Solution Beer
Chapter 2, Solution 1.
(a)
(b)
We measure:
R = 37 lb, α = 76° R = 37 lb
76° !
Vector Mechanics for Engineers: Staticsand Dynamics, 8/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 kNα = 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