Solucionario De Mecanica Vectotial Para Ingenieros: Estatica Ed Mcgraw Hill
Páginas: 80 (19797 palabras)
Publicado: 22 de noviembre de 2012
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 1.
(a)
(b)
We measure:
R = 37 lb, α = 76° R = 37 lb
76° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete OnlineSolutions Manual Organization System
Chapter 2, Solution 2.
(a)
(b)
We measure:
R = 57 lb, α = 86° R = 57 lb
86° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions ManualOrganization System
Chapter 2, Solution 3.
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 10.5 kN
α = 22.5°
R = 10.5 kN
22.5° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: CompleteOnline Solutions Manual Organization System
Chapter 2, Solution 4.
(a)
Parallelogram law: We measure:
R = 5.4 kN α = 12° R = 5.4 kN
12° !
(b)
Triangle rule:
We measure:
R =
5.4 kN α = 12° R = 5.4 kN
12° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J.Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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 andDynamics, 8/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 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
α = 25.1° !
(b)β + 45° + 25.104° = 180° β = 109.896°
Using the Law of Sines
Faa′ 200 N = sin β sin 45° Faa′ 200 N = sin109.896° sin 45°
or
Faa′ = 266 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete
OnlineSolutions Manual Organization System
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°
(a) (b)
α = 72.2° !
R = 1.391 kN !
Vector Mechanics forEngineers: Statics and Dynamics, 8/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 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution...
Chapter 2, Solution 1.
(a)
(b)
We measure:
R = 37 lb, α = 76° R = 37 lb
76° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete OnlineSolutions Manual Organization System
Chapter 2, Solution 2.
(a)
(b)
We measure:
R = 57 lb, α = 86° R = 57 lb
86° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions ManualOrganization System
Chapter 2, Solution 3.
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 10.5 kN
α = 22.5°
R = 10.5 kN
22.5° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: CompleteOnline Solutions Manual Organization System
Chapter 2, Solution 4.
(a)
Parallelogram law: We measure:
R = 5.4 kN α = 12° R = 5.4 kN
12° !
(b)
Triangle rule:
We measure:
R =
5.4 kN α = 12° R = 5.4 kN
12° !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J.Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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 andDynamics, 8/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 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
α = 25.1° !
(b)β + 45° + 25.104° = 180° β = 109.896°
Using the Law of Sines
Faa′ 200 N = sin β sin 45° Faa′ 200 N = sin109.896° sin 45°
or
Faa′ = 266 N !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete
OnlineSolutions Manual Organization System
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°
(a) (b)
α = 72.2° !
R = 1.391 kN !
Vector Mechanics forEngineers: Statics and Dynamics, 8/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 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution...
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