Cap 02

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Chapter 2, Solution 1.
(a)
(b)
We measure:
R = 37 lb, α = 76°
R = 37 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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76° !
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Chapter 2, Solution 2.
(a)
(b)
We measure:
R = 57 lb, α = 86°
R = 57 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.
86° !
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Chapter 2, Solution 3.
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 10.5 kN
α = 22.5°
R = 10.5 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.
22.5° !
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Chapter 2, Solution 4.
(a)
Parallelogram law:
We measure:
R = 5.4 kN α = 12°
(b)
R = 5.4 kN
12° !
R = 5.4 kN
12° !
Triangle rule:
We measure:
R = 5.4 kN α = 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.
Using the triangle rule and the Law of Sines
(a)
sin β
sin 45°
=
150 N 200 N
sin β = 0.53033
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.
Using the triangle rule and the Law of Sines
(a)
sin α
sin 45°
=
120 N 200 N
sin α = 0.42426
α = 25.104°
or
(b)
α = 25.1° !
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
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.
Faa′ = 266 N !
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Chapter 2, Solution 7.
Using the triangle rule and the Law of Cosines,
Have: β = 180° − 45°
β = 135°
Then:
R 2 = ( 900 ) + ( 600 ) − 2 ( 900 )( 600 ) cos 135°
2
2
or R = 1390.57 N
Using the Law of Sines,
600 1390.57
=
sin γ
sin135°
or γ = 17.7642°
and α = 90° − 17.7642°
α = 72.236°
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.
(a)
α = 72.2° !
(b)
R = 1.391 kN !
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Chapter 2, Solution 8.
By trigonometry: Law of Sines
F2
R
30
=
=
sin α
sin 38° sin β
α = 90° − 28° = 62°, β = 180° − 62° − 38° = 80°
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.
Using the Law of Sines
F1
R
20 lb
=
=
sin α
sin 38° sin β
α = 90° − 10° = 80°, β = 180° − 80° − 38° = 62°
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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Chapter 2, Solution 10.
Using the Law of Sines:
60 N
80 N
=
sin α sin10°
or α = 7.4832°
β = 180° − (10° + 7.4832° )
= 162.517°
Then:
R
80 N
=
sin162.517° sin10°
or R = 138.405 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.
(a)
α = 7.48° !
(b)
R = 138.4 N !
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Chapter 2, Solution 11.
Using the triangle rule and the Law of Sines
Have:
β = 180° − ( 35° + 25° )
= 120°
Then:
P
R
80 lb
=
=
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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Chapter 2, Solution 12.
Using the triangle rule and the Law of Sines
(a) Have:
80 lb
70 lb
=
sin α
sin 35°
sin α = 0.65552
α = 40.959°
or α = 41.0° !
β = 180 − ( 35° + 40.959° )
(b)
= 104.041°
Then:
R
70 lb
=
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.
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Chapter 2, Solution 13.
We observe that force P is minimum when α = 90°.
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.
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Chapter 2, Solution 14.
For TBC to be a minimum,
R and TBC must be perpendicular.
Thus
TBC = ( 70 N ) sin 4°
= 4.8829 N
And
R = ( 70 N ) cos 4°
= 69.829 N
(a)
(b)
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.
TBC = 4.88 N
6.00° !
R = 69.8 N !
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Chapter 2, Solution 15.
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
= 1380.33 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°
Then:
α + β + 75° = 180°
α = 76.588°
R = 37.2 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.
76.6° !
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Chapter 2, Solution 16.
Using the Law of Cosines and the Law of Sines,
R 2 = ( 45 lb ) + (15 lb ) − 2 ( 45 lb )(15 lb ) cos135°
2
2
or R = 56.609 lb
56.609 lb 15 lb
=
sin135°
sinθ
or θ = 10.7991°
R = 56.6 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.
85.8° !
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Chapter 2, Solution 17.
γ = 180° − 25° − 50°
γ = 105°
Using the Law of Cosines:
R 2 = ( 5 kN ) + ( 8 kN ) − 2 ( 5 kN )( 8 kN ) cos105°
2
2
or R = 10.4740 kN
Using the Law of Sines:
10.4740 kN 8 kN
=
sin105°
sin β
or β = 47.542°
and α = 47.542° − 25°
α = 22.542°
R = 10.47 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.
22.5° "
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Chapter 2, Solution 19.
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
= 1710.42 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°
Hence:
φ = α + 45° = 72.028°
R = 41.4 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.
72.0° !
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Chapter 2, Solution 19.
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
= 1710.42 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°
Hence:
φ = α + 45° = 72.028°
R = 41.4 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.
72.0° !
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Chapter 2, Solution 20.
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
= 1710.42 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°
Finally:
φ = α + 45° = 87.972°
R = 41.4 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.
88.0° !
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Chapter 2, Solution 21.
2.4 kN Force:
Fx = ( 2.4 kN ) cos 50°
Fx = 1.543 kN Fy = ( 2.4 kN ) sin 50°
Fy = 1.839 kN 1.85 kN Force:
Fx = (1.85 kN ) cos 20°
Fx = 1.738 kN Fy = (1.85 kN ) sin 20°
Fy = 0.633 kN 1.40 kN Force:
Fx = (1.40 kN ) cos 35°
Fx = 1.147 kN Fy = − (1.40 kN ) sin 35°
Fy = −0.803 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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Chapter 2, Solution 22.
Fx = ( 5 kips ) cos 40°
5 kips:
or Fx = 3.83 kips Fy = ( 5 kips ) sin 40°
or Fy = 3.21 kips 7 kips:
Fx = − ( 7 kips ) cos 70°
or Fx = −2.39 kips Fy = ( 7 kips ) sin 70°
or Fy = 6.58 kips 9 kips:
Fx = − ( 9 kips ) cos 20°
or Fx = −8.46 kips Fy = ( 9 kips ) sin 20°
or Fy = 3.08 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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Chapter 2, Solution 23.
Determine the following distances:
680 N Force:
dOA =
( −160 mm )2 + ( 300 mm )2
dOB =
( 600 mm )2 + ( 250 mm )2
dOC =
( 600 mm )2 + ( −110 mm )2
Fx = 680 N
= 340 mm
= 650 mm
= 610 mm
( −160 mm )
340 mm
Fx = − 320 N !
( 300 mm )
Fy = 680 N
340 mm
Fy = 600 N !
390 N Force:
Fx = 390 N
( 600 mm )
650 mm
Fx = 360 N !
Fy = 390 N
( 250 mm )
650 mm
Fy = 150 N !
610 N Force:
Fx = 610 N
( 600 mm )
610 mm
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
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Chapter 2, Solution 24.
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 = − ( 204 lb )
48
,
102
Fy = + ( 204 lb )
90
,
102
Fx = −96.0 lb Fy = 180.0 lb 212 lb Force:
Fx = + ( 212 lb )
56
,
106
Fx = 112.0 lb 90
,
106
Fy = 180.0 lb Fx = − ( 400 lb )
80
,
100
Fx = −320 lb Fy = − ( 400 lb )
60
,
100
Fy = −240 lb Fy = + ( 212 lb )
400 lb 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
© 2007 The McGraw-Hill Companies.
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Chapter 2, Solution 25.
(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 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 26.
(a)
P=
Px
cos 40°
P=
30 lb
cos 40°
or P = 39.2 lb !
(b)
Py = Px tan 40°
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.
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Chapter 2, Solution 27.
(a)
Py = 100 N
P=
P=
Py
sin 75°
100 N
sin 75°
or P = 103.5 N "
(b)
Px =
Px =
Py
tan 75°
100 N
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.
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Chapter 2, Solution 28.
We note:
CB exerts force P on B along CB, and the horizontal component of P is Px = 260 lb.
Then:
(a)
Px = P sin 50°
P=
Px
sin 50°
=
260 lb
sin 50°
= 339.40 lb
(b)
P = 339 lb !
Px = Py tan 50°
Py =
Px
tan 50°
=
260 lb
tan 50°
= 218.16 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.
Py = 218 lb !
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Chapter 2, Solution 29.
(a)
P=
45 N
cos 20°
or P = 47.9 N !
(b)
Px = ( 47.9 N ) sin 20°
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.
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Chapter 2, Solution 30.
(a)
P=
18 N
sin 20°
or P = 52.6 N !
(b)
Py =
18 N
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.
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Chapter 2, Solution 31.
From the solution to Problem 2.21:
F2.4 = (1.543 kN ) i + (1.839 kN ) j
F1.85 = (1.738 kN ) i + ( 0.633 kN ) j
F1.40 = (1.147 kN ) i − ( 0.803 kN ) j
R = ΣF = ( 4.428 kN ) i + (1.669 kN ) j
R=
( 4.428 kN )2 + (1.669 kN )2
= 4.7321 kN
tan α =
1.669 kN
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.
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Chapter 2, Solution 32.
From the solution to Problem 2.22:
F5 = ( 3.83 kips ) i + ( 3.21 kips ) j
F7 = − ( 2.39 kips ) i + ( 6.58 kips ) j
F9 = − ( 8.46 kips ) i + ( 3.08 kips ) j
R = ΣF = − ( 7.02 kips ) i + (12.87 ) j
R=
( − 7.02 kips )2 + (12.87 kips )2
= 14.66 kips
 12.87 
 = 61.4°
 − 7.02 
α = tan −1 
R = 14.66 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.
61.4° !
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Chapter 2, Solution 33.
From the solution to Problem 2.24:
FOA = − ( 48.0 lb ) i + ( 90.0 lb ) j
FOB = (112.0 lb ) i + (180.0 lb ) j
FOC = − ( 320 lb ) i − ( 240 lb ) j
R = ΣF = − ( 256 lb ) i + ( 30 lb ) j
R=
( − 256 lb )2 + ( 30 lb )2
= 257.75 lb
tan α =
30 lb
−256 lb
α = − 6.6839°
R = 258 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.
6.68° !
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Chapter 2, Solution 34.
From Problem 2.23:
FOA = − ( 320 N ) i + ( 600 N ) j
FOB = ( 360 N ) i + (150 N ) j
FOC = ( 600 N ) i − (110 N ) j
R = ΣF = ( 640 N ) i + ( 640 N ) j
R=
( 640 N )2 + ( 640 N )2
= 905.097 N
tan α =
640 N
640 N
α = 45.0°
R = 905 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.
45.0° !
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Chapter 2, Solution 35.
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
Thus:
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.
R = 45.2 lb
62.3° !
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Chapter 2, Solution 36.
(a)
Since R is to be horizontal, Ry = 0
Then, Ry = ΣFy = 0
90 lb + ( 70 lb ) sin α − (130 lb ) cos α = 0
(13) cosα = ( 7 ) sin α + 9
13 1 − sin 2 α = ( 7 ) sin α + 9
Squaring both sides:
(
)
169 1 − sin 2 α = ( 49 ) sin 2 α + (126 ) sin α + 81
( 218) sin 2 α + (126 ) sin α − 88 = 0
Solving by quadratic formula:
(b)
sin α = 0.40899
or
α = 24.1° !
or
R = 117.0 lb !
Since R is horizontal, R = Rx
Then, R = Rx = ΣFx
ΣFx = ( 70 ) cos 24.142° + (130 ) sin 24.142°
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 37.
300-N Force:
Fx = ( 300 N ) cos 20° = 281.91 N
Fy = ( 300 N ) sin 20° = 102.61 N
400-N Force:
Fx = ( 400 N ) cos85° = 34.862 N
Fy = ( 400 N ) sin 85° = 398.48 N
600-N Force:
Fx = ( 600 N ) cos 5° = 597.72 N
Fy = − ( 600 N ) sin 5° = −52.293 N
and
Rx = ΣFx = 914.49 N
Ry = ΣFy = 448.80 N
R=
( 914.49 N )2 + ( 448.80 N )2
= 1018.68 N
Further:
tan α =
α = tan −1
448.80
914.49
448.80
= 26.1°
914.49
R = 1019 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.
26.1° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 38.
ΣFx :
Rx = ΣFx
Rx = ( 600 N ) cos 50° + ( 300 N ) cos85° − ( 700 N ) cos 50°
Rx = − 38.132 N
ΣFy :
Ry = ΣFy
Ry = ( 600 N ) sin 50° + ( 300 N ) sin 85° + ( 700 N ) sin 50°
Ry = 1294.72 N
R=
( − 38.132 N )2 + (1294.72 N )2
R = 1295 N
tan α =
1294.72 N
38.132 N
α = 88.3°
R = 1.295 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.
88.3° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 39.
We have:
Rx = ΣFx = −
84
12
3
TBC + (156 lb ) − (100 lb )
116
13
5
Rx = −0.72414TBC + 84 lb
or
and
R y = ΣFy =
80
5
4
TBC − (156 lb ) − (100 lb )
116
13
5
Ry = 0.68966TBC − 140 lb
(a)
So, for R to be vertical,
Rx = −0.72414TBC + 84 lb = 0
TBC = 116.0 lb !
(b) Using
TBC = 116.0 lb
R = R y = 0.68966 (116.0 lb ) − 140 lb = −60 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
Chapter 2, Solution 40.
(a)
Since R is to be vertical, Rx = 0
Then, Rx = ΣFx = 0
( 600 N ) cosα + ( 300 N ) cos (α + 35°) − ( 700 N ) cos α = 0
Expanding: 3 ( cos α cos 35° − sin α sin 35° ) − cos α = 0
Then:
1
cos 35° −  
3
tan α =
sin 35°
α = 40.265°
α = 40.3° !
(b)
Since R is vertical, R = Ry
Then:
R = Ry = ΣFy
R = ( 600 N ) sin 40.265° + ( 300 N ) sin 75.265° + ( 700 N ) sin 40.265°
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
Chapter 2, Solution 41.
Selecting the x axis along aa′, we write
Rx = ΣFx = 300 N + ( 400 N ) cos α + ( 600 N ) sin α
(1)
R y = ΣFy = ( 400 N ) sin α − ( 600 N ) cos α
(2)
(a) Setting R y = 0 in Equation (2):
Thus
tan α =
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 = 1021.11 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
Chapter 2, Solution 42.
(a)
Require Ry = ΣFy = 0:
( 900 lb ) cos 25° + (1200 lb ) sin 35° − TAE sin 65° = 0
or TAE = 1659.45 lb
TAE = 1659 lb !
(b)
R = ΣFx
R = − ( 900 lb ) sin 25° − (1200 lb ) cos 35° − (1659.45 lb ) cos 65°
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
Chapter 2, Solution 43.
Free-Body Diagram
Force Triangle
Law of Sines:
FAC
TBC
400 lb
=
=
sin 25° sin 60° sin 95°
(a)
FAC =
400 lb
sin 25° = 169.691 lb
sin 95°
(b)
TBC =
400
sin 60° = 347.73 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.
FAC = 169.7 lb !
TBC = 348 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 44.
Free-Body Diagram:
ΣFx = 0:
4
21
− TCA +
TCB = 0
5
29
or
 29  4 
TCB =    TCA
 21  5 
ΣFy = 0:
3
20
TCA +
TCB − ( 3 kN ) = 0
5
29
Then
3
20  29 4

TCA +
× TCA  − ( 3 kN ) = 0

5
29  21 5

or
TCA = 2.2028 kN
(a) TCA = 2.20 kN !
(b) TCB = 2.43 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
Chapter 2, Solution 45.
Free-Body Diagram:
ΣFy = 0:
− FB sin 50° + FC sin 70° = 0
FC =
ΣFx = 0:
sin 50°
( FB )
sin 70°
− FB cos 50° − FC cos 70° + 940 N = 0

 sin 50°  
FB cos 50° + cos 70° 
  = 940
 sin 70°  

FB = 1019.96 N
FC =
sin 50°
(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
Chapter 2, Solution 46.
Free-Body Diagram:
ΣFx = 0:
− TAB cos 25° − TAC cos 40° + ( 70 lb ) cos10° = 0
(1)
ΣFy = 0:
TAB sin 25° − TAC sin 40° + ( 70 lb ) sin10° = 0
(2)
Solving Equations (1) and (2) simultaneously:
(a) TAB = 38.6 lb !
(b) TAC = 44.3 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
Chapter 2, Solution 47.
Free-Body Diagram:
(a)
ΣFx = 0:
− TAB cos 30° + R cos 65° = 0
R=
ΣFy = 0:
cos 30°
TAB
cos 65°
− TAB sin 30° + R sin 65° − ( 550 N ) = 0
cos 30°


TAB  − sin 30° +
sin 65°  − 550 = 0
°
cos
65


(b)
R=
or
TAB = 405 N !
or
R = 830 N !
cos30°
( 450 N )
cos 65°
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
Chapter 2, Solution 48.
Free-Body Diagram At B:
ΣFx = 0:
−
12
17
TBA +
TBC = 0
13
293
TBA = 1.07591 TBC
or
5
TBA +
13
ΣFy = 0:
2
TBC − 300 N = 0
293
5

 293
TBC =  300 − TBA 
13

 2
TBC = 2567.6 − 3.2918 TBA
TBC = 2567.6 − 3.2918 (1.07591TBC )
TBC = 565.34 N
or
Free-Body Diagram At C:
ΣFx = 0:
−
TCD =
17
24
TBC +
TCD = 0
25
293
17
25
( 565.34 N )  
293
 24 
TCD = 584.86 N
ΣFy = 0:
WC = −
−
2
7
TBC +
TCD − WC = 0
25
293
2
7
( 565.34 N ) + ( 584.86 N )
25
293
or
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.
WC = 97.7 N !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 49.
Free-Body Diagram:
ΣFx = 0:
− 8 kips + 15 kips − TD cos 40° = 0
TD = 9.1378 kips
TD = 9.14 kips !
ΣFy = 0:
( 9.1378 kips ) sin 40° − TC
=0
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
Chapter 2, Solution 50.
Free-Body Diagram:
ΣFy = 0:
− 9 kips + TD sin 40° = 0
TD = 14.0015 kips
TD = 14.00 kips ΣFx = 0:
− 6 kips + TB − (14.0015 kips ) cos 40° = 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
Chapter 2, Solution 51.
Free-Body Diagram:
ΣFx = 0:
FC + ( 2.3 kN ) sin15° − ( 2.1 kN ) cos15° = 0
or
ΣFy = 0:
FC = 1.433 kN FD − ( 2.3 kN ) cos15° + ( 2.1 kN ) sin15° = 0
or
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.
FD = 1.678 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 52.
Free-Body Diagram:
ΣFx = 0:
− FB cos15° + 2.4 kN + (1.9 kN ) sin15° = 0
or
FB = 2.9938 kN
FB = 2.99 kN ΣFy = 0:
FD − (1.9 kN ) cos15° + ( 2.9938 kN ) sin15° = 0
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 53.
From Similar Triangles we have:
L2 − ( 2.5 m ) = ( 8 − L ) − ( 5.45 m )
2
2
2
− 6.25 = 64 − 16 L − 29.7025
or
cos β =
And
or
Then
L = 2.5342 m
5.45 m
8 m − 2.5342 m
β = 4.3576°
cos α =
2.5 m
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
(a)
W = 786 N "
(b)
TABC = 3.26 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
Chapter 2, Solution 54.
From Similar Triangles we have:
L2 − ( 3 m ) = ( 8 − L ) − ( 4.95 m )
2
2
2
− 9 = 64 − 16 L − 24.5025
L = 3.0311 m
or
cos β =
Then
β = 4.9989°
or
cos α =
And
4.95 m
8 m − 3.0311 m
3m
3.0311 m
α = 8.2147°
or
Free-Body Diagram At B:
ΣFx = 0:
(a)
− TABC cos α − TDE cos α + TABC cos β = 0
or
TDE =
cos β − cos α
TABC
cos α
ΣFy = 0:
TABC sin α + TDE sin α + TABC sin β − ( 720 N ) = 0


 cos β − cos α 
TABC sin α + sin α 
 + sin β  = 720
cos α




TABC =
( 720 ) cosα
sin (α + β )
Substituting for α and β gives
TABC =
( 720 ) cos8.2147°
sin (8.2147° + 4.9989° )
TABC = 3117.5 N
or
(b)
TDE =
TABC = 3.12 kN "
cos 4.9989° − cos8.2147°
( 3117.5 N )
cos8.2147°
TDE = 20.338 N
or
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.
TDE = 20.3 N "
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 55.
Free-Body Diagram At C:
3
15
15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5
17
17
or
ΣFy = 0:
−
17
TAC + 5 TBC = 750
5
(1)
4
8
8
TAC + TBC − (150 lb ) − 190 lb = 0
5
17
17
17
TAC + 2 TBC = 1107.5
5
or
(2)
Then adding Equations (1) and (2)
7 TBC = 1857.5
or
TBC = 265.36 lb
Therefore
(a) TAC = 169.6 lb !
(b)
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.
TBC = 265 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 56.
Free-Body Diagram At C:
3
15
15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5
17
17
17
or
− TAC + 5 TBC = 750
5
4
8
8
ΣFy = 0: TAC + TBC − (150 lb ) − W = 0
5
17
17
17
17
or
TAC + 2 TBC = 300 + W
5
4
17
7 TBC = 1050 + W
Adding Equations (1) and (2) gives
4
17
or
TBC = 150 +
W
28
−
Using Equation (1)
or
Now for
T ≤ 240 lb ⇒
or
(2)
17
17 

TAC + 5 150 +
W = 750
5
28 

25
W
28
25
TAC : 240 =
W
28
W = 269 lb
TAC =
TBC : 240 = 150 +
or
(1)
17
W
28
W = 148.2 lb
Therefore
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.
0 ≤ W ≤ 148.2 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 57.
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)
(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 − LO ) m
or
LO = 758 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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 58.
Free-Body Diagram At A:
First Note ...
With LAB =
( 22 in.)2 + (16.5 in.)2
LAB = 27.5 in.
LAD =
( 30 in.)2 + (16 in.)2
LAD = 34 in.
Then FAB = k AB ( LAB − LO )
= ( 9 lb/in.)( 27.5 in. − 22.5 in.)
= 45 lb
FAD = k AD ( LAD − LO )
= ( 3 lb/in.)( 34 in. − 22.5 in.)
= 34.5 lb
(a)
ΣFx = 0:
−
4
7
15
( 45 lb ) + TAC + ( 34.5 lb ) = 0
5
25
17
or TAC = 19.8529 lb
TAC = 19.85 lb !
(b)
ΣFy = 0:
3
24
8
( 45 lb ) + (19.8529 lb ) + ( 34.5 lb ) − W = 0
5
25
17
W = 62.3 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
Chapter 2, Solution 59.
(a)
For TAB to be a minimum
TAB must be perpendicular to TAC
∴ α + 10° = 60°
(b)
or
α = 50.0° W
or
TAB = 35.0 lb W
Then TAB = ( 70 lb ) sin 30°
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
Chapter 2, Solution 60.
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
Force triangle is isoceles with
2 β = 180° − 85°
β = 47.5°
P = 2 ( 900 N ) cos 47.5° = 1216 N
Since P > 0, solution is correct
(a)
P = 1216 N !
(b)
α = 77.5° !
α = 180° − 55° − 47.5° = 77.5°
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 61.
Note: Refer to Note in Problem 2.60
Free-Body Diagram At C:
Force Triangle
(a) Law of Cosines
P 2 = (1400 N ) + ( 700 N ) − 2 (1400 N )( 700 N ) cos85°
2
2
or
P = 1510 N !
or
α = 57.5° !
(b) Law of Sines
sin β
sin 85°
=
1400 N 1510 N
sin β = 0.92362
β = 67.461°
α = 180° − 55° − 67.461°
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
Chapter 2, Solution 62.
Free-Body Diagram At C:
ΣFx = 0:
2Tx − 1200 N = 0
Tx = 600 N
(Tx )2 + (Ty )
2
= T2
( 600 N )2 + (Ty )
2
= ( 870 N )
2
Ty = 630 N
By similar triangles:
1.8 m
AC
=
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 63.
TBC must be perpendicular to FAC to be as small as possible.
Free-Body Diagram: C
Force Triangle is a Right Triangle
α = 55°
α = 55° !
(a) We observe:
(b)
TBC = ( 400 lb ) sin 60°
or TBC = 346.41 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.
TBC = 346 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 64.
At Collar A ...
Fs = k ( L′AB − LAB )
Have
For stretched length
L′AB =
(12 in.)2 + (16 in.)2
L′AB = 20 in.
For unstretched length
LAB = 12 2 in.
(
)
Fs = 4 lb/in. 20 − 12 2 in.
Then
Fs = 12.1177 lb
For the collar ...
ΣFy = 0
−W +
4
(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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 65.
At Collar A ...
ΣFy = 0:
− 9 lb +
or
h
2
12 + h 2
Fs = 0
hFs = 9 144 + h 2
Fs = k ( L′AB − LAB )
Now
Where the stretched length
L′AB =
(12 in.)2 + h2
LAB = 12 2 in.
Then
hFs = 9 144 + h 2
Becomes
h 3 lb/in.

or
( h − 3)
( 144 + h
2
)
− 12 2  = 9 144 + h 2

144 + h 2 = 12 2 h
Solving Numerically ...
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.
h = 16.81 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 66.
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 = ( 310 N ) sin (180° − 70° − 30° )
= 305.29 N
TBC = 305 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
Chapter 2, Solution 67.
Free-Body Diagram At C:
Since TAB = TBC = 140 lb, Force triangle is isosceles:
With
2β + 75° = 180°
β = 52.5°
Then
α = 90° − 52.5° − 30°
α = 7.50°
P
= (140 lb ) cos 52.5°
2
P = 170.453 lb
P = 170.5 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.
7.50° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 68.
Free-Body Diagram of Pulley
(a)
(
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
2
T = 1373 N (b)
(
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
2
T = 1373 N (c)
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
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 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
Chapter 2, Solution 69.
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 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
Chapter 2, Solution 70.
Free-Body Diagram: Pulley C
(a)
ΣFx = 0: TACB ( cos 30° − cos 50° ) − ( 800 N ) cos 50° = 0
Hence
TACB = 2303.5 N
TACB = 2.30 kN (b)
ΣFy = 0: TACB ( sin 30° + sin 50° ) + ( 800 N ) sin 50° − Q = 0
( 2303.5 N )( sin 30° + sin 50° ) + (800 N ) sin 50° − Q = 0
or
Q = 3529.2 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.
Q = 3.53 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 71.
Free-Body Diagram: Pulley C
ΣFx = 0: TACB ( cos 30° − cos 50° ) − P cos 50° = 0
P = 0.34730TACB
or
(1)
ΣFy = 0: TACB ( sin 30° + sin 50° ) + P sin 50° − 2000 N = 0
1.26604TACB + 0.76604 P = 2000 N
or
(2)
(a) Substitute Equation (1) into Equation (2):
1.26604TACB + 0.76604 ( 0.34730TACB ) = 2000 N
Hence:
TACB = 1305.41 N
TACB = 1305 N (b) Using (1)
P = 0.34730 (1305.41 N ) = 453.37 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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 72.
First replace 30 lb forces by their resultant Q:
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:
27.4° < α < 223° !
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
Chapter 2, Solution 73.
(a)
Fx = ( 950 lb ) sin 50° cos 40°
= 557.48 lb
Fx = 557 lb !
Fy = − ( 950 lb ) cos 50°
= − 610.65 lb
Fy = − 611 lb !
Fz = ( 950 lb ) sin 50° sin 40°
= 467.78 lb
Fz = 468 lb !
(b)
cosθ x =
557.48 lb
950 lb
or θ x = 54.1° !
cosθ y =
− 610.65 lb
950 lb
or θ y = 130.0° !
cosθ z =
467.78 lb
950 lb
or θ z = 60.5° !
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 74.
(a)
Fx = − ( 810 lb ) cos 45° sin 25°
= − 242.06 lb
Fx = −242 lb !
Fy = − ( 810 lb ) sin 45°
= − 572.76 lb
Fy = − 573 lb !
Fz = (810 lb ) cos 45° cos 25°
= 519.09 lb
Fz = 519 lb !
(b)
cosθ x =
−242.06 lb
810 lb
or θ x = 107.4° !
cosθ y =
− 572.76 lb
810 lb
or θ y = 135.0° !
cosθ z =
519.09 lb
810 lb
or θ z = 50.1° !
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
Chapter 2, Solution 75.
(a)
Fx = ( 900 N ) cos 30° cos 25°
= 706.40 N
Fx = 706 N !
Fy = ( 900 N ) sin 30°
= 450.00 N
Fy = 450 N !
Fz = − ( 900 N ) cos 30° sin 25°
= − 329.04 N
Fz = − 329 N !
(b)
cosθ x =
706.40 N
900 N
or θ x = 38.3° !
cosθ y =
450.00 N
900 N
or θ y = 60.0° !
cosθ z =
−329.40 N
900 N
or θ z = 111.5° !
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 76.
(a)
Fx = − (1900 N ) sin 20° sin 70°
= − 610.65 N
Fx = − 611 N !
Fy = (1900 N ) cos 20°
= 1785.42 N
Fy = 1785 N !
Fz = (1900 N ) sin 20° cos 70°
= 222.26 N
Fz = 222 N !
(b)
cosθ x =
−610.65 N
1900 N
or θ x = 108.7° !
cosθ y =
1785.42 N
1900 N
or θ y = 20.0° !
cosθ z =
222.26 N
1900 N
or θ z = 83.3° !
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
Chapter 2, Solution 77.
(a)
Fx = (180 lb ) cos 35° sin 20°
= 50.430 lb
Fx = 50.4 lb !
Fy = − (180 lb ) sin 35°
= −103.244 lb
Fy = −103.2 lb !
Fz = (180 lb ) cos 35° cos 20°
= 138.555 lb
Fz = 138.6 lb !
(b)
cosθ x =
50.430 lb
180 lb
or θ x = 73.7° !
cosθ y =
−103.244 lb
180 lb
or θ y = 125.0° !
cosθ z =
138.555 lb
180 lb
or θ z = 39.7° !
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
Chapter 2, Solution 78.
(a)
Fx = (180 lb ) cos 30° cos 25°
= 141.279 lb
Fx = 141.3 lb !
Fy = − (180 lb ) sin 30°
= − 90.000 lb
Fy = − 90.0 lb !
Fz = (180 lb ) cos 30° sin 25°
= 65.880 lb
Fz = 65.9 lb !
(b)
cosθ x =
141.279 lb
180 lb
or θ x = 38.3° !
cosθ y =
−90.000 lb
180 lb
or θ y = 120.0° !
cosθ z =
65.880 lb
180 lb
or θ z = 68.5° !
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 79.
(a)
Fx = − ( 220 N ) cos 60° cos 35°
= − 90.107 N
Fx = − 90.1 N W
Fy = ( 220 N ) sin 60°
= 190.526 N
Fy = 190.5 N W
Fz = − ( 220 N ) cos 60° sin 35°
= − 63.093 N
Fz = − 63.1 N W
(b)
cosθ x =
−90.107 Ν
220 N
θ x = 114.2° W
cosθ y =
190.526 N
220 N
θ y = 30.0° W
cosθ z =
−63.093 N
220 N
θ z = 106.7° 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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 80.
(a)
Fx = 180 N
With Fx = F cos 60° cos 35°
180 N = F cos 60° cos 35°
or F = 439.38 N
F = 439 N !
(b)
cosθ x =
180 N
439.48 N
θ x = 65.8° !
Fy = ( 439.48 N ) sin 60°
Fy = 380.60 N
cosθ y =
380.60 N
439.48 N
θ y = 30.0° !
Fz = − ( 439.48 N ) cos 60° sin 35°
Fz = −126.038 N
cosθ z =
−126.038 N
439.48 N
θ z = 106.7° !
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
Chapter 2, Solution 81.
F=
Fx2 + Fy2 + Fz2
F =
( 65 N )2 + ( − 80 N )2 + ( − 200 N )2
F = 225 N !
cosθ x =
Fx
65 N
=
F
225 N
θ x = 73.2° !
cosθ y =
Fy
F
=
− 80 N
225 N
θ y = 110.8° !
cosθ z =
Fz
− 200 N
=
F
225 N
θ z = 152.7° !
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
Chapter 2, Solution 82.
F=
Fx2 + Fy2 + Fz2
F =
( 450 N )2 + ( 600 N )2 + ( −1800 N )2
F = 1950 N !
cosθ x =
Fx
450 N
=
F 1950 N
θ x = 76.7° !
cosθ y =
Fy
F
=
600 N
1950 N
θ y = 72.1° !
cosθ z =
Fz
−1800 N
=
1950 N
F
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 83.
(a)
(
We have ( cosθ x ) + cosθ y
2
( cosθ y )
2
= 1 − ( cosθ x ) − ( cosθ z )
Since Fy < 0 we must have
Thus
2
) + ( cosθ z )2 = 1
2
2
cosθ y < 0
cosθ y = − 1 − ( cos 43.2° ) − cos ( 83.8° )
2
2
cosθ y = − 0.67597
θ y = 132.5° !
(b) Then:
F =
F=
Fy
cosθ y
− 50 lb
− 0.67597
F = 73.968 lb
And
Fx = F cosθ x
Fx = ( 73.968 lb ) cos 43.2°
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 84.
(a)
(
We have ( cosθ x ) + cosθ y
2
2
) + ( cosθ z )2 = 1
(
or ( cosθ z ) = 1 − ( cosθ x ) − cosθ y
2
Since Fz < 0 we must have
Thus
2
)
2
cosθ z < 0
cosθ z = − 1 − ( cos113.2° ) − cos ( 78.4° )
2
2
cosθ z = − 0.89687
θ z = 153.7° !
(b) Then:
F =
Fz
− 35 lb
=
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 = ( 39.025 lb ) cos 78.4°
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 85.
(a)
We have
Fy = F cosθ y
Fy = ( 250 N ) cos 72.4°
Fy = 75.592 N
Fy = 75.6 N !
Then
F 2 = Fx2 + Fy2 + Fz2
( 250 N )2 = (80 N )2 + ( 75.592 N )2 + Fz2
Fz = 224.47 N
Fz = 224 N !
(b)
cosθ x =
Fx
F
cosθ x =
80 N
250 N
θ x = 71.3° !
cosθ z =
Fz
F
cosθ z =
224.47 N
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 86.
(a)
Have
Fx = F cosθ x
Fx = ( 320 N ) cos104.5°
Fx = − 80.122 N
Fx = − 80.1 N !
Then:
F 2 = Fx2 + Fy2 + Fz2
( 320 N )2 = ( − 80.122 N )2 + Fy2 + ( −120 N )2
Fy = 285.62 N
Fy = 286 N !
(b)
cosθ y =
Fy
cosθ y =
285.62 N
320 N
F
θ y = 26.8° !
cosθ z =
Fz
F
cosθ z =
−120 N
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 87.
!!!"
DB = ( 36 in.) i − ( 42 in.) j − ( 36 in.) k
DB =
( 36 in.)2 + ( − 42 in.)2 + ( − 36 in.)2
TDB = TDBλDB = TDB
TDB =
= 66 in.
!!!"
DB
DB
55 lb
( 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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 88.
!!!"
EB = ( 36 in.) i − ( 45 in.) j + ( 48 in.) k
EB =
( 36 in.)2 + ( − 45 in.)2 + ( 48 in.)2
TEB = TEBλEB = TEB
TEB =
= 75 in.
!!!"
EB
EB
60 lb
( 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
(TEB ) z
= − 36.0 lb !
= 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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 89.
!!!"
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
!!!"
BA 2100 N
( 4 m ) i + ( 20 m ) j − ( 5 m ) k 
= F
=
21 m 
BA
F = ( 400 N ) i + ( 2000 N ) j − ( 500 N ) k
Fx = + 400 N, Fy = + 2000 N, Fz = − 500 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
Chapter 2, Solution 90.
!!!"
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
!!!"
DA 1260 N
( 4 m ) i + ( 20 m ) j + (14.8 m ) k 
= F
=
25.2 m 
DA
F = ( 200 N ) i + (1000 N ) j + ( 740 N ) k
Fx = + 200 N, Fy = + 1000 N, Fz = + 740 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
Chapter 2, Solution 91.
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
TBG = TBG λBG = TBG
TBG =
uuuv
BG
BG
450 N
 − (1 m ) i + (1.85 m ) j − ( 0.8 m ) k 
2.25 m 
= − ( 200 N ) i + ( 370 N ) j − (160 N ) k
∴ (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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 92.
uuuuv
BH = ( 0.75 m ) i + (1.5 m ) j − (1.5 m ) k
BH =
( 0.75 m )2 + (1.5 m )2 + ( −1.5 m )2
= 2.25 m
TBH = TBH λBH = TBH
TBH =
uuuuv
BH
BH
600 N
( 0.75 m ) i + (1.5 m ) j − (1.5 m ) k 
2.25 m 
= ( 200 N ) i + ( 400 N ) j − ( 400 N ) k
∴ (TBH ) x = 200 N (TBH ) y = 400 N (TBH ) z
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.
= − 400 N COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 93.
P = ( 4 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
= (1.18479 kips ) i − ( 2 kips ) j + ( 3.2552 kips ) k
Q = (8 kips ) [ − cos 45° sin15°i + sin 45°j − cos 45° cos15°k ]
= − (1.46410 kips ) i + ( 5.6569 kips ) j − ( 5.4641 kips ) k
R = P + Q = − ( 0.27931 kip ) i + ( 3.6569 kips ) j − ( 2.2089 kips ) k
R=
( − 0.27931 kip)2 + (3.6569 kips )2 + ( − 2.2089 kips)2
R = 4.2814 kips
cosθ x =
cos θ y =
cos θ z =
R = 4.28 kips or
Rx − 0.27931 kip
=
= − 0.065238
R
4.2814 kips
Ry
R
=
3.6569 kips
= 0.85414
4.2814 kips
Rz − 2.2089 kips
=
= − 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 94.
P = ( 6 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
= (1.77719 kips ) i − ( 3 kips ) j + ( 4.8828 kips ) k
Q = ( 7 kips ) [ − cos 45° sin15°i + sin 45°j − cos 45° cos15°k ]
= − (1.28109 kips ) i + ( 4.94975 kips ) j − ( 4.7811 kips ) k
R = P + Q = ( 0.49610 kip ) i + (1.94975 kips ) j + ( 0.101700 kip ) k
R=
( 0.49610 kip)2 + (1.94975 kips)2 + ( 0.101700 kip)2
R = 2.0144 kips
cos θ x =
cos θ y =
cos θ z =
or
R = 2.01 kips or
θ x = 75.7° Rx 0.49610 kip
=
= 0.24628
R
2.0144 kips
Ry
R
=
1.94975 kips
= 0.967906
2.0144 kips
Rz 0.101700 kip
=
= 0.050486
R
2.0144 kips
θ y = 14.56° θ z = 87.1° 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
Chapter 2, Solution 95.
uuur
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB =
( − 600 mm )2 + (360 mm )2 + ( 270 mm )2
AB = 750 mm
uuuv
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
AC =
( − 600 mm )2 + ( 320 mm )2 + ( −510 mm )2
AC = 850 mm
uuur
AB
510 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
TAB = TAB
=
AB 750 mm 
TAB = − ( 408 N ) i + ( 244.8 N ) j + (183.6 N ) k
uuur
AC
765 N
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
TAC = TAC
=
AC 850 mm 
TAC = − ( 540 N ) i + ( 288 N ) j − ( 459 N ) k
R = TAB + TAC = − ( 948 N ) i + ( 532.8 N ) j − ( 275.4 N ) k
Then
and
R = 1121.80 N
− 948 N
cos θ x =
1121.80 N
532.8 N
cos θ y =
1121.80 N
− 275.4 N
cos θ z =
1121.80 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.
R = 1122 N θ x = 147.7° θ y = 61.6° θ z = 104.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 96.
!!!"
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB =
( − 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2
= 750 mm
AB = 750 mm
!!!"
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
AC =
( − 600 mm )2 + ( 320 mm) 2 + ( − 510 mm) 2
= 850 mm
AC = 850 mm
!!!"
AB
765 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
TAB = TAB
=
AB 750 mm 
TAB = − ( 612 N ) i + ( 367.2 N ) j + ( 275.4 N ) k
!!!"
AC
510 N
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
TAC = TAC
=
AC 850 mm 
TAC = − ( 360 N ) i + (192 N ) j − ( 306 N ) k
R = TAB + TAC = − ( 972 N ) i + ( 559.2 N ) j − ( 30.6 N ) k
Then
R = 1121.80 N R = 1122 N !
− 972 N
θ x = 150.1° !
1121.80 N
559.2 N
cos θ y =
θ y = 60.1° !
1121.80 N
− 30.6 N
cos θ z =
θ z = 91.6° !
1121.80 N
cos θ x =
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
Chapter 2, Solution 97.
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 )
(a)
( RA ) x = 0
R A = TAB + TAC
∴ ( RA ) x = ΣFx = 0:
( 760 lb) sin 50° cos 40° − TAC cos 45° sin 25° = 0
TAC = 1492.41 lb
or
∴ TAC = 1492 lb (b)
Then
( 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
RA = 2038.1 lb
RA = 2040 lb cosθ x =
0
2038.1 lb
θ x = 90.0° cos θ y =
−1543.81 lb
2038.1 lb
θ y = 139.2° cos θ z =
1330.65 lb
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
© 2007 The McGraw-Hill Companies.
θ z = 49.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 98.
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 )
(a)
( RA ) x = 0
R A = TAB + TAC
∴ ( RA ) x = ΣFx = 0:
TAB sin 50° cos 40° − ( 980 lb ) cos 45° sin 25° = 0
TAB = 499.06 lb
or
∴ TAB = 499 lb (b)
Then
and
( 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
RA = 1338.35 lb
RA = 1338 lb 0
1338.35 lb
θ x = 90.0° cos θ x =
cos θ y =
cos θ z =
−1013.75 lb
1338.35 lb
873.78 lb
1338.35 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.
θ y = 139.2° θ z = 49.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 99.
!!!"
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
Cable AB:
AB =
( − 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2
TAB = TAB
!!!"
AB
600 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
=
AB 750 mm 
TAB = − ( 480 N ) i + ( 288 N ) j + ( 216 N ) k
!!!"
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
Cable AC:
AC =
( − 600 mm )2 + ( 320 mm) 2 + ( − 510 mm) 2
TAC = TAC
TAC = −
Load P:
= 750 mm
= 850 mm
!!!"
AC
TAC
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
=
AC 850 mm 
60
32
51
TAC i +
TAC j −
TAC k
85
85
85
P = − Pj
(a)
( RA ) z
= ΣFz = 0:
( 216 N ) −
51
TAC = 0
85
or
TAC = 360 N !
(b)
( RA ) y = ΣFy = 0:
( 288 N ) +
32
TAC − P = 0
85
or
P = 424 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
Chapter 2, Solution 100.
uuur
AB = − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k
Cable AB:
AB =
( − 4 m)2 + ( −20 m)2 + (5 m )2
TAB = TAB
= 21 m
uuur
AB
T
= AB  − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k 
AB 21 m
uuur
AC = (12 m ) i − ( 20 m ) j + ( 3.6 m ) k
Cable AC:
AC =
(12 m )2 + ( − 20 m )2 + ( 3.6 m )2
TAC = TAC
= 23.6 m
uuur
AC 1770 N
(12 m ) i − ( 20 m ) j + ( 3.6 m ) k 
=
AC 23.6 m 
= ( 900 N ) i − (1500 N ) j + ( 270 N ) k
uuur
AD = − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k
Cable AD:
AD =
( − 4 m )2 + ( − 20 m )2 + (14.8 m )2
TAD = TAD
=
= 25.2 m
uuur
AD
TAD
 − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k 
=
AD 25.2 m 
TAD
 − (10 m ) i − ( 50 m ) j − ( 37 m ) k 
63 m 
Now...
R = TAB + TAC + TAD and R = Rj; Rx = Rz = 0
4
10
TAB + 900 −
TAD = 0
21
63
5
37
ΣFy = 0:
TAB + 270 −
TAD = 0
21
63
Solving equations (1) and (2) simultaneously yields:
ΣFx = 0:
−
(1)
(2)
TAD = 1.775 kN !
TAB = 3.25 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
Chapter 2, Solution 101.
d AB =
( −450 mm) 2 + ( 600 mm )2
= 750 mm
d AC =
( 600 mm )2 + ( − 320 mm)2
= 680 mm
d AD =
( 500 mm)2 + ( 600 mm )2 + ( 360 mm )2
TAB =
= 860 mm
TAB
 − ( 450 mm ) i + ( 600 mm ) j
750 mm 
TAB = ( − 0.6 i + 0.8 j) TAB
TAC =
TAC
( 600 mm ) j − ( 320 mm ) k 
680 mm 
8 
 15
TAC =  j − k  TAC
 17
17 
TAD =
TAD
( 500 mm ) i + ( 600 mm ) j + ( 360 mm ) k 
860 mm 
30
18 
 25
TAD =  i +
j+
k  TAD
 43
43
43 
W = −W j
At point A:
ΣF = 0:
i component:
− 0.6 TAB
k component:
−
TAB + TAC + TAD + W = 0
25
+
TAD = 0
43
 5   25 
TAB =     TAD
or
 3   43 
18
18
TAC +
TAD = 0
17
43
or
j component:
(1)
 17   18 
TAC =     TAD
 8   43 
15
30
TAC +
TAD − W = 0
17
43
15  17 18
 30
TAD − W = 0
0.8 TAB +
 ⋅ TAD  +
17 8 43
43
255
TAD − W = 0
0.8 TAB +
172
(2)
0.8 TAB +
(3)
From Equation (1):
 5   25 
6 kN =     TAD
 3   43 
or
TAD = 6.1920 kN
From Equation (3):
0.8 ( 6 kN ) +
255
( 6.1920 kN ) − W = 0
172
∴ W = 13.98 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
Chapter 2, Solution 102.
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
 3   43 
(1)
 17   18 
TAC =     TAD
 8   43 
(2)
0.8 TAB +
255
TAD − W = 0
172
From Equation (1)
 5   25 
TAB =     ( 4.3 kN )
 3   43 
or
TAB = 4.1667 kN
From Equation (3)
0.8 ( 4.1667 kN ) +
255
( 4.3 kN ) − W = 0
172
∴ W = 9.71 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.
(3)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 103.
uuur
AB = − ( 4.20 m ) i − ( 5.60 m ) j
AB = ( − 4.20 m ) + ( − 5.60 m ) = 7.00 m
uuur
AC = ( 2.40 m ) i − ( 5.60 m ) j + ( 4.20 m ) k
2
2
AC = ( 2.40 m ) + ( − 5.60 m ) + ( 4.20 m ) = 7.40 m
uuur
AD = − ( 5.60 m ) j − ( 3.30 m ) k
2
2
AD = ( − 5.60 m ) + ( − 3.30 m ) = 6.50 m
uuur
AB
TAB
= TAB
=
( − 4.20i − 5.60j)
AB 7.00 m
2
TAB = TAB λ AB
2
2
4 
 3
TAB =  − i − j TAB
 5
5 
uuur
AC
TAC
TAC = TAC λ AC = TAC
=
( 2.40i − 5.60j + 4.20k )
AC 7.40 m
28
21 
 12
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
33 
 56
TAD =  − j −
k  TAD
 65
65 
P = Pj
For equilibrium at point A:
ΣF = 0
TAB + TAC + TAD + P = 0
i component:
3
12
− TAB +
TAC = 0
5
37
or
TAB =
20
TAC
37
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.
(1)
COSMOS: Complete Online Solutions Manual Organization System
4
28
56
− TAB −
TAC −
TAD + P = 0
5
37
65
j component:
4
28
56  65 7

− TAB −
TAC −
 ⋅ TAC  + P = 0
5
37
65 11 37
4
700
− TAB −
TAC + P = 0
5
407
(2)
21
33
TAC −
TAD = 0
37
65
k component:
or
 65   7 
TAD =     TAC
 11   37 
(3)
From Equation (1):
 20 
259 N =   TAC
 37 
or
From Equation (2):
−
TAC = 479.15 N
4
700
( 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 104.
See Problem 2.103 for the analysis leading to the linear algebraic Equations (1), (2), and (3)
TAB =
−
20
TAC
37
(1)
4
700
TAB −
TAC + P = 0 (2)
5
407
 65  7 
TAD =    TAC
 11  37 
(3)
Substituting for TAC = 444 N into Equation (1)
TAB =
Gives
20
( 444 N )
37
TAB = 240 N
or
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 105.
d BA =
( −11 in.)2 + ( 9.6 in.)2
= 14.6 in.
dCA =
( 9.6 in.)2 + ( − 7.2 in.)2
= 12.0 in.
d DA =
( 9.6 in.)2 + ( 9.6 in.)2 + ( 4.8 in.)2
FBA = FBAλBA =
= 14.4 in.
FBA
( −11 in.) i + ( 9.6 in.) j
14.6 in. 
  11 
 9.6  
= FBA  − 
i + 
 j
14.6

 14.6  
 
FCA = FCAλCA =
FCA
( 9.6 in.) j − ( 7.2 in.) k 
12.0 in. 
 4 
3 
= FCA   j −   k 
5 
 5 
FDA = FDAλDA =
FDA
( 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
At point A:
ΣF = 0: FBA + FCA + FDA + P = 0
i
component:
j
component:
k
component:
 11 
2
−
FBA +   FDA = 0

 14.6 
3
 9.6 
4
2
 14.6  FBA +  5  FCA +  3  FDA − P = 0


 
 
3
1
−   FCA +   FDA = 0
5
3
(1)
(2)
(3)
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
 14.6  2 
FBA = 
  FDA
 11  3 
 14.6  2 
29.2 lb = 
  FDA
 11  3 
From Equation (1)
FDA = 33 lb
or
Solving Eqn. (3) for FCA gives:
5
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 106.
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
 14.6 
3
(1)
 9.6 
4
2

 FBA +   FCA +   FDA − P = 0
 14.6 
5
3
(2)
 3
1
−   FCA +   FDA = 0 (3)
5
 3
From Equation (1):
 14.6  2 
FBA = 
  FDA
 11  3 
From Equation (3):
5
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 
Since P = 45 lb
 838 

 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 107.
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
AC =
(18 m )2 + ( −30 m )2 + ( 5.4 m )2
TAC = T λ AC = TAC
= 35.4 m
uuur
AC
TAC
(18 m ) i − ( 30 m ) j + ( 5.4 m ) k 
=
35.4 m 
AC
TAC = TAC ( 0.50847i − 0.84746 j + 0.152542k )
and
uuur
AB = − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k
AB =
( −6 m )2 + ( −30 m )2 + ( 7.5 m )2
TAB = T λ AB = TAB
= 31.5 m
uuur
AB
TAB
 − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k 
=
AB 31.5 m 
TAB = TAB ( −0.190476i − 0.95238j + 0.23810k )
uuur
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
uuur
AD
TAD
 − ( 6 m ) i − ( 30 m ) j − ( 22.2 m ) k 
=
AD 37.8 m 
TAD = TAD ( −0.158730i − 0.79365j − 0.58730k )
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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COSMOS: Complete Online Solutions Manual Organization System
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 : − 0.190476TAB + 0.50847TAC − 0.158730TAD = 0
(1)
j: − 0.95238TAB − 0.84746TAC − 0.79365TAD + P = 0
(2)
k : 0.23810TAB + 0.152542TAC − 0.58730TAD = 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 "
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
Chapter 2, Solution 108.
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 109.
!!!"
AB = − ( 6.5 ft ) i − (8 ft ) j + ( 2 ft ) k
AB =
TAB =
( −6.5 ft )2 + ( −8 ft )2 + ( 2 ft )2
= 10.5 ft
TAB
 − ( 6.5 ft ) i − ( 8 ft ) j + ( 2 ft ) k 
10.5 ft 
= TAB ( −0.61905i − 0.76190 j + 0.190476k )
!!!"
AC = (1 ft ) i − ( 8 ft ) j + ( 4 ft ) k
AC =
TAC =
(1 ft )2 + ( −8 ft )2 + ( 4 ft )2
= 9 ft
TAC
(1 ft ) i − ( 8 ft ) j + ( 4 ft ) k 
9 ft 
= TAC ( 0.111111i − 0.88889 j + 0.44444k )
!!!"
AD = (1.75 ft ) i − ( 8 ft ) j − (1 ft ) k
AD =
TAD =
(1.75 ft )2 + ( −8 ft )2 + ( −1 ft )2
= 8.25 ft
TAD
(1.75 ft ) i − ( 8 ft ) j − (1 ft ) k 
8.25 ft 
= TAD ( 0.21212i − 0.96970 j − 0.121212k )
At A ΣF = 0
ΣFx = 0:
−0.61905TAB + 0.111111TAC + 0.21212TAD = 0
(1)
ΣFy = 0:
−0.76190TAB − 0.88889TAC − 0.96970TAD + W = 0
(2)
ΣFz = 0:
0.190476TAB + 0.44444TAC − 0.121212TAD = 0
(3)
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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Chapter 2, Solution 110.
See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below.
− 0.61905 TAB + 0.111111TAC + 0.21212 TAD = 0
(1)
− 0.76190 TAB − 0.88889 TAC − 0.96970 TAD + W = 0
(2)
0.190476 TAB + 0.44444 TAC − 0.121212TAD = 0
(3)
Now substituting for TAD = 220 lb Gives:
− 0.61905 TAB + 0.111111TAC + 46.662 = 0
(4)
− 0.76190 TAB − 0.88889 TAC − 213.33 + W = 0
(5)
0.190476 TAB + 0.44444 TAC − 26.666 = 0
(6)
Solving Equations (4) and (6) simultaneously gives
TAB = 79.992 lb and TAC = 25.716 lb
Substituting into Equation (5) yields
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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Chapter 2, Solution 111.
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 + 8j − 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
Then, isolating the factors of i, j, and k, we obtain three algebraic equations:
TBE
T
T
cos 45° + CF cos 30° − DG cos15° + P = 0
65
65
65
i:
or
TBE cos 45° + TCF cos30° − TDG cos15° + P 65 = 0
8
8
8
+ TCF
+ TDG
−W = 0
65
65
65
j: TBE
or
TBE + TCF + TDG − W
65
=0
8
k: −
or
(1)
(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 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
Chapter 2, Solution 112.
See Problem 2.111 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
(1)
65
=0
8
(2)
k : − TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(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 = ( −1.729P + 0.668 ) lb
Hence, for TCF > 0
or
−1.729P + 0.668 > 0
P < 0.386 lb
∴ φ ≤ P < 0.386 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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Chapter 2, Solution 113.
d DA =
( 400 mm )2 + ( − 600 mm )2
d DB =
( − 200 mm )2 + ( − 600 mm )2 + (150 mm )2
d DC =
( − 200 mm )2 + ( − 600 mm )2 + ( −150 mm )2
= 721.11 mm
= 650 mm
= 650 mm
TDA = TDAλDA
=
TDA
( 400 mm ) i − ( 600 mm ) j
721.11 mm 
= TDA ( 0.55470i − 0.83205 j)
TDB = TDBλDB
=
TDB
 − ( 200 mm ) i − ( 600 mm ) j + (150 mm ) k 
650 mm 
12
3 
 4
= TDB  − i −
j + k
13
13
13


TDC = TDC λDC
TDC =
TDC
 − ( 200 mm ) i − ( 600 mm ) j − (150 mm ) k 
650 mm 
12
3 
 4
= TDC  − i −
j − k
13
13
13


W = Wj
At point D ΣF = 0: TDA + TDB + TDC + W = 0
4
4
TDB − TDC = 0
13
13
12
12
− TDB − TDC + W = 0
13
13
3
3
TDB − TDC = 0
13
13
i component:
0.55470 TDA −
(1)
j component:
−0.83205 TDA
(2)
k component:
(
)
Setting W = (16 kg ) 9.81 m/s 2 = 156.96 N
And Solving Equations (1), (2), and (3) simultaneously:
TDA = 62.9 N !
TDB = 56.7 N !
TDC = 56.7 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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Chapter 2, Solution 114.
d DA =
( 400 mm )2 + ( − 600 mm )2
d DB =
( − 200 mm )2 + ( − 600 mm )2 + ( 200 mm )2
d DC =
( − 200 mm )2 + ( − 600 mm )2 + ( − 200 mm )2
= 721.11 mm
= 663.32 mm
= 663.32 mm
TDA = TDAλDA
=
TDA
( 400 mm ) i − ( 600 mm ) j
721.11 mm 
= TDA ( 0.55470i − 0.83205 j)
TDB = TDBλDB
=
TDB
 − ( 200 mm ) i − ( 600 mm ) j + ( 200 mm ) k 
663.32 mm 
= TDB ( − 0.30151i − 0.90454 j + 0.30151k )
TDC = TDC λDC
=
TDC
 − ( 200 mm ) i − ( 600 mm ) j − ( 200 mm ) k 
663.32 mm 
= TDC ( − 0.30151i − 0.90454 j − 0.30151k )
At point D
ΣF = 0: TDA + TDB + TDC + W = 0
0.55470 TDA − 0.30151TDB − 0.30151TDC = 0
i component:
−0.83205 TDA − 0.90454 TDB − 0.90454 TDC + W = 0
j component:
0.30151TDB − 0.30151TDC = 0
k component:
(
)
Setting W = (16 kg ) 9.81 m/s 2 = 156.96 N
And Solving Equations (1), (2), and (3) simultaneously:
TDA = 62.9 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
(1)
(2)
(3)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 115.
From the solutions of 2.107 and 2.108:
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 116.
d BA =
( 6 m )2 + ( 6 m )2 + ( 3 m )2
d AC =
( −10.5 m )2 + ( − 6 m )2 + ( − 8 m )2
d AD =
( − 6 m )2 + ( − 6 m )2 + ( 7 m )2
d AE =
( 6 m )2 + ( − 4.5 m )2
FBA = FBAλBA =
=9m
= 14.5 mm
= 11 mm
= 7.5 m
FBA
( 6 m ) i + ( 6 m ) j + ( 3 m ) k 
9m
2
1 
2
= FBA  i + j + k 
3
3 
3
TAC = TAC λ AC =
TAC
 − (10.5 m ) i − ( 6 m ) j − ( 8 m ) k 
14.5 m 
12
16 
 21
= TAC  − i −
j−
k
29
29
29 

TAD = TAD λ AD =
TAD
 − ( 6 m ) i − ( 6 m ) j + ( 7 m ) k 
11 m 
6
7 
 6
= TAD  − i − j + k 
11
11 
 11
WAE = WAE λ AE =
W
( 6 m ) i − ( 4.5 m ) j
7.5 m 
= W ( 0.8i − 0.6 j)
WO = − W j
At point A: ΣF = 0: FBA + TAC + TAD + WAE + WO = 0
i component:
2
21
6
FBA −
TAC − TAD + 0.8W = 0
3
29
11
(1)
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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j component:
2
12
6
FBA −
TAC − TAD − 1.6W = 0
3
29
11
(2)
k component:
1
16
7
FBA −
TAC + TAD = 0
3
29
11
(3)
(
)
Setting W = ( 20 kg ) 9.81 m/s 2 = 196.2 N
And Solving Equations (1), (2), and (3) simultaneously:
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 117.
ΣFx = 0:
− TAD ( sin 30° )( sin 50° ) + TBD ( sin 30° )( cos 40° ) + TCD ( sin 30° )( cos 60° ) = 0
Dividing through by sin 30° and evaluating:
− 0.76604 TAD + 0.76604 TBD + 0.5 TCD = 0
(1)
ΣFy = 0:
− TAD ( cos 30° ) − TBD ( cos 30° ) − TCD ( cos 30° ) + 62 lb = 0
or TAD + TBD + TCD = 71.591 lb
(2)
ΣFz = 0:
TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0
or
0.64279 TAD + 0.64279 TBD − 0.86603TCD = 0
(3)
Solving Equations (1), (2), and (3) simultaneously:
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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 118.
From the solutions to Problems 2.111 and 2.112, have
(2′)
TBE + TCF + TDG = 0.2 65
−TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(3)
TBE cos 45° + TCF cos 30° − TDG cos15° − P 65 = 0 (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°
TCF = 0.94455 − 0.37137TDG
or
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 = 0.66790 − 0.62863TDG
Substituting (4) and (5) into (1′) :
1.29028 − 1.73205TDG − P 65 = 0
∴ TDG is taut for P <
1.29028
lb
65
or 0 ≤ P ≤ 0.1600 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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Chapter 2, Solution 119.
d AB =
( − 30 ft )2 + ( 24 ft )2 + ( 32 ft )2
d AC =
( − 30 ft )2 + ( 20 ft )2 + ( −12 ft )2
TAB = TAB λ AB =
= 50 ft
= 38 ft
TAB
 − ( 30 ft ) i + ( 24 ft ) j + ( 32 ft ) k 
50 ft 
= TAB ( − 0.6i + 0.48 j + 0.64k )
TAC = TAC λ AC =
TAC
 − ( 30 ft ) i + ( 20 ft ) j − (12 ft ) k 
38 ft 
20
12 
 30
= TAC  − i +
j−
k
38
38 
 38
N=
16
30
Ni +
Nj
34
34
W = − (175 lb ) j
At point A: ΣF = 0: TAB + TAC + N + W = 0
i component:
− 0.6 TAB −
30
16
TAC +
N=0
38
34
(1)
j component:
0.48 TAB +
20
30
TAC +
N − 175 lb = 0
38
34
(2)
12
TAC = 0
38
Solving Equations (1), (2), and (3) simultaneously:
k component:
0.64 TAB −
(3)
TAB = 30.9 lb TAC = 62.5 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 120.
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.
Now at point A: ΣF = 0: TAB + TAC + N + W + P = 0
i component:
− 0.6 TAB −
30
16
TAC +
N=0
38
34
(1)
j component:
0.48 TAB +
20
30
TAC +
N − 175 lb = 0
38
34
(2)
k component:
0.64 TAB −
12
TAC − 45 lb = 0
38
(3)
Solving (1), (2), and (3) simultaneously:
TAB = 81.3 lb TAC = 22.2 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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Chapter 2, Solution 121.
Note: BE shares the same unit vector as AB.
Thus:
λBE = λ AB =
( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k
201.56 mm
TBE = TBE λBE =
TBE
( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k 
201.56 mm 
TCF = TCF λCF =
TCF
( 25 mm ) cos 30°i + ( 200 mm ) j + ( 25 mm ) sin 30°k 
201.56 mm 
TDG = TDG λDG =
TDG
 − ( 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 122.
See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
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 = 0.5 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 123.
uuur
DA = − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k
( − 8 ft )2 + ( 40 ft )2 + (10 ft )2
DA =
TDA =
= 42 ft
TADB
 − ( 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
( 3 ft )2 + ( 36 ft )2 + ( − 8 ft )2
DB =
TDB =
= 37 ft
TADB
( 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
( a − 8 ft )2 + ( − 24 ft )2 + ( −3 ft )2
DC =
TDC
TDC =
At D
( a − 8)2 + 585
=
( a − 8)2 + 585 ft
( a − 8 ft ) i − ( 24 ft ) j − ( 3 ft ) k 
ΣF = 0:
ΣFx = 0: − 0.190476 TADB + 0.081081TADB +
ΣFz = 0: 0.23810 TADB − 0.21622 TADB −
( a − 8)
TDC
( a − 8)2 + 585
3
( a − 8) + 585
2
=0
TDC = 0
(1)
(2)
Dividing equation (1) by equation (2) gives
( 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:
ΣFx = 0:
− 0.109395 TADB + 0.52705 TDC = 0
(3)
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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COSMOS: Complete Online Solutions Manual Organization System
And writing ΣFy = 0 gives:
1.92535 TADB − 0.84327 TDC − W = 0
(4)
Substituting into equation (3) for TDC = 17 lb gives:
− 0.109395 TADB + 0.52705 (17 lb ) = 0
or
TADB = 81.9 lb Substituting into equation (4) for TDC = 17 lb and TADB = 81.9 lb gives:
1.92535 ( 81.9 lb ) − 0.84327 (17 lb ) − W = 0
or
W = 143.4 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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 124.
See Problem 2.123 for the analysis leading to the linear
algebraic Equations (3) and (4) below:
− 0.109395 TADB + 0.52705 TDC = 0
(3)
1.92535 TADB − 0.84327 TDC − W = 0
(4)
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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 125.
d AB =
( − 2.7 m )2 + ( 2.4 m )2 + ( − 3.6 m )2
d AC =
( 2.4 m )2 + (1.8 m )2
d AD =
(1.2 m )2 + ( 2.4 m )2 + ( − 0.3 m )2
= 2.7 m
d AE =
( − 2.4 m )2 + ( 2.4 m )2 + (1.2 m )2
= 3.6 m
= 5.1 m
=3m
TAB = TAB λ AB
=
TAB
 − ( 2.7 m ) i + ( 2.4 m ) j − ( 3.6 m ) k 
5.1 m 
8
12 
 9
= TAB  − i +
j − k
17
17 
 17
TAC = TAC λ AC
=
TAC
( 2.4 m ) j + (1.8 m ) k 
3m
= TAC ( 0.8 j + 0.6k )
TAD = 2TADE λ AD
=
2TADE
(1.2 m ) i + ( 2.4 m ) j − ( 0.3 m ) k 
2.7 m 
16
2 
8
= 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
1 
 2
= TADE  − i + j + k 
3
3 
 3
W = − Wj
At point A:
ΣF = 0:
TAB + TAC + TAD + TAE + W = 0
9
8
2
TAB + TADE − TADE = 0
17
9
3
i component:
−
j component:
8
16
2
TAB + 0.8 TAC + TADE + TADE − W = 0
17
9
3
k component:
−
12
2
1
TAB + 0.6 TAC − TADE + TADE = 0
17
9
3
(1)
(2)
(3)
Simplifying (1), (2), (3):
− 81TAB + 34 TADE = 0
(1′)
72 TAB + 122.4 TAC + 374 TADE = 153 W
(2′)
−108 TAB + 91.8 TAC + 17 TADE = 0
(3′)
Setting W = 1400 N and solving (1), (2), (3) simultaneously:
TAB = 203 N "
TAC = 149.6 N "
TADE = 485 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
Chapter 2, Solution 126.
See Problem 2.125 for the analysis leading to the linear algebraic Equations (1′ ) , ( 2′ ) , and ( 3′ ) below:
i component:
− 81 TAB + 34 TADE = 0
(1′)
j component:
72 TAB + 122.4 TAC + 37.4 TADE = 153 W
( 2′)
k component:
−108 TAB + 91.8 TAC + 17 TADE = 0
( 3′)
Setting TAB = 300 N and solving (1), (2), (3) simultaneously:
(a)
TAC = 221 N !
(b) TADE = 715 N !
(c)
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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W = 2060 N !
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Chapter 2, Solution 127.
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
uuur
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 x i + 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 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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COSMOS: Complete Online Solutions Manual Organization System
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)
or
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.
Q = 1.963 kN !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 128.
From the analysis of Problem 2.127, particularly the results:
y 2 + z 2 = 0.84 m 2
TAB =
680 N
y
Q=
680 N
z
y
With y = 550 mm = 0.55 m, we obtain:
z 2 = 0.84 m 2 − ( 0.55 m )
2
∴ z = 0.73314 m
and
TAB =
(a)
or
680 N
= 1236.36 N
0.55
TAB = 1.236 kN !
and
Q = 1236.36 ( 0.73314 ) N = 906 N
(b)
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 129.
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° (b)
β = 180° − ( 30° + 45.6° )
= 104.4°
Then:
R
14 lb
=
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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 130.
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:
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 131.
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:
P=
(a)
450 N
= 549.3 N
cos 35°
P = 549 N !
Px = ( 450 N ) tan 35°
(b)
= 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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 132.
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 !
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
Chapter 2, Solution 133.
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
0.104903TACB = 14.8629 kN
TACB = 141.682 kN
(b) TACB = 141.7 kN !
Now
ΣFy = 0: TACB ( sin 42° − sin 32° ) + ( 20 kN ) sin 42° − W = 0
or
(141.682 kN )( 0.139211) + ( 20 kN )( 0.66913) − W
=0
(a) W = 33.1 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
Chapter 2, Solution 134.
Free-Body Diagram: Pulley A
ΣFx = 0: 2P sin 25° − P cos α = 0
and
cos α = 0.8452
For
or
α = ±32.3°
α = +32.3°
ΣFy = 0: 2P cos 25° + P sin 32.3° − 350 lb = 0
or P = 149.1 lb
For
32.3° α = −32.3°
ΣFy = 0: 2P cos 25° + P sin − 32.3° − 350 lb = 0
or P = 274 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.
32.3° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 135.
Fx = F sin 30° sin 50° = 220.6 N (Given)
(a)
F =
220.6 N
= 575.95 N
sin30° sin50°
F = 576 N !
cosθ x =
(b)
Fx
220.6
=
= 0.38302
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° !
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 136.
Fz = F cosθ z = ( 600 lb ) cos136.8°
(a)
= −437.38 lb
Fz = −437 lb !
Then:
F 2 = Fx2 + Fy2 + Fz2
2
( ) + ( −437.38 lb )2
So: ( 600 lb ) = ( 200 lb ) + Fy
2
2
Hence: Fy = −
(b)
cosθ x =
( 600 lb )2 − ( 200 lb )2 − ( −437.38 lb )2
= −358.75 lb
Fy = −359 lb !
Fx
200
=
= 0.33333
F
600
θ x = 70.5° !
cosθ y =
Fy
F
=
−358.75
= −0.59792
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
© 2007 The McGraw-Hill Companies.
θ y = 126.7° !
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Chapter 2, Solution 137.
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 + 0.64278 j − 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° !
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
Chapter 2, Solution 138.
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
uuur
AB = − ( 0.72 m ) i + (1.2 m ) j − ( 0.54 m ) k,
AB = 1.5 m
uuur
AC = (1.2 m ) j + ( 0.64 m ) k,
AC = 1.36 m
uuur
AD = ( 0.8 m ) i + (1.2 m ) j − ( 0.54 m ) k,
AD = 1.54 m
uuur
AB
TAB = TABλ AB = TAB
= ( −0.48i + 0.8j − 0.36k ) TAB
and
AB
uuur
AC
TAC = TAC λ AC = TAC
= ( 0.88235j + 0.47059k ) TAC
AC
uuur
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
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 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
Chapter 2, Solution 139.
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
AB =
( −32 in.)2 + ( −48 in.)2 + ( 36 in.)2
TAB = T λ AB = TAB
= 68 in.
uuur
AB
T
= AB  − ( 32 in.) i − ( 48 in.) j + ( 36 in.) k 
68 in.
AB
TAB = TAB ( −0.47059i − 0.70588 j + 0.52941k )
uuur
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.
uuur
AC
T
= AC ( 45 in.) i − ( 48 in.) j + ( 36 in.) k 
75 in.
AC
TAC = TAC ( 0.60i − 0.64 j + 0.48k )
uuur
AD = ( 25 in.) i − ( 48 in.) j − ( 36 in.) k
Finally,
AD =
( 25 in.)2 + ( −48 in.)2 + ( −36 in.)2
= 65 in.
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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COSMOS: Complete Online Solutions Manual Organization System
TAD = T λ AD = TAD
uuur
AD
T
= AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k 
65 in.
AD
TAD = TAD ( 0.38461i − 0.73846 j − 0.55385k )
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 : − 0.47059TAB + 0.60TAC − 0.38461TAD = 0
(1)
j: − 0.70588TAB − 0.64TAC − 0.73846TAD + W = 0
(2)
k : 0.52941TAB + 0.48TAC − 0.55385TAD = 0
(3)
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 140.
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
AB =
( −0.48 m )2 + ( 0.72 m )2 + ( −0.16 m )2
TAB = T λ AB = TAB
= 0.88 m
uuur
AB
TAB
 − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k 
=
AB
0.88 m 
TAB = TAB ( −0.54545i + 0.81818 j − 0.181818k )
and
uuur
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
uuur
AC
TAC
( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k 
=
AC
0.77 m 
TAC = TAC ( 0.31169i + 0.93506 j − 0.16883k )
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:
i:
( −0.54545 + 0.31169 ) T
+P=0
P = 0.23376T
or
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
W = 1.75324T
or
k:
( −0.181818 − 0.16883) T
+Q =0
Q = 0.35065T
or
With W = 1200 N:
T =
1200 N
= 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.