# Geodesta

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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

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

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

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

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

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

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

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

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