Vector Mechanics For Engineers - Statics 8th Edition CH 2 Solutions
Vector Mechanics For Engineers - Statics 8th Edition CH 2 Solutions
Vector Mechanics For Engineers - Statics 8th Edition CH 2 Solutions
Chapter 2, Solution 1.
(a)
(b)
R = 37 lb 76° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Chapter 2, Solution 2.
(a)
(b)
R = 57 lb 86° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 3.
We measure:
R = 10.5 kN
α = 22.5°
R = 10.5 kN 22.5° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 4.
We measure:
R = 5.4 kN α = 12°
R = 5.4 kN 12° !
We measure:
R = 5.4 kN α = 12°
R = 5.4 kN 12° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 5.
β = 32.028°
α + β + 45° = 180°
α = 103.0° !
(b) Using the Law of Sines
Fbb′ 200 N
=
sin α sin 45°
Fbb′ = 276 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 6.
α = 25.104°
or α = 25.1° !
(b) β + 45° + 25.104° = 180°
β = 109.896°
Using the Law of Sines
Faa′ 200 N
=
sin β sin 45°
Faa′ 200 N
=
sin109.896° sin 45°
or Faa′ = 266 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 7.
or R = 1390.57 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Chapter 2, Solution 8.
F2 R 30
= =
sin α sin 38° sin β
Then:
F2 R 30 lb
= =
sin 62° sin 38° sin 80°
or (a) F2 = 26.9 lb !
(b) R = 18.75 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 9.
F1 R 20 lb
= =
sin α sin 38° sin β
Then:
F1 R 20 lb
= =
sin 80° sin 38° sin 62°
or (a) F1 = 22.3 lb !
(b) R = 13.95 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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60 N 80 N
Using the Law of Sines: =
sin α sin10°
or α = 7.4832°
= 162.517°
Then:
R 80 N
=
sin162.517° sin10°
or R = 138.405 N
(a) α = 7.48° !
(b) R = 138.4 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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= 120°
P R 80 lb
Then: = =
sin 35° sin120° sin 25°
or (a) P = 108.6 lb !
(b) R = 163.9 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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80 lb 70 lb
(a) Have: =
sin α sin 35°
sin α = 0.65552
α = 40.959°
or α = 41.0° !
= 104.041°
R 70 lb
Then: =
sin104.041° sin 35°
or R = 118.4 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Then:
(a) P = ( 80 lb ) sin 35°
or P = 45.9 lb !
And:
(b) R = ( 80 lb ) cos 35°
or R = 65.5 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Using the force triangle and the Laws of Cosines and Sines
We have:
= 135°
= 1380.33 lb2
or R = 37.153 lb
and
25 lb 37.153 lb
=
sin β sin135°
25 lb
sin β = sin135°
37.153 lb
= 0.47581
β = 28.412°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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or R = 56.609 lb
56.609 lb 15 lb
=
sin135° sinθ
or θ = 10.7991°
R = 56.6 lb 85.8° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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R 2 = ( 5 kN ) + ( 8 kN ) − 2 ( 5 kN )( 8 kN ) cos105°
2 2
or R = 10.4740 kN
or β = 47.542°
α = 22.542°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Using the force triangle and the Laws of Cosines and Sines
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2 2
Then:
= 1710.42 kN 2
R = 41.357 kN
and
20 kN 41.357 kN
=
sin α sin110°
20 kN
sin α = sin110°
41.357 kN
= 0.45443
α = 27.028°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Using the force triangle and the Laws of Cosines and Sines
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2 2
Then:
= 1710.42 kN 2
R = 41.357 kN
and
20 kN 41.357 kN
=
sin α sin110°
20 kN
sin α = sin110°
41.357 kN
= 0.45443
α = 27.028°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Using the force triangle and the Laws of Cosines and Sines
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2 2
Then:
= 1710.42 kN 2
R = 41.357 kN
and
30 kN 41.357 kN
=
sin α sin110°
30 kN
sin α = sin110°
41.357 kN
= 0.68164
α = 42.972°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Fx = 1.543 kN
Fy = 1.839 kN
Fx = 1.738 kN
Fy = 0.633 kN
Fx = 1.147 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
or Fx = 3.83 kips
or Fy = 3.21 kips
or Fx = −2.39 kips
or Fy = 6.58 kips
or Fx = −8.46 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Fx = − 320 N !
Fy = 680 N
( 300 mm )
340 mm
Fy = 600 N !
Fx = 360 N !
Fy = 390 N
( 250 mm )
650 mm
Fy = 150 N !
Fx = 600 N !
Fy = 610 N
( −110 mm )
610 mm
Fy = −110 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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OB = ( 56 )2 + ( 90 )2 = 106 in.
Then:
204 lb Force:
48
Fx = − ( 204 lb ) , Fx = −96.0 lb
102
90
Fy = + ( 204 lb ) , Fy = 180.0 lb
102
212 lb Force:
56
Fx = + ( 212 lb ) , Fx = 112.0 lb
106
90
Fy = + ( 212 lb ) , Fy = 180.0 lb
106
400 lb Force:
80
Fx = − ( 400 lb ) , Fx = −320 lb
100
60
Fy = − ( 400 lb ) , Fy = −240 lb
100
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Py
(a) P=
sin 35°
960 N
=
sin 35°
or P = 1674 N
Py
(b) Px =
tan 35°
960 N
=
tan 35°
or Px = 1371 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Px
(a) P=
cos 40°
30 lb
P=
cos 40°
or P = 39.2 lb !
Py = ( 30 lb ) tan 40°
or Py = 25.2 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
(a) Py = 100 N
Py
P=
sin 75°
100 N
P=
sin 75°
or P = 103.5 N "
Py
(b) Px =
tan 75°
100 N
Px =
tan 75°
or Px = 26.8 N "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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°
260 lb
=
sin 50°
= 339.40 lb P = 339 lb !
Px
Py =
tan 50°
260 lb
=
tan 50°
= 218.16 lb Py = 218 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
45 N
(a) P=
cos 20°
or P = 47.9 N !
or Px = 16.38 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
18 N
(a) P=
sin 20°
or P = 52.6 N !
18 N
(b) Py =
tan 20°
or Py = 49.5 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
R = ΣF = ( 4.428 kN ) i + (1.669 kN ) j
R= ( 4.428 kN )2 + (1.669 kN )2
= 4.7321 kN
1.669 kN
tan α =
4.428 kN
α = 20.652°
R = 4.73 kN 20.6° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
12.87
α = tan −1 = 61.4°
− 7.02
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
R = ΣF = − ( 256 lb ) i + ( 30 lb ) j
R= ( − 256 lb )2 + ( 30 lb )2
= 257.75 lb
30 lb
tan α =
−256 lb
α = − 6.6839°
R = 258 lb 6.68° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
R = ΣF = ( 640 N ) i + ( 640 N ) j
R= ( 640 N )2 + ( 640 N )2
= 905.097 N
640 N
tan α =
640 N
α = 45.0°
R = 905 N 45.0° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Cable BC Force:
84
Fx = − (145 lb ) = −105 lb
116
80
Fy = (145 lb ) = 100 lb
116
100-lb Force:
3
Fx = − (100 lb ) = −60 lb
5
4
Fy = − (100 lb ) = −80 lb
5
156-lb Force:
12
Fx = (156 lb ) = 144 lb
13
5
Fy = − (156 lb ) = −60 lb
13
and
Rx = ΣFx = −21 lb, Ry = ΣFy = −40 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
or α = 24.1° !
(b) Since R is horizontal, R = Rx
Then, R = Rx = ΣFx
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
300-N Force:
400-N Force:
600-N Force:
and
Rx = ΣFx = 914.49 N
Ry = ΣFy = 448.80 N
Further:
448.80
tan α =
914.49
448.80
α = tan −1 = 26.1°
914.49
R = 1019 N 26.1° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
ΣFx :
Rx = ΣFx
Rx = − 38.132 N
ΣFy :
Ry = ΣFy
Ry = 1294.72 N
R= ( − 38.132 N )2 + (1294.72 N )2
R = 1295 N
1294.72 N
tan α =
38.132 N
α = 88.3°
R = 1.295 kN 88.3° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
We have:
84 12 3
Rx = ΣFx = − TBC + (156 lb ) − (100 lb )
116 13 5
or Rx = −0.72414TBC + 84 lb
and
80 5 4
R y = ΣFy = TBC − (156 lb ) − (100 lb )
116 13 5
Ry = 0.68966TBC − 140 lb
Rx = −0.72414TBC + 84 lb = 0
TBC = 116.0 lb !
(b) Using
TBC = 116.0 lb
R = R = 60.0 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Then, Rx = ΣFx = 0
1
cos 35° −
Then: tan α = 3
sin 35°
α = 40.265°
α = 40.3° !
Then: R = Ry = ΣFy
R = 1130 N
R = 1.130 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
600
Thus tan α = = 1.5
400
α = 56.3° !
Rx = 1021.11 N
R = Rx = 1021 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
TAE = 1659 lb !
(b) R = ΣFx
R = 2060 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Law of Sines:
400 lb
(a) FAC = sin 25° = 169.691 lb FAC = 169.7 lb !
sin 95°
400
(b) TBC = sin 60° = 347.73 lb TBC = 348 lb !
sin 95°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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Free-Body Diagram:
4 21
ΣFx = 0: − TCA + TCB = 0
5 29
29 4
or TCB = TCA
21 5
3 20
ΣFy = 0: TCA + TCB − ( 3 kN ) = 0
5 29
3 20 29 4
Then TCA + × TCA − ( 3 kN ) = 0
5 29 21 5
or TCA = 2.2028 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
sin 50°
FC = ( FB )
sin 70°
sin 50°
FB cos 50° + cos 70° = 940
sin 70°
FB = 1019.96 N
sin 50°
FC = (1019.96 N )
sin 70°
or FC = 831 N !
FB = 1020 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
cos 30°
R= TAB
cos 65°
cos 30°
TAB − sin 30° + sin 65° − 550 = 0
cos 65°
or TAB = 405 N !
cos30°
(b) R= ( 450 N )
cos 65°
or R = 830 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram At B:
12 17
ΣFx = 0: − TBA + TBC = 0
13 293
5 2
ΣFy = 0: TBA + TBC − 300 N = 0
13 293
5 293
TBC = 300 − TBA
13 2
or TBC = 565.34 N
Free-Body Diagram At C:
17 24
ΣFx = 0: − TBC + TCD = 0
293 25
( 565.34 N )
17 25
TCD =
293 24
TCD = 584.86 N
2 7
ΣFy = 0: − TBC + TCD − WC = 0
293 25
2 7
WC = − ( 565.34 N ) + ( 584.86 N )
293 25
or WC = 97.7 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
ΣFx = 0:
TD = 9.1378 kips
TD = 9.14 kips !
TC = 5.87 kips !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
ΣFy = 0:
TD = 14.0015 kips
TD = 14.00 kips
ΣFx = 0:
TB = 16.73 kips
TB = 16.73 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
or FC = 1.433 kN
or FD = 1.678 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
or FB = 2.9938 kN
FB = 2.99 kN
FD = 1.060 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
L2 − ( 2.5 m ) = ( 8 − L ) − ( 5.45 m )
2 2 2
− 6.25 = 64 − 16 L − 29.7025
or L = 2.5342 m
5.45 m
And cos β =
8 m − 2.5342 m
or β = 4.3576°
2.5 m
Then cos α =
2.5342 m
or α = 9.4237°
Free-Body Diagram At B:
ΣFx = 0:
− TABC cos α − ( 35 N ) cos α + TABC cos β = 0
or TABC =
( 35) cos 9.4237°
cos 4.3576° − cos 9.4237°
TABC = 3255.9 N
ΣFy = 0:
TABC sin α + ( 35 N ) sin α + TABC sin β − W = 0
sin 9.4237° ( 3255.9 N + 35 N ) + ( 3255.9 N ) sin 4.3576° − W = 0
or W = 786.22 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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L2 − ( 3 m ) = ( 8 − L ) − ( 4.95 m )
2 2 2
− 9 = 64 − 16 L − 24.5025
or L = 3.0311 m
4.95 m
Then cos β =
8 m − 3.0311 m
or β = 4.9989°
3m
And cos α =
3.0311 m
or α = 8.2147°
Free-Body Diagram At B:
(a) ΣFx = 0:
cos β − cos α
or TDE = TABC
cos α
ΣFy = 0:
cos β − cos α
TABC sin α + sin α + sin β = 720
cos α
TABC =
( 720 ) cosα
sin (α + β )
TABC =
( 720 ) cos8.2147°
sin (8.2147° + 4.9989° )
TABC = 3117.5 N
TDE = 20.338 N
Free-Body Diagram At C:
3 15 15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5 17 17
17
or − TAC + 5 TBC = 750 (1)
5
4 8 8
ΣFy = 0: TAC + TBC − (150 lb ) − 190 lb = 0
5 17 17
17
or TAC + 2 TBC = 1107.5 (2)
5
7 TBC = 1857.5
or TBC = 265.36 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Free-Body Diagram At C:
3 15 15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5 17 17
17
or − TAC + 5 TBC = 750 (1)
5
4 8 8
ΣFy = 0: TAC + TBC − (150 lb ) − W = 0
5 17 17
17 17
or TAC + 2 TBC = 300 + W (2)
5 4
17
Adding Equations (1) and (2) gives 7 TBC = 1050 + W
4
17
or TBC = 150 + W
28
17 17
− TAC + 5 150 + W = 750
28
Using Equation (1)
5
25
or TAC = W
28
25
Now for T ≤ 240 lb ⇒ TAC : 240 = W
28
or W = 269 lb
17
TBC : 240 = 150 + W
28
or W = 148.2 lb
Therefore 0 ≤ W ≤ 148.2 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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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:
4 35 12
ΣFx = 0: − ( 3W ) + (W ) + Fs = 0
5 37 37
or
Fs = 4.4833W
and
3 12 35
ΣFy = 0: ( 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
(a) W = 62.8 N
(b) Have spring force
Fs = k ( LAB − LO )
Where
FAB = k AB ( LAB − LO )
and
So:
281.74 N = 800 N/m (1.110 − LO ) m
or
LO = 758 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram At A:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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∴ α + 10° = 60°
or α = 50.0° W
or TAB = 35.0 lb W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Note: In problems of this type, P may be directed along one of the cables, with T = Tmax in that cable and
T = 0 in the other, or P may be directed in such a way that T is maximum in both cables. The second
possibility is investigated first.
Free-Body Diagram At C:
Force Triangle
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Free-Body Diagram At C:
Force Triangle
or P = 1510 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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2Tx − 1200 N = 0
Tx = 600 N
(Tx )2 + (Ty )
2
= T2
Ty = 630 N
By similar triangles:
AC 1.8 m
=
870 N 630 N
AC = 2.4857 m
L = 2( AC )
L = 2 ( 2.4857 m )
L = 4.97 m
L = 4.97 m "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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LAB = 12 2 in.
Then (
Fs = 4 lb/in. 20 − 12 2 in. )
Fs = 12.1177 lb
ΣFy = 0
4
−W + (12.1177 lb ) = 0
5
W = 9.69 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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At Collar A ...
ΣFy = 0:
h
− 9 lb + Fs = 0
12 + h 2
2
or hFs = 9 144 + h 2
Becomes h 3 lb/in.
( 144 + h 2
)
− 12 2 = 9 144 + h 2
or ( h − 3) 144 + h 2 = 12 2 h
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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where magnitude and direction of TBD are known, and the direction
of FAB is known.
= 305.29 N
TBC = 305 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram At C:
β = 52.5°
α = 7.50°
P
= (140 lb ) cos 52.5°
2
P = 170.453 lb
P = 170.5 lb 7.50°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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(b) ( )
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
1
T = ( 2746.8 N )
2
T = 1373 N
( )
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
(c)
1
T = ( 2746.8 N )
3
T = 916 N
( )
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
(d) 1
T = ( 2746.8 N )
3
T = 916 N
( )
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
1
(e) T = ( 2746.8 N )
4
T = 687 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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(b)
( )
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
1
T = ( 2746.8 N )
3
T = 916 N
(d)
( )
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
1
T = ( 2746.8 N )
4
T = 687 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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(a) ΣFx = 0: TACB ( cos 30° − cos 50° ) − ( 800 N ) cos 50° = 0
TACB = 2.30 kN
(b) ΣFy = 0: TACB ( sin 30° + sin 50° ) + ( 800 N ) sin 50° − Q = 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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or P = 0.34730TACB (1)
TACB = 1305 N
P = 453 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Q = 2 ( 30 lb ) cos 25°
Q = 54.378 lb
Equivalent loading at A:
Law of Cosines:
(120 lb )2 = (100 lb )2 + ( 54.378 lb )2 − 2 (100 lb )( 54.378 lb ) cos (125° − α ) cos (125° − α ) = − 0.132685
This gives two values: 125° − α = 97.625°
α = 27.4°
125° − α = − 97.625°
α = 223°
Thus for R < 120 lb:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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467.78 lb
cosθ z =
950 lb
or θ z = 60.5° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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519.09 lb
cosθ z =
810 lb
or θ z = 50.1° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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−329.40 N
cosθ z =
900 N
or θ z = 111.5° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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= − 610.65 N
Fx = − 611 N !
= 1785.42 N
Fy = 1785 N !
= 222.26 N
Fz = 222 N !
−610.65 N
(b) cosθ x =
1900 N
or θ x = 108.7° !
1785.42 N
cosθ y =
1900 N
or θ y = 20.0° !
222.26 N
cosθ z =
1900 N
or θ z = 83.3° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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= −103.244 lb
Fy = −103.2 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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= − 90.107 N
Fx = − 90.1 N W
= 190.526 N
Fy = 190.5 N W
= − 63.093 N
Fz = − 63.1 N W
−90.107 Ν
(b) cosθ x =
220 N
θ x = 114.2° W
190.526 N
cosθ y =
220 N
θ y = 30.0° W
−63.093 N
cosθ z =
220 N
θ z = 106.7° W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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(a) Fx = 180 N
or F = 439.38 N
F = 439 N !
180 N
(b) cosθ x =
439.48 N
θ x = 65.8° !
Fy = 380.60 N
380.60 N
cosθ y =
439.48 N
θ y = 30.0° !
Fz = −126.038 N
−126.038 N
cosθ z =
439.48 N
θ z = 106.7° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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F = ( 65 N )2 + ( − 80 N )2 + ( − 200 N )2
F = 225 N !
Fx 65 N
cosθ x = =
F 225 N
θ x = 73.2° !
Fy − 80 N
cosθ y = =
F 225 N
θ y = 110.8° !
Fz − 200 N
cosθ z = =
F 225 N
θ z = 152.7° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Fx 450 N
cosθ x = =
F 1950 N
θ x = 76.7° !
Fy 600 N
cosθ y = =
F 1950 N
θ y = 72.1° !
Fz −1800 N
cosθ z = =
F 1950 N
θ z = 157.4° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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cosθ y = − 0.67597
θ y = 132.5° !
Fy
(b) Then: F =
cosθ y
− 50 lb
F=
− 0.67597
F = 73.968 lb
And Fx = F cosθ x
Fx = 53.9 lb !
Fz = F cosθ z
Fz = ( 73.968 lb ) cos83.8°
Fz = 7.99 lb !
F = 74.0 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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cosθ z = − 0.89687
θ z = 153.7° !
Fz − 35 lb
(b) Then: F = =
cosθ z − 0.89687
F = 39.025 lb
And Fx = F cosθ x
Fx = ( 39.025 lb ) cos113.2°
Fx = −15.37 lb !
Fy = F cosθ y
Fy = 7.85 lb !
F = 39.0 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Fy = 75.592 N
Fy = 75.6 N !
Fz = 224 N !
Fx
(b) cosθ x =
F
80 N
cosθ x =
250 N
θ x = 71.3° !
Fz
cosθ z =
F
224.47 N
cosθ z =
250 N
θ z = 26.1° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Fx = ( 320 N ) cos104.5°
Fx = − 80.122 N
Fx = − 80.1 N !
Fy = 286 N !
Fy
(b) cosθ y =
F
285.62 N
cosθ y =
320 N
θ y = 26.8° !
Fz
cosθ z =
F
−120 N
cosθ z =
320 N
θ z = 112.0° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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!!!"
DB = ( 36 in.) i − ( 42 in.) j − ( 36 in.) k
55 lb
TDB = ( 36 in.) i − ( 42 in.) j − ( 36 in.) k
66 in.
= ( 30 lb ) i − ( 35 lb ) j − ( 30 lb ) k
∴ (TDB ) x = 30.0 lb !
(TDB ) y = − 35.0 lb !
(TDB ) z = − 30.0 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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!!!"
EB = ( 36 in.) i − ( 45 in.) j + ( 48 in.) k
60 lb
TEB = ( 36 in.) i − ( 45 in.) j + ( 48 in.) k
75 in.
= ( 28.8 lb ) i − ( 36 lb ) j + ( 38.4 lb ) k
∴ (TEB ) x = 28.8 lb !
(TEB ) y = − 36.0 lb !
(TEB ) z = 38.4 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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!!!"
BA = ( 4 m ) i + ( 20 m ) j − ( 5 m ) k
BA = ( 4 m )2 + ( 20 m )2 + ( − 5 m )2 = 21 m
!!!"
BA 2100 N
F = F λ BA = F = ( 4 m ) i + ( 20 m ) j − ( 5 m ) k
BA 21 m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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!!!"
DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) k
DA = ( 4 m )2 + ( 20 m )2 + (14.8 m )2 = 25.2 m
!!!"
DA 1260 N
F = F λ DA = F = ( 4 m ) i + ( 20 m ) j + (14.8 m ) k
DA 25.2 m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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uuuv
BG = − (1 m ) i + (1.85 m ) j − ( 0.8 m ) k
BG = ( −1 m )2 + (1.85 m )2 + ( − 0.8 m )2
BG = 2.25 m
uuuv
BG
TBG = TBG λBG = TBG
BG
450 N
TBG = − (1 m ) i + (1.85 m ) j − ( 0.8 m ) k
2.25 m
∴ (TBG ) x = − 200 N
(TBG ) y = 370 N
(TBG ) z = −160.0 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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uuuuv
BH = ( 0.75 m ) i + (1.5 m ) j − (1.5 m ) k
∴ (TBH ) x = 200 N
(TBH ) y = 400 N
(TBH ) z = − 400 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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P = ( 4 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
Rx − 0.27931 kip
cosθ x = = = − 0.065238
R 4.2814 kips
Ry 3.6569 kips
cos θ y = = = 0.85414
R 4.2814 kips
Rz − 2.2089 kips
cos θ z = = = − 0.51593
R 4.2814 kips
or θ x = 93.7°
θ y = 31.3°
θ z = 121.1°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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P = ( 6 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
Rx 0.49610 kip
cos θ x = = = 0.24628
R 2.0144 kips
Ry 1.94975 kips
cos θ y = = = 0.967906
R 2.0144 kips
Rz 0.101700 kip
cos θ z = = = 0.050486
R 2.0144 kips
or θ x = 75.7°
θ y = 14.56°
θ z = 87.1°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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uuur
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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!!!"
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB = 750 mm
!!!"
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
AC = 850 mm
!!!"
AB 765 N
TAB = TAB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB 750 mm
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Have TAB = ( 760 lb )( sin 50° cos 40°i − cos 50°j + sin 50° sin 40°k )
TAC = TAC ( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25°k )
( 760 lb) sin 50° cos 40° − TAC cos 45° sin 25° = 0
or TAC = 1492.41 lb
∴ TAC = 1492 lb
(b) ( RA ) y = ΣFy = ( − 760 lb) cos 50° − (1492.41 lb) sin 45°
( RA ) y = −1543.81 lb
( RA ) z = ΣFz = ( 760 lb) sin 50° sin 40° + (1492.41 lb) cos 45° cos 25°
( RA ) z = 1330.65 lb
∴ R A = − (1543.81 lb ) j + (1330.65 lb ) k
Then RA = 2038.1 lb RA = 2040 lb
0
cosθ x = θ x = 90.0°
2038.1 lb
−1543.81 lb
cos θ y = θ y = 139.2°
2038.1 lb
1330.65 lb
cos θ z = θ z = 49.2°
2038.1 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Have TAB = TAB ( sin 50° cos 40°i − cos 50°j + sin 50° sin 40°k )
TAC = ( 980 lb )( − cos 45° sin 25°i − sin 45°j + cos 45° cos 25°k )
TAB sin 50° cos 40° − ( 980 lb ) cos 45° sin 25° = 0
or TAB = 499.06 lb
∴ TAB = 499 lb
(b) ( RA ) y = ΣFy = − ( 499.06 lb) cos 50° − (980 lb) sin 45°
( RA ) y = −1013.75 lb
( RA ) z = ΣFz = ( 499.06 lb) sin 50° sin 40° + (980 lb) cos 45° cos 25°
( RA ) z = 873.78 lb
∴ R A = − (1013.75 lb ) j + (873.78 lb ) k
Then RA = 1338.35 lb RA = 1338 lb
0
and cos θ x = θ x = 90.0°
1338.35 lb
−1013.75 lb
cos θ y = θ y = 139.2°
1338.35 lb
873.78 lb
cos θ z = θ z = 49.2°
1338.35 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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!!!"
Cable AB: AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
( RA ) z
51
(a) = ΣFz = 0: ( 216 N ) − TAC = 0 or TAC = 360 N !
85
( RA ) y = ΣFy = 0:
32
(b) ( 288 N ) + TAC − P = 0 or P = 424 N !
85
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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uuur
Cable AB: AB = − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k
AC = (12 m )2 + ( − 20 m )2 + ( 3.6 m )2
= 23.6 m
uuur
(12 m ) i − ( 20 m ) j + ( 3.6 m ) k
AC 1770 N
TAC = TAC =
AC 23.6 m
= ( 900 N ) i − (1500 N ) j + ( 270 N ) k
uuur
Cable AD: AD = − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k
AD = ( − 4 m )2 + ( − 20 m )2 + (14.8 m )2
= 25.2 m
uuur
− ( 4 m ) i − ( 20 m ) j + (14.8 m ) k
AD TAD
TAD = TAD =
AD 25.2 m
− (10 m ) i − ( 50 m ) j − ( 37 m ) k
TAD
=
63 m
Now...
R = TAB + TAC + TAD and R = Rj; Rx = Rz = 0
4 10
ΣFx = 0: −
TAB + 900 − TAD = 0 (1)
21 63
5 37
ΣFy = 0: TAB + 270 − TAD = 0 (2)
21 63
Solving equations (1) and (2) simultaneously yields:
TAD = 1.775 kN !
TAB = 3.25 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below.
5 25
TAB = TAD (1)
3 43
17 18
TAC = TAD (2)
8 43
255
0.8 TAB + TAD − W = 0 (3)
172
From Equation (1)
5 25
TAB = ( 4.3 kN )
3 43
or TAB = 4.1667 kN
From Equation (3)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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uuur
AB = − ( 4.20 m ) i − ( 5.60 m ) j
uuur
AC = ( 2.40 m ) i − ( 5.60 m ) j + ( 4.20 m ) k
uuur
AD = − ( 5.60 m ) j − ( 3.30 m ) k
uuur
AB TAB
TAB = TAB λ AB = TAB = ( − 4.20i − 5.60j)
AB 7.00 m
3 4
TAB = − i − j TAB
5 5
uuur
AC TAC
TAC = TAC λ AC = TAC = ( 2.40i − 5.60j + 4.20k )
AC 7.40 m
12 28 21
TAC = i − j+ k TAC
37 37 37
uuur
AD TAD
TAD = TAD λ AD = TAD = ( − 5.60 j − 3.30k )
AD 6.50 m
56 33
TAD = − j − k TAD
65 65
P = Pj
For equilibrium at point A: ΣF = 0
3 12
i component: − TAB + TAC = 0
5 37
20
or TAB = TAC (1)
37
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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4 28 56
j component: − TAB − TAC − TAD + P = 0
5 37 65
4 28 56 65 7
− TAB − TAC − ⋅ TAC + P = 0
5 37 65 11 37
4 700
− TAB − TAC + P = 0 (2)
5 407
21 33
k component: TAC − TAD = 0
37 65
65 7
or TAD = TAC (3)
11 37
or TAC = 479.15 N
4 700
From Equation (2): − ( 259 N ) − ( 479.15 N ) + P = 0
5 407
∴ P = 1031 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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See Problem 2.103 for the analysis leading to the linear algebraic Equations (1), (2), and (3)
20
TAB = TAC (1)
37
4 700
− TAB − TAC + P = 0 (2)
5 407
65 7
TAD = TAC (3)
11 37
Substituting for TAC = 444 N into Equation (1)
20
Gives TAB = ( 444 N )
37
or TAB = 240 N
And from Equation (3)
4 700
− ( 240 N ) − ( 444 N ) + P = 0
5 407
∴ P = 956 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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11 9.6
= FBA − i + j
14.6 14.6
FCA
FCA = FCAλCA = ( 9.6 in.) j − ( 7.2 in.) k
12.0 in.
4 3
= FCA j − k
5 5
FDA
FDA = FDAλDA = ( 9.6 in.) i + ( 9.6 in.) j + ( 4.8 in.) k
14.4 in.
2 2 1
= FDA i + j + k
3 3 3
P = −Pj
11 2
i component: − FBA + FDA = 0 (1)
14.6 3
3 1
k component: − FCA + FDA = 0 (3)
5 3
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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14.6 2
From Equation (1) FBA = FDA
11 3
14.6 2
29.2 lb = FDA
11 3
or FDA = 33 lb
5
Solving Eqn. (3) for FCA gives: FCA = FDA
9
5
FCA = ( 33 lb )
9
Substituting into Eqn. (2) for FBA , FDA, and FCA in terms of FDA gives:
9.6 4 5 2
14.6 ( 29.2 lb ) + 5 9 ( 33 lb ) + 3 ( 33 lb ) − P = 0
∴
P = 55.9 lb "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below.
11 2
− FBA + FDA = 0 (1)
14.6 3
9.6 4 2
FBA + FCA + FDA − P = 0 (2)
14.6 5 3
3 1
− FCA + FDA = 0 (3)
5 3
14.6 2
From Equation (1): FBA = FDA
11 3
5
From Equation (3): FCA = FDA
9
Substituting into Equation (2) for FBA and FCA gives:
9.6 14.6 2 4 5 2
FDA + FDA + FDA − P = 0
14.6 11 3 5 9 3
838
or FDA = P
495
838
Since P = 45 lb FDA = 45 lb
495
or FDA = 26.581 lb
14.6 2
and FBA = ( 26.581 lb )
11 3
or FBA = 23.5 lb
5
and FCA = ( 26.581 lb )
9
or FCA = 14.77 lb
and FDA = 26.6 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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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
uuur
AC = (18 m ) i − ( 30 m ) j + ( 5.4 m ) k
and
uuur
AB = − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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With P = Pj, at A:
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 "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Based on the results of Problem 2.107, 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 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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!!!"
AB = − ( 6.5 ft ) i − (8 ft ) j + ( 2 ft ) k
AB = ( −6.5 ft )2 + ( −8 ft )2 + ( 2 ft )2 = 10.5 ft
TAB
TAB = − ( 6.5 ft ) i − ( 8 ft ) j + ( 2 ft ) k
10.5 ft
AC = (1 ft )2 + ( −8 ft )2 + ( 4 ft )2 = 9 ft
TAC
TAC = (1 ft ) i − ( 8 ft ) j + ( 4 ft ) k
9 ft
AD = (1.75 ft )2 + ( −8 ft )2 + ( −1 ft )2 = 8.25 ft
TAD
TAD = (1.75 ft ) i − ( 8 ft ) j − (1 ft ) k
8.25 ft
At A ΣF = 0
Substituting for W = 320 lb and Solving Equations (1), (2), (3) simultaneously yields:
TAB = 86.2 lb !
TAC = 27.7 lb !
TAD = 237 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below.
W = 297 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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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.
− cos15°i + 8 j − sin15°k
TDG = TDG λ DG = TDG
65
At A: ΣF = 0: TBE + TCF + TDG + W + P = 0
TBE T T
i: cos 45° + CF cos 30° − DG cos15° + P = 0
65 65 65
8 8 8
j: TBE + TCF + TDG −W = 0
65 65 65
65
or TBE + TCF + TDG − W =0 (2)
8
TBE T T
k: − sin 45° + CF sin 30° − DG sin15° = 0
65 65 65
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
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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See Problem 2.111 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
65
j: TBE + TCF + TDG − W =0 (2)
8
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:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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TDB = 56.7 N !
TDC = 56.7 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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TDA = TDAλDA
TDA
= ( 400 mm ) i − ( 600 mm ) j
721.11 mm
TDB = TDBλDB
TDB
= − ( 200 mm ) i − ( 600 mm ) j + ( 200 mm ) k
663.32 mm
TDC
= − ( 200 mm ) i − ( 600 mm ) j − ( 200 mm ) k
663.32 mm
= TDC ( − 0.30151i − 0.90454 j − 0.30151k )
( )
Setting W = (16 kg ) 9.81 m/s 2 = 156.96 N
TDB = 57.8 N !
TDC = 57.8 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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TAB = 0.5409 P
TAC = 0.295P
TAD = 0.2959P
Using P = 8 kN:
TAB = 4.33 kN !
TAC = 2.36 kN !
TAD = 2.37 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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d BA = ( 6 m )2 + ( 6 m )2 + ( 3 m )2 =9m
d AC = ( −10.5 m )2 + ( − 6 m )2 + ( − 8 m )2 = 14.5 mm
d AD = ( − 6 m )2 + ( − 6 m )2 + ( 7 m )2 = 11 mm
d AE = ( 6 m )2 + ( − 4.5 m )2 = 7.5 m
FBA
FBA = FBAλBA = ( 6 m ) i + ( 6 m ) j + ( 3 m ) k
9m
2 2 1
= FBA i + j + k
3 3 3
TAC
TAC = TAC λ AC = − (10.5 m ) i − ( 6 m ) j − ( 8 m ) k
14.5 m
21 12 16
= TAC − i − j− k
29 29 29
TAD
TAD = TAD λ AD = − ( 6 m ) i − ( 6 m ) j + ( 7 m ) k
11 m
6 6 7
= TAD − i − j + k
11 11 11
W
WAE = WAE λ AE = ( 6 m ) i − ( 4.5 m ) j
7.5 m
= W ( 0.8i − 0.6 j)
WO = − W j
2 21 6
i component: FBA − TAC − TAD + 0.8W = 0 (1)
3 29 11
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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2 12 6
j component: FBA − TAC − TAD − 1.6W = 0 (2)
3 29 11
1 16 7
k component: FBA − TAC + TAD = 0 (3)
3 29 11
( )
Setting W = ( 20 kg ) 9.81 m/s 2 = 196.2 N
FBA = 1742 N
TAC = 1517 N
TAD = 403 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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ΣFx = 0:
− TAD ( sin 30° )( sin 50° ) + TBD ( sin 30° )( cos 40° ) + TCD ( sin 30° )( cos 60° ) = 0
ΣFy = 0:
ΣFz = 0:
TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0
TAD = 30.5 lb !
TBD = 10.59 lb !
TCD = 30.5 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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TCF ( sin 45° + sin 30° ) + TDG ( sin 45° − sin15° ) = 0.2 65 sin 45°
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°
1.29028 − 1.73205TDG − P 65 = 0
1.29028
∴ TDG is taut for P < lb
65
or 0 ≤ P ≤ 0.1600 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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d AB = ( − 30 ft )2 + ( 24 ft )2 + ( 32 ft )2 = 50 ft
d AC = ( − 30 ft )2 + ( 20 ft )2 + ( −12 ft )2 = 38 ft
TAB
TAB = TAB λ AB = − ( 30 ft ) i + ( 24 ft ) j + ( 32 ft ) k
50 ft
= TAB ( − 0.6i + 0.48 j + 0.64k )
TAC
TAC = TAC λ AC = − ( 30 ft ) i + ( 20 ft ) j − (12 ft ) k
38 ft
30 20 12
= TAC − i + j− k
38 38 38
16 30
N= Ni + Nj
34 34
W = − (175 lb ) j
At point A: ΣF = 0: TAB + TAC + N + W = 0
30 16
i component: − 0.6 TAB − TAC + N=0 (1)
38 34
20 30
j component: 0.48 TAB + TAC + N − 175 lb = 0 (2)
38 34
12
k component: 0.64 TAB −
TAC = 0 (3)
38
Solving Equations (1), (2), and (3) simultaneously:
TAB = 30.9 lb
TAC = 62.5 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Refer to the solution of problem 2.119 and the resulting linear algebraic Equations (1), (2), (3). Include force
P = − ( 45 lb ) k with other forces of Problem 2.119.
30 16
i component: − 0.6 TAB − TAC + N=0 (1)
38 34
20 30
j component: 0.48 TAB + TAC + N − 175 lb = 0 (2)
38 34
12
k component: 0.64 TAB − TAC − 45 lb = 0 (3)
38
TAB = 81.3 lb
TAC = 22.2 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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λBE = λ AB =
( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k
201.56 mm
TBE
TBE = TBE λBE = ( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k
201.56 mm
TCF
TCF = TCF λCF = ( 25 mm ) cos 30°i + ( 200 mm ) j + ( 25 mm ) sin 30°k
201.56 mm
TDG
TDG = TDG λDG = − ( 25 mm ) cos15°i + ( 200 mm ) j − ( 25 mm ) sin15°k
201.56 mm
W = − W j; P = Pk
At point A: ΣF = 0: TBE + TCE + TDG + W + P = 0
i component: 0.087704 TBE + 0.107415 TCF − 0.119806 TDG = 0 (1)
j component: 0.99226 TBE + 0.99226 TCF + 0.99226 TDG − W = 0 (2)
k component: − 0.087704 TBE + 0.062016 TCF − 0.032102 TDG + P = 0 (3)
Setting W = 10.5 N and P = 0, and solving (1), (2), (3) simultaneously:
TBE = 1.310 N !
TCF = 4.38 N !
TDG = 4.89 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
Setting W = 10.5 N and P = 0.5 N, and solving (1), (2), (3) simultaneously:
TBE = 4.84 N !
TCF = 1.157 N !
TDG = 4.58 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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uuur
DA = − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k
DA = ( − 8 ft )2 + ( 40 ft )2 + (10 ft )2 = 42 ft
TADB
TDA = − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k
42 ft
= TADB ( − 0.190476i + 0.95238 j + 0.23810k )
uuur
DB = ( 3 ft ) i + ( 36 ft ) j − ( 8 ft ) k
DB = ( 3 ft )2 + ( 36 ft )2 + ( − 8 ft )2 = 37 ft
TADB
TDB = ( 3 ft ) i + ( 36 ft ) j − ( 8 ft ) k
37 ft
= TADB ( 0.081081i + 0.97297 j − 0.21622k )
uuur
DC = ( a − 8 ft ) i − ( 24 ft ) j − ( 3 ft ) k
DC = ( a − 8 ft )2 + ( − 24 ft )2 + ( −3 ft )2 = ( a − 8)2 + 585 ft
TDC
TDC = ( a − 8 ft ) i − ( 24 ft ) j − ( 3 ft ) k
( a − 8)2 + 585
At D ΣF = 0:
( a − 8) = 0.190476 − 0.081081
−3 − 0.23810 + 0.21622
or a = 23 ft
Substituting into equation (1) for a = 23 ft and combining the coefficients for TADB gives:
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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or TADB = 81.9 lb
Substituting into equation (4) for TDC = 17 lb and TADB = 81.9 lb gives:
or W = 143.4 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Substituting for W = 120 lb and solving equations (3) and (4) simultaneously yields
TADB = 68.6 lb !
TDC = 14.23 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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TAB = TAB λ AB
TAB
= − ( 2.7 m ) i + ( 2.4 m ) j − ( 3.6 m ) k
5.1 m
9 8 12
= TAB − i + j − k
17 17 17
TAC = TAC λ AC
TAC
= ( 2.4 m ) j + (1.8 m ) k
3m
TAD = 2TADE λ AD
2TADE
= (1.2 m ) i + ( 2.4 m ) j − ( 0.3 m ) k
2.7 m
8 16 2
= TADE i + j − k
9 9 9
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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TAE = TAE λ AE
TADE
= − ( 2.4 m ) i + ( 2.4 m ) j + (1.2 m ) k
3.6 m
2 2 1
= TADE − i + j + k
3 3 3
W = − Wj
At point A: ΣF = 0: TAB + TAC + TAD + TAE + W = 0
9 8 2
i component: − TAB + TADE − TADE = 0 (1)
17 9 3
8 16 2
j component: TAB + 0.8 TAC + TADE + TADE − W = 0 (2)
17 9 3
12 2 1
k component: − TAB + 0.6 TAC − TADE + TADE = 0 (3)
17 9 3
Simplifying (1), (2), (3):
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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See Problem 2.125 for the analysis leading to the linear algebraic Equations (1′ ) , ( 2′ ) , and ( 3′ ) below:
(c) W = 2060 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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( AB )2 = x2 + y 2 + z 2
Here (1 m )2 = ( 0.40 m ) + y 2 + z 2
2
or y 2 + z 2 = 0.84 m 2
Thus, with y given, z is determined.
Now
uuur
AB 1
λ AB = = ( 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 x i + N zk + Pj + TAB λ AB = 0
P − yTAB = 0
With P = 680 N,
680 N
TAB =
y
Now, from the free body diagram of collar B:
ΣF = 0: N x i + N y j + Qk − TABλ AB = 0
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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680 N
Q = TAB z = 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
680 N
(a) TAB = = 2266.7 N
0.30
or TAB = 2.27 kN !
and
(b) Q = 2266.7 ( 0.866 ) = 1963.2 N
or Q = 1.963 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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y 2 + z 2 = 0.84 m 2
680 N
TAB =
y
680 N
Q= z
y
z 2 = 0.84 m 2 − ( 0.55 m )
2
∴ z = 0.73314 m
and
680 N
(a) TAB = = 1236.36 N
0.55
or TAB = 1.236 kN !
and
or Q = 0.906 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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20 lb 14 lb
(a) Have: =
sin α sin 30°
sin α = 0.71428
α = 45.6°
(b) β = 180° − ( 30° + 45.6° )
= 104.4°
R 14 lb
Then: =
sin104.4° sin 30°
R = 27.1 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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OA = ( 70 )2 + ( 240 )2 = 250 mm
500 N Force:
70
Fx = −500 N Fx = −140.0 N !
250
240
Fy = +500 N Fy = 480 N !
250
435 N Force:
210
Fx = +435 N Fx = 315 N !
290
200
Fy = +435 N Fy = 300 N !
290
510 N Force:
120
Fx = +510 N Fx = 240 N !
255
225
Fy = −510 N Fy = −450 N !
255
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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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:
450 N
(a) P= = 549.3 N
cos 35°
P = 549 N !
= 315.1 N
Px = 315 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Law of Sines:
TAC T 5 kN
= BC =
sin115° sin 5° sin 60°
5 kN
(a) TAC = sin115° = 5.23 kN TAC = 5.23 kN !
sin 60°
5 kN
(b) TBC = sin 5° = 0.503 kN TBC = 0.503 kN !
sin 60°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Free-Body Diagram
First, consider the sum of forces in the x-direction because there is only one unknown force:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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and
For α = +32.3°
or P = 149.1 lb 32.3°
For α = −32.3°
or P = 274 lb 32.3°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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220.6 N
F = = 575.95 N
sin30° sin50°
F = 576 N !
Fx 220.6
(b) cosθ x = = = 0.38302
F 575.95
θ x = 67.5° !
Fy = F cos 30° = 498.79 N
Fy 498.79
cosθ y = = = 0.86605
F 575.95
θ y = 30.0° !
Fz = − F sin 30° cos 50°
= −185.107 N
Fz −185.107
cosθ z = = = −0.32139
F 575.95
θ z = 108.7° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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= −437.38 lb Fz = −437 lb !
Then:
Fx 200
(b) cosθ x = = = 0.33333 θ x = 70.5° !
F 600
Fy −358.75
cosθ y = = = −0.59792 θ y = 126.7° !
F 600
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Q = ( 600 lb ) [ cos 40° cos 20°i + sin 40° j − cos 40° sin 20°k ]
R = 758 lb !
Rx 319.86 lb
cosθ x = = = 0.42199
R 757.98 lb
θ x = 65.0° !
Ry 635.67 lb
cosθ y = = = 0.83864
R 757.98 lb
θ y = 33.0° !
Rz 261.04 lb
cosθ z = = = 0.34439
R 757.98 lb
θ z = 69.9° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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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
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
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
uuur
AB = ( 32 in.) i − ( 48 in.) j + ( 36 in.) k
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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uuur
AD T
TAD = T λ AD = TAD = AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k
AD 65 in.
Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
In Equations (1), (2) and (3), set TAD = 120 lb, and, using conventional methods for solving Linear Algebraic
Equations (MATLAB or Maple, for example), we obtain:
TAB = 32.6 lb
TAC = 102.5 lb
W = 177.2 lb "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
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
uuur
AB = − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k
and
uuur
AC = ( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k
At A: ΣF = 0: TAB + TAC + P + Q + W = 0
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:
or P = 0.23376T
j: ( 0.81818 + 0.93506 ) T −W = 0
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
or W = 1.75324T
k: ( −0.181818 − 0.16883) T +Q =0
or Q = 0.35065T
With W = 1200 N:
1200 N
T = = 684.45 N
1.75324
P = 160.0 N !
Q = 240 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.