SOLUCIONARIO DE MECANICA PARA ING

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 Solutions 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 for Engineers: Statics and Dynamics, 8/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.

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 6

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