Vector Mechanics for Engineers - Statics 8th Edition Ch 2 Solutions
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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 © 2007 The McGraw-Hill Companies.
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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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° !
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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° !
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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.
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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.
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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.
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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.
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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 !
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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.
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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.
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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.
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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.
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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 !
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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.
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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.
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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
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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.
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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.
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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 "
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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 !
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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 !
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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 © 2007 The McGraw-Hill Companies.
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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.
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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.
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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.
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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.
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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.
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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 !
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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.
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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.
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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.
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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°
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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.
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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.
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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 © 2007 The McGraw-Hill Companies.
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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.
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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°
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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.
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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°
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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°
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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.
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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.
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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 © 2007 The McGraw-Hill Companies.
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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.
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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.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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.
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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.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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 © 2007 The McGraw-Hill Companies.
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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 !
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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.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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.
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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.
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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°
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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 © 2007 The McGraw-Hill Companies.
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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.
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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 © 2007 The McGraw-Hill Companies.
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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 © 2007 The McGraw-Hill Companies.
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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.
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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.
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