Solution manual vector mechanics engineers dynamics 8th beer chapter 02
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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° !
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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° !
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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° !
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Chapter 2, Solution 4.
(a)
Parallelogram law:
We measure:
R = 5.4 kN α = 12°
(b)
R = 5.4 kN
12° !
R = 5.4 kN
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.
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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 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-Hill Companies.
Faa′ = 266 N !
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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°
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.
(a)
α = 72.2° !
(b)
R = 1.391 kN !
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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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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 62°
or (a) F1 = 22.3 lb !
(b) R = 13.95 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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Chapter 2, Solution 10.
Using the Law of Sines:
60 N
80 N
=
sin α sin10°
or α = 7.4832°
β = 180° − (10° + 7.4832° )
= 162.517°
Then:
R
80 N
=
sin162.517° sin10°
or R = 138.405 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.
(a)
α = 7.48° !
(b)
R = 138.4 N !
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Chapter 2, Solution 11.
Using the triangle rule and the Law of Sines
Have:
β = 180° − ( 35° + 25° )
= 120°
Then:
P
R
80 lb
=
=
sin 35° sin120° sin 25°
or (a) P = 108.6 lb !
(b) R = 163.9 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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Chapter 2, Solution 12.
Using the triangle rule and the Law of Sines
(a) Have:
80 lb
70 lb
=
sin α
sin 35°
sin α = 0.65552
α = 40.959°
or α = 41.0° !
β = 180 − ( 35° + 40.959° )
(b)
= 104.041°
Then:
R
70 lb
=
sin104.041° sin 35°
or R = 118.4 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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Chapter 2, Solution 13.
We observe that force P is minimum when α = 90°.
Then:
(a)
P = ( 80 lb ) sin 35°
or P = 45.9 lb
!
And:
(b)
R = ( 80 lb ) cos 35°
or R = 65.5 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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Chapter 2, Solution 14.
For TBC to be a minimum,
R and TBC must be perpendicular.
Thus
TBC = ( 70 N ) sin 4°
= 4.8829 N
And
R = ( 70 N ) cos 4°
= 69.829 N
(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.
TBC = 4.88 N
6.00° !
R = 69.8 N !
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Chapter 2, Solution 15.
Using the force triangle and the Laws of Cosines and Sines
We have:
γ = 180° − (15° + 30° )
= 135°
Then:
R 2 = (15 lb ) + ( 25 lb ) − 2 (15 lb )( 25 lb ) cos135°
2
2
= 1380.33 lb2
or
R = 37.153 lb
and
25 lb 37.153 lb
=
sin β
sin135°
25 lb
sin β =
sin135°
37.153 lb
= 0.47581
β = 28.412°
Then:
α + β + 75° = 180°
α = 76.588°
R = 37.2 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.
76.6° !
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Chapter 2, Solution 16.
Using the Law of Cosines and the Law of Sines,
R 2 = ( 45 lb ) + (15 lb ) − 2 ( 45 lb )(15 lb ) cos135°
2
2
or R = 56.609 lb
56.609 lb 15 lb
=
sin135°
sinθ
or θ = 10.7991°
R = 56.6 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.
85.8° !
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Chapter 2, Solution 17.
γ = 180° − 25° − 50°
γ = 105°
Using the Law of Cosines:
R 2 = ( 5 kN ) + ( 8 kN ) − 2 ( 5 kN )( 8 kN ) cos105°
2
2
or R = 10.4740 kN
Using the Law of Sines:
10.4740 kN 8 kN
=
sin105°
sin β
or β = 47.542°
and α = 47.542° − 25°
α = 22.542°
R = 10.47 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° "
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Chapter 2, Solution 19.
Using the force triangle and the Laws of Cosines and Sines
We have:
Then:
γ = 180° − ( 45° + 25° ) = 110°
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2
2
= 1710.42 kN 2
R = 41.357 kN
and
20 kN
41.357 kN
=
sin α
sin110°
20 kN
sin α =
sin110°
41.357 kN
= 0.45443
α = 27.028°
Hence:
φ = α + 45° = 72.028°
R = 41.4 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.
72.0° !
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Chapter 2, Solution 19.
Using the force triangle and the Laws of Cosines and Sines
We have:
Then:
γ = 180° − ( 45° + 25° ) = 110°
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2
2
= 1710.42 kN 2
R = 41.357 kN
and
20 kN
41.357 kN
=
sin α
sin110°
20 kN
sin α =
sin110°
41.357 kN
= 0.45443
α = 27.028°
Hence:
φ = α + 45° = 72.028°
R = 41.4 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.
72.0° !
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Chapter 2, Solution 20.
Using the force triangle and the Laws of Cosines and Sines
We have:
Then:
γ = 180° − ( 45° + 25° ) = 110°
R 2 = ( 30 kN ) + ( 20 kN ) − 2 ( 30 kN )( 20 kN ) cos110°
2
2
= 1710.42 kN 2
R = 41.357 kN
and
30 kN
41.357 kN
=
sin α
sin110°
30 kN
sin α =
sin110°
41.357 kN
= 0.68164
α = 42.972°
Finally:
φ = α + 45° = 87.972°
R = 41.4 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.
88.0° !
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Chapter 2, Solution 21.
2.4 kN Force:
Fx = ( 2.4 kN ) cos 50°
Fx = 1.543 kN
Fy = ( 2.4 kN ) sin 50°
Fy = 1.839 kN
1.85 kN Force:
Fx = (1.85 kN ) cos 20°
Fx = 1.738 kN
Fy = (1.85 kN ) sin 20°
Fy = 0.633 kN
1.40 kN Force:
Fx = (1.40 kN ) cos 35°
Fx = 1.147 kN
Fy = − (1.40 kN ) sin 35°
Fy = −0.803 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 22.
Fx = ( 5 kips ) cos 40°
5 kips:
or Fx = 3.83 kips
Fy = ( 5 kips ) sin 40°
or Fy = 3.21 kips
7 kips:
Fx = − ( 7 kips ) cos 70°
or Fx = −2.39 kips
Fy = ( 7 kips ) sin 70°
or Fy = 6.58 kips
9 kips:
Fx = − ( 9 kips ) cos 20°
or Fx = −8.46 kips
Fy = ( 9 kips ) sin 20°
or Fy = 3.08 kips
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 23.
Determine the following distances:
680 N Force:
dOA =
( −160 mm )2 + ( 300 mm )2
dOB =
( 600 mm )2 + ( 250 mm )2
dOC =
( 600 mm )2 + ( −110 mm )2
Fx = 680 N
= 340 mm
= 650 mm
= 610 mm
( −160 mm )
340 mm
Fx = − 320 N !
( 300 mm )
Fy = 680 N
340 mm
Fy = 600 N !
390 N Force:
Fx = 390 N
( 600 mm )
650 mm
Fx = 360 N !
Fy = 390 N
( 250 mm )
650 mm
Fy = 150 N !
610 N Force:
Fx = 610 N
( 600 mm )
610 mm
Fx = 600 N !
Fy = 610 N
( −110 mm )
610 mm
Fy = −110 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.
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Chapter 2, Solution 24.
We compute the following distances:
OA =
( 48)2 + ( 90 )2
= 102 in.
OB =
( 56 )2 + ( 90 )2
= 106 in.
OC =
(80 )2 + ( 60 )2
= 100 in.
Then:
204 lb Force:
Fx = − ( 204 lb )
48
,
102
Fy = + ( 204 lb )
90
,
102
Fx = −96.0 lb
Fy = 180.0 lb
212 lb Force:
Fx = + ( 212 lb )
56
,
106
Fx = 112.0 lb
90
,
106
Fy = 180.0 lb
Fx = − ( 400 lb )
80
,
100
Fx = −320 lb
Fy = − ( 400 lb )
60
,
100
Fy = −240 lb
Fy = + ( 212 lb )
400 lb Force:
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 25.
(a)
P=
=
Py
sin 35°
960 N
sin 35°
or P = 1674 N
(b)
Px =
=
Py
tan 35°
960 N
tan 35°
or Px = 1371 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.
