Solution Beer

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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

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