2 ( 0.4 )   245.25 − P  
4 
7
Case 2
For motion:
FA + FD > ( FA )max + ( FD )max
Substituting
245.25 − P >
or
P > 35.0 N
From Case 1:
ND = N A
8
7 
 245.25 − P 
7
4 
FA + FD = ( 245.25 − P ) N
( FA )max + ( FD )max
= 2µ s N A
ΣM D = 0: − ( 0.7 m ) N A − (1 m) ( 245.25 N ) + (1.75 m ) P = 0
10  7

 P − 245.25  N
7 4

or
NA =
For motion:
FA + FD > ( FA )max + ( FD )max
Substituting:
10  7

245.25 − P > 2 ( 0.4 )   P − 245.25  

 7 4
or
P < 175.2 N
Therefore, have downward motion for
35.0 N < P < 175.2 N
PROBLEM 8.34
A collar B of weight W is attached to the spring AB and can move along
the rod shown. The constant of the spring is 1.5 kN/m and the spring is
unstretched when θ = 0. Knowing that the coefficient of static friction
between the collar and the rod is 0.40, determine the range of values of W
for which equilibrium is maintained when (a) θ = 20o , (b) θ = 30o.
SOLUTION
FBD collar:
Stretch of spring x = AB − a =
Impending motion down:
a
−a
cosθ
 a

 1

Fs = kx = k 
− a  = (1.5 kN/m )( 0.5 m ) 
− 1
θ
θ
cos
cos




 1

= ( 0.75 kN ) 
− 1
 cos θ

ΣFx = 0: N − Fs cosθ = 0
N = Fs cosθ = ( 0.75 kN )(1 − cosθ )
Impending motion up:
Impending slip:
F = µ s N = ( 0.4 )( 0.75 kN )(1 − cosθ )
= ( 0.3 kN )(1 − cosθ )
+ down, – up
ΣFy = 0: Fs sin θ ± F − W = 0
( 0.75 kN )( tan θ
or
(a) θ = 20°:
− sin θ ) ± ( 0.3 kN )(1 − cosθ ) − W = 0
W = ( 0.3 kN ) [ 2.5 ( tan θ − sin θ ) ± (1 − cosθ )]
( impossible )
Wup = −0.00163 kN
Wdown = 0.03455 kN
( OK )
Equilibrium if 0 ≤ W ≤ 34.6 N
(b) θ = 30°:
Wup = 0.01782 kN
Wdown = 0.0982 kN
( OK )
( OK )
Equilibrium if 17.82 N ≤ W ≤ 98.2 N
PROBLEM 8.35
A collar B of weight W is attached to the spring AB and can move along
the rod shown. The constant of the spring is 1.5 kN/m and the spring is
unstretched when θ = 0. Knowing that the coefficient of static friction
between the collar and the rod is 0.40, determine the range of values of W
for which equilibrium is maintained when (a) θ = 20o , (b) θ = 30o.
SOLUTION
FBD collar:
Stretch of spring x = AB − a =
a
−a
cosθ
 a

 1

Fs = k 
− a  = (1.5 kN/m )( 0.5 m ) 
− 1
 cos θ

 cos θ

 1

= ( 0.75 kN ) 
− 1 = ( 750 N )( sec θ − 1)
θ
cos


ΣFy = 0: Fs cosθ − W + N = 0
W = N + ( 750 N ) (1 − cosθ )
or
Impending slip:
F = µ s N (F must be +, but N may be positive or negative)
ΣFx = 0: Fs sin θ − F = 0
or
F = Fs sin θ = ( 750 N )( tan θ − sin θ )
(a) θ = 20°:
F = ( 750 N )( tan 20° − sin 20° ) = 16.4626 N
Impending motion:
(Note: for
N =
F
µs
=
16.4626 N
= 41.156 N
0.4
N < 41.156 N, motion will occur, equilibrium for
N > 41.156)
W = N + ( 750 N )(1 − cos 20° ) = N + 45.231 N
But
So equilibrium for W ≤ 4.07 N and W ≥ 86.4 N
(b) θ = 30°:
F = ( 750 N )( tan 30° − sin 30° ) = 58.013 N
Impending motion:
N =
F
µs
=
58.013
= 145.032 N
0.4
W = N + ( 750 N )(1 − cos 30° ) = N ± 145.03 N
= −44.55 N ( impossible ) , 245.51 N
Equilibrium for W ≥ 246 N
PROBLEM 8.36
The slender rod AB of length l = 30 in. is attached to a collar at B and
rests on a small wheel located at a horizontal distance a = 4 in. from the
vertical rod on which the collar slides. Knowing that the coefficient of
static friction between the collar and the vertical rod is 0.25 and
neglecting the radius of the wheel, determine the range of values of P for
which equilibrium is maintained when Q = 25 lb and θ = 30o.
SOLUTION
FBD rod + collar:
Note: d =
a
4 in.
=
= 8 in., so AC = 22 in.
sin θ
sin 30°
Neglect weights of rod and collar.
ΣM B = 0:
( 30 in.)( sin 30° )( 25 lb ) − (8 in.) C
=0
C = 46.875 lb
ΣFx = 0: N − C cos 30° = 0
N = ( 46.875 lb ) cos 30° = 40.595 lb
F = µ s N = 0.25 ( 40.595 lb )
Impending motion up:
= 10.149 lb
ΣFy = 0: − 25 lb + ( 46.875 lb ) sin 30° − P − 10.149 lb = 0
or
P = −1.563 lb − 10.149 lb = −11.71 lb
Impending motion down: Direction of F is now upward, but still have
F = µ s N = 10.149 lb
ΣFy = 0: − 25 lb + ( 46.875 lb ) sin 30° − P + 10.149 lb = 0
or
P = −1.563 lb + 10.149 lb = 8.59 lb
∴
Equilibrium for −11.71 lb ≤ P ≤ 8.59 lb
PROBLEM 8.37
The 4.5-kg block A and the 3-kg block B are connected by a slender rod
of negligible mass. The coefficient of static friction is 0.40 between all
surfaces of contact. Knowing that for the position shown the rod is
horizontal, determine the range of values of P for which equilibrium is
maintained.
SOLUTION
Note: φ s = tan −1 µ s = tan −1 0.4 = 21.801°
FBDs:
(b) Block A impending slip
(a) Block A impending slip
FAB = WA ctn ( 45° − φ s )
FAB = WA tan ( 45° − φ s )
(
(
)
= ( 4.5 kg ) 9.81 m/s tan ( 23.199° )
2
= 103.005 N
= 18.9193 N
Block B:
(
WB = ( 3 kg ) 9.81 m/s 2
)
= 29.43 N
From Block B:
)
= ( 4.5 kg ) 9.81 m/s 2 ctn ( 23.199° )
ΣFy′ = 0: N − WB cos 30° − FAB sin 30° = 0
PROBLEM 8.37 CONTINUED
N = ( 29.43 N ) cos 30° + (18.9193 N ) sin 30° = 34.947 N
Case (a)
F = µ s N = 0.4 ( 34.947 N ) = 13.979 N
Impending motion:
ΣFx′ = 0: FAB cos 30° − WB sin 30° − 13.979 N − P = 0
P = (18.9193 N ) cos 30° − ( 29.43 N ) sin 30° − 13.979 N
= −12.31 N
N = ( 29.43 N ) cos 30° + (103.005 N ) sin 30° = 76.9896 N
Case (b)
F = 0.4 ( 76.9896 N ) = 30.7958 N
Impending motion:
ΣFx′ = 0:
(103.005 N ) cos 30° − ( 29.43 N ) sin 30° + 30.7958 N − P = 0
P = 105.3 N
For equilibrium −12.31 N ≤ P ≤ 105.3 N
PROBLEM 8.38
Bar AB is attached to collars which can slide on the inclined rods shown.
A force P is applied at point D located at a distance a from end A.
Knowing that the coefficient of static friction µ s between each collar and
the rod upon which it slides is 0.30 and neglecting the weights of the bar
and of the collars, determine the smallest value of the ratio a/L for which
equilibrium is maintained.
SOLUTION
FBD bar + collars:
Impending motion
φ s = tan −1 µ s = tan −1 0.3 = 16.6992°
Neglect weights: 3-force FBD and ( ACB = 90°
So
AC =
a
= l sin ( 45° − φs )
cos ( 45° + φs )
a
= sin ( 45° − 16.6992° ) cos ( 45° + 16.6992° )
l
a
= 0.225
l
PROBLEM 8.39
The 6-kg slender rod AB is pinned at A and rests on the 18-kg cylinder C.
Knowing that the diameter of the cylinder is 250 mm and that the
coefficient of static friction is 0.35 between all surfaces of contact,
determine the largest magnitude of the force P for which equilibrium is
maintained.
SOLUTION
FBD rod:
FBD cylinder:
ΣM A = 0:
( 0.4 m ) N1 − ( 0.25 m ) Wr
=0
N1 = 0.625Wr = 36.7875 N
Cylinder:
ΣFy = 0: N 2 − N1 − WC = 0
ΣM D = 0:
or
N 2 = 0.625Wr + 3Wr = 3.625Wr = 5.8N1
( 0.165 m ) F1 − ( 0.085 m ) F2
=0
or
F2 = 1.941F1
Since µ s1 = µ s 2 , motion will impend first at top of the cylinder
So
F1 = µ s N1 = 0.35 ( 36.7875 N ) = 12.8756 N
and
F2 = 1.941 (12.8756 N ) = 24.992 N
[Check
F2 = 25 N < µ S N 2 = 74.7 N
OK ]
ΣFx = 0: P − F1 − F2 = 0
or
P = 12.8756 N + 24.992 N
or P = 37.9 N
PROBLEM 8.40
Two rods are connected by a collar at B. A couple M A of magnitude
12 lb ⋅ ft is applied to rod AB. Knowing that µ s = 0.30 between the
collar and rod AB, determine the largest couple M C for which
equilibrium will be maintained.
SOLUTION
FBD AB:
ΣM A = 0:
8 in 2 + 4 in 2 ( N ) − M A = 0
N =
Impending motion:
(12 lb ⋅ ft )(12 in./ft )
8.9443 in.
= 16.100 lb
F = µ s N = 0.3 (16.100 lb ) = 4.83 lb
(Note: For max, MC, need F in direction shown; see FBD BC.)
FBD BC + collar:
ΣM C = 0: M C − (17 in.)
or
MC =
1
2
2
N − ( 8 in.)
N − (13 in.)
F =0
5
5
5
17 in.
16 in.
26 in.
(16.100 lb ) +
(16.100 lb ) +
( 4.830 lb ) = 293.77 lb ⋅ in.
5
5
5
( MC )max
= 24.5 lb ⋅ ft
PROBLEM 8.41
In Problem 8.40, determine the smallest couple M C for which
equilibrium will be maintained.
SOLUTION
FBD AB:
ΣM A = 0: N
)
(
8 in 2 + 4 in 2 − M A = 0
N =
Impending motion:
(12 lb ⋅ ft )(12 in./ft )
8.9443 in.
= 16.100 lb
F = µ s N = 0.3 (16.100 lb )
= 4.830 lb
(Note: For min. MC, need F in direction shown; see FBD BC.)
FBD BC + collar:
ΣM C = 0: M C − (17 in.)
MC =
1
2
2
F =0
N − ( 8 in.)
N + (13 in.)
5
5
5
1
(17 in. + 16 in.)(16.100 lb ) − ( 26 in.)( 4.830 lb ) 
5
= 181.44 lb ⋅ in.
( MC )min
= 15.12 lb ⋅ ft
W
PROBLEM 8.42
Blocks A, B, and C having the masses shown are at rest on an incline.
Denoting by µ s the coefficient of static friction between all surfaces of
contact, determine the smallest value of µ s for which equilibrium is
maintained.
SOLUTION
For impending motion, C will start down and A will start up. Since, the
normal force between B and C is larger than that between A and B, the
corresponding friction force can be larger as well. Thus we assume that
motion impends between A and B.
FBD A:
ΣFy′ = 0: N AB − WA cos30° = 0; N AB =
FAB = µ s N AB =
Impending motion:
3
WA
2
3
WAµ s
2
ΣFx′ = 0: T − FAB − WA sin 30° = 0
T =
or
(
) W2
3µ s + 1
A
ΣFy′ = 0: NCD − N AB − (WB + WC ) cos30° = 0
FBD B + C:
NCD =
or
Impending motion:
3
(WA + WB + WC )
2
FCD = µ s NCD =
3
(WA + WB + WC ) µs
2
ΣFx′ = 0: T + FAB + FCD − (WB + WC ) sin 30° = 0
T =
Equating T’s:
µs =
WB + WC
3
−
µ s ( 2WA + WB + WC )
2
2
3µ s ( 3WA + WB + WC ) = WB + WC − WA
mB + mC − mA
1.5 kg + 4 kg − 2 kg
=
+
+
3
m
m
m
3
6
( A B C)
( kg + 1.5 kg + 4 kg ) 3
µ s = 0.1757 W
PROBLEM 8.42 CONTINUED
FBD B:
ΣFy′ = 0: N BC − N AB − WB cos30° = 0
N BC =
or
( FBC ) max
3
(WA + WB )
2
= µ s N BC = 0.1757
3
(WA + WB )
2
(
= 0.1522 ( mA + mB ) g = 0.1522 ( 3.5 kg ) 9.81 m/s 2
= 5.224 N
ΣFx′ = 0: FAB + FBC − WB sin 30° = 0
or
FBC = − FAB +
1
3
W
WB = −
WA ( 0.1757 ) + B
2
2
2
= ( −0.1522mA + 0.5mB ) g
=  −0.1522 ( 2 kg ) + 0.5 (1.5 kg ) ( 9.81 m/s 2 )
= 4.37 N
FBC < FBC max
OK
)
PROBLEM 8.43
A slender steel rod of length 9 in. is placed inside a pipe as shown.
Knowing that the coefficient of static friction between the rod and the
pipe is 0.20, determine the largest value of θ for which the rod will not
fall into the pipe.
SOLUTION
ΣM A = 0:
FBD rod:
3 in.
N B − ( 4.5 in.) cosθ  W = 0
cosθ
N B = (1.5cos 2 θ )W
or
Impending motion:
FB = µ s N B = (1.5µ s cos 2 θ )W
= ( 0.3cos 2 θ )W
ΣFx = 0: N A − N B sin θ + FB cosθ = 0
N A = (1.5cos 2 θ )W ( sin θ − 0.2 cosθ )
or
Impending motion:
FA = µ s N A
= ( 0.3cos 2 θ )W ( sin θ − 0.2 cosθ )
ΣFy = 0: FA + N B cosθ + FB sinθ − W = 0
(
FA = W 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ
or
)
Equating FA’s
0.3cos 2 θ ( sin θ − 0.2cosθ ) = 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ
0.6cos 2 θ sin θ + 1.44cos3 θ = 1
Solving numerically
θ = 35.8° W
PROBLEM 8.44
In Problem 8.43, determine the smallest value of
will not fall out of the pipe.
θ for which the rod
SOLUTION
ΣM A = 0:
FBD rod:
3 in.
N B − ( 4.5 in.) cosθ  W = 0
cosθ
N B = 1.5W cos 2 θ
or
Impending motion:
(
FB = µ s N B = 0.2 1.5W cos 2 θ
)
= 0.3W cos 2 θ
ΣFx = 0: N A − N B sin θ − FB cosθ = 0
N A = W cos 2 θ (1.5sin θ + 0.3cosθ )
or
Impending motion:
FA = µ s N A
= W cos 2 θ ( 0.3sin θ + 0.06 cosθ )
ΣFy = 0: N B cosθ − FB sin θ − W − FA = 0
or
FA = W cos 2 θ (1.5cosθ − 0.3sin θ ) − 1
Equating FA’s:
cos 2 θ (1.44cosθ − 0.6sin θ ) = 1
Solving numerically
θ = 20.5° W
PROBLEM 8.45
Two slender rods of negligible weight are pin-connected at C and
attached to blocks A and B, each of weight W. Knowing that θ = 70o
and that the coefficient of static friction between the blocks and the
horizontal surface is 0.30, determine the largest value of P for which
equilibrium is maintained.
SOLUTION
FBD pin C:
FAB = P sin10° = 0.173648P
FBC = P cos10° = 0.98481P
ΣFy = 0: N A − W − FAB sin 30° = 0
FBD block A:
N A = W + 0.173648P sin 30° = W + 0.086824P
or
ΣFx = 0: FA − FAB cos30° = 0
FA = 0.173648P cos30° = 0.150384P
or
FA = µ s N A
For impending motion at A:
Then
NA =
FA
µs
: W + 0.086824 P =
0.150384
P
0.3
P = 2.413W
or
ΣFy = 0: N B − W − FBC cos30° = 0
N B = W + 0.98481P cos30° = W + 0.85287 P
ΣFx = 0: FBC sin 30° − FB = 0
FBD block B:
FB = 0.98481P sin 30° = 0.4924 P
FB = µ s N B
For impending motion at B:
Then
or
NB =
FB
µs
: W + 0.85287 P =
0.4924P
0.3
P = 1.268W
Thus, maximum P for equilibrium
Pmax = 1.268W W
PROBLEM 8.46
A 40-lb weight is hung from a lever which rests against a 10° wedge at A
and is supported by a frictionless hinge at C. Knowing that the coefficient
of static friction is 0.25 at both surfaces of the wedge and that for the
position shown the spring is stretched 4 in., determine (a) the magnitude
of the force P for which motion of the wedge is impending, (b) the
components of the corresponding reaction at C.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
 4 in. 
Fs = kx = ( 240 lb/ft ) 
 = 80 lb
 12 in./ft 
FBD lever:
ΣM C = 0:
(12 in.)(80 lb ) − (16 in.)( 40 lb ) − ( 21 in.) RA cos (φs − 10° )
+ ( 2 in.) RA sin (φs − 10° ) = 0
(b)
or
RA = 15.3793 lb
ΣFx = 0:
(15.379 lb ) sin ( 4.036° ) − Cx
ΣFy = 0:
(15.379 lb ) cos ( 4.036° ) − 80 lb − 40 lb + C y
=0
C x = 1.082 lb
=0
C y = 104.7 lb
W
W
FBD wedge:
ΣFy = 0: RW cos14.036° − (15.3793 lb ) cos 4.036° = 0
or
(a)
RW = 15.8133 lb
ΣFx = 0: P − (15.3793 lb ) sin 4.036° − (15.8133 lb ) sin14.036° = 0
P = 4.92 lb W
PROBLEM 8.47
Solve Problem 8.46 assuming that force P is directed to the left.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
 4 in. 
Fs = kx = ( 240 lb/ft ) 
 = 80 lb
 12 in./ft 
FBD lever:
ΣM C = 0:
(12 in.)(80 lb ) − (16 in.)( 40 lb ) − ( 21 in.) RA cos 24.036°
− ( 2 in.) RA sin 24.036° = 0
RA = 16.005 lb
or
(b)
ΣFx = 0: C x − (16.005 lb ) sin 24.036° = 0
ΣFy = 0: C y − 80 lb − 40 lb + (16.005 lb ) cos ( 24.036° ) = 0
C x = 6.52 lb
W
C y = 105.4 lb W
FBD wedge:
ΣFy = 0: RW cos14.036° − (16.005 lb ) cos 24.036° = 0
(a)
or
RW = 15.067 lb
ΣFx = 0:
(16.005 lb ) sin 24.036° + (15.067 lb ) sin14.036° − P = 0
P = 10.17 lb W
PROBLEM 8.48
Two 8° wedges of negligible mass are used to move and position a
240-kg block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, determine the magnitude of the force P for which
motion of the block is impending.
SOLUTION
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
(
)
W = 240 kg 9.81 m/s 2 = 2354.4 N
FBD block:
R2
2354.4 N
=
sin 41.801° sin 46.398°
R2 = 2167.12 N
FBD wedge:
P
2167.12 N
=
sin 51.602° sin 60.199°
P = 1957 N
P = 1.957 kN W
PROBLEM 8.49
Two 8° wedges of negligible mass are used to move and position a
240-kg block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, determine the magnitude of the force P for which
motion of the block is impending.
SOLUTION
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
(
)
W = 240 kg 9.81 m/s 2 = 2354.4 N
FBD block + wedge:
R2
2354.4 N
=
sin 41.801° sin 38.398°
R2 = 2526.6 N
FBD wedge:
P
2526.6 N
=
sin 51.602° sin 68.199°
P = 2132.7 N
P = 2.13 kN W
PROBLEM 8.50
The elevation of the end of the steel beam supported by a concrete floor
is adjusted by means of the steel wedges E and F. The base plate CD has
been welded to the lower flange of the beam, and the end reaction of the
beam is known to be 150 kN. The coefficient of static friction is 0.30
between the two steel surfaces and 0.60 between the steel and the
concrete. If the horizontal motion of the beam is prevented by the force
Q, determine (a) the force P required to raise the beam, (b) the
corresponding force Q.
SOLUTION
φ s = tan −1 µ s = tan −1 0.3 = 16.70° for steel on steel
FBD AB + CD:
ΣFy = 0: N − 150 kN = 0
Impending motion:
N = 150 kN
F = µ s N = 0.3 (150 kN ) = 45 kN
ΣFx = 0: F − Q = 0
(b) Q = 45.0 kN
FBD top wedge:
W
Assume bottom wedge doesn’t move:
ΣFy = 0: RW cos (10° + 16.70° ) − 150 kN = 0
RW = 167.9 kN
ΣFx = 0: P − 45 kN − (167.9 kN ) sin 26.70° = 0
P = 120.44 kN
FBD bottom wedge:
( a ) P = 120.4 kN
W
Bottom wedge is two-force member, so φ = 26.70° for equilibrium, but
φ s = tan −1 µ s = tan −1 0.6 = 31.0° ( steel on concrete )
So
φ < φs
OK.
PROBLEM 8.51
The elevation of the end of the steel beam supported by a concrete floor
is adjusted by means of the steel wedges E and F. The base plate CD has
been welded to the lower flange of the beam, and the end reaction of the
beam is known to be 150 kN. The coefficient of static friction is 0.30
between the two steel surfaces and 0.60 between the steel and the
concrete. If the horizontal motion of the beam is prevented by the force
Q, determine (a) the force P required to raise the beam, (b) the
corresponding force Q.
SOLUTION
φ s = tan −1 µ s = tan −1 0.3 = 16.70° for steel on steel
FBD AB + CD + top wedge: Assume top wedge doesn’t move
Rw =
150 kN
= 167.90 kN
cos26.70°
Q = (150 kN ) tan 26.70° = 75.44 kN
(b) Q = 75.4 kN
FBD top wedge:
ΣFx = 0: 75.44 kN − 167.9 kN sin 26.70° − F = 0
F = 0 as expected.
PROBLEM 8.51 CONTINUED
FBD bottom wedge:
φ s = tan −1 µ s = tan −1 0.6 = 30.96° steel on concrete
P
167.90 kN
=
sin 57.66°
sin 59.04°
(a) P = 165.4 kN
PROBLEM 8.52
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that θ = 45°, determine the smallest force P required to raise
block A.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
750 lb
=
sin104.036° sin16.928°
R2 = 2499.0 lb
FBD wedge B:
P
2499.0
=
sin 73.072° sin 75.964°
P = 2464 lb
P = 2.46 kips
PROBLEM 8.53
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that θ = 45°, determine the smallest force P for which
equilibrium is maintained.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
750 lb
=
sin ( 75.964° ) sin ( 73.072° )
R2 = 760.56 lb
FBD wedge B:
P
760.56
=
sin16.928° sin104.036°
P = 228.3 lb
P = 228 lb
PROBLEM 8.41
In Problem 8.40, determine the smallest couple M C for which
equilibrium will be maintained.
SOLUTION
FBD AB:
ΣM A = 0: N
)
(
8 in 2 + 4 in 2 − M A = 0
N =
Impending motion:
(12 lb ⋅ ft )(12 in./ft )
8.9443 in.
= 16.100 lb
F = µ s N = 0.3 (16.100 lb )
= 4.830 lb
(Note: For min. MC, need F in direction shown; see FBD BC.)
FBD BC + collar:
ΣM C = 0: M C − (17 in.)
MC =
1
2
2
F =0
N − ( 8 in.)
N + (13 in.)
5
5
5
1
(17 in. + 16 in.)(16.100 lb ) − ( 26 in.)( 4.830 lb ) 
5
= 181.44 lb ⋅ in.
( MC )min
= 15.12 lb ⋅ ft
W
PROBLEM 8.42
Blocks A, B, and C having the masses shown are at rest on an incline.
Denoting by µ s the coefficient of static friction between all surfaces of
contact, determine the smallest value of µ s for which equilibrium is
maintained.
SOLUTION
For impending motion, C will start down and A will start up. Since, the
normal force between B and C is larger than that between A and B, the
corresponding friction force can be larger as well. Thus we assume that
motion impends between A and B.
FBD A:
ΣFy′ = 0: N AB − WA cos30° = 0; N AB =
FAB = µ s N AB =
Impending motion:
3
WA
2
3
WAµ s
2
ΣFx′ = 0: T − FAB − WA sin 30° = 0
T =
or
(
) W2
3µ s + 1
A
ΣFy′ = 0: NCD − N AB − (WB + WC ) cos30° = 0
FBD B + C:
NCD =
or
Impending motion:
3
(WA + WB + WC )
2
FCD = µ s NCD =
3
(WA + WB + WC ) µs
2
ΣFx′ = 0: T + FAB + FCD − (WB + WC ) sin 30° = 0
T =
Equating T’s:
µs =
WB + WC
3
−
µ s ( 2WA + WB + WC )
2
2
3µ s ( 3WA + WB + WC ) = WB + WC − WA
mB + mC − mA
1.5 kg + 4 kg − 2 kg
=
+
+
3
m
m
m
3
6
( A B C)
( kg + 1.5 kg + 4 kg ) 3
µ s = 0.1757 W
PROBLEM 8.42 CONTINUED
FBD B:
ΣFy′ = 0: N BC − N AB − WB cos30° = 0
N BC =
or
( FBC ) max
3
(WA + WB )
2
= µ s N BC = 0.1757
3
(WA + WB )
2
(
= 0.1522 ( mA + mB ) g = 0.1522 ( 3.5 kg ) 9.81 m/s 2
= 5.224 N
ΣFx′ = 0: FAB + FBC − WB sin 30° = 0
or
FBC = − FAB +
1
3
W
WB = −
WA ( 0.1757 ) + B
2
2
2
= ( −0.1522mA + 0.5mB ) g
=  −0.1522 ( 2 kg ) + 0.5 (1.5 kg ) ( 9.81 m/s 2 )
= 4.37 N
FBC < FBC max
OK
)
PROBLEM 8.43
A slender steel rod of length 9 in. is placed inside a pipe as shown.
Knowing that the coefficient of static friction between the rod and the
pipe is 0.20, determine the largest value of θ for which the rod will not
fall into the pipe.
SOLUTION
ΣM A = 0:
FBD rod:
3 in.
N B − ( 4.5 in.) cosθ  W = 0
cosθ
N B = (1.5cos 2 θ )W
or
Impending motion:
FB = µ s N B = (1.5µ s cos 2 θ )W
= ( 0.3cos 2 θ )W
ΣFx = 0: N A − N B sin θ + FB cosθ = 0
N A = (1.5cos 2 θ )W ( sin θ − 0.2 cosθ )
or
Impending motion:
FA = µ s N A
= ( 0.3cos 2 θ )W ( sin θ − 0.2 cosθ )
ΣFy = 0: FA + N B cosθ + FB sinθ − W = 0
(
FA = W 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ
or
)
Equating FA’s
0.3cos 2 θ ( sin θ − 0.2cosθ ) = 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ
0.6cos 2 θ sin θ + 1.44cos3 θ = 1
Solving numerically
θ = 35.8° W
PROBLEM 8.44
In Problem 8.43, determine the smallest value of
will not fall out of the pipe.
θ for which the rod
SOLUTION
ΣM A = 0:
FBD rod:
3 in.
N B − ( 4.5 in.) cosθ  W = 0
cosθ
N B = 1.5W cos 2 θ
or
Impending motion:
(
FB = µ s N B = 0.2 1.5W cos 2 θ
)
= 0.3W cos 2 θ
ΣFx = 0: N A − N B sin θ − FB cosθ = 0
N A = W cos 2 θ (1.5sin θ + 0.3cosθ )
or
Impending motion:
FA = µ s N A
= W cos 2 θ ( 0.3sin θ + 0.06 cosθ )
ΣFy = 0: N B cosθ − FB sin θ − W − FA = 0
or
FA = W cos 2 θ (1.5cosθ − 0.3sin θ ) − 1
Equating FA’s:
cos 2 θ (1.44cosθ − 0.6sin θ ) = 1
Solving numerically
θ = 20.5° W
PROBLEM 8.45
Two slender rods of negligible weight are pin-connected at C and
attached to blocks A and B, each of weight W. Knowing that θ = 70o
and that the coefficient of static friction between the blocks and the
horizontal surface is 0.30, determine the largest value of P for which
equilibrium is maintained.
SOLUTION
FBD pin C:
FAB = P sin10° = 0.173648P
FBC = P cos10° = 0.98481P
ΣFy = 0: N A − W − FAB sin 30° = 0
FBD block A:
N A = W + 0.173648P sin 30° = W + 0.086824P
or
ΣFx = 0: FA − FAB cos30° = 0
FA = 0.173648P cos30° = 0.150384P
or
FA = µ s N A
For impending motion at A:
Then
NA =
FA
µs
: W + 0.086824 P =
0.150384
P
0.3
P = 2.413W
or
ΣFy = 0: N B − W − FBC cos30° = 0
N B = W + 0.98481P cos30° = W + 0.85287 P
ΣFx = 0: FBC sin 30° − FB = 0
FBD block B:
FB = 0.98481P sin 30° = 0.4924 P
FB = µ s N B
For impending motion at B:
Then
or
NB =
FB
µs
: W + 0.85287 P =
0.4924P
0.3
P = 1.268W
Thus, maximum P for equilibrium
Pmax = 1.268W W
PROBLEM 8.46
A 40-lb weight is hung from a lever which rests against a 10° wedge at A
and is supported by a frictionless hinge at C. Knowing that the coefficient
of static friction is 0.25 at both surfaces of the wedge and that for the
position shown the spring is stretched 4 in., determine (a) the magnitude
of the force P for which motion of the wedge is impending, (b) the
components of the corresponding reaction at C.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
 4 in. 
Fs = kx = ( 240 lb/ft ) 
 = 80 lb
 12 in./ft 
FBD lever:
ΣM C = 0:
(12 in.)(80 lb ) − (16 in.)( 40 lb ) − ( 21 in.) RA cos (φs − 10° )
+ ( 2 in.) RA sin (φs − 10° ) = 0
(b)
or
RA = 15.3793 lb
ΣFx = 0:
(15.379 lb ) sin ( 4.036° ) − Cx
ΣFy = 0:
(15.379 lb ) cos ( 4.036° ) − 80 lb − 40 lb + C y
=0
C x = 1.082 lb
=0
C y = 104.7 lb
W
W
FBD wedge:
ΣFy = 0: RW cos14.036° − (15.3793 lb ) cos 4.036° = 0
or
(a)
RW = 15.8133 lb
ΣFx = 0: P − (15.3793 lb ) sin 4.036° − (15.8133 lb ) sin14.036° = 0
P = 4.92 lb W
PROBLEM 8.47
Solve Problem 8.46 assuming that force P is directed to the left.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
 4 in. 
Fs = kx = ( 240 lb/ft ) 
 = 80 lb
 12 in./ft 
FBD lever:
ΣM C = 0:
(12 in.)(80 lb ) − (16 in.)( 40 lb ) − ( 21 in.) RA cos 24.036°
− ( 2 in.) RA sin 24.036° = 0
RA = 16.005 lb
or
(b)
ΣFx = 0: C x − (16.005 lb ) sin 24.036° = 0
ΣFy = 0: C y − 80 lb − 40 lb + (16.005 lb ) cos ( 24.036° ) = 0
C x = 6.52 lb
W
C y = 105.4 lb W
FBD wedge:
ΣFy = 0: RW cos14.036° − (16.005 lb ) cos 24.036° = 0
(a)
or
RW = 15.067 lb
ΣFx = 0:
(16.005 lb ) sin 24.036° + (15.067 lb ) sin14.036° − P = 0
P = 10.17 lb W
PROBLEM 8.48
Two 8° wedges of negligible mass are used to move and position a
240-kg block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, determine the magnitude of the force P for which
motion of the block is impending.
SOLUTION
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
(
)
W = 240 kg 9.81 m/s 2 = 2354.4 N
FBD block:
R2
2354.4 N
=
sin 41.801° sin 46.398°
R2 = 2167.12 N
FBD wedge:
P
2167.12 N
=
sin 51.602° sin 60.199°
P = 1957 N
P = 1.957 kN W
PROBLEM 8.49
Two 8° wedges of negligible mass are used to move and position a
240-kg block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, determine the magnitude of the force P for which
motion of the block is impending.
SOLUTION
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
(
)
W = 240 kg 9.81 m/s 2 = 2354.4 N
FBD block + wedge:
R2
2354.4 N
=
sin 41.801° sin 38.398°
R2 = 2526.6 N
FBD wedge:
P
2526.6 N
=
sin 51.602° sin 68.199°
P = 2132.7 N
P = 2.13 kN W
PROBLEM 8.50
The elevation of the end of the steel beam supported by a concrete floor
is adjusted by means of the steel wedges E and F. The base plate CD has
been welded to the lower flange of the beam, and the end reaction of the
beam is known to be 150 kN. The coefficient of static friction is 0.30
between the two steel surfaces and 0.60 between the steel and the
concrete. If the horizontal motion of the beam is prevented by the force
Q, determine (a) the force P required to raise the beam, (b) the
corresponding force Q.
SOLUTION
φ s = tan −1 µ s = tan −1 0.3 = 16.70° for steel on steel
FBD AB + CD:
ΣFy = 0: N − 150 kN = 0
Impending motion:
N = 150 kN
F = µ s N = 0.3 (150 kN ) = 45 kN
ΣFx = 0: F − Q = 0
(b) Q = 45.0 kN
FBD top wedge:
W
Assume bottom wedge doesn’t move:
ΣFy = 0: RW cos (10° + 16.70° ) − 150 kN = 0
RW = 167.9 kN
ΣFx = 0: P − 45 kN − (167.9 kN ) sin 26.70° = 0
P = 120.44 kN
FBD bottom wedge:
( a ) P = 120.4 kN
W
Bottom wedge is two-force member, so φ = 26.70° for equilibrium, but
φ s = tan −1 µ s = tan −1 0.6 = 31.0° ( steel on concrete )
So
φ < φs
OK.
PROBLEM 8.51
The elevation of the end of the steel beam supported by a concrete floor
is adjusted by means of the steel wedges E and F. The base plate CD has
been welded to the lower flange of the beam, and the end reaction of the
beam is known to be 150 kN. The coefficient of static friction is 0.30
between the two steel surfaces and 0.60 between the steel and the
concrete. If the horizontal motion of the beam is prevented by the force
Q, determine (a) the force P required to raise the beam, (b) the
corresponding force Q.
SOLUTION
φ s = tan −1 µ s = tan −1 0.3 = 16.70° for steel on steel
FBD AB + CD + top wedge: Assume top wedge doesn’t move
Rw =
150 kN
= 167.90 kN
cos26.70°
Q = (150 kN ) tan 26.70° = 75.44 kN
(b) Q = 75.4 kN
FBD top wedge:
ΣFx = 0: 75.44 kN − 167.9 kN sin 26.70° − F = 0
F = 0 as expected.
PROBLEM 8.51 CONTINUED
FBD bottom wedge:
φ s = tan −1 µ s = tan −1 0.6 = 30.96° steel on concrete
P
167.90 kN
=
sin 57.66°
sin 59.04°
(a) P = 165.4 kN
PROBLEM 8.52
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that θ = 45°, determine the smallest force P required to raise
block A.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
750 lb
=
sin104.036° sin16.928°
R2 = 2499.0 lb
FBD wedge B:
P
2499.0
=
sin 73.072° sin 75.964°
P = 2464 lb
P = 2.46 kips
PROBLEM 8.53
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that θ = 45°, determine the smallest force P for which
equilibrium is maintained.
SOLUTION
φ s = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
750 lb
=
sin ( 75.964° ) sin ( 73.072° )
R2 = 760.56 lb
FBD wedge B:
P
760.56
=
sin16.928° sin104.036°
P = 228.3 lb
P = 228 lb
PROBLEM 8.54
A 16° wedge A of negligible mass is placed between two 80-kg blocks B
and C which are at rest on inclined surfaces as shown. The coefficient of
static friction is 0.40 between both the wedge and the blocks and block C
and the incline. Determine the magnitude of the force P for which motion
of the wedge is impending when the coefficient of static friction between
block B and the incline is (a) 0.40, (b) 0.60.
SOLUTION
φ s = tan −1 µ s = tan −1 0.4 = 21.8014°;
(a)
(
)
W = 80 kg 9.81 m/s 2 = 784.8 N
FBD wedge:
By symmetry:
R1 = R 2
ΣFy = 0: 2R2 sin ( 8° + 21.8014° ) − P = 0
P = 0.99400 R2
FBD block C:
R2
W
=
sin 41.8014° sin18.397°
R2 = 2.112 W
PROBLEM 8.54 CONTINUED
P = 0.994 R2 = ( 0.994 )( 2.112W )
P = 2.099 ( 784.8 N ) = 1647.5 N
(a) P = 1.648 kN
(b) Note that increasing the friction between block B and the incline has no effect on the above calculations.
The physical effect is that slip of B will not impend.
(b) P = 1.648 kN
PROBLEM 8.55
A 16° wedge A of negligible mass is placed between two 80-kg blocks B
and C which are at rest on inclined surfaces as shown. The coefficient of
static friction is 0.40 between both the wedge and the blocks and block C
and the incline. Determine the magnitude of the force P for which motion
of the wedge is impending when the coefficient of static friction between
block B and the incline is (a) 0.40, (b) 0.60.
SOLUTION
(a) φ s = tan −1 µ s = tan −1 0.4 = 21.801°
FBD wedge:
(
)
W = 80 kg 9.81 m/s 2 = 784.8 N
FBD block C:
Note that, since ( RCI ) y > ( RC ) y , while the horizontal components are equal,
20° + φ < 32.199°
φ < 12.199° < φ s
Therefore, motion of C is not impending; thus, motion of B up the incline is impending.
RB
P
=
sin 52.198° sin 59.603°
P = 1.0916RB
PROBLEM 8.55 CONTINUED
FBD block B:
RB
W
=
sin ( 20° + φ sB ) sin ( 68.199° − φsB )
RB =
or
W sin ( 20° + φsB )
sin ( 68.199° − φsB )
(a) Have φ sB = φs = 21.801°
Then
RB =
( 784.8 N ) sin ( 20° + 21.801° )
sin ( 68.199° − 21.801° )
= 722.37 N
P = 1.0916 ( 722.37 N )
and
or P = 789 N
(b) Have φ sB = tan −1 µ sB = tan −1 0.6 = 30.964°
Then
and
RB =
( 784.8 N ) sin ( 20° + 30.964° )
sin ( 68.199° − 30.964° )
P = 1.0916 (1007.45 N )
= 1007.45 N
or P = 1100 N
PROBLEM 8.56
A 10° wedge is to be forced under end B of the 12-lb rod AB. Knowing
that the coefficient of static friction is 0.45 between the wedge and the
rod and 0.25 between the wedge and the floor, determine the smallest
force P required to raise end B of the rod.
SOLUTION
FBD AB:
φ s1 = tan −1 ( µ s )1 = tan −1 0.45 = 24.228°
ΣM A = 0: rR1 cos (10° + 24.228° ) − rR1 sin (10° + 24.228° ) −
2r
π
(12 lb ) = 0
R1 = 28.902 lb
FBD wedge:
φ s 2 = tan −1 ( µ s )2 = tan −1 0.25 = 14.036°
P
28.902 lb
=
;
sin ( 38.264° ) sin 75.964°
P = 22.2 lb
PROBLEM 8.57
A small screwdriver is used to pry apart the two coils of a circular key
ring. The wedge angle of the screwdriver blade is 16° and the coefficient
of static friction is 0.12 between the coils and the blade. Knowing that a
force P of magnitude 0.8 lb was required to insert the screwdriver to the
equilibrium position shown, determine the magnitude of the forces
exerted on the ring by the screwdriver immediately after force P is
removed.
SOLUTION
FBD wedge:
By symmetry:
R1 = R2
ΣFy = 0: 2R1 sin ( 8° + φ s ) − P = 0
Have
So
φ s = tan −1 µ s = tan −1 0.12 = 6.843°
P = 0.8 lb
R1 = R2 = 1.5615 lb
When P is removed, the vertical components of R1 and R2 vanish,
leaving the horizontal components, R1 cos (14.843° ) , only
Therefore, side forces are 1.509 lb
But these will occur only instantaneously as the angle between the force
and the wedge normal is 8° > φ s = 6.84°, so the screwdriver will
slip out.
PROBLEM 8.58
A conical wedge is placed between two horizontal plates that are then
slowly moved toward each other. Indicate what will happen to the wedge
(a) if µ s = 0.20, (b) if µ s = 0.30.
SOLUTION
As the plates are moved, the angle θ will decrease.
(a)
φ s = tan −1 µ s = tan −1 0.2 = 11.31°. As θ decreases, the minimum angle at the contact approaches
12.5° > φs = 11.31°, so the wedge will slide up and out from the slot.
(b)
φ s = tan −1 µ s = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this
time the angle at the other contact is 25° − 16.7° = 8.3° < φ s ) The wedge binds in the slot.
PROBLEM 8.59
A 6° steel wedge is driven into the end of an ax handle to lock the handle
to the ax head. The coefficient of static friction between the wedge and
the handle is 0.35. Knowing that a force P of magnitude 250 N was
required to insert the wedge to the equilibrium position shown, determine
the magnitude of the forces exerted on the handle by the wedge after
force P is removed.
SOLUTION
FBD wedge:
By symmetry
R1 = R2
φ s = tan −1 µ s = tan −1 0.35 = 19.29°
ΣFy = 0: 2R sin (19.29° + 3° ) − P = 0
R1 = R2 = 329.56 N
When force P is removed, the vertical components of R1 and R2 vanish,
leaving only the horizontal components H1 = H 2 = R cos ( 22.29° )
H1 = H 2 = 305 N
Since the wedge angle 3° < φs = 19.3°, the wedge is “self-locking” and
will remain seated.
PROBLEM 8.60
A 15° wedge is forced under a 100-lb pipe as shown. The coefficient of
static friction at all surfaces is 0.20. Determine (a) at which surface
slipping of the pipe will first occur, (b) the force P for which motion of
the wedge is impending.
SOLUTION
FBD pipe:
ΣM C = 0: rFA − rFB = 0
(a)
FA = FB
or
But it is apparent that N B > N A , so since ( µ s ) A = ( µ s ) B ,
motion must first impend at A
FB = FA = µ s N A = 0.2 N A
and
ΣM B = 0:
(b)
( r sin15° )W
+ r (1 + sin15° ) FA − ( r cos15° ) N A = 0
0.2588 (100 lb ) + 1.2588 ( 0.2 N A ) − 0.9659 N A = 0
or
N A = 36.24 lb
and
FA = 7.25 lb
ΣFy′ = 0: N B − N A sin15° − FA cos15° − W cos15° = 0
N B = ( 36.24 lb ) sin15° + ( 7.25 lb + 100 lb ) cos15°
= 112.97 lb
FBD wedge:
( note N B
> N A as stated, and FB < µ s N B )
ΣFy = 0: NW + ( 7.25 lb ) sin15° − (112.97 lb ) cos15° = 0
NW = 107.24 lb
Impending slip:
FW = µ s NW = 0.2 (107.24 ) = 21.45 lb
ΣFx = 0: 21.45 lb + ( 7.25 lb ) cos15° + (112.97 lb ) sin15° − P = 0
P = 57.7 lb
PROBLEM 8.61
A 15° wedge is forced under a 100-lb pipe as shown. Knowing that the
coefficient of static friction at both surfaces of the wedge is 0.20,
determine the largest coefficient of static friction between the pipe and
the vertical wall for which slipping is impending at A.
SOLUTION
ΣM C = 0: rFA − rFB = 0
FBD pipe:
or
FA = FB
It is apparent that N B > N A , so if ( µ s ) A = ( µ s ) B , motion must impend
( )
first at A. As ( µ s ) A is increased to some µ s*
A
, motion will impend
simultaneously at A and B.
Then
FA = FB = µ sB N B = 0.2 N B
ΣFy = 0: N B cos15° − FB sin15° − FA − 100 lb = 0
N B cos15° − 0.2 N B sin15° − 0.2 N B = 100 lb
or
N B = 140.024 lb
So
FA = FB = 0.2 N B = 28.005 lb
ΣFx = 0: N A − N B sin15° − FB cos15° = 0
N A = 140.024sin15° + 28.005cos15° = 63.29 lb
Then
or
(µ )
*
s
A
=
FA
28.005 lb
=
NA
63.29 lb
(µ )
*
s
A
= 0.442
PROBLEM 8.62
Bags of grass seed are stored on a wooden plank as shown. To move the
plank, a 9° wedge is driven under end A. Knowing that the weight of the
grass seed can be represented by the distributed load shown and that the
coefficient of static friction is 0.45 between all surfaces of contact,
(a) determine the force P for which motion of the wedge is impending,
(b) indicate whether the plank will slide on the floor.
SOLUTION
FBD plank + wedge:
(a)
( 2.4 m ) N B − ( 0.45 m )( 0.64 kN/m )( 0.9 m )
ΣM A = 0:
− ( 0.6 m )
1
( 0.64 kN/m )( 0.9 m )
2
− (1.4 m )
1
(1.28 kN/m )(1.5 m ) = 0
2
N B = 0.740 kN = 740 N
or
ΣFy = 0: NW − ( 0.64 kN/m )( 0.9 m ) −
−
1
( 0.64 kN/m )( 0.9 m )
2
1
(1.28 kN/m )(1.5m ) = 0
2
NW = 1.084 kN = 1084 N
or
Assume impending motion of the wedge on the floor and the plank
on the floor at B.
So
and
FW = µ s NW = 0.45 (1084 N ) = 478.8 N
FB = µ s N B = 0.45 ( 740 N ) = 333 N
ΣFx = 0: P − FW − FB = 0
or
Check wedge:
(b)
ΣFy = 0:
or
ΣFx = 0:
or
P = 478.8 N + 333 N
P = 821 N
(1084 N ) cos 9° + (821 N − 479 N ) sin 9° − N A
=0
N A = 1124 N
(821 N − 479 N ) cos 9° − (1084 N ) sin 9° − FA
=0
FA = 168 N
FA < µ s N A = 0.45 (1124 N ) = 506 N
So, no impending motion at wedge/plank
∴ Impending motion of plank on floor at B
PROBLEM 8.63
Solve Problem 8.62 assuming that the wedge is driven under the plank at
B instead of at A.
SOLUTION
FBD plank:
(a)
ΣFx = 0: FA − Bx = 0
FA = Bx
ΣM A = 0:
( 2.4 m ) By − ( 0.45 m )( 0.64 kN/m )( 0.9 m )
− ( 0.6 m )
1
( 0.64 kN/m )( 0.9 m )
2
− (1.4 m )
1
(1.28 kN/m )(1.5 m ) = 0
2
By = 0.740 kN = 740 N
or
ΣFy = 0: N A − ( 0.64 kN/m )( 0.9 m ) −
−
1
( 0.64 kN/m )( 0.9 m )
2
1
(1.28 kN/m )(1.5 m ) = 0
2
N A = 1.084 kN = 1084 N
or
Since By < N A , assume impending motion of the wedge under
the plank at B.
FBD wedge:
( RB ) y
= By = 740 N and Bx = µ s By = 0.45 ( 740 N ) = 333 N
( RB ) x
= ( RB ) y tan ( 9° + φs )
φ s = tan −1 µ s = tan −1 0.45 = 24.228°
So
( RB ) x
= ( 740 N ) tan ( 9° + 24.228° ) = 485 N
ΣFx = 0: 485 N − 333 N − P = 0
P = 818 N
(b) Check:
FA = Bx = 333 N and
FA
333
=
= 0.307 < µ s
N A 1084
OK
No impending slip of plank at A
PROBLEM 8.64
The 20-lb block A is at rest against the 100-lb block B as shown. The
coefficient of static friction µ s is the same between blocks A and B and
between block B and the floor, while friction between block A and the
wall can be neglected. Knowing that P = 30 lb, determine the value of
µ s for which motion is impending.
SOLUTION
FBD’s:
FAB = µ s N AB
Impending motion at all surfaces
FB = µ s N B
A + B:
ΣFy = 0: N B − 30 lb − 20 lb − 100 lb = 0
N B = 150 lb
or
and
FB = µ s N B = (150 lb ) µ s
ΣFx = 0: N A − FB = 0
A:
so that
N A = (150 lb ) µ s
ΣFx′ = 0: N A cos 20° + ( 30 lb + 20 lb ) sin 20° − N AB = 0
or
N AB = 17.1010 lb + µ s (140.954 lb )
ΣFy′ = 0: FAB + N A sin 20° − ( 30 lb + 20 lb ) cos 20° = 0
or
But
FAB = 46.985 lb − µ s ( 51.303 lb )
FAB = µ s N AB : 46.985 − 51.303µs = 17.101µ s + 140.954µ s2
µ s2 + 0.4853µ s − 0.3333 = 0
µ s = −0.2427 ± 0.6263
µs > 0
so
µ s = 0.384
PROBLEM 8.65
Solve Problem 8.64 assuming that µ s is the coefficient of static friction
between all surfaces of contact.
SOLUTION
FBD’s:
A + B:
B:
FA = µ s N A
Impending motion at all surfaces, so
FB = µ s N B
FAB = µ s N AB
A + B:
ΣFx = 0: N A − FB = 0 or N A = FB = µ s N B
ΣFy = 0: FA − 30 lb − 20 lb − 100 lb + N B = 0 or
So
B:
or
NB =
150 lb
1 + µ s2
and
FB =
µ s N A + N B = 150 lb
µs
(150 lb )
1 + µ s2
ΣFx′ = 0: N AB + (100 lb − N B ) sin 20° − FB cos 20° = 0
N AB = N B sin 20° + FB cos 20° − (100 lb ) sin 20°
ΣFy′ = 0: − FAB + ( N B − 100 lb ) cos 20° − FB sin 20° = 0
or
FAB = N B cos 20° − FB sin 20° − (100 lb ) cos 20°
PROBLEM 8.65 CONTINUED
Now
FAB = µ s N AB :
=
150 lb
µs
cos 20° −
(150 lb ) sin 20° − (100 lb ) cos 20°
1 + µ s2
1 + µ s2
µs
µ s2
150
lb
sin
20
°
+
(
)
(150 lb ) cos 20° − µ s (100 lb ) sin 20°
1 + µ s2
1 + µ s2
2µ s3 − 5µ s2ctn 20° − 4µ s + ctn 20° = 0
Solving numerically:
µ s = 0.330
PROBLEM 8.66
Derive the following formulas relating the load W and the
force P exerted on the handle of the jack discussed in
Section 8.6. (a) P = (Wr/a ) tan (θ + φs ) , to raise the load;
(b) P = (Wr/a ) tan (φ s − θ ) , to lower the load if the screw is selflocking; (c) P = (Wr/a ) tan (θ − φ s ) , to hold the load if the screw is not
self-locking.
SOLUTION
FBD jack handle:
See Section 8.6
ΣM C = 0: aP − rQ = 0 or P =
r
Q
a
FBD block on incline:
(a) Raising load
Q = W tan (θ + φ s )
P=
r
W tan (θ + φs )
a
PROBLEM 8.66 CONTINUED
(b) Lowering load if screw is self-locking ( i.e.: if φs > θ )
Q = W tan (φs − θ )
P=
r
W tan (φ s − θ )
a
P=
r
W tan (θ − φ s )
a
(c) Holding load is screw is not self-locking ( i.e. if φs < θ )
Q = W tan (θ − φs )
PROBLEM 8.67
The square-threaded worm gear shown has a mean radius of 30 mm and a
lead of 7.5 mm. The larger gear is subjected to a constant clockwise
couple of 720 N ⋅ m. Knowing that the coefficient of static friction
between the two gears is 0.12, determine the couple that must be applied
to shaft AB in order to rotate the large gear counterclockwise. Neglect
friction in the bearings at A, B, and C.
SOLUTION
FBD large gear:
ΣM C = 0:
( 0.24 m )W
− 720 N ⋅ m = 0
W = 3000 N
Block on incline:
θ = tan −1
7.5 mm
= 2.2785°
2π ( 30 mm )
φ s = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = ( 3000 N ) tan 9.1213°
= 481.7 N
PROBLEM 8.67 CONTINUED
Worm gear:
r = 30 mm
= 0.030 m
ΣM B = 0: rQ − M = 0
M = rQ = ( 0.030 m )( 481.7 N )
M = 14.45 N ⋅ m
PROBLEM 8.68
In Problem 8.67, determine the couple that must be applied to shaft AB in
order to rotate the gear clockwise.
SOLUTION
FBD large gear:
ΣM C = 0:
( 0.24 m )W
− 720 N ⋅ m = 0
W = 3000 N
Block on incline:
θ = tan −1
7.5 mm
= 2.2785°
2π ( 30 mm )
φ s = tan −1 µ = tan −1 0.12
φ s = 6.8428°
φ s − θ = 4.5643°
PROBLEM 8.68 CONTINUED
Q = ( 3000 N ) tan 4.5643°
= 239.5 N
Worm gear:
ΣM B = 0: M − rQ = 0
M = rQ = ( 0.030 m )( 239.5 N ) = 7.18 N ⋅ m
PROBLEM 8.69
High-strength bolts are used in the construction of many steel structures.
For a 24-mm-nominal-diameter bolt the required minimum bolt tension is
210 kN. Assuming the coefficient of friction to be 0.40, determine the
required couple that should be applied to the bolt and nut. The mean
diameter of the thread is 22.6 mm, and the lead is 3 mm. Neglect friction
between the nut and washer, and assume the bolt to be square-threaded.
SOLUTION
FBD block on incline:
θ = tan −1
3 mm
= 2.4195°
( 22.6 mm )π
φ s = tan −1 µ s = tan −1 0.40
φ s = 21.8014°
Q = ( 210 kN ) tan ( 21.8014° + 2.4195° )
Q = 94.47 kN
Torque =
d
22.6 mm
Q=
( 94.47 kN )
2
2
= 1067.5 N ⋅ m
Torque = 1.068 kN ⋅ m
PROBLEM 8.70
The ends of two fixed rods A and B are each made in the form of a singlethreaded screw of mean radius 0.3 in. and pitch 0.1 in. Rod A has a righthanded thread and rod B a left-handed thread. The coefficient of static
friction between the rods and the threaded sleeve is 0.12. Determine the
magnitude of the couple that must be applied to the sleeve in order to
draw the rods closer together.
SOLUTION
Block on incline:
θ = tan −1
0.1 in.
= 3.0368°
2π ( 0.3 in.)
φ s = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = ( 500 lb ) tan 9.8796° = 87.08 lb
Couple on each side
M = rQ = ( 0.3 in.)( 87.08 lb ) = 26.12 lb ⋅ in.
Couple to turn = 2M = 52.2 lb ⋅ in.
PROBLEM 8.71
Assuming that in Problem 8.70 a right-handed thread is used on both rods
A and B, determine the magnitude of the couple that must be applied to
the sleeve in order to rotate it.
SOLUTION
Block on incline A:
θ = tan −1
0.1 in.
= 3.0368°
2π ( 0.3 in.)
φ s = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = ( 500 lb ) tan 9.8796°
= 87.08 lb
Couple at A = ( 0.3 in.)( 87.08 lb )
= 26.124 lb ⋅ in.
Block on incline B:
Q = ( 500 lb ) tan 3.806°
= 33.26 lb
Couple at B = ( 0.3 in.)( 33.26 lb )
= 9.979 lb ⋅ in.
Total couple = 26.124 lb ⋅ in. + 9.979 lb ⋅ in.
Couple to turn = 36.1 lb ⋅ in. W
PROBLEM 8.72
The position of the automobile jack shown is controlled by a screw ABC
that is single-threaded at each end (right-handed thread at A, left-handed
thread at C). Each thread has a pitch of 2 mm and a mean diameter of
7.5 mm. If the coefficient of static friction is 0.15, determine the
magnitude of the couple M that must be applied to raise the automobile.
SOLUTION
FBD joint D:
FAD = FCD
By symmetry:
ΣFy = 0: 2FAD sin 25° − 4 kN = 0
FAD = FCD = 4.7324 kN
FBD joint A:
FAE = FAD
By symmetry:
ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0
FAC = 8.5780 kN
Block and incline A:
θ = tan −1
2 mm
π ( 7.5 mm )
= 4.8518°
φ s = tan −1 µ s = tan −1 0.15 = 8.5308°
PROBLEM 8.72 CONTINUED
Q = ( 8.578 kN ) tan (13.3826° )
= 2.0408 kN
Couple at A:
M A = rQ
 7.5

=
mm  ( 2.0408 kN )
 2

= 7.653 N ⋅ m
By symmetry: Couple at C:
M C = 7.653 N ⋅ m
Total couple M = 2 ( 7.653 N ⋅ m )
M = 15.31 N ⋅ m W
PROBLEM 8.73
For the jack of Problem 8.72, determine the magnitude of the couple M
that must be applied to lower the automobile.
SOLUTION
FBD joint D:
FAD = FCD
By symmetry:
ΣFy = 0: 2FAD sin 25° − 4 kN = 0
FAD = FCD = 4.7324 kN
FBD joint A:
FAE = FAD
By symmetry:
ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0
FAC = 8.5780 kN
Block and incline at A:
θ = tan −1
2 mm
= 4.8518°
π ( 7.5 mm )
φ s = tan −1 µ s = tan −1 0.15
φ s = 8.5308°
PROBLEM 8.73 CONTINUED
φ s − θ = 3.679°
Q = ( 8.5780 kN ) tan 3.679°
Q = 0.55156 kN
Couple at A: M A = Qr
 7.5 mm 
= ( 0.55156 kN ) 

 2 
= 2.0683 N ⋅ m
By symmetry:
Couple at C : M C = 2.0683 N ⋅ m
Total couple M = 2 ( 2.0683 N ⋅ m )
M = 4.14 N ⋅ m W
PROBLEM 8.74
In the gear-pulling assembly shown, the square-threaded screw AB has a
mean radius of 22.5 mm and a lead of 6 mm. Knowing that the
coefficient of static friction is 0.10, determine the couple which must be
applied to the screw in order to produce a force of 4.5 kN on the gear.
Neglect friction at end A of the screw.
SOLUTION
Block on incline:
θ = tan −1
6 mm
= 2.4302°
2π ( 22.5 mm )
φ s = tan −1 µ s = tan −1 0.1
φ s = 5.7106°
Q = ( 4.5 kN ) tan 8.1408°
= 0.6437 kN
Couple M = rQ
= ( 22.5 mm )( 0.6437 kN )
= 14.483 N ⋅ m
M = 14.48 N ⋅ m W
NOTE FOR PROBLEMS 8.75–8.89
Note to instructors: In this manual, the singular sin (tan–1µ) ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ, there is
little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79, µ s = 0.40,
and the error made by using the approximation is about 7.7%.
PROBLEM 8.75
A 120-mm-radius pulley of mass 5 kg is attached to a 30-mm-radius shaft
which fits loosely in a fixed bearing. It is observed that the pulley will
just start rotating if a 0.5-kg mass is added to block A. Determine the
coefficient of static friction between the shaft and the bearing.
SOLUTION
FBD pulley:
ΣFy = 0: R − 103.005 N − 49.05 N − 98.1 N = 0
R = 250.155 N
ΣM O = 0:
( 0.12 m )(103.005 N − 98.1 N ) − rf ( 250.155 N ) = 0
rf = 0.0023529 m = 2.3529 mm
φ s = sin −1

µ s = tan φs = tan  sin −1

rf
rs
rf 
 −1 2.3529 mm 
 = tan  sin

rs 
30 mm 

µ s = 0.0787 W
PROBLEM 8.76
The double pulley shown is attached to a 0.5-in.-radius shaft which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
between the shaft and the poorly lubricated bearing is 0.40, determine the
magnitude of the force P required to start raising the load.
SOLUTION
FDB pulley:
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
*
)
rf = ( 0.5 in.) sin tan −1 0.40 = 0.185695 in.
ΣM C = 0:
( 4.5 in. + 0.185695 in.)( 40 lb )
− ( 2.25 in. − 0.185695 in.) P = 0
P = 90.8 lb
* See note before Problem 8.75.
PROBLEM 8.77
The double pulley shown is attached to a 0.5-in.-radius shaft which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
between the shaft and the poorly lubricated bearing is 0.40, determine the
magnitude of the force P required to start raising the load.
SOLUTION
FBD pulley:
(
)
(
rf = rs sin φs = rs sin tan −1 µ s = ( 0.5 in.) sin tan −1 0.4
)
*
rf = 0.185695 in.
ΣM C = 0:
( 4.5 in. − 0.185695 in.)( 40 lb )
− ( 2.25 in. − 0.185695 in.) P = 0
P = 83.6 lb
* See note before Problem 8.75.
PROBLEM 8.78
The double pulley shown is attached to a 0.5-in.-radius shaft which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
between the shaft and the poorly lubricated bearing is 0.40, determine the
magnitude of the force P required to maintain equilibrium.
SOLUTION
FBD pulley:
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
*
)
rf = ( 0.5 in.) sin tan −1 0.40 = 0.185695 in.
ΣM C = 0:
( 4.5 in. − 0.185695 in.)( 40 lb )
− ( 2.25 in. + 0.185695 in.)( P ) = 0
P = 70.9 lb
* See note before Problem 8.75.
PROBLEM 8.79
The double pulley shown is attached to a 0.5-in.-radius shaft which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
between the shaft and the poorly lubricated bearing is 0.40, determine the
magnitude of the force P required to maintain equilibrium.
SOLUTION
FBD pulley:
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
*
)
rf = ( 0.5 in.) sin tan −1 0.4 = 0.185695 in.
ΣM C = 0:
( 4.5 in. + 0.185695 in.)( 40 lb )
− ( 2.25 in. + 0.185695 in.) P = 0
P = 77.0 lb
* See note before Problem 8.75.
PROBLEM 8.80
Control lever ABC fits loosely on a 18-mm-diameter shaft at support B.
Knowing that P = 130 N for impending clockwise rotation of the lever,
determine (a) the coefficient of static friction between the pin and the
lever, (b) the magnitude of the force P for which counterclockwise
rotation of the lever is impending.
SOLUTION
(a) FBD lever (Impending CW
rotation):
ΣM C = 0:
( 0.2 m + rf ) ( 75 N ) − ( 0.12 m − rf ) (130 N ) = 0
rf = 0.0029268 m = 2.9268 mm
sin φ s =

µ s = tan φs = tan  sin −1

rf
rs
*
rf 
 −1 2.9268 mm 
 = tan  sin

18 mm 
rs 

= 0.34389
µ s = 0.344
(b) FBD lever (Impending CCW
rotation):
ΣM D = 0:
( 0.20 m − 0.0029268 m )( 75 N )
− ( 0.12 m + 0.0029268 m ) P = 0
P = 120.2 N
* See note before Problem 8.75.
PROBLEM 8.81
The block and tackle shown are used to raise a 600-N load. Each of the
60-mm-diameter pulleys rotates on a 10-mm-diameter axle. Knowing that
the coefficient of kinetic friction is 0.20, determine the tension in each
portion of the rope as the load is slowly raised.
SOLUTION
Pulley FBD’s:
rp = 30 mm
Left:
rf = raxle sin φk = raxle sin tan −1 µ k
(
)
*
(
= ( 5 mm ) sin tan −1 0.2
)
= 0.98058 mm
Left:
ΣM C = 0:
Right:
( rp − rf ) ( 600 lb ) − 2rpTAB = 0
TAB =
or
30 mm − 0.98058 mm
( 600 N ) = 290.19 N
2 ( 30 mm )
TAB = 290 N
ΣFy = 0: 290.19 N − 600 N + TCD = 0
TCD = 309.81 N
or
TCD = 310 N
Right:
ΣM G = 0:
or
( rp + rf ) TCD − ( rp − rf ) TEF
TEF =
=0
30 mm + 0.98058 mm
( 309.81 N ) = 330.75 N
30 mm − 0.98058 mm
TEF = 331 N
* See note before Problem 8.75.
PROBLEM 8.82
The block and tackle shown are used to lower a 600-N load. Each of the
60-mm-diameter pulleys rotates on a 10-mm-diameter axle. Knowing that
the coefficient of kinetic friction is 0.20, determine the tension in each
portion of the rope as the load is slowly lowered.
SOLUTION
rp = 30 mm
Pulley FBDs:
Left:
(
rf = raxle sin φk = raxle sin tan −1 µ k
)
*
(
= ( 5 mm ) sin tan −1 0.2
)
= 0.98058 mm
ΣM C = 0:
( rp + rf ) ( 600 N ) − 2rpTAB = 0
TAB =
or
30 mm + 0.98058 mm
( 600 N ) = 309.81 N
2 ( 30 mm )
TAB = 310 N
ΣFy = 0: TAB − 600 N + TCD = 0
Right:
TCD = 600 N − 309.81 N = 290.19 N
or
TCD = 290 N
ΣM H = 0:
or
( rp − rf ) TCD − ( rp + rf ) TEF
TEF =
=0
30 mm − 0.98058 mm
( 290.19 N )
30 mm + 0.98058 mm
TEF = 272 N
* See note before Problem 8.75.
PROBLEM 8.83
The link arrangement shown is frequently used in highway bridge
construction to allow for expansion due to changes in temperature. At
each of the 3-in.-diameter pins A and B the coefficient of static friction is
0.20. Knowing that the vertical component of the force exerted by BC on
the link is 50 kips, determine (a) the horizontal force which should be
exerted on beam BC to just move the link, (b) the angle that the resulting
force exerted by beam BC on the link will form with the vertical.
SOLUTION
FBD link AB:
Note that AB is a two force member. For impending motion, the pin
forces are tangent to the friction circles.
θ = sin −1
where
rf
25 in.
(
rf = rp sin φs = rp sin tan −1 µ s
(
)
*
)
= (1.5 in.) sin tan −1 0.2 = 0.29417 in.
Then
θ = sin −1
0.29417 in.
= 1.3485°
12.5 in.
(b) θ = 1.349°
Rvert = R cosθ
Rhoriz = R sin θ
Rhoriz = Rvert tan θ = ( 50 kips ) tan1.3485° = 1.177 kips
(a) Rhoriz = 1.177 kips
* See note before Problem 8.75.
PROBLEM 8.84
A gate assembly consisting of a 24-kg gate ABC and a 66-kg
counterweight D is attached to a 24-mm-diameter shaft B which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
is 0.20 between the shaft and the bearing, determine the magnitude of the
force P for which counterclockwise rotation of the gate is impending.
SOLUTION
FBD gate:
(
)
W = 24 kg ( 9.81 m/s ) = 235.44 N
= r sin φ = r sin ( tan µ )
= ( 0.012 m ) sin ( tan 0.2 ) = 0.0023534 m
W1 = 66 kg 9.81 m/s 2 = 647.46 N
2
2
rf
−1
s
s
s
s
−1
ΣM C = 0:
P=
( 0.6 m − rf )W1 + ( 0.15 m − rf ) P − (1.8 m + rf )W2 = 0
(1.80235 m )( 235.44 N ) − ( 0.59765 m )( 647.46 N )
( 0.14765 m )
= 253.2 N
P = 253 N
PROBLEM 8.85
A gate assembly consisting of a 24-kg gate ABC and a 66-kg
counterweight D is attached to a 24-mm-diameter shaft B which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
is 0.20 between the shaft and the bearing, determine the magnitude of the
force P for which counterclockwise rotation of the gate is impending.
SOLUTION
It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single combined weight of
( 90 kg ) ( 9.81 m/s2 ) = 882.9 N, located at a distance
right of B.
(1.8 m )( 24 kg ) − ( 0.6 m )( 24 kg )
x=
(
rf = rs sin φs = rs sin tan −1 µ s
90 kg
)
*
(
= ( 0.012 m ) sin tan −1 0.2
= 0.04 m to the
)
= 0.0023534 m
FBD pulley + gate:
β = sin −1
α = tan −1
0.04 m
= 14.931°
0.15 m
rf
0.0023534 m
= 0.8686°
0.15524 m
OB
= sin −1
OB =
0.15
= 0.15524 m
cos α
then
θ = α + β = 15.800°
P = W tan θ = 248.9 N
P = 250 N
* See note before Problem 8.75.
PROBLEM 8.86
A gate assembly consisting of a 24-kg gate ABC and a 66-kg
counterweight D is attached to a 24-mm-diameter shaft B which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
is 0.20 between the shaft and the bearing, determine the magnitude of the
force P for which clockwise rotation of the gate is impending.
SOLUTION
FBD gate:
(
)
(
)
W1 = 66 kg 9.81 m/s 2 = 647.46 N
W2 = 24 kg 9.81 m/s 2 = 235.44 N
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
*
)
= ( 0.012 m ) sin tan −1 0.2 = 0.0023534 m
ΣM C = 0:
P=
( 0.6 m + rf )W1 + ( 0.15 m + rf ) P − (1.8 m − rf )W2 = 0
(1.79765 m )( 235.44 N ) − ( 0.60235 m )( 647.46 N )
0.15235 m
= 218.19 N
P = 218 N
* See note before Problem 8.75.
PROBLEM 8.87
A gate assembly consisting of a 24-kg gate ABC and a 66-kg
counterweight D is attached to a 24-mm-diameter shaft B which fits
loosely in a fixed bearing. Knowing that the coefficient of static friction
is 0.20 between the shaft and the bearing, determine the magnitude of the
force P for which clockwise rotation of the gate is impending.
SOLUTION
It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single weight of
( 90 kg )( 9.81 N/kg ) = 882.9 N, located at a distance
x=
(1.8 m )( 24 kg ) − ( 0.15 m )( 66 kg )
90 kg
= 0.04 m to the
right of B.
FBD pulley + gate:
(
rf = rs sin φs = rs sin tan −1 µ s
)
*
(
= ( 0.012 m ) sin tan −1 0.2
)
rf = 0.0023534 m
α = tan −1
β = sin −1
rf
OB
0.04 m
= 14.931°
0.15 m
= sin −1
OB =
0.0023534 m
= 0.8686°
0.15524 m
0.15 m
= 0.15524 m
cos α
then
θ = α − β = 14.062°
P = W tan θ = 221.1 N
P = 221 N
* See note before Problem 8.75.
PROBLEM 8.88
A loaded railroad car has a weight of 35 tons and is supported by eight
32-in.-diameter wheels with 5-in.-diameter axles. Knowing that the
coefficients of friction are µ s = 0.020 and µ k = 0.015, determine the
horizontal force required (a) for impending motion of the car, (b) to keep
the car moving at a constant speed. Neglect rolling resistance between the
wheels and the track.
SOLUTION
FBD wheel:
Ww =
1
1
1
Wc = ( 35 ton ) = ( 70,000 ) lb
8
8
8
(
rf = ra sin φ = ra sin tan −1 µ
θ = sin
)
*
(
)
(
)
 ( 2.5 in.) sin tan −1µ 

= sin


rw
16 in.


−1 r f
−1 
= sin −1 0.15625sin tan −1µ 


(a) For impending motion use µ s = 0.02: then θ s = 0.179014°
(b) For steady motion use µ k = 0.15: then θ k = 0.134272°
Pw = Ww tan θ
(a)
Pc = Wc tan θ = 8Ww tan θ
Pc = ( 70,000 lb ) tan ( 0.179014° )
Pc = 219 lb
(b)
Pc = ( 70,000 lb ) tan ( 0.134272° )
Pc = 164.0 lb
* See note before Problem 8.75.
PROBLEM 8.89
A scooter is designed to roll down a 2 percent slope at a constant speed.
Assuming that the coefficient of kinetic friction between the 1-in.diameter axles and the bearing is 0.10, determine the required diameter of
the wheels. Neglect the rolling resistance between the wheels and the
ground.
SOLUTION
FBD wheel:
Note: The wheel is a two-force member in equilibrium, so R and W
must be collinear and tangent to friction circle.
2% slope ⇒ tan θ = 0.02
Also
But
sin θ =
rf
rw
(
)
sin tan −1 0.02 = 0.019996
(
rf = ra sin φk = ra sin tan −1 µ k
(
)
*
)
= (1 in.) sin tan −1 0.1 = 0.099504 in.
Then
and
rw =
rf
sin θ
=
0.099504
= 4.976 in.
0.019996
d w = 2rw
* See note before Problem 8.75.
d w = 9.95 in.
PROBLEM 8.90
A 25-kg electric floor polisher is operated on a surface for which the
coefficient of kinetic friction is 0.25. Assuming that the normal force per
unit area between the disk and the floor is uniformly distributed,
determine the magnitude Q of the horizontal forces required to prevent
motion of the machine.
SOLUTION
M H = dQ = ( 0.4 m ) Q
Couple exerted on handle
MF =
Couple exerted on floor
where
µ k = 0.25,
(
(Equation 8.9)
)
P = ( 25 kg ) 9.81 m/s 2 = 245.25 N,
R = 0.18 m
MH = MF,
For equilibrium
so
2
µ k PR
3
Q=
2
3
( 0.25)( 245.25 N )( 0.18 m )
0.4 m
Q = 18.39 N
PROBLEM 8.91
The pivot for the seat of a desk chair consists of the steel plate A, which
supports the seat, the solid steel shaft B which is welded to A and which
turns freely in the tubular member C, and the nylon bearing D. If a person
of weight W = 180 lb is seated directly above the pivot, determine the
magnitude of the couple M for which rotation of the seat is impending
knowing that the coefficient of static friction is 0.15 between the tubular
member and the bearing.
SOLUTION
For an annular bearing area
Since R =
Now
D
2
M =
2
R3 − R13
µ s P 22
3
R2 − R12
M =
1
D3 − D13
µ s P 22
3
D2 − D12
( Equation 8.8)
µ s = 0.15, P = W = 180 lb, D1 = 1.00 in., D2 = 1.25 in.
(1.25 in.) − ( 4 in.)
0.15
(180 lb )
3
(1.25 in.)2 − (1 in.)2
3
M =
3
M = 15.25 lb ⋅ in.
PROBLEM 8.92
As the surfaces of a shaft and a bearing wear out, the frictional resistance
of a thrust bearing decreases. It is generally assumed that the wear is
directly proportional to the distance traveled by any given point of the
shaft and thus to the distance r from the point to the axis of the shaft.
Assuming, then, that the normal force per unit area is inversely
proportional to r, show that the magnitude M of the couple required to
overcome the frictional resistance of a worn-out end bearing (with
contact over the full circular area) is equal to 75 percent of the value
given by formula (8.9) for a new bearing.
SOLUTION
Let the normal force on ∆A be ∆N , and
∆N
k
=
r
∆A
∆F = µ∆N , ∆M = r ∆F
As in the text
The total normal force
2π  R k

P = lim Σ∆N = ∫0  ∫0 rdr  dθ
∆A → 0
 r

(
)
P = 2π ∫0 kdr = 2π kR
The total couple
R
or
P
2π R
k
2π  R

M worn = lim Σ∆M = ∫0  ∫0 r µ rdr  dθ
∆A → 0
r


M worn = 2πµ k ∫0 rdr = 2πµ k
R
R2
P R2
= 2πµ
2
2π R 2
or
M worn =
1
µ PR
2
Now
M new =
2
µ PR
3
Thus
k =
M worn
=
M new
[Eq. (8.9)]
1
2
2
3
µ PR 3
= = 75%
µ PR 4
PROBLEM 8.93
Assuming that bearings wear out as indicated in Problem 8.92, show that
the magnitude M of the couple required to overcome the frictional
1
resistance of a worn-out collar bearing is M = µk P ( R1 + R2 )
2
where P = magnitude of the total axial force
R1, R2 = inner and outer radii of collar
SOLUTION
Let normal force on ∆A be ∆N , and
∆N
k
=
∆A
r
∆F = µ∆N , ∆M = r ∆F
As in the text
The total normal force P is
2π  R k

P = lim Σ∆N = ∫0  ∫R 2 rdr  dθ
1
∆A → 0
r


P = 2π ∫R 2 kdr = 2π k ( R2 − R1 )
R
k =
or
1
P
2π ( R2 − R1 )
k
2π  R

M worn = lim Σ∆M = ∫0  ∫R 2 r µ rdr  dθ
1
∆A → 0
r


The total couple is
M worn = 2πµ k ∫
R2
R1
( rdr ) = πµ k (
R22
−
R12
)=
(
πµ P R22 − R12
2π ( R2 − R1 )
M worn =
)
1
µ P ( R2 + R1 )
2
PROBLEM 8.94
Assuming that the pressure between the surfaces of contact is uniform,
show that the magnitude M of the couple required to overcome frictional
2 µk P R23 − R13
resistance for the conical bearing shown is M =
3 sin θ R22 − R12
SOLUTION
Let normal force on ∆A be ∆N , and
∆A = r ∆s∆φ
∆N = k ∆A
so
∆N
= k,
∆A
∆s =
∆r
sin θ
where φ is the azimuthal angle around the symmetry axis of rotation
∆Fy = ∆N sin θ = kr ∆r ∆φ
P = lim Σ∆Fy
Total vertical force
∆A → 0
2π
P = ∫0
(
(∫
R2
krdr
R1
P = π k R22 − R12
Friction force
)
) dφ = 2π k ∫
R2
rdr
R1
or
k =
Total couple
M = 2π
(
− R12
)
∆F = µ∆N = µ k ∆A
∆M = r ∆F = r µ kr
Moment
π
P
R22
∆r
∆φ
sin θ
2π  R µ k

M = lim Σ∆M = ∫0  ∫R 2
r 2dr  dφ
1
∆A → 0
sin θ


µ k R2 2
2 πµ
P
r dr =
∫
R1
2
sin θ
3 sin θ π R2 − R32
(
)
M =
(R
3
2
− R33
)
2 µ P R23 − R13
3 sin θ R22 − R12
PROBLEM 8.95
Solve Problem 8.90 assuming that the normal force per unit area between
the disk and the floor varies linearly from a maximum at the center to
zero at the circumference of the disk.
SOLUTION
Let normal force on ∆A be ∆N , and
r
∆N

= k 1 − 
R
∆A


r
r


∆F = µ∆N = µ k 1 −  ∆A = µ k 1 −  r ∆r ∆θ
R
R



r
2π  R 
P = lim Σ∆N = ∫0  ∫0 k 1 −  rdr  dθ
∆A → 0
R



 R 2 R3 
r
R
P = 2π k ∫0 1 −  rdr = 2π k 
−

3R 
R

 2
P=
1
π kR 2
3
k =
or
3P
π R2
r 
2π  R

M = lim Σr ∆F = ∫0  ∫0 r µ k 1 −  rdr  dθ
∆A → 0
R 


 R3 R 4  1
r3 
R
3
= 2πµ k ∫0  r 2 −  dr = 2πµ k 
−
 = πµ kR
R
3
4
R
6




=
where
πµ 3P 3 1
R = µ PR
6 π R2
2
µ = µk = 0.25
R = 0.18 m
(
)
P = W = ( 25 kg ) 9.81 m/s 2 = 245.25 N
Then
Finally,
M =
1
( 0.25)( 245.25 N )( 0.18 m ) = 5.5181 N ⋅ m
2
Q=
M
5.5181 N ⋅ m
=
d
0.4 m
Q = 13.80 N
PROBLEM 8.96
A 1-ton machine base is rolled along a concrete floor using a series of
steel pipes with outside diameters of 5 in. Knowing that the coefficient of
rolling resistance is 0.025 in. between the pipes and the base and
0.0625 in. between the pipes and the concrete floor, determine the
magnitude of the force P required to slowly move the base along the
floor.
SOLUTION
FBD pipe:
θ = sin −1
0.025 in. + 0.0625 in.
= 1.00257°
5 in.
P = W tan θ for each pipe, so also for total
P = ( 2000 lb ) tan (1.00257° )
P = 35.0 lb
PROBLEM 8.97
Knowing that a 120-mm-diameter disk rolls at a constant velocity down a
2 percent incline, determine the coefficient of rolling resistance between
the disk and the incline.
SOLUTION
FBD disk:
tan θ = slope = 0.02
b = r tan θ = ( 60 mm )( 0.02 )
b = 1.200 mm
PROBLEM 8.98
Determine the horizontal force required to move a 1-Mg automobile with
460-mm-diameter tires along a horizontal road at a constant speed.
Neglect all forms of friction except rolling resistance, and assume the
coefficient of rolling resistance to be 1 mm.
SOLUTION
FBD wheel:
r = 230 mm
b = 1 mm
θ = sin −1
b
r
b

P = W tan θ = W tan  sin −1  for each wheel, so for total
r

(
)
1 

P = (1000 kg ) 9.81 m/s 2 tan  sin −1

230 

P = 42.7 N
PROBLEM 8.99
Solve Problem 8.88 including the effect of a coefficient of rolling
resistance of 0.02 in.
SOLUTION
FBD wheel:
(
rf = ra sin φ = ra sin tan −1 µ
(
= ( 2.5 in.) sin tan −1 µ
)
)
P = W tan θ for each wheel, so also for total P = W tan θ
tan θ ≈
b + rf
rw
P = ( 70, 000 lb )
So
for small θ
( 0.02 in.) + rf
16 in.
(a) For impending motion, use µ s = 0.02:
Then
rf = 0.04999 in.
and
P = 306 lb
and
P = 252 lb
(b) For steady motion, use µ k = 0.015:
Then
rf = 0.037496 in.
PROBLEM 8.100
Solve Problem 8.89 including the effect of a coefficient of rolling
resistance of 0.07 in.
SOLUTION
FBD wheel:
The wheel is a two-force body, so R and W are colinear and tangent to the friction circle.
tan θ = slope = 0.02
tan θ ≈
Now
b + rf
rw
rw ≈
or
(
b + rf
tan θ
rf = ra sin φk = ra sin tan −1 µk
(
)
)
= ( 0.5 in.) sin tan −1 0.1
= 0.049752
∴ rw ≈
0.07 in. + 0.049752 in.
= 5.9876 in.
0.02
d = 2rw
d = 11.98 in.
PROBLEM 8.101
A hawser is wrapped two full turns around a bollard. By exerting a 320-N
force on the free end of the hawser, a dockworker can resist a force of
20 kN on the other end of the hawser. Determine (a) the coefficient of
static friction between the hawser and the bollard, (b) the number of times
the hawser should be wrapped around the bollard if a 80-kN force is to be
resisted by the same 320-N force.
SOLUTION
β = 4π rad
Two full turns of rope →
(a)
µ s β = ln
µs =
T2
T1
or
µs =
1
β
ln
T2
T1
1
20 000 N
ln
= 0.329066
4π
320 N
µ s = 0.329
(b)
β =
=
1
µs
ln
T2
T1
1
80 000 N
ln
0.329066
320 N
= 16.799 rad
β = 2.67 turns
PROBLEM 8.102
Blocks A and B are connected by a cable that passes over support C.
Friction between the blocks and the inclined surfaces can be neglected.
Knowing that motion of block B up the incline is impending when
WB = 16 lb, determine (a) the coefficient of static friction between the
rope and the support, (b) the largest value of WB for which equilibrium is
maintained. (Hint: See Problem 8.128.)
SOLUTION
FBD A:
FBD B:
ΣFt = 0: TA − 20 lb sin 30° = 0
ΣFx′ = 0: WB sin 30° − TB = 0
TA = 10 lb
TB =
From hint, β = 60° =
π
3
rad regardless of shape of support C
(a) For motion of B up incline when WB = 16 lb,
and
WB
2
µ s β = ln
TA
TB
or
TB =
µs =
1
β
ln
WB
= 8 lb
2
TA
3 10 lb
=
= 0.213086
ln
TB
π
8 lb
µ s = 0.213
(b) For maximum WB , motion of B impends down and TB > TA
So
Now
So that
TB = TAe µ s β = (10 lb ) e0.213086 π /3 = 12.500 lb
WB = 2TB
WB = 25.0 lb
PROBLEM 8.103
Blocks A and B are connected by a cable that passes over support C.
Friction between the blocks and the inclined surfaces can be neglected.
Knowing that the coefficient of static friction between the rope and the
support is 0.50, determine the range of values of WB for which
equilibrium is maintained. (Hint: See Problem 8.128.)
SOLUTION
FBD A:
FBD B:
ΣFt = 0: TA − 20 lbsin 30° = 0
TA = 10 lb
ΣFt ′ = 0: WB sin 30° − TB = 0
TB =
From hint, β = 60° =
π
3
WB
2
rad, regardless of shape of support C.
For impending motion of B up, TA > TB , so
TA = TBe µs β
or
TB = TAe − µs β = (10 lb ) e−0.5π /3 = 5.924 lb
WB = 2TB = 11.85 lb
For impending motion of B down, TB > TB , so
TB = TAe µs β = (10 lb ) e0.5π /3 = 16.881 lb
WB = 2TB = 33.76 lb
For equilibrium
11.85 lb ≤ WB ≤ 33.8 lb
PROBLEM 8.104
A 120-kg block is supported by a rope which is wrapped 1 12 times
around a horizontal rod. Knowing that the coefficient of static friction
between the rope and the rod is 0.15, determine the range of values of P
for which equilibrium is maintained.
SOLUTION
β = 1.5 turns = 3π rad
For impending motion of W up
P = We µs β = (1177.2 N ) e(
0.15 )3π
= 4839.7 N
For impending motion of W down
− 0.15 3π
P = We − µs β = (1177.2 N ) e ( )
= 286.3 N
For equilibrium
286 N ≤ P ≤ 4.84 kN
PROBLEM 8.105
The coefficient of static friction between block B and the horizontal
surface and between the rope and support C is 0.40. Knowing that
WA = 30 lb, determine the smallest weight of block B for which
equilibrium is maintained.
SOLUTION
Support at C:
FBD block B:
ΣFy = 0: N B − WB = 0
Impending motion
or
FB = µ s N B = 0.4 N B = 0.4WB
ΣFx = 0: FB − TB = 0
or
Now
so that
TB = FB = 0.4WB
WA = TBe µ s β
At support, for impending motion of WA down,
so
N B = WB
− 0.4 π /2
TB = WAe − µ s β = ( 30 lb ) e ( ) = 16.005 lb
WB =
TB
0.4
WB = 40.0 lb
PROBLEM 8.106
The coefficient of static friction µ s is the same between block B and
the horizontal surface and between the rope and support C. Knowing
that WA = WB , determine the smallest value of µ s for which
equilibrium is maintained.
SOLUTION
Support at C
FBD B:
ΣFy = 0: N B − W = 0
or
FB = µ s N B = µ sW
Impending motion:
ΣFx = 0: FB − TB = 0
or
1 = µ s e µs β
π
or
Solving numerically:
TB = FB = µ sW
W = TBe µ s β = µ sWe µ s β
Impending motion of rope on support:
or
NB = W
µ se 2
µs
=1
µ s = 0.475
PROBLEM 8.107
In the pivoted motor mount shown, the weight W of the 85-kg motor is
used to maintain tension in the drive belt. Knowing that the coefficient of
static friction between the flat belt and drums A and B is 0.40, and
neglecting the weight of platform CD, determine the largest torque which
can be transmitted to drum B when the drive drum A is rotating
clockwise.
SOLUTION
FBD motor + mount:
For impending slipping of belt,
ΣM D = 0:
T2 = T1e µs β = T1e0.4π = 3.5136T1
( 0.24 m )(833.85 N ) − ( 0.14 m ) T2 − ( 0.26 m ) T1 = 0
( 0.14 m )( 3.5136 ) + 0.26 m  T1 = 200.124 N ⋅ m
or
and
T1 = 266.16 N
T2 = 3.5136T1 = 935.18 N
FBD drum:
ΣM B = 0: M B − ( 0.06m )( 266.16 N − 935.18 N ) = 0
M B = 40.1 N ⋅ m
(Compare to M B = 81.7 N ⋅ m using V-belt, Problem 8.130.)
PROBLEM 8.108
Solve Problem 8.107 assuming that the drive drum A is rotating
counterclockwise.
SOLUTION
FBD motor + mount:
T1 = T2e µ s β = T1e0.4π = 3.5136T2
Impending slipping of belt:
ΣM D = 0:
( 0.24 m )W − ( 0.26 m ) T1 − ( 0.14 m ) T2
=0
( 0.26 m )( 3.5136 ) + 0.14 m  T2 = ( 0.24 m )( 833.85 N )
or
T2 = 189.95 N
and
T1 = 667.42 N
FBD drum:
ΣM B = 0:
( 0.06 m )( 667.42 N − 189.95 N ) − M B
=0
M B = 28.6 N ⋅ m
PROBLEM 8.109
A flat belt is used to transmit a torque from pulley A to pulley B. The
radius of each pulley is 3 in., and a force of magnitude P = 225 lb is
applied as shown to the axle of pulley A. Knowing that the coefficient of
static friction is 0.35, determine (a) the largest torque which can be
transmitted, (b) the corresponding maximum value of the tension in the
belt.
SOLUTION
FBD pulley A:
T2 = T1e µs β
Impending slipping of belt:
T2 = T1e0.35π = 3.0028T1
ΣFx = 0: T1 + T2 − 225 lb = 0
T1 (1 + 3.0028 ) = 225 lb
T2 = 3.0028T1
(a)
or
T1 = 56.211 lb
or
T2 = 168.79 lb
ΣM A = 0: M A + ( 6 in.)(T1 − T2 ) = 0
or
M A = ( 3 in.)(168.79 lb − 56.21 lb )
∴ max. torque: M A = 338 lb ⋅ in.
max. tension: T2 = 168.8 lb
(b)
(Compare with M A = 638 lb ⋅ in. with V-belt, Problem 8.131.)
PROBLEM 8.110
Solve Problem 8.109 assuming that the belt is looped around the pulleys
in a figure eight.
SOLUTION
FBDs pulleys:
θ = sin −1
3 in.
π
= 30° = rad.
6 in.
6
β =π +2
π
6
4π
3
=
T2 = T1e µs β
Impending belt slipping:
T2 = T1e(
0.35 ) 4π /3
= 4.3322T1
ΣFx = 0: T1 cos 30° + T2 cos 30° − 225 lb = 0
(T1 + 4.3322T1 ) cos 30° = 225 lb
T2 = 4.3322T1
(a)
so that
ΣM A = 0: M A + ( 3 in.)(T1 − T2 ) = 0
or
T1 = 48.7243 lb
T2 = 211.083 lb
or
M A = ( 3 in.)( 211.083 lb − 48.224 lb )
M max = M A = 487 lb ⋅ in.
(b)
Tmax = T2 = 211 lb
PROBLEM 8.111
A couple M B of magnitude 2 lb ⋅ ft is applied to the drive drum B of a
portable belt sander to maintain the sanding belt C at a constant speed.
The total downward force exerted on the wooden workpiece E is 12 lb,
and µ k = 0.10 between the belt and the sanding platen D. Knowing that
µ s = 0.35 between the belt and the drive drum and that the radii of
drums A and B are 1.00 in., determine (a) the minimum tension in the
lower portion of the belt if no slipping is to occur between the belt and
the drive drum, (b) the value of the coefficient of kinetic friction between
the belt and the workpiece.
SOLUTION
FBD lower portion of belt:
ΣFy = 0: N E − N D = 0
N D = N E = 12 lb
or
FD = ( µ k )belt/platen N D
Slipping:
FD = 0.1(12 lb ) = 1.2 lb
FE = ( µk )belt/wood N E
and
F = (12 lb ) ( µ k )belt/wood
(1)
ΣFx = 0: TB − TA − FD − FE = 0
(2)
FBD drum A: (assumed free to rotate)
ΣM A = 0: rA (TA − TT ) = 0
or
TT = TA
PROBLEM 8.111 CONTINUED
FBD drum B:
ΣM B = 0: M B + r (TT − TB ) = 0
TB − TT =
or
TB = TT e µs β = TT e0.35π
Impending slipping:
So
Now
From Equation (2):
From Equation (1):
Therefore
M B  2 lb ⋅ ft   12 in. 
=

 = 24 lb
r
 1 in.   ft 
(e
0.35π
)
− 1 TT = 24 lb
or
TT = 11.983 lb
TA = TT = 11.983 lb then TB = (11.983 lb ) e0.35π = 35.983 lb
35.983 lb − 11.983 lb − 1.2 lb = FE = 22.8 lb
( µk )belt/wood
=
FE
22.8 lb
=
= 1.900
12 lb
12 lb
(a) Tmin = TA = 11.98 lb
(b)
( µk )belt/wood
= 1.900
PROBLEM 8.112
A band belt is used to control the speed of a flywheel as shown.
Determine the magnitude of the couple being applied to the flywheel
knowing that the coefficient of kinetic friction between the belt and the
flywheel is 0.25 and that the flywheel is rotating clockwise at a constant
speed. Show that the same result is obtained if the flywheel rotates
counterclockwise.
SOLUTION
FBD wheel:
ΣM E = 0: − M E + ( 7.5 in.)(T2 − T1 ) = 0
M E = ( 7.5 in.)(T2 − T1 )
or
ΣM C = 0:
FBD lever:
T1 + T2 = 100 lb
or
Impending slipping:
or
So
( 4 in.)(T1 + T2 ) − (16 in.)( 25 lb ) = 0
T2 = T1e µs β
T2 = T1e
( )
0.25 32π
= 3.2482T1
T1 (1 + 3.2482 ) = 100 lb
T1 = 23.539 lb
and
M E = ( 7.5 in.)( 3.2482 − 1)( 23.539 lb ) = 396.9 lb ⋅ in.
M E = 397 lb ⋅ in.
Changing the direction of rotation will change the direction of M E and
will switch the magnitudes of T1 and T2 .
The magnitude of the couple applied will not change.
PROBLEM 8.113
The drum brake shown permits clockwise rotation of the drum but
prevents rotation in the counterclockwise direction. Knowing that the
maximum allowed tension in the belt is 7.2 kN, determine (a) the
magnitude of the largest counterclockwise couple that can be applied to
the drum, (b) the smallest value of the coefficient of static friction
between the belt and the drum for which the drum will not rotate
counterclockwise.
SOLUTION
FBD lever:
ΣM B = 0:
( 25 mm ) TC − ( 62.5 mm ) TA
=0
TC = 2.5TA
Impending ccw rotation:
FBD lever:
TC = Tmax = 7.2 kN
(a)
TC = 2.5TA
But
TA =
So
7.2 kN
= 2.88 kN
2.5
ΣM D = 0: M D + (100 mm ) (TA − TC ) = 0
M D = (100 mm )( 7.2 − 2.88 ) kN
M D = 432 N ⋅ m
(b) Also, impending slipping:
µs =
Therefore,
1
β
ln
µ s β = ln
TC
TA
TC
1
=
ln2.5 = 0.2187
π
4
TA
3
( µs )min
= 0.219
PROBLEM 8.114
A differential band brake is used to control the speed of a drum which
rotates at a constant speed. Knowing that the coefficient of kinetic
friction between the belt and the drum is 0.30 and that a couple of
magnitude is 150 N ⋅ m applied to the drum, determine the corresponding
magnitude of the force P that is exerted on end D of the lever when the
drum is rotating (a) clockwise, (b) counterclockwise.
SOLUTION
FBD lever:
ΣM B = 0:
( 0.34 m ) P + ( 0.04 m ) TC − ( 0.15 m ) TA
P=
=0
15TA − 4TC
34
(1)
FBD drum:
(a) For cw rotation, M E
( 0.14 m ) (TA − TC ) − M E
ΣM E = 0:
TA − TC =
Impending slipping:
=0
150 N ⋅ m
= 1071.43 N
0.14 m
TA = TC e µk β = TC e
( 0.3) 76π
TA = 3.00284TC
So
and
( 3.00284 − 1) TC
= 1071.43 N
TA = 1606.39 N
or
TC = 534.96 N
PROBLEM 8.114 CONTINUED
P=
From Equation (1):
15 (1606.39 N ) − 4 ( 534.96 N )
34
P = 646 N
(b) For ccw rotation,
Also, impending slip ⇒
ME
and
TC = 3.00284TA , so TA = 534.96 N
TC = 1606.39 N
and
And Equation (1) ⇒
ΣM E = 0 ⇒ TC − TA = 1071.43 N
P=
15 ( 534.96 N ) − 4 (1606.39 N )
34
P = 47.0 N
PROBLEM 8.115
A differential band brake is used to control the speed of a drum.
Determine the minimum value of the coefficient of static friction for
which the brake is self-locking when the drum rotates counterclockwise.
SOLUTION
FBD lever:
For self-locking P = 0
ΣM B = 0:
( 0.04 m ) TC − ( 0.15 m ) TA
=0
TC = 3.75TA
FBD drum:
TC = TAe µs β
For impending slipping of belt
or
Then
µ s β = ln
µs =
1
β
TC
TA
ln
TC
1
=
ln 3.75 = 0.3606
π
7
TA
6
( µs )req
= 0.361
PROBLEM 8.116
Bucket A and block C are connected by a cable that passes over drum B.
Knowing that drum B rotates slowly counterclockwise and that the
coefficients of friction at all surfaces are µ s = 0.35 and µ k = 0.25,
determine the smallest combined weight W of the bucket and its contents
for which block C will (a) remain at rest, (b) be about to move up the
incline, (c) continue moving up the incline at a constant speed.
SOLUTION
FBD block:
ΣFn = 0: NC − ( 200 lb ) cos 30° = 0; N = 100 3 lb
ΣFt = 0: TC − ( 200 lb ) sin 30° ∓ FC = 0
TC = 100 lb ± FC
(1)
where the upper signs apply when FC acts
FBD drum:
(a) For impending motion of block , FC
, and
(
)
FC = µ s NC = 0.35 100 3 lb = 35 3 lb
(
)
TC = 100 − 35 3 lb
So, from Equation (1):
TC = WAe µk β
But belt slips on drum, so
−0.25(
WA =  100 − 35 3 lb  e


(
)
2π
3
)
WA = 23.3 lb
and FC = µ s NC = 35 3 lb
(b) For impending motion of block , FC
From Equation (1):
Belt still slips, so
(
)
TC = 100 + 35 3 lb
−0.25(
WA = TC e − µk β =  100 + 35 3 lb  e


(
)
2π
3
)
WA = 95.1 lb
PROBLEM 8.116 CONTINUED
(c) For steady motion of block , FC
, and FC = µk NC = 25 3 lb
(
)
T = 100 + 25 3 lb.
Then, from Equation (1):
Also, belt is not slipping on drum, so
−0.35(
WA = TC e− µ s β =  100 + 25 3 lb  e


(
)
2π
3
)
WA = 68.8 lb
PROBLEM 8.117
Solve Problem 8.116 assuming that drum B is frozen and cannot
rotate.
SOLUTION
ΣFn = 0: NC − ( 200 lb ) cos 30° = 0; NC = 100 3 lb
FBD block:
ΣFt = 0: ± FC + ( 200 lb ) sin 30° − T = 0
T = 100 lb ± FC
(1)
where the upper signs apply when FC acts
(a) For impending motion of block , FC
So
(
)
FC = 0.35 100 3 lb = 35 3 lb
T = 100 lb − 35 3 lb = 39.375 lb
and
FBD drum:
Also belt slipping is impending
or
and FC = µ s NC
T = WAe µ s β
so
WA = Te− µ s β = ( 39.378 lb ) e
( )
−0.35 23π
WA = 18.92 lb
(b) For impending motion of block , FC
, and
FC = µ s NC = 35 3 lb
But
(
)
T = 100 + 35 3 lb = 160.622 lb.
Also belt slipping is impending
So
WA = Te+ µ s β = (160.622 lb ) e
( );
0.35 23π
WA = 334 lb
(c) For steady motion of block , FC
Then
(
, and FC = µk NC = 25 3 lb
)
T = 100 lb + 25 3 lb = 143.301 lb.
Now belt is slipping
So
WA = Te µk β = (143.301 lb ) e
( )
0.25 23π
WA = 242 lb
PROBLEM 8.118
A cable passes around three 30-mm-radius pulleys and supports two
blocks as shown. Pulleys C and E are locked to prevent rotation, and the
coefficients of friction between the cable and the pulleys are µ s = 0.20
and µ k = 0.15. Determine the range of values of the mass of block A for
which equilibrium is maintained (a) if pulley D is locked, (b) if pulley D
is free to rotate.
SOLUTION
θ = sin −1
Note:
βC = β D =
So
r
π
= 30° = rad
2r
6
2π
3
and
βE = π
(a) All pulleys locked ⇒ slipping impends at all surface simultaneously.
T2 = WAe µ s βC ; T1 = T2e µ s β D ; WB = T1e µs β E
If A impends ,
So
Then
If A impends ,
So
µ β +β +β
WB = WAe s ( C D E )
mA = mBe
− µ s ( βC + β D + β E )
or
= ( 8 kg ) e
WA = WBe
(
− µs ( βC + β D + β E )
−0.2 23π + 23π + π
)
= 1.847 kg
µ β +β +β
WA = T2e µ s βC = T1e µs β D e µ s βC = WBe s ( E D C )
0.2(π + 23π + 23π )
µ β +β +β
mA = mBe s ( E D C ) = ( 8 kg ) e
= 34.7 kg
Equilibrium for 1.847 kg ≤ mA ≤ 34.7 kg
PROBLEM 8.118 CONTINUED
(b) Pulleys C & E locked, pulley D free ⇒ T1 = T2 , other relations remain the same.
If A impends ,
So
T2 = WAe µs βC = T1
mA = mBe
− µs ( βC + β E )
If A impends slipping is reversed,
Then
µ β +β
WB = T1e µs β E = WAe s ( C E )
= ( 8 kg ) e
(
−0.2 23π + π
)
= 2.807 kg
WA = WBe
+ µ s ( βC + β E )
0.2( 5π )
µ β +β
mA = mBe s ( C E ) = ( 8 kg ) e 3 = 22.8 kg
Equilibrium for 2.81 kg ≤ mA ≤ 22.8 kg
PROBLEM 8.119
A cable passes around three 30-mm-radius pulleys and supports two
blocks as shown. Two of the pulleys are locked to prevent rotation, while
the third pulley is rotated slowly at a constant speed. Knowing that the
coefficients of friction between the cable and the pulleys are µ s = 0.20
and µ k = 0.15, determine the largest mass mA which can be raised (a) if
pulley C is rotated, (b) if pulley E is rotated.
SOLUTION
θ = sin −1
Note:
βC = β D =
r
π
= 30° = rad
2r
6
2π
3
βE = π
and
Mass A moves up
(a) C rotates , for maximum WA have no belt slipping on C, so
WA = T2e µ s βC
T1 = T2e µk β D
D and E are fixed, so
and
Thus
µ β +β
−µ β + β
WB = T1e µk β E = T2e k ( D E ) ⇒ T2 = WBe k ( D E )
mA g = mB ge
µs βC − µk ( β D + β E )
or
mA = ( 8 kg ) e
( 0.43π − 0.1π − 0.15π )
mA = 5.55 kg
PROBLEM 8.119 CONTINUED
T1 = WBe µs β E
(b) E rotates , no belt slip on E, so
µ β +β
T1 = T2e µk β D = WAe k ( C D )
C and D fixed, so
or
Then
mA g = T1e
− µk ( βC + β D )
mA = ( 8 kg ) e(
= mB ge
µ s β E − µk ( βC + β D )
0.2π − 0.1π − 0.1π )
= 8.00 kg
mA = 8.00 kg
PROBLEM 8.120
A cable passes around three 30-mm-radius pulleys and supports two
blocks as shown. Pulleys C and E are locked to prevent rotation, and the
coefficients of friction between the cable and the pulleys are µ s = 0.20
and µ k = 0.15. Determine the range of values of the mass of block A for
which equilibrium is maintained (a) if pulley D is locked, (b) if pulley D
is free to rotate.
SOLUTION
θ = sin −1
Note:
βC =
So
0.075 m
π
= 30° = rad
0.15 m
6
5
2
1
π , βD = π , βE = π
6
3
2
(a) All pulleys locked, slipping at all surfaces.
T1 = WAe µs βC ,
For mA impending ,
T2 = T1e µ s β D ,
WB = T2e µk β E ,
and
µ β +β +β
mB g = mA ge s ( C D E )
So
8 kg = mAe
(
)
0.2 56 + 23 + 12 π
mA = 2.28 kg
or
For mA impending down, all tension ratios are inverted, so
mA = ( 8 kg ) e
(
)
0.2 56 + 23 + 12 π
= 28.1 kg
Equilibrium for 2.28 kg ≤ mA ≤ 28.1 kg
(b) Pulleys C and E locked, D free ⇒ T1 = T2 , other ratios as in (a)
T1 = WAe µ s βC = T2
mA impending ,
and
So
µ β +β
WB = T2e µs β E = WAe s ( C E )
µ β +β
mB g = mA g e ( C E )
8 kg = mAe
or
(
)
0.2 56 + 12 π
mA = 3.46 kg
mA impending , all tension ratios are inverted, so
mA = 8 kg e
(
)
0.2 56 + 12 π
= 18.49 kg
Equilibrium for 3.46 kg ≤ mA ≤ 18.49 kg
PROBLEM 8.121
A cable passes around three 30-mm-radius pulleys and supports two
blocks as shown. Two of the pulleys are locked to prevent rotation, while
the third pulley is rotated slowly at a constant speed. Knowing that the
coefficients of friction between the cable and the pulleys are µ s = 0.20
and µ k = 0.15, determine the largest mass mA which can be raised (a) if
pulley C is rotated, (b) if pulley E is rotated.
SOLUTION
θ = sin −1
Note:
βC =
So
0.075 m
π
= 30° = rad
0.15 m
6
5
2
1
π , βD = π , βE = π
6
3
2
WA = T1e µ s βC . If D
(a) To raise maximum mA , with C rotating
and E are fixed, cable must slip there, so T2 = T1e µk β D
and
µ β +β
WB = T2e µk β E = T1e k ( D E )
µ β +β
= WAe− µ s βC e k ( D E )
(8 kg ) g
= mA ge
( )e0.15( 23 + 12 )π
−0.2 56 π
mA = 7.79 kg
(b) With E rotating , T2 = WBe µs β E . With C and D fixed.
T1 = WAe µk βC
so
and
µ β +β
T2 = T1e µk β D = WAe k ( C D )
µ β +β
WB = WAe k ( C D )e− µ s β E
(8 kg ) g
= mA ge
(
)
( )
0.15 56 + 23 π −0.2 12 π
e
mA = 5.40 kg
PROBLEM 8.122
A recording tape passes over the 1-in.-radius drive drum B and under
the idler drum C. Knowing that the coefficients of friction between the
tape and the drums are µ s = 0.40 and µ k = 0.30 and that drum C is free
to rotate, determine the smallest allowable value of P if slipping of the
tape on drum B is not to occur.
SOLUTION
FBD drive drum:
ΣM B = 0: r (TA − T ) − M = 0
M
2.7 lb ⋅ in.
=
= 2.7 lb
r
1 in.
TA − T =
Impending slipping:
TA = Te µ S β = Te0.4π
(
)
So
T e0.4π − 1 = 2.7 lb
or
T = 1.0742 lb
If C is free to rotate, P = T
P = 1.074 lb
PROBLEM 8.123
Solve Problem 8.122 assuming that the idler drum C is frozen and cannot
rotate.
SOLUTION
FBD drive drum:
ΣM B = 0: r (TA − T ) − M = 0
TA − T =
Impending slipping:
So
or
M
2.7 lb ⋅ in.
=
= 2.7 lb
r
1 in.
TA = Te µ s β = Te0.4π
(e
0.4π
)
− 1 T = 2.7 lb
T = 1.07416 lb
If C is fixed, the tape must slip
So
P = Te µk βC = 1.07416 lb e
0.3 π2
= 1.7208 lb
P = 1.721 lb
PROBLEM 8.124
For the band brake shown, the maximum allowed tension in either belt is
5.6 kN. Knowing that the coefficient of static friction between the belt
and the 160-mm-radius drum is 0.25, determine (a) the largest clockwise
moment M 0 that can be applied to the drum if slipping is not to occur,
(b) the corresponding force P exerted on end E of the lever.
SOLUTION
FBD pin B:
T1 = T2
(a) By symmetry:
 2 
ΣFy = 0: B − 2 
T =0
 2 1 


For impending rotation
or
B=
2T1 =
2T2
(1)
:
T3 > T1 = T2 > T4 , so T3 = Tmax = 5.6 kN
Then
T1 = T3e − µs β L = ( 5.6 kN ) e
or
T1 = 4.03706 kN = T2
(
−0.25 π4 + π6
and
T4 = T2e− µs β R = ( 4.03706 kN ) e
or
T4 = 2.23998 kN
)
( )
−0.25 34π
ΣM F = 0: M 0 + r (T4 − T3 + T2 − T1 ) = 0
or
M 0 = ( 0.16 m )( 5.6 kN − 2.23998 kN ) = 0.5376 kN ⋅ m
M 0 = 538 N ⋅ m
Lever:
(b) Using Equation (1)
B=
2T1 =
2 ( 4.03706 kN )
= 5.70927 kN
ΣM D = 0:
( 0.05 m )( 5.70927 kN ) − ( 0.25 m ) P = 0
P = 1.142 kN
PROBLEM 8.125
Solve Problem 8.124 assuming that a counterclockwise moment is
applied to the drum.
SOLUTION
FBD pin B:
T1 = T2
(a) By symmetry:
 2 
ΣFy = 0: B − 2 
T =0
 2 1 


For impending rotation
or
B=
(1)
2T1
:
T4 > T2 = T1 > T3 , so T4 = Tmax = 5.6 kN
Then
T2 = T4e− µs β R = ( 5.6 kN ) e
or
T2 = 3.10719 kN = T1
( )
−0.25 34π
and
T3 = T1e− µs β L = ( 3.10719 kN ) e
or
T3 = 2.23999 kN
(
−0.25 π4 + π6
)
ΣM F = 0: M 0 + r (T2 − T1 + T3 − T4 ) = 0
M 0 = (160 mm )( 5.6 kN − 2.23999 kN ) = 537.6 N ⋅ m
M 0 = 538 N ⋅ m
FBD Lever:
(b) Using Equation (1)
B=
2T1 =
2 ( 3.10719 kN )
B = 4.3942 kN
ΣM D = 0:
( 0.05 m )( 4.3942 kN ) − ( 0.25 m ) P = 0
P = 879 N
PROBLEM 8.126
The strap wrench shown is used to grip the pipe firmly without marring
the surface of the pipe. Knowing that the coefficient of static friction is
the same for all surfaces of contact, determine the smallest value of µ s
for which the wrench will be self-locking when a = 10 in., r = 1.5 in.,
and θ = 65o.
SOLUTION
For the wrench to be self-locking, friction must be sufficient to maintain equilibrium as P is increased from
zero to Pmax , as well as to prevent slipping of the belt on the pipe.
FBD wrench:
 10 in.

 10 in.

ΣM E = 0: 
− 1.5 in.  F − 
− 1.5 in.  T2 = 0
 sin 65°

 tan65°

9.5338F = 3.1631 T2
or
3.01406 =
T2
F
(1)
ΣFx = 0: − T2 + N sin 65° + F cos 65° = 0
N = F /µ s
Impending slipping:
So
or
 sin 65°

F
+ cos 65°  = T2
 µs

0.90631
µs
+ 0.42262 =
T2
F
Solving Equations (1) and (2) yields µ s = 0.3497; must still check belt on pipe.
(2)
PROBLEM 8.126 CONTINUED
Small portion of belt at A:
ΣFt = 0: 2 F − T1 = 0
T1 = 2F
or
ln
Belt impending slipping:
So
Using Equation (1)
µs =
1
β
µs =
T2
= µs β
T1
ln
T2
1 T
= ln 2
T1
β 2F
180
ln1.50703
295π
= 0.0797
∴ for self-locking, need µ s = 0.350
PROBLEM 8.127
Solve Problem 8.126 assuming that θ = 75o.
SOLUTION
For the wrench to be self-locking, friction must be sufficient to maintain equilibrium as P is increased from
zero to Pmax , as well as to prevent slipping of the belt on the pipe.
FBD wrench:
 10 in.

 10 in.

ΣM E = 0: 
− 1.5 in.  F − 
− 1.5 in.  T2 = 0
 sin 75°

 tan 75°

or
T2
= 7.5056
F
(1)
ΣFx = 0: − T2 + N sin 75° + F cos 75° = 0
Impending slipping:
N = F /µ s
So
 sin 75°

F
+ cos 75°  = T2
 µs

or
T2
0.96593
=
+ 0.25882
F
µs
Solving Equations (1) and (2): µ s = 0.13329; must still check belt on pipe.
(2)
PROBLEM 8.127 CONTINUED
Small portion of belt at A:
ΣFt = 0: 2 F − T1 = 0
T1 = 2F
or
Impending belt slipping:
So
Using Equation (1):
ln
µs =
1
β
µs =
T2
= µs β
T1
ln
T2
1 T
= ln 2
T1
β 2F
180
7.5056
ln
285π
2
= 0.2659
∴ for self-locking, µ s = 0.266
PROBLEM 8.128
Prove that Equatins (8.13) and (8.14) are valid for any shape of surface
provided that the coefficient of friction is the same at all points of
contact.
SOLUTION
∆θ
ΣFn = 0: ∆N − T + (T + ∆T )  sin
=0
2
∆N = ( 2T + ∆T ) sin
or
∆θ
2
∆θ
ΣFt = 0: (T + ∆T ) − T  cos
− ∆F = 0
2
∆F = ∆T cos
or
∆F = µ s ∆N
Impending slipping:
So
In limit as
So
and
∆T cos
∆θ
2
∆θ
∆θ
sin ∆θ
= µ s 2T sin
+ µ s ∆T
2
2
2
∆θ → 0: dT = µ sTdθ ,
or
dT
= µ s dθ
T
T2
β
∫T1 T = ∫0 µ s dθ ;
dT
ln
T2
= µs β
T1
or T2 = T1e µs β
Note: Nothing above depends on the shape of the surface, except it is
assumed smooth.
PROBLEM 8.129
Complete the derivation of Equation (8.15), which relates the tension
in both parts of a V belt.
SOLUTION
Small belt section:
ΣFy = 0: 2
∆N
α
∆θ
sin − T + (T + ∆T )  sin
=0
2
2
2
∆θ
ΣFx = 0: (T + ∆T ) − T  cos
− ∆F = 0
2
Impending slipping:
In limit as ∆θ → 0:
∆F = µ s ∆N ⇒ ∆T cos
dT =
µ sTdθ
α
sin
So
∆θ
2T + ∆T
∆θ
= µs
sin
α
2
2
sin
2
dT
µs
dθ
=
α
T
sin
2
or
2
µ
T2
β
s
∫T1 T = α ∫0 dθ
sin
dT
2
or
or
ln
T2
µβ
= s
α
T1
sin
2
T2 = T1e
µ s β /sin α2
PROBLEM 8.130
Solve Problem 8.107 assuming that the flat belt and drums are
replaced by a V belt and V pulleys with α = 36o. (The angle α is as
shown in Figure 8.15a.)
SOLUTION
ΣM D = 0:
FBD motor + mount:
Impending slipping:
( 0.24 m )W − ( 0.26 m ) T1 − ( 0.14 m ) T2
T2 = T1e
=0
µ S β /sin α2
0.4π
T2 = T1e sin18° = 58.356T1
Thus
( 0.24 m )(833.85 N ) − 0.26 m + ( 0.14 m )( 58.356 ) T1 = 0
T1 = 23.740 N
T2 = 1385.369 N
FBD Drum:
ΣM B = 0: M B + ( 0.06 m )( 23.740 N − 1385.369 N ) = 0
M B = 81.7 N ⋅ m
(Compare to M B = 40.1 N ⋅ m using flat belt, Problem 8.107.)
PROBLEM 8.131
Solve Problem 8.109 assuming that the flat belt and drums are
replaced by a V belt and V pulleys with α = 36o. (The angle α is as
shown in Figure 8.15a.)
SOLUTION
FBD pulley A:
Impending slipping:
T2 = T1e
µ s β /sin α2
T2 = T1e0.35π /sin18° = 35.1015T1
ΣFx = 0: T1 + T2 − 225 lb = 0
T1 (1 + 35.1015 ) = 225 lb
So
T1 = 6.2324 lb
T2 = 218.768 lb = Tmax
ΣM A = 0: M + ( 3 in.)(T1 − T2 ) = 0
M = ( 3 in.)( 218.768 lb − 6.232 lb )
(a)
M = 638 lb ⋅ in.
(Compare to 338 lb ⋅ in. with flat belt, Problem 8.109.)
(b)
Tmax = 219 lb
PROBLEM 8.132
Considering only values of θ less than 90°, determine the smallest value
of θ required to start the block moving to the right when (a) W = 75 lb,
(b) W = 100 lb.
SOLUTION
FBD block: (motion impending)
φ s = tan −1 µ s = 14.036°
30 lb
W
=
sinφs
sin (θ − φs )
sin (θ − φs ) =
W sin14.036°
30 lb
W
123.695 lb
or
sin (θ − 14.036° ) =
(a)
W = 75 lb: θ = 14.036° + sin −1
75 lb
123.695 lb
θ = 51.4°
(b)
W = 100 lb: θ = 14.036° + sin −1
100 lb
123.695 lb
θ = 68.0°
PROBLEM 8.133
The machine base shown has a mass of 75 kg and is fitted with skids at A
and B. The coefficient of static friction between the skids and the floor is
0.30. If a force P of magnitude 500 N is applied at corner C, determine
the range of values of θ for which the base will not move.
SOLUTION
FBD machine base (slip impending):
φ s = tan −1 µ s = tan −1 0.3 = 16.699°
W
P
=
sin ( 90° − φs − θ ) sin φs
sin ( 90° − φ s − θ ) =
W sin16.699°
P
 735.75 lb

90° − 16.699° − θ = sin −1 
( 0.28734 )
 500 lb

θ = 73.301° − 25.013°
θ = 48.3°
FBD machine base (tip about B impending):
PROBLEM 8.133 CONTINUED
ΣM B = 0:
( 0.2 m )( 735.75 N ) + ( 0.5 m ) ( 500 N ) cosθ 
− ( 0.4 m ) ( 500 N ) sinθ  = 0
0.8 sin θ − cosθ = 0.5886
Solving numerically
So, for equilibrium
θ = 78.7°
48.3° ≤ θ ≤ 78.7°
PROBLEM 8.134
Knowing that a couple of magnitude 30 N ⋅ m is required to start the
vertical shaft rotating, determine the coefficient of static friction between
the annular surfaces of contact.
SOLUTION
For annular contact regions, use Equation 8.8 with impending slipping:
M =
2
R3 − R13
µ s N 22
3
R2 − R12
( 0.06 m ) − ( 0.025 m )
2
µ s ( 4000 N )
3
( 0.06 m )2 − ( 0.025 m )2
3
So,
30 N ⋅ m =
3
µ s = 0.1670
PROBLEM 8.135
The 20-lb block A and the 30-lb block B are supported by an incline
which is held in the position shown. Knowing that the coefficient of static
friction is 0.15 between the two blocks and zero between block B and the
incline, determine the value of θ for which motion is impending.
SOLUTION
FBD’s
Block A:
A:
ΣFn = 0: N A − ( 20 lb ) cosθ = 0
B:
ΣFn = 0: N B − N A − ( 30 lb ) cosθ = 0
or
or
N A = ( 20 lb ) cosθ
N B = N A + ( 30 lb ) cosθ = ( 50 lb ) cosθ
Impending motion at all surfaces:
FA = µ s N A
= 0.15 ( 20 lb ) cosθ
= ( 3 lb ) cosθ
Block B:
A:
ΣFt = 0: FA + ( 20 lb ) sin θ − T = 0
B:
ΣFt = 0: − FA + ( 30 lb ) sin θ − T = 0
So
(10 lb ) sin θ
(10 lb ) sin θ
θ = tan −1
− 2 FA = 0
= 2 ( 3 lb ) cosθ
6 lb
= 30.96°
10 lb
θ = 31.0°
PROBLEM 8.136
The 20-lb block A and the 30-lb block B are supported by an incline
which is held in the position shown. Knowing that the coefficient of static
friction is 0.15 between all surfaces of contact, determine the value of θ
for which motion is impending.
SOLUTION
FBD’s
Block A:
A:
ΣFn = 0: N A − ( 20 lb ) cosθ = 0
B:
ΣFn = 0: N B − N A − ( 30 lb ) cosθ = 0
or
N A = ( 20 lb ) cosθ
N B = N A + ( 30 lb ) cosθ = ( 50 lb ) cosθ
or
Impending motion at all surfaces; B impends :
FA = µ s N A = ( 0.15 )( 20 lb ) cosθ = ( 3 lb ) cosθ
FB = µ s N B = ( 0.15 )( 50 lb ) cosθ = ( 7.5 lb ) cosθ
Block B:
A:
ΣFt = 0:
( 20 lb ) sin θ
+ FA − T = 0
B:
ΣFt = 0:
( 30 lb ) sin θ
− FA − FB − T = 0
So
(10 lb ) sin θ
(10 lb ) sin θ
tan θ =
− 2 FA − FB = 0
= 2 ( 3 lb ) cosθ + ( 7.5 lb ) cosθ
13.5 lb
= 1.35;
10 lb
θ = 53.5°
PROBLEM 8.137
Two cylinders are connected by a rope that passes over two fixed rods as
shown. Knowing that the coefficient of static friction between the rope
and the rods is 0.40, determine the range of values of the mass m of
cylinder D for which equilibrium is maintained.
SOLUTION
T = WAe µ s β B
For impending motion of A up:
and
µ β +β
WD = Te µs βC = WAe s ( B C )
or
mD g = ( 50 kg ) ge
(
0.4 π2 + π2
)
mD = 175.7 kg
For impending motion of A down, the tension ratios are inverted, so
µ β +β
WA = WDe s ( C B )
( 50 kg ) g
= mD ge
(
0.4 π2 + π2
)
mD = 14.23 kg
For equilibrium:
14.23 kg ≤ mD ≤ 175.7 kg
PROBLEM 8.138
Two cylinders are connected by a rope that passes over two fixed rods
as shown. Knowing that for cylinder D upward motion impends when
m = 20 kg, determine (a) the coefficient of static friction between the
rope and the rods, (b) the corresponding tension in portion BC of the
rope.
SOLUTION
(a) Motion of D impends upward, so
TBC = WDe µ s βC
(1)
µ β +β
WA = TBC e µs β B = WDe s ( C B )
So
 50 kg 
W
π π 
+  = ln A = ln 

2
2
W


 20 kg 
D
µs 
µ s = 0.29166
µ s = 0.292 W
(b) From Equation (1):
(
)
TBC = ( 20 kg ) 9.81 m/s 2 e0.29166 π /2
TBC = 310 N W
PROBLEM 8.139
A 10° wedge is used to split a section of a log. The coefficient of
static friction between the wedge and the log is 0.35. Knowing that a
force P of magnitude 600 lb was required to insert the wedge,
determine the magnitude of the forces exerted on the wood by the
wedge after insertion.
SOLUTION
FBD wedge (impending
motion ):
φ s = tan −1 µ s = tan −1 0.35 = 19.29°
R1 = R2
By symmetry:
ΣFy = 0: 2R1 sin ( 5° + φs ) − 600 lb = 0
or
R1 = R2 =
300 lb
= 729.30 lb
sin ( 5° + 19.29° )
When P is removed, the vertical components of R1 and R2 vanish, leaving
the horizontal components
R1x = R2 x = R1 cos ( 5° + φ s )
= ( 729.30 lb ) cos ( 5° + 19.29° )
R1x = R2 x = 665 lb W
(Note that φ s > 5°, so wedge is self-locking.)
PROBLEM 8.140
A flat belt is used to transmit a torque from drum B to drum A.
Knowing that the coefficient of static friction is 0.40 and that the
allowable belt tension is 450 N, determine the largest torque that can
be exerted on drum A.
SOLUTION
β A = 180° + 30° = π +
FBD’s drums:
β B = 180° − 30° = π −
π
6
π
6
=
7π
6
=
5π
6
Since β B < β A , slipping will impend first on B (friction coefficients
being equal)
So
T2 = Tmax = T1e µs β B
0.4 5π /6
450 N = T1e( )
or
T1 = 157.914 N
ΣM A = 0: M A + ( 0.12 m )(T1 − T2 ) = 0
M A = ( 0.12 m )( 450 N − 157.914 N ) = 35.05 N ⋅ m
M A = 35.1 N ⋅ m W
PROBLEM 8.141
Two 10-lb blocks A and B are connected by a slender rod of negligible
weight. The coefficient of static friction is 0.30 between all surfaces
of contact, and the rod forms an angle θ = 30° with the vertical.
(a) Show that the system is in equilibrium when P = 0. (b) Determine
the largest value of P for which equilibrium is maintained.
SOLUTION
FBD block B:
(b) For Pmax , motion impends at both surfaces
B:
ΣFy = 0: N B − 10 lb − FAB cos 30° = 0
N B = 10 lb +
3
FAB
2
(1)
FB = µ s N B = 0.3N B
Impending motion:
ΣFx = 0: FB − FAB sin 30° = 0
FAB = 2 FB = 0.6 N B
N B = 10 lb +
Solving (1) and (2)
FBD block A:
(2)
3
( 0.6 N B )
2
= 20.8166 lb
FAB = 0.6 N B = 12.4900 lb
Then
A:
ΣFx = 0: FAB sin 30° − N A = 0
NA =
Impending motion:
1
1
FAB = (12.4900 lb ) = 6.2450 lb
2
2
FA = µ s N A = 0.3 ( 6.2450 lb ) = 1.8735 lb
ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0
P = FA +
3
FAB − 10 lb
2
= 1.8735 lb +
3
(12.4900 lb ) − 10 lb = 2.69 lb
2
P = 2.69 lb W
(a)
Since P = 2.69 lb to initiate motion,
equilibrium exists with P = 0 W
PROBLEM 8.142
Determine the range of values of P for which equilibrium of the block
shown is maintained.
SOLUTION
FBD block (Impending motion down):
φ s = tan −1 µ s = tan −1 0.25
(
P = ( 500 lb ) tan 30° − tan −1 0.25
)
= 143.03 lb
(Impending motion up):
(
P = ( 500 lb ) tan 30° + tan −1 0.25
)
= 483.46 lb
Equilibrium for 143.0 lb ≤ P ≤ 483 lb W
PROBLEM 8.143
Two identical uniform boards, each of weight 40 lb, are temporarily
leaned against each other as shown. Knowing that the coefficient of static
friction between all surfaces is 0.40, determine (a) the largest magnitude
of the force P for which equilibrium will be maintained, (b) the surface at
which motion will impend.
SOLUTION
Board FBDs:
FC = µ s NC
Assume impending motion at C, so
= 0.4 NC
FBD II:
ΣM B = 0:
( 6 ft ) NC − (8 ft ) FC − ( 3 ft )( 40 lb ) = 0
6 ft − 0.4 ( 8 ft )  NC = ( 3 ft )( 40 lb )
or
NC = 42.857 lb
and
FC = 0.4 NC = 17.143 lb
ΣFx = 0: N B − FC = 0
N B = FC = 17.143 lb
ΣFy = 0: − FB − 40 lb + NC = 0
FB = NC − 40 lb = 2.857 lb
Check for motion at B:
FB
2.857 lb
=
= 0.167 < µ s , OK, no motion.
N B 17.143 lb
PROBLEM 8.143 CONTINUED
FBD I:
ΣM A = 0:
(8 ft ) N B + ( 6 ft ) FB − ( 3 ft )( P + 40 lb ) = 0
P=
(8 ft )(17.143 lb ) + ( 6 ft )( 2.857 lb ) − 40 lb = 11.429 lb
3 ft
Check for slip at A (unlikely because of P)
ΣFx = 0: FA − N B = 0
or
FA = N B = 17.143 lb
ΣFy = 0: N A − P − 40 lb + FB = 0
or
N A = 11.429 lb + 40 lb − 2.857 lb
= 48.572 lb
Then
Therefore,
FA
17.143 lb
=
= 0.353 < µ s ,
NA
48.572 lb
OK,
no slip ⇒ assumption is correct
(a) Pmax = 11.43 lb W
(b) Motion impends at C W
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
5
At
x = a, y = b: b = ka 2
or
k =
b
5
a2
∴ y =
b
a
5
2
5
x2
or
a
b
dI y =
=
Then
x=
2
2
5
y5
1 3
x dy
3
1 a3 65
y dy
3 b 65
Iy =
1 a3 b 65
∫ y dy
3 b 65 0
1 5 a3 115
=
y
3 11 b 65
=
b
0
5 a3 115
b
33 b 65
or I y =
5 3
a bW
33
PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At
x = 0, y = 0: 0 = ka 2 + c
k =−
x = a,
y =b:
∴k = −
Then
y =−
=−
Now
c
a2
b=c
b
a2
b
2
x − a) + b
2(
a
(
)
b 2
x − 2ax + a 2 + b
a2
2b 3
 b

dI y = x 2dA = x 2 ( ydx ) =  − 2 x 4 +
x − bx 2 + bx 2  dx
a
 a

2b 3 
 b
=  − 2 x4 +
x  dx
a 
a

Then
b
2b 3 
a 
I y = ∫ dI y = ∫0  − 2 x 4 +
x  dx
a 
 a
a
 1 x5 2 x 4 
= b − 2
+

a 4 0
 a 5
 a 3 2a 3 
1
3 1
= b 
+
 = ba  − 
4 
 2 5
 5
Iy =
3a3b
W
10
PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
y =h−
By observation
h
x
b
x

= h 1 − 
b


Now
dI y = x 2dA = x 2 ( h − y ) dx 

x 

= x 2  h − h 1 −   dx
b 


=
Then
hx3
dx
b
hx3
hx 4
I y = ∫ dI y = ∫
dx =
b
4b
b
b
0
=
0
hb 4
4b
Iy =
b 3h
W
4
PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
y = kx 2 + c
Have
x = 0, y = b: b = k ( 0 ) + c
At
or
c=b
At
x = 2a, y = 0:
or
k =−
y =−
Then
=
Then
I y = ∫ x 2dA,
2a
I y = ∫a x 2dA =
0 = k ( 2a ) + b
2
b
4a 2
b 2
x +b
4a 2
(
b
4a 2 − x 2
2
4a
dA = ydx =
)
(
)
b
4a 2 − x 2 dx
4a 2
(
)
b 2a 2
x 4a 2 − x 2 dx
2 ∫a
4a
2a
=
b
4a 2
 2 x3 x5 
− 
 4a
3
5 a

=
b
b
8a3 − a3 −
32a5 − a5
3
20a 2
(
)
=
7a3b 31a 3b
−
3
20
(
)
Iy =
47 3
a bW
60
PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
3
At x = a, y = b: b = ka 2
or k =
b
3
a2
∴y =
b 32
x
a 32
I x = ∫ y 2dA
dA = xdy
a 2 
b
= ∫0 y 2  2 y 5 dy 
 b 5

5 175
= 2 ×
y
b 5 17
b
17
0
5a b 5
=
17 b 25
a
or I x =
5 3
ab W
17
PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At
x = 0, y = 0: 0 = ka 2 + c
k =−
c
a2
x = a, y = b
k =−
y =b−
Then
Now
( x − a )2
b
2
x − a)
2(
a
=
a2
(b − y )
b
x − a = a2 1 −
Then
x = a2 1 −
and
and
b
a2
dI x = y 2dA = y 2 ( xdy )
From above
Then
b=c
y
b
y
+a
b

y
dI x = ay 2 1 + 1 −  dy

b 


y
b
I x = ∫ dI x = a ∫0 y 2 1 + 1 −  dy

b 

y3
=a
3
b
0

y
b
+ a ∫0 y 2  1 −  dy

6 

PROBLEM 9.6 CONTINUED
For the second integral use substitution
u =1−
y
1
⇒ du = − dy, y = b(1 − u )
b
b
y = 0 u =1
y=b u =0

y
1
b 2
b 2
2
∫0 y  1 − b  dy = −∫0 b (1 − u ) u 2 du


Now
0
=
0
−b3 1
∫
 u 12 − 2u 32 + u 52  du = −b3  2 u 23 − 4 u 52 + 2 u 72 




5
7 1


3
2 4 2
 70 − 84 + 30  16b3
= +b 3  − +  = b 3 
=
105
3 5 7

 105
Then
Ix = a
b3 16ab3
51 3
+
=
ab
3
105
105
or I x =
17 3
ab W
35
PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
y =h−
By observation
h
x
b
x

= h 1 − 
b

y

x = b 1 − 
h

or
Now
dI x = y 2dA = y 2 ( b − x ) dy
by 

= y2  b − b +
dy 
h


=
Then
by 4
dy =
h
4h
h by
0
Ix = ∫
3
by 3
dy
h
h
=
0
bh 4
4h
or I x =
bh3
W
4
PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
y = kx 2 + c
Have
At
x = 0, y = b: b = k (0) + c
or
c=b
At
x = 2a, y = 0: 0 = k (2a) 2 + b
k =−
or
Then
Now
y =
dI x =
=
b
4a 2
(
)
(
)
b
4a 2 − x 2
4a 2
1 3
y dx
3
3
1 b3
4a 2 − x 2 dx
6
3 64a
PROBLEM 9.8 CONTINUED
Then
I x = ∫ dI x
(
)
=
3
1 b3 2 a
4a 2 − x 2 dx
6 ∫a
3 64a
=
b3 2 a
6
4 2
2 4
6
∫ 64a − 48a x + 12a x − x dx
192a 6 a
(
b3
=
192a 6
=
2a

12 2 5 x 7 
6
4 3
64
a
x
16
a
x
a x −
−
+


5
7 a

b3 
64a 7( 2 − 1) − 16a 7 ( 8 − 1)
6 
192a
+
=
)
12 7
1

a ( 32 − 1) − (128 − 1) 
5
7

ab3 
372 127 
3
−
 64 − 112 +
 = 0.043006ab
192 
5
7 
I x = 0.0430ab3 W
PROBLEM 9.9
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
1
At
x = a, y = b: b = ka 2
k =
or
y =
Then
b
a
b 12
x
a
3
Now
dI x =
1 3
1  b  32
y dx = 
 x dx
3
3 a 
3
Then
3
a
a1 b 
2
I x = 2∫0 dI x = 2∫0 
 x dx
3 a 
3
2  b  2 52
= 
 x
3 a  5
a
=
0
4 b3 52
a
15 a 23
Ix =
4 3
ab W
15
PROBLEM 9.10
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At
x = 2a, y = b:
or
k =
2a = kb3
2a
b3
Then
x=
2a 3
y
b3
or
y =
b
Now
dI x =
( 2a )
1
1
3
x3
1 3
1 b3
y dx =
xdx
3
3 2a
2a
Then
1 b3 2a
1 b3 1 2
I x = ∫ dI x =
xdx
=
x
∫
3 2a a
6 a 2 a
=
(
b3
4a 2 − a 2
12a
)
Ix =
1 3
ab W
4
PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
x = 0, y = b: b = ke0 = k
At
y = be
Then
dI x =
Now
Then
a
I x = ∫ dI x = ∫0
− ax
1 3
b3  − ax 3
y dx =
 e  dx
3
3

b3  − ax 3
 e  dx
3

b3 − 3ax
b3  −a  − 3ax
=
e
dx
=
∫

e
3
3 3 
=
a
=−
0
(
b3a −3
e − e0
9
)
ab3
( 0.95021) = 0.10558ab3
9
or I x = 0.1056ab3
PROBLEM 9.12
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
1
At
x = a, y = b: b = ka 2
Then
y =
Now
dI y = x 2dA,
or
Then
I y = ∫ dI y = 2∫
=
b
a
b 12
x
a
dA = ydx
dI y = x 2 ydx =
a
0
k =
b 52
x dx
a
b 52
4 b 72
x dx =
x
7 a
a
a
0
4 b 72
a
7 a 12
or I y =
4 3
ab
7
PROBLEM 9.13
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At
x = 2a, y = b: 2a = kb3
y =
or
Now
b
( 2a )
I y = ∫ x 2dA
2a
=
1
x3
1
3
dA = ydx
I y = ∫a x 2
Then
s
2a 3
y
b3
x=
Then
or
b
b
( 2a )
1
1
3
x 3 dx
5
( 2a )
1
3
2a
∫a x 3 dx
2a
=
=
b
( 2a ) 3
1
3b
10 ( 2a )
=
1
3
3 103
x
10
a
 2a 103 − a 103 
( )

10
3ba3  103
2 − 1 3 
1 

10 ( 2 ) 3 
= 2.1619a3b
or I y = 2.16a3b
PROBLEM 9.14
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
x = 0, y = b: b = ke0 = k
At
y = be
Then
− ax
dI y = x 2dA = x 2 ydx
Now
−x
= x 2be a dx
a
−x
a
−x
I y = ∫ dI y = ∫0 bx 2e a dx = b∫0 x 2e a dx
Then
Use integration by parts
u = x2
du = 2 xdx
Then
−x
dv = e a dx
v = −ae

−x
−x
a
I y = ∫0 x 2e a dx = b  −ax 2e a

a
0
− ax

−x
a
− ∫0  −ae a  2 xdx 



−x
a
= b  −a3e −1 + 2a ∫0 xe a dx 


Again use integration by parts:
u = x
du = dx
−x
dv = e a dx
v = −ae
− ax
PROBLEM 9.14 CONTINUED
Then
−x
a
∫0 xe a dx = −axe
− ax
a
0
− ∫0  −ae

a
= −a 2e−1 − a 2e
− ax
a
− ax
 dx


= −a 2e−1 − a 2e−1 + a 2e0
0
= −2a 2e −1 + a 2
Finally,
(
)
(
I y = b  −a 3e −1 + 2a −2a 2e −1 + a 2  = ba3 −5e −1 + 2


)
= 0.1606ba3
or I y = 0.1606ba3
PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
x = a,
At
Then
Now
y1 = y2 = b
y1: b = ma
or
m=
b
a
y2: b = ka3
or
k =
b
a3
y1 =
b
x
a
y2 =
b 3
x
a3
 a
x2 =  1
 3
b
 1
 y 3

b
a
A = 2∫ dA = 2∫
y −
1
b
 b3
=
Now
or
a
y
b
 a 1 a 
dA = ( x2 − x1 ) dy =  1 y 3 − y  dy
 3
b 
b
b

0
Then
x1 =
or
a
1
3

 a 3 4 a 1 2
y  dy = 2  1 y 3 −
y 

b 2 

 b 3 4
0
3ab
1
− ab = ab
2
2
 a 1 a 
dI x = y 2dA = y 2  1 y 3 − y  dy
 3
b 
b
b
 y 73
 3 y 103
y 3 
y 4 

I x = 2∫ dI x = 2a ∫
dy
2
a
−
=
−
1
b 
4b 
10 13
3
b

 b
0
b

0
 3
b3 
1
 3
= 2a  b3 −  = 2ab3 
− 
10
4
10
4



or I x =
And
k x2 =
Ix
=
A
1 ab3
10
1 ab
2
=
1 3
ab
10
1 2
b
5
kx =
b
5
PROBLEM 9.16
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
x = a, y = b:
At
k1 =
or
y1 =
Then
x1 =
and
1
b = k1a 4
b = k2a 4
b
a4
k2 =
y2 =
a
x2 =
b
b
a
1
y4
1
a4
b 4
x
a4
1
4
b
1
4
1
x4
a 4
y
b4
 1
x4 
a x4
A = ∫ ( y2 − y1 ) dx = b∫0  1 − 4 dx
 a4
a 


Now
a
 4 x 54 1 x5 
 = 3 ab
= b
−
1
4
5
5 a4 5 a 

0
dA = ( x1 − x2 ) dy
I x = ∫ y 2dA
Then
Ix =
b
 a
2
∫0 y 
b
1
4
1
y4 −
a 4
y  dy
b 4 
b
 4 y 134
1 y 7 
= a
−
1
7 b4 
13 b 4

0
1
 4
= ab3  − 
13
7


or I x =
Now
kx =
Ix
=
A
15 ab3
91
3 ab
5
=
15 3
ab
91
25 2
b = 0.52414b
91
or k x = 0.524b
PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
x = a,
At
y1 = y2 = b
y1: b = ka3
or
k =
b
a3
y2: b = ma
or
m=
b
a
Then
y1 =
b 3
x
a3
y2 =
b
x
a
b 
b
dA = ( y2 − y1 ) dx =  x − 3 x3  dx
a
a


Now
a
b a
x3 
b 1
1

A = ∫ dA = 2 ∫0  x − 2 dx = 2  x 2 − 2 x 4 
a 
a 2
4a
a 
0
 1
b  a2
1
=2 
− 2 a 4  = ab
a 2
4a
 2
dI y = x 2dA =
Now
Then
b  3 x 5  
 x − 2  dx 
a 
a  
b a
x5 
a
I y = ∫0 dI y = 2 ∫0  x3 − 2  dx
a 
a 
a
b 1
1
b  a 4 1 a6 

= 2  x 4 − 2 x6  = 2 
−

a 4
a 4
6 a 2 
6a
0
=
And
1 3
ab
6
k y2 =
Iy
A
or I y =
=
1 a 3b
6
1 ab
2
=
1 2
a
3
1 3
ab
6
or k y =
a
3
PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At
x = a, y = b:
y1 =
Now
b 4
x
a4
b = k2a 4
b
a4
k2 =
and
y2 =
k1 =
or
Then
1
b = k1a 4
b
1
a4
b
a
A = ∫ ( y2 − y1 ) dx = b∫
 x 14
a

0
a
1
4
−
1
4
1
x4
x 4 
dx
a4 

a
 4 x 54 1 x5 
 = 3 ab
= b
−
1
4
5
5 a4 5 a 

0
dA = ( y2 − y1 ) dx
Now
I y = ∫ x 2dA
Then
 b 1

b
a
I y = ∫0 x 2  1 x 4 − 4 x 4  dx
 4

a
a

 9
x6 
a x4
= b ∫0  1 − 4  dx
 a4 a 


a
 4 x 134
1 x7 
= b
−
7 a4 
13 a 14

0
1 
 4
= b  a3 − a3 
7 
 13
or I y =
Now
ky =
Iy
A
=
15 a 3b
91
3 ab
5
=
15 3
ab
91
25
a = 0.52414a
91
or k y = 0.524a
PROBLEMS 9.19 AND 9.20
P 9.19 Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
P 9.20 Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
First determine constants m, b, and c
y1: at x = 2a,
y =0
0 = m ( 2a ) + b
x = a,
At
y =h
h = m(a) + b
m=−
Solving yields
y1 = −
Then
h
a
b = 2h
h
x + 2h
a
y2: at x = 2a,
y =0
0 = c sin k ( 2a )
At
x = a,
y = 2h
2h = c sin ka
Solving
c sin k ( 2a ) = 0
c≠0
sin k ( 2a ) = 0, k ( 2a ) = π , k =
Substitute k, 2h = c sin ka
Then
yields
2h = c sin
y2 = 2h sin
π
2a
π
2
π
2a
or
c = 2h
x
To calculate the area of shaded surface, a differential strip parallel to the
y axis is chosen to be dA.

π
 h

dA = ( y2 − y1 ) dx =  2h sin
x −  − x + 2h   dx
2
a
a



PROBLEMS 9.19 AND 9.20 CONTINUED
π
h 
2a 
A = ∫ dA = ∫0  2h sin
x − 2hx + x  dx
2a
a 

2a
 4a
π
x2 
= h  − cos
x − 2x +

2a
2a  a
 π
 4a
( 2a )2 
π
= h  − cos 2a − 2 ( 2a ) +
2a
2a 
 π

 4a
π
a2 
− h  − cos a − 2 ( a ) +

2a
2a 
 π
 4a

= h
− 4a + 2a  −
π


a

 4 1
h  −2a +  = ah  − 
2

π 2
A = 0.77324ah
PROBLEM 9.19
2a
I x = ∫a dI x
Moment of inertia
dI x =
where
dI x =
Now
(
)
1 3
y2 − y13 dx
3
3
3
1 
π  
h  
x  −  2h − x   dx
 2h sin
3 
2a 
a  

3
1 3 3 π
x 
3
= 8h sin
x − h  2 −   dx
3 
2a
a  

3
8h3 2a 3 π
h3 2 a 
x
−
sin
2 −  dx
Ix =
xdx
∫
∫
−a
a 
3
2a
3
a

Then
Now
π
π
π


3
2
∫ sin 2a xdx = ∫ sin 2a x 1 − cos 2a x  dx


π 
π
π 


x  dx − ∫  sin
x cos 2
x dx
= ∫  sin
2a 
2a
2a 


2a
2a
π
π
cos3
x+
x
= − cos
2a
3π
2a
π
Then
2a 
sin 3
a 
∫

2a
π 
2a 
π
1
π 
x dx = − cos
x − cos3
x
2 
π  2a
3
2a  a
=−
2a 
1  4a
 −1 +  =
3  3π
π 
PROBLEMS 9.19 AND 9.20 CONTINUED
And
3
2a 
2
a 
x
a
x
−  dx = −  2 − 
∫
a
a
4




4
2a
a
4
4
a
2a 
a
a
a
= − 2 −
 + 2 −  =
4
a 
4
a
4
Ix =
Then
8h3  4a  h3  a  h3a  32 1 
− 

−
 =

3  3π  3  4 
3  3π 4 
I x = 1.0484ah3
1.0484ah3
= 1.1644h
0.77324ah
Ix
=
A
kx =
and
k x = 1.164h
PROBLEM 9.20
I y = ∫ dI y
dI y = x 2dA
y1 = 2h −
From Problem 9.19
dA = ( y2 − y1 ) dx
h
x
a
y2 = 2h sin
π
2a
x
Now


π
x3  
dI y =  2hx 2 sin
x − h  2 x 2 −
  dx
a  
2a


Then
πx
x3 
2a
2a 
I y = ∫a dI y = h ∫a  2 x 2 sin
− 2x2 +
 dx
2a
a 

Now using integration by parts
Then
π
u = x2
dv = sin
du = 2 xdx
v=−
π
π
2a
2a
π
xdx
cos
π
2a
x
π



2
2
∫ x sin 2a xdx = x  − π cos 2a x  − ∫  − π cos 2a x  2 xdx




2a
2a
PROBLEM 9.20 CONTINUED
u = x
Now let
dv = cos
du = dx
Then
π
π
v=
4a 
2a
π
π
2a
sin
xdx
π
2a
x
π
π





2
2
∫ x sin 2a xdx = − π x cos 2a x + π  x  π sin 2a x  − ∫  π sin 2a x  dx 


 
 
2a
2a
2a
2a
Finally,
 2a 2
π
8a 2
π
4a 2 2a
π  1 3 1 4
I y = 2h  −
x cos
x + 2 x sin
x+ 2
cos
x − x +
x 
2a
2a
2a  3
8a 
π
π π
 π
a
3
 2a
16a3 ( 2a )
1
8a 2
a3 a 4 
2
= 2h  ( 2a ) − 3 −
+
( 2a )4 − 2 a + −  = 1.5231a3h
3
8a
3
8a 
π
π
 π

I y = 1.523a3h
and
k y2 =
Iy
A
=
1.5231a3h
= 1.4035a 2
0.77324
k y = 1.404a
PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the shaded
area with respect to the x axis.
SOLUTION
A = A1 + A2 + A3
First note that
= (1.2 in.)( 0.3 in.) + ( 2.4 in.)( 0.4 in.) + ( 2.4 in.)( 0.3 in.)
= ( 0.36 + 0.96 + 0.72 ) in 2
= 2.04 in 2
I x = ( I x )1 + ( I x )2 + ( I x )3
Now
where
Then
(
)
( I x )1 =
1
(1.2 in.)( 0.3 in.)3 + 0.36 in 2 (1.36 in.)2 = 0.6588 in 4
12
( I x )2
=
1
( 0.4 in.)( 2.4 in.)3 = 0.4608 in 4
12
( I x )3
=
1
( 2.4 in.)( 0.3 in.)3 + 0.72 in 2 (1.35 in.)2 = 1.3176 in 4
12
(
)
I x = 0.6588 in 4 + 0.4608 in 4 + 1.3176 in 4 = 2.4372 in 4
or I x = 2.44 in 4 W
and
k x2 =
Ix
2.4372 in 4
=
= 1.1947 in 2
A
2.04 in 2
or k x = 1.093 in. W
PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of the shaded
area with respect to the x axis.
SOLUTION
First note that
A = A1 − A2 − A3
= (100 mm )(120 mm ) − ( 80 mm )( 40 mm ) − ( 80 mm )( 20 mm ) = 7200 mm 2
= (12 000 − 3200 − 1600 ) mm 2 = 7200 mm
I x = ( I x )1 − ( I x )2 − ( I x )3
Now
where
Then
( I x )1 =
1
(100 mm )(120 mm )3 = 14.4 × 106 mm4
12
(
)
(
)
( I x )2
=
1
(80 mm )( 40 mm )3 + 3200 mm2 ( 40 mm )2 = 5.5467 × 106 mm 4
12
( I x )3
=
1
(80 mm )( 20 mm )2 + 1600 mm2 ( 30 mm )2 = 1.4933 × 106 mm 4
12
I x = (14.4 − 5.5467 − 1.4933) × 106 mm 4 = 7.36 × 106 mm 4
or I x = 7.36 × 106 mm 4 W
and
k x2 =
Ix
7.36 × 106
=
= 1022.2 mm 2
A
7200
or k x = 32.0 mm W
PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the shaded
area with respect to the y axis.
SOLUTION
First note that
A = A1 + A2 + A3
= (1.2 in.)( 0.3 in.) + ( 2.4 in.)( 0.4 in.) + ( 2.4 in.)( 0.3 in.)
= ( 0.36 + 0.96 + 0.72 ) in 2 = 2.04 in 2
Now
Where:
( )1 + ( I y )2 + ( I y )3
Iy = Iy
( I y )1 = 121 ( 0.3 in.)(1.2 in.)3 = 0.0432 in 4
( I y )2 = 121 ( 2.4 in.)( 0.4 in.)3 = 0.0128 in 4
( I y )3 = 121 ( 0.3 in.)( 2.4 in.)3 = 0.3456 in 4
Then
I y = ( 0.0432 + 0.0128 + 0.3456 ) in 4 = 0.4016 in 4
or I y = 0.402 in 4 W
And
k y2 =
Iy
A
=
0.4016
= 0.19686 in 2
2.04 in 2
or k y = 0.444 in. W
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration of the shaded
area with respect to the y axis.
SOLUTION
First note that
A = A1 − A2 − A3
= (100 mm )(120 mm ) − ( 80 mm )( 40 mm )
− ( 80 mm )( 20 mm ) = 7200 mm 2
= (12 000 − 3200 − 1600)mm 2 = 7200 mm 2
( )1 − ( I y )2 − ( I y )3
Iy = Iy
Now
where
( I y )1 = 121 (120 mm )(100 mm )3 = 10 × 106 mm4
( I y )2 = 121 ( 40 mm )(80 mm )3 = 1.7067 × 106 mm4
( I y )3 = 121 ( 20 mm )(80 mm )3 = 0.8533 × 106 mm4
Then
I y = (10 − 1.7067 − 0.8533) × 106 mm 4 = 7.44 × 106 mm 4
or I y = 7.44 × 106 mm 4 W
And
k y2 =
Iy
A
=
7.44 × 106 mm 4
= 1033.33 mm 2
7200 mm 2
k = 32.14550 mm
or k y = 32.1 mm W
PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to
the x and y axes.
SOLUTION
Have
I x = ( I x )1 + ( I x )2 + ( I x )3
3
3
1
1
=  ( 2a )( 4a )  +  ( a )( 3a ) 
3
3

 

2
2
  π
π 2  4a   π 2 
4a  
4
+   a − a    + a  3a +
 
4  3π   4 
3π  
 16

4
9π
4  4
 128 4   27 4   π
=
−
+
+2+
a + a +
a
3
3
16
9
π
4
9
π 

 
 
 161 37  4
4
=
+
 a = 60.9316a
3
π


or I x = 60.9a 4 W
Also
( )1 + ( I y )2 + ( I y )3
Iy = Iy
3
3
1
1
π

=  ( 4a )( 2a )  +  ( 3a )( a )  +  a 4 
3
 3
  16 
π 
 32
=
+ 1 +  a 4 = 11.8630a 4
16 
 3
or I y = 11.86a 4 W
PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with respect to
the x and y axes.
SOLUTION
Have
I x = ( I x )1 − ( I x )2 − ( I x )3
3
1
π  π 
=  ( 3a )( 2a )  −  a 4  −  a 4 
12
 8  8 
π π
π


=  2 − −  a4 =  2 −  a4
8 8
4


or I x = 1.215a 4 W
Also
( )1 − ( I y )2 − ( I y )3
Iy = Iy
2
1
3
a 
=  ( 2a )( 3a ) + ( 3a )( 2a )   
 2  
12
2
2
  π
π  4a   π 
4a  
−   a 4 − a 2    + a 2  2a −
 
2  3π   2 
3π  
  8

2
2
  π
π 2  4a   π 2 
4a  
4
−  a − a 
  + a a −
 
2  3π   2 
3π  
  8

8
8
8
9 3
π
=  +  a4 −  −
+ 2π − +
3 9π
2 2
 8 9π
 4
a

8
π 4
8  4 
11π  4
π
− −
+ − +
 a =  10 −
a
8
9
π
2
3
9
π
4 



= 1.3606a 4
or I y = 1.361a 4 W
PROBLEM 9.37
For the 6-in2 shaded area shown, determine the distance d 2 and the
moment of inertia with respect to the centroidal axis parallel to AA′
knowing that the moments of inertia with respect to AA′ and BB′ are
30 in4 and 58 in4, respectively, and that d1 = 1.25 in.
SOLUTION
Have
I AA′ = I + Ad12
and
I BB′ = I + Ad 22
(
I AA′ − I BB′ = A d12 − d 22
subtracting
or
( 30 − 58) in 4
(
)
)
2
= 6 in 2 (1.25 in.) − d 22 


Solve for d 2
(
)
d 22 = 1.252 + 4.6667 in 2 = 6.2292 in 2
Then
d 2 = 2.4958 in.
or d 2 = 2.50 in. W
and
(
)
I = I AA′ − Ad12 = 30 in 4 − 6 in 2 (1.25 in.) = 20.625 in 4
2
or I = 20.6 in 4 W
PROBLEM 9.38
Determine for the shaded region the area and the moment of inertia with
respect to the centroidal axis parallel to BB′ knowing that d1 = 1.25 in.
and d2 = 0.75 in. and that the moments of inertia with respect to AA′ and
BB′ are 20 in4 and 15 in4, respectively.
SOLUTION
Have
I AA′ = I + Ad12
and
I BB′ = I + Ad 22
subtracting
(
I AA′ − I BB′ = A d12 − d 22
)
2
2
20 in 4 − 15 in 4 = A (1.25 ) − ( 0.75 )  in 2


5 in 4 = A [1.5625 − 0.5625] in 2
or A = 5 in 2 W
and
(
)
I = I AA′ − Ad 2 = 20 in 4 − 5 in 2 (1.25 in.) = 12.1875 in 4
2
or I = 12.19 in 4 W
PROBLEM 9.39
The centroidal polar moment of inertia J C of the 15.5 × 103 mm 2
shaded region is 250 × 106 mm 4 . Determine the polar moments of
inertia J B and J D of the shaded region knowing that J D = 2 J B and
d = 100 mm.
SOLUTION
Have
2
J B = J C + AdCB
and
2
J D = J C + AdCD
J D = 2J B
Now
(
2
2
J C + AdCD
= 2 J C + AdCB
Then
2
dCB
= a2 + d 2
Now
a2 =
=
)
(
)

1  JC
+ d2 

2 A

1  250 × 106 mm 4
2
+ (100 mm )  = 13064.5 mm 2

3
2
2 15.5 × 10 mm

a = 114.300 mm
or
Then
2
A 4a 2 + d 2 = J C + 2 A a 2 + d 2
Substituting
or
2
dCD
= ( 2a ) + d 2
and
(
)
(
)
2
2
J B = 250 × 106 mm 4 + 15.5 × 103 mm 2 (114.300 mm ) + (100 mm ) 


(
)
= 250 × 106 + 357.5 × 106 mm 4 = 607.5 × 106 mm 4
or J B = 608 × 106 mm 4 W
And
(
)
2
2
J D = 250 × 106 mm 4 + 15.5 × 103 mm 2 ( 228.60 mm ) + (100 mm ) 


(
)
= 250 × 106 + 964.99 × 10 mm 4 = 1214.99 × 106 mm 4
or J D = 1215 × 106 mm 4 W
PROBLEM 9.40
Determine the centroidal polar moment of inertia J C of the 10 × 103 mm 2
shaded area knowing that the polar moments of inertia of the area with
respect to points A, B, and D are J A = 45 × 106 mm 4 , J B = 130 × 106 mm 4 ,
and J D = 252 × 106 mm 4 , respectively.
SOLUTION
2
J A = J C + AdCA
Have
J A = J C + Aa 2
Then
2
J B = J C + AdCB
Have
(1)
2
dCB
= a2 + d 2
where
(
J B = JC + A a2 + d 2
Then
2
J D = J C + AdCD
Have
)
(
)
(3)
J D − J B = 3 Aa 2
Then Equation (3) − Equation(2):
and Equation(4) − 3[ Equation(1)]:
JC = J A −
(2)
2
dCD
= 4a 2 + d 2
where
J D = J C + A 4a 2 + d 2
Then
or
2
dCA
= a2
where
(4)
( J D − J B ) − 3J A
= −3 J C
1
( JD − JB )
3
= 45 × 106 mm 4 −
(
)
1
252 × 106 − 130 × 106 mm 4 = 4.3333 × 106 mm 4
3
or J C = 4.33 × 106 mm 4 W
Note
a = 63.77 mm
and
d = 92.195 mm
PROBLEM 9.41
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First calculate the centroid C of the area
Y = 0.6 in. + 0.9 in. = 1.5 in.
From symmetry
XA = ΣAx
To compute X use the equation
or
X =
1
2
 2 (1.8 × 3.6 ) in  × ( 2.3 in.)
1
( 3 × 5.4 ) in 2 − (1.8 × 3.6 ) in 2
2
( 3 × 5.4 ) in 2  × ( 2.7 in.) −


= 2.8 in.
The moment of inertia of the composite area is obtained by subtracting the moment of inertia of the triangle
from the moment of inertia of the rectangle
I x = ( I x )1 − ( I x )2
( I x )1 =
where
and
Then
( I x )2
1
( 5.4 in.)( 3 in.)3 = 12.15 in 4
12
3
1
= 2  ( 3.6 in.)( 0.9 in.)  = 0.4374 in 4
12

I x = (12.15 − 0.4374 ) in 4 = 11.7126 in 4
or I x = 11.71 in 4 W
( )1 − ( I y )2
Iy = Iy
Similarly,
where
( I y )1 = 121 ( 3 in.)( 5.4 in.)3 + (3 × 5.4) in 2  ( 2.8 in. − 2.7 in.)2
= 39.582 in 4
PROBLEM 9.41 CONTINUED
and
( I y )2 = 361 (1.8 in.)( 3.6 in.)3 +  12 (1.8)( 3.6) in 2  ( 2.8 in. − 2.3 in.)2
= 3.1428 in 4
Then
I y = ( 39.582 − 3.1428 ) in 4 = 36.4392 in 4
or I y = 36.4 in 4 W
PROBLEM 9.42
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
X = 0.9 in.
By symmetry
Have
AY = ΣyA
Where
A = (1.8 in.)(1.4 in.) +
1
(1.8 in.)( 2.1 in.)
2
= ( 2.52 + 1.89 ) in 2 = 4.41 in 2
Then
( 4.41 in ) Y
2
(
)
(
= ( 0.7 in.) 2.52 in 2 + ( 2.1 in.) 1.89 in 2
)
= 5.733 in 3
or
Y = 1.3 in.
I x = ( I x )1 + ( I x )2
Now
where
( I x )1 =
1
(1.8 in.)(1.4 in.)3
12
(
)
+ 2.52 in 2 (1.3 in. − 0.7 in.) = 1.3188 in 4
And
( I x )2
=
2
1
(1.8 in.)( 2.1 in.)3
36
(
)
+ 1.89 in 2 ( 2.1 in. − 1.3 in.) = 1.67265 in 4
Then
2
I x = (1.3188 + 1.67265 ) in 4 = 2.99145 in 4
or I x = 2.99 in 4 W
Also
where
( )1 + ( I y )2
Iy = Iy
( I y )1 = 121 (1.4 in.)(1.8 in.)3 = 0.6804 in 4
PROBLEM 9.42 CONTINUED
and
( I y )2 = 2  361 ( 2.1 in.)( 0.9 in.)3
2
1

+  × 1.89 in 2  ( 0.3 in.)  = 0.25515 in 4
2


Then
I y = ( 0.6804 + 0.25515 ) in 4
or I y = 0.936 in 4 W
PROBLEM 9.43
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
A = A1 − A2
= (100 mm )(160 mm ) − ( 40 mm )(100 mm )
= (16 000 − 4000 ) mm 2
= 12 000 mm 2
First locate the centroid:
AX = ΣAx
Have
or
(12 000 mm ) X = (16 000)(50) − ( 4000 )( 38) mm
2
X =
or
or
= 648 000 mm3
648 000 mm3
= 54 mm
12 000 mm 2
AY = ΣAy
And
or
3
(12 000 mm ) Y
2
= (16 000 )( 86 ) − ( 4000 )( 86 )  mm3 = 936 000 mm3
Y =
936 000 mm3
= 78 mm
12 000 mm 2
PROBLEM 9.43 CONTINUED
I x = ( I x )1 − ( I x )2
Now
where
( I x )1 =
(
)
1
( 40 mm )(100 mm )3 + 16 000 mm2 (80 mm − 78 mm )2
12
= 34.197 × 106 mm 4
( I x )2
=
(
)
1
( 40 mm )(100 mm )3 + 4000 mm2 ( 80 mm − 78 mm )2
12
= 3.5893 × 106 mm 4
Then
I x = ( 34.197 − 3.5893) × 106 mm 4
or I x = 30.6 × 106 mm 4 W
Also
where
( )1 − ( I y )2
Iy = Iy
( I y )1 = 121 (160 mm )(100 mm )3 + (16 000 mm2 ) ( 54 mm − 50 mm )2 = 13.589 × 106 mm4
( I y )2 = 121 (100 mm )( 40 mm )3 + ( 4000 mm2 ) ( 54 mm − 38 mm )2 = 1.5573 × 106 mm4
Then
I y = (13.589 − 1.5573) × 106 mm 4
or I y = 12.03 × 106 mm 4 W
PROBLEM 9.44
Determine the moments of inertia I x and I y of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
x1 = 160 mm
First locate centroid
y1 = 60 mm
A1 = 320 mm × 120 mm = 38 400 mm 2
x2 = −30 mm
y2 = 40 mm
A2 = 60 mm × 80 mm = 4800 mm 2
x3 = 280 mm
y3 = −105 mm
A3 = 80 mm × 210 mm = 16 800 mm 2
Then
X =
ΣxA
ΣA
160 ( 38 400 ) − 30 ( 4800 ) + 280 (16 800 )  mm3
= 
( 38 400 + 4800 + 16 800 ) mm 2
= 178.4 mm
And
Y =
ΣyA
ΣA
60 ( 38 400 ) + 40 ( 4800 ) − 105 (16 800 )  mm3
= 
( 38 400 + 4800 + 16 800 ) mm 2
= 12.20 mm
PROBLEM 9.44 CONTINUED
Then
I x = ( I x )1 + ( I x )2 + ( I x )3
(
)
3
2
1
=  ( 320 mm )(120 mm ) + 38 400 mm 2 ( 60 mm − 12.2 mm ) 
12

(
)
3
2
1
+  ( 60 mm )( 80 mm ) + 4800 mm 2 ( 40 mm − 12.2 mm ) 
12


(
)
3
2
1
+  ( 80 mm )( 210 mm ) + 16 800 mm 2 (105 mm + 12.2 mm ) 
12


= ( 46.080 + 87.7379 ) + ( 2.5600 + 3.7096 ) + ( 61.7400 + 230.7621)  × 106 mm 4
= 432.5896 × 106 mm 4
or I x = 433 × 106 mm 4 W
And
( )1 + ( I y )2 + ( I y )3
Iy = Iy
(
)
3
2
1
=  (120 mm )( 320 mm ) + 38 400 mm 2 (178.4 mm − 160 mm ) 
12


(
)
3
2
1
+  ( 80 mm )( 60 mm ) + 4800 mm 2 ( 30 mm + 178.4 mm ) 
12


(
)
3
2
1
+  ( 210 mm )( 80 mm ) + 16 800 mm 2 ( 280 mm − 178.4 mm ) 

12
= ( 327.6800 + 13.0007 ) + (1.4400 + 208.4667 ) + ( 8.9600 + 173.4190 )  × 106 mm 4
= 732.9664 × 106 mm 4
or I y = 733 × 106 mm 4 W
PROBLEM 9.45
Determine the polar moment of inertia of the area shown with respect to
(a) point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the area
A, in 2
1
2
π
2
−
( 2.7 )(1.8) = 7.6341
π
2
( 0.9 )2
Σ
yA, in 3
0.76394
5.8319
0.38197
0.4860
6.3618
5.3460
Y =
Y ΣA = ΣyA:
Then
J O = ( J O )1 − ( J O )2 =
(a)
= −1.2723
y , in.
π
8
5.3460 in 2
= 0.84033 in.
6.3618 in 2
π
( 2.7 in.)(1.8 in.) ( 2.7 )2 + (1.8)2  in 2 − ( 0.9 in.)4
4
= 19.5814 in 4
or J O = 19.58 in 4 W
JO = JC + A ( y )
(b)
or
(
)
2
J C = 19.5814 in 4 − 6.3618 in 2 ( 0.84033 in.) = 15.0890 in 4
2
J C = 15.09 in 4 W
PROBLEM 9.46
Determine the polar moment of inertia of the area shown with respect to
(a) point O, (b) the centroid of the area.
SOLUTION
First locate centroid
Y =0
Symmetry implies
A1 = π ( 20 in.)(12 in.)
x1 = 0
= ( 240π ) in 2
x2 =
4 ( 9 in.)
= (12π ) in.
3π
A2 = −
π
2
x3 = −
A3 = −
Then
ΣxA
X =
=
ΣA
=
( 9 in.)2
= − ( 40.5π ) in 2
4 ( 6 in.)
8
= − in.
π
3π
π
2
( 6 in.)2
( 0 ) ( 240π in 2 ) + (12π
= − (18π ) in 2
(
)
in.) −40.5π in 2 −
8 in.
π
240π in 2 − 40.5π in 2 − 18π in 2
−486 in 3 + 144 in 3
−342 in 3
=
= −0.59979 in.
181.5π in 2
181.5π in 2
( −18π in )
2
PROBLEM 9.46 CONTINUED
(a) Have
J O = ( J O )1 − ( J O )2 − ( J O )3
=
π
4
π
π
( 20 in.)(12 in.) ( 20 in.)2 + (12 in.)2  −  ( 9 in.)4  −  ( 6 in.)4 
4

4

= π ( 32640 − 1640.25 − 324.00 ) in 4 = 96,371 in 4
or J O = 96.4 × 103 in 4 W
(b) Have
Then
J O = J C + Ax 2
(
)
J C = 96,371 in 4 − 181.5π in 2 ( −0.59979 in.)
2
= 96,371 in 4 − 204.5629 in 4 = 96,166.4379 in 4
or J C = 96.2 × 103 in 4 W
PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect to
(a) point O, (b) the centroid of the area.
SOLUTION
A, mm 2
π
1
2
Σ
Now
2
−
(120 )2
= 22 619.5
1
( 240 )( 90 ) = −10 800
2
y , mm
yA, mm3
50.9296
1.1520 × 106
30
−0.324 × 106
0.828 × 106
11 819.5
Y =
ΣAY
0.828 × 106 mm3
=
= 70.054 mm
ΣA
11819.5 mm 2
J O = ( J O )1 − ( J O )2
(a)
where
and
( J O )1 =
π
(120 mm ) = 162.86 × 10
4
( J O )2 = ( I x′ ) 2 + ( I y ′ ) 2
4
=
6
mm 4
1
1
( 240 mm )( 90 mm )3 + 2  ( 90 mm )(120 mm )3 
12
12


= 40.5 × 106 mm 4
Then
J O = (162.86 − 40.5 ) × 106 mm 4 = 122.36 × 106 mm 4
or J O = 122.4 × 106 mm 4 W
PROBLEM 9.47 CONTINUED
J O = J C + Ay 2
(b)
or
(
)
J C = 122.36 × 106 mm 4 − 11 819.5 mm 2 ( 70.054 mm )
2
= (122.36 − 58.005 )106 mm 4
or J C = 64.4 × 106 mm 4 W
PROBLEM 9.48
Determine the polar moment of inertia of the area shown with respect to
(a) point O, (b) the centroid of the area.
SOLUTION
First locate centroid
x1 = 100 mm
y1 = −28.75 mm
A1 = ( 400 mm )( 57.5 mm ) = 23 000 mm 2
x2 = 100 mm
A2 =
1
( 300 mm )( 240 mm ) = 36 000 mm2
2
x3 = 50 mm
A3 = −
Then
y3 = 40 mm
1
(150 mm )(120 mm ) = −9000 mm2
2
(
)
(
)
(
2
2
2
ΣxA (100 mm ) 23 000 mm + (100 mm ) 36 000 mm + ( 50 mm ) −9000 mm
=
X =
ΣA
23 000 mm 2 + 36 000 mm 2 − 9000 mm 2
=
And
y2 = 80 mm
( 2.3 + 3.6 − 0.45) × 106 mm3
50 × 103 mm 2
(
)
= 109.0 mm
)
(
)
(
2
2
2
ΣyA ( −28.75 mm ) 23 000 mm + ( 80 mm ) 36 000 mm + ( 40 mm ) −9000 mm
Y =
=
ΣA
50 × 103 mm 2
=
( −661.25 + 2880 − 360 ) × 103 mm3
50 × 103 mm 2
= 37.175 mm
)
PROBLEM 9.48 CONTINUED
JO = I x + I y
(a) Now
I x = ( I x )1 + ( I x )2 − ( I x )3
where
1
3
( I x )1 = ( 400 mm )( 57.5 mm )3
Then
= 25.3479 × 106 mm 4
( I x )2
=
1
( 300 mm )( 240 mm )3 = 345.6000 × 106 mm 4
12
( I x )3
=
1
(150 mm )(120 mm )3 = 21.6000 × 106 mm4
12
I x = ( 25.3479 + 345.6000 − 21.6000 ) × 106 mm 4
= 349.348 × 106 mm 4
Also
where
( )1 + ( I y )2 − ( I y )3
Iy = Iy
( I y )1 = 121 ( 57.5 mm )( 400 mm )3 + ( 23 000 mm2 ) (100 mm )2
= ( 306.6667 + 230.0000 ) × 106 mm 4 = 536.6667 × 106 mm 4
( I y )2 = 121 ( 240 mm )( 300 mm )3 = 540.0000 × 106 mm4
( I y )3 = 121 (150 mm )(120 mm )3 = 33.7500 × 106 mm4
Then
Finally,
I y = ( 536.6667 + 540 − 33.75 ) × 106 mm 4 = 1042.917 × 106 mm 4
J O = ( 349.348 + 1042.917 ) × 106 mm 4 = 1392.265 × 106 mm 4
or J O = 1392 × 106 mm 4 W
(b) Have
Then
J O = J C + Ad 2
where
(
2
d2 = X + Y
2
)
2
2
J C = 1392.265 × 106 mm 4 − 50 × 103 mm 2 (109.0 mm ) + ( 37.175 mm ) 


= (1392.265 − 594.050 − 69.099 ) × 106 mm 4
= 729.1660 × 106 mm 4
or J C = 729 × 106 mm 4 W
PROBLEM 9.49
Two 1-in. steel plates are welded to a rolled S section as shown. Determine
the moments of inertia and the radii of gyration of the section with respect
to the centroidal x and y axes.
SOLUTION
S-section
A = 14.7 in 2
I x = 305 in 4
I y = 15.7 in 4
A = AS + 2 Aplate
= 14.7 in 2 + 2 ( 8 in.)(1 in.) = 30.7 in 2
( )S + 2 ( I x )plate
Ix = Ix
 ( 8 in.)(1 in.)3

2
= 305 in 4 + 2 
+ ( 8 in.)(1 in.)( 6.5 in.) 
12


= ( 305 + 677.33) in 4 = 982.33 in 4
I x = 9.82 in 4 W
or
and
k x2 =
Ix
982.33 in 4
=
= 31.998 in 2
4
A
30.7 in
k x = 5.66 in. W
or
Also
( )S
Iy = Iy
( )plate
+ 2 Iy
 (1 in.)( 8 in.)3 

= 15.7 in 4 + 2 
12


= (15.7 + 85.333) in 4 = 101.03 in 4
or I y = 101.0 in 4 W
and
k y2 =
Iy
A
=
101.03 in 4
= 3.29098 in 2
30.7 in 2
or k y = 1.814 in. W
PROBLEM 9.50
To form a reinforced box section, two rolled W sections and two plates are
welded together. Determine the moments of inertia and the radii of gyration
of the combined section with respect to the centroidal axes shown.
SOLUTION
W-section
A = 7.08 in 2
I x = 18.3 in 4
I y = 82.8 in 4
A = 2 AW + 2 Aplate
= 2 7.08 in 2 + ( 7.93 in.)( 0.3 in.) 
= 18.918 in 2
Now
( )W
Ix = 2 Ix
( )plate
+ 2 Ix
2

 6.495 in.  
= 2 18.3 in 4 + 7.08 in 2 
 
2

 

(
)
 ( 7.93 in.)( 0.3 in.)3

2
+ 2
+ ( 7.93 in.)( 0.3 in.)  ( 6.495 in. + 0.15 in.) 
12


= 2 92.967 in 4  + 2 105.07 in 4  = 396.07 in 4
I x = 396 in 4 W
or
and
or
k x2 =
Ix
396.07 in 4
=
= 20.936 in 2
2
A
18.918 in
k x = 4.58 in. W
PROBLEM 9.50 CONTINUED
Also
( )W + 2 ( I y )plate
Iy = 2 Iy
(
= 2 82.8 in
4
)
 ( 0.3 in.)( 7.93 in.)3 
 = (165.60 + 24.9339 ) in 4
+ 2
12


= 190.53 in 4
or I y = 190.5 in 4 W
and
k y2 =
Iy
A
=
190.53 in 4
= 10.072
18.918 in 2
or k x = 3.17 in. W
PROBLEM 9.51
Two C250 × 30 channels are welded to a 250 × 52 rolled S section as
shown. Determine the moments of inertia and the radii of gyration of the
combined section with respect to its centroidal x and y axes.
SOLUTION
Use Figure 9.13B (textbook) properties of rolled-steel shapes (SI units) to get the values for C250 and S250
shapes
S250 × 52 section:
A = 6670 mm 2
I x = 61.2 × 106 mm 4
I y = 3.59 × 106 mm 4
A = 3780 mm 2
C250 × 30 section:
I x = 32.6 × 106 mm 4
I y = 1.14 × 106 mm 4
A = AS + 2 AC
How, for the combined section:
= 6670 + 2 ( 3780 )  mm 2
= 14 230 mm 2
I x = ( I x ) S + 2 ( I x )C
= 61.2 × 106 + 2 32.6 × 106  mm 4


(
)
( )S + 2 ( I y )C + AC d 2 
or I x = 126.4 × 106 mm 4
Iy = Iy
where d is the distance from the centroid of the C section to the centroid C of the combined section
Now
2


 126

I y = 3.59 × 106 mm 4 + 2  1.14 × 106 mm 4 + 3780 mm 2 
+ 69 − 15.3  mm 2 
 2



(
) (
)
= ( 3.59 + 2.28 + 102.9588 ) × 106 mm 4
or I y = 108.8 × 106 mm 4
Also
kx =
=
Ix
A
126.4 × 106 mm 4
14 230 mm 2
or k x = 94.2 mm
PROBLEM 9.51 CONTINUED
And
ky =
=
Iy
A
108.8 × 106 mm 4
14 230 mm 2
or k y = 87.5 mm
PROBLEM 9.52
Two channels are welded to a d × 300-mm steel plate as shown.
Determine the width d for which the ratio I x / I y of the centroidal
moments of inertia of the section is 16.
SOLUTION
A = 3780 mm 2
Channel:
I x = 32.6 × 106 mm 4
I y = 1.14 × 106 mm 4
( )C + ( I x )plate
Ix = 2 Ix
Now
(
)
= 2 32.6 × 106 mm 4 +
d
( 300 mm )3
12
(
)
= 65.2 × 106 + 2.25d × 106 mm 4
And
( )channel + ( I y )plate
Iy = 2 Iy
2

d

= 2 1.14 × 106 mm 4 + 3780 mm 2  + 15.3 mm 
2


(
)
 ( 300 mm )d 3
+
12

(
)
=  2.28 × 106 + 1890d + 115.668 × 103 d + 1.7697 × 106 + 25d 3  mm 4


(
)
= 25d 3 + 1890d 2 + 115.67 × 103 d + 4.0497 × 106 mm 4
I x = 16I y
Given
Then 65.2 × 106 + 2.25d × 106
(
= 16 25d 3 + 1890d 2 + 115.67 × 103 d + 4.0497 × 106
or
)
25d 3 + 1890d 2 − 24.955d − 25300 = 0
Solving numerically
d = 12.2935 mm
or d = 12.29 mm
PROBLEM 9.53
Two L3 × 3 × 14 -in. angles are welded to a C10 × 20 channel. Determine
the moments of inertia of the combined section with respect to centroidal
axes respectively parallel and perpendicular to the web of the channel.
SOLUTION
A = 1.44 in 2
Angle:
I x = I y = 1.24 in 4
A = 5.88 in 2
Channel:
I x = 2.81 in 4
I y = 78.9 in 4
Locate the centroid
X =0
Y =
=
Now
(
)
(
)
2  1.44 in 2 ( 0.842 in.)  + 5.88 in 2 ( −0.606 in.)
ΣAy

= 
2
ΣA
2 1.44 in + 5.88 in 2
(
( 2.42496 − 3.5638) in 3
8.765 in 4
)
= −0.12995 in.
( I x ) = 2 ( I x )L + ( I x )C = 2 1.24 in 4 + (1.44 in 2 ) ( 0.842 in. + 0.12995 in.)2 
(
)
2
+  2.81 in 4 + 5.88 in 2 ( 0.606 in. − 0.12995 in.) 


= 2 ( 2.6003) in 4 + 4.1425 in 4 = 9.3431 in 4
or I x = 9.34 in 4
Also
( I y ) = 2 ( I y )L + ( I y )C = 2 2.14 in 4 + 1.44 in 2 ( 5 in. − 0.842 in.)2  + 7.89 in 4
= 2 ( 26.136 ) in 4 + 78.9 in 4 = 131.17 in 4
or I y = 131.2 in 4
PROBLEM 9.54
To form an unsymmetrical girder, two L3 × 3 × 14 -in. angles and two
L6 × 4 × 12 -in. angles are welded to a 0.8-in. steel plate as shown.
Determine the moments of inertia of the combined section with respect
to its centroidal x and y axes.
SOLUTION
L3 × 3 ×
Angle:
A = 1.44 in 2
L6 × 4 ×
A = 4.75 in 2
1
:
4
I x = I y = 1.24 in 4
1
:
2
I x = 6.27 in 4
I y = 17.4 in 4
Plate:
A = ( 27 in.)( 0.8 in.) = 21.6 in 2
Ix =
1
( 0.8 in.)( 27 in.)3 = 1312.2 in 4
12
Iy =
1
( 27 in.)( 0.8 in.)3 = 1.152 in 4
12
PROBLEM 9.54 CONTINUED
X =0
Centroid:
Y =
or
(
(
)
)
(
)
2
2  1.44 in 2 ( 27 in. − 0.84 in.)  + 2  4.75 in 2 ( 0.987 in.)  + 21.6 in 2 (13.5 in.)




Y =
2 1.44 in 2 + 4.75 in 2 + 21.6 in 2
=
Now
)
ΣAy
ΣA
(
376.31 in 3
= 11.0745 in.
33.98 in 2
I x = 2 ( I x )1 + 2 ( I x )3 + ( I x )2
2
2
= 2 6.25 + 4.75 (11.075 − 0.987 )  in 4 + 2 1.24 + 1.44 ( 27 − 0.842 − 11.075)  in 4




2
+ 1312.2 + 21.6 (13.5 − 11.075)  in 4


= 2 ( 489.67 ) in 4 + 2 ( 328.84 ) in 4 + 1439.22 in 4 = 3076.24 in 4
or I x = 3076 in 4
Also
( I y ) = 2 ( I y )1 + 2 ( I y )3 + ( I y )2
2
2
= 2 17.4 + 4.75 ( 0.4 + 1.99 )  in 4 + 2 1.24 + 1.44 ( 0.4 + 0.842 )  in 4 + 1.152 in 4




= 2 ( 44.532 ) in 4 + 2 ( 3.4613) in 4 + 1.152 in 4
= 97.139 in 4
or I y = 97.1 in 4
PROBLEM 9.55
Two L127 × 76 × 12.7-mm angles are welded to a 10-mm steel plate.
Determine the distance b and the centroidal moments of inertia I x and I y
of the combined section knowing that I y = 3I x .
SOLUTION
A = 2420 mm 2
Angle:
I x = 3.93 × 106 mm 4
I y = 1.074 × 106 mm 4
A = ( 200 mm )(10 mm ) = 2000 mm 2
Plate:
Ix =
1
( 200 mm )(10 mm )3 = 0.01667 × 106 mm4
12
Iy =
1
(10 mm )( 200 mm )3 = 6.6667 × 106 mm 4
12
X =0
Centroid
Y =
or
Y =
(
ΣAy
ΣA
)
2 2420 mm 2 ( 44.5 mm ) + 2000 mm 2 ( −5 mm )
 2 ( 2420 ) + 2000  mm 2
=
205.380 mm3
6840 mm 2
= 30.026 mm
Now
I x = 2 ( I x )angle + ( I x )plate
2
= 2 3.93 × 106 + ( 2420 )( 44.5 − 30.026 )  mm 4


2
+ 0.01667 × 106 + ( 2000 )( 30.026 + 5 )  mm 4


(
)
= 2 4.43698 × 106 mm 4 + 2.4703 × 106 mm 4
= 11.344 × 106 mm 4
or I x = 11.34 × 106 mm 4
PROBLEM 9.55 CONTINUED
Also
Where
( )angle + ( I y )plate
Iy = 2 Iy
( I y )angle = 1.074 × 106 mm4 + ( 2420 mm2 ) ( b − 19.05 mm )2
(
)
= 1.074 × 106 + ( 2420 ) b 2 − 38.1b + 362.9  mm 4


=  2420b 2 − 92202b + 1.9522 × 106  mm 4
and
Now
Then
or
( I y )plate = 6.6667 × 106 mm4
( )
I y = 3 Ix
(
)
2  2420b 2 − 92202b + 1.9522 × 106  mm 4 + 6.6667 × 106 mm 4 = 3  11.34 × 106 mm 4 


2420b 2 − 9.2202b + 1.9522 × 106 − 13.6767 × 106 = 0
b 2 − 38.1b − 4844.8 = 0
b = 91.2144 mm
or b = 91.2 mm
PROBLEM 9.56
A channel and an angle are welded to an a × 20-mm steel plate. Knowing
that the centroidal y axis is located as shown, determine (a) the width a,
(b) the moments of inertia with respect to the centroidal x and y axes.
SOLUTION
(a) Using Figure 9.13B
From the geometry of L152 × 152 × 19, C150 × 15.6, plate a × 20 mm and how they are welded
x A = 44.9 mm AA = 5420 mm 2
xC = −12.5 mm AC = 1980 mm 2
a

xP = −  − 152  mm
2

AP = ( 20a ) mm 2
X =
From the condition
ΣxA
=0
ΣA
 a


− 152  mm  20a mm 2 = 0

 2

( 44.9 mm ) ( 5420 mm2 ) − (12.5 mm ) (1980 mm2 ) − 
or
a 2 − 304a − 21860.8 = 0
(
)
a = 364.05 mm
or a = 364 mm
And
AP = ( 20 mm )( 364 mm )
= 7280 mm 2
PROBLEM 9.56 CONTINUED
(b) Locate the centroid
Y =
ΣAy
ΣA

mm  + ( 7280 mm ) ( −10 mm )
(5420 mm ) ( 44.9 mm ) + (1980 mm )  152
2

2
=
2
2
( 5420 + 1980 + 7280 ) mm 2
= 21.867 mm
Now
I x = ( I x ) A + ( I x )C + ( I x ) P
(
)
2
= 11.6 × 106 mm 4 + 5420 mm 2 ( 44.9 mm − 21.867 mm ) 


(
)
2
+ 6.21 × 106 mm 4 + 1980 mm 2 ( 76 mm − 21.867 mm ) 


(
)
3
2
1
+  ( 364.05 mm )( 20 mm ) + 7281 mm 2 (10 mm + 21.867 mm ) 
12

= (11.6 + 2.8754 ) + ( 6.21 + 5.8022 ) + ( 0.2427 + 7.3939 )  × 106 mm 4
= (14.4754 + 12.0122 + 7.6366 ) × 106 mm 4
= 34.1242 × 106 mm 4
or I x = 34.1 × 106 mm 4
And
( ) A + ( I y )C + ( I y )P
Iy = Iy
(
)
2
= 11.6 × 106 mm 4 + 5420 mm 2 ( 44.9 mm ) 


(
)
2
+ 0.347 × 106 mm 4 + 1980 mm 2 (12.5 mm ) 


2
1
3
 364.05 mm
 
+  ( 20 mm )( 364.05 mm ) + 7821 mm 2 
− 152 mm  
2

 
12
(
)
= (11.6 + 10.9268 ) × 106 mm 4 + ( 0.347 + 0.3094 ) × 106 mm 4
+ ( 80.4140 + 6.5638 ) × 106 mm 4
= ( 22.5268 + 0.6564 + 86.9778 ) × 106 mm 4
= 110.161 × 10−6 mm 4
or I y = 110.2 × 106 mm 4
PROBLEM 9.57
The panel shown forms the end of a trough which is filled with water to
the line AA′ . Referring to Sec. 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel
(the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text have
yP =
I AA′ =
For a quarter circle
and
I AA′
yA
y =
yP =
16
r4
4r
π
, A = r2
3π
4
π
Then
π
r4
16
 4r  π 2 

 r 
 3π  4 
or yP =
3π
r
16
PROBLEM 9.58
The panel shown forms the end of a trough which is filled with water to
the line AA′ . Referring to Sec. 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel
(the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text have
yP =
I AA′
yA
I AA′ =
For a semiellipse
y =
π
8
ab3
4b
π
, A = ab
3π
2
π
Then
ab3
8
yP =
 4b  π 

 ab 
 3π  2 
or yP =
3π
b
16
PROBLEM 9.59
The panel shown forms the end of a trough which is filled with water to
the line AA′. Referring to Sec. 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel
(the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text
yP =
Have
I AA′
yA
YA = ΣyA
Now
=
h
4 1
( 2b × h ) + h  × 2b × h 
2
3 2

=
7 2
bh
3
I AA′ = ( I AA′ )1 + ( I AA′ )2
And
where
1
3
( I AA′ )1 = ( 2b )( h )3
=
( I AA′ )2 = I x + Ad 2 =
=
Then
Finally,
2 3
bh
3
1
1
4 
( 2b )( h )3 +  × 2b × h 
 h 
36
2
3



2
11 3
bh
6
I AA′ =
2 3 11 3 5 3
bh + bh = bh
3
6
2
5 3
bh
yP = 2
7 2
bh
3
or yP =
15
h
14
PROBLEM 9.60
The panel shown forms the end of a trough which is filled with water to
the line AA′. Referring to Sec. 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel
(the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text
yP =
Have
where
I AA′
yA
I AA′ = ( I AA′ )1 + ( I AA′ )2
2
2
  π
π  4r   π 2 
4r  
3
1
=  ( 2r )( r )  +   r 4 − r 2 
+
+
r
r



 
2  3π   2 
3π  
3
   8

=
And
2 4 π
8
π 4
9  4 
5π  4
+ + +
r + −
r = 2 +
r
3
2 3 8π 
8 
 8 9π

4r  π 2  
r
 
YA = ΣyA =  ( 2r × r )  +  r +
 r 
3π  2  
2
 
π 2

5 π 
= 1 + +  r 3 =  +  r 3
2 3

3 2
Then
5π  4

2 +
r
8 
yP = 
= 1.2242r
5 π  3
 + r
3 2
or yP = 1.224r
PROBLEM 9.61
The cover for a 10 × 22-in. access hole in an oil storage tank is attached to
the outside of the tank with four bolts as shown. Knowing that the specific
weight of the oil is 57.4 lb/ft3 and that the center of the cover is located
10 ft below the surface of the oil, determine the additional force on each
bolt because of the pressure of the oil.
SOLUTION
Using the equation developed on page 491 of the text have
yP =
I AA′
yA
R = γ yA
R = 57.4 lb/ft 3 × 10 ft × ( 22 × 10 ) in 2 ×
Then
1 ft 2
144 in 2
= 876.94 lb
and
I AA′ =
1
( 22 in.)(10 in.)3 + ( 22 in.)(10 in.) (120 in.)2
12
= 3.169833 × 106 in 4 = 152.8662 ft 4
yA = 10 ft × ( 22 × 10 ) in 2 ×
and
yp =
Then
Now symmetry implies
1 ft 2
= 15.27778 ft 3
2
144 in
152.8662 ft 4
= 10.00579 ft
15.27778 ft 2
FA = FB
and
FC = FD
Equilibrium
ΣM CD = 0:
(1 ft )( 2FAa ) − ( 0.5 − 0.00579 ) ft × 876.94 lb = 0
FA = 216.70 lb
or FA = FB = 217 lb
Also
or
ΣFx = 0:
−2 ( 216.70 ) + 876.94 − 2FC = 0
or FC = FD = 222 lb
PROBLEM 9.62
A vertical trapezoidal gate that is used as an automatic valve is held shut
by two springs attached to hinges located along edge AB. Knowing that
each spring exerts a couple of magnitude 8.50 kN ⋅ m , determine the
depth d of water for which the gate will open.
SOLUTION
d = ( h + 1.58 ) m
I ss′
yA
From page 491
yp =
Now
yA = ΣyA
γ = ρg
R = γ yA
1

1

= ( h + 0.34 m )   × 2.4 m × 1.02 m  + ( h + 0.68 m )   × 1.68 m × 1.02 m 
2

2

= ( 2.0808h + 0.99878 ) m3
(
)(
)
R = 103 kg/m3 9.81 m/s 2 ( 2.0808h + 0.99878 ) m3
Also,
= 20 413 ( h + 0.480 ) N
1

3
2
1

And I ss′ = ( I ss′ )1 + ( I ss′ )2 =  ( 2.4 m )(1.02 m ) +  ( 2.4 m )(1.02 m )  ( h + 0.34 ) m 2 
36
2




1

3
2
1

+  (1.68 m )(1.08 m ) +  (1.68 m )(1.02 m )  ( h + 0.68 ) m 2 
2

 36

2
2
= 0.07075 + 1.224 ( h + 0.34 ) + 0.04952 + 0.8568 ( h + 0.68 )  m 4


(
)
(
)
= 0.12027 + 1.224 h 2 + 0.68h + 0.1156 + 0.8568 h 2 + 1.36h + 0.4624  m 4


(
)
= 2.0808h 2 + 1.9976h + 0.65795 m 4
PROBLEM 9.62 CONTINUED
Then
yp
( 2.0808h
=
)
+ 1.9976h + 0.65795 m 4
( 2.0808h + 0.99878) m
=
For gate to open
2
3
h 2 + 0.960h + 0.3162
m
h + 0.480
(
)
ΣM AB = 0: M open − y p − h R = 0
 h 2 + 0.960h + 0.3162
 
2 ( 8500 N ⋅ m ) − 
− h  m  ( 20 413)( h + 0.48 )  N = 0
h + 0.480
 

or
or
(
17 000 = 20 413 h 2 + 0.96h + 0.3162 − h 2 − 0.480h
)
0.48h − 0.5166 = 0
h = 1.0763 m
Now
d = h + 1.58 m = (1.0763 + 1.58 ) m = 2.6563 m
or d = 2.66 m
PROBLEM 9.63
Determine the x coordinate of the centroid of the volume shown. (Hint:
The height y of the volume is proportional to the x coordinate; consider an
analogy between this height and the water pressure on a submerged surface.)
SOLUTION
x dV
x = ∫ EL
∫ dV
Have
where
dV = ydA
Now
y =
and
xEL = x
60
1
x= x
300
5
1 


x =
1 
∫  5 x  dA


∫ x  5 x  dA
Then
( I z )A
x 2dA
= ∫
=
∫ xdA ( xA) A
where ( I z ) A is the moment of inertia of the area with respect to the z axis, and x is analogous to y p
Now
( Iz )A
=
1
1
( 240 mm )( 300 mm )3 + ( 240 mm )( 300 mm ) ( 200 mm )2
36
2
= 1.620 × 109 mm 4
and
Then
1

xA = ( 200 mm )  ( 240 mm )( 300 mm )  = 7.20 × 106 mm3
2

x =
1.620 × 109 mm 4
7.20 × 106 mm3
or x = 225 mm
PROBLEM 9.64
Determine the x coordinate of the centroid of the volume shown; this
volume was obtained by intersecting an elliptic cylinder with an oblique
plane. (Hint: The height y of the volume is proportional to the x coordinate;
consider an analogy between this height and the water pressure on a
submerged surface.)
SOLUTION
y =
Have
x ∫ dV = ∫ xEL dV ,
h
x
a
where
xEL = x
h 
dV = ydA =  x  dA
a 
And
h 
( Iz )A
x 2dA


= ∫
=
x =
h 
∫ xdA ( xA) A
∫  a x  dA


∫ x  a x  dA
Now
For the given volume
( Iz )A
=
π
4
( 2 in.)( 3.5 in.)3 + π ( 3.5 in.)( 2 in.) ( 3.5 in.)2
= ( 21.4375π + 85.7500π ) in 4 = 336.74 in 4
and
Then
( xA) A
= 3.5 in. π ( 3.5 in.)( 2 in.)  = 76.969 in 3
x =
336.74 in 4
= 4.375 in.
76.969 in 3
or x = 4.38 in.
PROBLEM 9.65
Show that the system of hydrostatic forces acting on a submerged plane
area A can be reduced to a force P at the centroid C of the area and two
couples. The force P is perpendicular to the area and is of magnitude
P = γ Ay sin θ , where γ is the specific weight of the liquid, and the
couples are M x′ = ( γ I x′ sin θ ) i and M y′ = γ I x′y′ sin θ j , where
I x′y′ = ∫ x′y ′ dA (see Sec. 9.8). Note that the couples are independent
of the depth at which the area is submerged.
(
)
SOLUTION
The pressure p at an arbitrary depth ( y sin θ ) is
p = γ ( y sin θ )
so that the hydrostatic force dF exerted on an infinitesimal area dA is
dF = (γ y sin θ ) dA
Equivalence of the force P and the system of infinitesimal forces dF requires
ΣF : P = ∫ dF = ∫ γ y sin θ dA = γ sin θ ∫ ydA
or P = γ Ay sin θ
Equivalence of the force and couple
( P, M x′ + M y′ ) and
the system of infinitesimal hydrostatic forces
requires
ΣM x: − yP − M x′ = ∫ ( − ydF )
Now
−∫ ydF = −∫ y (γ y sin θ ) dA = −γ sin θ ∫ y 2dA
= − (γ sin θ ) I x
Then
or
− yP − M x′ = − (γ sin θ ) I x
M x′ = (γ sin θ ) I x − y (γ Ay sin θ )
(
= γ sin θ I x − Ay 2
)
or M x′ = γ I x′ sin θ
ΣM y : xP + M y′ = ∫ xdF
Now
∫ xdF = ∫ x (γ y sin θ ) dA = γ sin θ ∫ xydA
= (γ sin θ ) I xy
( Equation 9.12 )
PROBLEM 9.65 CONTINUED
Then
or
xP + M y′ = (γ sin θ ) I xy
M y′ = (γ sin θ ) I xy − x (γ Ay sin θ )
(
= γ sin θ I xy − Ax y
or, using Equation 9.13,
)
or M y′ = γ I x′y′ sin θ
PROBLEM 9.66
Show that the resultant of the hydrostatic forces acting on a submerged
plane area A is a force P perpendicular to the area and of magnitude
P = γ Ay sin θ = pA , where γ is the specific weight of the liquid and p
is the pressure at the centroid C of the area. Show that P is applied at a
point C p , called the center of pressure, whose coordinates are x p = I xy / Ay
and y p = I x / Ay , where I xy = ∫ xy dA (see Sec. 9.8). Show also that the
difference of ordinates y p − y is equal to k x2′ / y and thus depends upon
the depth at which the area is submerged.
SOLUTION
The pressure p at an arbitrary depth ( y sin θ ) is
p = γ ( y sin θ )
so that the hydrostatic force dP exerted on an infinitesimal area dA is
dP = (γ y sin θ ) dA
The magnitude P of the resultant force acting on the plane area is then
P = ∫ dP = ∫ γ y sin θ dA = γ sin θ ∫ ydA
= γ sin θ ( yA )
Now
p = γ y sin θ
∴ P = pA
Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then
requires
ΣM x: − yP P = − ∫ ydP
Now
2
∫ ydP = ∫ y (γ y sin θ ) dA = γ sin θ ∫ y dA
= ( γ sin θ ) I x
Then
or
yP P = (γ sin θ ) I x
yP =
(γ sin θ ) I x
γ sin θ ( yA)
or yP =
Ix
Ay
ΣM y : xP P = ∫ xdP
Now
∫ xdP = ∫ x (γ y sin θ ) dA = γ sin θ ∫ xydA
= (γ sin θ ) I xy
( Equation 9.12 )
PROBLEM 9.66 CONTINUED
Then
or
xP P = (γ sin θ ) I xy
xP =
(γ sin θ ) I xy
γ sin θ ( yA)
or xP =
Now
Ay
I x = I x′ + Ay 2
From above
I x = ( Ay ) yP
By definition
I x′ = k x′ A
Substituting
I xy
2
( Ay ) yP
= k x2′ A + Ay 2
yP − y =
Rearranging yields
k x2′
y
Although k x′ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on
the depth.
∴ ( yP − y ) = f ( depth )
PROBLEM 9.67
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
y = a 1−
First note
=
1
4a 2 − x 2
2
dI xy = dI x′y′ + xEL yELdA
Have
dI x′y′ = 0
where
yEL =
( symmetry )
xEL = x
1
1
y =
4a 2 − x 2
2
4
dA = ydx =
Then
x2
4a 2
1
4a 2 − x 2dx
2
2a  1
 1

4a 2 − x 2 
4a 2 − x 2  dx
I xy = ∫ dI xy = ∫0 x 
4
 2

(
)
2a
1 2a
1
1 
= ∫0 4a 2 x − x3 dx =  2a 2 x 2 − x 4 
8
8
4 0
=
a4
8
1
2
4

2 ( 2) − 4 ( 2) 


or I xy =
1 4
a
2
PROBLEM 9.68
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
At
x = a, y = b: a = kb 2
or
k =
a
b2
Then
x=
a 2
y
b2
Have
where
dI xy = dI x′y′ + xEL yELdA
dI x′y′ = 0
yEL = y
Then
( symmetry )
dA = xdy =
xEL =
1
a
x = 2 y2
2
2b
a 2
y dy
b2
b a

 a

I xy = ∫ dI xy = ∫0  2 y 2  ( y )  2 y 2dy 
 2b

b

b
a2 b
a2  1 
= 4 ∫0 y 5dy = 4  y 6 
2b
2b  6  0
or I xy =
1 2 2
ab
12
PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
First note that
y =−
Now
h
x
b
dI xy = dI x′y′ + xEL yELdA
dI x′y′ = 0 ( symmetry )
where
xEL = x
yEL =
1
1h
y =−
x
2
2b
h
dA = ydx = − xdx
b
Then
0  1 h  h

I xy = ∫ dI xy = ∫−b x  −
x  − xdx 
 2 b  b

0
=
1 h2 0 3
1 h2  1 4 
x
dx
x
=
∫
2 b 2 −b
2 b 2  4  −b
1
or I xy = − b 2h 2
8
PROBLEM 9.70
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
At
π 
x = a, y = b: b = c sin  a 
 2a 
y = b sin
Now
dI x′y′ = 0
Have
dI xy
Now
or
π
2a
or
c=b
x
(symmetry)
dA = ydx
0
= dI x′y′ + xEL yELdA
1 a
π
a 1 
I xy = ∫0 x  y  ( ydx ) = ∫0 xb 2 sin 2
xdx
2
2a
2 

b  x 2 x sin πa x cos πa
=
−
−
2 4
4 2πa
8 2πa

2
( )
=
a

x
2
 0
( )
b 2  a 2 4a 2 4a 2 
+ 2 +


2  4
8π
8π 2 
or I xy =
(
a 2b 2
4 + π2
8π 2
)
The following table is provided for the convenience of the instructor, as many problems in this and the next
lesson are related.
Type of
Problem
Compute
I x and I y
Fig. 9.12
Fig. 9.13B
Fig. 9.13A
Compute
I xy
9.67
9.72
9.73
9.74
9.75
9.78
9.79
9.80
9.81
9.83
9.82
9.84
9.85
9.86
9.87
9.89
9.88
9.90
9.91
9.92
9.93
9.95
9.94
9.96
9.97
9.98
9.100
9.101
9.103
9.106
I x′, I y′, I x′y′
by equations
Principal
axes by
equations
I x′, I y′, I x′y′
by
Mohr's circle
Principal
axes by
Mohr’s
circle
PROBLEM 9.71
Using the parallel-axis theorem, determine the product of inertia of the area
shown with respect to the centroidal x and y axes.
SOLUTION
( )1 + ( I xy )2 + ( I xy )3
I xy = I xy
Have
( I xy )2 = 0
Symmetry implies
I xy = I x′y′ + x yA
For the other rectangles
I x′y′ = 0
Where symmetry implies
A in 2
x , in.
y , in.
Ax y in 4
1
4 ( 0.5 ) = 2
−2.75
1.0
−5.5
3
4 ( 0.5 ) = 2
2.75
−1.0
−5.5
Σ
−11.00
or I xy = −11.00 in 4
PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
Note: Orientation of A3 corresponding to a 180° rotation of the axes. Equation 9.20 then yields
I x′y′ = I xy
( I xy )1 = 0
Symmetry implies
( I x′y′ )2 = − 721 ( 9 in.)2 ( 4.5 in.)2 = −22.78125 in 4
Using Sample Problem 9.6
and
X 2 = 9 in.
Then
and
Therefore,
A2 =
1
( 9 in.)( 4.5 in.) = 20.25 in 2
2
( I x′y′ )3 = − 721 ( 9 in.)2 ( 4.5 in.)2 = −22.78125 in 4
Similarly,
and
Y2 = 1.5 in.
X 3 = −9 in.
Y2 = −1.5 in.
A3 =
0
( )1 + ( I xy )2 + ( I xy )3
I xy = I xy
1
( 9 in.)( 4.5 in.) = 20.25 in 2
2
with
( I xy )2 = ( I xy )3
I xy = I x′y′ + x y A
I xy = 2  −22.78125 + ( 9 )(1.5 )( 20.25 )  in 4
= 501.1875 in 4
or I xy = 501 in 4 W
PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
( )1 + ( I xy )2
Have
I xy = I xy
For each semicircle
I xy = I x′y′ + x y A
Thus
I xy = Σx y A
A, mm 2
1
2
Σ
π
2
π
2
x , mm
(120 )2
= 7200π
− 60
(120 )2
= 7200π
60
I x′y′ = 0 (symmetry)
and
y , mm
Ax y , mm 4
160
69.12 × 106
−
π
160
π
69.12 × 106
138.24 × 106
or I xy = 138.2 × 106 mm 4 W
PROBLEM 9.74
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
( )1 + ( I xy )2
I xy = I xy
Have
I xy = I x′y′ + Ax y
For each rectangle
and
I x′y′ = 0 (symmetry)
I xy = Σ x y A
Thus
A, mm 2
x , mm
y , mm
Ax y , mm 4
1
76 ( 6.4 ) = 486.4
−12.9
9.4
−58 980.86
2
44.6 ( 6.4 ) = 285.44
21.9
−16.1
−100 643.29
Σ
−159 624.15
or I xy = −0.1596 × 106 mm 4 W
PROBLEM 9.75
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
Have
Now symmetry implies
and for the other rectangles
Thus
( )1 + ( I xy )2 + ( I xy )3
I xy = I xy
( I xy )1 = 0
I xy = I x′y′ + x y A
where
I x′y′ = 0 (symmetry)
I xy = ( x y A)2 + ( x y ) A3
= ( −69 mm )( −25 mm ) (12 mm )( 38 mm ) 
+ ( 69 mm )( 25 mm ) (12 mm )( 38 mm ) 
= ( 786 600 + 786 600 ) mm 4 = 1 573 200 mm 4
or I xy = 1.573 × 106 mm 4 W
PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
( I xy )1 = 0
Symmetry implies
Using Sample Problem 9.6 and Equation 9.20, note that the orientation of A2 corresponds to a 90° rotation of
1 2 2
the axes; thus
I x′y′ =
b h
2
72
( )
( I x′y′ )3 = 721 b2h2
Also, the orientation of A3 corresponds to a 270° rotation of the axes; thus
( I x′y′ )2 = 721 ( 9 in.)2 ( 6 in.)2 = 40.5 in 4
Then
and
x2 = 6 in.,
x3 = −6 in.,
and
Then
A2 =
1
( 9 in.)( 6 in.) = 27 in 2
2
( I x′y′ )3 = ( I x′y′ )2 = 40.5 in 4
Also
Now
y2 = −2 in.,
0
y3 = 2 in.,
( )1 − ( I xy )2 − ( I xy )3
I xy = I xy
and
A3 = A2 = 27 in 2
I xy = I x′y′ + x y A
(
( I xy )2 = ( I xy )3
)
I xy = −2  40.5 in 4 + ( 6 in.)( −2 in.) 27 in 2 


= −2 ( 40.5 − 324 ) in 4
or I xy = 567 in 4 W
PROBLEM 9.77
Using the parallel-axis theorem, determine the product of inertia of the area
shown with respect to the centroidal x and y axes.
SOLUTION
I xy = I x′y′ + x y A
Have
Where I x′y′ = 0 for each rectangle
( )1 + ( I xy )2 + ( I xy )3
I xy = I xy
Then
= Σ x yA
Now
x1 = − (178.4 mm − 160 mm ) = −18.4 mm
y1 = 60 mm − 12.2 mm = 47.8 mm
A1 = 320 mm × 120 mm = 38400 mm 2
and
x2 = − (178.4 mm + 30 mm ) = −208.4 mm
y2 = 40 mm − 12.2 mm = 27.8 mm
A2 = 60 mm × 80 mm = 4800 mm 2
and
x3 = ( 320 mm − 178.4 mm ) − 40 mm = 101.6 mm
y3 = − (12.2 mm + 105 mm ) = −117.2 mm
A3 = ( 80 mm × 210 mm ) = 16800 mm 2
Then
(
)
(
)
I xy = ( −18.4 mm )( 47.8 mm ) 38400 mm 2  + ( −208.4 mm )( 27.8 mm ) 4800 mm 2 


 
(
)
+ (101.6 mm )( −117.2 mm ) 16800 mm 2 


= − ( 33.7736 + 27.8089 + 200.0463) × 106 mm 4
= −261.6288 × 106 mm 4
or I xy = −262 × 106 mm 4 W
PROBLEM 9.78
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
( )1 + ( I xy )2
I xy = I xy
Have
For each rectangle
I xy = I x′y′ + x yA
Then
and
I x′y′ = 0 (symmetry)
I xy = Σx yA = ( −0.75 in.)( −1.5 in.) ( 3 in.)( 0.5 in.) 
+ ( 0.5 in.)(1.00 in.) ( 4.5 in.)( 0.5 in.) 
= (1.6875 + 1.125 ) in 4 = 2.8125 in 4
or I xy = 2.81 in 4 W
PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of inertia
and the product of inertia with respect to new axes obtained by rotating
the x and y axes about O (a) through 45o counterclockwise, (b) through
30o clockwise.
SOLUTION
π
Ix =
From Figure 9.12:
16
π
=
8
a4
π
Iy =
16
=
π
2
( 2a )( a )3
( 2a )3 ( a )
a4
I xy =
From Problem 9.67:
1 4
a
2
1
1 π
π  5
I x + I y =  a4 + a4  =
π a4
2
28
2  16
(
First note
)
1
1 π
π 
3
I x − I y =  a4 − a4  = − π a4
2
2 8
2 
16
(
)
Now use Equations (9.18), (9.19), and (9.20).
Equation (9.18):
I x′ =
=
Equation (9.19):
I y′ =
=
Equation (9.20):
I x′y′ =
(
)
(
)
1
1
Ix + I y +
I x − I y cos 2θ − I xy sin 2θ
2
2
5
3
1
π a 4 − π a 4 cos 2θ − a 4 sin 2θ
16
16
2
(
)
(
)
1
1
Ix + I y −
I x − I y cos 2θ + I xy sin 2θ
2
2
5
3
1
π a 4 + π a 4 cos 2θ + a 4 sin 2θ
16
16
2
(
)
1
I x − I y sin 2θ + I xy cos 2θ
2
=−
3
1
π a 4 sin 2θ + a 4 cos 2θ
16
2
PROBLEM 9.79 CONTINUED
(a)
θ = +45°:
I x′ =
5
3
1
π a 4 − π a 4 cos90° − a 4 sin 90°
16
16
2
or I x′ = 0.482a 4 W
I y′ =
5
3
1
π + π a 4 cos 90° + a 4
16
16
2
or I y′ = 1.482a 4 W
I x′y′ = −
3
1
π a 4 sin 90° + a 4 cos 90°
16
2
or I x′y′ = −0.589a 4 W
(b) θ = −30° :
I x′ =
5
3
1
π a 4 − π a 4 cos ( −60° ) − a 4 sin ( −60° )
16
16
2
or I x′ = 1.120a 4 W
I y′ =
5
3
1
π a 4 + π a 4 cos ( −60° ) + a 4 sin ( −60° )
16
16
2
or I y′ = 0.843a 4 W
I x′y′ = −
3
1
π a 4 sin ( −60° ) + a 4 cos ( −60° )
16
2
or I x′y′ = 0.760a 4 W
PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the area of
Problem 9.72 with respect to new centroidal axes obtained by rotating the
x and y axes 45° clockwise.
SOLUTION
From the solution to Problem 9.72
I xy = 501.1875 in 4
A2 = A3 = 20.25 in 2
First compute the moment of inertia
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )2 = ( I x )3
with
3
3
1
1
=  (12 in.)( 9 in.)  + 2  ( 9 in.)( 4.5in.) 
12

12

= ( 729 + 136.6875 ) in 4 = 865.6875 in 4
and
( )1 + ( I y )2 + ( I y )3
Iy = Iy
with
( I y )2 = ( I y )3
(
)
3
3
2
1
1
=  ( 9 in.)(12 in.)  + 2  ( 4.5 in.)( 9 in.) + 20.25 in 2 ( 9 in.) 
12

 36

= (1296 + 182.25 + 3280.5 ) in 4 = 4758.75 in 4
From Equation 9.18
I x′ =
=
Ix + I y
2
+
Ix − I y
2
cos 2θ − I xy sin 2θ
865.6875 in 4 + 4758.75 in 4 865.6875 in 4 − 4758.75 in 4
cos  2 ( −45° ) 
+
2
2
−501.1875 in 4 sin  2 ( −45° ) 
= ( 2812.21875 + 501.1875 ) in 4 = 3313.4063 in 4
or I x′ = 3.31 × 103 in 4
PROBLEM 9.80 CONTINUED
Similarly
I y′ =
Ix + I y
2
−
Ix − I y
2
cos 2θ + I xy sin 2θ
= ( 2812.21875 − 501.1875 ) in 4 = 2311.0313 in 4
or I y′ = 2.31 × 103 in 4
and
I x′y′ =
=
Ix − I y
2
sin 2θ + I xy cos 2θ
865.6875 in 4 − 4758.75 in 4
sin  2 ( −45° ) 
2
+ 501.1875cos  2 ( −45° ) 
= ( −1946.53125 )( −1) in 4 = 1946.53125 in 4
or I x′y′ = 1.947 × 103 in 4
PROBLEM 9.81
Determine the moments of inertia and the product of inertia of the area of
Problem 9.73 with respect to new centroidal axes obtained by rotating the
x and y axes through 30o clockwise.
SOLUTION
I xy = 138.24 × 106 mm 4
From Problem 9.73,
I x = ( I x )1 + ( I x )2
( I x )1 = ( I x )2
4
π
= 2  (120 mm ) 
8

= 51.84π × 106 mm 4
( )1 + ( I y )2
( I y )1 = ( I y )2
Iy = Iy
π
4
2
2
π
= 2  (120 mm ) + (120 mm ) ( 60 mm ) 
2
8

= 103.68π × 106 mm 4
(
)
(
)
Have
I x = 2 25.92π × 106 = 51.84π × 106 mm 4
and
I y = 2 51.84π × 106 = 103.68π × 106 mm 4
Then
1
I x + I y = 77.76π × 106 mm 4
2
and
(
)
(
)
1
I x − I y = −25.92π × 106 mm 4
2
PROBLEM 9.81 CONTINUED
Now, from Equations 9.18, 9.19, and 9.20
Equation 9.18:
I x′ =
(
)
(
)
1
1
Ix + I y +
I x − I y cos 2θ − I xy sin 2θ
2
2
= 77.76π × 106 − 25.92π × 106 cos ( −60° ) − 138.24 × 106 sin ( −60° )  mm 4
= 323.29 × 106 mm 4
or I x = 323 × 106 mm 4
Equation 9.19:
I y′ =
(
)
(
)
1
1
Ix + I y −
I x − I y cos 2θ + I xy sin 2θ
2
2
= 77.76π × 106 + 25.92π × 106 cos ( −60° ) + 138.24 × 106 sin ( −60° )  mm 4
= 165.29 × 106 mm 4
or I y′ = 165.29 × 106 mm 4
Equation 9.20:
I x′y′ =
(
)
1
I x − I y sin 2θ + I xy cos 2θ
2
=  −25.92π × 106 sin ( −60° ) + 138.24 × 106 cos ( −60° )  mm 4
= 139.64 × 106 mm 4
or I x′y′ = 139.6 × 104 mm 4
PROBLEM 9.82
Determine the moments of inertia and the product of inertia of the area of
Problem 9.75 with respect to new centroidal axes obtained by rotating the
x and y axes through 60o counterclockwise.
SOLUTION
From Problem 9.75
I xy = 1.5732 × 106 mm 4
Now
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 =
where
and
( I x )2 = ( I x )3
1
(150 mm )(12 mm )3 = 21 600 mm4
12
1
(12 mm )( 38 mm )3 + (12 mm )( 38 mm ) ( 25 mm )2
12
=
= 339 872 mm 4
Then
I x =  21 600 + 2 ( 339 872 )  mm 4 = 701 344 mm 4 = 0.70134 × 106 mm 4
( )1 + ( I y )2 + ( I y )3
Iy = Iy
Also
where
and
( I y )1 = 121 (12 mm )(150 mm )3 = 3.375 × 106 mm4
( I y )2 = ( I y )3 = 121 ( 38 mm )(12 mm )3 + (12 mm )( 38 mm ) ( 69 mm )2
= 2.1765 × 106 mm 4
Then
I y = ( 3.375 + 2 ( 2.1765 )  × 106 mm 4 = 7.728 × 106 mm 4
(
)
(
)
Now
1
I x + I y = 4.2146 × 106 mm 4
2
and
1
I x − I y = −3.5133 × 106 mm 4
2
PROBLEM 9.82 CONTINUED
Using Equations 9.18, 9.19, and 9.20
From Equation 9.18:
I x′ =
Ix + I y
2
Ix − I y
+
2
cos 2θ − I xy sin 2θ
(
)
=  4.2147 × 106 + −3.5133 × 106 cos (120° ) − 1.5732 × 106 sin (120° )  mm 4


= 4.6089 × 106 mm 4
or I x′ = 4.61 × 106 mm 4
From Equation 9.19:
I y′ =
Ix + I y
2
−
Ix − I y
2
cos 2θ + I xy sin 2θ
(
)
=  4.2147 × 106 − −3.5133 × 106 cos (120° ) + 1.5732 × 106 sin (120° )  mm 4


= 3.8205 × 106 mm 4
or I y′ = 3.82 × 106 mm 4
From Equation 9.20:
I x′y′ =
Ix − I y
2
sin 2θ + I xy cos 2θ
=  −3.5133 × 106 sin (120° ) + 1.5732 × 106 cos (120° )  mm 4
= −3.8292 × 106 mm 4
or I x′y′ = −3.83 × 106 mm 4
PROBLEM 9.83
Determine the moments of inertia and the product of inertia of the
L76 × 51 × 6.4-mm angle cross section of Problem 9.74 with respect to
new centroidal axes obtained by rotating the x and y axes through 45o
clockwise.
SOLUTION
From Problem 9.74
I xy = −0.1596 × 106 mm 4
From Figure 9.13
I x = 0.166 × 106 mm 4
I y = 0.453 × 106 mm 4
(
)
(
)
1
I x + I y = 0.3095 × 106 mm 4
2
Now
1
I x − I y = −0.1453 × 106 mm 4
2
Using Equations (9.18), (9.19), and (9.20)
Equation (9.18):
I x′ =
Ix + I y
2
+
Ix − I y
2
(
cos 2θ − I xy sin 2θ
)
(
)
= 0.3095 × 106 + −0.1435 × 106 cos ( −90° ) − −0.1596 × 106 sin ( −90° )  mm 4


= 0.1499 × 106 mm 4
or I x′ = 0.1499 × 106 mm 4
PROBLEM 9.83 CONTINUED
Equation (9.19):
I y′ =
Ix + I y
2
−
Ix − I y
cos 2θ + I xy sin 2θ
2
(
)
(
)
= 0.3095 × 106 − −0.1435 × 106 cos ( −90° ) + −0.1596 × 106 sin ( −90° )  mm 4


= 0.4691 × 106 mm 4
or I y′ = 0.469 × 106 mm 4
Equation (9.20):
I x′y′ =
Ix − I y
2
sin 2θ + I xy cos 2θ
=  −0.1435 × 106 sin ( −90° ) + 0.1596 × 106 cos ( −90° )  mm 4
= 0.1435 × 106 mm 4
or I x′y′ = 0.1435 × 106 mm 4
PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the
L5 × 3 × 12 -in. angle cross section of Problem 9.78 with respect to new
centroidal axes obtained by rotating the x and y axes through 30o
counterclockwise.
SOLUTION
From Problem 9.78
I xy = 2.8125 in 4
From Figure 9.13
I x = 9.45 in 4 ,
I y = 2.58 in 4
(
)
(
)
1
I x + I y = 6.015 in 4
2
Now
1
I x − I y = 3.435 in 4
2
Using Equations (9.18), (9.19), and (9.20)
Equation (9.18):
Ix + I y
I x′ =
+
2
Ix − I y
2
cos 2θ − I xy sin 2θ
= 6.015 + 3.435cos ( 60° ) − 2.8125sin ( 60° )  in 4 = 5.2968 in 4
or I x′ = 5.30 in 4
Equation (9.19):
I y′ =
Ix + I y
2
−
Ix − I y
2
cos 2θ + I xy sin 2θ
= 6.015 − 3.435cos ( 60° ) + 2.8125sin ( 60° )  in 4 = 6.7332 in 4
or I y′ = 6.73 in 4
Equation (9.20):
I x′y′ =
Ix − I y
2
sin 2θ + I xy cos 2θ
= 3.435sin ( 60° ) + 2.8125cos ( 60° )  in 4 = 4.3810 in 4
or I x′y′ = 4.38 in 4
PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of the
principal axes at the origin and the corresponding values of the moments
of inertia.
SOLUTION
From Problem 9.79:
π
Ix =
8
I xy =
Problem 9.67:
Now, Equation (9.25):
Iy =
a4
tan 2θ m = −
Then
2
a4
1 4
a
2
2 I xy
Ix − I y
=−
π
8
=
π
1 
2  a4 
2 
a4 −
π
2
a4
8
= 0.84883
3π
2θ m = 40.326°
and
220.326°
or θ m = 20.2° and 110.2°
Also, Equation (9.27):
I max, min =
=
Ix + I y
2
2
 Ix − I y 
2
± 
 + I xy
2


1π 4 π 4
 a + a 
2 8
2 
2
1 π
π 
1 
±   a4 − a4  +  a4 
2 
2 
2 8
2
= ( 0.981748 ± 0.772644 ) a 4
or I max = 1.754a 4
and I min = 0.209a 4
By inspection, the a axis corresponds to Imin and the b axis corresponds to
Imax.
PROBLEM 9.86
For the area indicated, determine the orientation of the principal axes at
the origin and the corresponding values of the moments of inertia.
Area of Problem 9.72
SOLUTION
From the solutions to Problem 9.72 and 9.80
(
(
)
1
I x + I y = 2812.21875 in 4
2
I xy = 501.1875 in 4
)
1
I x − I y = −1946.53125 in 4
2
Then Equation (9.25):
or
tan 2θ m = −
2 I xy
Ix − I y
2θ m = 14.4387°
=−
501.1875
= 0.257477
−1946.53125
194.4387°
and
or θ m = 7.22° and 97.2°
Equation (9.27):
I max, min =
Ix + I y
2
2
 Ix − I y 
± 
+ I xy2
 2 


= 2812.21875 ±
( −1946.53125 )2 + ( 501.1875 )2
= ( 2812.21875 ± 2010.0181) in 4
or I max = 4.82 × 103 in 4
and I min = 802 in 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.87
For the area indicated, determine the orientation of the principal axes at
the origin and the corresponding values of the moments of inertia.
Area of Problem 9.73
SOLUTION
From Problems 9.73 and 9.81
I x = 51.84π × 106 mm 4
I y = 103.68π × 106 mm 4
I xy = 138.24 × 106 mm 4
tan 2θ m = −
Equation (9.25):
2 I xy
Ix − I y
=−
(
2 138.24 × 106
)
51.84π × 10 − 103.68π × 106
6
= 1.69765
2θ m = 59.500°
and
239.500°
or θ m = 29.7° and 119.7°
Then
I max, min
2
 Ix − I y 
1
=
Ix + I y ± 
+ I xy2
 2 
2


(
=
)
( 51.84 + 103.68) π × 106
2
 ( 51.84 − 103.68 ) π × 106 
6

 + 138.24 × 10
2


2
±
(
)
2
= ( 244.29 ± 160.44 ) × 106 mm 4
or I max = 405 × 106 mm 4
and I min = 83.9 × 106 mm 4
Note: By inspection the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.88
For the area indicated, determine the orientation of the principal axes at
the origin and the corresponding values of the moments of inertia.
Area of Problem 9.75
SOLUTION
From Problems 9.75 and 9.82
I x = 0.70134 × 106 mm 4
I y = 7.728 × 106 mm 4
I xy = 1.5732 × 106 mm 4
Then
(
)
(
)
1
I x + I y = 4.2147 × 106 mm 4
2
1
I x − I y = −3.5133 × 106 mm 4
2
Equation (9.25):
tan 2θ = −
2 I xy
Ix − I y
=−
(
2 1.5732 × 106
)
0.70134 × 10 − 7.728 × 106
6
= 0.44778
Then
2θ m = 24.12°
204.12°
and
or θ m = 12.06° and 102.1°
Also, Equation (9.27):
I max, min =
Ix + I y
2
2
 Ix − I y  2
± 
I
 2  xy


= 4.2147 × 106 ±
( −3.5133 × 10 ) + (1.5732 × 10 )
6
2
6
2
= ( 4.2147 ± 3.8494 ) × 106 mm 4
or I max = 8.06 × 106 mm 4
and I min = 0.365 × 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.89
For the angle cross section indicated, determine the orientation of the
principal axes at the origin and the corresponding values of the moments
of inertia.
The L76 × 51 × 6.4-mm angle cross section of Problem 9.74
SOLUTION
From Problems 9.74 and 9.83
I x = 0.166 × 106 mm 4
I y = 0.453 × 106 mm 4
I xy = −0.1596 × 106 mm 4
Then
(
)
(
)
1
I x + I y = 0.3095 × 106 mm 4
2
1
I x − I y = −0.1435 × 106 mm 4
2
Equation (9.25):
Then
tan 2θ m = −
2 I xy
Ix − I y
=−
(
2 −0.1596 × 106
)
( 0.166 − 0.453) × 106
2θ m = −48.041°
and
= −1.1122
131.96°
θ m = −24.0° and 66.0°
or
Also, Equation (9.27):
I max, min =
( Ix + I y ) ±
2
2
 Ix − I y 
2

 + I xy
2


= 0.3095 × 106 ±
( −0.1435 × 10 ) + ( −0.1596 × 10 )
6
2
6
2
= ( 0.3095 ± 0.21463) × 106 mm 4
or
I max = 0.524 × 106 mm 4
I min = 0.0949 × 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.90
For the angle cross section indicated, determine the orientation of the
principal axes at the origin and the corresponding values of the moments
of inertia.
The L5 × 3 ×
1
-in. angle cross section of Problem 9.78
2
SOLUTION
From Problems 9.78 and 9.84
I xy = 2.81 in 4
I x = 9.45 in 4
I y = 2.58 in 4
Then
(
)
(
)
1
I x + I y = 6.015 in 4
2
1
I x − I y = 3.435 in 4
2
2 I xy
Equation (9.25):
tan 2θ m = −
Then
2θ m = −39.2849
Ix − I y
=−
and
2 ( 2.81)
9.45 − 2.58
= −0.8180
140.7151
or θ m = −19.64 and 70.36
Also, Equation (9.27):
I max, min
( Ix + I y ) ±
=
2
2
 Ix − I y 
2

 + I xy
2


= 6.015 ± 3.4352 − 2.812
= ( 6.015 ± 4.438 ) in 4
or I max = 10.45 in 4
and I min = 1.577 in 4
Note: By inspection, the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.91
Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67 the
moments of inertia and the product of inertia with respect to new axes
obtained by rotating the x and y axes about O (a) through 45o
counterclockwise, (b) through 30o clockwise.
SOLUTION
π
Ix =
From Problem 9.79:
8
π
Iy =
2
a4
1 4
a
2
I xy =
Problem 9.67:
a4
The Mohr’s circle is defined by the diameter XY, where
1 
π
X  a4, a4 
8
2 

Now
I ave =
1
1 π
π 
5
I x + I y =  a 4 + a 4  = π a 4 = 0.98175a 4
2
2 8
2  16
(
)
2
and
R=
1 
π
Y  a4, − a4 
2
2 

and
 Ix − I y 
2

 + I xy =
 2 
2
 1  π 4 π 4 
1 4
 2  8 a − 2 a  +  2 a 



 
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
=−
2 I xy
Ix − I y
π
8
1 
2  a4 
2 
a4 −
= 0.84883
or
2θ m = 40.326°
π
2
a4
2
PROBLEM 9.91 CONTINUED
Then
α = 90° − 40.326°
= 49.674°
β = 180° − ( 40.326° + 60° )
= 79.674°
(a)
I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674°
or I x′ = 0.482a 4
I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674°
or I y′ = 1.482a 4
I x′y′ = − R sin α = −0.77264a 4 sin 49.674°
or I x′y′ = −0.589a 4
(b)
I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674°
or I x′ = 1.120a 4
I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674°
or I y′ = 0.843a 4
I x′y′ = R sin β = 0.77264a 4 sin 79.674°
or I x′y′ = 0.760a 4
PROBLEM 9.92
Using Mohr’s circle, determine the moments of inertia and the product of
inertia of the area of Problem 9.72 with respect to new centroidal axes
obtained by rotating the x and y axes 45° clockwise.
SOLUTION
From the solution to
Problem 9.72:
I xy = 501.1875 in 4
Problem 9.80:
I x = 865.6875 in 4
I y = 4758.75 in 4
(
)
(
)
1
I x + I y = 2812.21875 in 4
2
Now
1
I x − I y = −1946.53125 in 4
2
The Mohr’s circle is defined by the points X and Y where
( I x , I xy )
X:
I ave =
Now
Y:
(
( I y , −I xy )
)
1
I x + I y = 2812.2 in 4
2
2
and
R=
 Ix − I y 
2

 + I xy =
2


( −1946.53125)2 + 501.18752
= 2010.0 in 4
Also,
tan 2θ m =
I xy
Ix − I y
=
501.1875
= 0.2575
1946.53125
2
or
Then
2θ m = 14.4387°
α = 180° − (14.4387° + 90° ) = 75.561°
in 4
PROBLEM 9.92 CONTINUED
Then
I x′ , I y′ = I ave ± R cos α = 2812.2 ± 2010.0cos 75.561°
or I x′ = 3.31 × 103 in 4
and I y′ = 2.31 × 103 in 4
and
I x′y′ = R sin α = 2010.0sin 75.561°
or I x′y′ = 1.947 × 103 in 4
PROBLEM 9.93
Using Mohr’s circle, determine the moments of inertia and the product of
inertia of the area of Problem 9.73 with respect to new centroidal axes
obtained by rotating the x and y axes through 30o clockwise.
SOLUTION
From Problems 9.73 and 9.81
I xy = 138.24 × 106 mm 4
I x = 51.84π × 106 mm 4
= 162.86 × 106 mm 4
I y = 103.68π × 106 mm 4
= 325.72 × 106 mm 4
I ave =
Now
(
1
Ix + I y
2
)
= 244.29 × 106 mm 4
2
R=
 Ix − I y 
2

 + I xy
2


= 160.4405 × 106 mm 4
2θ m = 59.5°
From Problem 9.87
Then
Then
α = 180 − 60° − 2θ m = 60.5°
I x′ = I ave + R cos α = 244.29 + 160.4405cos 60.5°
= 323.29 × 106 mm 4
or I x′ = 323 × 106 mm 4
I y′ = I ave − R cos α = 244.24 − 160.4405cos 60.5°
= 165.29 × 106 mm 4
or I y′ = 165.3 × 106 mm 4
I x′y′ = R sin α = 160.44sin 60.5° = 139.6 × 106 mm 4
PROBLEM 9.94
Using Mohr’s circle, determine the moments of inertia and the product of
inertia of the area of Problem 9.75 with respect to new centroidal axes
obtained by rotating the x and y axes through 60o counterclockwise.
SOLUTION
From Problems 9.75 and 9.82
I x = 0.70134 × 106 mm 4
I y = 7.728 × 106 mm 4
I xy = 1.5732 × 106 mm 4
Now
I ave =
(
)
1
I x + I y = 4.2147 × 106 mm 4
2
2
and
Then
and
Then
R=
 Ix − I y 
2
6
4

 + I xy = 3.8494 × 10 mm
2


 −2 (1.5732 ) 
2θ m = tan −1 
 = 24.12°
 0.70134 − 7.728 
α = 120° − 24.12° − 90 = 5.88°
I x′ = I ave + R sin α = ( 4.2147 + 3.8494sin 5.88° ) × 106 mm 4
= 4.6091 × 106 mm 4
or I x′ = 4.61 × 106 mm 4
I y′ = I ave − R sin α = ( 4.2147 − 3.8494sin 5.88° ) × 106 mm 4
= 3.8203 × 106 mm 4
or I y′ = 3.82 × 106 mm 4
I x′y′ = − R cos α = −3.8494 cos 5.88° = −3.8291 × 106 mm 4
or I x′y′ = −3.83 × 106 mm 4
PROBLEM 9.95
Using Mohr’s circle, determine the moments of inertia and the product of
inertia of the L76 × 51 × 6.4-mm angle cross section of Problem 9.74
with respect to new centroidal axes obtained by rotating the x and y axes
through 45o clockwise.
SOLUTION
From Problems 9.74 and 9.83
I x = 0.166 × 106 mm 4
I y = 0.453 × 106 mm 4
I xy = −0.1596 × 106 mm 4
Now
I ave =
(
)
1
I x + I y = 0.3095 × 106 mm 4
2
2
and
R=
 Ix − I y 
2

 + I xy
2


= 0.21463 × 106 mm 4
Then
and
 −2 ( −0.1596 ) 
2θ m = tan −1 
 = −48.04°
 0.166 − 0.453 
α + 90° − 2θ = 90°; α = 2θ m
Then
I x′ = I ave − R sin α = ( 0.3095 − 0.21463sin 48.04° ) × 106 mm 4
= 0.14989 × 106 mm 4
or I x′ = 0.1499 × 106 mm 4
and
I y′ = I ave + R sin α = ( 0.3095 + 0.21463sin 48.04° ) × 106 mm 4
= 0.46910 × 106 mm 4
or I y′ = 0.4690 × 106 mm 4
and
I x′y′ = R cos α = 0.21463cos 48.04° = 0.1435 × 106 mm 4
or I x′y′ = 0.1435 × 106 mm 4
PROBLEM 9.96
Using Mohr’s circle, determine the moments of inertia and the product of
inertia of the L5 × 3 × 12 -in. angle cross section of Problem 9.78 with
respect to new centroidal axes obtained by rotating the x and y axes
through 30o counterclockwise.
SOLUTION
I x = 9.45 in 4
Have
I y = 2.58 in 4
I xy = 2.8125 in 4
From Problem 9.78
I ave =
Now
2
R=
and
Ix + I y
 Ix − I y 

 + I xy
 2 
( )
2
= 6.015 in 4
2
= 4.43952 in 4
Then
 −2 ( 2.8125 ) 
2θ m = tan −1 
 = −39.31°
 9.45 − 2.58 
2θ m + 60 + α = 180°,
Then
α = 80.69°
I x′ = I ave − R cos α = 6.015 in 4 − ( 4.43952 in 4 ) cos80.69°
= 5.29679 in 4
or I x′ = 5.30 in 4
I y′ = I ave + R cos α = 6.015 in 4 + ( 4.43952 in 4 ) cos80.69°
= 6.73321 in 4
or I y′ = 6.73 in 4
I x′y′ = R sin α = ( 4.43952 in 4 ) sin 80.69° = 4.38104 in 4
or I x′y′ = 4.38 in 4
PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine
the orientation of the principal axes at the origin and the corresponding
values of the moments of inertia.
SOLUTION
From Problem 9.79:
Ix =
Problem 9.67:
I xy =
π
8
a4
Iy =
π
2
a4
1 4
a
2
The Mohr’s circle is defined by the diameter XY, where
1 
π
X  a4, a4 
8
2 

I ave =
Now
1 
π
Y  a4, − a4 
2
2 

and
1
1π
π 
I x + I y =  a 4 + a 4  = 0.98175a 4
2
2 8
2 
(
)
and
R=
2
1

2
 2 I x − I y  + I xy =


(
)
2
 1  π 4 π 4 
1 4
  a − a  +  a 
2 
2 
2  8
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
=−
2 I xy
Ix − I y
π
8
1 
2  a4 
2 
a4 −
π
2
a4
= 0.84883
or
and
2θ m = 40.326°
θ m = 20.2°
2
PROBLEM 9.97 CONTINUED
∴ The principal axes are obtained by rotating the xy axes through
20.2° counterclockwise
About O.
Now
I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4
or I max = 1.754a 4
and I min = 0.209a 4
From the Mohr’s circle it is seen that the a axis corresponds to I min and
the b axis corresponds to I max .
PROBLEM 9.98
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.72
SOLUTION
From the solution to Problem 9.72:
I xy = 501.1875 in 4
From the solution to Problem 9.80:
I x = 865.6875 in 4
I y = 4758.75 in 4
(
)
(
)
1
I x + I y = 2812.21875 in 2
2
1
I x − I y = −1946.53125 in 4
2
X:
The Mohr’s circle is defined by the point
I ave =
Now
(
( I x , I xy ) ,
Y:
( I y,
− I xy
)
1
I x + I y = 2812.2 in 4
2
2
and
R=
 Ix − I y 

 + I xy =
 2 
( −1946.53125)2 + 501.18752
= 2010.0 in 4
)
PROBLEM 9.98 CONTINUED
tan 2θ m = −
I xy
 Ix − I y 


 2 
=−
501.1875
= 0.2575,
−1946.53125
2θ m = 14.4387°
or θ m = 7.22° counterclockwise
Then
I max, min = I ave ± R = ( 2812.2 ± 2010.0 ) in 4
or I max = 4.82 × 103 in 4
and I min = 802 in 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.99
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.76
SOLUTION
I xy = 576 in 4
From the solution to Problem 9.76
Now
I x = ( I x )1 − ( I x )2 − ( I x )3 , where
=
π
( I x )2 = ( I x )3
(15 in.)4 − 2  ( 9 in.)( 6 in.)3  = ( 39761 − 324 ) in 4
1
12
4

= 39, 437 in 4
( )1 − ( I y )2 − ( I y )3 ,
Iy = Iy
and
=
π
4
where
( I y )2 = ( I y )3
(15 in.)4 − 2  ( 6 in.)( 9 in.)3 + ( 9 in.)( 6 in.)( 6 in.)2 
1
 36
1
2

= ( 39, 761 − 243 − 1944 ) in 4 = 37,574 in 4
The Mohr’s circle is defined by the point (X, Y) where
X:
I ave =
Now
and
R=
Y:
( I y , −I xy )
)
1
1
I x + I y = ( 39, 437 + 37,574 ) in 4 = 38,506 in 4
2
2
Ix − I y
2
(
( I x , I xy )
2
+ I xy2 =
1

2
4
 2 ( 39, 437 − 37,574 )  + 567 = 1090.5 in


PROBLEM 9.99 CONTINUED
tan 2θ m =
− I xy
Ix − I y
2
=
−567
= −0.6087
1
( 39, 437 − 37,574 )
2
or θ m = −15.66° clockwise
Then
I max, min = I ave ± R = ( 38,506 ± 1090.50 ) in 4
or I max = 39.6 × 103 in 4
and I min = 37.4 × 103 in 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to the I max and the b axis corresponds to
I min .
PROBLEM 9.100
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.73
SOLUTION
I x = 162.86 × 106 mm 4
From Problems 9.73 and 9.81
I y = 325.72 × 106 mm 4
I xy = 138.24 × 106 mm 4
X (162.86,138.24 ) × 106 mm 4
Define points
I ave =
Now
(
Y ( 325.72, −138.24 ) × 106 mm 4
)
1
1
I x + I y = (162.86 + 325.72 ) × 106 mm 4
2
2
= 244.29 × 106 mm 4
2
and
R=
 Ix − I y 
2

 + I xy =
 2 
 (162.86 − 325.72 )

× 106  + 138.24 × 106

2


2
(
)
2
= 160.44 × 106 mm 4
and
 −2 (138.24 ) × 106

2θ m = tan −1 
= 59.4999°
6
 (162.86 − 325.72 ) × 10 
or θ m = 29.7° counterclockwise
Then
(
)
I max, min = I ave ± R = 244.29 × 106 ± 160.44 × 106 mm 4
or I max = 405 × 106 mm 4
and I min = 83.9 × 106 mm 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to
I max .
PROBLEM 9.101
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.74
SOLUTION
From Problems 9.74 and 9.83
I x = 0.166 × 106 mm 4 ,
I y = 0.453 × 106 mm 4 ,
X ( 0.166, −0.1596 ) × 106 mm 4
Define points
I ave =
Now
(
and
I xy = −0.1596 × 106 mm 4
Y ( 0.453, −0.1596 ) × 106 mm 4
)
1
1
I x + I y = ( 0.166 + 0.453) × 106 mm 4
2
2
= 0.3095 × 106 mm 4
2
and
R=
 Ix − I y 
2

 + I xy =
2


 ( 0.166 − 0.453)106 
6

 + −0.1596 × 10
2


2
(
)
2
= 0.21463 × 106 mm 4
Also
 −2 I xy
2θ m = tan −1 
 Ix − I y


 −2 ( −0.1596 ) 
 = tan −1 
 = −48.04°

 0.166 − 0.453 

θ m = −24.02°
or θ = −24.0° clockwise
Then
I max,
min
= I ave ± R = ( 0.3095 ± 0.21463) × 106 mm 4
or I max = 0.524 × 106 mm 4
and I min = 0.0949 × 106 mm 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to
I max .
PROBLEM 9.102
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.75
SOLUTION
From Problems 9.75 and 9.82
I x = 0.70134 × 106 mm 4 ,
Now
I ave =
(
I y = 7.728 × 106 mm 4 ,
)
1
1
I x + I y = ( 0.70134 + 7.728 ) × 106 mm 4 = 4.2147 × 106 mm 4
2
2
2
and
R=
I xy = 1.5732 × 106 mm 4
 Ix − I y 
2

 + I xy =
 2 
 ( 0.70134 − 7.728 ) × 106 
6

 + 1.5732 × 10
2


2
(
)
2
= 3.8495 × 106 mm 4
X ( 0.70134, 15732 ) × 106 mm
Define points
Y ( 7.728, − 1.5732 ) × 106 mm
Also
 −2 (1.5732 ) 
2θ m = tan −1 
 = 24.122°, θ m = 12.06°
 0.70134 − 7.728 
or θ m = 12.06° counterclockwise
Then
I max,
min
= I ave ± R = ( 4.2147 ± 3.8495 ) × 106 mm 4
or I max = 8.06 × 106 mm 4
and I min = 0.365 × 106 mm 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.103
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.71
SOLUTION
I xy = −11.0 in 4
From Problem 9.71
Compute I x and I y for area of Problem 9.71
5 in. × ( 0.5 in.)
3
Ix =
12
 ( 0.5 in.)( 4 in.)3

2
+ 2
+ ( 4 in. × 0.5 in.)(1.0 in.) 
12


= 9.38542 in 4
 ( 0.5 in.)3 ( 4 in.)
 0.5 in. × ( 5 in.)3
2
Iy = 2
+ ( 4 in. × 0.5 in.)( 2.75 in.)  +
12
12


= 35.54167 in 4
X ( 9.38542, − 11) ,
Define points
I ave =
Now
Ix + I y
2
2
and
R=
=
and
Y ( 35.54167, 11)
9.38542 in 4 + 35.54167 in 4
= 22.46354 in 4
2
 Ix − I y 

 + I xy
 2 
( )
2
2
=
2
 9.38542 − 35.54167 

 + (11.0 )
2


= 17.08910 in 4
Also
Then


−2 ( −11.0 )
2θ m = tan −1 
 = −40.067
 9.38542 − 35.54167 
or θ m = −20.033° clockwise
I max, min = I ave ± R = 22.46354 ± 17.08910
= 39.55264, 5.37444
or I max = 39.55 in 4
I min = 5.37 in 4
Note: The a axis corresponds to I min and b axis corresponds to I max .
PROBLEM 9.104
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.77
SOLUTION
From Problems 9.44 and 9.77
I x = 432.59 × 106 mm 4 ,
I y = 732.97 × 106 mm 4 ,
I xy = −261.63 × 106 mm 4
X ( 432.59, − 261.63) × 106 mm 4
Define points
Y ( 732.97, 261.63) × 106 mm 4
Now
I ave =
(
)
1
1
I x + I y = ( 432.59 + 732.97 ) × 106 = 582.78 × 106 mm 4
2
2
2
and
R=
2
 Ix − I y 
1  432.59 − 732.97
2
6
× 10  + −261.63 × 106

 + I xy =

2 
2

 2 
(
)
2
= 301.67 × 106 mm 4
Also
tan 2θ m = −
Ix − I y
=
−2 ( −261.63) × 106
( 432.59 − 732.97 ) × 106
= −60.14°
θ m = −30.1° clockwise
or
Then
2 I xy
I max,
min
= I ave ± R = ( 582.78 ± 301.67 ) × 106 mm 4
or
I max = 884 × 106 mm 4
and
I min = 281 × 106 mm 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.105
The moments and product of inertia for an L102 × 76 × 6.4-mm angle
cross section with respect to two rectangular axes x and y through C are,
I x = 0.166 × 106 mm 4 ,
I y = 0.453 × 106 mm 4 , and
respectively,
I xy < 0 , with the minimum value of the moment of inertia of the area
with respect to any axis through C being I min = 0.051 × 106 mm 4 . Using
Mohr’s circle, determine (a) the product of inertia I xy of the area, (b) the
orientation of the principal axes, (c) the value of I max .
SOLUTION
I x = 0.166 × 106 mm 4 , I y = 0.453 × 106 mm 4 and I xy < 0
Given:
Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when
the 102 mm leg of the angle is parallel to the x axis. For I xy < 0 the angle must be oriented as shown.
(a) Now
I ave =
(
)
1
1
I x + I y = ( 0.166 + 0.453) × 106 mm 4
2
2
= 0.3095 × 106 mm 4
I min = I ave − R
Now
R = I ave − I min
or
R = ( 0.3095 − 0.051) × 106 mm 4
Then
= 0.2585 × 106 mm 4
2
 Ix − I y 
R =
+ I xy
 2 


2
From
I xy =
( )
2
2

2
 0.166 − 0.453  
6
4
( 0.2585 ) − 
  × 10 mm
2

 

I xy = ±0.21501 × 106 mm 4
Since
I xy < 0,
I xy = −0.21501 × 106 mm 4
or I xy = −0.215 × 106 mm 4
PROBLEM 9.105 CONTINUED
(b)
 −2 ( −0.21501) 
2θ m = tan −1 
 = −56.28°
 0.166 − 0.453 
or θ m = −28.1 clockwise
(c)
I max = I ave + R = ( 0.3095 + 0.2585 ) × 106 mm 4
or I max = 0.568 × 106 mm 4
PROBLEM 9.106
Using Mohr’s circle, determine for the cross section of the rolled-steel
angle shown the orientation of the principal centroidal axes and the
corresponding values of the moments of inertia. (Properties of the cross
sections are given in Fig. 9.13.)
SOLUTION
I x = 9.45 in 4
From Figure 9.13
I y = 2.58 in 4
I xy = −2.81 in 4
From Problem 9.78
The Mohr’s circle is defined by the diameter XY where
X ( 9.45, − 2.81) in 4
Y ( 2.58, 2.81) in 4
Now
I ave =
(
(
)
1
1
Ix + I y =
9.45 in 4 + 2.58 in 4
2
2
)
= 6.015 in 4
and
R=
=
2
1

2
 2 I x − I y  + I xy


(
)
(
1
9.45 in 4 − 2.58 in 4
2
) + ( 2.81 in )
2
4
2
= 5.612 in 4
tan 2θ m =
−2 I xy
Ix − I y
=
(
−2 −2.81 in 4
)
9.45 in − 2.58 in 4
4
= 0.81805
2θ m = 32.285°
or
or θ m = 19.643° counterclockwise
About C.
Now
I max,
min
= I ave ± R = ( 6.015 ± 5.612 ) in 4
or I max = 11.63 in 4
and I min = 0.403 in 4
From the Mohr’s circle it is seen that the a axis corresponds to I max and
the b axis corresponds to I min .
PROBLEM 9.107
Using Mohr’s circle, determine for the cross section of the rolled-steel
angle shown the orientation of the principal centroidal axes and the
corresponding values of the moments of inertia. (Properties of the cross
sections are given in Fig. 9.13.)
SOLUTION
I x = 7.20 × 106 mm 4 ,
From Figure 9.13B:
( )1 + ( I xy )2 ,
I xy = I xy
Have
x1 =
Now
where
102
− 25.3 = 25.7 mm,
2
I y = 2.64 × 106 mm 4
I xy = I x′y′ + x yA
y1 = 50.3 −
and
I x′y′ = 0
12.7
= 43.95 mm
2
A1 = 102 × 12.7 = 1295.4 mm 2
x2 = −25.3 −
1

y2 = −  (152 − 12.7 ) − ( 50.3 − 12.7 )  = 32.05 mm
2


12.7
= −18.95 mm
2
A2 = (12.7 )(152 − 12.7 ) = 1769.11 mm 2
Then
{
(
)
(
= (1.46317 + 1.07446 ) × 106 mm 4 = 2.5376 × 106 mm 4
The Mohr’s circle is defined by points X and Y, where
(
) (
X I x , I xy , Y I y , − I xy
Now
)}
I xy = ( 25.7 mm )( 43.95 mm ) 1295.4 mm 2  + ( −18.95 mm )( −32.05 mm ) 1769.11 mm 2  × 106

 

I ave =
(
)
)
1
1
I x + I y = ( 7.20 + 2.64 ) × 106 mm 4 = 4.92 × 106 mm 4
2
2
PROBLEM 9.107 CONTINUED
and
R=
 Ix − I y 
 1
2
2
2
6
4

 + I xy =  ( 7.20 − 2.64 ) + 2.5376  × 10 mm
2
2






= 3.4114 × 106 mm 4
tan θ m = −
2 I xy
Ix − I y
=−
2 ( 2.5376 )
( 7.20 − 2.64 )
= −1.11298,
2θ = −48.0607°
θ = −24.0° clockwise
or
Now
or
and
I max, min = I ave ± R = ( 4.92 ± 3.4114 ) × 106 mm 4
I max = 8.33 × 106 mm 4
I min = 1.509 × 106 mm 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.108
For a given area the moments of inertia with respect to two rectangular
centroidal x and y axes are I x = 640 in 4 and I y = 280 in 4 , respectively.
Knowing that after rotating the x and y axes about the centroid 60°
clockwise the product of inertia relative to the rotated axes is −180 in 4 ,
use Mohr’s circle to determine (a) the orientation of the principal axes,
(b) the centroidal principal moments of inertia.
SOLUTION
I ave =
Have
(
(
(
)
)
1
1
640 in 4 + 280 in 4 = 460 in 4
Ix + I y =
2
2
(
)
)
1
1
640 in 4 − 280 in 4 = 180 in 4
Ix − I y =
2
2
2θ = −120°,
I x′y′ = −180 in 4 ,
Also have
(
)
(
Ix > I y
)
Letting the points I x , I xy and I x′ , I x′y′ be denoted by X an X ′, respectively, three possible Mohr’s
circles can be constructed
Assume the first case applies
Then
Also
Ix − I y
2
= R cos 2θ m
I x′y′ = R cos α
R cos 2θ m = 180 in 4
or
or
R cos α = 180 in 4
∴ α = ±2θ m
120° = 2θ m + ( 90° − α )
Also have
∴ α = −2θ m
Note
and
2 ( 2θ m ) = 30°
2θ m − α = 30°
or
or
2θ m = α = 15°
2θ m > 0 
 implies case 2 applies
α < 0
PROBLEM 9.108 CONTINUED
θ m = 7.5° clockwise
(a) Therefore,
(b) Have
Then
R cos15° = 180
I max,
min
or
R = 186.35 in 4
= I ave ± R = 460 ± 186.350
or
I max = 646 in 4
and
I min = 274 in 4
Note: From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.109
It is known that for a given area I y = 300 in 4 and I xy = −125 in 4 ,
where the x and y axes are rectangular centroidal axes. If the axis
corresponding to the maximum product of inertia is obtained by rotating
the x axis 67.5o counterclockwise about C, use Mohr’s circle to
determine (a) the moment of inertia I x of the area, (b) the principal
centroidal moments of inertia.
SOLUTION
Ix > I y
First assume
( )max is
(Note: Assuming I x < I y is not consistent with the requirement that the axis corresponding to the I xy
obtained by rotating the x axis through 67.5° counterclockwise)
2θ m = 2 ( 67.5° ) − 90° = 45°
From Mohr’s circle have
tan 2θ m =
(a) From
Have
Ix = I y + 2
I xy
tan 2θ m
2 I xy
Ix − I y
= 300 in 4 + 2
125 in 4
= 550 in 4
tan 45°
or I x = 550 in 4
(b) Now
and
Then
I ave =
1
550 + 300 4
in = 425 in 4
Ix + I y =
2
2
(
R=
I max,
)
I xy
sin 2θ m
min
=
125 in 4
= 176.78 in 4
sin 45°
= I ave ± R = ( 425 ± 176.76 ) in 4
= ( 601.78, 248.22 ) in 4
or I max = 602 in 4
and I min = 248 in 4
PROBLEM 9.110
Using Mohr’s circle, show that for any regular polygon (such as a
pentagon) (a) the moment of inertia with respect to every axis through the
centroid is the same, (b) the product of inertia with respect to every pair
of rectangular axes through the centroid is zero.
SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y.
Because the y axis is an axis of symmetry, it follows that I xy = 0. Since
I xy = 0, the x and y axes must be principal axes. Assuming
I x = I max and I y = I min , the Mohr’s circle is then drawn as shown.
Now rotate the coordinate axes through an angle α as shown; the
resulting moments of inertia, I x′ and I y′ , and product of inertia, I x′y′ , are
indicated on the Mohr’s circle. However, the x′ axis is an axis of
symmetry, which implies I x′y′ = 0. For this to be possible on the Mohr’s
circle, the radius R must be equal to zero (thus, the circle degenerates into
a point). With R = 0, it immediately follows that
(a)
I x = I y = I x′ = I y′ = I ave (for all moments of
inertia with respect to an axis through C)
(b)
I xy = I x′y′ = 0 (for all products of inertia with respect to
all pairs of rectangular axes with origin at C)
PROBLEM 9.121
The parabolic spandrel shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Using direct integration,
express the moment of inertia of the solid with respect to the x axis in
terms of m and b.
SOLUTION
At
x = a, y = b:
b = ka 2
y =
Then
or
k =
b
a2
b 2
x
a2
( )
dm = ρ π r 2 dx
Now
2
 b

= πρ  2 x 2  dx
a

Then
m = πρ
b2 a 4
∫ x dx
a4 0
a
1
b2
= πρ 4 x5
5
a
0
=
Now
1
πρ ab 2
5
or
πρ =
5m
ab 2
2
2

1 b
1 
   b

d I x =  r 2  dm =  2 x 2  πρ  2 x 2  dx 
2 a
2 
   a


=
5m 1 b 2 4 b 2 4
5 b2 8
×
x
×
x
dx
=
m x dx
2 a9
ab 2 2 a 4
a4
a
Then..
5 b2 a
5 b2 1
I x = m 9 ∫0 x8dx = m 9 × x9
2 a
2 a
9 0
or I x =
5
mb 2
18
PROBLEM 9.122
Determine by direct integration the moment of inertia with respect to the
z axis of the right circular cylinder shown assuming that it has a uniform
density and a mass m.
SOLUTION
m = ρV = ρπ a 2 L
For the cylinder
dm = ρπ a 2dx
For the element shown
=
dI z = dI z + x 2dm
and
=
Then
m
dx
L
1 2
a dm + x 2dm
4
L
L
0 
1 2
1 
 m  m  1
I z = ∫ dI z = ∫
a + x 2  dx  =  a 2 x + x3 
3 0
4
 L  L  4
=
m1 2
1 3
 a L+ L
L4
3 
or I z =
(
1
m 3a 2 + 4 L2
12
)
PROBLEM 9.123
The area shown is revolved about the x axis to form a homogeneous solid
of revolution of mass m. Determine by direct integration the moment of
inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your
answers in terms of m and a.
SOLUTION
2a = k ( 2a )
3
x = 2a
At
y =
Then
or
1
4a 2
1 3
x
4a 2
(
dm = ρ π r 2dx
Now
k =
)
2
πρ 6
 1

= πρ  2 x3  dx =
x dx
16a 4
 4a

πρ
m=
Then
16a 4
=
or πρ =
2a
6
∫a x dx
πρ 1
4
16a 7
2a
x
=
7
a
πρ 
127
7
7
2a ) − ( a )  =
πρ a3
4 (
112a 

112
112m
127a3
2
(a) Now
1 1
1 
  πρ 6 
d I x =  r 2  dm =  2 x3  
x dx 
4
2
2


 4a
  16a

=
x6
1 6 112m
7m
x
dx =
x12dx
×
×
4
3
4
32a
127a
16a
4064a11
Then
2a
Ix =
=
7m
7m 1 13
2 a 12
x dx =
x
11 ∫a
4064a
4064a11 13
a
7m 
57337 2
ma = 1.0853ma 2
( 2a )13 − ( a )13  =
11 

52832
52832a
or I x = 1.085ma 2
PROBLEM 9.123 CONTINUED
2
1  1
 πρ 6
1


dI y =  r 2 + x 2  dm =   2 x3  + x 2 
x dx
4
4


 16a
 4  4a
Have
=
1
112m  1 12

x + x8  dx
×

4
16a
127a3  64a 4

2a
Then
Iy =
7m 2a  1 12
7m  1
1 

x + x8  dx =
x13 + x8 
7 ∫a 
4
7
4
9
127a
127a  832a
 64a

a
=
7m  1
1
1
1
13
9
13
9
2a ) + ( 2a ) −
a − (a) 
7 
4(
4( )
9
9
127a  832a
832a

=
7m  8191 9 511 9 
a +
a  = 3.67211ma 2

9
127a 7  832

or I y = 3.67ma 2
PROBLEM 9.124
Determine by direct integration the moment of inertia with respect to the
x axis of the tetrahedron shown assuming that it has a uniform density and
a mass m.
SOLUTION
Have
x=
a
y

y + a = a 1 + 
h
h


and
z =
b
y

y + b = b 1 + 
h
h

2
y
1
 1

dm = ρ  xzdy  = ρ ab  1 +  dy
h
2
 2

For the element shown
2
1
y

m = ∫ dm = ∫
ρ ab 1 +  dy
2
h

0
−h
Then
0
3
1
h 
y 
= ρ ab × 1 +  
2
3 
h  
−h
Now, for the element
Then
=
1
3
3
ρ abh (1) − (1 − 1) 


6
=
1
ρ abh
6
I AA′,area
1 3
1 3
y
=
xz =
ab 1 + 
36
36
h

4
1
y 

dI AA′, mass = ρ tI AA′, area = ρ ( dy )  ab3 1 +  
h  

 3
4
PROBLEM 9.124 CONTINUED
Now
dI x = dI AA′,mass
 2  1 2 
+  y +  z   dm
 3  

4
y
1

=
ρ ab3 1 +  dy
h
36

2
2

1 
y     1
y


+  y 2 +  b 1 +     ρ ab  1 +  dy 
h     2
h

3 



=
4

y
y3 y 4 
1
1

+ 2  dy
ρ ab3 1 +  dy + ρ ab  y 2 + 2
h
h
12
2
h 


m=
Now
1
ρ abh
6
4
 1 b2 
y
3m  2
y3
y 4 
dI x =  m 1 +  +
+ 2   dy
 y + 2
h
h 
h
h  
 2 h 
Then
and
0
I x = ∫ dI x = ∫−h
4

m  2
y
y3
y 4 
+ 2  dy
b 1 +  + 6  y 2 + 2
2h  
h
h
h  

0
5
1
m  2 h
y
1 y4
y5 
=
+ 2  
b × 1 +  + 6  y 3 +
2h 
5
h
2 h
5h  
3
−h
=

m 1 2
1
1
5
3
1
( −h )4 + 2 ( −h )5  
 b h (1) − 6  ( −h ) +
2h  5
2h
5h
3

or I x =
(
1
m b2 + h2
10
)
PROBLEM 9.125
Determine by direct integration the moment of inertia with respect to the
y axis of the tetrahedron shown assuming that it has a uniform density and
a mass m.
SOLUTION
Have
x=
a
y

y + a = a 1 + 
h
h


and
z =
b
y

y + b = b 1 + 
h
h

2
For the element shown
y
1
 1

dm = ρ  xzdy  = ρ ab  1 +  dy
h
2
 2

2
1
y

m = ∫ dm = ∫
ρ ab 1 +  dy
2
h

0
−h
Then
0
3
1
h 
y 
= ρ ab × 1 +  
2
3 
h  
−h
=
1
3
3
ρ abh (1) − (1 − 1) 


6
=
1
ρ abh
6
I BB′,area =
Also
Then, using
1 3
xz
12
I DD′,area =
1 3
zx
12
I mass = ρ tI area have
 1

dI BB′,mass = ρ ( dy )  xz 3 
 12

 1

dI DD′,mass = ρ ( dy )  zx3 
 12

PROBLEM 9.125 CONTINUED
Now
dI y = dI BB′,mass + dI DD′,mass
Have
(
)
=
1
ρ xz x 2 + z 2 dy
12
=
2
1
y 

ρ ab 1 +   a 2 + b 2
12
h  

m=
(
)
2
y 

1
+

  dy
h  

(
1
m 2
ρ abh ⇒ dI y =
a + b2
6
2h
)
4
y

1 +  dy
h

Then
(
4
)
m 2
y

I y = ∫ dI y = ∫
a + b 2 1 +  dy
2h
h

0
−h
0
5
m 2
h 
y 
2
a + b ×  1 +  
=
2h
5 
h  
−h
=
(
)
(
)
m 2
5
5
a + b 2 (1) − (1 − 1) 


10
or I y =
(
1
m a 2 + b2
10
)
PROBLEM 9.126
Determine by direct integration the moment of inertia with respect to the
z axis of the semiellipsoid shown assuming that it has a uniform density
and a mass m.
SOLUTION
First note that when
1
z = 0:

x2  2
y = b 1 − 2 
a 

y = 0:

x2  2
z = c 1 − 2 
a 

1

x2 
dm = ρ (π yzdx ) = πρ bc 1 − 2  dx
a 

For the element shown
m = ∫ dm =
Then

a
πρ bc 1
0
∫

x2 
 dx
a 2 
−
a
1
2


= πρ bc  x − 2 x3  = πρ abc
3
3a

0
I AA′,area =
For the element
4
zy 3
π

dI AA′,mass = ρ tI AA′,area = ρ ( dx )  zy 3 
4

Then
Now
π
dI z = dI AA′,mass + x 2dm
=
=
π
4

ρ b3c 1 −

2


x2 
x2  
2
πρ
1
dx
x
bc
+
−


 dx 
2

a 
a 2  


3m  b 2 
x2
x4   2 x4 
 1 − 2 2 + 4  +  x − 2   dx
2a  4 
a
a  
a  
PROBLEM 9.126 CONTINUED
Finally
I z = ∫ dI z
=
3m a  b 2 
x2
x4   2 x4 
1
2
−
+


 +  x − 2   dx
∫
2a 0  4 
a 2 a 4  
a  
a
3m  b 2 
2 x3 1 x5   1 3 1 x5  
=
+
  x −
+ x −

2a  4 
3 a 2 5 a 4   3
5 a 2  
0
=
3  b2 
2 1
 1 1 
m  1 − +  + a 2  −  
2 4
3 5
 3 5 
or I z =
(
1
m a 2 + b2
5
)
PROBLEM 9.127
A thin steel wire is bent into the shape shown. Denoting the mass per unit
length of the wire by m ′, determine by direct integration the moment of
inertia of the wire with respect to each of the coordinate axes.
SOLUTION
1
2
2 2
dy
−1
= − x 3  a 3 − x 3 
dx


First note
2
2
2
−2
 dy 
1 +   = 1 + x 3  a 3 − x 3 


 dx 
Then
2
 a 3
= 
x
2
 dy 
dm = m′dL = m′ 1 +   dx
 dx 
For the element shown
1
 a 3
= m′   dx
x
1
Then
m = ∫ dm =
∫ 0a
m′
a3
x
Now
dx =
a  23
a
0
I x = ∫ y dm = ∫
2
1
3

1
2 a
3
3
m′a 3  x 3  = m′a
2
2
 0
1
3
a 3 


− x  m′ 1 dx
  x3 


2
3
5
 a2
1
4 1
2
= m′a 3 ∫ 0a  1 − 3a 3 x 3 + 3a 3 x − x 3  dx
 3

x

a
1 3
2
9 4 4 3 2
3 8
= m′a 3  a 2 x 3 − a 3 x 3 + a 3 x 2 − x 3 
4
2
8 0
2
 3 9 3 3 3
= m′a3  − + −  = m′a3
 2 4 2 8 8
Symmetry implies
or I x =
1 2
ma
4
Iy =
1 2
ma
4
PROBLEM 9.127 CONTINUED
Alternative Solution
I y = ∫ x dm =
2
1
= m′a 3 ×
=
Also
∫ 0a

x
2



5
1
 = m′a 3 ∫ 0a x 3 dx
dx
1
x 3 
1
m′
a3
3  83  a 3
x
= m′a3
8   0 8
1 2
ma
4
(
)
I z = ∫ x 2 + y 2 dm = I y + I x
or I z =
1 2
ma
2
PROBLEM 9.128
A thin triangular plate of mass m is welded along its base AB to a block
as shown. Knowing that the plate forms an angle θ with the y axis,
determine by direct integration the mass moment of inertia of the plate
with respect to (a) the x axis, (b) the y axis, (c) the z axis.
SOLUTION
ζ =−
For line BC
=
h
x+h
a
2
h
( a − 2x )
a
1 
m = ρV = ρ t  ah 
2 
Also
=
1
ρ tah
2
2
(a) Have
dI x =
=
1 2
ζ dm′
3
dm′ = ρ tζ dx
where
Then
1 2
ζ 
ζ dm′ +   dm′
12
2
a 1
I x = ∫ dI x = 2 ∫ 02 ζ 2 ( ρ tζ dx )
3
3
=
a h
2

ρ t ∫ 02  ( a − 2 x )  dx
3
a


a
2 h3 1  1 
4
= ρ t 3 ×  −  ( a − 2 x )  2

0
3 a
4 2
=−
=
1
h3
4
4
ρ t 3 ( a − a ) − ( a ) 

12 a 
1
ρ tah3
12
or I x =
1 2
mh W
6
PROBLEM 9.128 CONTINUED
Iζ = ∫ x 2dm
Now
Iζ = ∫ x 2dm′ = 2 ∫ 02 x 2 ( ρ tζ dx )
a
and
a
h

= 2 ρ t ∫ 02 x 2  ( a − 2 x )  dx
a

a
h a
1 2
= 2 ρ t  x3 − x 4 
a 3
4 0
h a  a 
1a
   −  
a  3  2 
4 2
3
= 2ρt
=
(b) Have
4


1
1
ρ ta3h =
ma 2
48
24
2
I y = ∫ ry2dm = ∫  x 2 + (ζ sin θ )  dm


= ∫ x 2dm + sin 2 θ ∫ ζ 2dm
Now
I x = ∫ ζ 2dm ⇒ I y = Iζ + I x sin 2 θ
=
(c) Have
1
1
ma 2 + mh 2 sin 2 θ
24
6
(
)
or I y =
m 2
a + 4h 2 sin 2 θ W
24
(
)
I z = ∫ rz2dm = ∫ x 2 + y 2 dm
2
= ∫  x 2 + (ζ cosθ )  dm


= ∫ x 2dm + cos 2 θ ∫ ζ 2dm
= Iζ + I x cos 2 θ
=
1
1
ma 2 + mh 2 cos 2 θ
24
6
or I z =
(
)
m 2
a + 4h 2 cos 2 θ W
24
PROBLEM 9.129
Shown is the cross section of a molded flat-belt pulley. Determine its
mass moment of inertia and its radius of gyration with respect to the axis
AA′. (The specific weight of brass is 0.306 lb/in3 and the specific weight
of the fiber-reinforced polycarbonate used is 0.0433 lb/in3.)
SOLUTION
First note for the cylindrical ring shown that
m = ρV = ρ t ×
π
4
(d
2
2
− d12
)
and, using Figure 9.28, that
2
I AA′ =
1  d2 
1 d 
m2   − m1  1 
2  2 
2  2
2
=
1 
π 2 2 
π 2  2
 ρ t × d 2  d 2 −  ρ t × d1  d1 

8 
4 
4  

=
1π  4
4
 ρ t  d 2 − d1
8 4 
=
1π  2
2
2
2
 ρ t  d 2 − d1 d 2 + d1
8 4 
=
1
m d12 + d 22
8
(
(
(
)
)(
)
)
Now treat the pulley as four concentric rings and, working from the brass
outward, have
m=
π  0.306 lb/in 3

4  32.2 ft/s 2
+

π 1.0433 lb/in 3
4
32.2 ft/s

( 0.875 in.) ( 0.55 in.)2 − ( 0.25 in.)2  
2
{(0.875 in.) (0.85 in.)
2
2
+ ( 0.10 in.) (1.1 in.) − ( 0.85 in.)

2
2
+ ( 0.475 in.) (1.4 in.) − (1.1 in.) 


=
π
128.8
2
2
− ( 0.55 in.) 

}
( 0.06426 + 0.01593 + 0.00211 + 0.015426 ) lb ⋅ s 2/ft
PROBLEM 9.129 CONTINUED
Now
(
m = 1567.38 ⋅ 10−6 + 388.553 ⋅ 10−6 + 51.465 ⋅ 10−6
)
+ 376.259 ⋅ 10−6 lb ⋅ s 2 /ft = 2383.657 ⋅ 10−6 lb ⋅ s 2 /ft
Then
I AA′ =
 0.25 2  0.55  2 
1 
−6
2
⋅
⋅
1567.38
10
lb
s
/ft
ft  + 
ft  


8
12
12



 



 0.55 2  0.85 2 
ft  + 
ft  
+ 388.553 ⋅ 10−6 lb ⋅ s 2 /ft 

 12
 
 12
 0.85 2  1.1  2 
ft  + 
ft  
+ 51.465 ⋅ 10−6 lb ⋅ s 2 /ft 
  12  
 12
+ 376.259 ⋅ 10
=
−6
 1.1 2  1.4  2  
lb ⋅ s /ft 
ft  + 
ft   
 12   
 12 
2
1
( 3.9728 + 2.7657 + 0.69067 + 8.2829 ) ⋅ 10−6 lb ⋅ ft ⋅ s2
8
= 1.96401 × 10−6 lb ⋅ ft ⋅ s 2
or I AA′ = 1.964 ⋅10−6 lb ⋅ ft ⋅ s 2 W
and
2
k AA
′ =
I AA′
1.96401 × 10−6 lb ⋅ ft ⋅ s 2
=
m
2383.657 × 10−6 lb ⋅ s 2 /ft
= 8.23947 × 10− 4 ft 2
k AA′ = 2.87044 ⋅ 10− 2 ft = 0.34445 in.
or k AA′ = 0.344 in. W
PROBLEM 9.130
Shown is the cross section of an idler roller. Determine its moment of
inertia and its radius of gyration with respect to the axis AA′. (The
density of bronze is 8580 kg/m3; of aluminum, 2770 kg/m3; and of
neoprene, 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m = ρV = ρ t ×
π
(d
4
2
2
)
− d12 =
π
4
(
ρ t d 22 − d12
)
and, using Figure 9.28, that
2
I AA′
1 d 
1 d 
= m2  2  − m1  1 
2  2 
2  2
2
=
1 
π 2 2 
π 2  2
 ρ t × d 2  d 2 −  ρ t × d1  d1 

8 
4 
4  

=
1π  4
4
 ρ t  d 2 − d1
8 4 
=
1π  2
2
2
2
 ρ t  d 2 − d1 d 2 + d1
8 4 
=
1
m d12 + d 22
8
(
(
(
)
)(
)
)
Now treat the roller as three concentric rings and, working from the
bronze outward, have
Have
m=
(8580 kg/m ) ( 0.0195 m ) ( 0.009 m )
4{
π
3
(
)
(
)
2
2
− ( 0.006 m ) 

2
2
+ 2770 kg/m3 ( 0.0165 m ) ( 0.012 m ) − ( 0.009 m ) 


}
2
2
+ 1250 kg/m3 ( 0.0165 m ) ( 0.027 m ) − ( 0.012 m ) 


π
[7.52895 + 2.87942 + 12.06563] × 10−3 kg
4
= 5.9132 × 10−3 kg + 2.26149 × 10−3 kg
=
+ 9.47632 × 10−3 kg
= 17.6510 × 10−3 kg
PROBLEM 9.130 CONTINUED
And
I AA′ =
{(
)
(
)
(
)
1
2
2
5.9132 × 10−3 kg ( 0.006 ) + ( 0.009 )  m 2


8
2
2
+ 2.26149 × 10−3 kg ( 0.009 ) + ( 0.012 )  m 2


2
2
+ 9.47632 × 10−3 kg ( 0.012 ) + ( 0.027 )  m 2


=
}
1
( 691.844 + 508.835 + 8272.827 )10−9 kg ⋅ m 2
8
= 1.18419 × 10−6 kg ⋅ m 2
or I AA′ = 1.184 × 10−6 kg ⋅ m 2 W
Now
2
k AA
′ =
I AA′ 1.18419 × 10−6 kg m 2
=
m
17.6510 × 10−3 kg
= 67.08902 × 10−6 m 2
k AA′ = 8.19079 × 10−3 m
or k AA′ = 8.19 mm W
PROBLEM 9.131
Given the dimensions and the mass m of the thin conical shell shown,
determine the moment of inertia and the radius of gyration of the shell
with respect to the x axis. (Hint: Assume that the shell was formed by
removing a cone with a circular base of radius a from a cone with a
circular base of radius a + t . In the resulting expressions, neglect terms
containing t2, t3, etc. Do not forget to account for the difference in the
heights of the two cones.)
SOLUTION
First note
h′
h
=
a+t
a
or
h′ =
h
(a + t )
a
For a cone of height H whose base has a radius r, have
Ix =
where
3
mr 2
10
m = ρV
= ρ×
Ix =
Then
π
3
r 2H
3 π
2  2
 ρr H  r
10  3

=
π
10
ρ r 4H
Now, following the hint have
mshell = mouter − minner =
=
=
π
3
π
3


ρ ( a + t ) ×
2

ρ a 2 h  1 +

π
3
ρ ( a + t ) h′ − a 2h 
2


h
( a + t ) − a 2h 
a

3
 π
t
t

2 
 − 1 = ρ a h 1 + 3 + ... − 1
a
3
a



Neglecting the t2 and t3 terms obtain
mshell ≈ πρ aht
PROBLEM 9.131 CONTINUED
Also
( I x )shell = ( I x )outer − ( I x )inner
π
=
10
π
=
10
π
=
10
=
π
10
ρ ( a + t ) h′ − a 4h 
4




ρ ( a + t ) ×
4

ρ a 4h  1 +



h
( a + t ) − a 4h 
a

5

t
 − 1
a

ρ a 4h 1 + 5
t

+ ... − 1
a

Neglecting t2 and higher order terms, obtain
( I x )shell
≈
π
2
ρ a3ht
or I x =
Now
1 2
ma W
2
1 2
ma
I
k x2 = x = 2
m
m
or k x =
a
W
2
PROBLEM 9.132
A portion of an 8-in.-long steel rod of diameter 1.50 in. is turned to form
the conical section shown. Knowing that the turning process reduces the
moment of inertia of the rod with respect to the x axis by 20 percent,
determine the height h of the cone.
SOLUTION
8-in. rod
and
( I x )0 = 12 m0a 2
a = 0.75 in.
where
and
(
m0 = ρV0 = ρ π a 2 L
L = 8 in.
( I x )0 = 12 ρπ a4 L
Therefore,
( )cyl + ( I x )cone = 12 mcyla2 + 103 mconea 2
Rod and cone
Ix = Ix
where
mcyl = ρVcyl = ρ π a 2 ( L − h ) 
and
1

mcone = ρVcone = ρ  π a 2h 
3

Then
Ix =
( )0
1
1
1

ρπ a 4 ( L − h ) + ρπ a 4h = 0.8  ρπ a 4 L 
2
10
2


5
5
1
4
L−
h+
h=
L
10
10
10
10
or
or
1
1
ρπ a 4 ( L − h ) + ρπ a 4h
2
10
I x = 0.8 I x
Given
Then
)
h=
1
1
L = ( 8.00 in.) = 2.00 in.
4
4
or h = 2.00 in. W
PROBLEM 9.133
The steel machine component shown is formed by machining a
hemisphere into the base of a truncated cone. Knowing that the density of
steel is 7850 kg/m3, determine the mass moment of inertia of the
component with respect to the y axis.
SOLUTION
First note
and
Now
L
L − 240
=
120
80
or
L = 720 mm
m = ρ stV
2
1
 π
m1 = ρ st  π a12h1  = × 7850 kg/m3 × ( 0.120 m ) ( 0.720 m )
3
 3
= 85.230 kg
2
2
 2
m2 = ρ st  π a22  = π × 7850 kg/m3 × ( 0.090 m ) = 11.9855 kg
3
3


2
1
 π
m3 = ρ st  π a32h3  = × 7850 kg/m3 × ( 0.080 m ) ( 0.720 − 0.240 ) m
3
3


= 25.253 kg
Now
( )1 − ( I y )2 − ( I y )3
Iy = Iy
PROBLEM 9.133 CONTINUED
where (using Figure 9.28)
( I y )1 = 103 m1a12 = 103 (85.230 kg )( 0.120 m )2 = 0.36819 kg ⋅ m2
( I y )2 = 12 ( I y )sphere = 12  52 mspherea22 
=
where
msphere = 2mhemisphere
12
2

 2 × 11.9855 kg  ( 0.090 m )
25

= 0.038833 kg ⋅ m 2
( I y )3 = 103 m3a32 = 103 ( 25.253 kg )( 0.080 m )2 = 0.048486 kg ⋅ m2
Then
I y = ( 0.36819 − 0.038833 − 0.048486 ) kg ⋅ m 2 = 0.28087 kg ⋅ m 2
= 0.281 kg ⋅ m 2
or I y = 281 × 10−3 kg ⋅ m 2 W
PROBLEM 9.134
After a period of use, one of the blades of a shredder has been worn to the
shape shown and is of weight 0.4 lb. Knowing that the moments of inertia
of the blade with respect to the AA′ and BB′ axes are
0.6 × 10−3 lb ⋅ ft ⋅ s 2 and 1.26 × 10−3 lb ⋅ ft ⋅ s 2 , respectively, determine
(a) the location of the centroidal axis GG′, (b) the radius of gyration with
respect to axis GG′.
SOLUTION
Have
dB =
(a)
and
4
− d A = ( 0.33333 − d A ) ft
12
I AA′ = I GG′ + md A2
I BB′ = I GG′ + md B2
Then
(
I BB′ − I AA′ = m d B2 − d A2
)
)
2
= m ( 0.33333 − d A ) −d A2 


= m ( 0.11111 − 0.66666d A )
Then
(1.26 − 0.6 ) × 10−3 lb ⋅ ft ⋅ s2
=
or
0.40 lb
( 0.11111 − 0.66666d A ) ft 2
32.2 ft/s 2
d A = 0.08697 ft
d A = 1.044 in. W
or
I AA′ = I GG′ + md A2
(b)
or
I GG′ = 0.6 × 10−3 lb ⋅ ft ⋅ s 2
−
0.4 lb
( 0.08697 ft )2
32.2 ft /s 2
= 0.50604 × 10−3 lb ⋅ ft ⋅ s 2
Then
2
kGG
′ =
I GG′
0.50604 × 10−3 lb ⋅ ft ⋅ s 2
=
0.4 lb
m
32.2 ft/s 2
= 0.04074 ft 2
kGG′ = 0.20183 ft = 2.4219 in.
or kGG′ = 2.42 in. W
PROBLEM 9.135
The cups and the arms of an anemometer are fabricated from a material
of density ρ. Knowing that the moment of inertia of a thin, hemispherical
shell of mass m and thickness t with respect to its centroidal axis GG′, is
5ma2/12, determine (a) the moment of inertia of the anemometer with
respect to the axis AA′, (b) the ratio of a to l for which the centroidal
moment of inertia of the cups is equal to 1 percent of the moment of
inertia of the cups with respect to the axis AA′.
SOLUTION
marm = ρVarm = ρ ×
(a) First note
π
4
d 2l
dmcup = ρ dVcup
and
= ρ ( 2π a cosθ )( t )( adθ ) 
π
mcup = ∫ dmcup = ∫ 02 2πρ a 2t cosθ dθ
Then
π
= 2πρ a 2t [sin θ ]02
= 2πρ a 2t
( I AA′ )anem. = ( I AA′ )cups + ( I AA′ )arms
Now
Using the parallel-axis theorem and assuming the arms are slender
rods, have
( I AA′ )anem.
2 
= 3 ( I GG′ )cup + mcup d AG


+ 3  I arm + marm d AGarm 
2
 5

2
 a   
= 3  mcup a 2 + mcup ( l + a ) +    
 2   
12

2
1
l 
2
+3  marml + marm   
 2  
 2
5

= 3mcup  a 2 + 2la + l 2  + marml 2
3

5
 π

= 3 2πρ a 2t  a 2 + 2la + l 2  +  ρ d 2l  l 2
3
 4

(
or
( I AA′ )anem
)
( )

 5 a2
 d 2l 
a
= πρ l 2 6a 2t  2 + 2 + 1 +
W
l
4 
3 l


PROBLEM 9.135 CONTINUED
( IGG′ )cup
( I AA′ )cup
(b) Have
or
= 0.01
5
5

mcup a 2 = 0.01mcup  a 2 + 2la + l 2 
12
3


Now let ζ =
( From Part a )
a
l
Then
5

5ζ 2 = 0.12  ζ 2 + 2ζ + 1
3

or
40ζ 2 − 2ζ − 1 = 0
Then
or
ζ =
2±
ζ = 0.1851
( −2 )2 − 4 ( 40 )( −1)
2 ( 40 )
and
ζ = − 0.1351
∴
a
= 0.1851 W
l
PROBLEM 9.136
A square hole is centered in and extends through the aluminum machine
component shown. Determine (a) the value of a for which the mass
moment of inertia of the component with respect to the axis AA′, which
bisects the top surface of the hole, is maximum, (b) the corresponding
values of the mass moment of inertia and the radius of gyration with
respect to the axis AA′. (The density of aluminum is 2800 kg / m3 .)
SOLUTION
First note
m1 = ρV1 = ρ b 2 L
And
m2 = ρV2 = ρ a 2 L
(a) Using Figure 9.28 and the parallel-axis theorem, have
I AA′ = ( I AA′ )1 − ( I AA′ )2
2
1
a 
=  m1 b 2 + b 2 + m1   
 2  
12
(
)
2
1
a 
−  m2 a 2 + a 2 + m2   
 2  
12
(
(
)
)
(
)
1 
1
 5 
= ρ b2 L  b2 + a 2  − ρ a 2 L  a 2 
4 
6
 12 
=
ρL
( 2b
12
4
+ 3b 2a 2 − 5a 4
)
dI AA′
ρL
=
6b 2a − 20a3 = 0
da
12
(
Then
a=0
or
)
a=b
and
3
10
d 2 I AA′
ρL
1
6b 2 − 60a 2 = ρ L b 2 − 10a 2
=
2
12
2
da
Also..
(
)
(
and for
a=b
)
Now, for
a = 0,
∴
d 2 I AA′
>0
da 2
( I AA′ )max
occurs when
a = 84
or
a=b
3 d 2 I AA′
,
<0
10
da 2
3
10
3
= 46.009 mm
10
a = 46.0 mm
PROBLEM 9.136 CONTINUED
(b) From part (a)
( I AA′ )mass
ρ L 
2
4

3 
3  
2b + 3b  b
=
 − 5  b

12 
10 
10  




=
2
4
(
)
49
49
4
ρ Lb 4 =
2800 kg/m3 ( 0.3 m )( 0.4 m )
240
240
= 8.5385 × 10−3 kg ⋅ m 2
or
( I AA′ )mass
2
k AA
′ =
and
= 8.54 × 10−3 kg ⋅ m 2
( I AA′ )mass
m
where
(
m = m1 − m2 = ρ L b − a
2
2
)
2


 
3
7
2
ρ Lb 2
= ρ L b −  b
  =


10
10

 

Then
2
k AA
′
49
ρ Lb 4
7 2
7
240
=
=
b =
(84 mm )2 = 2058 mm 2
7
24
24
2
ρ Lb
10
k AA′ = 45.3652 mm
or k AA′ = 45.4 mm
To the instructor:
The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at
this time for use in the solutions of Problems 9.137–9.142
Thin rectangular plate
( I x )m
( )m + md 2
= I x′
=
 b  2  h  2 
1
m b 2 + h 2 + m   +   
12
 2  
 2 
=
1
m b2 + h2
3
(
)
(
)
( I y )m = ( I y′ )m + md 2
=
1
b
mb 2 + m  
12
2
=
1 2
mb
3
2
( )m + md 2
I z = I z′
=
1
h
mh 2 + m  
12
2
=
1 2
mh
3
2
Thin triangular plate
Have
1

m = ρV = ρ  bht 
2

and
I z , area =
Then
I z , mass = ρ tI z , area
1 3
bh
36
= ρt ×
=
1 3
bh
36
1
mh 2
18
1
mb 2
18
1
+ I z , mass =
m b2 + h2
18
I y, mass =
Similarly,
I x, mass = I y, mass
Now
(
)
Thin semicircular plate
Have
π

m = ρV = ρ  a 2t 
2


And
I y, area = I z, area =
Then
π
8
a4
I y, mass = I z, mass = ρ tI y, area
= ρt ×
π
8
a4
1 2
ma
4
=
I x, mass = I y, mass + I z , mass =
Now
1 2
ma
2
Also
I x, mass = I x′, mass + my 2
or
16  2
1
I x′, mass = m  −
a
2
 2 9π 
And
I z , mass = I z′, mass + my 2
or
16  2
1
I z′, mass = m  −
a
2
 4 9π 
Thin Quarter-Circular Plate
y = z =
Have
π

m = ρV = ρ  a 2t 
4


and
I y, area = I z , area =
Then
I y, mass = I z, mass = ρ tI y, area
π
16
a4
= ρt ×
=
π
16
1 2
ma
4
a4
4a
3π
Now
Also
or
and
or
I x, mass = I y, mass + I z , mass =
1 2
ma
2
(
I x, mass = I x′, mass + m y 2 + z 2
32  2
1
I x′, mass = m  −
a
2
 2 9π 
I y, mass = I y′, mass + mz 2
16  2
1
I y′, mass = m  −
a
2
 4 9π 
)
PROBLEM 9.137
A 0.1-in-thick piece of sheet metal is cut and bent into the machine
component shown. Knowing that the specific weight of steel is
0.284 lb / in 3 , determine the moment of inertia of the component with
respect to each of the coordinate axes.
SOLUTION
m1 = ρV =
=
γ
g
tA
0.284 lb/in 3
( 0.10 in.)( 38 in.)( 24 in.)
32.2 ft/s 2
= 0.80438 lb ⋅ s 2 /ft
m2 =
 1

0.284
( 0.1 in.)   ( 38 in.)( 24 in.)
32.2
 2 

= 0.402186 lb ⋅ s 2 /ft
m3 =
 π

0.284
( 0.1 in.)   ( 24 in.)2  = 0.39900 lb ⋅ s 2/ft
32.2
 4 

Using Fig. 9.28 for component 1 and the equations derived for components 2 and 3, have
I x = ( I x )1 + ( I x )2 + ( I x )3
 1  24 2  24  2  2
2
= 0.80438 lb ⋅ s 2 /ft    + 
  ft + 0.402186 lb ⋅ s /ft
12
12
2
×
12






(
)
(
)
 1  24  2  24  2  2
   +
  ft
 3 × 12  
18  12 
 1  24 2 
+ 0.39900 lb ⋅ s 2 /ft     ft 2
 2  12  
(
)
= (1.0725 + 0.26812 + 0.7980 ) lb ⋅ ft ⋅ s 2
= 2.13862 lb ⋅ ft ⋅ s 2
or I x = 2.14 lb ⋅ ft ⋅ s 2
PROBLEM 9.137 CONTINUED
Also
( )1 + ( I y )2 + ( I y )3
Iy = Iy
 1  38 2  24 2 
 38  2  24  2  2 
= 0.80438 lb ⋅ s 2 /ft    +    ft 2 + 
 +
  ft 
 12  
 2 × 12   
 2 × 12 
12  12 
(
)
 1  38 2  38 2  2
+ 0.402186 lb ⋅ s 2/ft    + 
  ft
 3 × 12  
18  12 
(
)
 1  24 2 
+ 0.39900 lb ⋅ s 2 /ft     ft 2
 4  12  
(
)
= ( 3.76122 + 0.672172 + 0.3990 ) lb ⋅ ft ⋅ s 2
= 4.83234 lb ⋅ ft ⋅ s 2
or I y = 4.83 lb ⋅ ft ⋅ s 2
I z = ( I z )1 + ( I z )2 + ( I z )3
 1  38 2  38  2  2
= 0.80438 lb ⋅ s 2 /ft    + 
  ft
 2 × 12  
12  12 
(
)
 1  38 2  24  2   38  2  24  2  
2
+ 0.401286 lb ⋅ s 2/ft    +    + 
 +
   ft
×
×
18
12
12
3
12
3
12












 
 
 
(
)
 1  24 2 
+ 0.3990 lb ⋅ s 2 /ft     ft 2
 4  12  
(
)
= ( 2.68871 + 0.940296 + 0.3990 ) lb ⋅ ft ⋅ s 2
= 4.0280 lb ⋅ ft ⋅ s 2
or I z = 4.03 lb ⋅ ft ⋅ s 2
PROBLEM 9.138
A 3-mm-thick piece of sheet metal is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg / m3 ,
determine the moment of inertia of the component with respect to each of
the coordinate axes.
SOLUTION
First compute the mass of each component.
Have m = ρ stV = ρ st tA
(
)
m1 = 7850 kg/m3 ( 0.003 m )( 0.70 m )( 0.780 m ) = 12.858 kg
 π 
2
m2 = 7850 kg/m3 ( 0.003 m )   ( 0.39 m )  = 5.6265 kg
2
 

(
)
 1 

m3 = 7850 kg/m3 ( 0.003 m )   ( 0.780 m )( 0.3 m )  = 2.7554 kg
 2 

(
)
Using Fig. 9-28 for component 1 and the equations derived above for components 2 and 3 have
I x = ( I x )1 + ( I x )2 + ( I x )3
2
1
2
 0.78   2
= (12.858 kg )  ( 0.78 ) + 
 m
 2  
12
2

 1
16 
2  4 × 0.39 
2  2
+ ( 5.6265 kg )  −
+
+
0.39
0.39
m


(
)
(
)



2
 2 9π 
 3π 
 
2
2
2
2   0.78 
1 
 0.30    2
+ ( 2.7554 kg )  ( 0.78 ) + ( 0.30 )  + 
+
  m
  3 
18 
 3   

= ( 2.6076 + 1.2836 + 0.3207 ) kg ⋅ m 2
= 4.2119 kg ⋅ m 2
or I x = 4.21 kg ⋅ m 2
PROBLEM 9.138 CONTINUED
And
( )1 + ( I y )2 + ( I y )3
Iy = Iy
1
2
2
2
2 
1
= (12.858 kg )  ( 0.7 ) + ( 0.78 )  + ( 0.7 ) + ( 0.78 )   m 2




12
4

2
2
1
+ ( 5.6265 kg )  ( 0.39 ) + ( 0.39 )  m 2
4


2
 1

2
2
 0.78    4
+ ( 2.7554 kg )  ( 0.78 ) + ( 0.7 ) + 
  m
18
 3   


= ( 4.7077 + 1.0697 + 1.6295 ) kg ⋅ m 2
= 7.4069 kg ⋅ m 2
or I y = 7.41 kg ⋅ m 2
And
I z = ( I z )1 + ( I z )2 + ( I z )3

1 
2 1
= (12.858 kg ) ( 0.7 ) 
+  m2
 12 4  

2
1
+ ( 5.6265 kg )  ( 0.39 )  m 2
4

2
 1

2
2
 0.30    2
+ ( 2.7554 kg )  ( 0.3) + ( 0.70 ) + 
  m
 3   

18
= ( 2.1001 + 0.21395 + 1.39145 ) kg ⋅ m 2 = 3.7055 kg ⋅ m 2
or I z = 3.71 kg ⋅ m 2
PROBLEM 9.139
The cover of an electronic device is formed from sheet aluminum that is
2 mm thick. Determine the mass moment of inertia of the cover with
respect to each of the coordinate axes. (The density of aluminum is
2770 kg / m3 .)
SOLUTION
m = ρV
Have
(
)
m1 = 2770 kg/m3 ( 0.002 m )( 0.048 m )( 0.06 m )
Now
= 0.015955 kg
(
)
m2 = 2770 kg/m3 ( 0.002 m )( 0.06 m )( 0.124 m )
= 0.041218 kg
(
)
m3 = 2770 kg/m3 ( 0.002 m )( 0.048 m )( 0.124 m )
= 0.032974 kg
Using Fig. 9.28 and the parallel axis theorem, have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
and
( I x )3 = ( I x ) 4


1 
1 
2 1
2 1
= ( 0.015955 kg ) ( 0.048 ) 
+   m 2 + ( 0.041218 kg ) ( 0.124 ) 
+  m2
 12 4  
 12 4  



1 
2
2  1
+ 2 ( 0.032974 kg )  ( 0.048 ) + ( 0.124 )  
+  m2


 12 4  

(
)
= 0.012253 × 10−3 + 0.211256 × 10−3 + 0.388654 × 10−3 kg ⋅ m 2 = 0.61216 × 10−3 kg ⋅ m 2
or I x = 0.612 × 10−3 kg ⋅ m 2 W
PROBLEM 9.139 CONTINUED
( )1 + ( I y )2 + ( I y )3 + ( I y )4
Iy = Iy

2
2
2  1 
1
= ( 0.015955 kg )  ( 0.06 )  m 2 + ( 0.041218 kg )  ( 0.06 ) + ( 0.124 )     m 2

  3 
3




1
2  1 
2
2
2
1
+ ( 0.032974 kg ) ( 0.124 )    m 2 + ( 0.032974 kg )  ( 0.124 ) + ( 0.06 ) + ( 0.124 )  m 2
3
12
4
 



= ( 0.019146 + 0.260718 + 0.16900 + 0.287709 ) × 10−3  kg ⋅ m 2 = 0.73657 × 10−3 kg ⋅ m 2
or I y = 0.737 × 10−3 kg ⋅ m 2 W
I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4


2
2  1 
2  1 
= ( 0.015955 kg )  ( 0.06 ) + ( 0.048 )     m 2 + ( 0.041218 kg ) ( 0.06 )    m 2


 3 
 3 



 1 
2  1 
2
2
+ ( 0.032974 kg ) ( 0.048 )    m 2 + ( 0.032974 kg )   ( 0.048 ) + ( 0.06 )  m 2
3
3
 

 

= ( 0.031399 + 0.049462 + 0.025324 + 0.14403) × 10−3  kg ⋅ m 2
= 0.250215 × 103 kg ⋅ m 2
or I z = 0.250 × 10−3 kg ⋅ m 2 W
PROBLEM 9.140
A framing anchor is formed of 2-mm-thick galvanized steel. Determine
the mass moment of inertia of the anchor with respect to each of the
coordinate exes. (The density of galvanized steel is 7530 kg/m3.)
SOLUTION
First compute the mass of each component
m = ρV = ρ At
Have
Now
(
)
= ( 7530 kg/m ) ( 0.002 m )( 0.045 m )( 0.020 m ) = 0.013554 kg
m1 = 7530 kg/m3 ( 0.002 m )( 0.045 m )( 0.070 m ) = 0.047439 kg
m2
3
(
)
m3 = 7530 kg/m3 ( 0.002 m ) ×
1
( 0.04 m )( 0.095 m ) = 0.028614 kg
2
Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3, have
I x = ( I x )1 + ( I x )2 + ( I x )3
2
2
2
2 
1
1
= ( 0.047439 kg )  ( 0.07 )  m 2 + ( 0.013554 kg )  ( 0.020 ) + ( 0.07 ) + ( 0.01)   m 2


3

12
1
2
2
2
2 
1
+ ( 0.028614 kg )  ( 0.095 ) + ( 0.04 )  + ( 2 × 0.095 ) + ( 0.040 )   m 2
 9

18 
= ( 0.077484 + 0.068222 + 0.136751) × 10−3  kg ⋅ m 2 = 0.282457 × 10 −3 kg ⋅ m 2
or I x = 0.2825 × 10−3 kg ⋅ m 2 W
PROBLEM 9.140 CONTINUED
( )1 + ( I y )2 + ( I y )3
Iy = Iy

2
2
2  1 
1
= ( 0.047439 kg )  ( 0.045 )  m 2 + ( 0.013554 kg )  ( 0.045 ) + ( 0.02 )     m 2

  3 
3



1
2
2
2
1
+ ( 0.028614 kg )  ( 0.04 ) ( 0.045 ) + ( 0.04 )  m 2
9
18

= ( 0.03202 + 0.010956 + 0.065574 ) × 10−3  kg ⋅ m 2 = 0.10855 × 103 kg ⋅ m 2
or I y = 0.1086 × 103 kg ⋅ m 2 W
I z = ( I z )1 + ( I z )2 + ( I z )3


2
2  1 
21
2
= ( 0.047439 kg )  ( 0.045 ) + ( 0.070 )     m 2 + ( 0.013554 kg ) ( 0.045 )   + ( 0.070 )  m 2
  3 
 3



2
1
2
2
  2
2
+ ( 0.028614 kg )  ( 0.095 ) + 0.045 +  0.095   m
3
 
18
= ( 0.0109505 + 0.075564 + 0.187064 ) × 10−3  kg ⋅ m3
= 0.37213 × 10−3 kg ⋅ m 2
or I z = 0.372 × 10−3 kg ⋅ m 2 W
PROBLEM 9.141
A 2-mm-thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the moment of inertia of the component with respect to each of
the coordinate axes.
SOLUTION
m = ρstV = ρsttA
Have
Then
m1 = 7850 kg/m3 × 0.002 m × ( 0.320 × 0.240 ) m 2
= 1.20576 kg
m2 = 7850 kg/m3 × 0.002 m × ( 0.320 × 0.180 ) m 2
= 0.90432 kg
1

m3 = 7850 kg/m3 × 0.002 m ×  × 0.180 × 0.240  m 2
2


= 0.33912 kg
2
π
m4 = 7850 kg/m3 × 0.002 m ×  ( 0.100 m ) 
2

= 0.24662 kg
Using Fig. 9-2B for components 1 and 2 and the equations derived above
for components 3 and 4, have
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3 − ( I x )4
1
3
( I x )1 = (1.20576 kg )( 0.240 m )2
= 23.151 × 10−3 kg ⋅ m 2
( I x )2
=
1
( 0.90432 kg )( 0.180 m )2
3
= 9.7667 × 10−3 kg ⋅ m 2
( I x )3
=
1
( 0.33912 kg ) ( 0.180 )2 + ( 0.240 )2  m 2
6
= 5.0868 × 10−3 kg ⋅ m 2
PROBLEM 9.141 CONTINUED
( I x )4
=
1
( 0.24662 kg )( 0.100 m )2
4
= 0.61655 × 10−3 kg ⋅ m 2
Then
I x = ( 23.151 + 9.7667 + 5.0868 − 0.61655 ) × 10−3  kg ⋅ m 2
or I x = 37.4 × 10−3 kg ⋅ m 2
( )1 + ( I y )2 + ( I y )3 − ( I y )4
Iy = Iy
And
( I y )1 = 13 (1.20576 kg ) ( 0.320)2 + ( 0.240 )2  m2
where
= 64.307 × 10−3 kg ⋅ m 2
( I y )2 = 13 ( 0.90432 kg )( 0.320 m )2
= 30.867 × 10−3 kg ⋅ m 2
( I y )3 = 16 ( 0.33912 kg )( 0.240 m )2
= 3.2556 × 10−3 kg ⋅ m 2

( I y )4 = ( 0.24662 kg )  12 − 916π 2  ( 0.100 m )2 


2

2
 4 × 0.100   2 
+ ( 0.160 ) + 
 m 
3π

 


= 7.5466 × 10−3 kg ⋅ m 2
Then
I y = ( 64.307 + 30.067 + 3.2556 − 7.5466 )10−3  kg ⋅ m 2
or I y = 90.9 × 10−3 kg ⋅ m 2
I z = ( I z )1 + ( I z )2 + ( I z )3 − ( I z )4
And
where
1
3
( I z )1 = (1.20576 kg )( 0.320 m )2
( I z )2
=
= 41.157 × 10−3 kg ⋅ m 2
1
( 0.90432 kg ) ( 0.320 )2 + ( 0.180 )2  m 2
3
= 40.634 × 10−3 kg ⋅ m 2
( I z )3
=
1
( 0.33912 kg )( 0.180 m )2 = 1.83125 × 10−3 kg ⋅ m2
6
PROBLEM 9.141 CONTINUED
 1
 4

( I z )4 = ( 0.24662 kg )   ( 0.100 m )2  + ( 0.160 m )2  


= 6.9300 × 10−3 kg ⋅ m 2
Then
( I z )z
= ( 41.157 + 40.634 + 1.83125 − 6.9300 ) × 10−3  kg ⋅ m 2
or I z = 76.7 × 10−3 kg ⋅ m 2
PROBLEM 9.142
The piece of roof flashing shown is formed from sheet copper that is
0.032 in. thick. Knowing that the specific weight of copper is 558 lb/ft3,
determine the mass moment of inertia of the flashing with respect to each
of the coordinate axes.
SOLUTION
Have
m = ρcopperV
=
Then
γ copper
g
tA
m1 = m2
=
558 lb/ft 3
 1 ft 
× 0.032 in. × ( 48 × 6 ) in 2 × 

2
32.2 ft/s
 12 in. 
3
= 0.092422 lb ⋅ s 2 /ft
Using Fig. 9-2B for components 1 and 2, have
Now
Then
I x = ( I x )1 + ( I x )2
and
( I x )1 = ( I x )2
2
1
 6  
ft   = 1.54037 lb ⋅ ft ⋅ s 2
I x = 2  0.092422 lb ⋅ s 2 /ft 
 12  
 3
(
)
or I x = 1.54 × 10−3 lb ⋅ ft ⋅ s 2
( )1 + ( I y )2
Iy = Iy

where
48 
( I y )1 = 13 ( 0.092422 lb ⋅s2/ft )  12


2

= 500.62 × 10−3 lb ⋅ ft ⋅ s 2
and
( I y )2 = ∫ ry2dm
where
ry2 = x 2 + z 2
= x 2 + (ζ cos30 )
2
2
 6 
+    ft 2
 12  
PROBLEM 9.142 CONTINUED
dm = ρ dV =
and
Then
( I y )2 =
=
γ copper
g
γ copper
g
γ copper
g
tdζ dx
(
)
t ∫ 0L ∫ 0a x 2 + ζ 2 cos 2 30° dζ dx
1


t ∫ 0L  ax 2 + a3 cos 2 30°  dx
3


=
1 γ copper
t aL3 + a3 L cos 2 30°
3 g
=
1
m2 L2 + a 2 cos 2 30°
3
=
1
0.092422 lb ⋅ s 2 /ft
3
(
(
(
)
A = aL
)

) ( 4)
2

1

+  cos 30° 
2

2
 ft 2

= 498.69 × 10−3 lb ⋅ ft ⋅ s 2
Finally,
I y = ( 500.62 + 498.69 ) × 10−3 lb ⋅ ft ⋅ s 2
or I y = 999 × 10−3 lb ⋅ ft ⋅ s 2
I z = ( I z )1 + ( I z )2
where
( I z )1 =
(
)
1
 48 
0.092422 lb ⋅ s 2/ft 
ft 
3
 12 
= 492.92 × 10−3 lb ⋅ ft ⋅ s 2
and
( I z )2
where
rz2 = x 2 + y 2
and
y = ζ sin 30°
Then
( I z )2
= ∫ rz2dm
2
= ∫  x 2 + (ζ sin 30° )  dm


2
PROBLEM 9.142 CONTINUED
( )2
Similarly, as I y
( I z )2
=
(
1
m2 L2 + a 2 sin 2 30°
3
(
1
= 0.092422 lb ⋅ s 2 /ft
3
)
)
 2  1 2 2

( 4 ) +   sin 30°  ft 2
2


= 494.84 × 10−3 lb ⋅ ft ⋅ s 2
Then
I x = ( 492.92 + 494.84 ) × 10−3 lb ⋅ ft ⋅ s 2
or I z = 988 × 10−3 lb ⋅ ft ⋅ s 2
PROBLEM 9.143
The machine element shown is fabricated from steel. Determine the mass
moment of inertia of the assembly with respect to (a) the x axis, (b) the y
axis, (c) the z axis. (The specific weight of steel is 0.284 lb / in 3. )
SOLUTION
m = ρV =
Have
γ
g
V =