Solucionario mecanica vectorial para ingenieria estatica

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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 Online SolutionsManual 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 9....
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