Beer Jhonson
Chapter 2, Solution 1.
(a)
(b)
We measure:
R = 37 lb, α = 76°
R = 37 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.
76° !
COSMOS: CompleteOnline Solutions Manual Organization System
Chapter 2, Solution 2.
(a)
(b)
We measure:
R = 57 lb, α = 86°
R = 57 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.
86° !
COSMOS: Complete Online SolutionsManual Organization System
Chapter 2, Solution 3.
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 10.5 kN
α = 22.5°
R = 10.5 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.
22.5° !COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 4.
(a)
Parallelogram law:
We measure:
R = 5.4 kN α = 12°
R = 5.4 kN
R = 5.4 kN
(b)
12° !
12° !
Triangle rule:
We measure:
R = 5.4 kN α = 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 !
VectorMechanics 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 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
(b)
α = 25.1° !
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
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-HillCompanies.
Faa′ = 266 N !
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)
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.
α = 72.2° !
R = 1.391 kN !
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....
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