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
© 2007 The McGraw-Hill Companies.
76° !
COSMOS: Complete Online 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 Solutions Manual 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°
(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.
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 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 !
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°
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 !
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.
COSMOS: Complete Online Solutions Manual Organization System
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.
COSMOS: Complete Online Solutions Manual Organization System
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 !
COSMOS: Complete Online Solutions Manual Organization System
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.
COSMOS: Complete Online Solutions Manual Organization System
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.
COSMOS: Complete Online Solutions Manual Organization System
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.
COSMOS: Complete Online Solutions Manual Organization System
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 !
COSMOS: Complete Online Solutions Manual Organization System
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° !
COSMOS: Complete Online Solutions Manual Organization System
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° !
COSMOS: Complete Online Solutions Manual Organization System
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° "
COSMOS: Complete Online Solutions Manual Organization System
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° !
COSMOS: Complete Online Solutions Manual Organization System
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° !
COSMOS: Complete Online Solutions Manual Organization System
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° !
COSMOS: Complete Online Solutions Manual Organization System
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 = 340 mm
dOB =
( 600 mm )2 + ( 250 mm )2 = 650 mm
dOC =
( 600 mm )2 + ( −110 mm )2 = 610 mm
Fx = 680 N
( −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.
COSMOS: Complete Online Solutions Manual Organization System
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
Fx = −96.0 lb 90
,
102
Fy = 180.0 lb Fy = + ( 204 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 26.
(a)
P=
Px
cos 40°
P=
30 lb
cos 40°
or P = 39.2 lb !
(b)
Py = Px tan 40°
Py = ( 30 lb ) tan 40°
or Py = 25.2 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 27.
(a)
Py = 100 N
P=
P=
Py
sin 75°
100 N
sin 75°
or P = 103.5 N "
(b)
Px =
Px =
Py
tan 75°
100 N
tan 75°
or Px = 26.8 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 28.
We note:
CB exerts force P on B along CB, and the horizontal component of P is Px = 260 lb.
Then:
(a)
Px = P sin 50°
P=
Px
sin 50°
=
260 lb
sin 50°
= 339.40 lb
(b)
P = 339 lb !
Px = Py tan 50°
Py =
Px
tan 50°
=
260 lb
tan 50°
= 218.16 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.
Py = 218 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 29.
(a)
P=
45 N
cos 20°
or P = 47.9 N !
(b)
Px = ( 47.9 N ) sin 20°
or Px = 16.38 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 30.
(a)
P=
18 N
sin 20°
or P = 52.6 N !
(b)
Py =
18 N
tan 20°
or Py = 49.5 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 31.
From the solution to Problem 2.21:
F2.4 = (1.543 kN ) i + (1.839 kN ) j
F1.85 = (1.738 kN ) i + ( 0.633 kN ) j
F1.40 = (1.147 kN ) i − ( 0.803 kN ) j
R = ΣF = ( 4.428 kN ) i + (1.669 kN ) j
R=
( 4.428 kN )2 + (1.669 kN )2
= 4.7321 kN
tan α =
1.669 kN
4.428 kN
α = 20.652°
R = 4.73 kN
20.6° !
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 32.
From the solution to Problem 2.22:
F5 = ( 3.83 kips ) i + ( 3.21 kips ) j
F7 = − ( 2.39 kips ) i + ( 6.58 kips ) j
F9 = − ( 8.46 kips ) i + ( 3.08 kips ) j
R = ΣF = − ( 7.02 kips ) i + (12.87 ) j
R=
( − 7.02 kips )2 + (12.87 kips )2 = 14.66 kips
 12.87 
 = 61.4°
 − 7.02 
α = tan −1 
R = 14.66 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.
61.4° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 33.
From the solution to Problem 2.24:
FOA = − ( 48.0 lb ) i + ( 90.0 lb ) j
FOB = (112.0 lb ) i + (180.0 lb ) j
FOC = − ( 320 lb ) i − ( 240 lb ) j
R = ΣF = − ( 256 lb ) i + ( 30 lb ) j
R=
( − 256 lb )2 + ( 30 lb )2
= 257.75 lb
tan α =
30 lb
−256 lb
α = − 6.6839°
R = 258 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.
6.68° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 34.
From Problem 2.23:
FOA = − ( 320 N ) i + ( 600 N ) j
FOB = ( 360 N ) i + (150 N ) j
FOC = ( 600 N ) i − (110 N ) j
R = ΣF = ( 640 N ) i + ( 640 N ) j
R=
( 640 N )2 + ( 640 N )2
= 905.097 N
tan α =
640 N
640 N
α = 45.0°
R = 905 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.
45.0° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 35.
Cable BC Force:
Fx = − (145 lb )
Fy = (145 lb )
84
= −105 lb
116
80
= 100 lb
116
100-lb Force:
Fx = − (100 lb )
3
= −60 lb
5
Fy = − (100 lb )
4
= −80 lb
5
156-lb Force:
Fx = (156 lb )
12
= 144 lb
13
Fy = − (156 lb )
5
= −60 lb
13
and
Rx = ΣFx = −21 lb,
R=
Ry = ΣFy = −40 lb
( −21 lb )2 + ( −40 lb )2 = 45.177 lb
Further:
tan α =
α = tan −1
40
21
40
= 62.3°
21
Thus:
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.
R = 45.2 lb
62.3° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 36.
(a)
Since R is to be horizontal, Ry = 0
Then, Ry = ΣFy = 0
90 lb + ( 70 lb ) sin α − (130 lb ) cos α = 0
(13) cosα = ( 7 ) sin α + 9
13 1 − sin 2 α = ( 7 ) sin α + 9
Squaring both sides:
(
)
169 1 − sin 2 α = ( 49 ) sin 2 α + (126 ) sin α + 81
( 218) sin 2 α + (126 ) sin α − 88 = 0
Solving by quadratic formula:
(b)
sin α = 0.40899
or
α = 24.1° !
or
R = 117.0 lb !
Since R is horizontal, R = Rx
Then, R = Rx = ΣFx
ΣFx = ( 70 ) cos 24.142° + (130 ) sin 24.142°
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 37.
300-N Force:
Fx = ( 300 N ) cos 20° = 281.91 N
Fy = ( 300 N ) sin 20° = 102.61 N
400-N Force:
Fx = ( 400 N ) cos85° = 34.862 N
Fy = ( 400 N ) sin 85° = 398.48 N
600-N Force:
Fx = ( 600 N ) cos 5° = 597.72 N
Fy = − ( 600 N ) sin 5° = −52.293 N
and
Rx = ΣFx = 914.49 N
Ry = ΣFy = 448.80 N
R=
( 914.49 N )2 + ( 448.80 N )2 = 1018.68 N
Further:
tan α =
α = tan −1
448.80
914.49
448.80
= 26.1°
914.49
R = 1019 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.
26.1° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 38.
ΣFx :
Rx = ΣFx
Rx = ( 600 N ) cos 50° + ( 300 N ) cos85° − ( 700 N ) cos 50°
Rx = − 38.132 N
ΣFy :
Ry = ΣFy
Ry = ( 600 N ) sin 50° + ( 300 N ) sin 85° + ( 700 N ) sin 50°
Ry = 1294.72 N
R=
( − 38.132 N )2 + (1294.72 N )2
R = 1295 N
tan α =
1294.72 N
38.132 N
α = 88.3°
R = 1.295 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.
88.3° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 39.
We have:
Rx = ΣFx = −
84
12
3
TBC + (156 lb ) − (100 lb )
116
13
5
Rx = −0.72414TBC + 84 lb
or
and
R y = ΣFy =
80
5
4
TBC − (156 lb ) − (100 lb )
116
13
5
Ry = 0.68966TBC − 140 lb
(a)
So, for R to be vertical,
Rx = −0.72414TBC + 84 lb = 0
TBC = 116.0 lb !
(b) Using
TBC = 116.0 lb
R = R y = 0.68966 (116.0 lb ) − 140 lb = −60 lb
R = R = 60.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 40.
(a)
Since R is to be vertical, Rx = 0
Then, Rx = ΣFx = 0
( 600 N ) cosα + ( 300 N ) cos (α + 35°) − ( 700 N ) cos α = 0
Expanding: 3 ( cos α cos 35° − sin α sin 35° ) − cos α = 0
Then:
1
cos 35° −  
3
tan α =
sin 35°
α = 40.265°
α = 40.3° !
(b)
Since R is vertical, R = Ry
Then:
R = Ry = ΣFy
R = ( 600 N ) sin 40.265° + ( 300 N ) sin 75.265° + ( 700 N ) sin 40.265°
R = 1130 N
R = 1.130 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 41.
Selecting the x axis along aa′, we write
Rx = ΣFx = 300 N + ( 400 N ) cos α + ( 600 N ) sin α
(1)
R y = ΣFy = ( 400 N ) sin α − ( 600 N ) cos α
(2)
(a) Setting R y = 0 in Equation (2):
Thus
tan α =
600
= 1.5
400
α = 56.3° !
(b) Substituting for α in Equation (1):
Rx = 300 N + ( 400 N ) cos 56.3° + ( 600 N ) sin 56.3°
Rx = 1021.11 N
R = Rx = 1021 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 42.
(a)
Require Ry = ΣFy = 0:
( 900 lb ) cos 25° + (1200 lb ) sin 35° − TAE sin 65° = 0
or TAE = 1659.45 lb
TAE = 1659 lb !
(b)
R = ΣFx
R = − ( 900 lb ) sin 25° − (1200 lb ) cos 35° − (1659.45 lb ) cos 65°
R = 2060 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 43.
Free-Body Diagram
Force Triangle
Law of Sines:
FAC
TBC
400 lb
=
=
sin 25° sin 60° sin 95°
(a)
FAC =
400 lb
sin 25° = 169.691 lb
sin 95°
FAC = 169.7 lb !
(b)
TBC =
400
sin 60° = 347.73 lb
sin 95°
TBC = 348 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 44.
Free-Body Diagram:
ΣFx = 0:
4
21
− TCA +
TCB = 0
5
29
or
 29  4 
TCB =    TCA
 21  5 
ΣFy = 0:
3
20
TCA +
TCB − ( 3 kN ) = 0
5
29
Then
3
20  29 4

TCA +
× TCA  − ( 3 kN ) = 0

5
29  21 5

or
TCA = 2.2028 kN
(a) TCA = 2.20 kN !
(b) TCB = 2.43 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 45.
Free-Body Diagram:
ΣFy = 0:
− FB sin 50° + FC sin 70° = 0
FC =
ΣFx = 0:
sin 50°
( FB )
sin 70°
− FB cos 50° − FC cos 70° + 940 N = 0

 sin 50°  
FB cos 50° + cos 70° 
  = 940
 sin 70°  

FB = 1019.96 N
FC =
sin 50°
(1019.96 N )
sin 70°
or
FC = 831 N !
FB = 1020 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 46.
Free-Body Diagram:
ΣFx = 0:
− TAB cos 25° − TAC cos 40° + ( 70 lb ) cos10° = 0
(1)
ΣFy = 0:
TAB sin 25° − TAC sin 40° + ( 70 lb ) sin10° = 0
(2)
Solving Equations (1) and (2) simultaneously:
(a) TAB = 38.6 lb !
(b) TAC = 44.3 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 47.
Free-Body Diagram:
(a)
ΣFx = 0:
− TAB cos 30° + R cos 65° = 0
R=
ΣFy = 0:
cos 30°
TAB
cos 65°
− TAB sin 30° + R sin 65° − ( 550 N ) = 0
cos 30°


TAB  − sin 30° +
sin 65°  − 550 = 0
°
cos
65


(b)
R=
or
TAB = 405 N !
or
R = 830 N !
cos30°
( 450 N )
cos 65°
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 48.
Free-Body Diagram At B:
ΣFx = 0:
−
12
17
TBA +
TBC = 0
13
293
TBA = 1.07591 TBC
or
5
TBA +
13
ΣFy = 0:
2
TBC − 300 N = 0
293
5

 293
TBC =  300 − TBA 
13

 2
TBC = 2567.6 − 3.2918 TBA
TBC = 2567.6 − 3.2918 (1.07591TBC )
TBC = 565.34 N
or
Free-Body Diagram At C:
ΣFx = 0:
−
17
24
TBC +
TCD = 0
25
293
TCD =
17
25
( 565.34 N )  
293
 24 
TCD = 584.86 N
ΣFy = 0:
WC = −
−
2
7
TBC +
TCD − WC = 0
25
293
2
7
( 565.34 N ) + ( 584.86 N )
25
293
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.
WC = 97.7 N !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 49.
Free-Body Diagram:
ΣFx = 0:
− 8 kips + 15 kips − TD cos 40° = 0
TD = 9.1378 kips
TD = 9.14 kips !
ΣFy = 0:
( 9.1378 kips ) sin 40° − TC = 0
TC = 5.87 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 50.
Free-Body Diagram:
ΣFy = 0:
− 9 kips + TD sin 40° = 0
TD = 14.0015 kips
TD = 14.00 kips ΣFx = 0:
− 6 kips + TB − (14.0015 kips ) cos 40° = 0
TB = 16.73 kips
TB = 16.73 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 51.
Free-Body Diagram:
ΣFx = 0:
FC + ( 2.3 kN ) sin15° − ( 2.1 kN ) cos15° = 0
or
ΣFy = 0:
FC = 1.433 kN FD − ( 2.3 kN ) cos15° + ( 2.1 kN ) sin15° = 0
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.
FD = 1.678 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 52.
Free-Body Diagram:
ΣFx = 0:
− FB cos15° + 2.4 kN + (1.9 kN ) sin15° = 0
or
FB = 2.9938 kN
FB = 2.99 kN ΣFy = 0:
FD − (1.9 kN ) cos15° + ( 2.9938 kN ) sin15° = 0
FD = 1.060 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 53.
From Similar Triangles we have:
L2 − ( 2.5 m ) = ( 8 − L ) − ( 5.45 m )
2
2
2
− 6.25 = 64 − 16 L − 29.7025
or
cos β =
And
or
Then
L = 2.5342 m
5.45 m
8 m − 2.5342 m
β = 4.3576°
cos α =
2.5 m
2.5342 m
or α = 9.4237°
Free-Body Diagram At B:
ΣFx = 0:
− TABC cos α − ( 35 N ) cos α + TABC cos β = 0
or
TABC =
( 35) cos 9.4237°
cos 4.3576° − cos 9.4237°
TABC = 3255.9 N
ΣFy = 0:
TABC sin α + ( 35 N ) sin α + TABC sin β − W = 0
sin 9.4237° ( 3255.9 N + 35 N ) + ( 3255.9 N ) sin 4.3576° − W = 0
or
W = 786.22 N
(a)
W = 786 N "
(b)
TABC = 3.26 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 54.
From Similar Triangles we have:
L2 − ( 3 m ) = ( 8 − L ) − ( 4.95 m )
2
2
2
− 9 = 64 − 16 L − 24.5025
L = 3.0311 m
or
cos β =
Then
β = 4.9989°
or
cos α =
And
4.95 m
8 m − 3.0311 m
3m
3.0311 m
α = 8.2147°
or
Free-Body Diagram At B:
ΣFx = 0:
(a)
− TABC cos α − TDE cos α + TABC cos β = 0
or
TDE =
cos β − cos α
TABC
cos α
ΣFy = 0:
TABC sin α + TDE sin α + TABC sin β − ( 720 N ) = 0


 cos β − cos α 
TABC sin α + sin α 
 + sin β  = 720
cos α




TABC =
( 720 ) cosα
sin (α + β )
Substituting for α and β gives
TABC =
( 720 ) cos8.2147°
sin (8.2147° + 4.9989° )
TABC = 3117.5 N
or
(b)
TDE =
TABC = 3.12 kN "
cos 4.9989° − cos8.2147°
( 3117.5 N )
cos8.2147°
TDE = 20.338 N
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.
TDE = 20.3 N "
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 55.
Free-Body Diagram At C:
3
15
15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5
17
17
or
ΣFy = 0:
−
17
TAC + 5 TBC = 750
5
(1)
4
8
8
TAC + TBC − (150 lb ) − 190 lb = 0
5
17
17
17
TAC + 2 TBC = 1107.5
5
or
(2)
Then adding Equations (1) and (2)
7 TBC = 1857.5
or
TBC = 265.36 lb
Therefore
(a) TAC = 169.6 lb !
(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 = 265 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 56.
Free-Body Diagram At C:
3
15
15
ΣFx = 0: − TAC + TBC − (150 lb ) = 0
5
17
17
17
or
− TAC + 5 TBC = 750
5
4
8
8
ΣFy = 0: TAC + TBC − (150 lb ) − W = 0
5
17
17
17
17
or
TAC + 2 TBC = 300 + W
5
4
17
7 TBC = 1050 + W
Adding Equations (1) and (2) gives
4
17
or
TBC = 150 +
W
28
−
Using Equation (1)
or
Now for
T ≤ 240 lb ⇒
or
(2)
17
17 

TAC + 5 150 +
W = 750
5
28 

25
W
28
25
TAC : 240 =
W
28
W = 269 lb
TAC =
TBC : 240 = 150 +
or
(1)
17
W
28
W = 148.2 lb
Therefore
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.
0 ≤ W ≤ 148.2 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 57.
Free-Body Diagram At A:
First note from geometry:
The sides of the triangle with hypotenuse AD are in the ratio 12:35:37.
The sides of the triangle with hypotenuse AC are in the ratio 3:4:5.
The sides of the triangle with hypotenuse AB are also in the ratio
12:35:37.
Then:
ΣFx = 0: −
4
35
12
( 3W ) + (W ) + Fs = 0
5
37
37
or
Fs = 4.4833W
and
ΣFy = 0:
3
12
35
( 3W ) + (W ) + Fs − 400 N = 0
5
37
37
Then:
3
12
35
( 3W ) + (W ) + ( 4.4833W ) − 400 N = 0
5
37
37
or
W = 62.841 N
and
Fs = 281.74 N
or
W = 62.8 N (a)
(b) Have spring force
Fs = k ( LAB − LO )
Where
FAB = k AB ( LAB − LO )
and
LAB =
( 0.360 m )2 + (1.050 m )2 = 1.110 m
So:
281.74 N = 800 N/m (1.110 − LO ) m
or
LO = 758 mm 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 58.
Free-Body Diagram At A:
First Note ...
With LAB =
( 22 in.)2 + (16.5 in.)2
LAB = 27.5 in.
LAD =
( 30 in.)2 + (16 in.)2
LAD = 34 in.
Then FAB = k AB ( LAB − LO )
= ( 9 lb/in.)( 27.5 in. − 22.5 in.)
= 45 lb
FAD = k AD ( LAD − LO )
= ( 3 lb/in.)( 34 in. − 22.5 in.)
= 34.5 lb
(a)
ΣFx = 0:
−
4
7
15
( 45 lb ) + TAC + ( 34.5 lb ) = 0
5
25
17
or TAC = 19.8529 lb
TAC = 19.85 lb !
(b)
ΣFy = 0:
3
24
8
( 45 lb ) + (19.8529 lb ) + ( 34.5 lb ) − W = 0
5
25
17
W = 62.3 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 59.
(a)
For TAB to be a minimum
TAB must be perpendicular to TAC
∴ α + 10° = 60°
(b)
or
α = 50.0°
or
TAB = 35.0 lb
Then TAB = ( 70 lb ) sin 30°
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 60.
Note:
In problems of this type, P may be directed along one of the cables, with T = Tmax in that cable and
T = 0 in the other, or P may be directed in such a way that T is maximum in both cables. The second
possibility is investigated first.
Free-Body Diagram At C:
Force Triangle
Force triangle is isoceles with
2 β = 180° − 85°
β = 47.5°
P = 2 ( 900 N ) cos 47.5° = 1216 N
Since P > 0, solution is correct
(a)
P = 1216 N !
(b)
α = 77.5° !
α = 180° − 55° − 47.5° = 77.5°
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: Complete Online Solutions Manual Organization System
Chapter 2, Solution 61.
Note: Refer to Note in Problem 2.60
Free-Body Diagram At C:
Force Triangle
(a) Law of Cosines
P 2 = (1400 N ) + ( 700 N ) − 2 (1400 N )( 700 N ) cos85°
2
2
or
P = 1510 N !
or
α = 57.5° !
(b) Law of Sines
sin β
sin 85°
=
1400 N 1510 N
sin β = 0.92362
β = 67.461°
α = 180° − 55° − 67.461°
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 62.
Free-Body Diagram At C:
ΣFx = 0:
2Tx − 1200 N = 0
Tx = 600 N
(Tx )2 + (Ty ) = T 2
2
( 600 N )2 + (Ty ) = (870 N )2
2
Ty = 630 N
By similar triangles:
1.8 m
AC
=
870 N 630 N
AC = 2.4857 m
L = 2( AC )
L = 2 ( 2.4857 m )
L = 4.97 m
L = 4.97 m "
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 63.
TBC must be perpendicular to FAC to be as small as possible.
Free-Body Diagram: C
Force Triangle is a Right Triangle
α = 55°
α = 55° !
(a) We observe:
(b)
TBC = ( 400 lb ) sin 60°
or TBC = 346.41 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.
TBC = 346 lb !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 64.
At Collar A ...
Fs = k ( L′AB − LAB )
Have
For stretched length
L′AB =
(12 in.)2 + (16 in.)2
L′AB = 20 in.
For unstretched length
LAB = 12 2 in.
(
)
Fs = 4 lb/in. 20 − 12 2 in.
Then
Fs = 12.1177 lb
For the collar ...
ΣFy = 0
−W +
4
(12.1177 lb ) = 0
5
W = 9.69 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 65.
At Collar A ...
ΣFy = 0:
− 9 lb +
h
2
12 + h 2
Fs = 0
or
hFs = 9 144 + h 2
Now
Fs = k ( L′AB − LAB )
Where the stretched length
L′AB =
(12 in.)2 + h2
LAB = 12 2 in.
Then
hFs = 9 144 + h 2
Becomes
h 3 lb/in.

or
( h − 3) 144 + h2 = 12 2 h
( 144 + h − 12 2 ) = 9 144 + h
2
Solving Numerically ...
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.
2
h = 16.81 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 66.
Free-Body Diagram: B
TBD + FAB + TBC = 0
(a) Have:
where magnitude and direction of TBD are known, and the direction
of FAB is known.
Then, in a force triangle:
α = 90.0° By observation, TBC is minimum when
(b) Have
TBC = ( 310 N ) sin (180° − 70° − 30° )
= 305.29 N
TBC = 305 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 67.
Free-Body Diagram At C:
Since TAB = TBC = 140 lb, Force triangle is isosceles:
With
2β + 75° = 180°
β = 52.5°
Then
α = 90° − 52.5° − 30°
α = 7.50°
P
= (140 lb ) cos 52.5°
2
P = 170.453 lb
P = 170.5 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.
7.50° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 68.
Free-Body Diagram of Pulley
(a)
(
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
2
T = 1373 N (b)
(
)
ΣFy = 0: 2T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
2
T = 1373 N (c)
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
3
T = 916 N (
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
(d)
T =
1
( 2746.8 N )
3
T = 916 N (
)
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
(e)
T =
1
( 2746.8 N )
4
T = 687 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 69.
Free-Body Diagram of Pulley and
Crate
(b)
(
)
ΣFy = 0: 3T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
3
T = 916 N (d)
(
)
ΣFy = 0: 4T − ( 280 kg ) 9.81 m/s 2 = 0
T =
1
( 2746.8 N )
4
T = 687 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 70.
Free-Body Diagram: Pulley C
(a)
ΣFx = 0: TACB ( cos 30° − cos 50° ) − ( 800 N ) cos 50° = 0
Hence
TACB = 2303.5 N
TACB = 2.30 kN (b)
ΣFy = 0: TACB ( sin 30° + sin 50° ) + ( 800 N ) sin 50° − Q = 0
( 2303.5 N )( sin 30° + sin 50° ) + (800 N ) sin 50° − Q = 0
or
Q = 3529.2 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.
Q = 3.53 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 71.
Free-Body Diagram: Pulley C
ΣFx = 0: TACB ( cos 30° − cos 50° ) − P cos 50° = 0
P = 0.34730TACB
or
(1)
ΣFy = 0: TACB ( sin 30° + sin 50° ) + P sin 50° − 2000 N = 0
1.26604TACB + 0.76604 P = 2000 N
or
(2)
(a) Substitute Equation (1) into Equation (2):
1.26604TACB + 0.76604 ( 0.34730TACB ) = 2000 N
Hence:
TACB = 1305.41 N
TACB = 1305 N (b) Using (1)
P = 0.34730 (1305.41 N ) = 453.37 N
P = 453 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 72.
First replace 30 lb forces by their resultant Q:
Q = 2 ( 30 lb ) cos 25°
Q = 54.378 lb
Equivalent loading at A:
Law of Cosines:
(120 lb )2 = (100 lb )2 + ( 54.378 lb )2 − 2 (100 lb )( 54.378 lb ) cos (125° − α ) cos (125° − α ) = − 0.132685
This gives two values:
125° − α = 97.625°
α = 27.4°
125° − α = − 97.625°
α = 223°
Thus for R < 120 lb:
27.4° < α < 223° !
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 73.
(a)
Fx = ( 950 lb ) sin 50° cos 40°
= 557.48 lb
Fx = 557 lb !
Fy = − ( 950 lb ) cos 50°
= − 610.65 lb
Fy = − 611 lb !
Fz = ( 950 lb ) sin 50° sin 40°
= 467.78 lb
Fz = 468 lb !
(b)
cosθ x =
557.48 lb
950 lb
or θ x = 54.1° !
cosθ y =
− 610.65 lb
950 lb
or θ y = 130.0° !
cosθ z =
467.78 lb
950 lb
or θ z = 60.5° !
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: Complete Online Solutions Manual Organization System
Chapter 2, Solution 74.
(a)
Fx = − ( 810 lb ) cos 45° sin 25°
= − 242.06 lb
Fx = −242 lb !
Fy = − ( 810 lb ) sin 45°
= − 572.76 lb
Fy = − 573 lb !
Fz = (810 lb ) cos 45° cos 25°
= 519.09 lb
Fz = 519 lb !
(b)
cosθ x =
−242.06 lb
810 lb
or θ x = 107.4° !
cosθ y =
− 572.76 lb
810 lb
or θ y = 135.0° !
cosθ z =
519.09 lb
810 lb
or θ z = 50.1° !
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 75.
(a)
Fx = ( 900 N ) cos 30° cos 25°
= 706.40 N
Fx = 706 N !
Fy = ( 900 N ) sin 30°
= 450.00 N
Fy = 450 N !
Fz = − ( 900 N ) cos 30° sin 25°
= − 329.04 N
Fz = − 329 N !
(b)
cosθ x =
706.40 N
900 N
or θ x = 38.3° !
cosθ y =
450.00 N
900 N
or θ y = 60.0° !
cosθ z =
−329.40 N
900 N
or θ z = 111.5° !
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: Complete Online Solutions Manual Organization System
Chapter 2, Solution 76.
(a)
Fx = − (1900 N ) sin 20° sin 70°
= − 610.65 N
Fx = − 611 N !
Fy = (1900 N ) cos 20°
= 1785.42 N
Fy = 1785 N !
Fz = (1900 N ) sin 20° cos 70°
= 222.26 N
Fz = 222 N !
(b)
cosθ x =
−610.65 N
1900 N
or θ x = 108.7° !
cosθ y =
1785.42 N
1900 N
or θ y = 20.0° !
cosθ z =
222.26 N
1900 N
or θ z = 83.3° !
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 77.
(a)
Fx = (180 lb ) cos 35° sin 20°
= 50.430 lb
Fx = 50.4 lb !
Fy = − (180 lb ) sin 35°
= −103.244 lb
Fy = −103.2 lb !
Fz = (180 lb ) cos 35° cos 20°
= 138.555 lb
Fz = 138.6 lb !
(b)
cosθ x =
50.430 lb
180 lb
or θ x = 73.7° !
cosθ y =
−103.244 lb
180 lb
or θ y = 125.0° !
cosθ z =
138.555 lb
180 lb
or θ z = 39.7° !
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 78.
(a)
Fx = (180 lb ) cos 30° cos 25°
= 141.279 lb
Fx = 141.3 lb !
Fy = − (180 lb ) sin 30°
= − 90.000 lb
Fy = − 90.0 lb !
Fz = (180 lb ) cos 30° sin 25°
= 65.880 lb
Fz = 65.9 lb !
(b)
cosθ x =
141.279 lb
180 lb
or θ x = 38.3° !
cosθ y =
−90.000 lb
180 lb
or θ y = 120.0° !
cosθ z =
65.880 lb
180 lb
or θ z = 68.5° !
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: Complete Online Solutions Manual Organization System
Chapter 2, Solution 79.
(a)
Fx = − ( 220 N ) cos 60° cos 35°
= − 90.107 N
Fx = − 90.1 N
Fy = ( 220 N ) sin 60°
= 190.526 N
Fy = 190.5 N
Fz = − ( 220 N ) cos 60° sin 35°
= − 63.093 N
Fz = − 63.1 N
(b)
cosθ x =
−90.107 Ν
220 N
θ x = 114.2°
cosθ y =
190.526 N
220 N
θ y = 30.0°
cosθ z =
−63.093 N
220 N
θ z = 106.7°
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 80.
(a)
Fx = 180 N
With Fx = F cos 60° cos 35°
180 N = F cos 60° cos 35°
or F = 439.38 N
F = 439 N !
(b)
cosθ x =
180 N
439.48 N
θ x = 65.8° !
Fy = ( 439.48 N ) sin 60°
Fy = 380.60 N
cosθ y =
380.60 N
439.48 N
θ y = 30.0° !
Fz = − ( 439.48 N ) cos 60° sin 35°
Fz = −126.038 N
cosθ z =
−126.038 N
439.48 N
θ z = 106.7° !
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 81.
F=
Fx2 + Fy2 + Fz2
F =
( 65 N )2 + ( − 80 N )2 + ( − 200 N )2
F = 225 N !
cosθ x =
Fx
65 N
=
F
225 N
θ x = 73.2° !
cosθ y =
Fy
F
=
− 80 N
225 N
θ y = 110.8° !
cosθ z =
Fz
− 200 N
=
F
225 N
θ z = 152.7° !
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 82.
F=
Fx2 + Fy2 + Fz2
F =
( 450 N )2 + ( 600 N )2 + ( −1800 N )2
F = 1950 N !
cosθ x =
Fx
450 N
=
F 1950 N
θ x = 76.7° !
cosθ y =
Fy
F
=
600 N
1950 N
θ y = 72.1° !
cosθ z =
Fz
−1800 N
=
1950 N
F
θ z = 157.4° !
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 83.
(a)
(
We have ( cosθ x ) + cosθ y
2
) + ( cosθ z )2 = 1
2
( cosθ y ) = 1 − ( cosθ x )2 − ( cosθ z )2
2
Since Fy < 0 we must have
Thus
cosθ y < 0
cosθ y = − 1 − ( cos 43.2° ) − cos ( 83.8° )
2
2
cosθ y = − 0.67597
θ y = 132.5° !
(b) Then:
F =
F=
Fy
cosθ y
− 50 lb
− 0.67597
F = 73.968 lb
And
Fx = F cosθ x
Fx = ( 73.968 lb ) cos 43.2°
Fx = 53.9 lb !
Fz = F cosθ z
Fz = ( 73.968 lb ) cos83.8°
Fz = 7.99 lb !
F = 74.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 84.
(a)
(
We have ( cosθ x ) + cosθ y
2
) + ( cosθ z )2 = 1
2
(
or ( cosθ z ) = 1 − ( cosθ x ) − cosθ y
2
Since Fz < 0 we must have
Thus
2
)
2
cosθ z < 0
cosθ z = − 1 − ( cos113.2° ) − cos ( 78.4° )
2
2
cosθ z = − 0.89687
θ z = 153.7° !
(b) Then:
F =
Fz
− 35 lb
=
cosθ z
− 0.89687
F = 39.025 lb
And
Fx = F cosθ x
Fx = ( 39.025 lb ) cos113.2°
Fx = −15.37 lb !
Fy = F cosθ y
Fy = ( 39.025 lb ) cos 78.4°
Fy = 7.85 lb !
F = 39.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 85.
(a)
We have
Fy = F cosθ y
Fy = ( 250 N ) cos 72.4°
Fy = 75.592 N
Fy = 75.6 N !
Then
F 2 = Fx2 + Fy2 + Fz2
( 250 N )2 = (80 N )2 + ( 75.592 N )2 + Fz2
Fz = 224.47 N
Fz = 224 N !
(b)
cosθ x =
Fx
F
cosθ x =
80 N
250 N
θ x = 71.3° !
cosθ z =
Fz
F
cosθ z =
224.47 N
250 N
θ z = 26.1° !
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 86.
(a)
Have
Fx = F cosθ x
Fx = ( 320 N ) cos104.5°
Fx = − 80.122 N
Fx = − 80.1 N !
Then:
F 2 = Fx2 + Fy2 + Fz2
( 320 N )2 = ( − 80.122 N )2 + Fy2 + ( −120 N )2
Fy = 285.62 N
Fy = 286 N !
(b)
cosθ y =
Fy
cosθ y =
285.62 N
320 N
F
θ y = 26.8° !
cosθ z =
Fz
F
cosθ z =
−120 N
320 N
θ z = 112.0° !
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 87.
!!!"
DB = ( 36 in.) i − ( 42 in.) j − ( 36 in.) k
DB =
( 36 in.)2 + ( − 42 in.)2 + ( − 36 in.)2 = 66 in.
!!!"
DB
TDB = TDBλDB = TDB
DB
TDB =
55 lb
( 36 in.) i − ( 42 in.) j − ( 36 in.) k 
66 in. 
= ( 30 lb ) i − ( 35 lb ) j − ( 30 lb ) k
∴ (TDB ) x = 30.0 lb !
(TDB ) y = − 35.0 lb !
(TDB ) z = − 30.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 88.
!!!"
EB = ( 36 in.) i − ( 45 in.) j + ( 48 in.) k
EB =
( 36 in.)2 + ( − 45 in.)2 + ( 48 in.)2 = 75 in.
!!!"
EB
TEB = TEBλEB = TEB
EB
TEB =
60 lb
( 36 in.) i − ( 45 in.) j + ( 48 in.) k 
75 in. 
= ( 28.8 lb ) i − ( 36 lb ) j + ( 38.4 lb ) k
∴ (TEB ) x = 28.8 lb !
(TEB ) y = − 36.0 lb !
(TEB ) z = 38.4 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 89.
!!!"
BA = ( 4 m ) i + ( 20 m ) j − ( 5 m ) k
BA =
( 4 m )2 + ( 20 m )2 + ( − 5 m )2 = 21 m
!!!"
BA 2100 N
( 4 m ) i + ( 20 m ) j − ( 5 m ) k 
F = F λ BA = F
=
21 m 
BA
F = ( 400 N ) i + ( 2000 N ) j − ( 500 N ) k
Fx = + 400 N, Fy = + 2000 N, Fz = − 500 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 90.
!!!"
DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) k
DA =
( 4 m )2 + ( 20 m )2 + (14.8 m )2 = 25.2 m
!!!"
DA 1260 N
( 4 m ) i + ( 20 m ) j + (14.8 m ) k 
F = F λ DA = F
=
25.2 m 
DA
F = ( 200 N ) i + (1000 N ) j + ( 740 N ) k
Fx = + 200 N, Fy = + 1000 N, Fz = + 740 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 91.
uuuv
BG = − (1 m ) i + (1.85 m ) j − ( 0.8 m ) k
BG =
( −1 m )2 + (1.85 m )2 + ( − 0.8 m )2
BG = 2.25 m
uuuv
BG
TBG = TBG λBG = TBG
BG
TBG =
450 N
 − (1 m ) i + (1.85 m ) j − ( 0.8 m ) k 
2.25 m 
= − ( 200 N ) i + ( 370 N ) j − (160 N ) k
∴ (TBG ) x = − 200 N (TBG ) y = 370 N (TBG ) z = −160.0 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
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Chapter 2, Solution 92.
uuuuv
BH = ( 0.75 m ) i + (1.5 m ) j − (1.5 m ) k
BH =
( 0.75 m )2 + (1.5 m )2 + ( −1.5 m )2
= 2.25 m
uuuuv
BH
TBH = TBH λBH = TBH
BH
TBH =
600 N
( 0.75 m ) i + (1.5 m ) j − (1.5 m ) k 
2.25 m 
= ( 200 N ) i + ( 400 N ) j − ( 400 N ) k
∴ (TBH ) x = 200 N (TBH ) y = 400 N (TBH ) z = − 400 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 93.
P = ( 4 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
= (1.18479 kips ) i − ( 2 kips ) j + ( 3.2552 kips ) k
Q = (8 kips ) [ − cos 45° sin15°i + sin 45°j − cos 45° cos15°k ]
= − (1.46410 kips ) i + ( 5.6569 kips ) j − ( 5.4641 kips ) k
R = P + Q = − ( 0.27931 kip ) i + ( 3.6569 kips ) j − ( 2.2089 kips ) k
R=
( − 0.27931 kip)2 + (3.6569 kips )2 + ( − 2.2089 kips)2
R = 4.2814 kips
cosθ x =
cos θ y =
cos θ z =
R = 4.28 kips or
Rx − 0.27931 kip
=
= − 0.065238
R
4.2814 kips
Ry
R
=
3.6569 kips
= 0.85414
4.2814 kips
Rz − 2.2089 kips
=
= − 0.51593
R
4.2814 kips
or
θ x = 93.7° θ y = 31.3° θ z = 121.1° 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 94.
P = ( 6 kips ) [ cos 30° sin 20°i − sin 30°j + cos 30° cos 20°k ]
= (1.77719 kips ) i − ( 3 kips ) j + ( 4.8828 kips ) k
Q = ( 7 kips ) [ − cos 45° sin15°i + sin 45°j − cos 45° cos15°k ]
= − (1.28109 kips ) i + ( 4.94975 kips ) j − ( 4.7811 kips ) k
R = P + Q = ( 0.49610 kip ) i + (1.94975 kips ) j + ( 0.101700 kip ) k
R=
( 0.49610 kip)2 + (1.94975 kips)2 + ( 0.101700 kip)2
R = 2.0144 kips
cos θ x =
cos θ y =
cos θ z =
or
R = 2.01 kips or
θ x = 75.7° Rx 0.49610 kip
=
= 0.24628
R
2.0144 kips
Ry
R
=
1.94975 kips
= 0.967906
2.0144 kips
Rz 0.101700 kip
=
= 0.050486
R
2.0144 kips
θ y = 14.56° θ z = 87.1° 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 95.
uuur
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB =
( − 600 mm )2 + (360 mm )2 + ( 270 mm )2
AB = 750 mm
uuuv
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
AC =
( − 600 mm )2 + ( 320 mm )2 + ( −510 mm )2
AC = 850 mm
uuur
AB
510 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
TAB = TAB
=
AB 750 mm 
TAB = − ( 408 N ) i + ( 244.8 N ) j + (183.6 N ) k
uuur
AC
765 N
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
TAC = TAC
=
AC 850 mm 
TAC = − ( 540 N ) i + ( 288 N ) j − ( 459 N ) k
R = TAB + TAC = − ( 948 N ) i + ( 532.8 N ) j − ( 275.4 N ) k
Then
and
R = 1121.80 N
− 948 N
cos θ x =
1121.80 N
532.8 N
cos θ y =
1121.80 N
− 275.4 N
cos θ z =
1121.80 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.
R = 1122 N θ x = 147.7° θ y = 61.6° θ z = 104.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 96.
!!!"
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
AB =
( − 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2 = 750 mm
AB = 750 mm
!!!"
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
AC =
( − 600 mm )2 + ( 320 mm) 2 + ( − 510 mm) 2 = 850 mm
AC = 850 mm
!!!"
AB
765 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
TAB = TAB
=
AB 750 mm 
TAB = − ( 612 N ) i + ( 367.2 N ) j + ( 275.4 N ) k
!!!"
AC
510 N
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
TAC = TAC
=
AC 850 mm 
TAC = − ( 360 N ) i + (192 N ) j − ( 306 N ) k
R = TAB + TAC = − ( 972 N ) i + ( 559.2 N ) j − ( 30.6 N ) k
Then
R = 1121.80 N R = 1122 N !
− 972 N
θ x = 150.1° !
1121.80 N
559.2 N
cos θ y =
θ y = 60.1° !
1121.80 N
− 30.6 N
cos θ z =
θ z = 91.6° !
1121.80 N
cos θ x =
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 97.
Have
TAB = ( 760 lb )( sin 50° cos 40°i − cos 50°j + sin 50° sin 40°k )
TAC = TAC ( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25°k )
(a)
( RA ) x = 0
R A = TAB + TAC
∴ ( RA ) x = ΣFx = 0:
( 760 lb) sin 50° cos 40° − TAC cos 45° sin 25° = 0
TAC = 1492.41 lb
or
∴ TAC = 1492 lb (b)
Then
( RA ) y = ΣFy = ( − 760 lb) cos 50° − (1492.41 lb) sin 45°
( RA ) y = −1543.81 lb
( RA ) z = ΣFz = ( 760 lb) sin 50° sin 40° + (1492.41 lb) cos 45° cos 25°
( RA ) z = 1330.65 lb
∴ R A = − (1543.81 lb ) j + (1330.65 lb ) k
RA = 2038.1 lb
RA = 2040 lb cosθ x =
0
2038.1 lb
θ x = 90.0° cos θ y =
−1543.81 lb
2038.1 lb
θ y = 139.2° cos θ z =
1330.65 lb
2038.1 lb
θ z = 49.2° 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 98.
Have
TAB = TAB ( sin 50° cos 40°i − cos 50°j + sin 50° sin 40°k )
TAC = ( 980 lb )( − cos 45° sin 25°i − sin 45°j + cos 45° cos 25°k )
(a)
( RA ) x = 0
R A = TAB + TAC
∴ ( RA ) x = ΣFx = 0:
TAB sin 50° cos 40° − ( 980 lb ) cos 45° sin 25° = 0
TAB = 499.06 lb
or
∴ TAB = 499 lb (b)
Then
and
( RA ) y = ΣFy = − ( 499.06 lb) cos 50° − (980 lb) sin 45°
( RA ) y = −1013.75 lb
( RA ) z = ΣFz = ( 499.06 lb) sin 50° sin 40° + (980 lb) cos 45° cos 25°
( RA ) z = 873.78 lb
∴ R A = − (1013.75 lb ) j + (873.78 lb ) k
RA = 1338.35 lb
RA = 1338 lb 0
1338.35 lb
θ x = 90.0° −1013.75 lb
1338.35 lb
θ y = 139.2° 873.78 lb
1338.35 lb
θ z = 49.2° cos θ x =
cos θ y =
cos θ z =
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 99.
!!!"
AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k
Cable AB:
AB =
( − 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2 = 750 mm
!!!"
AB
600 N
 − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k 
TAB = TAB
=
AB 750 mm 
TAB = − ( 480 N ) i + ( 288 N ) j + ( 216 N ) k
!!!"
AC = − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k
Cable AC:
AC =
( − 600 mm )2 + ( 320 mm) 2 + ( − 510 mm) 2 = 850 mm
!!!"
AC
TAC
 − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k 
TAC = TAC
=
AC 850 mm 
TAC = −
Load P:
60
32
51
TAC i +
TAC j −
TAC k
85
85
85
P = − Pj
(a)
( RA ) z = ΣFz = 0:
( 216 N ) −
51
TAC = 0
85
or
TAC = 360 N !
(b)
( RA ) y = ΣFy = 0:
( 288 N ) +
32
TAC − P = 0
85
or
P = 424 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 100.
uuur
AB = − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k
Cable AB:
AB =
( − 4 m)2 + ( −20 m)2 + (5 m )2 = 21 m
uuur
AB
T
TAB = TAB
= AB  − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k 
AB 21 m
uuur
AC = (12 m ) i − ( 20 m ) j + ( 3.6 m ) k
Cable AC:
AC =
(12 m )2 + ( − 20 m )2 + ( 3.6 m )2 = 23.6 m
uuur
AC 1770 N
(12 m ) i − ( 20 m ) j + ( 3.6 m ) k 
TAC = TAC
=
AC 23.6 m 
= ( 900 N ) i − (1500 N ) j + ( 270 N ) k
uuur
AD = − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k
Cable AD:
AD =
( − 4 m )2 + ( − 20 m )2 + (14.8 m )2 = 25.2 m
uuur
AD
TAD
 − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k 
TAD = TAD
=
AD 25.2 m 
=
TAD
 − (10 m ) i − ( 50 m ) j − ( 37 m ) k 
63 m 
Now...
R = TAB + TAC + TAD and R = Rj; Rx = Rz = 0
4
10
TAB + 900 −
TAD = 0
21
63
5
37
ΣFy = 0:
TAB + 270 −
TAD = 0
21
63
Solving equations (1) and (2) simultaneously yields:
ΣFx = 0:
−
(1)
(2)
TAD = 1.775 kN !
TAB = 3.25 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 101.
d AB =
( −450 mm) 2 + ( 600 mm )2 = 750 mm
d AC =
( 600 mm )2 + ( − 320 mm)2 = 680 mm
d AD =
( 500 mm)2 + ( 600 mm )2 + ( 360 mm )2 = 860 mm
TAB =
TAB
 − ( 450 mm ) i + ( 600 mm ) j
750 mm 
TAB = ( − 0.6 i + 0.8 j) TAB
TAC =
TAC
( 600 mm ) j − ( 320 mm ) k 
680 mm 
8 
 15
TAC =  j − k  TAC
 17
17 
TAD =
TAD
( 500 mm ) i + ( 600 mm ) j + ( 360 mm ) k 
860 mm 
30
18 
 25
TAD =  i +
j+
k  TAD
 43
43
43 
W = −W j
At point A:
i component:
k component:
ΣF = 0:
TAB + TAC + TAD + W = 0
25
− 0.6 TAB +
TAD = 0
43
 5   25 
TAB =     TAD
or
 3   43 
−
18
18
TAC +
TAD = 0
17
43
or
j component:
(1)
 17   18 
TAC =     TAD
 8   43 
15
30
TAC +
TAD − W = 0
17
43
15  17 18
 30
TAD − W = 0
0.8 TAB +
 ⋅ TAD  +
17 8 43
43
255
TAD − W = 0
0.8 TAB +
172
(2)
0.8 TAB +
(3)
From Equation (1):
 5   25 
6 kN =     TAD
 3   43 
or
TAD = 6.1920 kN
From Equation (3):
0.8 ( 6 kN ) +
255
( 6.1920 kN ) − W = 0
172
∴ W = 13.98 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 102.
See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below.
 5   25 
TAB =     TAD
 3   43 
(1)
 17   18 
TAC =     TAD
 8   43 
(2)
0.8 TAB +
255
TAD − W = 0
172
From Equation (1)
 5   25 
TAB =     ( 4.3 kN )
 3   43 
or
TAB = 4.1667 kN
From Equation (3)
0.8 ( 4.1667 kN ) +
255
( 4.3 kN ) − W = 0
172
∴ W = 9.71 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.
(3)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 103.
uuur
AB = − ( 4.20 m ) i − ( 5.60 m ) j
AB = ( − 4.20 m ) + ( − 5.60 m ) = 7.00 m
uuur
AC = ( 2.40 m ) i − ( 5.60 m ) j + ( 4.20 m ) k
2
2
AC = ( 2.40 m ) + ( − 5.60 m ) + ( 4.20 m ) = 7.40 m
uuur
AD = − ( 5.60 m ) j − ( 3.30 m ) k
2
2
2
AD = ( − 5.60 m ) + ( − 3.30 m ) = 6.50 m
uuur
AB
TAB
TAB = TAB λ AB = TAB
=
( − 4.20i − 5.60j)
AB 7.00 m
2
2
4 
 3
TAB =  − i − j TAB
 5
5 
uuur
AC
TAC
TAC = TAC λ AC = TAC
=
( 2.40i − 5.60j + 4.20k )
AC 7.40 m
28
21 
 12
TAC =  i −
j+
k  TAC
 37
37
37 
uuur
AD
TAD
TAD = TAD λ AD = TAD
=
( − 5.60 j − 3.30k )
AD 6.50 m
33 
 56
TAD =  − j −
k  TAD
 65
65 
P = Pj
For equilibrium at point A:
ΣF = 0
TAB + TAC + TAD + P = 0
i component:
3
12
− TAB +
TAC = 0
5
37
or
TAB =
20
TAC
37
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.
(1)
COSMOS: Complete Online Solutions Manual Organization System
4
28
56
− TAB −
TAC −
TAD + P = 0
5
37
65
j component:
4
28
56  65 7

− TAB −
TAC −
 ⋅ TAC  + P = 0
5
37
65 11 37
4
700
− TAB −
TAC + P = 0
5
407
(2)
21
33
TAC −
TAD = 0
37
65
k component:
or
 65   7 
TAD =     TAC
 11   37 
(3)
From Equation (1):
 20 
259 N =   TAC
 37 
or
From Equation (2):
−
TAC = 479.15 N
4
700
( 259 N ) −
( 479.15 N ) + P = 0
5
407
∴ P = 1031 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 104.
See Problem 2.103 for the analysis leading to the linear algebraic Equations (1), (2), and (3)
TAB =
−
20
TAC
37
(1)
4
700
TAB −
TAC + P = 0 (2)
5
407
 65  7 
TAD =    TAC
 11  37 
(3)
Substituting for TAC = 444 N into Equation (1)
TAB =
Gives
20
( 444 N )
37
TAB = 240 N
or
And from Equation (3)
−
4
700
( 240 N ) −
( 444 N ) + P = 0
5
407
∴ P = 956 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 105.
d BA =
( −11 in.)2 + ( 9.6 in.)2 = 14.6 in.
dCA =
( 9.6 in.)2 + ( − 7.2 in.)2 = 12.0 in.
d DA =
( 9.6 in.)2 + ( 9.6 in.)2 + ( 4.8 in.)2 = 14.4 in.
FBA = FBAλBA =
FBA
( −11 in.) i + ( 9.6 in.) j
14.6 in. 
  11 
 9.6  
= FBA  − 
i + 
 j
14.6

 14.6  
 
FCA = FCAλCA =
FCA
( 9.6 in.) j − ( 7.2 in.) k 
12.0 in. 
 4 
3 
= FCA   j −   k 
5 
 5 
FDA = FDAλDA =
FDA
( 9.6 in.) i + ( 9.6 in.) j + ( 4.8 in.) k 
14.4 in. 
 2 
2
1 
= FDA   i +   j +   k 
3
 3 
 3 
P = −Pj
At point A:
ΣF = 0: FBA + FCA + FDA + P = 0
i
component:
j
component:
k
component:
 11 
2
−
FBA +   FDA = 0

 14.6 
3
 9.6 
4
2
 14.6  FBA +  5  FCA +  3  FDA − P = 0


 
 
3
1
−   FCA +   FDA = 0
5
3
(1)
(2)
(3)
continued
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
 14.6  2 
FBA = 
  FDA
 11  3 
 14.6  2 
29.2 lb = 
  FDA
 11  3 
From Equation (1)
FDA = 33 lb
or
Solving Eqn. (3) for FCA gives:
5
FCA =   FDA
9
5
FCA =   ( 33 lb )
9
Substituting into Eqn. (2) for FBA , FDA, and FCA in terms of FDA gives:
 9.6 
 4  5 
2
 14.6  ( 29.2 lb ) +  5  9  ( 33 lb ) +  3  ( 33 lb ) − P = 0


  
 
∴
P = 55.9 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 106.
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below.
 11 
2
−
 FBA +   FDA = 0
 14.6 
3
(1)
 9.6 
4
2

 FBA +   FCA +   FDA − P = 0
 14.6 
5
3
(2)
 3
1
−   FCA +   FDA = 0 (3)
5
 3
From Equation (1):
 14.6  2 
FBA = 
  FDA
 11  3 
From Equation (3):
5
FCA =   FDA
9
Substituting into Equation (2) for FBA and FCA gives:
 9.6  14.6  2 
 4  5 
2


  FDA +    FDA +   FDA − P = 0
 14.6  11  3 
 5  9 
3
 838 
or 
 FDA = P
 495 
Since P = 45 lb
 838 

 FDA = 45 lb
 495 
or FDA = 26.581 lb
 14.6  2 
and FBA = 
  ( 26.581 lb )
 11  3 
or FBA = 23.5 lb 5
and FCA =   ( 26.581 lb )
9
or FCA = 14.77 lb and FDA = 26.6 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 107.
The force in each cable can be written as the product of the magnitude of
the force and the unit vector along the cable. That is, with
uuur
AC = (18 m ) i − ( 30 m ) j + ( 5.4 m ) k
AC =
(18 m )2 + ( −30 m )2 + ( 5.4 m )2 = 35.4 m
uuur
AC
TAC
(18 m ) i − ( 30 m ) j + ( 5.4 m ) k 
TAC = T λ AC = TAC
=
35.4 m 
AC
TAC = TAC ( 0.50847i − 0.84746 j + 0.152542k )
and
uuur
AB = − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k
AB =
( −6 m )2 + ( −30 m )2 + ( 7.5 m )2 = 31.5 m
uuur
AB
TAB
 − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k 
TAB = T λ AB = TAB
=
AB 31.5 m 
TAB = TAB ( −0.190476i − 0.95238j + 0.23810k )
uuur
AD = − ( 6 m ) i − ( 30 m ) j − ( 22.2 m ) k
Finally
AD =
( −6 m )2 + ( −30 m )2 + ( −22.2 m )2 = 37.8 m
TAD = T λ AD = TAD
uuur
AD
TAD
 − ( 6 m ) i − ( 30 m ) j − ( 22.2 m ) k 
=
AD 37.8 m 
TAD = TAD ( −0.158730i − 0.79365j − 0.58730k )
continued
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
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With P = Pj, at A:
ΣF = 0: TAB + TAC + TAD + Pj = 0
Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations:
i : − 0.190476TAB + 0.50847TAC − 0.158730TAD = 0
(1)
j: − 0.95238TAB − 0.84746TAC − 0.79365TAD + P = 0
(2)
k : 0.23810TAB + 0.152542TAC − 0.58730TAD = 0
(3)
In Equations (1), (2) and (3), set TAB = 3.6 kN, and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain:
TAC = 1.963 kN
TAD = 1.969 kN
P = 6.66 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.
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Chapter 2, Solution 108.
Based on the results of Problem 2.107, particularly Equations (1), (2) and (3), we substitute TAC = 2.6 kN and
solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations
(MATLAB or Maple, for example), to obtain
TAB = 4.77 kN
TAD = 2.61 kN
P = 8.81 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 109.
!!!"
AB = − ( 6.5 ft ) i − (8 ft ) j + ( 2 ft ) k
AB =
TAB =
( −6.5 ft )2 + ( −8 ft )2 + ( 2 ft )2 = 10.5 ft
TAB
 − ( 6.5 ft ) i − ( 8 ft ) j + ( 2 ft ) k 
10.5 ft 
= TAB ( −0.61905i − 0.76190 j + 0.190476k )
!!!"
AC = (1 ft ) i − ( 8 ft ) j + ( 4 ft ) k
AC =
(1 ft )2 + ( −8 ft )2 + ( 4 ft )2 = 9 ft
TAC =
TAC
(1 ft ) i − ( 8 ft ) j + ( 4 ft ) k 
9 ft 
= TAC ( 0.111111i − 0.88889 j + 0.44444k )
!!!"
AD = (1.75 ft ) i − ( 8 ft ) j − (1 ft ) k
AD =
(1.75 ft )2 + ( −8 ft )2 + ( −1 ft )2 = 8.25 ft
TAD =
TAD
(1.75 ft ) i − ( 8 ft ) j − (1 ft ) k 
8.25 ft 
= TAD ( 0.21212i − 0.96970 j − 0.121212k )
At A ΣF = 0
ΣFx = 0:
−0.61905TAB + 0.111111TAC + 0.21212TAD = 0
(1)
ΣFy = 0:
−0.76190TAB − 0.88889TAC − 0.96970TAD + W = 0
(2)
ΣFz = 0:
0.190476TAB + 0.44444TAC − 0.121212TAD = 0
(3)
Substituting for W = 320 lb and Solving Equations (1), (2), (3) simultaneously yields:
TAB = 86.2 lb !
TAC = 27.7 lb !
TAD = 237 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
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Chapter 2, Solution 110.
See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below.
− 0.61905 TAB + 0.111111TAC + 0.21212 TAD = 0
(1)
− 0.76190 TAB − 0.88889 TAC − 0.96970 TAD + W = 0
(2)
0.190476 TAB + 0.44444 TAC − 0.121212TAD = 0
(3)
Now substituting for TAD = 220 lb Gives:
− 0.61905 TAB + 0.111111TAC + 46.662 = 0
(4)
− 0.76190 TAB − 0.88889 TAC − 213.33 + W = 0
(5)
0.190476 TAB + 0.44444 TAC − 26.666 = 0
(6)
Solving Equations (4) and (6) simultaneously gives
TAB = 79.992 lb and TAC = 25.716 lb
Substituting into Equation (5) yields
W = 297 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
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Chapter 2, Solution 111.
Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all
have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the
generators of the cone.
Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB.
Hence:
It follows that:
λ AB = λ BE =
cos 45°i + 8j − sin 45°k
65
 cos 45°i + 8j − sin 45°k 
TBE = TBE λ BE = TBE 

65


 cos 30°i + 8j + sin 30°k 
TCF = TCF λ CF = TCF 

65


 − cos15°i + 8 j − sin15°k 
TDG = TDG λ DG = TDG 

65


At A:
ΣF = 0: TBE + TCF + TDG + W + P = 0
Then, isolating the factors of i, j, and k, we obtain three algebraic equations:
TBE
T
T
cos 45° + CF cos 30° − DG cos15° + P = 0
65
65
65
i:
or
TBE cos 45° + TCF cos30° − TDG cos15° + P 65 = 0
8
8
8
+ TCF
+ TDG
−W = 0
65
65
65
j: TBE
or
TBE + TCF + TDG − W
65
=0
8
k: −
or
(1)
(2)
TBE
T
T
sin 45° + CF sin 30° − DG sin15° = 0
65
65
65
−TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(3)
With P = 0 and the tension in cord BE = 0.2 lb:
Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination,
matrix methods or iteration – with MATLAB or Maple, for example), we obtain:
TCF = 0.669 lb
TDG = 0.746 lb
W = 1.603 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
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Chapter 2, Solution 112.
See Problem 2.111 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
i : TBE cos 45° + TCF cos 30° − TDG cos15° + 65 P = 0
(1)
65
=0
8
(2)
k : − TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(3)
j: TBE + TCF + TDG − W
With W = 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1),
(2), and (3) for the tension TCF as a function of P and requiring it to be positive (> 0).
Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrix
methods or iteration – with MATLAB or Maple, for example), we obtain:
TCF = ( −1.729P + 0.668 ) lb
Hence, for TCF > 0
or
−1.729P + 0.668 > 0
P < 0.386 lb
∴ φ ≤ P < 0.386 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
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Chapter 2, Solution 113.
d DA =
( 400 mm )2 + ( − 600 mm )2 = 721.11 mm
d DB =
( − 200 mm )2 + ( − 600 mm )2 + (150 mm )2 = 650 mm
d DC =
( − 200 mm )2 + ( − 600 mm )2 + ( −150 mm )2 = 650 mm
TDA = TDAλDA
=
TDA
( 400 mm ) i − ( 600 mm ) j
721.11 mm 
= TDA ( 0.55470i − 0.83205 j)
TDB = TDBλDB
=
TDB
 − ( 200 mm ) i − ( 600 mm ) j + (150 mm ) k 
650 mm 
12
3 
 4
= TDB  − i −
j + k
13
13
13


TDC = TDC λDC
TDC =
TDC
 − ( 200 mm ) i − ( 600 mm ) j − (150 mm ) k 
650 mm 
12
3 
 4
= TDC  − i −
j − k
13
13
13


W = Wj
At point D ΣF = 0: TDA + TDB + TDC + W = 0
4
4
TDB − TDC = 0
13
13
12
12
−0.83205 TDA − TDB − TDC + W = 0
13
13
3
3
TDB − TDC = 0
13
13
0.55470 TDA −
i component:
j component:
k component:
(
)
Setting W = (16 kg ) 9.81 m/s 2 = 156.96 N
And Solving Equations (1), (2), and (3) simultaneously:
TDA = 62.9 N !
TDB = 56.7 N !
TDC = 56.7 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
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(1)
(2)
(3)
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Chapter 2, Solution 114.
d DA =
( 400 mm )2 + ( − 600 mm )2 = 721.11 mm
d DB =
( − 200 mm )2 + ( − 600 mm )2 + ( 200 mm )2 = 663.32 mm
d DC =
( − 200 mm )2 + ( − 600 mm )2 + ( − 200 mm )2 = 663.32 mm
TDA = TDAλDA
=
TDA
( 400 mm ) i − ( 600 mm ) j
721.11 mm 
= TDA ( 0.55470i − 0.83205 j)
TDB = TDBλDB
=
TDB
 − ( 200 mm ) i − ( 600 mm ) j + ( 200 mm ) k 
663.32 mm 
= TDB ( − 0.30151i − 0.90454 j + 0.30151k )
TDC = TDC λDC
=
TDC
 − ( 200 mm ) i − ( 600 mm ) j − ( 200 mm ) k 
663.32 mm 
= TDC ( − 0.30151i − 0.90454 j − 0.30151k )
At point D
ΣF = 0: TDA + TDB + TDC + W = 0
i component:
0.55470 TDA − 0.30151TDB − 0.30151TDC = 0
(1)
j component:
−0.83205 TDA − 0.90454 TDB − 0.90454 TDC + W = 0
(2)
0.30151TDB − 0.30151TDC = 0
k component:
(
)
Setting W = (16 kg ) 9.81 m/s 2 = 156.96 N
And Solving Equations (1), (2), and (3) simultaneously:
TDA = 62.9 N !
TDB = 57.8 N !
TDC = 57.8 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
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Chapter 2, Solution 115.
From the solutions of 2.107 and 2.108:
TAB = 0.5409 P
TAC = 0.295P
TAD = 0.2959P
Using P = 8 kN:
TAB = 4.33 kN !
TAC = 2.36 kN !
TAD = 2.37 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
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Chapter 2, Solution 116.
d BA =
( 6 m )2 + ( 6 m )2 + ( 3 m )2 = 9 m
d AC =
( −10.5 m )2 + ( − 6 m )2 + ( − 8 m )2 = 14.5 mm
d AD =
( − 6 m )2 + ( − 6 m )2 + ( 7 m )2 = 11 mm
d AE =
( 6 m )2 + ( − 4.5 m )2 = 7.5 m
FBA = FBAλBA =
FBA
( 6 m ) i + ( 6 m ) j + ( 3 m ) k 
9m
2
1 
2
= FBA  i + j + k 
3
3 
3
TAC = TAC λ AC =
TAC
 − (10.5 m ) i − ( 6 m ) j − ( 8 m ) k 
14.5 m 
12
16 
 21
= TAC  − i −
j−
k
29
29
29 

TAD = TAD λ AD =
TAD
 − ( 6 m ) i − ( 6 m ) j + ( 7 m ) k 
11 m 
6
7 
 6
= TAD  − i − j + k 
11
11 
 11
WAE = WAE λ AE =
W
( 6 m ) i − ( 4.5 m ) j
7.5 m 
= W ( 0.8i − 0.6 j)
WO = − W j
At point A: ΣF = 0: FBA + TAC + TAD + WAE + WO = 0
i component:
2
21
6
FBA −
TAC − TAD + 0.8W = 0
3
29
11
(1)
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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j component:
2
12
6
FBA −
TAC − TAD − 1.6W = 0
3
29
11
(2)
k component:
1
16
7
FBA −
TAC + TAD = 0
3
29
11
(3)
(
)
Setting W = ( 20 kg ) 9.81 m/s 2 = 196.2 N
And Solving Equations (1), (2), and (3) simultaneously:
FBA = 1742 N TAC = 1517 N TAD = 403 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
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Chapter 2, Solution 117.
ΣFx = 0:
− TAD ( sin 30° )( sin 50° ) + TBD ( sin 30° )( cos 40° ) + TCD ( sin 30° )( cos 60° ) = 0
Dividing through by sin 30° and evaluating:
− 0.76604 TAD + 0.76604 TBD + 0.5 TCD = 0
(1)
ΣFy = 0:
− TAD ( cos 30° ) − TBD ( cos 30° ) − TCD ( cos 30° ) + 62 lb = 0
or TAD + TBD + TCD = 71.591 lb
(2)
ΣFz = 0:
TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0
or
0.64279 TAD + 0.64279 TBD − 0.86603TCD = 0
(3)
Solving Equations (1), (2), and (3) simultaneously:
TAD = 30.5 lb !
TBD = 10.59 lb !
TCD = 30.5 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
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Chapter 2, Solution 118.
From the solutions to Problems 2.111 and 2.112, have
(2′)
TBE + TCF + TDG = 0.2 65
−TBE sin 45° + TCF sin 30° − TDG sin15° = 0
(3)
TBE cos 45° + TCF cos 30° − TDG cos15° − P 65 = 0 (1′ )
Applying the method of elimination to obtain a desired result:
Multiplying (2′) by sin 45° and adding the result to (3):
TCF ( sin 45° + sin 30° ) + TDG ( sin 45° − sin15° ) = 0.2 65 sin 45°
TCF = 0.94455 − 0.37137TDG
or
Multiplying (2′) by sin 30° and subtracting (3) from the result:
TBE ( sin 30° + sin 45° ) + TDG ( sin 30° + sin15° ) = 0.2 65 sin 30°
or
TBE = 0.66790 − 0.62863TDG
Substituting (4) and (5) into (1′) :
1.29028 − 1.73205TDG − P 65 = 0
∴ TDG is taut for P <
1.29028
lb
65
or 0 ≤ P ≤ 0.1600 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
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Chapter 2, Solution 119.
d AB =
( − 30 ft )2 + ( 24 ft )2 + ( 32 ft )2 = 50 ft
d AC =
( − 30 ft )2 + ( 20 ft )2 + ( −12 ft )2 = 38 ft
TAB = TAB λ AB =
TAB
 − ( 30 ft ) i + ( 24 ft ) j + ( 32 ft ) k 
50 ft 
= TAB ( − 0.6i + 0.48 j + 0.64k )
TAC = TAC λ AC =
TAC
 − ( 30 ft ) i + ( 20 ft ) j − (12 ft ) k 
38 ft 
20
12 
 30
= TAC  − i +
j−
k
38
38 
 38
N=
16
30
Ni +
Nj
34
34
W = − (175 lb ) j
At point A: ΣF = 0: TAB + TAC + N + W = 0
i component:
− 0.6 TAB −
30
16
TAC +
N=0
38
34
(1)
j component:
0.48 TAB +
20
30
TAC +
N − 175 lb = 0
38
34
(2)
12
TAC = 0
38
Solving Equations (1), (2), and (3) simultaneously:
k component:
0.64 TAB −
(3)
TAB = 30.9 lb TAC = 62.5 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 120.
Refer to the solution of problem 2.119 and the resulting linear algebraic Equations (1), (2), (3). Include force
P = − ( 45 lb ) k with other forces of Problem 2.119.
Now at point A: ΣF = 0: TAB + TAC + N + W + P = 0
i component:
− 0.6 TAB −
30
16
TAC +
N=0
38
34
(1)
j component:
0.48 TAB +
20
30
TAC +
N − 175 lb = 0
38
34
(2)
k component:
0.64 TAB −
12
TAC − 45 lb = 0
38
(3)
Solving (1), (2), and (3) simultaneously:
TAB = 81.3 lb TAC = 22.2 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
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Chapter 2, Solution 121.
Note: BE shares the same unit vector as AB.
Thus:
λBE = λ AB =
( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k
201.56 mm
TBE = TBE λBE =
TBE
( 25 mm ) cos 45°i + ( 200 mm ) j − ( 25 mm ) sin 45°k 
201.56 mm 
TCF = TCF λCF =
TCF
( 25 mm ) cos 30°i + ( 200 mm ) j + ( 25 mm ) sin 30°k 
201.56 mm 
TDG = TDG λDG =
TDG
 − ( 25 mm ) cos15°i + ( 200 mm ) j − ( 25 mm ) sin15°k 
201.56 mm 
W = − W j;
P = Pk
At point A: ΣF = 0: TBE + TCE + TDG + W + P = 0
i component:
0.087704 TBE + 0.107415 TCF − 0.119806 TDG = 0
(1)
j component:
0.99226 TBE + 0.99226 TCF + 0.99226 TDG − W = 0
(2)
k component:
− 0.087704 TBE + 0.062016 TCF − 0.032102 TDG + P = 0
(3)
Setting W = 10.5 N and P = 0, and solving (1), (2), (3) simultaneously:
TBE = 1.310 N !
TCF = 4.38 N !
TDG = 4.89 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
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Chapter 2, Solution 122.
See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
i component: 0.087704 TBE + 0.107415 TCF − 0.119806 TDG = 0
(1)
j component: 0.99226 TBE + 0.99226 TCF + 0.99226 TDG − W = 0
(2)
k component: − 0.087704 TBE + 0.062016 TCF − 0.032102 TDG + P = 0
(3)
Setting W = 10.5 N and P = 0.5 N, and solving (1), (2), (3) simultaneously:
TBE = 4.84 N !
TCF = 1.157 N !
TDG = 4.58 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
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Chapter 2, Solution 123.
uuur
DA = − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k
DA =
( − 8 ft )2 + ( 40 ft )2 + (10 ft )2 = 42 ft
TDA =
TADB
 − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k 
42 ft 
= TADB ( − 0.190476i + 0.95238 j + 0.23810k )
uuur
DB = ( 3 ft ) i + ( 36 ft ) j − ( 8 ft ) k
DB =
( 3 ft )2 + ( 36 ft )2 + ( − 8 ft )2 = 37 ft
TDB =
TADB
( 3 ft ) i + ( 36 ft ) j − ( 8 ft ) k 
37 ft 
= TADB ( 0.081081i + 0.97297 j − 0.21622k )
uuur
DC = ( a − 8 ft ) i − ( 24 ft ) j − ( 3 ft ) k
( a − 8 ft )2 + ( − 24 ft )2 + ( −3 ft )2 = ( a − 8)2 + 585 ft
DC =
TDC
TDC =
At D
( a − 8)2 + 585
( a − 8 ft ) i − ( 24 ft ) j − ( 3 ft ) k 
ΣF = 0:
ΣFx = 0: − 0.190476 TADB + 0.081081TADB +
ΣFz = 0: 0.23810 TADB − 0.21622 TADB −
( a − 8)
TDC = 0
( a − 8)2 + 585
3
( a − 8) + 585
2
TDC = 0
(1)
(2)
Dividing equation (1) by equation (2) gives
( a − 8) = 0.190476 − 0.081081
−3
− 0.23810 + 0.21622
or
a = 23 ft
Substituting into equation (1) for a = 23 ft and combining the coefficients for TADB gives:
ΣFx = 0:
− 0.109395 TADB + 0.52705 TDC = 0
(3)
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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And writing ΣFy = 0 gives:
1.92535 TADB − 0.84327 TDC − W = 0
(4)
Substituting into equation (3) for TDC = 17 lb gives:
− 0.109395 TADB + 0.52705 (17 lb ) = 0
or
TADB = 81.9 lb Substituting into equation (4) for TDC = 17 lb and TADB = 81.9 lb gives:
1.92535 ( 81.9 lb ) − 0.84327 (17 lb ) − W = 0
or
W = 143.4 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 124.
See Problem 2.123 for the analysis leading to the linear
algebraic Equations (3) and (4) below:
− 0.109395 TADB + 0.52705 TDC = 0
(3)
1.92535 TADB − 0.84327 TDC − W = 0
(4)
Substituting for W = 120 lb and solving equations (3) and (4) simultaneously yields
TADB = 68.6 lb !
TDC = 14.23 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 125.
d AB =
( − 2.7 m )2 + ( 2.4 m )2 + ( − 3.6 m )2 = 5.1 m
d AC =
( 2.4 m )2 + (1.8 m )2 = 3 m
d AD =
(1.2 m )2 + ( 2.4 m )2 + ( − 0.3 m )2 = 2.7 m
d AE =
( − 2.4 m )2 + ( 2.4 m )2 + (1.2 m )2 = 3.6 m
TAB = TAB λ AB
=
TAB
 − ( 2.7 m ) i + ( 2.4 m ) j − ( 3.6 m ) k 
5.1 m 
8
12 
 9
= TAB  − i +
j − k
17
17 
 17
TAC = TAC λ AC
=
TAC
( 2.4 m ) j + (1.8 m ) k 
3m
= TAC ( 0.8 j + 0.6k )
TAD = 2TADE λ AD
=
2TADE
(1.2 m ) i + ( 2.4 m ) j − ( 0.3 m ) k 
2.7 m 
16
2 
8
= TADE  i +
j − k
9
9
9


continued
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
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COSMOS: Complete Online Solutions Manual Organization System
TAE = TAE λ AE
=
TADE
 − ( 2.4 m ) i + ( 2.4 m ) j + (1.2 m ) k 
3.6 m 
2
1 
 2
= TADE  − i + j + k 
3
3 
 3
W = − Wj
At point A:
ΣF = 0:
TAB + TAC + TAD + TAE + W = 0
i component:
−
9
8
2
TAB + TADE − TADE = 0
17
9
3
(1)
j component:
8
16
2
TAB + 0.8 TAC + TADE + TADE − W = 0
17
9
3
(2)
k component:
−
12
2
1
TAB + 0.6 TAC − TADE + TADE = 0
17
9
3
(3)
− 81TAB + 34 TADE = 0
(1′)
72 TAB + 122.4 TAC + 374 TADE = 153 W
(2′)
−108 TAB + 91.8 TAC + 17 TADE = 0
(3′)
Simplifying (1), (2), (3):
Setting W = 1400 N and solving (1), (2), (3) simultaneously:
TAB = 203 N "
TAC = 149.6 N "
TADE = 485 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 126.
See Problem 2.125 for the analysis leading to the linear algebraic Equations (1′ ) , ( 2′ ) , and ( 3′ ) below:
i component:
− 81 TAB + 34 TADE = 0
(1′)
j component:
72 TAB + 122.4 TAC + 37.4 TADE = 153 W
( 2′)
k component:
−108 TAB + 91.8 TAC + 17 TADE = 0
( 3′)
Setting TAB = 300 N and solving (1), (2), (3) simultaneously:
(a)
TAC = 221 N !
(b) TADE = 715 N !
(c)
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.
W = 2060 N !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 127.
Free-Body Diagrams of collars
For both Problems 2.127 and 2.128:
( AB )2 = x2 + y 2 + z 2
Here
(1 m )2 = ( 0.40 m )2 + y 2 + z 2
or
y 2 + z 2 = 0.84 m 2
Thus, with y given, z is determined.
Now
uuur
AB
1
λ AB =
=
( 0.40i − yj + zk ) m = 0.4i − yk + zk
AB 1 m
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A:
ΣF = 0: N x i + N zk + Pj + TAB λ AB = 0
Setting the j coefficient to zero gives:
P − yTAB = 0
With P = 680 N,
TAB =
680 N
y
Now, from the free body diagram of collar B:
ΣF = 0: N x i + N y j + Qk − TABλ AB = 0
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Setting the k coefficient to zero gives:
Q − TAB z = 0
And using the above result for TAB we have
Q = TAB z =
680 N
z
y
Then, from the specifications of the problem, y = 300 mm = 0.3 m
z 2 = 0.84 m 2 − ( 0.3 m )
2
∴ z = 0.866 m
and
TAB =
(a)
680 N
= 2266.7 N
0.30
TAB = 2.27 kN !
or
and
Q = 2266.7 ( 0.866 ) = 1963.2 N
(b)
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.
Q = 1.963 kN !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 128.
From the analysis of Problem 2.127, particularly the results:
y 2 + z 2 = 0.84 m 2
TAB =
680 N
y
Q=
680 N
z
y
With y = 550 mm = 0.55 m, we obtain:
z 2 = 0.84 m 2 − ( 0.55 m )
2
∴ z = 0.73314 m
and
TAB =
(a)
or
680 N
= 1236.36 N
0.55
TAB = 1.236 kN !
and
Q = 1236.36 ( 0.73314 ) N = 906 N
(b)
or
Q = 0.906 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 129.
Using the triangle rule and the Law of Sines
(a)
Have:
20 lb
14 lb
=
sin α
sin 30°
sin α = 0.71428
α = 45.6° (b)
β = 180° − ( 30° + 45.6° )
= 104.4°
Then:
R
14 lb
=
sin104.4° sin 30°
R = 27.1 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 130.
We compute the following distances:
OA =
( 70 )2 + ( 240 )2 = 250 mm
OB =
( 210 )2 + ( 200 )2 = 290 mm
OC =
(120 )2 + ( 225)2 = 255 mm
500 N Force:
 70 
Fx = −500 N 

 250 
Fx = −140.0 N !
 240 
Fy = +500 N 

 250 
Fy = 480 N !
 210 
Fx = +435 N 

 290 
Fx = 315 N !
 200 
Fy = +435 N 

 290 
Fy = 300 N !
 120 
Fx = +510 N 

 255 
Fx = 240 N !
 225 
Fy = −510 N 

 255 
Fy = −450 N !
435 N Force:
510 N 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 131.
Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC
is 450 N.
Then:
P=
(a)
450 N
= 549.3 N
cos 35°
P = 549 N !
Px = ( 450 N ) tan 35°
(b)
= 315.1 N
Px = 315 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
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Chapter 2, Solution 132.
Free-Body Diagram
Force Triangle
Law of Sines:
TAC
T
5 kN
= BC =
sin115° sin 5° sin 60°
(a)
TAC =
5 kN
sin115° = 5.23 kN
sin 60°
TAC = 5.23 kN !
(b)
TBC =
5 kN
sin 5° = 0.503 kN
sin 60°
TBC = 0.503 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 133.
Free-Body Diagram
First, consider the sum of forces in the x-direction because there is only one unknown force:
ΣFx = 0: TACB ( cos 32° − cos 42° ) − ( 20 kN ) cos 42° = 0
or
0.104903TACB = 14.8629 kN
TACB = 141.682 kN
(b) TACB = 141.7 kN !
Now
ΣFy = 0: TACB ( sin 42° − sin 32° ) + ( 20 kN ) sin 42° − W = 0
or
(141.682 kN )( 0.139211) + ( 20 kN )( 0.66913) − W = 0
(a) W = 33.1 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 134.
Free-Body Diagram: Pulley A
ΣFx = 0: 2P sin 25° − P cos α = 0
and
cos α = 0.8452
For
or
α = ±32.3°
α = +32.3°
ΣFy = 0: 2P cos 25° + P sin 32.3° − 350 lb = 0
or P = 149.1 lb
For
32.3° α = −32.3°
ΣFy = 0: 2P cos 25° + P sin − 32.3° − 350 lb = 0
or P = 274 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.
32.3° COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 135.
Fx = F sin 30° sin 50° = 220.6 N (Given)
(a)
F =
220.6 N
= 575.95 N
sin30° sin50°
F = 576 N !
cosθ x =
(b)
Fx
220.6
=
= 0.38302
F
575.95
θ x = 67.5° !
Fy = F cos 30° = 498.79 N
cosθ y =
Fy
F
=
498.79
= 0.86605
575.95
θ y = 30.0° !
Fz = − F sin 30° cos 50°
= − ( 575.95 N ) sin 30° cos 50°
= −185.107 N
cosθ z =
Fz
−185.107
=
= −0.32139
F
575.95
θ z = 108.7° !
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 136.
Fz = F cosθ z = ( 600 lb ) cos136.8°
(a)
= −437.38 lb
Fz = −437 lb !
Then:
F 2 = Fx2 + Fy2 + Fz2
( ) + ( −437.38 lb )2
So: ( 600 lb ) = ( 200 lb ) + Fy
2
2
2
Hence: Fy = − ( 600 lb ) − ( 200 lb ) − ( −437.38 lb )
2
(b)
cosθ x =
2
2
= −358.75 lb
Fy = −359 lb !
Fx
200
=
= 0.33333
F
600
θ x = 70.5° !
cosθ y =
Fy
F
=
−358.75
= −0.59792
600
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.
θ y = 126.7° !
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Chapter 2, Solution 137.
P = ( 500 lb ) [ − cos 30° sin15°i + sin 30° j + cos 30° cos15°k ]
= ( 500 lb ) [ −0.2241i + 0.50 j + 0.8365k ]
= − (112.05 lb ) i + ( 250 lb ) j + ( 418.25 lb ) k
Q = ( 600 lb ) [ cos 40° cos 20°i + sin 40° j − cos 40° sin 20°k ]
= ( 600 lb ) [ 0.71985i + 0.64278 j − 0.26201k ]
= ( 431.91 lb ) i + ( 385.67 lb ) j − (157.206 lb ) k
R = P + Q = ( 319.86 lb ) i + ( 635.67 lb ) j + ( 261.04 lb ) k
R=
( 319.86 lb )2 + ( 635.67 lb )2 + ( 261.04 lb )2 = 757.98 lb
R = 758 lb !
cosθ x =
Rx
319.86 lb
=
= 0.42199
R
757.98 lb
θ x = 65.0° !
cosθ y =
Ry
R
=
635.67 lb
= 0.83864
757.98 lb
θ y = 33.0° !
cosθ z =
Rz
261.04 lb
=
= 0.34439
R
757.98 lb
θ z = 69.9° !
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 138.
The forces applied at A are:
TAB , TAC , TAD and P
where P = Pj . To express the other forces in terms of the unit vectors
i, j, k, we write
uuur
AB = − ( 0.72 m ) i + (1.2 m ) j − ( 0.54 m ) k,
AB = 1.5 m
uuur
AC = (1.2 m ) j + ( 0.64 m ) k,
AC = 1.36 m
uuur
AD = ( 0.8 m ) i + (1.2 m ) j − ( 0.54 m ) k,
AD = 1.54 m
uuur
AB
TAB = TABλ AB = TAB
= ( −0.48i + 0.8j − 0.36k ) TAB
and
AB
uuur
AC
TAC = TAC λ AC = TAC
= ( 0.88235j + 0.47059k ) TAC
AC
uuur
AD
TAD = TADλ AD = TAD
= ( 0.51948i + 0.77922 j − 0.35065k ) TAD
AD
Equilibrium Condition with W = −Wj
ΣF = 0: TAB + TAC + TAD − Wj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and
factoring i, j, and k:
( −0.48TAB + 0.51948TAD ) i + ( 0.8TAB + 0.88235TAC + 0.77922TAD − W ) j
+ ( −0.36TAB + 0.47059TAC − 0.35065TAD ) k = 0
Equating to zero the coefficients of i, j, k:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAB = 3 kN in Equations (1), (2) and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives
TAC = 4.3605 kN
TAD = 2.7720 kN
W = 8.41 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 139.
The (vector) force in each cable can be written as the product of the
(scalar) force and the unit vector along the cable. That is, with
uuur
AB = ( 32 in.) i − ( 48 in.) j + ( 36 in.) k
AB =
( −32 in.)2 + ( −48 in.)2 + ( 36 in.)2 = 68 in.
TAB = T λ AB = TAB
uuur
AB
T
= AB  − ( 32 in.) i − ( 48 in.) j + ( 36 in.) k 
68 in.
AB
TAB = TAB ( −0.47059i − 0.70588 j + 0.52941k )
uuur
AC = ( 45 in.) i − ( 48 in.) j + ( 36 in.) k
and
AC =
( 45 in.)2 + ( −48 in.)2 + ( 36 in.)2 = 75 in.
uuur
AC
T
TAC = T λ AC = TAC
= AC ( 45 in.) i − ( 48 in.) j + ( 36 in.) k 
75 in.
AC
TAC = TAC ( 0.60i − 0.64 j + 0.48k )
uuur
AD = ( 25 in.) i − ( 48 in.) j − ( 36 in.) k
Finally,
AD =
( 25 in.)2 + ( −48 in.)2 + ( −36 in.)2 = 65 in.
continued
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
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uuur
AD
T
TAD = T λ AD = TAD
= AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k 
65 in.
AD
TAD = TAD ( 0.38461i − 0.73846 j − 0.55385k )
With W = Wj, at A we have:
ΣF = 0: TAB + TAC + TAD + Wj = 0
Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
i : − 0.47059TAB + 0.60TAC − 0.38461TAD = 0
(1)
j: − 0.70588TAB − 0.64TAC − 0.73846TAD + W = 0
(2)
k : 0.52941TAB + 0.48TAC − 0.55385TAD = 0
(3)
In Equations (1), (2) and (3), set TAD = 120 lb, and, using conventional methods for solving Linear Algebraic
Equations (MATLAB or Maple, for example), we obtain:
TAB = 32.6 lb
TAC = 102.5 lb
W = 177.2 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 2, Solution 140.
The (vector) force in each cable can be written as the product of the
(scalar) force and the unit vector along the cable. That is, with
uuur
AB = − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k
AB =
( −0.48 m )2 + ( 0.72 m )2 + ( −0.16 m )2 = 0.88 m
uuur
AB
TAB
 − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k 
TAB = T λ AB = TAB
=
AB
0.88 m 
TAB = TAB ( −0.54545i + 0.81818 j − 0.181818k )
and
uuur
AC = ( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k
AC =
( 0.24 m )2 + ( 0.72 m )2 − ( 0.13 m )2 = 0.77 m
uuur
AC
TAC
( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k 
TAC = T λ AC = TAC
=
AC
0.77 m 
TAC = TAC ( 0.31169i + 0.93506 j − 0.16883k )
At A:
ΣF = 0: TAB + TAC + P + Q + W = 0
Noting that TAB = TAC because of the ring A, we equate the factors of
i, j, and k to zero to obtain the linear algebraic equations:
i:
( −0.54545 + 0.31169 ) T + P = 0
P = 0.23376T
or
j:
( 0.81818 + 0.93506 ) T − W = 0
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
W = 1.75324T
or
k:
( −0.181818 − 0.16883) T + Q = 0
Q = 0.35065T
or
With W = 1200 N:
T =
1200 N
= 684.45 N
1.75324
P = 160.0 N !
Q = 240 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 3, Solution 1.
Resolve 90 N force into vector components P and Q
where Q = ( 90 N ) sin 40°
= 57.851 N
Then M B = − rA/BQ
= − (0.225 m )(57.851 N )
= −13.0165 N ⋅ m
M B = 13.02 N ⋅ m
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 2.
Fx = ( 90 N ) cos 25°
= 81.568 N
Fy = ( 90 N ) sin 25°
= 38.036 N
x = ( 0.225 m ) cos 65°
= 0.095089 m
y = (0.225 m ) sin 65°
= 0.20392 m
M B = xFy − yFx
= ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N )
= −13.0165 N ⋅ m
M B = 13.02 N ⋅ m
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 3, Solution 3.
Px = ( 3 lb ) sin 30°
= 1.5 lb
Py = ( 3 lb ) cos 30°
= 2.5981 lb
M A = xB/ A Py + yB/ A Px
= ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb )
= 16.0335 lb ⋅ in.
M A = 16.03 lb ⋅ in.
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 3, Solution 4.
For P to be a minimum, it must be perpendicular to the line joining points
A and B
with rAB =
( 3.4 in.)2 + ( 4.8 in.)2
= 5.8822 in.
 y
 
α = θ = tan −1  
x
 4.8 in. 
= tan −1 

 3.4 in. 
= 54.689°
Then
M A = rAB Pmin
or
Pmin =
M A 19.5 lb ⋅ in.
=
rAB
5.8822 in.
= 3.3151 lb
∴ Pmin = 3.32 lb
54.7°
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.
Pmin = 3.32 lb
35.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 5.
M A = rB/ A P sin θ
By definition
where
θ = φ + ( 90° − α )
and
φ = tan −1 
 4.8 in. 

 3.4 in. 
= 54.689°
Also
rB/ A =
( 3.4 in.)2 + ( 4.8 in.)2
= 5.8822 in.
Then
(17 lb ⋅ in.) = ( 5.8822 in.)( 2.9 lb ) sin ( 54.689° + 90° − α )
or
sin (144.689° − α ) = 0.99658
or
144.689° − α = 85.260°; 94.740°
∴ α = 49.9°, 59.4°
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 3, Solution 6.
(a)
(a) M A = rB/ A × TBF
M A = xTBFy + yTBFx
= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°
= 386.41 N ⋅ m
or M A = 386 N ⋅ m
(b)
(b) For FC to be a minimum, it must be perpendicular to the line
joining A and C.
∴ M A = d ( FC )min
d =
with
( 2 m )2 + (1.35 m )2
= 2.4130 m
Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min
( FC )min = 160.137 N
and
 1.35 m 
 = 34.019°
 2m 
φ = tan −1 
θ = 90 − φ = 90° − 34.019° = 55.981°
∴ ( FC )min = 160.1 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.
56.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 7.
(a)
M A = xTBFy + yTBFx
= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°
= 386.41 N ⋅ m
or M A = 386 N ⋅ m
(b)
Have
or
M A = xFC
FC =
MA
386.41 N ⋅ m
=
2m
x
= 193.205 N
(c)
∴ FC = 193.2 N
For FB to be minimum, it must be perpendicular to the line joining A
and B
∴ M A = d ( FB )min
with
Then
d =
( 2 m )2 + ( 0.40 m )2 = 2.0396 m
386.41 N ⋅ m = ( 2.0396 m ) ( FC )min
( FC )min = 189.454 N
and
 2m 
 = 78.690°
 0.4 m 
θ = tan −1 
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.
( FC )min = 189.5 N
78.7° COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 8.
(a)
(
)
M B = rA/B cos15° W
= (14 in.)( cos15° )( 5 lb )
= 67.615 lb ⋅ in.
or
M B = 67.6 lb ⋅ in.
(b)
M B = rD/B P sin 85°
67.615 lb ⋅ in. = ( 3.2 in.) P sin 85°
or
(c)
P = 21.2 lb For ( F )min, F must be perpendicular to BC.
Then,
M B = rC/B F
67.615 lb ⋅ in. = (18 in.) F
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.
F = 3.76 lb
75.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 9.
Slope of line EC =
(a)
Then
and
Then
TABx =
35 in.
5
=
76 in. + 8 in. 12
12
(TAB )
13
=
12
( 260 lb ) = 240 lb
13
TABy =
5
( 260 lb ) = 100 lb
13
M D = TABx ( 35 in.) − TABy ( 8 in.)
= ( 240 lb )( 35 in.) − (100 lb )( 8 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in. (b) Have
M D = TABx ( y ) + TABy ( x )
= ( 240 lb )( 0 ) + (100 lb )( 76 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in. 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 3, Solution 10.
Slope of line EC =
35 in.
7
=
112 in. + 8 in. 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
M D = TABx ( y ) + TABy ( x )
Have
∴ 7840 lb ⋅ in. =
24
7
TAB ( 0 ) +
TAB (112 in.)
25
25
TAB = 250 lb
or TAB = 250 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 11.
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y ( d )
M D = 1152 N ⋅ m
where
(TAB max ) y = TAB max sin θ = ( 2880 N ) sin θ
Now
sin θ =
1.05 m
( d + 0.24 )2 + (1.05)2 m

∴ 1152 N ⋅ m = 2880 N 



 (d )
2
2 
( d + 0.24 ) + (1.05) 
1.05
or
( d + 0.24 )2 + (1.05)2 = 2.625d
or
( d + 0.24 )2 + (1.05)2 = 6.8906d 2
or
5.8906d 2 − 0.48d − 1.1601 = 0
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm
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 3, Solution 12.
with d AB =
( 42 mm )2 + (144 mm )2
= 150 mm
sin θ =
42 mm
150 mm
cosθ =
144 mm
150 mm
and FAB = − FAB sin θ i − FAB cosθ j
=
2.5 kN
( − 42 mm ) i − (144 mm ) j
150 mm 
= − ( 700 N ) i − ( 2400 N ) j
Also rB/C = − ( 0.042 m ) i + ( 0.056 m ) j
Now M C = rB/C × FAB
= ( − 0.042 i + 0.056 j) × ( − 700 i − 2400 j) N ⋅ m
= (140.0 N ⋅ m ) k
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.
M C = 140.0 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 13.
( 42 mm )2 + (144 mm )2
with d AB =
= 150 mm
sin θ =
42 mm
150 mm
cosθ =
144 mm
150 mm
FAB = − FAB sin θ i − FAB cosθ j
=
2.5 kN
( − 42 mm ) i − (144 mm ) j
150 mm 
= − ( 700 N ) i − ( 2400 N ) j
Also rB/C = − ( 0.042 m ) i − ( 0.056 m ) j
Now M C = rB/C × FAB
= ( − 0.042 i − 0.056 j) × ( − 700i − 2400 j) N ⋅ m
= ( 61.6 N ⋅ m ) k
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.
M C = 61.6 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 14.
ΣM D :
 88

 105

M D = ( 0.090 m ) 
× 80 N  − ( 0.280 m ) 
× 80 N 
137
137




= −12.5431 N ⋅ m
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.
M D = 12.54 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 15.
Note: B = B ( cos β i + sin β j)
B′ = B ( cos β i − sin β j)
C = C ( cos α i + sin α j)
By definition:
B × C = BC sin (α − β )
(1)
B′ × C = BC sin (α + β )
(2)
Now ... B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α − sin β cos α ) k
and
(3)
B′ × C = B ( cos β i − sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α + sin β cos α ) k
(4)
Equating the magnitudes of B × C from equations (1) and (3) yields:
BC sin (α − β ) = BC ( cos β sin α − sin β cos α )
(5)
Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields:
BC sin (α + β ) = BC ( cos β sin α + sin β cos α )
(6)
Adding equations (5) and (6) gives:
sin (α − β ) + sin (α + β ) = 2cos β sin α
or
sin α cos β =
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.
1
1
sin (α + β ) + sin (α − β )
2
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 16.
Have d = λ AB × rO/ A
where λ AB =
rB/ A
rB/ A
and rB/ A = ( −210 mm − 630 mm ) i
+ ( 270 mm − ( −225 mm ) ) j
= − ( 840 mm ) i + ( 495 mm ) j
rB/ A =
( −840 mm )2 + ( 495 mm )2
= 975 mm
Then λ AB =
=
− ( 840 mm ) i + ( 495 mm ) j
975 mm
1
( −56i + 33j)
65
Also rO/ A = ( 0 − 630 ) i + ( 0 − (−225) ) j
= − ( 630 mm ) i + ( 225 mm ) j
∴d =
1
( −56i + 33j) × − ( 630 mm ) i + ( 225 mm ) j
65
= 126.0 mm
d = 126.0 mm W
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 3, Solution 17.
(a)
where
λ =
A×B
A×B
A = 12i − 6 j + 9k
B = − 3i + 9 j − 7.5k
Then
i
j
k
A × B = 12 − 6 9
− 3 9 − 7.5
= ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 ) k
= 9 ( − 4i + 7 j + 10k )
And A × B = 9 (− 4) 2 + (7)2 + (10)2 = 9 165
∴λ =
9 ( − 4i + 7 j + 10k )
9 165
or λ =
(b)
where
λ =
A×B
A×B
A = −14i − 2 j + 8k
B = 3i + 1.5j − k
Then
j k
i
A × B = −14 − 2 8
3 1.5 −1
= ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k
= 5 ( −2i + 2 j − 3k )
and
A × B = 5 (−2)2 + (2)2 + (−3)2 = 5 17
∴λ =
or λ =
5 ( −2i + 2 j − 3k )
5 17
1
( − 2i + 2 j − 3k ) 17
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.
1
( − 4i + 7 j + 10k ) 165
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 18.
(a)
Have A = P × Q
i j k
P × Q = 3 7 −2 in.2
−5 1 3
= [ (21 + 2)i + (10 − 9) j + (3 + 35)k ] in.2
(
) (
) (
)
= 23 in.2 i + 1 in.2 j + 38 in.2 k
∴A=
(23)2 + (1)2 + (38) 2 = 44.430 in.2
or A = 44.4 in.2 W
(b)
A = P×Q
i j k
P × Q = 2 − 4 3 in.2
6 −1 5
= [ (−20 − 3)i + (−18 − 10) j + (−2 + 24)k ] in.2
(
) (
) (
)
= − 23 in.2 i − 28 in.2 j + 22 in.2 k
∴A=
(− 23)2 + (−28)2 + (22) 2 = 42.391 in.2
or A = 42.4 in.2 W
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 3, Solution 19.
(a)
Have
MO = r × F
i
j
k
= − 6 3 1.5 N ⋅ m
7.5 3 − 4.5
= [ (−13.5 − 4.5)i + (11.25 − 27) j + (−18 − 22.5)k ] N ⋅ m
= ( −18.00i − 15.75 j − 40.5k ) N ⋅ m
or M O = − (18.00 N ⋅ m ) i − (15.75 Ν ⋅ m ) j − ( 40.5 N ⋅ m ) k W
(b)
Have
MO = r × F
i
j
k
= 2 − 0.75 −1 N ⋅ m
7.5
3
− 4.5
= [ (3.375 + 3)i + (−7.5 + 9) j + (6 + 5.625)k ] N ⋅ m
= ( 6.375i + 1.500 j + 11.625k ) N ⋅ m
or M O = ( 6.38 N ⋅ m ) i + (1.500 Ν ⋅ m ) j + (11.63 Ν ⋅ m ) k W
(c)
Have
MO = r × F
i
j k
= − 2.5 −1 1.5 N ⋅ m
7.5 3 4.5
= [ (4.5 − 4.5)i + (11.25 − 11.25) j + (−7.5 + 7.5)k ] N ⋅ m
or M O = 0 W
This answer is expected since r and F are proportional ( F = −3r ) . Therefore, vector F has a line of action
passing through the origin at O.
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 3, Solution 20.
(a)
Have
MO = r × F
i
j k
= − 7.5 3 − 6 lb ⋅ ft
3 −6 4
= [ (12 − 36)i + (−18 + 30) j + (45 − 9)k ] lb ⋅ ft
or M O = − ( 24.0 lb ⋅ ft ) i + (12.00 lb ⋅ ft ) j + ( 36.0 lb ⋅ ft ) k W
(b)
Have
MO = r × F
i
j k
= − 7.5 1.5 −1 lb ⋅ ft
3 −6 4
= [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5)k ] lb ⋅ ft
or M O = 0 W
(c)
Have
MO = r × F
i
j
k
= − 8 2 −14 lb ⋅ ft
3 −6 4
= [ (8 − 84)i + (−42 + 32) j + (48 − 6)k ] lb ⋅ ft
or M O = − ( 76.0 lb ⋅ ft ) i − (10.00 lb ⋅ ft ) j + ( 42.0 lb ⋅ ft ) k W
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 3, Solution 21.
With
TAB = − ( 369 N ) j
JJJG
AD
TAB = TAD
= ( 369 N )
AD
( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k
( 2.4 m )2 + ( −3.1 m )2 + ( −1.2 m )2
TAD = ( 216 N ) i − ( 279 N ) j − (108 N ) k
Then
R A = 2 TAB + TAD
= ( 216 N ) i − (1017 N ) j − (108 N ) k
Also
rA/C = ( 3.1 m ) i + (1.2 m ) k
Have
M C = rA/C × R A
i
j
k
= 0
3.1
1.2 N ⋅ m
216 −1017 −108
= ( 885.6 N ⋅ m ) i + ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k
M C = ( 886 N ⋅ m ) i + ( 259 N ⋅ m ) j − ( 670 N ⋅ m ) k W
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 3, Solution 22.
Have
M A = rC/ A × F
where
rC/ A = ( 215 mm ) i − ( 50 mm ) j + (140 mm ) k
Fx = − ( 36 N ) cos 45° sin12°
Fy = − ( 36 N ) sin 45°
Fz = − ( 36 N ) cos 45° cos12°
∴ F = − ( 5.2926 N ) i − ( 25.456 N ) j − ( 24.900 N ) k
and
i
j
k
M A = 0.215
− 0.050
0.140 N ⋅ m
− 5.2926 − 25.456 − 24.900
= ( 4.8088 N ⋅ m ) i + ( 4.6125 N ⋅ m ) j − ( 5.7377 N ⋅ m ) k
M A = ( 4.81 N ⋅ m ) i + ( 4.61 N ⋅ m ) j − ( 5.74 N ⋅ m ) k W
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 3, Solution 23.
Have
M O = rA/O × R
where
rA/D = ( 30 ft ) j + ( 3 ft ) k
T1 = − ( 62 lb ) cos10°  i − ( 62 lb ) sin10°  j
= − ( 61.058 lb ) i − (10.766 lb ) j
JJJG
AB
T2 = T2
AB
= ( 62 lb )
( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k
( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2
= (10 lb ) i − ( 60 lb ) j + (12 lb ) k
∴ R = − ( 51.058 lb ) i − ( 70.766 lb ) j + (12 lb ) k
i
j
k
MO =
0
30
3 lb ⋅ ft
− 51.058 −70.766 12
= ( 572.30 lb ⋅ ft ) i − (153.17 lb ⋅ ft ) j + (1531.74 lb ⋅ ft ) k
M O = ( 572 lb ⋅ ft ) i − (153.2 lb ⋅ ft ) j + (1532 lb ⋅ ft ) k W
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 3, Solution 24.
(a)
Have
M O = rB/O × TBD
where
rB/O = ( 2.5 m ) i + ( 2 m ) j
JJJG
BD
TBD = TBD
BD
 − (1 m ) i − ( 2 m ) j + ( 2 m ) k 
= ( 900 N ) 
( −1 m ) 2 + ( − 2 m ) 2 + ( 2 m ) 2
= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k
Then
i
j
k
M O = 2.5
2
0 N⋅m
− 300 − 600 600
M O = (1200 N ⋅ m ) i − (1500 N ⋅ m ) j − ( 900 N ⋅ m ) k W
continued
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
(b)
Have
M O = rB/O × TBE
where
rB/O = ( 2.5 m ) i + ( 2 m ) j
JJJG
BE
TBE = TBE
BE
 − ( 0.5 m ) i − ( 2 m ) j − ( 4 m ) k 
= ( 675 N ) 
( 0.5 m )2 + ( −2 m )2 + ( − 4 m )2
= − ( 75 N ) i − ( 300 N ) j − ( 600 N ) k
Then
i
j
k
M O = 2.5
2
0 N⋅m
− 75 − 300 − 600
M O = − (1200 N ⋅ m ) i + (1500 N ⋅ m ) j − ( 600 N ⋅ m ) k W
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 3, Solution 25.
Have M C = rA/C × P
where
rA/C = rB/C + rA/B
= (16 in.)( − cos80° cos15°i − sin 80° j − cos80° sin15°k )
+ (15.2 in.)( − sin 20° cos15°i + cos 20° j − sin 20° sin15°k )
= − ( 7.7053 in.) i − (1.47360 in.) j − ( 2.0646 in.) k
and
P = (150 lb )( cos 5° cos 70°i + sin 5° j − cos 5° sin 70°k )
= ( 51.108 lb ) i + (13.0734 lb ) j − (140.418 lb ) k
Then
i
j
k
M C = −7.7053 −1.47360 −2.0646 lb ⋅ in.
51.108 13.0734 −140.418
= ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k
or M C = (19.49 lb ⋅ ft ) i − ( 99.0 lb ⋅ ft ) j − ( 2.12 lb ⋅ ft ) k W
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 3, Solution 26.
Have
M C = rA/C × FBA
where
rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k
and
FBA = λ BA FBA


− ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k 

=
228 N )

(
( 0.1)2 + (1.8)2 + ( 0.6 )2 m 

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
i
j
k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k
or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k W
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 3, Solution 27.
Have
where
M C = TAD d
JJJG
d = Perpendicular distance from C to line AD
with
M C = rA/C TAD
and
rA/C = ( 3.1 m ) j + (1.2 m ) k
JJJG
AD
TAD = TAD
AD
( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k 
TAD = ( 369 N ) 
( 2.4 m )2 + ( − 3.1 m )2 + ( −1.2 m )2
= ( 216 N ) i − ( 279 N ) j − (108 N ) k
Then
i
j
k
3.1 1.2 N ⋅ m
MC = 0
216 − 279 −108
= ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k
and
MC =
( 259.2 N ⋅ m )2 + ( −669.6 N ⋅ m )2
= 718.02 N ⋅ m
∴ 718.02 N ⋅ m = ( 369 N ) d
or d = 1.946 m W
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 3, Solution 28.
Have
where
M O = TAC d
d = Perpendicular distance from O to rope AC
with
M O = rA/O × TAC
and
rA/O = ( 30 ft ) j + ( 3 ft ) k
TAC = − ( 62 lb ) cos10° i − ( 62 lb ) sin10° j
= − ( 61.058 lb ) i − (10.766 lb ) j
Then
i
j
k
0
30
3 lb ⋅ ft
MO =
− 61.058 −10.766 0
= ( 32.298 lb ⋅ ft ) i − (183.174 lb ⋅ ft ) j + (1831.74 lb ⋅ ft ) k
and
MO =
( 32.298 lb ⋅ ft )2 + ( −183.174 lb ⋅ ft )2 + (1831.74 lb ⋅ ft )2
= 1841.16 lb ⋅ ft
∴ 1841.16 lb ⋅ ft = ( 62 lb ) d
or d = 29.7 ft W
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 3, Solution 29.
M O = TAB d
Have
where
d = Perpendicular distance from O to rope AB
with
M O = rA/O × TAB
and
rA/O = ( 30 ft ) j + ( 3 ft ) k
JJJG
AB
TAB = TAB
AB
( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k 
= ( 62 lb ) 
( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2
= (10 lb ) i − ( 60 lb ) j + (12 lb ) k
Then
i
j k
M O = 0 30 3 lb ⋅ ft
10 − 60 12
= ( 540 lb ⋅ ft ) i + ( 30 lb ⋅ ft ) j − ( 300 lb ⋅ ft ) k
and
MO =
( 540 lb ⋅ ft )2 + ( 30 lb ⋅ ft )2 + ( −300 lb ⋅ ft )2
= 618.47 lb ⋅ ft
∴ 618.47 lb ⋅ ft = ( 62 lb ) d
or d = 9.98 ft W
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 3, Solution 30.
Have
where
M C = TBD d
d = Perpendicular distance from C to cable BD
with
M C = rB/C × TB/D
and
rB/C = ( 2 m ) j
JJJG
BD
TBD = TBD
BD
 − (1 m ) i − ( 2 m ) j + ( 2 m ) k 
= ( 900 N ) 
( −1 m )2 + ( − 2 m )2 + ( 2 m )2
= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k
Then
i
j
k
2
0 N⋅m
MC = 0
−300 − 600 600
= (1200 N ⋅ m ) i + ( 600 N ⋅ m ) k
and
MC =
(1200 N ⋅ m )2 + ( 600 N ⋅ m )2
= 1341.64 N ⋅ m
∴ 1341.64 = ( 900 N ) d
or d = 1.491 m W
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 3, Solution 31.
Have M C = Pd
From the solution of problem 3.25
M C = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k
Then
MC =
( 233.91)2 + ( −1187.48)2 + ( −25.422 )2
= 1210.57 lb ⋅ in.
and
d =
MC
1210.57 lb.in.
=
150 lb
P
or d = 8.07 in. W
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 3, Solution 32.
| M D | = FBAd
Have
where
d = perpendicular distance from D to line AB.
M D = rA/D × FBA
rA/D = − ( 0.12 m ) j + ( 0.72 m ) k
FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2 m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
i
j
k
∴ MD = 0
−0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k
and
|MD | =
(146.88)2 + (8.64 )2 + (1.44 )2 = 147.141 N ⋅ m
∴ 147.141 N ⋅ m = ( 228 N ) d
d = 0.64536 m
or d = 0.645 m W
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 3, Solution 33.
Have
| M C | = FBAd
d = perpendicular distance from C to line AB.
where
M C = rA/C × FBA
rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k
FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2 m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
i
j
k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k
and
| MC | =
(146.88)2 + ( 60.48)2 + ( 205.92 )2 = 260.07 N ⋅ m
∴ 260.07 N ⋅ m = ( 228 N ) d
d = 1.14064 m
or d = 1.141 m W
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 3, Solution 34.
(a) Have d = rC/ A sin θ = λ AB × rC/ A
where d = Perpendicular distance from C to pipe AB
with λ AB =
AB
=
AB
=
7i + 4 j − 32k
( 7 )2 + ( 4 )2 + ( −32 )2
1
( 7i + 4 j − 32k )
33
and rC/ A = − (14 ft ) i + ( 5 ft ) j + ( L − 22 ) ft  k
Then
i j
k
1
7 4 − 32 ft
λ AB × rC/ A =
33
−14 5 L − 22
{
}
1
 4 ( L − 22 ) + 32 ( 5 )  i + 32 (14 ) − 7 ( L − 22 )  j +  7 ( 5 ) + 4 (14 )  k ft
33 
1
( 4L + 72 ) i + ( −7 L + 602 ) j + 91k  ft
=
33 
=
and
d =
1
33
( 4L + 72 )2 + ( −7 L + 602 )2 + ( 91)2
or
or
dd 2
1
= 2  2 ( 4 )( 4L + 72 ) + 2 ( −7 )( −7 L + 602 )  = 0
dL
33
65L − 3926 = 0
L = 60.400 ft
But
L > Lgreenhouse
For
(b) with
(d ) min ,
L = 30 ft,
so
d =
1
33
L = 30.0 ft W
( 4 × 30 + 72 )2 + ( −7 × 30 + 602 )2 + ( 91)2
or d = 13.51 ft W
Note:
with
L = 60.4 ft,
d =
1
33
( 4 × 60.4 + 72 )2 + ( −7 × 60.4 + 602 )2 + ( 91)2 = 11.29 ft
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 3, Solution 35.
P ⋅ Q = ( − 4i + 8j − 3k ) ⋅ ( 9i − j − 7k )
= ( − 4 )( 9 ) + ( 8 )( −1) + ( − 3)( − 7 )
= − 23
or P ⋅ Q = −23 W
P ⋅ S = ( − 4i + 8 j − 3k ) ⋅ ( 5i − 6 j + 2k )
= ( − 4 )( 5 ) + ( 8 )( − 6 ) + ( − 3)( 2 )
= − 74
or P ⋅ S = −74 W
Q ⋅ S = ( 9i − j − 7k ) ⋅ ( 5i − 6 j + 2k )
= ( 9 )( 5 ) + ( −1)( − 6 ) + ( − 7 )( 2 )
= 37
or Q ⋅ S = 37 W
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 3, Solution 36.
By definition
B ⋅ C = BC cos (α − β )
where
B = B ( cos β ) i + ( sin β ) j
C = C ( cos α ) i + ( sin α ) j
∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β )
cos β cos α + sin β sin α = cos (α − β )
or
(1)
By definition
B′⋅ C = BC cos (α + β )
where
B′ = ( cos β ) i − ( sin β ) j
∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β )
or
cos β cos α − sin β sin α = cos (α + β )
(2)
Adding Equations (1) and (2),
2 cos β cos α = cos (α − β ) + cos (α + β )
or cos α cos β =
1
1
cos (α + β ) + cos (α − β ) W
2
2
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 3, Solution 37.
First note:
rB/ A = ( 0.56 m ) i + ( 0.9 m ) j
rC/ A = ( 0.9 m ) j − ( 0.48 m ) k
rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k
By definition
rB/ A =
( 0.56 m )2 + ( 0.9 m )2 = 1.06 m
rC/ A =
( 0.9 m )2 + ( − 0.48 m )2 = 1.02 m
rD/ A =
( − 0.52 m )2 + ( 0.9 m )2 + ( 0.36 m )2 = 1.10 m
rB/ A ⋅ rD/ A = rB/ A rD/ A cosθ
or ( 0.56i + 0.9 j) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.06 )(1.10 ) cosθ
( 0.56 )( − 0.52 ) + ( 0.9 )( 0.9 ) + ( 0 )( 0.36 ) = 1.166 cosθ
cosθ = 0.44494
θ = 63.6° W
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 3, Solution 38.
From the solution to problem 3.37
rC/ A = 1.02 m with rC/ A = ( 0.9 m ) i − ( 0.48 m ) j
rD/ A = 1.10 m with rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k
Now by definition
rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ
or ( 0.9 j − 0.48k ) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.02 )(1.10 ) cosθ
0 ( − 0.52 ) + ( 0.9 )( 0.9 ) + ( − 0.48)( 0.36 ) = 1.122cosθ
cosθ = 0.56791
or θ = 55.4° W
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 3, Solution 39.
(a) By definition
λ BC + λ EF = (1) (1) cosθ
where
λ BC =
=
λ EF =
=
Therefore
( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k
( 32 )2 + ( − 9 )2 + ( − 24 )2 ft
1
( 32i − 9 j − 24k )
41
− (14 ft ) i − (12 ft ) j + (12 ft ) k
( −14 )2 + ( −12 )2 + ( 12 )2 ft
1
( −7i − 6 j + 6k )
11
( 32i − 9 j − 24k ) ⋅ ( −7i − 6 j + 6k ) = cosθ
41
11
( 32 )( −7 ) + ( −9 )( −6 ) + ( −24 )( 6 ) = ( 41)(11) cosθ
cosθ = − 0.69623
or
(b) By definition
θ = 134.1° W
(TEG ) BC = (TEF ) cosθ
= (110 lb )( −0.69623)
= −76.585 lb
or (TEF ) BC = −76.6 lb W
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 3, Solution 40.
(a) By definition
λ BC ⋅ λ EG = (1) (1) cosθ
where
λ BC =
=
λ EG =
=
Therefore
( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k
( 32 )2 + ( − 9 )2 + ( − 24 )2 ft
1
( 32i − 9 j − 24k )
41
(16 ft ) i − (12 ft ) j + ( 9.75) k
(16 )2 + ( −12 )2 + ( 9.75)2 ft
1
(16i − 12 j + 9.75k )
22.25
( 32i − 9 j − 24k ) ⋅ (16i − 12 j + 9.75k ) = cosθ
41
22.25
( 32 )(16 ) + ( −9 )( −12 ) + ( −24 )( 9.75) = ( 41)( 22.25) cosθ
cosθ = 0.42313
or
(b) By definition
θ = 65.0° W
(TEG )BC = (TEG ) cosθ
= (178 lb )( 0.42313)
= 75.317 lb
or (TEG ) BC = 75.3 lb W
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 3, Solution 41.
First locate point B:
d
3.5
=
22
14
d = 5.5 m
or
d BA =
(a)
( 5.5 + 0.5)2 + ( −22 )2 + ( −3)2 = 23 m
Locate point D:
( −3.5 − 7.5sin 45° cos15° ) , (14 + 7.5cos 45° ) ,
( 0 + 7.5sin 45° sin15° ) m
or ( −8.6226 m, 19.3033 m, 1.37260 m )
Then
d BD =
( −8.6226 + 5.5)2 + (19.3033 − 22 )2 + (1.37260 − 0 )2 m
= 4.3482 m
and
cosθ ABD =
( 6i − 22 j − 3k ) ⋅ ( −3.1226i − 2.6967 j + 1.37260k )
d BA ⋅ d BD
=
d BA d BD
( 23)( 4.3482 )
= 0.36471
or
(b)
θ ABD = 68.6° W
(TBA )BD = TBA cosθ ABD
= ( 230 N )( 0.36471)
or (TBA ) BD = 83.9 N W
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 3, Solution 42.
First locate point B:
d
3.5
=
22
14
d = 5.5 m
or
(a) Locate point D:
( −3.5 − 7.5sin 45° cos10° ) , (14 + 7.5cos 45° ) ,
( 0 + 7.5sin 45° sin10° ) m
or ( −8.7227 m, 19.3033 m, 0.92091 m )
Then
d DC = ( 5.2227 m ) i − ( 5.3033 m ) j − ( 0.92091 m ) k
and
d DB =
( −5.5 + 8.7227 )2 + ( 22 − 19.3033)2 + ( 0 − 0.92091)2 m
= 4.3019 m
and
cos θ BDC =
( 3.2227i + 2.6967 j − 0.92091k ) ⋅ ( 5.2227i − 5.3033j − 0.92091k )
d DB ⋅ d DC
=
d DB d DC
( 4.3019 )( 7.5)
= 0.104694
(b)
or
θ BDC = 84.0° W
or
(TBD )DC = 26.2 N W
(TBD )DC = TBD cosθ BDC = ( 250 N )( 0.104694 )
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 3, Solution 43.
Volume of parallelopiped is found using the mixed triple product
(a)
Vol = P ⋅ ( Q × S )
3 −4 1
= − 7 6 − 8 in.3
9 −2 −3
= ( −54 + 288 + 14 − 48 + 84 − 54 ) in.3
= 230 in.3
or Volume = 230 in.3 W
(b)
Vol = P ⋅ ( Q × S )
−5 −7 4
= 6 − 2 5 in.3
−4 8 −9
= ( −90 + 140 + 192 + 200 − 378 − 32 ) in.3
= 32 in.3
or Volume = 32 in.3 W
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 3, Solution 44.
For the vectors to all be in the same plane, the mixed triple product is zero.
P ⋅(Q × S ) = 0
−3 −7 5
∴ 0 = −2 1 −4
8 Sy −6
0 = 18 + 224 − 10S y − 12S y + 84 − 40
So that
22 S y = 286
S y = 13
or S y = 13.00 W
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 3, Solution 45.
Have
rC = ( 2.25 m ) k
JJJG
CE
TCE = TCE
CE
( 0.90 m ) i + (1.50 m ) j − ( 2.25 m ) k 
TCE = (1349 N ) 
( 0.90 )2 + (1.50 )2 + ( −2.25 )2 m
= ( 426 N ) i + ( 710 N ) j − (1065 N ) k
Now
M O = rC × TCE
i
j
k
= 0
0
2.25 N ⋅ m
426 710 −1065
= − (1597.5 N ⋅ m ) i + ( 958.5 N ⋅ m ) j
∴ M x = −1598 N ⋅ m, M y = 959 N ⋅ m, M z = 0 W
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 3, Solution 46.
Have
rE = ( 0.90 m ) i + (1.50 m ) j
JJJG
DE
TDE = TDE
DE
 − ( 2.30 m ) i + (1.50 m ) j − ( 2.25 m ) k 
= (1349 N ) 
( − 2.30 )2 + (1.50 )2 + ( − 2.25)2 m
= − ( 874 N ) i + ( 570 N ) j − ( 855 N ) k
Now
M O = rE × TDE
i
j
k
= 0.90 1.50 0 N ⋅ m
− 874 570 − 855
= − (1282.5 N ⋅ m ) i + ( 769.5 N ⋅ m ) j + (1824 N ⋅ m ) k
∴ M x = −1283 N ⋅ m, M y = 770 N ⋅ m, M z = 1824 N ⋅ m
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 3, Solution 47.
Have
M z = k ⋅ ( rB ) y × TBA  + k ( rC ) y × TCD 




where
M z = − ( 48 lb ⋅ ft ) k
( rB ) y = ( rC ) y = ( 3 ft ) j
JJJG
( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
BA
TBA = TBA
= (14 lb ) 
BA
( 4.5)2 + ( − 3)2 + ( 9 )2 ft
= ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k
JJJG
( 6 ft ) i − ( 3 ft ) j − ( 6 ft ) k 
CD
TCD = TCD
= TCD 
CD
( 6 )2 + ( − 3)2 + ( − 6 )2 ft
=
Then
TCD
(2i − j − 2k )
3
{
}
− ( 48 lb ⋅ ft ) = k ⋅ ( 3 ft ) j × ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k 

T

+ k ⋅ ( 3 ft ) j ×  CD ( 2 i − j − 2 k )  
3



or
− 48 = −18 − 2TCD
TCD = 15.00 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 48.
Have
M y = j ⋅ ( rB ) z × TBA  × j⋅ ( rC ) z × TCD 
where
M y = 156 lb ⋅ ft
( rB ) z = ( 24 ft ) k;
( rC ) z = ( 6 ft ) k
JJJG
( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
BA
TBA = TBA
= TBA 
BA
( 4.5)2 + ( − 3)2 + ( 9 )2 ft
TBA
( 4.5i − 3j + 9k )
10.5
JJJG
( 6 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
CD
TCD = TCD
= ( 7.5 lb ) 
CD
( 6 )2 + ( − 3)2 + ( 9 )2 ft
=
= ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k
T


Then (156 lb ⋅ ft ) = j ⋅ ( 24 ft ) k × BA ( 4.5i − 3j + 9k ) 
10.5


{
}
+ j ⋅ ( 6 ft ) k × ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k 
or
156 =
108
TBA + 30
10.5
TBA = 12.25 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 49.
Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 
By
(1)
M y = − ( P cos φ )( 0.125 m )
(2)
M z = − ( P sin φ )( 0.125 m )
(3)
Equation ( 3) M z
− ( P sin φ )( 0.125 )
:
=
Equation ( 2 ) M y
− ( P cos φ )( 0.125 )
or
−4
= tan φ ∴
− 23
φ = 9.8658°
or
φ = 9.87°
From Equation (2)
− 23 N ⋅ m = − ( P cos 9.8658° )( 0.125 m )
P = 186.762 N
or P = 186.8 N
From Equation (1)
26 N ⋅ m = (186.726 N ) cos 9.8658° ( 0.225 m ) sin θ 
− (186.726 N ) sin 9.8658°  ( 0.225 m ) cosθ 
or 0.98521sin θ − 0.171341cosθ = 0.61885
Solving numerically,
θ = 48.1°
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 3, Solution 50.
Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 
(1)
M y = − ( P cos φ )( 0.125 m )
(2)
M z = − ( P sin φ )( 0.125 m )
By
Equation ( 3) M z
− ( P sin φ )( 0.125 )
:
=
Equation ( 2 ) M y
− ( P cos φ )( 0.125 )
or
− 3.5
= tan φ ; φ = 9.9262°
− 20
From Equation (3):
− 3.5 N ⋅ m = − ( P sin 9.9262° )( 0.125 m )
P = 162.432 N
From Equation (1):
M x = (162.432 N )( 0.225 m )( cos 9.9262° sin 60° − sin 9.9262° cos 60° )
= 28.027 N ⋅ m
or M x = 28.0 N ⋅ m
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 3, Solution 51.
First note:
JJJG
BA
TBA = TBA
BA
= ( 70 lb )
= ( 70 lb )
( 4 ) i + 1.5 − ( LBC + 1) j + ( − 6 ) k
2
( 4 )2 + 1.5 − ( LBC + 1) + ( − 6 )2
4 i + ( 0.5 − LBC ) j − 6 k
52 + ( 0.5 − LBC )
2
rA = ( 4 ft ) i + (1.5 ft ) j − (12 ft ) k
Have M O = rA × TBA
=
For the i components: − 763 lb ⋅ ft =
70
52 + ( 0.5 − LBC )
i
4 ft
2
52 + ( 0.5 − LBC ) 4
70 lb
1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft
2 
or
10.9 52 + ( 0.5 − LBC ) = 3 + 12 LBC
or
(10.9 )2 52 + ( 0.5 − LBC )2  = 9 + 72LBC + 144L2BC
or
25.19L2BC + 190.81LBC − 6198.8225 = 0
Then
j
k
1.5 ft
−12 ft
( 0.5 − LBC ) − 6
2
LBC =
−190.81 ±
(190.81)2 − 4 ( 25.19 )( − 6198.8225)
2 ( 25.19 )
Taking the positive root
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.
LBC = 12.35 ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 52.
First note:
JJJG
BA
TBA = TBA
BA
= ( 70 lb )
= ( 70 lb )
( 4 ) i + 1.5 − ( LBC + 1) j + ( − 6 ) k
2
( 4 )2 + 1.5 − ( LBC + 1) + ( − 6 )2
4 i + ( 0.5 − LBC ) j − 6 k
52 + ( 0.5 − LBC )
2
rA = ( 4 ft ) i + (1.5 ft ) j − (12 ft ) k
Have M O = rA × TBA
i
4 ft
2
52 + ( 0.5 − LBC ) 4
TBA
=
For the i components: − 900 lb ⋅ ft =
300 =
or
315 =
Then,
or
(1) ⇒
( 2)
52 + ( 0.5 − LBC )
TBA
52 + ( 0.5 − LBC )
For the k components: − 315 lb ⋅ ft =
or
TBA
2
(1 + 4LBC )
52 + ( 0.5 − LBC )
52 + ( 0.5 − LBC )
2
1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft
2 
TBA
4TBA
j
k
−12 ft
1.5 ft
( 0.5 − LBC ) − 6
(1)
 4 ( 0.5 − LBC ) − 1.5 ( 4 )  lb ⋅ ft
2 
(1 + LBC )
(2)
300
1 + 4LBC
=
315 4 (1 + LBC )
LBC =
59
ft
4
LBC = 14.75 ft
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 3, Solution 53.
(
Have
M AD = λ AD ⋅ rB/ A × TBH
where
λ AD =
)
( 0.8 m ) i − ( 0.6 m ) k = 0.8 i − 0.6 k
( 0.8 m )2 + ( − 0.6 m )2
rB/ A = ( 0.4 m ) i
JJJJG
( 0.3 m ) i + ( 0.6 m ) j − ( 0.6 m ) k 
BH
TBH = TBH
= (1125 N ) 
BH
( 0.3)2 + ( 0.6 )2 + ( − 0.6 )2 m
Then
0.8 0 − 0.6
M AD = 0.4 0
0 = −180 N ⋅ m
375 750 − 750
or M AD = −180.0 N ⋅ m
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 3, Solution 54.
(
)
Have
M AD = λ AD ⋅ rB/ A × TBG
where
λ AD = ( 0.8 m ) i − ( 0.6 m ) k
rB/ A = ( 0.4 m ) i
JJJG
 − ( 0.4 m ) i + ( 0.74 ) j − ( 0.32 m ) k 
BG
TBG = TBG
= (1125 N ) 
BG
( − 0.4 m )2 + ( 0.74 m )2 + ( − 0.32 m )2
= − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k
Then
0.8
0 − 0.6
M AD = 0.4
0
0
− 500 925 − 400
or M AD = − 222 N ⋅ m
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 3, Solution 55.
Have
(
M AD = λ AD ⋅ rE/ A × FEF
where
)
JJJG
AD
λ AD =
AD
λ AD =
( 7.2 m ) i + ( 0.9 m ) j
( 7.2 m )2 + ( 0.9 m )2
= 0.99228 i + 0.124035 j
rE/ A = ( 2.1 m ) i − ( 0.9 m ) j
JJJG
( 0.3 m ) i + (1.2 m ) j + ( 2.4 m ) k 
EF
FEF = FEF
= ( 24.3 kN ) 
EF
( 0.3 m )2 + (1.2 m )2 + ( 2.4 m )2
= ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k
Then
0.99228 0.124035
M AD =
2.1
2.7
− 0.9
10.8
0
0 kN ⋅ m
21.6
= −19.2899 − 5.6262
= − 24.916 kN ⋅ m
or M AD = − 24.9 kN ⋅ m
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 3, Solution 56.
(
Have
M AD = λ AD ⋅ rG/ A × E EF
Where
λ AD =
)
( 7.2 m ) i + ( 0.9 m ) j
( 7.2 m )2 + ( 0.9 m )2
= 0.99228 i + 0.124035 j
rG/ A = ( 6 m ) i − (1.8 m ) j
JJJJG
( −1.2 m ) i + ( 2.4 m ) j + ( 2.4 m ) k 
GH
FGH = FGH
= ( 21.3 kN ) 
GH
( −1.2 m )2 + ( 2.4 m )2 + ( 2.4 m )2
= − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k
Then
0.99228 0.124035 0
6
0 kN ⋅ m
M AD =
−1.8
14.2
14.2
− 7.1
= − 25.363 − 10.5678
= − 35.931 kN ⋅ m
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.
M AD = − 35.9 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 57.
(
M OA = λ OA ⋅ rC/O × P
Have
)
where
From triangle OBC
a
2
( OA) x =
( OA) z = ( OA) x tan 30° =
a 1 
a
=


2 3  2 3
Since
( OA)2 = ( OA)2x + ( OA)2y + ( OAz )2
or
 a 
2
a
a =   + ( OA ) y + 

2
2 3
2
2
∴
( OA) y =
2
a2 −
a2 a2
2
−
=a
4
12
3
Then
rA/O =
a
2
a
i +a j+
k
2
3
2 3
and
λ OA =
1
i+
2
2
1
j+
k
3
2 3
P = λ BC P
=
( a sin 30°) i − ( a cos30° ) k P
( )
=
P
i − 3k
2
a
(
)
rC/O = ai
1
2
∴ M OA =
1
1
=
=
2
1
3 2 3
P
( a )  
0
0
2
0
− 3
aP  2 
−
 (1) − 3
2  3 
(
aP
aP
M OA =
2
2
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 3, Solution 58.
(a) For edge OA to be perpendicular to edge BC,
uuur uuur
OA ⋅ BC = 0
where
From triangle OBC
( OA) x =
a
2
( OA) z = ( OA) x tan 30° =
a 1 
a
=
2  3  2 3
uuur  a 
 a 
∴ OA =   i + ( OA )y j + 
k
2
2 3
and
Then
or
so that
uuur
BC = ( a sin 30° ) i − ( a cos 30° ) k
=
a
a 3
i−
k
2
2
=
a
i − 3k
2
(
)
a
a
 a  
=0
 i + ( OA) y j + 
 k  ⋅ i − 3k
2
2
2 3 

(
)
a2
a2
+ ( OA )y ( 0 ) −
=0
4
4
uuur uuur
∴ OA ⋅ BC = 0
uuur
uuur
OA is perpendicular to BC. (b) Have M OA = Pd , with P acting along BC and d the
uuur uuur
perpendicular distance from OA to BC.
From the results of Problem 3.57,
M OA =
∴
Pa
2
Pa
= Pd
2
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.
d =
a
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 59.
(
Have
M DI = λ DI ⋅ rF /I × TEF
where
JJJG
DI
λ DI =
=
DI
)
( 4.8 ft ) i − (1.2 ft ) j
( 4.8 ft )2 + ( −1.2 ft )2
= 0.97014 i − 0.24254 j
rF /I = (16.2 ft ) k
JJJG
( 3.6 ft ) i − (10.8 ft ) j + (16.2 ft ) k 
EF
TEF = TEF
= ( 29.7 lb ) 
EF
( 3.6 ft )2 + ( −10.8 ft )2 + (16.2 ft )2
= ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k
Then
0.97014 − 0.24254 0
0
0
16.2 lb ⋅ ft
M DI =
5.4
−16.2 24.3
= − 21.217 + 254.60
= 233.39 lb ⋅ ft
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.
M DI = 233 lb ⋅ ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 60.
(
Have
M DI = λ DI ⋅ rG/I × TEG
where
JJJG
DI
λ DI =
=
DI
)
( 4.8 ft ) i − (1.2 ft ) j
( 4.8 ft )2 + ( −1.2 ft )2
= 0.97014 i − 0.24254 j
rG/I = − ( 35.1 ft ) k
JJJG
( 3.6 ft ) i − (10.8 ft ) j − ( 35.1 ft ) k 
EG
TEG = TEG
= ( 24.6 lb ) 
EG
( 3.6 ft )2 + ( −10.8 ft )2 + ( − 35.1 ft )2
= ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k
Then
0.97014 − 0.24254
0
0
0
M DI =
− 35.1 lb ⋅ ft
2.4
− 7.2
− 23.4
= 20.432 − 245.17
= − 224.74 lb ⋅ ft
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.
M DI = − 225 lb ⋅ ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 61.
F1 = F1λ1
First note that
and
F2 = F2λ 2
Let M1 = moment of F2 about the line of action of M1
and M 2 = moment of F1 about the line of action of M 2
Now, by definition
(
)
(
)
(
)
(
)
M1 = λ1 ⋅ rB/ A × F2 = λ1 ⋅ rB/ A × λ 2 F2
M 2 = λ 2 ⋅ rA/B × F1 = λ 2 ⋅ rA/B × λ1 F1
Since
F1 = F2 = F
rA/B = −rB/ A
and
(
)
M1 = λ1 ⋅ rB/ A × λ 2 F
(
)
M 2 = λ 2 ⋅ −rB/ A × λ1 F
Using Equation (3.39)
(
)
(
λ1 ⋅ rB/ A × λ 2 = λ 2 ⋅ −rB/ A × λ1
so that
(
)
)
M 2 = λ1 ⋅ rB/ A × λ 2 F
∴ M12 = M 21
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 3, Solution 62.
From the solution of Problem 3.53:
λ AD = 0.8i − 0.6k
TBH = ( 375 N ) i + ( 750 N ) j − ( 750 N ) k;
TBH = 1125 N
M AD = −180 N ⋅ m
Only the perpendicular component of TBH contributes to the moment of TBH about line AD. The
parallel component of TBH will be used to find the perpendicular component.
Have
( TBH )Parallel = λ AD ⋅ TBH
= [ 0.8i − 0.6k ] ⋅ ( 375 N ) i + ( 750 N ) j − ( 750 N ) k 
= ( 300 + 450 ) N
= 750 N
Since
TBH = ( TBH )Perpendicular + ( TBH )Parallel
Then
(TBH )Perpendicular = ( TBH )2 − ( TBH )2Parallel
=
(1125 N )2 − ( 750 N )2
= 838.53 N
and
M AD = ( TBH )Perpendicular d
180 N ⋅ m = ( 838.53 N ) d
d = 0.21466 m
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.
d = 215 mm
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 63.
From the solution of Problem 3.54:
λ AD = 0.8i − 0.6k
TBG = − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k
TBG = 1125 N
M AD = − 222 N ⋅ m
Only the perpendicular component of TBG contributes to the moment of TBG about line AD. The
parallel component of TBG will be used to find the perpendicular component.
Have
( TBG )Parallel = λ AD ⋅ TBG
= [ 0.8i − 0.6k ] ⋅  − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k 
= ( − 400 + 240 ) N
= −160 N
Since
TBG = ( TBG )Perpendicular + ( TBG )Parallel
Then
(TBG )Perpendicular = (TBG )2 − (TBG )2Parallel
=
(1125 N )2 − ( −160 N )2
= 1113.56 N
and
M AD = (TBG )Perpendicular d
222 N ⋅ m = (1113.56 N ) d
d = 0.199361 m
or d = 199.4 mm
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 3, Solution 64.
From the solution of Problem 3.59:
λ DI = 0.97014 i − 0.24254 j
TEF = ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k
TEF = 29.7 lb
M DI = 233.39 lb ⋅ ft
Only the perpendicular component of TEF contributes to the moment of TEF about line DI. The
parallel component of TEF will be used to find the perpendicular component.
Have
( TEF )Parallel = λ DI ⋅ TEF
= [ 0.97014 i − 0.24254 j] ⋅ ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k 
= ( 5.2388 + 3.9291)
= 9.1679 lb
Since
TEF = ( TEF )Perpendicular + ( TEF )Parallel
Then
(TEF )Perpendicular = ( TEF )2 − ( TEF )2Parallel
=
( 29.7 )2 − ( 9.1679 )2
= 28.250 lb
and
M DI = ( TEF )Perpendicular d
233.39 lb ⋅ ft = ( 28.250 lb ) d
d = 8.2616 ft
or d = 8.26 ft
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 3, Solution 65.
From the solution of Problem 3.60:
λ DI = 0.97014 i − 0.24254 j
TEG = ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k
TEG = 24.6 lb
M DI = − 224.74 lb ⋅ ft
Only the perpendicular component of TEG contributes to the moment of TEG about line DI. The
parallel component of TEG will be used to find the perpendicular component.
Have
( TEG )Parallel = λ DI ⋅ TEG
= [ 0.97014 i − 0.24254 j] ⋅ ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k 
= ( 2.3283 + 1.74629 )
= 4.0746 lb
Since
TEG = ( TEG )Perpendicular + (TEG )Parallel
Then
(TEG )Perpendicular = (TEG )2 − (TEG )2Parallel
=
( 24.6 )2 − ( 4.0746 )2
= 24.260 lb
and
M DI = (TEG )Perpendicular d
224.74 lb ⋅ ft = ( 24.260 lb ) d
d = 9.2638 ft
or d = 9.26 ft
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 3, Solution 66.
From the solution of Prob. 3.55:
λ AD = 0.99228 i + 0.124035 j
FEF = ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k
FEF = 24.3 kN
M AD = − 24.916 kN ⋅ m
Only the perpendicular component of FEF contributes to the moment of FEF about edge AD. The
parallel component of FEF will be used to find the perpendicular component.
Have
( FEF )Parallel = λ AD ⋅ FEF
= [ 0.99228 i + 0.124035 j] ⋅ ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k 
= 4.0187 kN
Since
FEF = ( FEF )Perpendicular + ( FEF )Parallel
Then
( FEF )Perpendicular = ( FEF )2 − ( FEF )2Parallel
=
( 24.3)2 − ( 4.0187 )2
= 23.965 kN
and
M AD = ( FEF )Perpendicular d
24.916 kN ⋅ m = ( 23.965 kN ) d
d =1.039683m
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.
or
d = 1.040 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 67.
From the solution of Prob. 3.56:
λ AD = 0.99228 i + 0.124035 j
FGH = − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k
FGH = 21.3 kN
M AD = − 35.931 kN ⋅ m
Only the perpendicular component of FGH contributes to the moment of FGH about edge AD. The
parallel component of FGH will be used to find the perpendicular component.
Have
( FGH )Parallel = λ AD ⋅FGH
= ( 0.99228 i + 0.124035 j) ⋅  − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k 
= − 5.2839 kN
Since
FGH = ( FGH )Perpendicular + ( FGH )Parallel
Then
( FGH )Perpendicular = ( FGH )2 − ( FGH )2Parallel
=
( 21.3)2 − ( 5.2839 )2
= 20.634 kN
and
M AD = ( FGH )Perpendicular d
35.931 kN ⋅ m = ( 20.634 kN ) d
d = 1.741349m
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.
or
d = 1.741 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 68.
(a)
Have
M1 = d1F1
Where
d1 = 0.6 m and F1 = 40 N
∴ M1 = ( 0.6 m )( 40 N )
or M1 = 24.0 N ⋅ m
(b)
Have
M Total = M1 + M 2
8 N ⋅ m = 24.0 N ⋅ m − ( 0.820 m )( cos α )( 24 N )
∴ cos α = 0.81301
(c)
Have
or
α = 35.6°
or
d 2 = 1.000 m
M1 + M 2 = 0
24 N ⋅ m − d 2 ( 24 N ) = 0
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 3, Solution 69.
(a)
M = Fd
12 N ⋅ m = F ( 0.45 m )
or F = 26.7 N
(b)
M = Fd
12 N ⋅ m = F ( 0.24 m )
or F = 50.0 N
(c)
M = Fd
Where d =
( 0.45 m )2 + ( 0.24 m )2
= 0.51 m
12 N ⋅ m = F ( 0.51 m )
or F = 23.5 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 3, Solution 70.
(a)
Note when a = 8 in., rC/F is perpendicular to the inclined 10 lb forces.
Have
M = ΣFd (
)
= − (10 lb )  a + 8 in. + 2 (1 in.)  − (10 lb )  2a 2 + 2 (1 in.) 


a = 8 in.,
For
M = − (10 lb )(18 in. + 24.627 in.)
= − 426.27 lb ⋅ in.
or M = 426 lb ⋅ in.
(b)
Have
M = 480 lb ⋅ in.
Also
M = Σ ( M + Fd ) (
)
= Moment of couple due to horizontal forces at A and D +
Moment of force-couple systems at C and F about C.
Then
− 480 lb ⋅ in. = −10 lb  a + 8 in. + 2 (1 in.)  +  M C + M F + FX ( a + 8 in.) + Fy ( 2a ) 
Where M C = − (10 lb )(1 in.) = −10 lb ⋅ in.
M F = M C = −10 lb ⋅ in.
Fx =
−10
lb
2
Fy =
−10
lb
2
∴ − 480 lb ⋅ in. = −10 lb ( a + 10 in.) − 10 lb ⋅ in. − 10 lb ⋅ in.
−
10 lb
10 lb
( a + 8 in.) −
( 2a )
2
2
303.43 = 31.213 a
or a = 9.72in.
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 3, Solution 71.
(a)
Have
M = ΣFd (
)
= ( 9 lb )(13.8 in.) − ( 2.5 lb )(15.2 in.)
= ( 86.2 lb ⋅ in.)
M = 86.2 lb ⋅ in.
(b)
Have
M = Td = 86.2 lb ⋅ in.
For T to be a minimum, d must be maximum.
∴ Tmin must be perpendicular to line AC.
tan θ =
15.2 in.
11.4 in.
θ = 53.130°
or θ = 53.1°
(c)
Have
M = Tmin d max Where M = 86.2 lb ⋅ in.
d max =
(15.2 in.)2 + (11.4 in.)2 + 2 (1.2 in.)
= 21.4 in.
∴ 86.2 lb ⋅ in. = Tmin ( 21.4 in.)
Tmin = 4.0280 lb
or Tmin = 4.03 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 72.
Based on M = M1 + M 2
M1 = (18 N ⋅ m ) k
M 2 = ( 7.5 N ⋅ m ) i
∴ M = ( 7.5 N ⋅ m ) i + (18 N ⋅ m ) k
and
M =
( 7.5 N ⋅ m )2 + (18 N ⋅ m )2
= 19.5 N ⋅ m
or M = 19.50 N ⋅ m
With
λ =
=
Then
( 7.5 N ⋅ m ) i + (18 N ⋅ m ) k
M
=
M
19.5 N ⋅ m
5
12
i+ k
13
13
cos θ x =
5
13
cosθ y = 0
cosθ z =
12
13
∴ θ x = 67.380°
∴ θ y = 90°
∴ θ z = 22.620°
or θ x = 67.4°,
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.
θ y = 90.0°,
θ z = 22.6°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 73.
Have
M = M1 + M 2
Where M1 = rC/B × PIC
rC/B = ( 38.4 in.) i − (16 in.) j
PIC = − ( 25 lb ) k
i
j
k
∴ M1 = 38.4 −16 0 lb ⋅ in.
0
0 −25
= ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j
and
M 2 = rD/ A × PZE
rD/ A = ( 8 in.) j − ( 22 in.) k
JJJG
 − (19.2 in.) i + ( 22 in.) k 
ED
PZ E = PZE
= ( 36.5 lb ) 
ED
( −19.2 in.)2 + ( 22 in.)2
= − ( 24 lb ) i + ( 27.5 lb ) k
i j k
∴ M 2 = 0 8 −22 lb ⋅ in.
−24 0 27.5
M 2 = ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k
and
M = M1 + M 2
= ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j + ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k 
= ( 620 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (192 lb ⋅ in.) k
continued
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
2
2
2

M =  ( 620 ) + (1488 ) + (192 )  lb ⋅ in.


= 1623.39 lb ⋅ in.
or M = 1.623 kip ⋅ in.
λ=
( 620 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (192 lb ⋅ in.) k
M
=
M
1623.39 lb ⋅ in.
= 0.38192 i + 0.91660 j + 0.118271k
cosθ x = 0.38192
or θ x = 67.5°
cosθ y = 0.91660
or θ y = 23.6°
cosθ z = 0.118271
or θ z = 83.2°
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 3, Solution 74.
Have
M = M1 + M 2
Where M1 = rE/D × FD
= − ( 0.7 m ) k × ( 80 N ) j
= ( 56.0 N ⋅ m ) i
And
M 2 = rG/F × FB
Now
d BF =
( −0.300 m )2 + ( 0.540 m )2 + ( 0.350 m )2
= 0.710 m
Then
FB = λBF FB
=
( −0.300 m ) i + ( 0.540 m ) j + ( 0.350 m ) k 71 N
(
)
0.710 m
= − ( 30 N ) i + ( 54 N ) j + ( 35 N ) k
∴
M 2 = ( 0.54 m ) j ×  − ( 30 N ) i + ( 54 N ) j + ( 35 N ) k 
= (18.90 N ⋅ m ) i + (16.20 N ⋅ m ) k
Finally M = ( 56.0 N ⋅ m ) i + (18.90 N ⋅ m ) i + (16.20 N ⋅ m ) k 
= ( 74.9 N ⋅ m ) i + (16.20 N ⋅ m ) k
and
M =
( 74.9 N ⋅ m )2 + (16.20 N ⋅ m )2
= 76.632 N ⋅ m
cosθ x =
74.9
76.632
or M = 76.6 N ⋅ m
cosθ y =
0
76.632
cosθ z =
16.20
76.632
or θ x = 12.20°
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.
θ y = 90.0°
θ z = 77.8°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 75.
M = ( M1 + M 2 ) + M P
Have
From the solution to Problem 3.74
( M1 + M 2 ) = ( 74.9 N ⋅ m ) i + (16.20 N ⋅ m ) k
Now
M P = rD / E × PE
= ( 0.54 m ) j + ( 0.70 m ) k  × ( 90 N ) i
= ( 63.0 N ⋅ m ) j − ( 48.6 N ⋅ m ) k
∴
M = ( 74.9 i + 16.20 k ) + ( 63.0 j − 48.6 k )
= ( 74.9 N ⋅ m ) i + ( 63.0 N ⋅ m ) j − ( 32.4 N ⋅ m ) k
and
M =
( 74.9 N ⋅ m )2 + ( 63.0 N ⋅ m )2 + ( − 32.4 N ⋅ m )2
= 103.096 N ⋅ m
or M = 103.1 N ⋅ m
and
cosθ x =
74.9
103.096
cosθ y =
63.0
103.096
cosθ z =
− 32.4
103.096
or θ x = 43.4°
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.
θ y = 52.3°
θ z = 108.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 76.
Have
M = M1 + M 2 + M P
From Problem 3.73 solution:
M1 = ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j
M 2 = ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k
Now
M P = rE/ A × PE
rE/ A = (19.2 in.) i + ( 8 in.) j − ( 44 in.) k
PE = ( 52.5 lb ) j
Therefore
i
j
k
M P = 19.2 8 − 44
0 52.5 0
= ( 2310 lb. in.) i + (1008 lb. in.) k
and
M = M1 + M 2 + M P
= [(400 + 220 + 2310)i + (960 + 528)j + (192 + 1008)k ] lb ⋅ in.
= ( 2930 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (1200 lb ⋅ in.) k
M =
( 2930 )2 + (1488)2 + (1200 )2
= 3498.4 lb ⋅ in.
or M = 3.50 kip ⋅ in.
continued
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
λ =
M
2930i + 1488j + 1200k
=
M
3498.4
= 0.83753i + 0.42534 j + 0.34301k
cosθ x = 0.83753
or
cosθ y = 0.42534
or θ y = 64.8°
cosθ z = 0.34301
or θ z = 69.9°
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.
θ x = 33.1°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 77.
Have
M = M1 + M 2 + M 3
Where
M1 = − (1.2 lb ⋅ ft ) cos 25° j − (1.2 lb ⋅ ft ) sin 25°k
M 2 = − (1.3 lb ⋅ ft ) j
M 3 = − (1.4 lb ⋅ ft ) cos 20° j + (1.4 lb ⋅ ft ) sin 20°k
∴
M = ( −1.08757 − 1.3 − 1.31557 ) j + ( − 0.507142 + 0.478828 ) k
= − ( 3.7031 lb ⋅ ft ) j − ( 0.028314 lb ⋅ ft ) k
and
M =
( −3.7031)2 + ( − 0.028314 )2 = 3.7032 lb ⋅ ft
or M = 3.70 lb ⋅ ft
λ =
M
−3.7031j − 0.028314k
=
M
3.7032
= − 0.99997 j − 0.0076458k
cosθ x = 0
or
θ x = 90°
or
θ y = 179.6°
or
θ z = 90.4°
cosθ y = − 0.99997
cosθ z = −0.0076458
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 3, Solution 78.
(a)
FB = P :
∴ FB = 160.0 N
50.0°
M B = −rBA P cos10°
= − ( 0.355 m )(160 N ) cos10°
= −55.937 N ⋅ m
or M B = 55.9 N ⋅ m
(b)
FC = P :
∴ FC = 160.0 N
50.0°
M C = M B − rCB ( FB )⊥
= M B − rCB FB sin 55°
= −55.937 N ⋅ m − ( 0.305 m )(160 N ) sin 55°
or M C = 95.9 N ⋅ m
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 3, Solution 79.
(a)
ΣF :
FB = 135 N
or FB = 135 N
ΣM :
M B = P dB
= (135 N )( 0.125 m )
= 16.875 N ⋅ m
or M B = 16.88 N ⋅ m
(b)
ΣM B :
M B = FC d
16.875 N ⋅ m = FC ( 0.075 m )
FC = 225 N
or FC = 225 N
ΣF :
0 = − FB + FC
FB = FC = 225 N
or FB = 225 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 3, Solution 80.
ΣF : PC = P = 700 N
(a) Based on
or PC = 700 N
60°
ΣM C : M C = − Px dCy + Py dCx
Px = ( 700 N ) cos60° = 350 N
where
Py = ( 700 N ) sin 60° = 606.22 N
dCx = 1.6 m
dCy = 1.1 m
∴ M C = − ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m )
= −385 N ⋅ m + 969.95 N ⋅ m
= 584.95 N ⋅ m
or M C = 585 N ⋅ m
ΣFx : PDx = P cos60°
(b) Based on
= ( 700 N ) cos 60°
= 350 N
ΣM D :
( P cos 60°)( d DA ) = PB ( d DB )
( 700 N ) cos 60° ( 0.6 m ) = PB ( 2.4 m )
PB = 87.5 N
or PB = 87.5 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
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COSMOS: Complete Online Solutions Manual Organization System
ΣFy : P sin 60° = PB + PDy
( 700 N ) sin 60° = 87.5 N + PDy
PDy = 518.72 N
PD =
=
( PDx )2 + ( PDy )
2
( 350 )2 + ( 518.72 )2 = 625.76 N
P 
 PDx 
 518.72 
 = 55.991°
 350 
θ = tan −1  Dy  = tan −1 
or PD = 626 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.
56.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 81.
ΣFx :
2.8cos 65° = FA cos θ + FC cos θ
= ( FA + FC ) cosθ
ΣFy :
(1)
2.8sin 65° = FA sin θ + FC sin θ
= ( FA + FC ) sin θ
Then
(2)
(2)
⇒ tan 65° = tan θ
(1)
or θ = 65.0°
ΣM A :
( 27 m )( 2.8 kN ) sin 65° = ( 72 m ) ( FC ) sin 65°
or FC = 1.050 kN
From Equation (1): 2.8 kN = FA + 1.050 kN
or FA = 1.750 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.
∴ FA = 1.750 kN
65.0°
FC = 1.050 kN
65.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 82.
Based on
ΣFx : − ( 54 lb ) cos 30° = − FB cos α − FC cos α
( FB + FC ) cosα = ( 54 lb ) cos 30°
(1)
( 54 lb ) sin 30° = FB sin α + FC sin α
ΣFy :
or ( FB + FC ) sin α = ( 54 lb ) sin 30°
From
(2)
Eq ( 2 )
: tan α = tan 30°
Eq (1)
∴ α = 30°
Based on
ΣM C :
( 54 lb ) cos ( 30° − 20° )  (10 in.) = ( FB cos10° )( 24 in.)
∴ FB = 22.5 lb
From Eq. (1),
or FB = 22.5 lb
30°
or FC = 31.5 lb
30°
( 22.5 + FC ) cos 30° = ( 54 ) cos 30°
FC = 31.5 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 83.
(a) Based on
− ( 54 lb ) cos 30° = − FC cos 30°
ΣFx :
∴
FC = 54 lb
or FC = 54.0 lb
30°
( 54 lb ) cos10°  (10 in.) = M C
ΣM C :
∴
M C = 531.80 lb ⋅ in.
or
M C = 532 lb ⋅ in.
(b) Based on
ΣFy :
( 54 lb ) sin 30° = FB sin α
FB sin α = 27
or
ΣM B :
(1)
531.80 lb ⋅ in. − ( 54 lb ) cos10° ( 24 in.)
= − FC ( 24 in.) cos 20° 
FC = 33.012 lb
or
And
ΣFx :
FC = 33.0 lb
− ( 54 lb ) cos 30° = − 33.012 lb − FB cos α
FB cos α = 13.7534
From
Eq (1)
:
Eq ( 2 )
From Eq. (1),
FB =
tan α =
27
13.7534
(2)
∴ α = 63.006°
27
= 30.301 lb
sin ( 63.006° )
or FB = 30.3 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.
63.0°
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Chapter 3, Solution 84.
(a) Have ΣFy :
FC + FD + FE = F
F = −200 lb + 150 lb − 150 lb
F = − 200 lb
or F = 200 lb ΣM G :
Have
FC ( d − 4.5 ft ) − FD ( 6 ft ) = 0
( 200 lb )( d − 4.5 ft ) − (150 lb )( 6 ft ) = 0
d = 9 ft
or d = 9.00 ft (b) Changing directions of the two 150-lb forces only changes the sign of
the couple.
∴ F = − 200 lb
or F = 200 lb And
ΣM G :
FC ( d − 4.5 ft ) + FD ( 6 ft ) = 0
( 200 lb )( d − 4.5 ft ) + (150 lb )( 6 ft ) = 0
d =0
or d = 0 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
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Chapter 3, Solution 85.
(a)
Based on ΣFz :
− 200 N + 200 N + 240 N = FA
FA = 240 N
or FA = ( 240 N ) k
Based on ΣM A :
(b)
( 200 N )( 0.7 m ) − ( 200 N )( 0.2 m ) = M A
M A = 100 N ⋅ m
or M A = (100.0 N ⋅ m ) j
Based on ΣFz :
− 200 N + 200 N + 240 N = F
F = 240 N
or F = ( 240 N ) k
Based on ΣM A :
(c)
100 N ⋅ m = ( 240 N )( x )
x = 0.41667 m
or x = 0.417 m From A along AB
Based on ΣM B :
− ( 200 N )( 0.3 m ) + ( 200 N )( 0.8 m ) − P (1 m ) = R ( 0 )
P = 100 N
or P = 100.0 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 3, Solution 86.
Let R be the single equivalent force...
ΣF :
R = FA + FC
= ( 260 N )( cos10° i − sin10° k ) + ( 320 N )( − cos8° i − sin 8° k )
= − ( 60.836 N ) i − ( 89.684 N ) k
or R = − ( 60.8 N ) i − ( 89.7 N ) k
ΣM A :
rAD Rx = rAC FC cos8°
rAD ( 60.836 N ) = ( 0.690 m )( 320 N ) cos8°
rAD = 3.5941 m
∴ R Would have to be applied 3.59 m to the right of A
on an extension of handle ABC.
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
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Chapter 3, Solution 87.
(a)
Have
ΣF :
FB + FC + FD = FA
Since
FB = − FD
∴
FA = FC = 22 lb
20°
or FA = 22.0 lb
Have
ΣM A :
20°
− FBT ( r ) − FCT ( r ) + FDT ( r ) = M A
− ( 28 lb ) sin15°  ( 8 in.) − ( 22 lb ) sin 25°  ( 8 in.)
+ ( 28 lb ) sin 45°  ( 8 in.) = M A
M A = 26.036 lb ⋅ in.
or M A = 26.0 lb ⋅ in.
(b)
Have
ΣF :
FA = FE
or FE = 22.0 lb
ΣM :
M A = [ FE cos 20°] ( a )
∴
26.036 lb ⋅ in. = ( 22 lb ) cos 20°  ( a )
20°
a = 1.25941 in.
or a = 1.259 in. Below A
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
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Chapter 3, Solution 88.
(a)
Let R be the single equivalent force. Then
R = (120 N ) k
ΣM B :
R = 120 N
− a (120 N ) = − ( 0.165 m )( 90 N ) cos15° + ( 0.201 m )( 90 N ) sin15°
a = 0.080516 m
∴ The line of action is y =
201
mm − 80.516 mm = 19.984 mm
2
or y = 19.98 mm
(b)
ΣM B :
Then
− ( 0.201 − 0.040 ) m  (120 N ) = − ( 0.165 m )( 90 N ) cosθ + ( 0.201 m )( 90 N ) sin θ
or
cosθ − 1.21818sin θ = 1.30101
or
cos 2 θ = (1.30101 + 1.21818sin θ )
or
1 − sin 2 θ = 1.69263 + 3.1697sin θ + 1.48396sin 2 θ
or
2.48396sin 2 θ + 3.1697 sin θ + 0.69263 = 0
sin θ =
−3.1697 ±
2
( 3.1697 )2 − 4 ( 2.48396 )( 0.69263)
2 ( 2.48396 )
or θ = −16.26°
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.
and
θ = −85.0°
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Chapter 3, Solution 89.
(a)
First note that F = P and that F must be equivalent to (P, MD) at point D,
Where
M D = 57.6 N ⋅ m
For F = ( F )min
F must act as far from D as possible
∴ Point of application is at point B
(b) For ( F )min
Now
F must be perpendicular to BD
d DB =
( 630 mm )2 + ( −160 mm )2
= 650 mm
tan α =
63
16
α = 75.7°
Then
M D = d DB F
57.6 N ⋅ m = ( 0.650 m ) F
F = 88.6 N
or F = 88.6 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
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75.7°
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Chapter 3, Solution 90.
Have
ΣF :
− ( 250 kN ) j = F
or F = − ( 250 kN ) j
Also have
ΣM G :
rP × P = M
i
j
k
− 0.030
0
0.060 kN ⋅ m = M
0
− 250
0
∴ M = (15 kN ⋅ m ) i + ( 7.5 kN ⋅ m ) k
or M = (15.00 kN ⋅ m ) i + ( 7.50 kN ⋅ m ) k
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
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Chapter 3, Solution 91.
Have ΣF :
TAB = F
JJJG
AB
where TAB = TAB
AB
= ( 54 lb )
2.25i − 18 j + 9k
( 2.25)2 + ( −18)2 + ( 9 )2
= ( 6 lb ) i − ( 48 lb ) j + ( 24 lb ) k
F = ( 6.00 lb ) i − ( 48.0 lb ) j + ( 24.0 lb ) k
So that
Have ΣM E :
rA/E × TAB = M
i
j
k
0 22.5 0 lb ⋅ ft = M
6 − 48 24
∴ M = ( 540 lb ⋅ ft ) i − (135 lb ⋅ ft ) k
or
M = ( 540 lb ⋅ ft ) i − (135.0 lb ⋅ ft ) k
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
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Chapter 3, Solution 92.
Have ΣF :
TCD = F
JJJG
CD
where TCD = TCD
CD
= ( 61 lb )
− 0.9i − 16.8j + 7.2k
( − 0.9 )2 + ( −16.8)2 + ( 7.2 )2
= − ( 3 lb ) i − ( 56 lb ) j + ( 24 lb ) k
So that
F = − ( 3.00 lb ) i − ( 56.0 lb ) j + ( 24.0 lb ) k
Have ΣM O = rC/D × TCD = M
i
j
k
0 22.5 0 lb ⋅ ft = M
− 3 − 56 24
∴ M = ( 540 lb ⋅ ft ) i + ( 67.5 lb ⋅ ft ) k
M = ( 540 lb ⋅ ft ) i + ( 67.5 lb ⋅ ft ) k
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
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Chapter 3, Solution 93.
Have ΣF :
TAB = F
JJJG
AB
where TAB = TAB
AB
= (10.5 kN )
− i − 4.75j + 2k
( −1)2 + ( − 4.75)2 + ( 2 )2
= − ( 2 kN ) i − ( 9.5 kN ) j + ( 4 kN ) k
F = − ( 2.00 kN ) i − ( 9.50 kN ) j + ( 4.00 kN ) k
So that
Have ΣM O :
rA × TAB = M
i
j k
3 4.75 0 kN ⋅ m = M
− 2 − 9.5 4
∴ M = (19 kN ⋅ m ) i − (12 kN ⋅ m ) j − (19 kN ⋅ m ) k
M = (19.00 kN ⋅ m ) i − (12.00 kN ⋅ m ) j − (19.00 kN ⋅ m ) k
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
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Chapter 3, Solution 94.
Let ( R, M O ) be the equivalent force-couple system
Then R = ( 220 N )( − sin 60° j − cos 60° k )
(
= (110 N ) − 3 j − k
)
or R = − (190.5 N ) j − (110 N ) k
Now
ΣM O :
M O = rOC × R
Where
rOC = ( 0.2 m ) i + ( 0.1 − 0.4sin 20° ) m  j + ( 0.4 cos 20° m ) k
Then
i
j
k
M O = − ( 0.1)(110 N ) 2 (1 − 4sin 20° ) 4 cos 20° ( m )
0
3
1
{
= − (11 N ⋅ m ) (1 − 4sin 20° )(1) − ( 4cos 20° )

or
( 3 ) i − 2 j + 2 3 k}
M O = ( 75.7 N ⋅ m ) i + ( 22.0 N ⋅ m ) j − ( 38.1 N ⋅ m ) k
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Chapter 3, Solution 95.
Have ΣF :
F = FD
JJG
AI
where F = F
AI
= ( 63 lb )
14.4i − 4.8j + 7.2k
(14.4 )2 + ( − 4.8)2 + ( 7.2 )2
F = ( 54.0 lb ) i − (18.00 lb ) j + ( 27.0 lb ) k
So that
Have ΣM D :
M + rI /O × F = M D
JJJG
AC
where M = M
AC
= ( 560 lb ⋅ in.)
9.6 i − 7.2 k
( 9.6 )2 + ( − 7.2 )2
= ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k
i
j
k
Then M D = ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k + 0 0 14.4 lb ⋅ in.
54 −18 27
= ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k + ( 259.2 lb ⋅ in.) i + ( 777.6 lb ⋅ in.) j
or
M D = ( 707 lb ⋅ in.) i + ( 778 lb ⋅ in.) j − ( 336 lb ⋅ in.) k
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
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Chapter 3, Solution 96.
First assume that the given force W and couples M1 and M 2 act at the
origin.
Now
W = − Wj
and
M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k
Note that since W and M are perpendicular, it follows that they can be
replaced with a single equivalent force.
F =W
(a) Have
or
F = − Wj = − ( 2.4 N ) j
or F = − ( 2.40 N ) j W
(b) Assume that the line of action of F passes through point P (x, 0, z).
Then for equivalence
M = rP/O × F
rP/O = xi + zk
where
∴
− ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k
i
j
k
= x 0 z = (Wz ) i − (Wx ) k
0 −W 0
Equating the i and k coefficients,
z =
(b) For
−M z cos 25°
W
and
 M − M 2 sin 25° 
x = − 1

W


W = 2.4 N, M1 = 0.068 N ⋅ m, M 2 = 0.065 N ⋅ m
x=
0.068 − 0.065sin 25°
= − 0.0168874 m
− 2.4
or x = −16.89 mm W
z =
− 0.065cos 25°
= − 0.024546 m
2.4
or z = − 24.5 mm W
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
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Chapter 3, Solution 97.
ΣM Bz : M 2 z = 0
(a) Have
(
)
k ⋅ rH /B × F1 + M1z = 0
(1)
rH /B = ( 31 in.) i − ( 2 in.) j
where
F1 = λ EH F1
=
( 6 in.) i + ( 6 in.) j − ( 7 in.) k 20 lb
(
)
=
20 lb
( 6i + 6 j − 7k )
11.0
11.0 in.
M1z = k ⋅ M1
M1 = λ EJ M1
=
−di + ( 3 in.) j − ( 7 in.) k
d 2 + 58 in.
( 480 lb ⋅in.)
Then from Equation (1),
0
0
1
31 −2 0
6 6 −7
20 lb ⋅ in. ( −7 )( 480 lb ⋅ in.)
+
=0
11.0
d 2 + 58
continued
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
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Solving for d, Equation (1) reduces to
20 lb ⋅ in.
3360 lb ⋅ in.
=0
(186 + 12 ) − 2
11.0
d + 58
d = 5.3955 in.
From which
or d = 5.40 in.
(b)
F2 = F1 =
20 lb
( 6i + 6 j − 7k )
11.0
= (10.9091i + 10.9091j − 12.7273k ) lb
or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k
M 2 = rH /B × F1 + M1
i j k
20 lb ⋅ in. ( − 5.3955 ) i + 3j − 7k
= 31 − 2 0
+
( 480 lb ⋅ in.)
11.0
9.3333
6 6 −7
= ( 25.455i + 394.55j + 360k ) lb ⋅ in.
+ ( − 277.48i + 154.285j − 360k ) lb ⋅ in.
M 2 = − ( 252.03 lb ⋅ in.) i + ( 548.84 lb ⋅ in.) j
or M 2 = − ( 21.0 lb ⋅ ft ) i + ( 45.7 lb ⋅ ft ) j
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
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Chapter 3, Solution 98.
(a) a : ΣFy :
Ra = − 400 N − 600 N
or R a = 1000 N
ΣM B :
M a = ( 2 kN ⋅ m ) + ( 2 kN ⋅ m ) + ( 5 m )( 400 N )
or M a = 6.00 kN ⋅ m
b : ΣFy :
Rb = −1200 N + 200 N
or R b = 1000 N
ΣM B :
M b = ( 0.6 kN ⋅ m ) + ( 5 m )(1200 N )
or M b = 6.60 kN ⋅ m
c : ΣFy :
Rc = 200 N − 1200 N
or R c = 1000 N
ΣM B :
M c = − ( 4 kN ⋅ m ) − (1.6 kN ⋅ m ) − ( 5 m )( 200 N )
or M c = 6.60 kN ⋅ m
d : ΣFy :
Rd = − 800 N − 200 N
or R d = 1000 N
ΣM B :
M d = − (1.6 kN ⋅ m ) + ( 4.2 kN ⋅ m ) + ( 5 m )( 800 N )
or M d = 6.60 kN ⋅ m
continued
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
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e : ΣFy :
Re = − 500 N − 400 N
or R e = 900 N
ΣM B :
M e = ( 3.8 kN ⋅ m ) + ( 0.3 kN ⋅ m ) + ( 5 m )( 500 N )
or M e = 6.60 kN ⋅ m
f : ΣFy :
R f = 400 N − 1400 N
or R f = 1000 N
ΣM B :
M f = ( 8.6 kN ⋅ m ) − ( 0.8 kN ⋅ m ) − ( 5 m )( 400 N )
or M f = 5.80 kN ⋅ m
g : ΣFy :
Rg = −1200 N + 300 N
or R g = 900 N
ΣM B :
M g = ( 0.3 kN ⋅ m ) + ( 0.3 kN ⋅ m ) + ( 5 m )(1200 N )
or M g = 6.60 kN ⋅ m
h : ΣFy :
Rh = − 250 N − 750 N
or R h = 1000 N
ΣM B :
M h = − ( 0.65 kN ⋅ m ) + ( 6 kN ⋅ m ) + ( 5 m )( 250 N )
or M h = 6.60 kN ⋅ m
(b) The equivalent loadings are (b), (d), (h)
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
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Chapter 3, Solution 99.
The equivalent force-couple system at B is...
ΣFy :
R = − 650 N − 350 N
or R = 1000 N
ΣM B :
M = (1.6 m )( 800 N ) + (1.27 kN ⋅ m ) + ( 5 m )( 650 N )
or M = 5.80 kN ⋅ m
∴ The equivalent loading of Problem 3.98 is (f)
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
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Chapter 3, Solution 100.
Equivalent force system...
(a) ΣFy :
R = − 400 N − 200 N
or R = 600 N
ΣM A :
− d ( 600 N ) = − ( 200 N ⋅ m ) + (100 N ⋅ m ) − ( 4 m )( 200 N )
or d = 1.500 m
(b) ΣFy :
R = − 400 N + 100 N
or R = 300 N
ΣM A :
− d ( 300 N ) = − ( 200 N ⋅ m ) − ( 600 N ⋅ m ) + ( 4 m )(100 N )
or d = 1.333 m
(c) ΣFy :
R = − 400 N − 100 N
or R = 500 N
ΣM A :
− d ( 500 N ) = − ( 200 N ⋅ m ) − ( 200 N ⋅ m ) − ( 4 m )(100 N )
or d = 1.600 m
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
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Chapter 3, Solution 101.
The equivalent force-couple system at A for each of the five force-couple systems will be determined and
compared to
F = ( 2 lb ) j
M = ( 48 lb ⋅ in.) i + ( 32 lb ⋅ in.) k
To determine if they are equivalent
Force-couple system at B:
Have ΣF :
F = ( 2 lb ) j
and ΣM A :
M = ΣM B + rB/ A × FB
(
)
M = ( 32 lb ⋅ in.) i + (16 lb ⋅ in.) k + ( 8 in.) i × ( 2 lb ) j
= ( 32 lb ⋅ in.) i + ( 32 lb ⋅ in.) k
∴ is not equivalent
Force-couple system at C:
Have ΣF :
F = ( 2 lb ) j
And ΣM A :
M = M C + rC/ A × FC
(
)
M = ( 68 lb ⋅ in.) i + ( 8 in.) i + (10 in.) k  × ( 2 lb ) j
= ( 48 lb ⋅ in.) i + (16 lb ⋅ in.) k
∴ is not equivalent
continued
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
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Force-couple system at E:
Have ΣF :
F = ( 2 lb ) j
and ΣM A :
M = M E + rE/ A × FE
(
)
M = ( 48 lb ⋅ in.) i + (16 in.) i − ( 3.2 in.) j × ( 2 lb ) j
= ( 48 lb ⋅ in.) i + ( 32 lb ⋅ in.) k
∴ is equivalent
Force-couple system at G:
Have ΣF :
F = ( 2 lb ) i + ( 2 lb ) j
F has two force components
∴ is not equivalent
Force-couple system at I:
Have ΣF :
F = ( 2 lb ) j
and ΣM A :
ΣM I + rI / A × FI
(
)
M = ( 80 lb ⋅ in.) i − (16 in.) k
+ (16 in.) i − ( 8 in.) j + (16 in.) k  × ( 2 lb ) j
M = ( 48 lb ⋅ in.) i + (16 lb ⋅ in.) k
∴ is not equivalent
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
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Chapter 3, Solution 102.
WA = mA g = ( 38 kg ) g
First
WB = mB g = ( 29 kg ) g
WC = mC g = ( 27 kg ) g
(a)
For resultant weight to act at C,
Then
ΣM C = 0
( 38 kg ) g  ( 2 m ) − ( 27 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0
∴ d =
76 − 58
= 0.66667 m
27
or d = 0.667 m
WC = mC g = ( 24 kg ) g
(b)
For resultant weight to act at C,
Then
ΣM C = 0
( 38 kg ) g  ( 2 m ) − ( 24 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0
∴ d =
76 − 58
= 0.75 m
24
or d = 0.750 m
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
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Chapter 3, Solution 103.
(a) Have ΣF :
− WC − WD − WE = R
∴ R = − 200 lb − 175 lb − 135 lb
= − 510 lb
or R = 510 lb
Have ΣM A :
− ( 200 lb )( 4.5 ft ) − (175 lb )( 7.8 ft ) − (135 lb )(12.75 ft ) = − R ( d )
∴ − 3986.3 lb ⋅ ft = ( − 510 lb ) d
or d = 7.82 ft (b) For equal reactions at A and B,
The resultant R must act at midspan.
From
L
ΣM A = − R  
2
∴ − ( 200 lb )( 4.5 ft ) − (175 lb )( 4.5 ft + a ) − (135 lb )( 4.5 ft + 2.5 a )
= − ( 510 lb )( 9 ft )
or 2295 + 512.5 a = 4590
and a = 4.48 ft Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Chapter 3, Solution 104.
Have ΣF :
−12 kN − WL − 18 kN = − 40 kN − 40 kN
WL = 50 kN
or WL = 50.0 kN
ΣM B :
(12 kN )( 5 m ) + ( 50 kN ) d = ( 40 kN )( 5 m )
d = 2.8 m
or heaviest load ( 50 kN ) is located
2.80 m from front axle
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
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Chapter 3, Solution 105.
(a) ΣF :
R = ( 80 N ) i − ( 40 N ) j − ( 60 N ) j + ( 90 N )( − sin 50°i − cos 50° j)
= (11.0560 N ) i − (157.851 N ) j
R=
(11.0560 N )2 + ( −157.851 N )2
= 158.2 N
tan θ =
−157.851
11.0560
θ = 86.0°
or R = 158.2 N
86.0°
(b)
ΣM F : d − (157.851 N ) = ( 0.32 m )( 80 N ) − ( 0.15 m )( 40 N ) − ( 0.35 m )( 60 N )
− ( 0.61 m )( 90 N ) cos 50° − ( 0.16 m )( 90 N ) sin 50°
or d = 302 mm to the right of F
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
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Chapter 3, Solution 106.
(a) ΣM I :
0 = ( 0.32 m )( 80 N ) + ( 0.1 m )( 40 N ) − ( 0.1 m )( 60 N ) − ( 0.36 m )( 90 N ) cos α
− ( 0.16 m )( 90 N ) sin α
or 4sin α + 9cos α = 6.5556
( 9cosα )2 = ( 6.5556 − 4sin α )2
(
)
81 1 − sin 2 α = 42.976 − 52.445sin α + 16sin 2 α
97sin 2 α − 52.445sin α − 38.024 = 0
Solving by the quadratic formula gives for the positive root
sin α = 0.95230
α = 72.233°
or α = 72.2° Note: The second root (α = − 24.3° ) is rejected since 0 < α < 90°.
(b) ΣF :
R = ( 80 N ) i − ( 40 N ) j − ( 60 N ) j
+ ( 90 N )( − sin 72.233°i − cos 72.233° j)
= − ( 5.7075 N ) i − (127.463 N ) j
R=
( − 5.7075 N )2 + ( −127.463 N )2
= 127.6 N
tan θ =
−127.463
− 5.7075
θ = 87.4°
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.
R = 127.6 N
87.4° COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 107.
(a) Have
ΣM D :
0 = M − ( 0.8 in.)( 40 lb ) − ( 2.9 in.)( 20 lb ) cos 30° − ( 3.3 in.)( 20 lb ) sin 30°
or M = 115.229 lb ⋅ in.
or
M = 115.2 lb ⋅ in.
Now, R is oriented at 45° as shown (since its line of action
passes through B and D).
0 = ( 40 lb ) cos 45° − ( 20 lb ) cos15°
Have ΣFx′ :
− ( 90 lb ) cos (α + 45° )
or α = 39.283°
or
(b) ΣFx :
α = 39.3°
Rx = 40 − 20sin 30° − 90cos 39.283°
= − 39.663 lb
Now
R=
2 Rx
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.
R = 56.1 lb
45.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 108.
(a) Reduce system to a force and couple at B:
Have R = ΣF = − (10 lb ) j + ( 25 lb ) cos 60°i + ( 25 lb ) sin 60° j − ( 40 lb ) i
= − ( 27.5 lb ) i + (11.6506 lb ) j
or R =
( − 27.5 lb )2 + (11.6506 lb )2 = 29.866 lb
 11.6506 
 = 22.960°
 27.5 
θ = tan −1 
or
R = 29.9 lb
Also M B = ΣM B = ( 80 lb ⋅ in.) k − (12 in.) i × ( −10 lb ) j − ( 8 in.) j × ( − 40 lb ) i
= − (120 lb ⋅ in.) k
(b)
Have M B = − (120 lb ⋅ in.) k = − ( u ) i × (11.6506 lb ) j
− (120 lb ⋅ in.) k = − (11.6506 lb )( u ) k
u = 10.2999 in. and x = 12 in. − 10.2999 in.
= 1.7001 in.
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.
23.0°
COSMOS: Complete Online Solutions Manual Organization System
Have M B = − (120 lb ⋅ in.) k = − ( v ) j × ( − 27.5 lb ) i
− (120 lb ⋅ in.) k = − ( 27.5 lb )( v ) k
v = 4.3636 in.
and y = 8 in. − 4.3636 in. = 3.6364 in.
or
1.700 in. to the right of A and 3.64 in. above C
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 3, Solution 109.
(a)
Position origin along centerline of sheet metal at the intersection with
line EF.
(a)
ΣF = R
Have
R =  − 0.52 j − 1.05 j − 2.1( sin 45° i + cos 45° j) − 0.64 i  kips
R = − ( 2.1249 kips ) i − ( 3.0549 kips ) j
( − 2.1249 )2 + ( − 3.0549 )2
R=
= 3.7212 kips
 − 3.0549 
 = 55.179°
 − 2.1249 
θ = tan −1 
or
Have
R = 3.72 kips
55.2° M EF = ΣM EF
Where M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.)
− ( 2.1 kips )( 0.8 in.) − ( 0.64 kip ) (1.6 in.) sin 45° + 1.6 in.
= 0.123923 kip ⋅ in.
To obtain distance d left of EF,
Have
M EF = dRy = d ( − 3.0549 kips )
d =
0.123923 kip ⋅ in.
= −0.040565 in.
− 3.0549 kips
or
d = 0.0406 in. left of EF (b)
Have
M EF = ΣM EF = 0
M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.)
− ( 2.1 kips )( 0.8 in.)
− ( 0.64 kip ) (1.6 in.) sin α + 1.6 in.
∴
(1.024 kip ⋅ in.) sin α = 0.848 kip ⋅ in.
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.
α = 55.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 110.
(a)
Have
ΣF = R
R =  − 0.52 j − 1.05j − 2.1( sin α i + cos α j) − 0.64i  kips
= − 0.64 kip + ( 2.1 kips )( sin α ) i  − 1.57 kips + ( 2.1 kips ) cos α  j
Then
tan α =
Rx
0.64 + 2.1sin α
=
Ry 1.57 + 2.1cos α
1.57 tan α + 2.1sin α = 0.64 + 2.1sin α
tan α =
0.64
1.57
α = 22.178°
or
(b)
From α = 22.178°
Rx = − 0.64 kip − ( 2.1 kips ) sin 22.178°
= −1.43272 kips
Ry = −1.57 kips − ( 2.1 kips ) cos 22.178°
= − 3.5146 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.
α = 22.2°
COSMOS: Complete Online Solutions Manual Organization System
R=
( −1.43272 )2 + ( − 3.5146 )2
= 3.7954 kips
or
Then
R = 3.80 kips
67.8°
M EF = ΣM EF
Where M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.) − ( 2.1 kips )( 0.8 in.)
− ( 0.64 kip ) (1.6 in.) sin 22.178° + 1.6 in.
= 0.46146 kip ⋅ in.
To obtain distance d left of EF,
Have
M EF = dRy
= d ( − 3.5146 kips )
d =
0.46146 kip ⋅ in.
− 3.5146 kips
= − 0.131298 in.
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.
d = 0.1313 in. left of EF
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 111.
Equivalent force-couple at A due to belts on pulley A
ΣF : −120 N − 160 N = RA
Have
∴ R A = 280 N
ΣM A : − 40 N ( 0.02 m ) = M A
Have
∴ M A = 0.8 N ⋅ m
Equivalent force-couple at B due to belts on pulley B
ΣF :
Have
( 210 N + 150 N )
∴ R B = 360 N
25° = R B
25°
ΣM B : − 60 N ( 0.015 m ) = M B
Have
∴ M B = 0.9 N ⋅ m
Equivalent force-couple at F
Have
ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j)
= ( 326.27 N ) i − (127.857 N ) j
R = RF =
2
2
RFx
+ RFy
=
( 326.27 )2 + (127.857 )2 = 350.43 N
 RFy 
−1  −127.857 
 = tan 
 = −21.399°
 326.27 
 RFx 
θ = tan −1 
or R F = R = 350 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.
21.4°
COSMOS: Complete Online Solutions Manual Organization System
Have
ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m
− ( 360 N ) cos 25° ( 0.010 m )
+ ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m
M F = − ( 3.5056 N ⋅ m ) k
To determine where a single resultant force will intersect line FE,
M F = dR y
∴ d =
MF
−3.5056 N ⋅ m
=
= 0.027418 m = 27.418 mm
Ry
−127.857 N
or d = 27.4 mm
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 3, Solution 112.
(a)
Have
R = ΣF
R = ( 25 N )( cos 40°i + sin 40° j) − (15 N ) i − (10 N ) j
= ( 4.1511 N ) i + ( 6.0696 N ) j
R =
( 4.1511)2 + ( 6.0696 )2
= 7.3533 N
 6.0696 

 4.1511 
θ = tan −1 
= 55.631°
R = 7.35 N
or
55.6° (a) and (b)
(b)
From
M B = ΣM B = dRy
where
M B = − ( 25 N ) cos 40° ( 0.375 m ) sin 50°
− ( 25 N ) sin 40° ( 0.375 m ) cos 50°
+ (15 N ) ( 0.150 m ) sin 50° − (10 N )( 0.150 m )
+ 6.25 N ⋅ m
∴ M B = −2.9014 N ⋅ m
and
d =
=
MB
Ry
−2.9014 N ⋅ m
6.0696 N
= 0.47802 m
or
(c)
d = 478 mm to the left of B From M B = rD/B × R
− ( 2.9014 N ⋅ m ) k = ( −d1 cos 50°i + d1 sin 50° j)
× ( 4.1511 N ) i + ( 6.096 N ) j
− ( 2.9014 N ⋅ m ) k = − ( 7.0814 d1 ) k
∴ d1 = 0.40972 m
or d1 = 410 mm from B along line AB
or 34.7 mm above and to left of A 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 3, Solution 113.
ΣFx = 0
Based on
P cos α − 15 N = 0
∴ P cos α = 15 N
(1)
ΣFy = 0
and
P sin α − 10 N = 0
∴ P sin α = 10 N
(2)
Dividing Equation (2) by Equation (1),
tan α =
10
15
∴ α = 33.690°
Substituting into Equation (1),
P=
15 N
= 18.0278 N
cos33.690°
or
P = 18.03 N
(a) Based on
ΣM B = 0
33.7°
− (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50°
− (18.0278 N ) sin 33.690° ( d + 0.150 m ) cos 50°
+ (15 N ) ( 0.150 m ) sin 50° − (10 N )( 0.150 m ) + 6.25 N ⋅ m = 0
−17.9186d = −3.7858
∴ d = 0.21128 m
or d = 211 mm
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
(b) Based on
ΣM D = 0
− (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50°
− (18.0278 N ) sin 33.690°  ( d + 0.150 m ) cos50° + 0.150 m 
+ (15 N ) ( 0.150 m ) sin 50° + 6.25 N ⋅ m = 0
−17.9186d = −3.7858
∴ d = 0.21128 m
or d = 211 mm
This result is expected, since R = 0 and M RB = 0 for
d = 211 mm implies that R = 0 and M = 0 at any other point for
the value of d found in part a.
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 3, Solution 114.
(a)
Let ( R, M D ) be the equivalent force-couple system at D.
First note...
At
x = b; y = h
For
y = k x2
We have h = k b 2
h
or k = 2
b
 h 
∴ y =  2  x2
b 
 h 
y =  2  x2
For any contact point c alone the surface
b 
dy
h
=2 2x
dx
b
 b2 
tan −1 
  2hx 
R = F
and
ΣΜ D : M D = − ( x ) F sin θ + ( h − y ) F cosθ

 
b2
2h x
 +  h − h x 2  F
= −x F 
2
4
2
2
4
 b + 4h x  
b

b + 4h 2 x 2


or


h 2

2
 − xb +  h − 2 x  ( 2hx ) 
b



MD = F 
4
2 2
b + 4h x

2h 2 x 3 
2
2
 − xb + 2h x −

b2 

= F
b 4 + 4h 2 x 2

2h 2 x3 
2
2
 2h − b x −

b2 
or M D = F 


b 4 + 4h 2 x 2


(
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
b = 1 ft, h = 2 ft
(b) With
 7 x − 8x3 
MD = 
F
2 
 1 + 16 x 
For M D to be a maximum
Then

 7 − 24 x 2
dM D
= 0 = F

dx


(
)
1
− 
1
1 + 16 x 2 − 7 x − 8x3   ( 32 x ) 1 + 16 x 2 2 
2

2

1 + 16 x


(
(
)
)
(
)
For the non-trivial solution:
(
)(
)
(
0 = 7 − 24 x 2 1 + 16 x 2 − 16 x 7 x − 8 x3
)
0 = 256 x 4 + 24 x 2 − 7
Solving by the quadratic formula gives for the positive root.
x = 0.354 ft 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 3, Solution 115.
For equivalence
ΣF : FB + FC + FD = R A
R A = − ( 240 N ) j − (125 N ) k − ( 300 N ) i + (150 N ) k
∴ R A = − ( 300 N ) i − ( 240 N ) j + ( 25 N ) k
Also for equivalence
ΣΜ Α : rB/ A × FB + rC/ A × FC + rD/ A × FD = M A
i
j
k
i
j
k
i
j
k
0
+ 0.06 m 0.03 m − 0.075 m + 0.06 m 0.08 m − 0.75 m
or M A = 0 0.12 m
0 − 240 N −125 N
−300 N
0
0
0
0
150 N
=  − (15 N ⋅ m ) i  + ( 22.5 N ⋅ m ) j + ( 9 N ⋅ m ) k  + (12 N ⋅ m ) i − ( 9 N ⋅ m ) j
or M A = − ( 3 N ⋅ m ) i + (13.5 N ⋅ m ) j + ( 9 N ⋅ m ) k
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 3, Solution 116.
Let ( R, M D ) be the equivalent force-couple system at O.
Now ΣF : R = ΣF
= (1.8 lb )( − sin 40°i − cos 40°k ) + (11 lb )( − sin12° j − cos12°k ) + (18 lb )( − sin15° j − cos15°k )
or R = − (1.157 lb ) i − ( 6.95 lb ) j − ( 29.5 lb ) k
Note that each belt force may be replaced by a force-couple that is equivalent to the same force plus the
moment of the force about the shaft (x axis) of the sander. Then ...
ΣM O : M O = ΣM O
i
j
k
= (1.8 lb )
0
0.75 in. 2.2 in.
− sin 40°
0
− cos 40°
− ( 2.5 in.)(11 lb )  i − ( 9 in.) i × (11 lb )( − sin12° j − cos12°k )
+ ( 2.5 in.)(18 lb )  i − ( 9 in.) i × (18 lb )( − sin15° j − cos15°k )
= (1.8 )( −0.75cos 40°i − 2.2sin 40° j + 0.75sin 40°k ) − 27.5i
+ ( 99 )( sin12°k − cos12° j) + 45i + (162 )( sin15°k − cos15° j)  ( lb ⋅ in.)
= ( −1.03416 − 27.5 + 45 ) i + ( −2.5454 − 96.837 − 156.480 ) j
+ ( 0.86776 + 20.583 + 41.929 ) k  ( lb ⋅ in.)
or M O = (16.47 lb ⋅ in.) i − ( 256 lb ⋅ in.) j + ( 63.4 lb ⋅ in.) k
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 3, Solution 117.
Have
ΣFx :
−10 N = Ax + Bx
ΣFy :
0 = Ay + By
⇒ By = − Ay
ΣFz :
6 N = Az + Bz
⇒ Bz = 6 − Az
and
ΣM O :
Now
d BA :
⇒ Bx = −10 − Ax
M O = rO/ A × A + rO/B × B
372 mm =
( 60 mm )2 + ( −72 mm )2 + ( d BA )2z
or ( d BA ) z = 360 mm
rO/ A = (135 mm ) i − ( 72 mm ) j + ( 310 mm ) k
Then
rO/B = ( 75 mm ) i − ( 50 mm ) k
( 60 N ⋅ m ) i + ( 0.05 N ⋅ m ) j − (10 N ⋅ m ) k
i
j
k
i
j
k
= 0.135 − 0.72 0.310 ( N ⋅ m ) + 0.075 0 − 0.050 ( N ⋅ m )
Ax
Ay
Az
Bx By
Bz
i:
(
)
60 = −0.072 Az − 0.310 Ay + ( 0.050 ) By
or 60 = −0.072 Az − 0.360 Ay
j:
0.05 = ( 0.310 Ax − 0.135 Az ) + ( −0.050 Bx − 0.075 Bz )
= 0.310 Ax − 0.050 ( −10 − Ax ) − 0.135 Az − 0.075 ( 6 − Az )
or 0 = 0.360 Ax − 0.060 Az
Az = 6 Ax
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.
(1)
COSMOS: Complete Online Solutions Manual Organization System
Now
Ax = 2 N
From equation (1)
∴ Az = 12.00 N
60 = −0.072 (12.00 ) − 0.360 Ay
or Ay = −169.1 N
Then
Bx = −12.00 N
By = 169.1 N
Bz = −6.00 N
∴ A = ( 2.00 N ) i − (169.1 N ) j + (12.00 N ) k
B = − (12.00 N ) i + (169.1 N ) j − ( 6.00 N ) k
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 3, Solution 118.
ΣF : B + C = R
Have
ΣFx : Bx + C x = 3.9 lb
or
ΣFy : C y = Ry
(2)
ΣFz : C z = −1.1 lb
(3)
Bx = 3.9 lb − Cx
(1)
ΣM A : rB/ A × B + rC/ A × C + M B = M RA
Have
i j k
i
j
k
1
1
4 0 2.0 + ( 2 lb ⋅ ft ) i = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k
∴
x 0 4.5 +
12
12
Bx 0 0
C x C y −1.1
( 2 − 0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k
= M xi + (1.5 ) j − (1.1) k
From
(a)
i - coefficient
2 − 0.166667C y = M x
(4)
j - coefficient
0.375Bx + 0.166667Cx + 0.36667 = 1.5
(5)
k - coefficient
0.33333C y = −1.1
(6)
or
C y = −3.3 lb
From Equations (1) and (5):
0.375 ( 3.9 − Cx ) + 0.166667Cx = 1.13333
Cx =
0.32917
= 1.58000 lb
0.20833
From Equation (1):
Bx = 3.9 − 1.58000 = 2.32 lb
∴ B = ( 2.32 lb ) i
C = (1.580 lb ) i − ( 3.30 lb ) j − (1.110 lb ) k
(b)
From Equation (2):
or R y = − ( 3.30 lb )
From Equation (4):
M x = −0.166667 ( −3.30 ) + 2.0 = 2.5500 lb ⋅ ft
or M x = ( 2.55 lb ⋅ ft )
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.
Ry = C y = −3.30 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 119.
(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:
(
)
R
M By
= j ⋅ ΣM BR = j ⋅ rF /B × FF + rE/B × FE = 0
where
rF /B = (1.125 m ) i − ( 0.575 m ) j + ( 0.7 m ) k
rE/B = (1.35 m ) i − ( 0.85 m ) j + ( 0.7 m ) k
FF = 50 N ( sin α ) j + ( cos α ) k 
FE = − ( 25 N ) k
∴
ΣM RB =
j
k
i
j
k
i
( 50 N ) 1.125 m −0.575 m 0.7 m + ( 25 N ) 1.35 m −0.85 m 0.70
0
sin α
cos α
0
0
−1
= ( −28.75cos α − 35sin α + 21.25 ) i − ( 56.25cos α − 33.75 ) j + ( 56.25sin α ) k  N ⋅ m
R
M By
= −56.25cos α + 33.75 = 0
Thus,
cos α = 0.60
α = 53.130°
or α = 53.1° (b) R = FE + FF
where
FE = − ( 25 N ) k
FF = ( 50 N )( sin 53.130° j + cos 53.130°k ) = ( 40 N ) j + ( 30 N ) k
∴ R = ( 40 N ) j + ( 5 N ) k and
M = ΣM RB = −  28.75 ( 0.6 ) + 35 ( 0.8 ) − 21.25 i − 56.25 ( 0.6 ) − 33.75 j + 56.25 ( 0.8 )  k
= − ( 24 N ⋅ m ) i − ( 0 ) j + ( 45 N ⋅ m ) k
or M = − ( 24.0 N ⋅ m ) i + ( 45.0 N ⋅ m ) k 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 3, Solution 120.
R = ΣF = FF + FE
(a) Have
FF = 50 N ( sin 60° ) j + ( cos 60° ) k  = ( 43.301 N ) j + ( 25 N ) k
where
FE = − ( 25 N ) k
∴ R = ( 43.301 N ) j or R = ( 43.3 N ) j
M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE
Have
rF /C = ( 0.225 m ) i − ( 0.050 m ) j
where
rE/C = ( 0.450 m ) i − ( 0.325 m ) j
i
∴
M CR
j
k
i
j
k
= 0.225 −0.050 0 N ⋅ m + 0.450 −0.325 0 N ⋅ m
0
43.301 25
0
0
−25
= ( 6.875 N ⋅ m ) i + ( 5.625 N ⋅ m ) j + ( 9.7427 N ⋅ m ) k
or M CR = ( 6.88 N ⋅ m ) i + ( 5.63 N ⋅ m ) j + ( 9.74 N ⋅ m ) k
(b) To determine which direction duct section CD has a tendency to turn, have
R
M CD
= λ DC ⋅ M CR
where
λ DC =
Then
− ( 0.45 m ) i + ( 0.1 m ) j
( −0.45)2 + ( 0.1)2
= −0.97619i + 0.21693j
R
M CD
= ( −0.97619i + 0.21693j) ⋅ ( 6.875i + 5.625 j + 9.7427k ) N ⋅ m
= ( −6.7113 + 1.22023) N ⋅ m
= −5.4911 N ⋅ m
Since λ DC ⋅ M CR < 0, duct DC tends to rotate counterclockwise relative to elbow C as viewed from D to C.
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 3, Solution 121.
ΣF : R = R A = Rλ
λ BC
Have
λ BC =
where
( 42 in.) i − ( 96 in.) j − (16 in.) k
∴ RA =
106 in.
21.2 lb
( 42i − 96j − 16k )
106
or R A = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k ΣM A : rC/ A × R + M = M A
Have
where
rC/ A = ( 42 in.) i + ( 48 in.) k =
1
( 42i + 48k ) ft
12
= ( 3.5 ft ) i + ( 4.0 ft ) k
R = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k
M = −λ BC M
=
−42i + 96 j + 16k
(13.25 lb ⋅ ft )
106
= − ( 5.25 lb ⋅ ft ) i + (12 lb ⋅ ft ) j + ( 2 lb ⋅ ft ) k
Then
i
j
k
3.5
0
4.0 lb ⋅ ft + ( −5.25i + 12 j + 2k ) lb ⋅ ft = M A
8.40 −19.20 −3.20
∴ M A = ( 71.55 lb ⋅ ft ) i + ( 56.80 lb ⋅ ft ) j − ( 65.20 lb ⋅ ft ) k
or M A = ( 71.6 lb ⋅ ft ) i + ( 56.8 lb ⋅ ft ) j − ( 65.2 lb ⋅ ft ) k 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 3, Solution 122.
From


− ( 0.6 ft ) i + ( 4.2 ft ) j − (1.5 ft ) k 
R C = R = ( 60 lb ) λ AB = 60 lb 

2
2
2 
 ( − 0.6 ft ) + ( 4.2 ft ) + ( −1.5 ft ) 
R C = − ( 8.00 lb ) i + ( 56.0 lb ) j − ( 20.0 lb ) k
From
M C = rA/C × R + M
where
rA/C = ( 7.8 ft ) i + (1.5 ft ) k

M = ( 22.5 lb ⋅ ft ) λ BA = ( 22.5 lb ⋅ ft ) 



( 0.6 ft ) i − ( 4.2 ft ) j + (1.5 ft ) k 

( 0.6 ft )2 + ( − 4.2 ft )2 + (1.5 ft )2 
= ( 3 lb ⋅ ft ) i − ( 21 lb ⋅ ft ) j + ( 7.5 lb ⋅ ft ) k
i
j k
∴ M C = 7.8 0 1.5 lb ⋅ ft + ( 3 i − 21 j + 7.5 k ) lb ⋅ ft
−8 56 −20
= ( − 84 + 3) lb ⋅ ft  i + (144 − 21) lb ⋅ ft  j + ( 436.8 + 7.5 ) lb ⋅ ft  k
or M C = − ( 81.0 lb ⋅ ft ) i + (123.0 lb ⋅ ft ) j + ( 444 lb ⋅ ft ) k
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 3, Solution 123.
Have:
ΣF :
FA + FC + FD + FE = R
R = − ( 80 kN ) j − ( 40 kN ) j − (100 kN ) j − ( 60 kN ) j
= − ( 280 kN ) j
or R = 280 kN
Have:
ΣM x :
FA ( z A ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( zG )
(80 kN )( 0 ) + ( 40 kN ) ( 3 m ) sin 60° + 60 kN ( 0 )
+ ( 60 kN )  − ( 3 m ) sin 60° = ( 280 kN ) Z G
∴
Z G = − 0.185577 m
or Z G = − 0.1856 m
ΣM z :
FA ( x A ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xG )
(80 kN )  − ( 3 m ) cos 60° − 1.5 m  + ( 40 kN )(1.5 m ) + 60 kN (1.5 m )
+ (100 kN ) ( 3 m ) cos 60° + 1.5 m  = ( 280 kN ) xG
or xG = 0.750 m
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 3, Solution 124.
Have:
ΣM x :
FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) + FF ( z F ) = R ( zG )
(80 kN )( 0 ) + FB ( 3 m ) sin 60° + ( 40 kN ) ( 3 m ) sin 60° + (100 kN )( 0 )
+ ( 60 kN )  − ( 3 m ) sin 60° + FF  − ( 3 m ) sin 60° = R ( 0 )
FB − FF = 20 kN
Also
ΣM z :
(1)
FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) + FF ( xF ) = R ( xG )
(80 kN ) − ( 3 m ) cos 60° − 1.5 m  + FB ( −1.5 m ) + ( 40 kN )(1.5 m )
+ (100 kN ) ( 3 m ) cos 60° + 1.5 m  + ( 60 kN ) (1.5 m ) + FF ( − 1.5 m ) = R ( 0 )
FB + FF = 140 kN
Solving equations (1) and (2):
(2)
FB = 80.0 kN
FF = 60.0 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 3, Solution 125.
Have
ΣF : FA + FB + FC + FD = R
− (116 kips ) j − ( 470 kips ) j − ( 66 kips ) j − ( 28 kips ) j = R
∴ R = − ( 680 kips ) j R = 680 kips
Have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )
(116 kips )( 24 ft ) + ( 470 kips )( 48 ft ) + ( 66 kips )(18 ft ) + ( 28 kips )(100.5 ft ) = ( 680 kips )( zE )
∴ z E = 43.156 ft or z E = 43.2 ft
Have
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )
(116 kips )( 30 ft ) + ( 470 kips )( 96 ft ) + ( 66 kips )(162 ft ) + ( 28 kips )( 96 ft ) = ( 680 kips )( xE )
∴ xE = 91.147 or xE = 91.1 ft
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 3, Solution 126.
Have
ΣF : FB + FC + FD + FE = R
− ( 470 kips ) j − ( 66 kips ) j − ( 28 kips ) j − (116 kips ) j = R
∴ R = − ( 680 kips ) j
Have
ΣM x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )
( 470 kips )( 48 ft ) + ( 66 kips )(18 ft ) + ( 28 kips )(100.5 ft ) + (116 kips )( b ) = ( 680 kips )( 48 ft )
∴ b = 52.397 ft or b = 52.4 ft
Have
ΣM z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )
( 470 kips )( 96 ft ) + ( 66 kips )(162 ft ) + ( 28 kips )( 96 ft ) + (116 kips )( a ) = ( 680 kips )( 96 ft )
∴ a = 58.448 ft or a = 58.4 ft
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 3, Solution 127.
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
Have
ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R
∴ R = − ( 780 N ) j
Have
ΣM z :
( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x )
∴ x = 1.34103 m
Have
ΣM x :
( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z )
∴ z = 0.93846 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
( 0.3 m ≤ x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m )
continued
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
Let x = 0.3 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0
∴ W = 380 N
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0
∴ z = 3.5684 m < 3.7 m
Let z = 3.7 m,
∴ acceptable
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0
∴ W = 395.29 N > 380 N
Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the
smallest weight W.
or W = 380 N at ( 0.3 m, 0, 3.57 m )
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 3, Solution 128.
For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of
the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box
touching the z-axis, satisfies this condition.
Let x = 0.6 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0
∴ W = 665 N
and
ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0
∴ z = 2.8105 m
( 2 m < z < 4 m)
∴ acceptable
or W = 665 N at ( 0.6 m, 0, 2.81 m )
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 3, Solution 129.
First, reduce the given force system to a force-couple system at the origin.
Have
ΣF :
( 2P ) i − ( P ) j + ( P ) j = R
∴ R = ( 2P ) i
Have
ΣM O : Σ ( rO × F ) = M OR
i
j k
i j k
2 2.5 + 0 0 4 = Pa ( −1.5i + 5j − 6k )
2 −1 0
0 1 0
M OR = Pa 2
R = 2Pi
(a)
or Magnitude of R = 2P
Direction of R : θ x = 0°, θ y = −90°, θ z = 90°
(b) Have
M1 = λ R ⋅ M OR
λR =
R
R
= i ⋅ ( −1.5Pai + 5Paj − 6Pak )
= −1.5Pa
and pitch
P=
M1
−1.5Pa
=
= −0.75a
2P
R
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.
or P = −0.75a
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M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( 5Pa ) j − ( 6Pa ) k
Require
M 2 = rQ/O × R
( 5Pa ) j − ( 6Pa ) k = ( yj + zk ) × ( 2Pi ) = − ( 2Py ) k + ( 2Pz ) j
From
i : 5Pa = 2Pz
∴ z = 2.5a
From
k : − 6Pa = −2Py
∴ y = 3a
∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at y = 3a, z = 2.5a
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 3, Solution 130.
First, reduce the given force system to a force-couple at the origin.
ΣF : Pi − Pi − Pk = R
Have
∴ R = − Pk
ΣM O : − P ( 3a ) k − P ( 3a ) j + P ( −ai + 3aj) = MOR
Have
∴ MOR = Pa ( −i − 3k )
Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same.
R = − Pk or Magnitude of R = P
(a)
Direction of R : θ x = 90°, θ y = 90°, θ z = −180°
M1 = λ R ⋅ M OR
(b) Have
= −k ⋅  Pa ( −i − 3k ) 
= 3Pa
and pitch
(c) Have
P=
M1 3Pa
=
= 3a
R
P
or P = 3a
M OR = M1 + M 2
∴ M 2 = M OR − M1 = Pa ( −i − 3k ) − ( −3Pak ) = − Pai
Require
M 2 = rQ/O × R
− Pai = ( xi + yj) × ( − P ) k = Pxj − Pyi
From
i : − Pa = − Py
or
y =a
j: x = 0
∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at x = 0, y = a
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 3, Solution 131.
First, reduce the given force system to a force-couple at the origin.
Have ΣF : − (10 N ) j − (11 N ) j = R
∴ R = − ( 21 N ) j
Have ΣM O : Σ ( rO × F ) + ΣM C = M OR
M OR =
i j
k
i j
k
0 0 0.5 N ⋅ m + 0 0 −0.375 N ⋅ m − (12 N ⋅ m ) j
0 −10 0
0 −11
0
= ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
R = − ( 21 N ) j
(a)
R = − ( 21.0 N ) j
Have M1 = λ R ⋅ M OR
(b)
or
λR =
R
R
= ( − j) ⋅ ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
= 12 N ⋅ m
and pitch P =
(c)
and
M1 = − (12 N ⋅ m ) j
M1 12 N ⋅ m
=
= 0.57143 m
R
21 N
Have
or P = 0.571 m
M OR = M1 + M 2
∴ M 2 = M OR − M1 = ( 0.875 N ⋅ m ) i
Require
∴
( 0.875 N ⋅ m ) i = ( xi + zk ) × − ( 21 N ) j
0.875i = − ( 21x ) k + ( 21z ) i
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.
M 2 = rQ/O × R
COSMOS: Complete Online Solutions Manual Organization System
From i:
0.875 = 21z
∴ z = 0.041667 m
From k:
0 = −21x
∴ z =0
∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 41.7 mm
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 3, Solution 132.
(a)
Have
First, reduce the given force system to a
force-couple system.
ΣF :
− ( 50 N ) k − ( 50 N ) j + ( 50 N ) k = R
R = − ( 50 N ) j;
R = 50 N
R = − ( 50.0 N ) j
Have
ΣM O :
( 0.1 N ⋅ m ) k − ( 0.1 N ⋅ m ) j + ( 0.1 N ⋅ m ) k = MOR
M OR = − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k
(b)
Have
M1 = λ R × MOR =
R
⋅ M OR
R
= − j ⋅  − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k 
= 0.1 N ⋅ m
and pitch
P=
M1
0.1 N ⋅ m
=
= 0.002 m
R
50 N
or P = 2.00 mm
(c)
Have
M1 = PR = ( 0.002 m )  − ( 50 N ) j
= − ( 0.1 N ⋅ m ) j
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
Note that because M z = ( 0.2 N ⋅ m ) k , the line of action of the wrench must pass through the x-axis to
compensate for M z as shown above:
With
M1 + ( r × R ) = M OR
Then
− ( 0.1 N ⋅ m ) j +  − ( d ) i × − ( 50N ) j
= − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k
or ( 50 N )( d )  k = ( 0.2 N ⋅ m ) k
and
d = 0.004 m
x = − d = − 0.004 m
or x = − 4.00 mm, z = 0
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 3, Solution 133.
(
)
First replace the given couples with an equivalent force-couple system R, M OB at the origin.
ΣF :
ΣM O :
R = − ( 35 lb ) i − (12 lb ) k
M OR = − ( 200 lb ⋅ in.) i + ( 8 in.) j + ( 8 in.) k  ×  − ( 35 lb ) i 
− (140 lb ⋅ in.) k + (10 in.) i + ( 4 in.) j ×  − (12 lb ) k 
= ( − 200 − 48 ) i + ( − 280 + 120 ) j + ( 280 − 140 ) k
= − ( 248 lb ⋅ in.) i − (1600 lb ⋅ in.) j + (140 lb ⋅ in.) k
Now
R=
( − 35 lb )2 + ( −12 lb )2
= 37 lb
Then
λ axis =
1
( − 35 i − 12 k )
37
R = − ( 35.0 lb ) i − (12.00 ) k
(a)
M1 = λ axis ⋅ M OR
(b)
Then
=
1
( − 35 i − 12 k ) ⋅ ( − 248 i − 160 j + 140 k )( lb ⋅ in.)
37
=
1
( 35 × 248 − 12 × 140 )( lb ⋅ in.)
37
=
7000
lb ⋅ in.
37
M
P= 1 =
R
7000
lb ⋅ in.
37
37 lb
or P = 5.11 in.
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
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COSMOS: Complete Online Solutions Manual Organization System
(c)
Have
M1 = λ axis M1
=
7000 lb ⋅ in.
( − 35 i − 12 k )
37 2
Then
M OR = M1 + M 2
or
M 2 = ( − 248 i − 160 j + 140 k ) −
7000
( − 35 i − 12 k )
37 2
= − ( 69.037 lb ⋅ in.) i − (160 lb ⋅ in.) j + ( 201.36 lb ⋅ in.) k
M z = rO/P × R
Require
i j k
or − 69.037i − 160 j + 201.36k = 0 y z
−35 0 −12
j:
−160 = − 35 z
z = 4.57 in.
or
k:
201.36 = 35 y
or
y = 5.75 in.
∴
The point of intersection is defined by
y = 5.75 in.
z = 4.57 in.
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 3, Solution 134.
First reduce the given force system to a force-couple at the origin at B.
15 
 8
Have ΣF : − ( 79.2 lb ) k − ( 51 lb )  i +
j = R
17 
 17
(a)
∴ R = − ( 24.0 lb ) i − ( 45.0 lb ) j − ( 79.2 lb ) k
and
R = 94.2 lb
Have
ΣM B : rA/B × FA + M A + M B = M RB
M RB =
i j
k
15 
 8
0 −20
0 − 660k − 714  i +
j  = 1584i − 660k − 42 ( 8i + 15 j)
17
17


0 0 −79.2
∴ M RB = (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k
(b)
Have
M1 = λ R ⋅ M OR
=
λR =
R
R
−24.0i − 45.0 j − 79.2k
⋅ (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k 
94.2
= 537.89 lb ⋅ in.
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
M1 = M1λ R
and
= − (137.044 lb ⋅ in.) i − ( 256.96 lb ⋅ in.) j − ( 452.24 lb ⋅ in.) k
Then pitch p =
or p = 5.71 in.
(c)
M1 537.89 lb ⋅ in.
=
= 5.7101 in.
R
94.2 lb
Have
M RB = M1 + M 2
∴ M 2 = M RB − M1 = (1248i − 630 j − 660k ) − ( −137.044i − 256.96 j − 452.24k )
= (1385.04 lb ⋅ in.) i − ( 373.04 lb ⋅ in.) j − ( 207.76 lb ⋅ in.) k
Require
M 2 = rQ/B × R
i
j
k
1385.04i − 373.04 j − 207.76k = x
0
z
−24 −45 −79.2
= ( 45 z ) i − ( 24 z ) j + ( 79.2 x ) j − ( 45 x ) k
From i:
1385.04 = 45z
From k:
−207.76 = −45x
∴ z = 30.779 in.
∴ x = 4.6169 in.
∴ The axis of the wrench intersects the xz-plane
at
x = 4.62 in., z = 30.8 in.
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 3, Solution 135.
(a)
First reduce the given force system to a force-couple at the origin.
ΣF : Pλ BA + Pλ DC + Pλ DE = R
Have
 4
12 
3  3
4  −9
4
R = P  j − k  +  i − j + 
i − j+
k 
5
25 
5  5
5   25
 5
∴ R =
R=
3P
25
3P
( 2i − 20 j − k )
25
( 2 )2 + ( 20 )2 + (1)2 =
27 5
P
25
Have
ΣM : Σ ( rO × P ) = M OR
−4 P
3P 
4P 
4P
12 P 
 3P
 −9 P
j−
k  + ( 20a ) j × 
i−
j  + ( 20a ) j × 
i−
j+
k  = M OR
5 
5 
5
25 
 5
 5
 25
( 24a ) j × 
∴ M OR =
(b)
24 Pa
( −i − k )
5
Have
M1 = λ R ⋅M OR
where λ R =
3P
25
1
R
=
=
( 2i − 20 j − k )
( 2i − 20 j − k )
R
25
27 5 P 9 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
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COSMOS: Complete Online Solutions Manual Organization System
Then M 1 =
and pitch
1
9 5
( 2i − 20 j − k ) ⋅
p=
24 Pa
−8Pa
( −i − k ) =
5
15 5
M1
−8Pa  25  −8a
=

=
R
81
15 5  27 5 P 
or p = −0.0988a
(c)
M1 = M 1λ R =
−8Pa  1 
8Pa
( −2i + 20 j + k )

 ( 2i − 20 j − k ) =
675
15 5  9 5 
Then
M 2 = M OR − M1 =
24Pa
8Pa
8Pa
( −i − k ) −
( −2i + 20 j + k ) =
( −403i − 20 j − 406k )
5
675
675
M 2 = rQ/O × R
Require
 8Pa 
 3P 

 ( −403i − 20 j − 406k ) = ( xi + zk ) × 
 ( 2i − 20 j − k )
 675 
 25 
 3P 
=
  20 zi + ( x + 2 z ) j − 20 xk 
 25 
From i:
8 ( −403)
Pa
 3P 
= 20 z 

675
 25 
∴ z = −1.99012a
From k:
8 ( −406 )
Pa
 3P 
= −20 x 

675
 25 
∴ x = 2.0049a
∴ The axis of the wrench intersects the xz-plane at
x = 2.00a, z = −1.990a
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 3, Solution 136.
First reduce the given force-couple system to an equivalent force-couple system ( R, M B ) at point B.
d BD =
( − 480 mm )2 + ( 560 mm )2 + ( − 480 mm )2
= 880 mm
FBD = FBDλBD =
132 N
( − 480i + 560 j − 480k )
880
= (12 N )( − 6i + 7 j − 6k )
d EB =
( 240 mm )2 + ( − 220 mm )2 + ( 480 mm )2
= 580 mm
FEB = FEBλEB =
145 N
( 240i − 220 j + 480k )
580
= ( 5 N )(12i − 11j + 24k )
ΣF :
R = FBD + FEB
= (12 N )( − 6i + 7 j − 6k ) + 5 N (12i − 11j + 24k )
= − (12 N ) i + ( 29 N ) j + ( 48 N ) k
d BF =
( 340 mm )2 + ( 240 mm )2 + ( − 60 mm )2
= 20 442 mm
Then
MB =
=
20 N ⋅ m
( 340i + 240 j − 60k )
20 442
20 N ⋅ m
(17i + 12 j − 3k )
442
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
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Now determine whether R and M B are perpendicular
R ⋅ M B = ( −12 j + 29 j + 48k ) ⋅
20
(17i + 12 j − 3k )
442
20
( −12 × 17 + 29 × 12 − 48 × 3)
442
=
=0
∴ R and M B are perpendicular so that ( R, M B ) can be reduced to the single equivalent force
R = − (12.00 N ) i + ( 29.0 N ) j + ( 48.0 N ) k
Now require M B = rB/P × R
or
j:
20 × 12
= −12 ( z − 0.480 ) + 0.480 ( 48 )
442
or
k:
i
j
k
20 N ⋅ m
(17i + 12 j − 3k ) = − 0.480 y z − 0.480 ( N ⋅ m )
442
−12 29
48
z = 1.449 m
− 20 × 3
= − 0.480 ( 29 ) + 12 y
442
or
y = 0.922 m
∴ The point of intersection is defined by
y = 0.922 m
z = 1.449 m
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 3, Solution 137.
First, reduce the given force system to a force-couple at the origin.
ΣF : FA + FG = R
Have
 ( 4 in.) i + ( 6 in.) j − (12 in.) k 
∴ R = (10 lb ) k + 14 lb 
 = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k
14 in.


R=
and
Have
56 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = MOR
{
}
M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k 
 (16 in.) i − (12 in.) j 
 ( 4 in.) i − (12 in.) j + ( 6 in.) k 
+ ( 84 lb ⋅ in.) 
 + ( 120 lb ⋅ in.) 

20 in.
14 in.




∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k
= (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k
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
The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR .
To be perpendicular R ⋅ M OR = 0
Have
R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0?
= 73.829 + 19.3716 − 24.572
≠0
∴ System cannot be reduced to a single equivalent force.
To reduce to an equivalent wrench, the moment component along the line of action of P is found.
M1 = λ R ⋅ M OR
λR =
R
R
 ( 4i + 6 j − 2k ) 
=
 ⋅ (18.4572i + 3.2286 j + 12.2858k )
56


= 9.1709 lb ⋅ ft
and
M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k )
And pitch
p=
M1 9.1709 lb ⋅ ft
=
= 1.22551 ft
R
56 lb
or p = 1.226 ft
Have
M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k )
= (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k
Require
M 2 = rQ/O × R
(13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6 j − 2k )
= − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k
From j:
−4.1244 = 4 z
or
z = −1.0311 ft
From k:
14.7368 = −4 y
or
y = −3.6842 ft
∴ line of action of the wrench intersects the yz plane at
y = −3.68 ft, z = 1.031 ft
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 3, Solution 138.
FA = ( FA ) x i + ( FA ) y j
Define
FB = ( FB ) x i + ( FB ) y j
ΣFx :
Then
( FA ) x + ( FB ) x = 0
( FA ) x = − ( FB ) x
ΣFy :
( FA ) y + ( FB ) y = R
( FA ) y = R − ( FB ) y
bk × ( FB ) x i + ( FB ) y j = − ak + R j + M j


ΣM A :
and
i:
− b ( FB ) y = aR
a
R
b
or
( FB ) y = −
Then
( FA ) y = R −  −
a 
R
b


a

= R 1 + 
b

j:
b ( FB ) x = M
M
b
or
( FB ) x =
Then
( FA ) x = −
M
b
∴ FA = −
M
a

i + R 1 +  j
b
b

FB =
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.
M
a
i − Rj
b
b
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 139.
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rxi + Ry j + Rzk
M = M xi + M y j + M zk
and
while the unknown forces A and B can be expressed as
A = Axi + Ay j + Azk
and
B = Bxi + Bzk
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position
vector rP are also known.
Then, for equivalence of the two systems
ΣFx : Rx = Ax + Bx
ΣFy : Ry = Ay
(1)
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
continued
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
(
)
Based on the above six independent equations for the six unknowns Ax , Ay , Az , Bx , Bz , b , there exists a
unique solution for A and B.
Ay = Ry
From Equation (2)
Equation (6)
1
Ax =   xRy − M z
 y
Equation (1)
1
Bx = Rx −   xRy − M z
 y
Equation (4)
1
Az =   M x + zRy
 y
Equation (3)
1
Bz = Rz −   M x + zRy
 y
Equation (5)
b=
(
)
(
(
)
)
(
)
( xM x + yM y + zM z )
( M x − yRz + zRy )
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 3, Solution 140.
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
Have
R = Rj
M = Mj
and
and are known.
The unknown forces A and B can be expressed as
A = Axi + Ay j + Azk
B = Bxi + By j + Bzk
and
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence
∑ Fx : 0 = Ax + Bx
(1)
∑ Fy : R = Ay + By
(2)
∑ Fz : 0 = Az + Bz
(3)
∑ M x : 0 = − zBy (4)
∑ M y : M = −aAz − xBz + zBx
(5)
∑ M z : 0 = aAy + xBy
(6)
Since A and B are made perpendicular,
A⋅B = 0
There are eight unknowns:
or
Ax Bx + Ay By + Az Bz = 0
(7)
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
continued
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
0 = − zBy
Next consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3) this equation becomes
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
By ≠ 0, so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become.
0 = Bx
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = −aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0 (7)′
Then
Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,
 R − By 
M = −aAz −  −a
 ( − Az )


B
y


or
Az = −
M
By (8)
aR
Substituting into Equation (7)′,
M
 M

By 
By  = 0
( R − By ) By +  − aR
aR


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
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COSMOS: Complete Online Solutions Manual Organization System
By =
or
a 2 R3
a2R2 + M 2
Then from Equations (2), (8), and (3)
Ay = R −
Az = −
Bz =
a 2 R3
RM 2
=
a2R2 + M 2
a2R2 + M 2

M 
a 2 R3
aR 2 M
=
−
 2 2
2
2
aR  a R + M 
a R2 + M 2
aR 2 M
a R2 + M 2
2
In summary
A=
RM
( Mj − aRk )
a R2 + M 2
B=
aR 2
( aRj + Mk )
a R2 + M 2
2
2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a
given point.
Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0.
From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane
parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to
the left of the origin, as shown in the figure below.
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 3, Solution 141.
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action ( AA′ ) . Note that it has been assumed that the line of
action of force B intersects the xz plane at point P ( x, 0, z ) . Denoting the known direction of line AA′ by
λ A = λxi + λ y j + λzk
it follows that force A can be expressed as
(
A = Aλ A = A λxi + λ y j + λz k
)
Force B can be expressed as
B = Bxi + By j + Bzk
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows
that the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
ΣFx : 0 = Aλx + Bx (1)
ΣFy : R = Aλ y + By (2)
ΣFz : 0 = Aλz + Bz (3)
ΣM x : 0 = − zBy
Since there are six unknowns
(4)
ΣM y : M = −aAλz + zBx − xBz
(5)
ΣM z : 0 = aAλ y + xBy
(6)
( A, Bx , By , Bz , x, z ) and six independent equations, it will be possible to
obtain a solution.
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
Case 1: Let z = 0 to satisfy Equation (4)
Now Equation (2)
Aλ y = R − By
Equation (3)
Bz = − Aλz
x=−
Equation (6)
aAλ y
By
 a 
= −
 R − By
 By 


(
)
Substitution into Equation (5)
  a 

M = −aAλz −  − 
 R − By ( − Aλz ) 
  By 

(
∴ A=−
)
1 M 
B
λz  aR  y
Substitution into Equation (2)
R=−
1 M 
By λ y + By
λz  aR 
∴ By =
Then
λz aR 2
λz aR − λ y M
R
aR
λy −
λz
M
λx MR
Bx = − Aλx =
λz aR − λ y M
A=−
MR
λz aR − λ y M
Bz = − Aλz =
=
λz MR
λz aR − λ y M
In summary
A=
B=
and
P
λA
aR
λy −
λz
M
R
( λ Mi + λz aRj + λz Mk )
λz aR − λ y M x


 λz aR − λ y M  
R 
x = a 1 −
 = a 1 − R 
 


By 
λz aR 2

 


or x =
Note that for this case, the lines of action of both A and B intersect the x axis.
continued
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.
λy M
λz R
COSMOS: Complete Online Solutions Manual Organization System
Case 2: Let By = 0 to satisfy Equation (4)
A=
Now Equation (2)
R
λy
Equation (1)
λ 
Bx = − R  x 
 λy 
 
Equation (3)
λ 
Bz = − R  z 
 λy 
 
aAλ y = 0
Equation (6)
which requires a = 0
Substitution into Equation (5)
  λ 
  λ 
M = z −R  x  − x −R  z 
  λ y  
  λ y  
or
M 
 λy
 R
λz x − λx z = 
This last expression is the equation for the line of action of force B.
In summary
 R
A =  λ A
 λy 
 
 R
B =   ( −λ x i − λ z k )
 λy 
 
Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below.
Note that the component of A in the xz plane is parallel to 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 142.
(a) Have
M B = rC/B FN
= ( 0.1 m )( 800 N )
= 80.0 N ⋅ m
or M B = 80.0 N ⋅ m
(b) By definition
M B = rA/B P sin θ
where
θ = 90° − ( 90° − 70° ) − α
= 90° − 20° − 10°
= 60°
∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60°
P = 205.28 N
or
P = 205 N
(c) For P to be minimum, it must be perpendicular to the line joining
points A and B. Thus, P must be directed as shown.
Thus
M B = dPmin = rA/B Pmin
or
80.0 N ⋅ m = ( 0.45 m ) Pmin
∴ Pmin = 177.778 N
or Pmin = 177.8 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.
20°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 143.
M C = rB/C × FB
Have
Noting the direction of the moment of each force component about C is
clockwise,
M C = xFBy + yFBx
where
x = 144 mm − 78 mm = 66 mm
y = 86 mm + 108 mm = 194 mm
and
FBx =
FBy =
78
( 78)2 + (86 )2
86
( 78) + (86 )2
2
( 580 N ) = 389.65 N
( 580 N ) = 429.62 N
∴ M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N )
= 103947 N ⋅ mm
= 103.947 N ⋅ m
or M C = 103.9 N ⋅ m
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 3, Solution 144.
(a) Have
M A = rE/ A × TDE
where
rE/ A = ( 92 in.) j
TDE = λ DETDE
=
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 0 92
0 lb ⋅ in. = − ( 22,080 lb ⋅ in.) i − ( 4416 lb ⋅ in ) k
48 264 −240
or M A = − (1840 lb ⋅ ft ) i − ( 368 lb ⋅ ft ) k
M A = rG/ A × TCG
(b) Have
rG/ A = (108 in.) i + ( 92 in.) j
where
TCG = λ CGTCG =
− ( 24 in.) i + (132 in.) j − (120 in.) k
( 24 )2 + (132 )2 + (120 )2 in.
( 360 lb )
= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 108 92
0 lb ⋅ in.
−48 264 −240
= − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k
or M A = − (1840 lb ⋅ ft ) i + ( 2160 lb ⋅ ft ) j + ( 2740 lb ⋅ ft ) k
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 3, Solution 145.
First note
and
AC = rC/ A =
( −2.4 )2 + (1.8)2 m = 3 m
AD = rD/ A =
(1.2 )2 + ( −2.4 )2 + ( 0.3)2 m = 2.7 m
rC/ A = − ( 2.4 m ) j + (1.8 m ) k
rD/ A = (1.2 m ) i − ( 2.4 m ) j + ( 0.3 m ) k
By definition
rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ
or
( −2.4 j + 1.8k ) ⋅ (1.2i − 2.4 j + 0.3k ) = ( 3)( 2.7 ) cosθ
( 0 )(1.2 ) + ( −2.4 )( −2.4 ) + (1.8)( 0.3) = 8.1cosθ
and
cosθ =
6.3
= 0.77778
8.1
θ = 38.942°
or θ = 38.9°
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 3, Solution 146.
M O = rA/O × TBA
Based on
where
M O = M xi + M y j + M zk
= M xi + (100 lb ⋅ ft ) j − ( 400 lb ⋅ ft ) k
rA/O = ( 6 ft ) i + ( 4 ft ) j
TBA = λ BATBA
=
( 6 ft ) i − (12 ft ) j − ( a ) k T
BA
d BA
i
j
k
T
∴ M xi + 100 j − 400k = 6 4 0 BA
d
6 −12 −a BA
=
TBA
 − ( 4a ) i + ( 6a ) j − ( 96 ) k 
d BA 
100
d BA
6a
From j-coefficient:
100d AB = 6aTBA or TBA =
From k -coefficient:
−400d AB = −96TBA or TBA =
Equating Equations (1) and (2) yields
a=
400
d BA
96
100 ( 96 )
6 ( 400 )
or a = 4.00 ft
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.
(1)
(2)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 147.
(
M DB = λ DB ⋅ rC/D × TCF
Have
λ DB =
where
)
( 48 in.) i − (14 in.) j = 0.96i − 0.28j
50 in.
rC/D = ( 8 in.) j − (16 in.) k
( 24 in.) i − ( 36 in.) j − (8 in.) k 132 lb
(
)
TCF = λ CF TCF =
44 in.
= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k
0.96 −0.28
∴ M DB =
0
72
0
8
−16 lb ⋅ in.
−108 −24
= 0.96 ( 8 )( −24 ) − ( −16 )( −108 )  + ( −0.28 ) ( −16 )( 72 ) − 0 
= −1520.64 lb ⋅ in.
or M DB = −1521 lb ⋅ in.
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 3, Solution 148.
(a) Based on
ΣF : FA = T = 1000 N
or FA = 1000 N
20°
ΣM A : M A = (T sin 50° )( dA )
= (1000 N ) sin 50° ( 2.25 m )
= 1723.60 N ⋅ m
or M A = 1724 N ⋅ m
(b) Based on
ΣF : FB = T = 1000 N
or FB = 1000 N
20°
ΣMB : M B = (T sin 50° )( d B )
= (1000 N ) sin 50° (1.25 m )
= 957.56 N ⋅ m
or M B = 958 N ⋅ m
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 3, Solution 149.
Require the equivalent forces acting at A and C be parallel and at an angle
of α with the vertical.
Then for equivalence,
ΣFx :
( 250 lb ) sin 30° = FA sin α + FB sin α
(1)
ΣFy : − ( 250 lb ) cos 30° = − FA cos α − FB cos α (2)
Dividing Equation (1) by Equation (2),
( 250 lb ) sin 30° = ( FA + FB ) sin α
− ( 250 lb ) cos 30° − ( FA + FB ) cos α
Simplifying yields α = 30°
Based on
ΣM C : ( 250 lb ) cos 30° (12 ft ) = ( FA cos 30° )( 32 ft )
∴ FA = 93.75 lb
or FA = 93.8 lb
60°
Based on
ΣM A : − ( 250 lb ) cos 30° ( 20 ft ) = ( FC cos 30° ) ( 32 ft )
∴ FC = 156.25 lb
or FC = 156.3 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.
60°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 150.
Have
ΣF : PAB = FC
where
PAB = λ AB PAB
=
( 2.0 in.) i + ( 38 in.) j − ( 24 in.) k 45 lb
(
)
44.989 in.
or FC = ( 2.00 lb ) i + ( 38.0 lb ) j − ( 24.0 lb ) k
Have
ΣM C : rB/C × PAB = M C
i
j
k
M C = 2 29.5 −33
0 lb ⋅ in.
1 19 −12
= ( 2 lb ⋅ in.) {( −33)( −12 ) i − ( 29.5 )( −12 ) j
+ ( 29.5 )(19 ) − ( −33)(1)  k
}
or M C = ( 792 lb ⋅ in.) i + ( 708 lb ⋅ in.) j + (1187 lb ⋅ in.) k
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 3, Solution 151.
For equivalence
ΣFx : − ( 90 N ) sin 30° + (125 N ) cos 40° = Rx
or Rx = 50.756 N
ΣFy : − ( 90 N ) cos30° − 200 N − (125 N ) sin 40° = Ry
or Ry = −358.29 N
and
( 50.756 )2 + ( −358.29 )2 = 361.87 N
R=
Then
tan θ =
Ry
Rx
=
−358.29
= −7.0591
50.756
∴ θ = −81.937°
or R = 362 N
81.9°
Also
ΣM A : M − ( 90 N ) sin 35° ( 0.6 m ) − ( 200 N ) cos 25°  ( 0.85 m ) − (125 N ) sin 65°  (1.25 m ) = 0
∴ M = 326.66 N ⋅ m
or M = 327 N ⋅ m
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 3, Solution 152.
For equivalence
Σ F: FA + FB + FC + FD = R C
R C = − ( 5 lb ) j − ( 3 lb ) j − ( 4 lb ) k − ( 7 lb ) i
∴ R C = ( −7 lb ) i − ( 8 lb ) j − ( 4 lb ) k
Also for equivalence
ΣM C : rA′/C × FA + rB′/C × FB + rD′/C × FD = M C
or
i j
k
i
j
k
i
j
k
M C = 0 0 −1.5 in. + 1 in. 0 −1.5 in. + 0 1.5 in. 1.5 in.
0 5 lb
0
0 −3 lb
0
−7 lb
0
0
= ( −7.50 lb ⋅ in. − 0 ) i  + ( 0 − 4.50 lb ⋅ in.) i + ( −3.0 lb ⋅ in. − 0 ) k 
+ (10.5 lb ⋅ in. − 0 ) j + ( 0 + 10.5 lb ⋅ in.) k 
or M C = − (12.0 lb ⋅ in.) i + (10.5 lb ⋅ in.) j + ( 7.5 lb ⋅ in.) k
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 3, Solution 153.
Have
ΣF : FA + FB + FC + FD = R
− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R
∴ R = − ( 330 lb ) j
Have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft )
∴ z D = 3.5523 ft
Have
or z D = 3.55 ft
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft )
∴ xD = 7.0263 ft or xD = 7.03 ft
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 4, Solution 1.
Free-Body Diagram:
(a)
ΣM B = 0:
− Ay ( 3.6 ft ) − (146 lb )(1.44 ft ) − ( 63 lb )( 3.24 ft ) − ( 90 lb )( 6.24 ft ) = 0
Ay = − 271.10 lb
(b)
ΣM A = 0 :
or A y = 271 lb
or B y = 570 lb
By (3.6 ft ) − (146 lb)(5.04 ft ) − (63 lb)(6.84 ft ) − (90 lb)(9.84 ft ) = 0
By = 570.10 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 2.
Free-Body Diagram:
(a)
ΣM C = 0:
( 3.5 kips ) (1.6 + 1.3 + 19.5cos15o ) ft  − 2FB (1.6 + 1.3 + 14 ) ft  + ( 9.5 kips )(1.6 ft ) = 0
2FB = 5.4009 kips
or FB = 2.70 kips
(b)
ΣM B = 0:
( 3.5 kips ) (19.5cos15o − 14 ) ft  − ( 9.5 kips ) (14 + 1.3) ft  + 2 FC (14 + 1.3 + 1.6 ) ft  = 0
2FC = 7.5991 kips, or
or FC = 3.80 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 4, Solution 3.
Free-Body Diagram:
(a)
ΣM K = 0:
( 25 kN )( 5.4 m ) + ( 3 kN )( 3.4 m ) − 2FH ( 2.5 m ) + ( 50 kN )( 0.5 m ) = 0
2FH = 68.080 kN
(b)
ΣM H = 0:
or FH = 34.0 kN
( 25 kN )( 2.9 m ) + ( 3 kN )( 0.9 m ) − ( 50 kN )( 2.0 m ) + 2FK ( 2.5 m ) = 0
2FK = 9.9200 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.
or FK = 4.96 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 4.
Free-Body Diagram: (boom)
(a)
ΣM B = 0:
( 25 kN )( 2.6 m ) + ( 3 kN )( 0.6 m ) − ( 25 kN )( 0.4 m ) − TCD ( 0.7 m ) = 0
TCD = 81.143 kN
(b)
ΣFx = 0:
Bx = 0 so that B = By
ΣFy = 0:
( −25 − 3 − 25 − 81.143) kN + B = 0
B = 134.143 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.
or TCD = 81.1 kN or B = 134.1 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 5.
Free-Body Diagram:
a1 = ( 20 in.) sin α − ( 8 in.) cos α
a2 = ( 32 in.) cos α − ( 20 in.) sin α
b = ( 64 in.) cos α
From free-body diagram of hand truck
ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0
(1)
ΣFy = 0: P − 2w + 2B = 0
(2)
α = 35°
For
a1 = 20sin 35° − 8cos 35° = 4.9183 in.
a2 = 32 cos 35° − 20sin 35° = 14.7413 in.
b = 64cos 35° = 52.426 in.
(a)
From Equation (1)
P ( 52.426 in.) − 80 lb (14.7413 in.) + 80 lb ( 4.9183 in.) = 0
∴ P = 14.9896 lb
(b)
or P = 14.99 lb From Equation (2)
14.9896 lb − 2 ( 80 lb ) + 2 B = 0
∴ B = 72.505 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.
or
B = 72.5 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 6.
a1 = ( 20 in.) sin α − ( 8 in.) cos α
Free-Body Diagram:
a2 = ( 32 in.) cos α − ( 20 in.) sin α
b = ( 64 in.) cos α
From free-body diagram of hand truck
ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 (1)
ΣFy = 0: P − 2w + 2B = 0
(2)
α = 40°
For
a1 = 20sin 40° − 8cos 40° = 6.7274 in.
a2 = 32 cos 40° − 20sin 40° = 11.6577 in.
b = 64cos 40° = 49.027 in.
(a)
From Equation (1)
P ( 49.027 in.) − 80 lb (11.6577 in.) + 80 lb ( 6.7274 in.) = 0
P = 8.0450 lb
or P = 8.05 lb (b)
From Equation (2)
8.0450 lb − 2 (80 lb ) + 2 B = 0
B = 75.9775 lb
or B = 76.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 7.
Free-Body Diagram:
(a) a = 2.9 m
ΣFx = 0:
Ax = 0
ΣM B = 0:
− (12 m ) Ay + (12 − 2.9 ) m  ( 3.9 kN ) + (12 − 2.9 − 2.6 ) m  ( 6.3 kN )
+ ( 2.8 + 1.45 ) m  ( 7.9 kN ) + (1.45 m )( 7.3 kN ) = 0
or
ΣFy = 0:
or
Ay = 10.0500 kN
or A = 10.05 kN
or B = 15.35 kN
10.0500 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0
By = 15.3500 kN
(b) a = 8.1 m
ΣM B = 0:
− (12 m ) Ay + (12 − 8.1) m  ( 3.9 kN ) + (12 − 8.1 − 2.6 ) m  ( 6.3 kN )
+ ( 2.8 + 4.05 ) m  ( 7.9 kN ) + ( 4.05 m )( 7.3 kN ) = 0
or
ΣFy = 0:
or
Ay = 8.9233 kN
or A = 8.92 kN
8.9233 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0
By = 16.4767 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.
or B = 16.48 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 8.
Free-Body Diagram:
(a)
ΣFx = 0:
Ax = 0
ΣM B = 0:
− (12 m ) Ay + (12 m − a )( 3.9 kN ) + (12 − 2.6 ) m − a  ( 6.3 kN )
a
a

+  2.8 m +  ( 7.9 kN ) + ( 7.3 kN ) = 0
2
2

or
(12 m ) Ay = 128.14 kN ⋅ m − (10.2 kN ) a + (15.2 kN ) a2
(12 m ) Ay = 128.14 kN ⋅ m − ( 2.6 kN ) a
Thus Ay is maximum for the smallest possible value of a:
a =0
(b) The corresponding value of Ay is
( Ay )max = 10.6783 kN, and
ΣFy = 0:
or A = 10.68 kN
or B = 14.72 kN
10.6783 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0
By = 14.7217 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 4, Solution 9.
Free-Body Diagram:
For (TC )max , TB = 0
ΣM O = 0:
(TC )max ( 4.8 in.) − (80 lb )( 2.4 in.) = 0
(TC )
= 40 lb  > [Tmax = 36 lb ]
max

(TC )max = 36.0 lb
For (TC )min , TB = Tmax = 36 lb
ΣM O = 0:
(TC )min ( 4.8 in.) + ( 36 lb )(1.6 in.) − (80 lb )( 2.4 in.) = 0
(TC )min = 28.0 lb
Therefore:
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.
28.0 lb ≤ TC ≤ 36.0 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 10.
Free-Body Diagram:
For Qmin , TD = 0
ΣM B = 0:
( 7.5 kN )( 0.5 m ) − Qmin ( 3 m ) = 0
Qmin = 1.250 kN
For Qmax , TB = 0
ΣM D = 0:
( 7.5 kN )( 2.75 m ) − Qmax ( 0.75 m ) = 0
Qmax = 27.5 kN
Therefore:
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.
1.250 kN ≤ Q ≤ 27.5 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 11.
Free-Body Diagram:
ΣM D = 0:
( 7.5 kN )( 2.75 m ) − TB ( 2.25 m ) + ( 5 kN )(1.5 m ) − Q ( 0.75 m ) = 0
Q = ( 37.5 − 3TB ) kN
ΣM B = 0:
(1)
( 7.5 kN )( 0.5 m ) − ( 5 kN )( 0.75 m ) + TD ( 2.25 m ) − Q ( 3 m ) = 0
Q = ( 0.75 TD ) kN
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
Thus, making 0 ≤ TB ≤ 12 kN in. (1), we have
1.500 kN ≤ Q ≤ 37.5 kN
(3)
And making 0 ≤ TD ≤ 12 kN in. (2), we have
0 ≤ Q ≤ 9.00 kN
(3) and (4) now give:
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.
(4)
1.500 kN ≤ Q ≤ 9.00 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 12.
Free-Body Diagram:
For (WA )min , E = 0
ΣM F = 0:
(WA )min ( 7.5 ft ) + ( 9 lb )( 4.8 ft ) + ( 28 lb )( 3 ft ) − ( 90 lb )(1.8 ft ) = 0
(WA )min = 4.6400 lb
For (WA ) max , F = 0
ΣM E = 0:
(WA )max (1.5 ft ) − ( 9 lb )(1.2 ft ) − ( 28 lb )( 3 ft ) − ( 90 lb )( 7.8 ft ) = 0
(WA )max = 531.20 lb
Thus
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.
4.64 lb ≤ WA ≤ 531 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 13.
Free-Body Diagram:
ΣM D = 0:
( 750 N )( 0.1 m − a ) − ( 750 N )( a + 0.075 m − 0.1 m ) − (125 N )( 0.05 m ) + B ( 0.2 m ) = 0
 87.5 N + 0.2 B 
a=

1500 N


(1)
Using the bounds on B:
B = − 250 N (i.e. 250 N downward) in (1) gives amin = 0.0250 m
B = 500 N (i.e. 500 N upward) in (1) gives amax = 0.1250 m
Therefore:
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.
25.0 mm ≤ a ≤ 125.0 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 14.
Free-Body Diagram:
Note that W = mg is the weight of the crate in the free-body diagram, and that
0 ≤ E y ≤ 2.5 kN
ΣFx = 0:
Ax = 0
ΣM A = 0:
− (1.2 m )(1.2 kN ) − ( 2.0 m )(1.6 kN ) − ( 3.8 m ) E y + ( 6 m )W = 0
or
ΣFy = 0:
or
6W = 4.64 kN + 3.8E y
(1)
Ay − 1.2 kN − 1.6 kN − E y + W = 0
Ay = 2.8 kN + E y − W
(2)
Considering the smallest possible value of E y :
For
E y = 0, W = Wmin = 0.77333 kN
From (2) the corresponding value of Ay is:
Ay = 2.02667 kN ≤ 2.5 kN, which satisfies the constraint on Ay .
For the largest allowable value of E y :
E y = 2.5 kN , W = Wmax = 2.3567 kN
From (2) the corresponding value of Ay is:
Ay = 2.9433 kN ≥ 2.5 kN which violates the constraint on Ay .
Thus
( Ay )max = 2.5 kN. Solving (1) and (2) for W with ( Ay )max = 2.5 kN,
W = Wmax = 1.59091 kN
Therefore:
773.33 N ≤ W ≤ 1590.91 N, or
773.33 N ≤ m(9.81 m/s 2 ) ≤ 1590.91 N, and
78.8 kg ≤ m ≤ 162.2 kg 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 4, Solution 15.
Free-Body Diagram:
Calculate lengths of vectors BD and CD:
BD = (11.2) 2 + (21.0) 2 ft = 23.8 ft
CD =
(a)
(11.2) + (8.4)2 ft = 14.0 ft
 11.2 ft 
 11.2 ft 
(221 lb )(24 ft ) + 
TCD (11.4 ft ) = 0
− (161 lb )(24 ft ) + 
 23.8 ft 
 14.0 ft 
ΣM A = 0 :
TCD = 150.000 lb
(b)
ΣFx = 0:
 11.2 ft 
 11.2 ft 
161 lb − 
 ( 221 lb ) − 
 (150 lb ) + Ax = 0
 23.8 ft 
 14.0 ft 
Ax = 63.000 lb
ΣFy = 0:
or
A x = 63.000 lb
 21.0 ft 
 11.2 ft 
(221 lb) − 
Ay − 

 (150 lb) = 0
 23.8 ft 
 14.0 ft 
Ay = 285.00 lb
A=
TCD = 150.0 lb Ax2 + Ay2 =
or
A y = 285.00 lb
(63)2 + (285) 2 = 291.88 lb
( 63 )
θ = tan −1 285 = 77.535°
Therefore
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 = 292 lb
77.5° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 16.
Free-Body Diagram:
(a)
Equilibrium for ABCD:
ΣM C = 0:
( A cos 60° )(1.6 in.) − ( 6 lb )(1.6 in.) + ( 4 lb )( 0.8 in.) = 0
A = 8.0000 lb
(b)
ΣFx = 0:
or
C x = 8.0000 lb
C y − 6 lb + ( 8 lb ) sin 60° = 0
or C y = −0.92820 lb
C =
60° Cx + 4 lb + ( 8 lb ) cos 60° = 0
or C x = − 8.0000 lb
ΣFy = 0:
A = 8.00 lb
C x2 + C y2 =
or
C y = 0.92820 lb
(8)2 + ( 0.92820 )2 = 8.0537 lb
 − 0.92820 
 = 6.6182°
−8


θ = tan −1 
Therefore:
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.
C = 8.05 lb
6.62° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 17.
Free-Body Diagram:
Equations of equilibrium:
ΣΜ Α = 0:
− ( 330 N )( 0.25 m ) + B sin α ( 0.3 m ) + B cos α ( 0.5 m ) = 0
(1)
ΣFx = 0:
Ax − B sin α = 0
(2)
ΣFy = 0:
Ay − ( 330 N ) + B cos α = 0
(3)
(a) Substitution α = 0 into (1), (2), and (3) and solving for A and B:
B = 165.000 N, Ax = 0, Ay = 165.0 N
or A = 165.0 N , B = 165.0 N
(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:
B = 275.00 N, Ax = 275.00 N, Ay = 330.00 N
A=
Ax2 + Ay2 =
θ = tan −1
Ay
Ax
(275)2 + (330) 2 = 429.56 N
= tan −1
330
= 50.194°
275
∴ A = 430 N
50.2°, B = 275 N
(c) Substituting α = 30° into (1), (2), and (3) and solving for A and B:
B = 141.506 N, Ax = 70.753 N, Ay = 207.45 N, ⇒
A=
Ax2 + Ay2 =
θ = tan −1
Ay
Ax
(70.753) 2 + (207.45) 2 = 219.18 N
= tan −1
207.45
= 71.168°
70.753
∴ A = 219 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.
71.2°, B = 141.5 N
60° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 18.
Free-Body Diagram:
Equations of equilibrium:
ΣΜ Α = 0 :
− (82.5 N ⋅ m ) + B sin α (0.3 m ) + B cos α (0.5 m ) = 0
(1)
ΣFx = 0:
Ax − B sin α = 0
(2)
ΣFy = 0:
Ay + B cos α = 0
(3)
(a) Substituting α = 0 into (1), (2), and (3) and solving for A and B:
B = 165.000 N, Ax = 0, Ay = −165.0 N
or A = 165.0 N , B = 165.0 N
∴ A = 275 N
(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:
B = 275.00 N, Ax = 275.00 N, Ay = 0
, B = 275 N
(c) Substituting α = 30° into (1), (2), and (3) and solving for A and B:
B = 141.506 N, Ax = 70.753 N, Ay = −122.548 N
A = Ax2 + A y2 = (70.753) 2 + (−122.548) 2 = 141.506 N
θ = tan −1
Ay
Ax
= tan −1
122.548
= 60.000°
70.753
∴ A = 141.5 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.
60.0°, B = 141.5 N
60° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
∴ TAB = 300
(b) From free-body diagram of lever BCD
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
∴ C x = −380 N
or
C x = 380 N
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
∴ C y = −240 N
C x2 + C y2 =
Then
C =
and
θ = tan −1 
or
C y = 240 N
( 380 )2 + ( 240 )2 = 449.44 N
 Cy 
− 240 
 = tan −1 
 = 32.276°

 − 380 
 Cx 
or C = 449 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.
32.3° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 20.
Free-Body Diagram:
From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − P ( 75 mm ) = 0
∴ TAB = 1.5P
(1)
ΣFx = 0: 0.6TAB + P − C x = 0
∴ C x = P + 0.6TAB
(2)
Cx = P + 0.6 (1.5P ) = 1.9 P
From Equation (1)
ΣFy = 0: 0.8TAB − C y = 0
∴ C y = 0.8TAB
(3)
C y = 0.8 (1.5P ) = 1.2 P
From Equation (1)
From Equations (2) and (3)
C = C x2 + C y2 =
(1.9 P )2 + (1.2 P )2 = 2.2472P
Since Cmax = 500 N,
∴ 500 N = 2.2472Pmax
or
Pmax = 222.49 lb
or P = 222 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 21.
Free-Body Diagram:
(a)
ΣΜ Βx = 0 :
or Fsp =
 2.4 in. 
−
 A − (0.9 in.)Fsp = 0
 cosα 
8
lb = kx = k (1.2 in.)
cos 30°
Solving for k:
k = 7.69800 lb/in.
k = 7.70 lb/in. (b)
8 lb 
=0
 cos30° 
( 3 lb ) sin 30° + Bx + 
ΣFx = 0:
Bx = −10.7376 lb
or
− ( 3 lb ) cos 30° + B y = 0
ΣFy = 0:
By = 2.5981 lb
or
B=
( −10.7376 )2 + ( 2.5981)2 = 11.0475 lb, and
θ = tan −1
2.5981
= 13.6020°
10.7376
Therefore:
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.
B = 11.05 lb
13.60° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 22.
Free-Body Diagram:
(a)
ΣΜ Βx = 0:
or
 2.4 in. 

 ( 3.6 lb ) − ( 0.9 in.)(12 lb ) = 0
 cosα 
α = 36.9° cosα = 0.80000, or α = 36.870°
(b)
ΣFx = 0:
( 3 lb ) sin 36.870° + Bx + (12 lb ) = 0
or
Bx = −14.1600 lb
ΣFy = 0:
− ( 3.6 lb ) cos 36.870° + By = 0
or
By = 2.8800 lb
B=
( −14.1600 )2 + ( 2.8800 )2 = 14.4499 lb, and
θ = tan −1
2.8800
= 11.4966°
14.1600
Therefore:
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.
B = 14.45 lb
11.50° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 23.
Free-Body Diagram:
From free-body diagram for (a):
− B ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0
ΣΜ A = 0:
Β = −187.50 Ν
ΣFx = 0:
or B = 187.5 N
−187.5 N − 50 N + Ax = 0
Ax = 237.50 N
ΣFy = 0:
Ay − 100 N = 0
Ay = 100.000 N
A=
and:
Ax2 + Ay2 =
θ = tan −1
Ay
Ax
( 237.5)2 + (100 )2 = 257.69 N
= tan −1
100
= 22.834°
237.5
∴ A = 258 N
22.8° From For (b)
ΣΜ A = 0:
− B cos 45° ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0
Β = 265.17 Ν
or B = 265.17 N
45° ∴ A = 253 N
20.2° − ( 265.17 N ) cos 45° − 50 N + Ax = 0
ΣFx = 0:
Ax = 237.50 N
Ay + ( 265.17 ) sin 45° − 100 N = 0
ΣFy = 0:
Ay = −87.504 N
and:
A=
Ax2 + Ay2 =
θ = tan −1
Ay
Ax
(237.50)2 + (−87.504)2 = 253.11 N
= tan −1
87.504
= 20.226°
237.50
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 4, Solution 24.
Free-Body Diagram:
From free-body diagram for (a):
− (100 N )( 0.3 m ) + A ( 0.2 m ) − ( 50 N )( 0.15 m ) − 10 N ⋅ m = 0
ΣΜ B = 0:
A = 237.50 Ν
ΣFx = 0:
or A = 238 N
Bx + 237.5 N − 50 N = 0
Bx = −187.50 N
ΣFy = 0:
By − 100 N = 0
By = 100.000 N
and:
B=
Bx2 + By2 =
θ = tan −1
By
( −187.5)2 + (100 )2 = 212.50 N
= tan −1
Bx
100
= 28.072°
187.5
∴ B = 213 N
28.1° From free-body diagram or (b):
− (100 N )( 0.3 m ) + A cos 45° ( 0.2 m ) − ( 50 N )( 0.15 m ) − 10 N ⋅ m = 0
ΣΜ B = 0:
A = 335.88 Ν
ΣFx = 0:
or A = 336 N
45° ∴ B = 233 N
36.3° Bx + ( 335.88 N ) cos 45° − 50 N = 0
Bx = −187.503 N
ΣFy = 0:
B y + ( 335.88 N ) sin 45° − 100 N = 0
By = −137.503 N
and:
B=
Bx2 + B y2 =
θ = tan −1
By
Bx
(−187.503) 2 + (−137.503)2 = 232.52 N
= tan −1
137.503
= 36.254°
187.503
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 4, Solution 25.
Free-Body Diagram:
Geometry:
x AC = ( 8 in.) cos 20° = 7.5175 in.
y AC = ( 8 in.) sin 20° = 2.7362 in.
⇒ yDA = 9.6 in. − 2.7362 in. = 6.8638 in.
 yDA 
−1  6.8638 
 = tan 
 = 42.397°
x
 7.5175 
 AC 
α = tan −1 
β = 90° − 20° − 42.397° = 27.603°
Equilibrium for lever:
(a)
TAD cos 27.603° ( 8 in.) − ( 60 lb ) (12 in.) cos 20° = 0
ΣM C = 0:
TAD = 95.435 lb
(b)
TAD = 95.4 lb Cx + ( 95.435 lb ) cos 42.397° = 0
ΣFx = 0:
C x = −70.478 lb
C y − 60 lb − ( 95.435 lb ) sin 42.397° = 0
ΣFy = 0:
C y = 124.348 lb
Cx2 + C y2 =
Thus:
C =
and
θ = tan −1
Cy
Cx
(−70.478) 2 + (124.348) 2 = 142.932 lb
= tan −1
124.348
= 60.456°
70.478
∴ C = 142.9 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.
60.5° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 26.
Free-Body Diagram:
(a) a = 2 in.
ΣΜ A = 0:
( 2 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0
E = −3.5698 lb
ΣFx = 0:
or E = 3.57 lb
60.0° A = 27.1 lb
58.4° Ax − 16 lb + ( 3.5698 lb ) cos 60° = 0
Ax = 14.2151 lb
ΣFy = 0:
Ay − 20 lb − ( 3.5698 lb ) sin 60° = 0
Ay = 23.092 lb
A=
(14.2151)2 + ( 23.092 )2 = 27.117 lb
θ = tan −1
23.092
= 58.384°
14.2151
Therefore:
(b) a = 7.5 in.
ΣΜ A = 0:
( 7.5 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0
E = 20.973 lb
ΣFx = 0:
or E = 21.0 lb
60.0° A = 26.6 lb
3.97° Ax − 16 lb − ( 20.973 lb ) cos 60° = 0
Ax = 26.487 lb
ΣFy = 0:
Ay − 20 lb + ( 20.973 lb ) sin 60° = 0
Ay = 1.83685 lb
A=
( 26.487 )2 + (1.83685)2 = 26.551 lb
θ = tan −1
1.83685
= 3.9671°
26.487
Therefore:
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 4, Solution 27.
Free-Body Diagram:
Geometry:
( 2.52 )2 + ( 0.39 )2 = 2.55 m
Distance BC =
Equilibrium for mast:
(a)
 2.52 

 TBC  ( 0.75 m ) − (135 N )( 2.16 m ) − ( 225 N )( 0.66 m ) = 0

 2.55 

ΣΜ A = 0:
TBC = 593.79 N
(b)
or TBC = 594 N  2.52 
Ax − 
 ( 593.79 N ) − 225 N − 135 Ν = 0
 2.55 
ΣFx = 0:
Ax = 586.80 N
 0.39 
Ay + 
 ( 593.79 N ) − 225 N − 135 Ν = 0
 2.55 
ΣFy = 0:
Ay = 269.19 N
Ax2 + Ay2 =
Thus:
A=
and
θ = tan −1
Ay
Ax
( 586.80 )2 + ( 269.19 )2 = 645.60 N
= tan −1
269.19
= 24.643°
586.80
∴ A = 646 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.
24.6° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 28.
Free-Body Diagram:
Geometry:
( 2.52 )2 + ( 0.462 )2 = 2.562 m
Distance BC =
Equilibrium for mast:
(a)
 2.52 

 TBC  ( 0.75 m ) − ( 90 N )( 2.16 m ) − (135 N )( 0.66 m ) = 0

 2.562 

ΣΜ A = 0:
TBC = 384.30 N
(b)
or
TBC = 384 N  2.52 
Ax − 
 ( 384.30 N ) = 0
 2.562 
ΣFx = 0:
Ax = 378.00 N
ΣFy = 0:
 0.462 
Ay + 
 ( 384.30 N ) − 135 N − 90 Ν = 0
 2.562 
Ay = 155.700 N
Ax2 + Ay2 =
Thus:
A=
and
θ = tan −1
Ay
Ax
( 378.00 )2 + (155.700 )2 = 408.81 N
= tan −1
155.700
= 22.387°
378.00
∴ A = 409 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.
22.4° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 29.
Free-Body Diagram:
Geometry:
AB =
Distance
(0.3)2 + (0.125)2 = 0.325 m
Equilibrium for bracket:
0.3 
 0.125 
T  ( 0.175 m ) − 
T  ( 0.225 m ) + T ( 0.075 m ) = 0
 0.325 
 0.325 
(150 N )( 0.225 m ) − 
ΣΜ C = 0:
T = 195.000 N
T = 195.0 N  0.3 
Cx + 
T  (195 N ) = 0
 0.325 
ΣFx = 0:
C x = −180.000 N
ΣFy = 0:
 0.125 
C y − 150 N + 
T  (195 N ) + 195 N = 0
 0.325 
C y = −120.000 N
C x2 + C y2 =
Thus:
C =
and
θ = tan −1
Cy
Cx
( −180 )2 + ( −120 )2 = 216.33 N
= tan −1
120
= 33.690°
180
∴ C = 216 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.
33.7° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 30.
Free-Body Diagram:
Geometry:
Distance BC =
( 4 )2 + ( 3)2 = 5 in.
Distance CD =
(14 )2 + ( 3)2 = 14.3178 in.
Equilibrium for bracket:
ΣΜ A = 0:
4
5


3
5


14
3



T  ( 4 in.) + 
T  ( 9 in.) = 0
14.3178
14.3178




(10 lb )( 9 in.) −  T  ( 4 in.) −  T  ( 9 in.) + 
T = 32.108 lb
or T = 32.1 lb 4
 14 
Ax +   ( 32.108 lb ) − 
 ( 32.108 lb ) = 0
5
 14.3178 
ΣFx = 0:
Ax = 5.7089 lb
ΣFy = 0:
Ay +
3
3

( 32.108 lb ) + 
 ( 32.108 lb ) + 10 lb = 0
5
 14.3178 
Ay = −35.992 lb
Ax2 + Ay2 =
Thus:
A=
and
θ = tan −1
Ay
Ax
( 5.7089 )2 + ( −35.992 )2 = 36.442 lb
= tan −1
35.992
= 80.987°
5.7089
∴ A = 36.4 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.
81.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 31.
Free-Body Diagram:
Geometry:
Distance BC = (7.2)2 + (3)2 = 7.8 in.
Distance CD = (10.8)2 + (3)2 = 11.2089 in.
Equilibrium for bracket:
7.2 
 3 
 10.8

T  ( 4 in.) − 
T  ( 9 in.) + 
T  ( 4 in.)
 7.8 
 7.8 
 11.2089 
(10 lb )( 9 in.) − 
ΣM A = 0:
3


+
T  ( 9 in.) = 0
 11.2089 
or T = 101.0 lb T = 101.014 lb
 7.2 
 10.8 
Ax + 
 (101.014 lb ) − 
 (101.014 lb ) = 0
7.8


 11.2089 
ΣFx = 0:
Ax = 4.0853 lb
ΣFy = 0:
3
 3 


Ay + 
 (101.014 lb ) + 
 (101.014 lb ) + 10 lb = 0
 7.8 
 11.2089 
Ay = − 75.887 lb
Ax2 + Ay2 =
Thus:
A=
and
θ = tan −1
Ay
Ax
( 4.0853)2 + ( − 75.887 )2 = 75.997 lb
= tan −1
75.887
= 86.919°
4.0853
∴ A = 76.0 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.
86.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 32.
Free-Body Diagram:
Geometry:
Distance AD = (0.9)2 + (0.375)2 = 0.975 m
Distance BD = (0.5)2 + (0.375)2 = 0.625 m
Equilibrium for beam:
(a)
ΣM C = 0:
0.375 
 0.375 
T  ( 0.9 m ) − 
T  ( 0.5 m ) = 0
 0.975 
 0.625 
(135 N )( 0.7 m ) − 
T = 146.250 N
(b)
ΣFx = 0:
or T = 146.3 N  0.9 
 0.5 
Cx + 
T  (146.250 N ) + 
T  (146.250 N ) = 0
0.975


 0.625 
C x = − 252.00 N
ΣFy = 0:
 0.375 
 0.375 
Cy + 
T  (146.250 N ) + 
T  (146.250 N ) − 135 N = 0
 0.975 
 0.625 
C y = − 9.0000 N
Thus:
C = C x2 + C y2 =
and
θ = tan −1
Cy
Cx
( − 252 )2 + ( − 9 )2 = 252.16 N
= tan −1
9
= 2.0454°
252
C = 252 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.
2.05° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 33.
Free-Body Diagram:
For both parts (a) and (b)
ΣM D = 0:
− RP − RC x = 0
Cx = −P
ΣFx = 0:
B cosθ − P = 0
B=
ΣFy = 0:
(1)
P
cosθ
(2)
 P 
Cy − 
 sin θ + P = 0
 cosθ 
C y = P ( tanθ − 1)
(a)
(3)
The magnitudes of the forces at B and C are equal:
B = C x2 + C y2
2
2
2
 P 

 = ( − P ) +  P ( tan θ − 1) 
 cosθ 
or
1
= 1 + tan 2 θ − 2 tan θ + 1
cos 2 θ
(
)
continued
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
1
, this gives
cos 2 θ
1
1
=1+
− 2 tan θ , or
2
cos θ
cos 2 θ
1
tan θ = , so
2
tan 2 θ + 1 =
Noting that
θ = 26.565°
θ = 26.6° (b) Using (2)
B=
P
, or
2/ 5
∴ B=
5
P
2
26.6° ∴ C=
5
P
2
26.6° and using (1) and (3)
C x = − P,
C =
Cy = −
P
2
2
P
5
P
 =
2
 2
( −P )2 +  −
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 4, Solution 34.
Free-Body Diagram:
For both parts (a) and (b)
ΣM D = 0:
− RP − RC x = 0
Cx = −P
B cosθ − P = 0
ΣFx = 0:
P
cosθ
B=
ΣFy = 0:
(1)
(2)
 P 
Cy − 
 sin θ + P = 0
 cosθ 
C y = P ( tanθ − 1)
(3)
(a) The magnitude of the reaction at C:
C =
C x2 + C y2
C =
( −P )2 +  P ( tan θ − 1)
2
which is smallest when tan θ = 1 , or
θ = 45.0° (b) Using (2)
B=
P
cos 45°
or B = 2P
45.0o and
C x = − P,
Cy = 0
or C = P
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 4, Solution 35.
Free-Body Diagram:
Equilibrium for bracket:
ΣM C = 0:
− T ( a ) − P ( a ) + (T sin 40° )( 2a sin 40° ) + (T cos 40° )( a + 2a cos 40° ) = 0
T = 0.56624P
ΣFx = 0:
or T = 0.566P Cx + ( 0.56624 P ) sin 40° = 0
C x = 0.36397 P
ΣFy = 0:
C y + 0.56624 P − P + ( 0.56624 P ) cos 40° = 0
Cy = 0
or C = 0.364P
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 4, Solution 36.
Free-Body Diagram:
(a)
ΣM C = 0:
 2a

− aTABD − aP + 
+ a  TABD cosθ = 0
 cosθ

or with TABD = 3P/4 :
3
 2
3
− a P − aP + 
+ 1 P cosθ = 0
4
 cosθ
4
cosθ =
(b)
ΣFx = 0:
ΣFy = 0:
1
, and θ = 70.529°
3
Cx −
83 
 P = 0
3 4 
Cx =
2
P
2
θ = 70.5° 3
1 3 
P + Cy − P +  P  = 0
4
3 4 
Cy = 0
Therefore:
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.
C=
1
P
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 37.
Free-Body Diagram:
Equilibrium for bracket:
(a)
ΣFy = 0:
− 600 N + T = 0
T = 600 N
(b)
ΣM C = 0:
− ( 600 N )( 0.6 m ) + A ( 0.09 m ) = 0
A = 4000 N
ΣFx = 0:
or T = 600 N or A = 4 kN
or B = 4 kN
B − 4000 N = 0
B = 4000 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 4, Solution 38.
Free-Body Diagram:
(
)
W = − ( 52 kg ) ( 9.81 m/s ) = 510.12 N
WC = − ( 80 kg ) 9.81 m/s 2 = 784.80 N
Note that
D
2
(a)
ΣM A = 0:
− ( 784.80 N )( 0.8 m ) cos 30° − ( 510.12 N )( 2.2 m ) cos 30° + B ( 3.5 m ) cos 5° = 0
B = 434.69 N
or B = 435 N
55° (b)
ΣFy = 0:
A cos10° − 784.80 N − 510.12 N + (434.69 N)cos35° = 0
A = 953.33 N
ΣFx = 0:
or A = 953 N
80° P − ( 953.33 N ) sin10° − (434.69 N)sin35° = 0
P = 414.87 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.
or P = 415 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 39.
Free-Body Diagram:
Equilibrium for rod:
(a)
ΣM E = 0:
( 6 lb ) cos 60° ( dOE ) − (T cos 45°) ( dOE ) = 0
T = 4.2426 lb
ΣFx = 0:
(b)
T = 4.24 lb ( 4.2426 lb ) cos 45° − ( 6 lb ) cos 60° − N A sin 45° + N D cos 45° = 0
N A = ND
ΣFy = 0:
(1)
− ( 6 lb ) sin 60° − ( 4.2426 lb ) sin 45° + N A cos 45° + N D cos 45° = 0
N A + N D = 11.5911 lb
(2)
Solving (1) and (2) gives:
N A = N D = 5.7956 lb
Therefore:
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.
N A = 5.80 lb
45° N D = 5.80 lb
45° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 40.
Free-Body Diagram:
Equilibrium for rod:
(a)
ΣM E = 0:
( 6 lb ) cos 60° ( dOE ) − (T cosθ ) ( dOE ) = 0
T =
3
lb
cosθ
(1)
Thus T is minimum when cos θ is maximum:
(b)
(c)
With θ = 0°, (1) gives: T = 3 lb
ΣFx = 0:
θ = 0° T = 3.00 lb 3 lb − ( 6 lb ) cos 60° − N A sin 45° − N D sin 45° = 0
N A = ND
ΣFy = 0:
(2)
− ( 6 lb ) sin 60° + N A cos 45° + N D cos 45° = 0
N A + N D = 7.3485 lb
(3)
Solving (2) and (3) gives:
N A = N D = 3.6742 lb
Therefore:
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.
N A = 3.67 lb
45° N D = 3.67 lb
45° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 41.
Free-Body Diagram:
Equilibrium for bracket:
ΣFy = 0:
T sin 20° − 270 N = 0
T = 789.43 N
Tx = ( 789.43 N ) cos 20° = 741.82 N, and
Note that:
Ty = ( 789.43 N ) sin 20° = 270 N
Thus Ty and the 270-N force form a couple:
270 N ( 0.25 m ) = 67.5 N ⋅ m clockwise
ΣM B = 0:
( 741.82 N )( 0.125 m ) − 67.5 N ⋅ m + FCD ( 0.2 m ) = 0
FCD = −126.138 N
ΣFy = 0:
or
FCD = 126.138 N
FAB − 126.138 N − 741.82 N = 0
FAB = 867.96 N
or
FAB = 867.96 N
Thus, FCD acts to the left, while FAB acts to the right, i.e. these forces are exerted by rollers B and C,
respectively. Rollers A and B exert no force. The forces exerted on the post are the opposites of the forces
exerted by the rollers:
A = D=0
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.
B = 868 N
C = 126.1 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 42.
Free-Body Diagram:
Equilibrium for bracket:
ΣFy = 0:
T sin 30° − 270 N = 0
T = 540 N
Tx = ( 540 N ) cos 30° = 467.65 N, and
Note that:
Ty = ( 540 N ) sin 30° = 270 N
Thus Ty and the 270-N force form a couple:
270 N ( 0.25 m ) = 67.5 N ⋅ m clockwise
ΣM B = 0:
( 467.65 N )( 0.125 m ) − 67.5 N ⋅ m + FCD ( 0.2 m ) = 0
FCD = 45.219 N or FCD = 45.219 N
ΣFy = 0:
FAB + 45.219 N − 467.65 N = 0
FAB = 422.43 N or FAB = 422.43 N
Thus, both FCD and FAB act to the right, i.e. these forces are exerted on the bracket by rollers D and B,
respectively. Rollers A and C exert no force. The forces exerted on the post are the opposites of the forces
exerted on the bracket:
A =C=0
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.
B = 422 N
D = 45.2 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 43.
Free-Body Diagram:
Geometry:
Equation of the slot: y =
x2
4
 2x 
 dy 
= 1.20000
  = slope of slot at C =  4 
dx
  ( x = 2.4 in.)
 C
It follows for the angles that:
α = tan −1 (1.2 ) = 50.194°
θ = 90° − α = 90° − 50.194° = 39.806°
 4.8 − 2.64 
 = 12.6804°
9.6


β = tan −1 
Coordinates for C, D, and E:
( 2.4 )2 = 2.44 in.
xC = 2.4 in.,
yC =
xD = 2.4 in.,
yD = 1.84 in. + (1.6 in.) tan β
4
= 1.84 in. + (1.6 in.) tan12.6804° = 2.20000 in.
xE = 0,
yE = yC + ( 2.4 in.) tan θ
= 1.44 in. + ( 2.4 in.) tan 39.806° = 3.4400 in.
continued
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
With P = 1 lb:
ΣM E = 0:
P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0
(1 lb )( 3.44 in.) − ( Q sin12.6804° )(1.24 in.) − ( Q cos12.6804° )( 2.4 in.) = 0
Q = 1.31616 lb
ΣFx = 0:
P − N C cosθ − Q sin β = 0
1 lb − NC cos 39.806° − (1.31616 lb ) sin12.6804° = 0
NC = 0.92563 lb
ΣFy = 0:
N B + NC sin θ − Q cos β = 0
N B + ( 0.92563 lb ) sin 39.806° − (1.31616 lb ) cos12.6804° = 0
N B = 0.69148 lb
(a) N = 0.691 lb , N = 0.926 lb
B
C
39.8° Q = 1.316 lb
77.3° (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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 44.
Free-Body Diagram:
Geometry:
Equation of the slot: y =
x2
4
 2x 
 dy 
= 1.20000
  = slope of slot at C =  4 
  ( x = 2.4 in.)
 dx C
It follows for the angles that:
α = tan −1 (1.2 ) = 50.194°
θ = 90° − α = 90° − 50.194° = 39.806°
 4.8 − 2.64 
 = 12.6804°
9.6


β = tan −1 
Coordinates for C, D, and E:
2
2.4 )
(
=
= 2.44 in.
xC = 2.4 in.,
yC
xD = 2.4 in.,
yD = 1.84 in. + (1.6 in.) tan β
4
= 1.84 in. + (1.6 in.) tan12.6804° = 2.20000 in.
xE = 0,
yE = yC + ( 2.4 in.) tan θ
= 1.44 in. + ( 2.4 in.) tan 39.806° = 3.4400 in.
continued
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
With Q = 2 lb:
(a)
ΣM E = 0:
P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0
P ( 3.44 in.) − ( 2 lb ) sin12.6804° (1.24 in.) − ( 2 lb ) cos12.6804° ( 2.4 in.) = 0
P = 1.51957 lb
(b)
ΣFx = 0:
or P = 1.520 lb
P − NC cosθ − Q sin β = 0
1.51957 lb − NC cos 39.806° − ( 2 lb ) sin12.6804° = 0
NC = 1.40656 lb
ΣFy = 0:
or NC = 1.407 lb
39.8° N B + NC sin θ − Q cos β = 0
N B + (1.40656 lb ) sin 39.806° − ( 2 lb ) cos12.6804° = 0
N B = 1.05075 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.
or N B = 1.051 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 45.
Note: Weight of block is W = (10 kg)(9.81 m/s2) = 98.1 N
(a) Free-Body Diagram:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 98.1 N = 0
Ay = 98.1 N
or A = 98.1 N Therefore:
ΣM A = 0:
M A − ( 98.1 N )( 0.45 m ) = 0
M A = 44.145 N ⋅ m
or M A = 44.1 N ⋅ m
(b) Free-Body Diagram:
ΣFx = 0:
Ax − 98.1 N = 0
Ax = 98.1 N
ΣFy = 0:
Ay − 98.1 N = 0
Ay = 98.1 N
continued
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
Thus:
A=
Ax2 + Ay2 =
( 98.1)2 + ( 98.1)2 = 138.734 N
or A = 138.7 N
ΣM A = 0:
45° M A + ( 98.1 N )( 0.45 m + 0.1 m ) = 0
M A = 44.145 N ⋅ m
or M A = 44.1 N ⋅ m
(c) Free-Body Diagram:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 98.1 N − 98.1 N = 0
Ay = 196.2 N
ΣM A = 0:
or A = 196.2 N M A − ( 98.1 N )( 0.45 m − 0.1 m ) − ( 98.1 N )( 0.45 m + 0.1 m ) = 0
M A = 88.290 N ⋅ m
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.
or M A = 88.3 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 46.
Free-Body Diagram:
With M = 0 and Ti = T0 = 12 lb
ΣFx = 0:
C x − 12 lb = 0
Cx = 12 lb
ΣFy = 0:
C y − 12 lb = 0
CY = 12 lb
Thus: C =
C x2 + C y2 =
ΣM C = 0:
(12) 2 + (12)2 = 16.9706 lb
or
C = 16.97 lb
or
M C = 2.40 lb ⋅ in.
45° MC – (12 lb)[(1.8 – 1) in.] + (12 lb)[(2 + 1 − 2.4) in.] = 0
M C = 2.40 lb ⋅ in.
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 4, Solution 47.
Free-Body Diagram:
With M = 8 lb.in. and Ti = 16 lb, To = 8 lb
ΣFx = 0:
Cx − 16 lb = 0
Cx = 16 lb
ΣFy = 0:
Cy – 8 lb = 0
Cy = 8 lb
Thus: C =
C x2 + C y2 =
and θ = tan −1
Cy
Cx
(16)2 + (8)2 = 17.8885 lb
 8
= tan −1   = 26.565°
 16 
∴
ΣM C = 0:
C = 17.89
26.6° MC – (16 lb)[(1.8 – 1) in.] + (8 lb)[(2 + 1 –2.4) in.] – 8 lb ⋅ in. = 0
MC = 16.00 lb ⋅ in.
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.
or
MC = 16.00 lb ⋅ in.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 48.
(a)
Free-Body Diagram:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 2 lb − 1lb = 0
Ay = 3 lb
or A = 3 lb ΣM A = 0:
MA – (2 lb)(8 in.) – (1 lb)(12 in.) = 0
M A = 28 lb ⋅ in.
(b)
or M A = 28 lb ⋅ in.
Free Body Diagram:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 2 lb − 1 lb + 1.2 lb = 0
Ay = 1.8 lb
or A = 1.8 lb ΣM A = 0:
MA – (2 lb)(8 in.) – (1 lb)(12 in.) + (1.2 lb)(16 in.) = 0
M A = 8.8 lb ⋅ in.
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.
or M A = 8.8 lb ⋅ in.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 49.
Free-Body Diagram:
Set M A = 20 lb ⋅ in. counter-clockwise to find Fmin:
ΣM A = 0:
20 lb ⋅ in. − (2 lb)(8 in.) – (1 lb)(12 in.) + Fmin (16 in.) = 0
Fmin = 0.5 lb
Set M A = 20 lb ⋅ in. clockwise to find Fmax :
ΣM A = 0:
− 20 lb ⋅ in. − (2 lb)(8 in.) – (1 lb)(12 in.) + Fmin (16 in.) = 0
Fmax = 3 lb
Therefore:
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.
0.5 lb ≤ FE ≤ 3 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 50.
(a)
Free-Body Diagram:
ΣFx = 0:
Ex = 0
ΣFy = 0:
E y − 16.2 kN − 5.4 kN − 18 kN = 0
Ey = 39.6 kN
ΣM E = 0:
M E + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) − (18 kN)(1.5 m) = 0
M E = − 64.8 kN ⋅ m
(b)
or E = 39.6 kN or ME = 64.8 kN. m
Free-Body Diagram:
ΣFx = 0:
Ex = 0
ΣFy = 0:
E y − 16.2 kN − 5.4 kN = 0
Ey = 21.6 kN
ΣM E = 0:
or E = 21.6 kN ME + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) = 0
M E = − 91.8 kN ⋅ m
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.
or ME = 91.8 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 51.
Free-Body Diagram:
ΣM E = 0:
ME + (16.2 kN)x + (5.4 kN)(2.6 m) – T(1.5 m) = 0
ME = (1.5 T − 16.2 x – 14.04) kN ⋅ m
(1)
For x = 0.6 m, (1) gives:
(ME )1 = (1.5 T − 23.76) kN ⋅ m
For x = 7 m, (1) gives:
(ME )2 = (1.5 T − 127.44) kN ⋅ m
(a)
The maximum absolute value of ME is obtained when (ME )1 = − (ME ) and
1.5 T − 23.76 kN = − (1.5 T – 127.44 kN)
T = 50.400 kN
(b)
or T = 50.4 kN For this value of T:
ME = 1.5(50.400) kN ⋅ m − 23.76 kN ⋅ m
= 51.84 kN ⋅ m
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.
or ME = 51.8 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 52.
Free-Body Diagram:
Geometry:
Distance BD =
(1.8)2 + (4)2 = 4.3863 m
Note also that: W = mg = (160 kg)(9.81 m/s2) = 1569.60 N
With MA = 360 N ⋅ m clockwise: (i.e. corresponding to Tmax )
ΣM A = 0:
 1.8 

− 360 N ⋅ m – [(540 N) cos 15o](5.6 m) + 
 Tmax  (4 m) = 0
 4.3863 

Tmax = 1998.79 N
or Tmax = 1.999 kN With MA = 360 N ⋅ m counter-clockwise: (i.e. corresponding to Tmin )
ΣM A = 0:
 1.8 

360 N ⋅ m – [(540 N) cos 15o](5.6 m) + 
 Tmin  (4 m) = 0
 4.3863 

Tmin = 1560.16 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.
or Tmin = 1.560 kN COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 53.
Free-Body Diagram:
(a) Using W = mg:
ΣFx = 0:
− A cos 45° + B sin 45° = 0
B= A
ΣFy = 0:
(1)
A sin 45° + B sin 45° − mg = 0
A+ B =
2mg
2 mg
and
(2)
From (1) and (2) it follows that
2A =
ΣM B = 0:
A=
1
mg
2
 l 

 1

mg   cosθ  + M − 
mg  [l cos(45° − θ )] = 0
2
2


 

(3)
Using that cos(α − β ) = cos α cos β + sin α sin β , (3) gives
 mgl 
 mgl 

 cosθ + M − 
 ( cosθ + sin θ ) = 0
 2 
 2 
 mgl 
M −
 sin θ = 0, and
 2 
sin θ =
(b)
2M
mgl


2(2.7 N ⋅ m)
 = 20.122°
2
 (2kg)(9.81 m/s )(0.8 m) 
θ = sin −1 
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.
 2M 
or θ = sin −1 
  mgl 
or θ = 20.1° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 54.
Free-Body Diagram:
For both parts (a) and (b)
(
)
o 

l cos (θ + 30° ) cos 30° W − l cos θ + 30  T
ΣM D = 0:
(
)
(
)
+ l sin θ + 30o cos 60o  W + l sin θ + 30o  T = 0




or W cos(θ + 60°) + T sin (θ + 30° ) − cos (θ + 30° )  = 0
(a)
For T = 0, (1) gives
cos (θ + 60° ) = 0
(b)
(1)
or θ = 30.0° For T = W , (1) gives:
cosθ cos 60° − sin θ sin 60° + sin θ cos 30° + cosθ sin 30°
− cosθ cos 30° + sin θ sin 30° = 0
or tan θ sin 30° + ( cos 60° + sin 30° − cos 30° ) = 0
Solving for θ :
tan θ = 2 ( cos 30° − 1)
or θ = −150000°
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.
or θ = −15.00° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 55.
Free-Body Diagram:
Using W = mg :
(a)
P( R cosθ + R cosθ ) − mg ( R sin θ ) = 0
ΣM C = 0:
2P = mg tan θ
tan θ =
(b)
2P
mg
or
 2P 

 mg 
θ = tan −1 
With m = 0.7 kg and P = 3 N:


2(3 N)
2 
 (0.7 kg)(9.81 m/s ) 
θ = tan −1 
= 41.145°
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.
or
θ = 41.1° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 56.
Free-Body Diagram:
Using W = mg, and h =
l
tan α
2
M − (mg )(h sin θ ) = 0
ΣM C = 0:
or sin θ =
=
M
mgh
M 2

 cot α 
mg  l


cot α 
or θ = sin −1  2M

mgl 

Note: θ ≤ 90° − α for cord BC to remain taut, (i.e. for TBC > 0).
With l = 1 m, m = 2 kg, and M = 3 N ⋅ m:

2(3 N ⋅ m) cot 30° 

2
 (2 kg)(9.81 m/s )(1 m) 
θ = sin −1 
= 31.984°
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.
or θ = 32.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 57.
Free-Body Diagram:
First note
T = tension in spring = ks
s = elongation of spring
where
( )θ − ( AB )θ
= AB
= 90°
θ 
 90° 
= 2l sin   − 2l sin 

2
 2 
  θ   1 
= 2l sin   − 

  2   2 
  θ   1 
∴ T = 2kl sin   − 

  2   2 
(1)
(a) From free-body diagram of rod BC

 θ 
ΣM C = 0: T l cos    − P ( l sin θ ) = 0
 2 

Substituting T From Equation (1)
  θ   1  
 θ 
2kl sin   − 
 l cos    − P ( l sin θ ) = 0

 2 
  2   2  
  θ   1 

θ 
θ 
 θ 
2kl 2 sin   − 
 cos   − Pl  2sin   cos    = 0

2
2
 2 

  2   2 
Factoring out
θ 
2l cos   , leaves
2
continued
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
  θ   1 
θ 
kl sin   − 
 − P sin   = 0

2
  2   2 
or
1  kl 
θ 
sin   =


2  kl − P 
2


kl
∴ θ = 2sin −1 

 2 ( kl − P ) 
(b) P =
kl
4


 kl  4  
kl
−1  4 
 = 2sin −1 

  = 2sin 

kl
 2 kl − 4 
3 2 
 2  3 kl  
θ = 2sin −1 
(
)
= 2sin −1 ( 0.94281)
= 141.058°
or θ = 141.1° 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 4, Solution 58.
Free-Body Diagrams:
(
)
Note that WE = mg = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣM A = 0:
M − rAT = 0
T =
( 58 N ⋅ m )φ
M
=
rA
0.035 m
Since the torsion spring is unstretched when θ = 0:
( 70 mm )φ = ( 35 mm )θ
φ =
1
θ
2
Therefore:
T =
( 58 N ⋅ m )θ
2(0.035 m)
ΣM B = 0:
rBT − lWE cosθ = 0
( 0.070 m )
( 58 N⋅ m )θ − 0.090 m 98.1N cosθ = 0
(
)(
)
2(0.035 m)
θ = 0.60890 cosθ
Solving for θ numerically:
θ = 0.52645 rad
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.
or θ = 30.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 59.
Free-Body Diagram:
Geometry:
Triangle ABC is isosceles. Thus distance CD = l cos
θ
2
,
Elongation of spring is equal to distance AB:
x = 2l sin
θ
2
,
and T = kx = 2kl sin
(a)
θ
2
.
Equilibrium for rod:
ΣM C = 0:
θ

P ( l cosθ ) − T  l cos  = 0
2

Pl cosθ − kl 2 (2sin
θ
2
θ
cos ) = 0
2
P cosθ − kl sin θ = 0
tan θ =
(b)
P
kl
P
or θ = tan −1    kl 
For p = 2kl :
 2kl 
−1
 = tan (2) = 63.435°
 kl 
θ = tan −1 
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.
or θ = 63.4° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 60.
Free-Body Diagram:
Spring force: Fs = ks = k ( l − l cosθ ) = kl (1 − cosθ )
(a)
l

Fs ( l sin θ ) − W  cosθ  = 0
2

ΣM D = 0:
kl (sin θ − cosθ sin θ ) −
kl (tan θ − sin θ ) −
(b)
W
cosθ = 0
2
W
=0
2
or tan θ − sin θ =
W
2kl
For given values of W = 4 lb, l = 30 in., k = 1.8 lb/ft = 0.15 lb/in.
tan θ − sin θ =
Solving numerically:
4 lb
= 0.44444
2(0.15 lb/in.)(30 in.)
θ = 50.584°
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.
or θ = 50.6° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 61.
Free-Body Diagram:
1. Three non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Determinate (c)
Equilibrium From free-body diagram of bracket:
ΣM A = 0: B (1 m ) − (100 N )( 0.6 m ) = 0
∴ B = 60.0 N
ΣFx = 0: Ax − 60 N = 0
∴ A x = 60.0 N
ΣFy = 0: Ay − 100 N = 0
∴ A y = 100 N
( 60.0 )2 + (100 )2 = 116.619 N
Then
A=
and
θ = tan −1 
 100 
 = 59.036°
 60.0 
∴ A = 116.6 N
59.0° 2. Four concurrent reactions through A
(a)
Improperly constrained (b)
Indeterminate (c)
No equilibrium 3. Two reactions
(a)
Partially constrained (b)
Determinate (c)
Equilibrium 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
From free-body diagram of bracket
ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0
∴ C = 50.0 N ΣFy = 0: A − 100 N + 50 N = 0
∴ A = 50.0 N 4. Three non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Determinate (c)
Equilibrium From free-body diagram of bracket
 1.0 
 = 39.8°
 1.2 
θ = tan −1 
BC =
(1.2 )2 + (1.0 )2 = 1.56205 m
 1.2  
ΣM A = 0: 
 B  (1 m ) − (100 N )( 0.6 m ) = 0
 1.56205  
39.8° ∴ B = 78.1 N
ΣFx = 0: C − ( 78.102 N ) cos 39.806° = 0
∴ C = 60.0 N
ΣFy = 0: A + ( 78.102 N ) sin 39.806° − 100 N = 0
∴ A = 50.0 N 5. Four non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Indeterminate (c)
Equilibrium From free-body diagram of bracket
ΣM C = 0:
(100 N )( 0.6 m ) − Ay (1.2 m ) = 0
∴ Ay = 50 N
or A y = 50.0 N 6. Four non-concurrent non-parallel reactions
(a)
Completely constrained (b)
Indeterminate (c)
Equilibrium continued
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
From free-body diagram of bracket
ΣM A = 0: − Bx (1 m ) − (100 N )( 0.6 m ) = 0
∴ Bx = −60.0 N
or B x = 60.0 N
or A x = 60.0 N
ΣFx = 0: − 60 + Ax = 0
∴ Ax = 60.0 N
7. Three non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Determinate (c)
Equilibrium From free-body diagram of bracket
ΣFx = 0: Ax = 0
ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0
∴ C = 50.0 N
or C = 50.0 N ΣFy = 0: Ay − 100 N + 50.0 N = 0
∴ Ay = 50.0 N
∴ A = 50.0 N 8. Three concurrent, non-parallel reactions
(a)
Improperly constrained (b)
Indeterminate (c)
No equilibrium 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 4, Solution 62.
Free-Body Diagram:
1. Three non-concurrent, non-parallel reactions
Completely constrained (a)
(b)
Determinate (c)
Equilibrium From free-body diagram of plate
ΣM A = 0: C ( 30 in.) − 50 lb (15 in.) = 0
C = 25.0 lb ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 50 lb + 25 lb = 0
Ay = 25 lb
A = 25.0 lb 2. Three non-current, non-parallel reactions
Completely constrained (a)
(b)
Determinate (c)
Equilibrium From free-body diagram of plate
B = 0
ΣFx = 0:
ΣM B = 0:
( 50 lb )(15 in.) − D ( 30 in.) = 0
D = 25.0 lb ΣFy = 0: 25.0 lb − 50 lb + C = 0
C = 25.0 lb continued
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
3. Four non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Indeterminate (c)
Equilibrium From free-body diagram of plate
ΣM D = 0: Ax ( 20 in.) − ( 50 lb )(15 in.)
∴ A x = 37.5 lb
∴ D x = 37.5 lb
ΣFx = 0: Dx + 37.5 lb = 0
4. Three concurrent reactions
(a)
Improperly constrained (b)
Indeterminate (c)
No equilibrium 5. Two parallel reactions
(a)
Partial constraint (b)
Determinate (c)
Equilibrium From free-body diagram of plate
ΣM D = 0: C ( 30 in.) − ( 50 lb )(15 in.) = 0
C = 25.0 lb ΣFy = 0: D − 50 lb + 25 lb = 0
D = 25.0 lb 6. Three non-concurrent, non-parallel reactions
(a)
Completely constrained (b)
Determinate (c)
Equilibrium From free-body diagram of plate
ΣM D = 0: B ( 20 in.) − ( 50 lb )(15 in.) = 0
B = 37.5 lb
ΣFx = 0: Dx + 37.5 lb = 0
D x = 37.5 lb
ΣFy = 0: Dy − 50 lb = 0
D y = 50.0 lb
or D = 62.5 lb
53.1° continued
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
7. Two parallel reactions
(a)
Improperly constrained (b)
Reactions determined by dynamics (c)
No equilibrium 8. Four non-concurrent, non-parallel reactions
Completely constrained (a)
(b)
Indeterminate (c)
Equilibrium From free-body diagram of plate
ΣM D = 0: B ( 30 in.) − ( 50 lb )(15 in.) = 0
B = 25.0 lb ΣFy = 0: Dy − 50 lb + 25.0 lb = 0
D y = 25.0 lb ΣFx = 0: Dx + C = 0
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 4, Solution 63.
Free-Body Diagram:
Note that the wheel is a three-force body, and let point D be the intersection of the three forces.
With a = 75 mm, it follows from the force triangle that
A
P
90 N
=
=
125 100
75
Then:
P = 100 ( 90 N ) = 120 N
75
or P = 120.0 N A = 125 ( 90 N ) = 150 N, and
75
(100 )
θ = tan −1 75 = 36.870°
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.
or A = 150.0 N
36.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 64.
Free-Body Diagram:
Note that the wheel is a three-force body, and let point D be the intersection of the three forces.
From the force triangle it follows that
90 N
=
a
A = 90
A
(100) + ( a )
2
(100)
2
(a )
2
2
+1
(1)
Setting A = 180 N and solving for a:
90 N
=
a
180 N
( )
2
2
a =
a=
( )
(100)2 + a 2
(100) + a
2
4
(100)
3
2
= 57.735 mm
From (1) it follows that A will decrease as a increases. Therefore the value of a
calculated is a lower limit:
a ≥ 57.7 mm 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 4, Solution 65.
Free-Body Diagram:
Geometry:
EF = ( 2.4 tan 30° + 0.9 ) in.
AF = EF tan 30°
tan φ =
0.9
( 2.4 tan30° + 0.9 ) tan 30° + 2.4
φ = 13.6019°
Equilibrium: force triangle
Using the law of sines on the force triangle:
Fsp
3 lb
B
=
=
sin120° sin ( 60° − φ ) sinφ
B = 11.05 lb
Fsp = 9.2376 lb
(a) Fsp = kx
9.2376 lb = k (1.2 in.)
Solving for k:
k = 7.698 lb/in.
(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.
or k = 7.70 lb/in. B = 11.05 lb
13.60° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 66.
Free-Body Diagram:
Note that the bent rod is a three-force body. D is the point where the lines of action of the three
forces intersect.
(a) The requirement B = C means that the force triangle must be isosceles. Therefore θ = φ .
Which leads to the force triangle shown.
From the geometry it follows that
tan θ = 1
2
or θ = 26.6° θ = 26.565°
(b) From the force triangle:
2B sin θ = P, or with sin θ =
B=C =
P
=
 1 
2

 5
1
5
5P
2
Therefore:
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.
B=
5P
2
26.6° C=
5P
2
26.6° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 67.
(a) Free-Body Diagram: (α = 90° )
The bracket is a three-force body and A is the intersection of the lines of action of the three forces.
 6
 = 26.565°
 12 
θ = tan −1 
From the force triangle:
A = (75 lb) cot θ
= (75 lb) cot 26.565°
= 150.000 lb
C =
75 lb
75 lb
=
= 167.705 lb
sin θ
sin 26.565°
or A = 150.0 lb
or C = 167.7 lb
63.4° continued
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
(b) Free-Body Diagram: (α = 45° )
Let E be the intersection of the lines of action of the three
forces acting on the bracket.
Triangle ABE is isosceles and therefore
AE = AB = 16 in.
From triangle CEF
 CF 
−1  6 
 = tan   = 12.0948°
 EF 
 28 
θ = tan −1 
From force triangle:
β = 180° − 135° − θ
= 180° − 135° − 12.0948°
= 32.905°
Using the law of sines:
A
C
75 lb
=
=
sin 32.905° sin135° sin12.0948°
Solving for A and C:
A = 194.452 lb
C = 253.10 lb
or A = 194.5 lb
or C = 253 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.
77.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 68.
Free-Body Diagram:
Let C be the intersection of the lines of action of the three
forces acting on the girder
From triangle BCD:
h = ( 40 ft ) cot 30° = 69.282 ft
From triangle ACD:
 69.282 ft 
 = 73.8979°
 20 ft 
α = tan −1 
or α = 73.9° From force triangle:
β = 90° − α = 90° − 73.8979° = 16.1021°
γ = 180° − 30° − β = 180° − 30° − 16.1021° = 133.898°
Using the law of sines:
TA
TB
6000 lb
=
=
sin 30° sin16.1021° sin133.898°
Solving for TA and TB :
TA = 4163.3 lb,
TB = 2309.4 lb
or TA = 4160 lb and TB = 2310 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 69.
Free-Body Diagram:
From the free-body diagram:
 (900 mm)sin 50° 
 = 82.726°
88 mm


θ = tan −1 
From the force triangle:
FN = (130 N ) tan θ = (130 N ) tan 82.726° = 1018.48 N
Force on nail is therefore
RB =
or FN = 1.018 kN
130 N
130 N
=
= 1026.74 N
cosθ
cos82.726°
or R B = 1.027 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.
82.7° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 70.
Free-Body Diagram:
From force triangle:
 2600 N 
 = 83.636°
 290 N 
θ = tan −1 
From free-body diagram:
tan θ =
l =
(900 mm)sin50°
l
( 900 mm ) sin 50° = 76.894 mm
tan 83.636°
or l = 76.9 mm 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 4, Solution 71.
Free-Body Diagram:
We note from the free-body diagram that the ladder is a three-force body. Point C in the free-body diagram is
the intersection between the lines of action of the three forces.
It then follows that:
sin θ =
1.75 m
( 9.2 − 1.8) m
θ = 13.6793°
Also:
 9.2

− 1.8  m = 2.8 m
AG = 
 2

 9.2 
BD = 
m  cosθ = ( 4.6 m ) cosθ
 2 
CD = CG + GD
=
=
AG
sin θ
2.8
sin θ
+ BG sin θ
+
9.2
2
sin θ
continued
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
Then
2.8
+ 4.6sin13.6793°
CD
tan φ =
= sin13.6793°
BD
4.6cos13.6793°
φ = 70.928°
Now using the law of sines on the force triangle:
FW
B
W
=
=
sin(90° − φ ) sin θ
sin [φ + (90° − θ )]
FW
B
W
=
=
cos φ
sin θ
sin(φ − θ )
(a) From the law of sines and noting that
W = ( 53 kg ) (9.81 m/s 2 ) = 519.93 N
FW
519.93 N
=
cos 70.928° cos(70.928° − 13.6793°)
FW = 314.03 N
or FW = 314 N
76.3° or B = 227 N
70.9° (b) In the same way
B
519.93 N
=
sin13.6793° cos(70.928° − 13.6793°)
B = 227.28 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 4, Solution 72.
Free-Body Diagram:
We note from the free-body diagram that the ladder is a three-force body. Point D in the free-body diagram is
the intersection between the lines of action of the three forces.
It then follows that:
BD =
9.2 m
cosθ = ( 4.6 m ) cosθ =, but also that
2
1.75 

BD = 1.75 tan θ +
m
tan θ 

Therefore
BD =
9.2 m
cosθ = ( 4.6 m ) cosθ
2
This implies:
4.6 cosθ = 1.75 tan θ +
1.75
tan θ
92sin θ = 35(1 + tan 2 θ ) = 35 sec2 θ
92sin θ cos 2 θ = 35
continued
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
Solving numerically for the smallest possible root:
θ = 31.722°
Then
sin 31.722° =
1.75
9.2 − a
a = 5.8717 m
or a = 5.87 m (b) From the force triangle
FW =
W
cos 31.722°
FW = 611.24 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.
or FW = 611 N
58.3° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 73.
Free-Body Diagram:
Let D be the intersection of the lines of action
of the three forces acting on the tool.
From the free-body diagram:
yDC =
(14.4 in.) cos 35° = 32.409 in.
xBC
=
tan 20°
tan 20°
yBC = (14.4 in.) sin 35° = 8.2595 in.


3.6 in.

 yDC − yBC − 1.8 in. 
α = tan −1 


3.6 in.
= tan −1 

 ( 32.409 − 8.2595 − 1.8 ) in. 
= 9.1505°
From the force triangle, using the law of sines:
20 lb
A
=
sin α sin 20°
or A = 43.0 lb
80.8° on tool, and A = 43.0 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.
80.8° on rim of can. COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 74.
Free-Body Diagram:
Let E be the intersection of the lines of action
of the three forces acting on the tool.
From the free-body diagram, using law of sines:
6 in. + ( 0.76 in.) tan 35°
L
=
sin 95°
sin 30°
L = 13.0146 in.
Also:
yBD = L − y AE − 0.88 in.
= 13.0146 in. −
0.76 in.
− 0.88 in.
cos 35°
= 11.2068 in.
And
 1.8 in. 

 yBD 
α = tan −1 
 1.8 in. 
= tan −1 
 = 9.1247°
 11.2086 in. 
Then from the force triangle and using the law of sines:
B
14 lb
=
sin150° sin 9.1247°
Solving for B:
B = 44.141 lb, or
on the member B = 44.141 lb
80.9°, and on the lid B = 44.1 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.
80.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 75.
Free-Body Diagram:
Based on the roller having impending motion to the left, the only contact
between the roller and floor will be at the edge of the tile.
First note
(
)
W = mg = ( 20 kg ) 9.81 m/s 2 = 196.2 N
From the geometry of the three forces acting on the roller
 92 mm 
 = 23.074°
 100 mm 
α = cos −1 
and
θ = 90° − 30° − α
= 60° − 23.074
= 36.926°
Applying the law of sines to the force triangle,
W
P
=
sin θ
sin α
or
196.2 N
P
=
sin 36.926° sin 23.074°
∴ P = 127.991 N
or P = 128.0 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.
30° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 76.
Free-Body Diagram:
Based on the roller having impending motion to the right, the only
contact between the roller and floor will be at the edge of the tile.
First note
(
W = mg = ( 20 kg ) 9.81 m/s 2
)
= 196.2 N
From the geometry of the three forces acting on the roller
 92 mm 
 = 23.074°
 100 mm 
α = cos −1 
and
θ = 90° + 30° − α
= 120° − 23.074°
= 96.926°
Applying the law of sines to the force triangle,
W
P
=
sin θ
sin α
or
196.2 N
P
=
sin 96.926° sin 23.074
∴ P = 77.460 N
or P = 77.5 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.
30° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 77.
Free-Body Diagram:
Note that the clamp is a three-force body. D is the intersection
of the lines of action of the three forces.
From the free-body diagram it follows that:
y AD = (4.2 in.) tan 78o = 19.7594 in.
yBD = y AD − 2.8 in.
= (19.7594 − 2.8) in. = 16.9594 in.
Then
 yBD 

 7.8 in. 
θ = tan −1 
 16.9594 in. 
= tan −1 
 = 65.3013° , and
 7.8 in. 
α = 90° − θ − 12°
= 90° − 65.3013° − 12° = 12.6987°
(a) Using the maximum allowable compressive force on
the clamp:
( RB ) y = RB sin θ = 40 lb
or RB =
40 lb
= 44.028 lb
sin 65.301°
or R B = 44.0 lb
(b)
65.3° Using the law of sines for the force triangle:
RB
NA
T
=
=
sin12° sin α
sin(90° + θ )
44.028 lb
NA
T
=
=
sin12°
sin12.6987° sin155.301°
which gives:
N A = 46.551 lb
(c)
T = 88.485 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.
or N A = 46.6 lb
or T = 88.5 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 78.
Free-Body Diagram: (for hoist AD)
Note that the hoist AD is a three-force body.
E is the intersection between the lines of action of the three forces
acting on the hoist.
From the free-body diagram:
x AE = (48 in.) cos 30° = 41.5692 in.
y AD = (48 in.)sin 30° = 24 in.
yBE = x AD tan 75° = (41.5692 in.)tan75°
= 155.1384 in.
Then:
 yBE − 16 in. 
−1  139.1384 
 = tan 

x
 41.5692 
AD


= 73.36588°
α = tan −1 
β = 75° − α = 75° − 73.36588° = 1.63412°
θ = 180° − 15° − β = 165° − 1.63412° = 163.366°
From the force triangle and using the law of sines:
260 lb
B
A
=
=
sinβ
sin θ
sin15°
260 lb
B
A
=
=
sin 1.63412° sin 163.366° sin 15°
Solving for A and B:
(a)
(b)
B = 2609.9 lb
or B = 2.61kips
75.0° or A = 2.36 kips
73.4° A = 2359.8 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 79.
Free-Body Diagram:
Note that the member is a three-force body. In the free-body diagram, D is the intersection between the lines
of action of the three forces.
(a)
From the force triangle:
T − 110 N 3
=
T
4
3T = 4T − 440 N
T = 440 N (b)
From the force triangle:
C
5
=
T
4
C =
5
5
T = (440 N) = 550 N
4
4
or C = 550 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.
36.9° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 80.
Free-Body Diagram:
Note that the member is a three-force body. In the free-body diagram, E is the intersection between the lines
of action of the three forces.
From the free-body diagram:
 15 
 = 61.928°
 8
α = tan −1 
 8
 = 33.690°
 12 
β = tan −1 
From the force triangle:
α − β = 61.928° − 33.690° = 28.238°
180° − α = 180° − 61.928° = 118.072°
Using the law of sines:
T − 18 lb
T
C
=
=
sin(α − β ) sin β
sin(180° − α )
T − 18 lb
T
C
=
=
sin ( 22.238° ) sin ( 33.690° ) sin(118.072°)
Then:
(T − 18 lb ) sin ( 33.690° ) = T sin ( 28.238° )
T = 122.414 lb
and
or T = 122.4 lb (122.414 lb ) sin (118.072°) = C sin ( 33.690° )
C = 194.723 lb
or C = 194.7 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.
33.7° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 81.
Free-Body Diagram:
Note that the peavey is a three-force body.
In the free-body diagram, D is the intersection of the lines
of action of the three forces acting on the peavey.
It then follows:


44 in.
 = 40.2364°
 44 in. + 8 in. 
β = tan −1 
α = 45° − β = 45° − 40.2364° = 4.7636°
From the force triangle, using the law of sines:
W
C
A
=
=
sin β
sin α
sin135°
80 lb
C
A
=
=
sin 40.236° sin 4.7636° sin135°
Solving for C and A:
(a)
(b)
C = 10.2852 lb
or C = 10.29 lb
45.0° or A = 87.6 lb
85.2° A = 87.576 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 82.
Free-Body Diagram:
Note that the peavey is a three-force body.
In the free-body diagram, D is the intersection of the lines
of action of the three forces acting on the peavey.
It then follows:

44 in. 

 DC + 8 in. 
β = tan −1 
where DC = ( 44 in. + a ) tan 30°
 R 
a=
−R
 tan 30° 
 4 in. 
=
 − 4 in.
 tan 30° 
= 2.9282 in.
Then:
DC = ( 46.9282 in.) tan 30° = 27.0940 in.
and

44 in. 
 = 51.4245°
 35.0940 in. 
β = tan −1 
α = 60° − β = 60° − 51.4245° = 8.5755°
Now from the force triangle, using the law of sines:
W
C
A
=
=
sin β sin α sin120°
80 lb
C
A
=
=
sin 51.424° sin 8.5755° sin120°
Solving for C and A:
(a)
C = 15.2587 lb
or C = 15.26 lb
(b)
A = 88.621 lb
or A = 88.6 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.
81.4° 30.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 83.
Free-Body Diagram:
From the free-body diagram, the member AB is a threeforce body. Let D be the intersection of the lines of action of
the three forces acting on AB. Then, using triangle BCD:
CD = ( 250 mm ) tan 60° = 433.01 mm
Also:
AF = AE + EF = AE + CD
= (300 + 433.01) mm
= 733.01 mm
FD
250
= tan −1
AF
733.01
= 18.8324°
θ = tan −1
From the force triangle
α = 180° − 30° − 18.8324°
= 131.168°
Using the law of sines
A
B
330 N
=
=
sin 30° sin18.8324° sin131.168°
Solving for A and B:
A = 219.19 N,
B = 141.507 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.
or A = 219 N
71.2° or B = 141.5 N
60.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 84.
Free-Body Diagram:
From the free-body diagram it follows that
tan θ =
9.6 − 8sin20°
8cos20°
θ = 42.397°
Also:
BE = 20sin 20° + ( 20cos 20° ) tan 42.397° = 12sin 20° (12cos 20° ) tan φ
φ = 60.456°
Then using the law of sines on the force triangle:
TAD
C
60 lb
=
=
sin ( 90° − φ ) sin ( 90° + θ ) sin (φ − θ )
TAD
C
60 lb
=
=
cos 60.456° cos 42.397° sin18.059°
(a)
TAD = 95.438 lb
(b)
C = 142.935 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.
or TAD = 95.41 lb or C = 142.935 lb
60.5° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 85.
Free-Body Diagram:
(a)
Using the law of cosines on triangle ABC:
2
2
R 2 = ( 2R ) + ( 2R ) − 2 ( 2R )( 2R ) cosθ
1 = 8 − 8cosθ
cosθ =
7
8
θ = 28.955°
Also,
2R cos (θ + α ) = R cos α
2R ( cosθ cos α − sin θ sin α ) = R cos α
tan α =
2 cosθ − 1 2cos 28.955° − 1
=
2 sin θ
2 sin 28.955°
or α = 37.8° α = 37.761°
(b)
From the free-body diagram:
2R
R
=
sin φ sin θ
sin φ = 2sin θ = 2 sin 28.955°
φ = 75.522°
Now using the law of sines on the force triangle:
NA
NB
W
=
=
sin [90° − (φ − α )] sin 90° − (θ + α )  sin (θ + α ) + (φ − α ) 
NA
NB
mg
=
=
cos(φ − α ) cos (θ + α ) sin (θ + φ )
NA
NB
mg
=
=
cos 37.762° cos 66.716° sin104.478°
Solving for N A and N 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.
N A = 0.816 mg
66.7° N B = 0.408 mg
37.8° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 86.
Free-Body Diagram:
(a)
Note that the rod is a three-force body. Using the law of cosines on triangle ABC:
2
R 2 = ( 2R ) + L2 − 2 ( 2 R ) L cosθ
cosθ =
3R 2 + L2
4RL
(1)
Also,
L
cos 45° = 2 R cos(θ + 45°)
2
L
cos 45° = 2R(cosθ cos 45° − sin θ sin 45°)
2
L
= cosθ − sin θ
4R
Using (1) and that sin θ = 1 − cos 2 θ
sin θ =
( 4 RL )2 − ( 3R 2 + L2 )
4RL
2
, this gives
continued
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
L
3R 2 + L2
=
−
4R
4RL
(
( 4RL )2 − ( 3R 2 + L2 )
2
4RL
)
( 4RL )2 − ( 3R 2 + L2 )
L2 − 3R 2 + L2 = −
2
Squaring both sides and simplifying:
(
)
9R 4 = 16R 2 L2 − 9R 4 + 6R 2 L2 + L4 , or
L4 − 10R 2 L2 + 18R 4 = 0
Solving for L2 :
(
)
(
)
L2 = 5 ± 7 R 2 , and taking the largest root
L2 = 5 + 7 R 2
(b)
or L = 2.77 R Using the value of L obtained in (a) and (1)
cosθ =
(
)
3R 2 + 5 + 7 R 2
(
4R 5 + 7
) R
1/2
θ = 15.7380°
Now using the law of sines on triangle ABC in the free body diagram:
2R
R
=
sin φ sin θ
sin φ = 2sin θ = 2 sin15.7380°
φ = 32.852°
Now using the law of sines on the force triangle:
NA
NB
mg
=
=
cos (135° − 32.852° ) sin ( 45° − 15.7380° ) sin (15.7380° + 32.852° )
Solving for N A and N 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.
N A = 1.303 mg
60.7° N B = 0.652 mg
12.15° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 87.
Free-Body Diagram:
Note that the wheel is a two-force body and therefore the
force at C is directed along CA and perpendicular to the
incline.
The wheelbarrow is a three-force body. Let D be the
intersection of the lines of action of the three forces acting
on the wheelbarrow. Then, using the triangle DEG
DE = EG tan 72° = ( 8 in.) tan 72° = 24.6215 in.
DF = DE − EF = 24.6215 in. − 9 in. = 15.6215 in.
Using triangle DFB:
φ = tan −1
FB
 40 
= tan −1 
 = 68.667°
DF
 15.6215 
From the force triangle:
α = φ − 18° = 68.667° − 18° = 50.667°
β = 180° − 50.667° − 18° = 111.333°
Using the law of sines:
B
C
120 lb
=
=
sin18° sin 50.667° sin111.333°
B = 39.809 lb,
C = 9.644 lb
(a)
(b)
Noting that the force on each handle is B/2:
1
B = 19.90 lb
2
39.3° C = 99.6 lb
72.0° Reaction at C:
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 4, Solution 88.
Free-Body Diagram:
From the free-body diagram:
a = ( 40 in.) tan θ − 32 in. =
8 in. − ( 23 in.) tan18°
tan18°
θ = 40.048°
Then,
φ = 90° − 18° − 40.048° = 31.952°
Now, using the law of sines on the force triangle:
2( B/2)
C
120 lb
=
=
sin18° sin 31.952° sin (162° − 31.952° )
and solving for B/2 and C:
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)
1
B = 24.2 lb
2
58.0° (b)
C = 83.0 lb
72.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 89.
Free-Body Diagram:
Note that the rod is a three-force body. In the free-body
diagram, E is the intersection between the lines of action
of the three forces.
Using triangle ACF in the free-body diagram:
yCF = d tan θ
From triangle CEF:
xFE = yCF tan θ = d tan 2 θ
and from triangle AGE:
cosθ =
Noting that 1 + tan 2 θ = sec 2 θ =
d + xFE
( L2 )
=
d + d tan 2 θ
( L2 )
(1)
1
cos 2 θ
(1) gives
cosθ =
2d  1 
, or
L  cos 2 θ 
cos3 θ =
2d
L
Using the given values of d = 2.8 in., and L = 10 in.
cos3 θ =
2(2.8 in.)
= 0.56
10 in.
cosθ = 0.82426
θ = 34.486°
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.
or θ = 34.5° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 90.
Free-Body Diagram:
As shown in the free-body diagram of the slender rod AB, the three forces
intersect at C. From the force geometry
tan β =
xGB
y AB
where
y AB = L cosθ
and
xGB =
1
L sin θ
2
1 L sin θ
1
∴ tan β = 2
= tan θ
L cosθ
2
or tan θ = 2 tan β 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 4, Solution 91.
Free-Body Diagram
(a) As shown in the free-body diagram. of the slender rod AB, the three
forces intersect at C. From the geometry of the forces
xCB
yBC
tan β =
where
xCB =
1
L sin θ
2
yBC = L cosθ
and
∴ tan β =
1
tan θ
2
or
tan θ = 2 tan β
For
β = 25°
tan θ = 2 tan 25° = 0.93262
∴ θ = 43.003°
or θ = 43.0° (
)
W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N
(b)
From force triangle
A = W tan β
= ( 98.1 N ) tan 25°
= 45.745 N
or A = 45.7 N
and
B=
W
98.1 N
=
= 108.241 N
cos β
cos 25°
or B = 108.2 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.
65.0° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 92.
Free-Body Diagram:
Note that the rod is a three-force body. In the free-body diagram, D is
the intersection between the lines of action of the three forces.
Using triangle BCE:
a = BE = BC sin θ
and from triangle BCD
BC = BD sin θ
Then a = BD sin 2 θ
Also from triangle ABD
BD = L sin θ , so
a = L sin 3 θ
1
−1  a  3
or θ = sin   L
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 4, Solution 93.
Free-Body Diagram:
Note that the athlete is a three-force body. From the free-body diagram
1
a tan θ
4a
2
=
tan θ
tan (θ + φ )
3a +
tan (θ + φ ) =
4 tan θ
1
3 + tan θ
2
(1)
From the force triangle, using the law of sines
FH
W
=
or, using FH = 0.8W
sin θ
sin 180° − (θ + φ ) 
sin (θ + φ ) = 1.25sin θ
Now, using (1)


4 tan θ 
−1
θ + φ = tan 
 = sin (1.25sin θ )
1
 3 + tan θ 
2


−1 
Solving numerically for θ
θ = 15.04° 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 4, Solution 94.
Free-Body Diagram:
Note that the rod is a three-force body. In the free-body diagram, E is
the intersection of the lines of action of the three forces.
(a)
Using triangle DBC which is isosceles
DB = a
and using triangle BDE
ED = DB tan 2θ = a tan 2θ
From triangle GED
ED =
a tan 2θ =
( L − a ) , and therefore
tan θ
L−a
, or
tan θ
a (tan θ tan 2θ + 1) = L
(1)
From triangle BCD:
a=
L
a
1.25 L
2
cos θ
= 1.6 cos θ
(2)
Using (2) in (1):
1.6 cos θ = 1 + tan θ tan 2θ
(3)
(
2
)
2 1 − cos θ
sin θ sin 2θ
sin θ 2sin θ cosθ
, (3) gives
=
=
Noting that tan θ tan 2θ =
cosθ cos 2θ
cosθ 2cos 2 θ − 1
2cos 2 θ − 1
1.6 cosθ = 1 +
(
2 1 − cos 2 θ
2
) , or
2 cos θ − 1
3
3.2 cos θ − 1.6 cos θ − 1 = 0
or θ = 23.5° Solving numerically, θ = 23.515°
(b) Substituting into (2) for L = 8 in.,
a=
5 ( 8 in.)
= 5.4528°
8 cos 23.515°
or a = 5.45 in. 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 4, Solution 95.
Free-Body Diagram:
The forces acting on the three-force member intersect at D.
(a) From triangle ACO
 r 
−1  1 
 = tan   = 18.4349°
 3r 
3
θ = tan −1 
tan θ =
(b) From triangle DCG
∴ DC =
and
or θ = 18.43° r
DC
r
r
=
= 3r
tan θ
tan18.4349°
DO = DC + r = 3r + r = 4r
 yDO 

 x AG 
α = tan −1 
where
yDO = ( DO ) cosθ = ( 4r ) cos18.4349°
= 3.4947r
and
x AG = ( 2r ) cosθ = ( 2r ) cos18.4349°
= 1.89737r
 3.4947r 
∴ α = tan −1 
 = 63.435°
 1.89737r 
where
90° + (α − θ ) = 90° + 45° = 135.00°
Applying the law of sines to the force triangle,
mg
R
= A
sin 90° + (α − θ )  sin θ
∴ RA = ( 0.44721) mg
Finally,
P = RA cos α
= ( 0.44721mg ) cos 63.435°
= 0.20000mg
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.
or P =
mg
5
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 96.
Free-Body Diagram:
ΣM A = 0:
( 450 mm ) i  × D + (150 mm ) i  × ( −180 N ) j + ( 250 mm ) i  × ( − 300 N ) k  = 0
( 450 mm ) Dyk − (450 mm)Dz j − (150 mm)(180 N)k + (250 mm)(300 N) j = 0
Setting the coefficients of the unit vectors equal to zero:
k:
Dy (450 mm) − (180 N )(150 mm ) = 0,
or
Dy = 60.000 N
i:
( 300 N )( 250 mm ) − Dz ( 450 mm ) = 0,
or
Dz = 166.667 N
D = ( 60.0 N ) j + (166.7 N ) k ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay + 60.000 N − 180 N = 0, or Ay = 120.000 N
ΣFz = 0:
Az + 166.667 N − 300 N = 0, or Az = 133.333 N
A = (120.0 N ) j + (133.3 N ) k 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 4, Solution 97.
Free-Body Diagram:
ΣFx = 0:
Dx = 0
ΣM D = 0:
( − 7 in.) i  × C + ( 2 in.) i + ( 3 in.) k  × ( 530 lb ) j + ( −192 lb ) k 
+ ( −3 in.) i + ( 6 in.) j × ( −96 lb ) j + ( 265 lb ) k  = 0
or − ( 7 in.) C yk + ( 7 in.) C z j + ( 2 in.)( 530 lb ) k + ( 2 in.)(192 lb ) j − ( 3 in.)( 530 lb ) i
+ ( 3 in.)( 96 lb ) k + ( 3 in.)( 265 lb ) j + ( 6 in.)( 265 lb ) i = 0
Setting the coefficients of the unit vectors to zero:
k:
− C y ( 7 in.) + ( 96 lb )( 3 in.) + ( 530 lb )( 2 in.) = 0,
or
C y = 192.571 lb
j:
C z ( 7 in.) + ( 265 lb )( 3 in.) + (192 lb )( 2 in.) = 0,
or
C z = −168.429 lb
C = (192.6 lb ) j − (168.4 lb ) k Then:
ΣFy = 0:
192.571 lb − 96 lb + Dy + 530 lb = 0,
ΣFz = 0:
−168.429 lb + 265 lb + Dz − 192 lb = 0,
or
Dy = −626.57 lb
or
Dz = 95.429 lb
D = − ( 626 lb ) j + ( 95.4 lb ) k 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 4, Solution 98.
Free-Body Diagram:
ΣFx = 0:
Dx = 0
ΣM D = 0:
( − 7 in.) i  × C + ( 2 in.) i + ( 3 in.) k  × ( 530 lb ) j + ( − 192 lb ) k 
+ ( − 3 in.) i + ( 6 in.) j × ( 265 lb ) j + ( − 96 lb ) k  = 0
or − ( 7 in.) C yk + ( 7 in.) C z j + ( 2 in.)( 530 lb ) k + ( 2 in.)(192 lb ) j − ( 3 in.)( 530 lb ) i
− ( 3 in.)( 265 lb ) k − ( 3 in.)( 96 lb ) j − ( 6 in.)( 96 lb ) i = 0
Setting the coefficients of the unit vectors equal to zero:
k:
− C y ( 7 in.) − ( 265 lb )( 3 in.) + ( 530 lb )( 2 in.) = 0,
or
j:
C z ( 7 in.) + ( 96 lb )( 3 in.) + (192 lb )( 2 in.) = 0, or
C z = − 96.000 lb
C y = 37.857 lb
C = ( 37.9 lb ) j − ( 96.0 lb ) k Then:
ΣFy = 0:
37.857 lb + 265 lb + Dy + 530 lb = 0, or
Dy = − 832.86 lb
ΣFz = 0:
− 96 lb + 96 lb + Dz − 192 lb = 0,
Dz = 192.000 lb
or
D = − ( 833 lb ) j + (192.0 lb ) k 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 4, Solution 99.
Free-Body Diagram:
(
)
Note that W = mg = (18 kg ) 9.81 m/s 2 = 176.580 N
Moment equilibrium:
ΣM A = 0:
(
)
rB/ A × By j + Bz k + rC/ A × C zk + rG/ A × ( −176.580 N ) j = 0
or
i
j k
i
j
k
i
j
k
1.5 0 0 + 1.2 1.2sin 60° −1.2cos 60° + 0.6 0.6sin 60° − 0.6cos 60° = 0
Cz
0 B y Bz
0
0
0 −176.580
0
or
(
)
1.2C z sin 60° − (105.948 N ) cos 60° i + ( −1.5Bz − 1.2C z ) j + 1.5By − 105.948 N k = 0
Solving the equation one component at a time:
From i component:
1.2C z sin 60° − (105.948 N ) cos 60° = 0,
From j component:
−1.5Bz − 1.2C z = 0,
From k component:
1.5By − 105.948 N = 0,
or
or
C z = 50.974 N
Bz = − 0.8 ( 50.974 N ) = − 40.779 N
or
By = 70.632 N
Force equations:
ΣFy = 0:
Ay − 176.580 N + 70.632 N = 0,
or
Ay = 105.948 N
ΣFz = 0:
Az + 50.974 N − 40.779 N = 0,
or
Az = −10.195 N
Therefore:
A = (105.9 N ) j − (10.20 N ) k B = ( 70.6 N ) j − ( 40.8 N ) k C = ( 51.0 N ) k 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 4, Solution 100.
Free-Body Diagram:
(a)
( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k
ΣM C = 0:
+ (150 mm ) i + ( − 50 mm ) j × T k + ( 325 mm ) i + ( 55 mm ) j × (− FE ) j = 0
or
( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j − ( 50 mm )(150 N ) i
− (150 mm ) T j − ( 50 mm ) T j − ( 325 mm ) FE k = 0
continued
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
Setting the coefficients of the unit vectors equal to zero:
i:
(150 N )( 50 mm ) − T ( 50 mm ) = 0 ,
j:
( −150 N )(150 mm ) − (150 N )(150 mm ) − Dz ( 250 mm ) = 0,
k:
( Dy ) ( 250 mm ) − FE ( 325 mm ) = 0 , or
or
T = 150 N
Dy =
T = 150 N or
Dz = −180 N
325
FE
250
Spring force FE = kx, where
k = 366 N/m
elongation of spring x = ( yE )θ =180° − ( yE )θ = 0°
= ( 300 + 55 ) mm − ( 300 − 55 ) mm
= 110 mm
So,
FE = ( 366 N/m )( 0.110 m ) = 40.26 N
Substituting into expression for Dy:
Dy = 52.338 N
Force equations:
ΣFx = 0:
Dx = 0
ΣFy = 0:
C y + 52.338 N − 40.26 N = 0,
or
ΣFz = 0:
C z + 150 N + 150 N − 180
N = 0,
C y = −12.078 N
or
C z = −120.000 N
Therefore:
C = − (12.08 N ) j − (120.0 N ) k D = ( 52.3 N ) j − (180.0 N ) k 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 4, Solution 101.
Free-Body Diagram:
Start by determining the spring force, FE :
FE = − FE cosθ j + FE sin θ k
The magnitude FE = kx, where
k = 366 N/m, and
elongation of spring x = ( yE )θ = 90° − ( yE )θ = 0°
=
( 300 mm )2 + ( 55 mm )2 − ( 300 mm − 55mm )
= 305 mm − 245 mm = 60 mm
So, FE = ( 366 N/m )( 0.06 m ) = 21.96 N. Note that the length of the spring at θ = 90° is therefore 305 mm.
Then
 300 
 55 
FE = − ( 21.96 N ) 
 j + ( 21.96 N ) 
k
305


 305 
or FE = − ( 21.6 N ) j + ( 3.96 N ) k
continued
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
( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k
+ (150 mm ) i + ( −50 mm ) j × T k
ΣM C = 0:
+ ( 325 mm ) i + ( −55 mm ) k  ×  − ( 21.6 N ) j + ( 3.96 N ) k  = 0
( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j
or
− ( 50 mm )(150 N ) i − (150 mm ) T j − ( 50 mm ) T j
− ( 325 mm )( 21.6 N ) k − ( 325 mm )( 3.96 N ) j − ( 55 mm )( 21.6 N ) i = 0
Setting the coefficients of the unit vectors equal to zero:
(a)
i:
(150 N )( 50 mm ) − T ( 50 mm ) − ( 21.6 N )( 55 mm ) = 0,
or
T = 126.240 N
T = 126.2 N j:
( −150 N )(150 mm ) − (126.24 N )(150 mm ) − ( 3.96 N )( 325 mm ) − Dz ( 250 mm ) = 0,
or Dz = −170.892 N
k:
( Dy ) ( 250 mm ) − ( 21.6 N )( 325 mm ) = 0, or
Dy = 28.080 N
Force equations:
ΣFx = 0:
Dx = 0
ΣFy = 0:
C y + 28.080 N − 21.6 N = 0,
ΣFz = 0:
C z + 126.240 N + 150 N − 170.892 N + 3.96 N = 0,
or
C y = −6.4800 N
or
C z = −109.308 N
Therefore:
C = − ( 6.48 N ) j − (109.3 N ) k D = ( 28.1 N ) j − (170.9 N ) k 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 4, Solution 102.
Free-Body Diagram:
(
)
The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N
ΣM C = 0:
or
rCA × N A + rCB × N B + rCG × W = 0
( −0.3 m ) i + (1.2 m ) k  × N A j + (1.8 m ) i + ( 0.9 m ) k  × N B j
+ ( 0.6 m ) i + ( 0.6 m ) k  × ( −W ) j = 0
or
− ( 0.3 m ) N Ak − (1.2 m ) N Ai + (1.8 m ) N Bk − ( 0.9 m ) N Bi − ( 0.6 m )Wk + ( 0.6 m )Wi = 0
Equating the coefficients of the unit vectors to zero:
i:
−1.2 N A − 0.9 N B + 0.6W = 0
4 N A + 3 N B 0 = 2W
j:
(1)
− 0.3 N A + 1.8 N B − 0.6W = 0
− N A + 6 N B 0 = 2W
− 2 ×  Eq. (1)  + Eq. ( 2 )
(2)
gives
−9 N A = 2W
NA =
2
W = 370.60 N
9
continued
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
Now (2) gives
NB =
1
2  10
W
 2W + W  =
6
9  27
N B = 617.67 N,
and from (1)
ΣFy = 0:
N A + N B + NC − W = 0
370.60 N + 617.67 N + NC − 1667.7 N = 0
NC = 679.43 N
Therefore the forces on the blocks are:
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.
N A = 371 N
N B = 618 N
NC = 679 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 103.
Free-Body Diagram:
The location of the bucket of sand will be ( xS , c, zS ).
(
)
The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N
ΣFy = 0:
N =
3N − W − WS = 0
1
(W + WS )
3
ΣM O = 0:
rOA × N + rOB × N + rOC × N + rOG × W − rOS × WS = 0
or
( 0.3 m ) i + (1.2 m ) k  × Nj + ( 2.4 m ) i + ( 0.9 m ) k  × Nj
(1)
+ ( 0.6 m ) i × Nj + (1.2 m ) i + ( 0.6 m ) k  × ( −W ) j + ( xS i + zS k ) × ( −WS ) j = 0
or
( 0.3 m ) N k − (1.2 m ) N i + ( 2.4 m ) N k − ( 0.9 m ) N i + ( 0.6 m ) N k
− (1.2 m )W k + ( 0.6 m )W i − xSWS k + zSWS i = 0
Equating the coefficients of the unit vectors to zero:
i:
− (1.2 ) N − ( 0.9 ) N + ( 0.6 )W + zSWS = 0
continued
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
or, using (1)
1

−2.1  (W + WS )  + 0.6W + zSWS = 0
3

WS =
k:
0.1W
zS − 0.7 m
0.3 N + 2.4 N + 0.6 N − 1.2W − xSWS = 0
or, using (1)
1

3.3  (W + WS )  − 1.2W − xSWS = 0
3


WS =
0.1W
1.1 m − xS
(2)
For (WS )min , (1) and (2) imply that xS, should be chosen as small as possible and that zS should be chosen as
large as possible with the constraint that
(1.1 m − xS ) = ( zS − 0.7 m )
or
xS + zS = 1.8 m.
The smallest xS and the largest zS that satisfy this condition are
xS = 0.6 m zS = 1.2 m The corresponding value of WS is:
WS =
0.1(1667.7 N )
= 333.54 N
1.1 m − 0.6 m
Therefore the smallest mass of the bucket of sand is
( mS )min =
333.54 N
= 34.000kg
9.81 m/s 2
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.
or ( mS )min = 34.0 kg COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 104.
Free-Body Diagram:
WAB = ( 5 lb/ft )( 2 ft ) = 10 lb
First note
WBC = ( 5 lb/ft )( 4 ft ) = 20 lb
W = WAB + WBC = 30 lb
To locate the equivalent force of the pipe assembly weight
rG/B × W = Σ ( ri × Wi ) = rG ( AB ) × WAB + rG ( BC ) × WBC
( xGi + zGk ) × ( −30 lb ) j = (1 ft ) k × ( −10 lb ) j + ( 2 ft ) i × ( −20 lb ) j
or
∴
− ( 30 lb ) xGk + ( 30 lb ) zG i = (10 lb ⋅ ft ) i − ( 40 lb ⋅ ft ) k
From i-coefficient
zG =
10 lb ⋅ ft 1
= ft
30 lb
3
k-coefficient
xG =
40 lb ⋅ ft
1
= 1 ft
30 lb
3
From free-body diagram. of piping
ΣM x = 0:
W ( zG ) − TA ( 2 ft ) = 0
1 
1 
∴ TA =  ft  30 lb  ft  = 5 lb
2 
3 
ΣFy = 0:
or
TA = 5.00 lb
5 lb + TD + TC − 30 lb = 0
∴ TD + TC = 25 lb
(1)
continued
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
ΣM z = 0:
4 
TD (1.25 ft ) + TC ( 4 ft ) − 30 lb  ft  = 0
3 
∴ 1.25TD + 4TC = 40 lb ⋅ ft
−4  Equation (1) 
−4TD − 4TC = −100
(3)
−2.75TD = −60
Equation (2) + Equation (3)
∴ TD = 21.818 lb
From Equation (1)
(2)
or
TD = 21.8 lb
TC = 25 − 21.818 = 3.1818 lb
or
Results:
TC = 3.18 lb
TA = 5.00 lb TC = 3.18 lb TD = 21.8 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 105.
Free-Body Diagram:
First note
W AB = (5 lb/ft )(2 ft ) = 10 lb
WBC = ( 5 lb/ft )( 4 ft ) = 20 lb
From free-body diagram. of pipe assembly
ΣFy = 0: TA + TC + TD − 10 lb − 20 lb = 0
∴ TA + TC + TD = 30 lb
(1)
ΣM x = 0: (10 lb )(1 ft ) − TA ( 2 ft ) = 0
or
From Equations (1) and (2)
TA = 5.00 lb
TC + TD = 25 lb
(2)
(3)
ΣM z = 0: TC ( 4 ft ) + TD ( amax ) − 20 lb ( 2 ft ) = 0
or
( 4 ft ) TC + TDamax = 40 lb ⋅ ft
(4)
continued
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
Using Equation (3) to eliminate TC
4 ( 25 − TD ) + TD amax = 40
amax = 4 −
or
60
TD
By observation, a is maximum when TD is maximum. From Equation (3), (TD )max occurs when TC = 0.
Therefore, (TD )max = 25 lb and
amax = 4 −
60
25
= 1.600 ft
Results: (a)
amax = 1.600 ft (b)
TA = 5.00 lb TC = 0 TD = 25.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 106.
Free-Body Diagram:
The free-body diagram indicates the forces on the camera and tripod slid along their lines of action to the
plane ABCD.
Note that the x-coordinate of the center of mass of the camera is:
xCAM = − ( 2.4 in. − 1 in.) = −1.4 in.
ΣM B = 0:
or
( − 3 in.) k × C y j + ( −1.5 in.) k − (1.4 in.) i  × ( − 0.44 lb ) j
+ ( −1.5 in.) k − ( 2.8 in.) i  × ( − 0.53 lb ) j + ( −1.5 in.) k − ( 3.2 in.) i  × Ay j = 0
( 3 in.) C yi − (1.5 in.)( 0.44 lb ) i + (1.4 in.)( 0.44 lb ) k − (1.5 in.)( 0.53 lb ) i
+ ( 2.8 in.)( 0.53 lb ) k + (1.5 in.) Ay i − ( 3.2 in.) Ayk = 0
Setting the coefficients of the unit vectors equal to zero:
k:
− Ay ( 3.2 in.) + ( 0.53 lb )( 2.8 in.) + ( 0.44 lb )(1.4 in.) = 0
Ay = 0.65625 lb
or A y = 0.656 lb
continued
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
i:
( 3 in.) C y − (1.5 in.)( 0.44 lb ) − (1.5 in )( 0.53 lb ) + (1.5 in.) (0.65625 lb) = 0
C y = 0.156875 lb
or C y = 0.1569 lb
or B y = 0.1569 lb
ΣFx = By − 0.53 lb − 0.44 lb + 0.156875 lb + 0.156875 lb = 0
By = 0.156875 lb
(b) Free-Body Diagram:
Condition for no tipping: By > 0
ΣM A = 0:
( 3.2 in.) i + ( −1.5 in.) k  × C y j + ( 3.2 in.) i + (1.5 in.) k  × B y j
{
}
+ (1.8 in.) i × ( − 0.44 lb ) j + 1.8 in. − (1.4 in.) cosθ  i − (1.4 in.) sinθ k × ( − 0.53 lb ) j = 0
or
( 3.2 in.) C yk + (1.5 in.) C yi + ( 3.2 in.) By k − (1.5 in.) Byi − (1.8 in.)( 0.44 lb ) k
+ 1.8 in. − (1.4 in.) cosθ  ( − 0.53 lb ) k + (1.4 in.) sinθ ( − 0.53 lb ) i = 0
Setting the coefficients of the unit vectors equal to zero:
i:
C y (1.5 in.) − By (1.5 in.) − ( 0.53 lb )(1.4 in.) sin θ = 0
C y = B y + ( 0.53 lb )
k:
1.4
sin θ
1.5
( By + C y ) ( 3.2 in.) − ( 0.44 lb )(1.8 in.) − ( 0.53 lb ) 1.8 in. − (1.4 in ) cosθ  = 0
or
 1.4 

2By ( 3.2 in.) + ( 0.53 lb ) 
 ( 3.2 in.) sin θ + (1.4 in.) cosθ  − ( 0.44 lb + 0.53 lb )(1.8 in.) = 0
 1.5 

continued
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
Solving for By :
By =

 1.4 
 
1
( 0.44 lb + 0.53 lb )1.8 in. − 0.53 lb 
 ( 3.2 in.) sin θ + (1.4 in.) cosθ  
2 ( 3.2 in.) 
 1.5 
 
and By > 0
2.3531 > 2.1333sin θ + cosθ
To solve for θ :
2.1333sin θ + cosθ = A cos (θ + α )
= A ( cosθ cos α − sin θ sin α )
where
A=
( 2.1333)2 + (1)2 = 2.3561, and
 − 2.1333 
 , which (noting that cos α > 0) gives
1


α = tan −1 
α = − 64.885°
The inequality for By becomes:
2.3531 > 2.3561cos(θ − 64.885°)
or cos (θ − 64.885° ) < 0.99873
or θ − 64.885° < cos −1 ( 0.99873)
or θ < 64.885° ± 2.8879°
θ max = 62.00° 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 4, Solution 107.
Free-Body Diagram:
ΣM B = 0:
or
rBA × TA + rBC × TC + rBO × WAB + rBF × WBD + rBG × WAD = 0
( )
3


L
L
3
 1 
− Li × TA j +  − 4
i + Lk   × TC j − i ×  − Wj 


4
2
 tan 60°

 3 


 L
  1   3L
  1 
3
3
+  − i +
Lk  ×  − Wj  +  − i +
Lk  ×  − Wj  = 0
4
4
 4
  3   4
  3 
or
− LTAk −
3
3
L
L
3
L
3
LTC k − LTC i + Wk + Wk +
LWi − Wk +
LWi = 0
4
4
6
12
12
4
12
Equating the coefficients of the unit vectors to zero:
i:
W 3
W 3
3
L+
L − TC L = 0
3 4
3 4
4
TC =
k:
2W
3 3
− TA L +
TA =
W 3L W L W L 2W 3 L
+
+
−
=0
3 4
3 4
3 2 3 34 3
W
3
continued
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
ΣFy = 0:
TB +
W
2W
+
−W = 0
3
3 3
TB =
2
1 
1 −
W
3
3
Therefore:
TA =
TB =
2
1 
1 −
W 3
3
TC =
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.
W
3
2W
3 3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 108.
Free-Body Diagram:
Note that:
xC =
L
1
−
zC
2
3
(a) Setting: TA = TB = T
ΣM O = 0:
or
rOA × TA + rOB × TB + rOC × TC + rOH × W + rOF × WBD + rOG × WAD = 0
−
( )
3
 L
L 
L
L
3 
 i + Lk  × TC j
i × TA j + i × TB j +  − 4
2
2
4 
 2 tan 60° 



L
  1   L
  1 
3
3
Lk  ×  − Wj  +  − i +
Lk  ×  − Wj  + ( xC i + zC k ) × ( − Wj) = 0
+  i +
4
4
4
  3   4
  3 
or
( )
3 L 
L
L
L
3
 T k + LTC i
− TAk + TAk +  − 4
 2 tan 60°  C
2
2
4


−
1L
1 3
1L
1 3
Wk +
LWi +
Wk +
LWi − xCWk + zCWi = 0
34
3 4
34
3 4
continued
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
Equating the coefficients of the unit vectors to zero:
i:
W 3
W 3
3L
L+
L − TC
+ WzC = 0
3 4
3 4
4
(1)
or, using the relation between xC, and zC:
 L

3
3 
− LTC + 
+  − 3 xC +
L  W = 0
4
2  
 2 3 
k:
(2)
1
L
L
3
1
1
LW +
LW − xCW = 0
− T + T + L  −
 TC −
2
2
2
4
12
12


1
3
L  −
 TC = xCW
4 
2
(3)
Substituting (3) into (2)
 1
 1
3
3
3 
− LTC + 
+
 LW − 3  L  −
 TC  = 0

4
2 
4  
  2
2 3
4
W
3
4
ΣFy = 0:
2T + W − 2W = 0
3
W
T =
3
TC =
TA = TB =
Therefore:
TC =
(b) Using TC in (1):
W
3
4
W
3
W 3
W 3
4 3
L+
L −  W  L + WzC = 0
3 4
3 4
3 4
1 

zC = 1 −
L
2 3

and from geometry
xC =
L
1
1
zC =
2− 3 L
−
2
3
3
(
)
Therefore:
xC =
1
2− 3 L
3
(
)

1 
zC = 1 −
L 2 3

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 4, Solution 109.
Free-Body Diagram:
(a) ΣFy = 0:
3 ( R ) − 135 N = 0
R = 45.0 N
ΣM A = 0:
rAB × R B + rAC × R C + rAG × FW = 0
or
( 0.9 m ) i + l k  × R j + l i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( − FW j) = 0
or
( 0.9 m ) R k − lR i + lR k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0
Equating the coefficients of the i unit vector to zero:
i:
− lR − ( 0.9 m ) R + ( 0.45 m ) FW = 0
Using that FW = 3R
− l − ( 0.9 m ) + ( 0.45 m )( 3) = 0
l = 0.450 m
or l = 450 mm continued
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
(b) Free-Body Diagram:
ΣM A = 0:
rAB × R B + rAC × R C + rAG × FW = 0
or
( 0.9 m ) i + ( 0.5 m ) k  × R j + ( 0.5 m ) i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( −135 N ) j = 0
( 0.9 m ) R k − ( 0.5 m ) R i + ( 0.5 m ) R k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0
Equating the coefficients of the unit vectors to zero
i:
− 0.5RB − 0.9 RC + 60.75 = 0
(1)
k:
0.9 RB + 0.5RC − 60.75 = 0
(2)
0.5 × [ Eq. (1)] + 0.9 × [ Eq. (2) ]
gives
 − 0.5 ( 0.5 ) + ( 0.9 )( 0.9 )  RB + ( 0.5 − 0.9 ) 60.75 = 0
RB = 43.393 N
Now using (1)
− 0.5 ( 43.393) − 0.9 RC + 60.67 = 0
RC = 43.393 N
ΣFy = 0:
RA + 43.393 + 43.393 − 135 = 0
RA = 48.2 N RB = 43.4 N RC = 43.4 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 4, Solution 110.
Free-Body Diagram:
ΣM O = 0:
( 0.75 ft ) i × ( Ay j + Azk ) + ( 3.75 ft ) i × ( By j + Bzk ) + ( 2.25 ft ) i + ( 3 ft ) k  × ( − 27 lb ) j
+ ( 4.5 ft ) i + ( 5.25 ft ) k  × C y j = 0
or
( 0.75 ft ) Ayk − ( 0.75 ft ) Az j + ( 3.75 ft ) Byk − ( 3.75 ft ) Bz j − ( 2.25 ft )( 27 lb ) k
+ ( 3 ft )( 27 lb ) i + ( 4.5 ft ) C yk − ( 5.25 ft ) C y i = 0
Setting the coefficients of the unit vectors equal to zero:
i:
( 27 lb )( 3 ft ) − C y ( 5.25 ft ) = 0
C y = 15.4286 lb
k:
( 27 lb )(1.5 ft ) + By ( 3 ft ) − (15.4286 lb )( 3.75 ft ) = 0
By = − 5.78575 lb
ΣFy = 0:
Ay − 5.78575 lb − 27 lb + 15.4286 lb = 0
Ay = 17.3572 lb
Therefore:
(a)
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 y = 17.36 lb
(b)
B y = 5.79 lb (c)
C y = 15.43 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 111.
Free-Body Diagram:
ΣM O = 0:
( 0.75 ft ) i × ( Ay j + Azk ) + ( 3.75 ft ) i × ( By j + Bzk ) + ( 2.25 ft ) i + ( 3 ft ) k  × ( − 27 lb ) j
+ ( 3.75 ft ) i + ( 6 ft ) k  × C y j = 0
or
( 0.75 ft ) Ayk − ( 0.75 ft ) Az j + ( 3.75 ft ) Byk − ( 3.75 ft ) Bz j − ( 2.25 ft )( 27 lb ) k
+ ( 3 ft )( 27 lb ) i + ( 3.75 ft ) C yk − ( 6.0 ft ) C y i = 0
Setting the coefficients of the unit vectors equal to zero:
ΣM x = 0:
( 27 lb )( 3 ft ) − C y ( 6 ft ) = 0
C y = 13.5000 lb
ΣM z = 0:
− ( 27 lb )(1.5 ft ) + By ( 3 ft ) − (13.5000 lb )( 3 ft ) = 0
By = 0
ΣFy = 0:
Ay + 13.5000 lb − 27 lb = 0
Ay = 13.5000 lb
Therefore:
A y = 13.50 lb
(a)
(b)
(c)
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.
By = 0 C y = 13.50 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 112
Free-Body Diagram:
Express all forces in terms of rectangular components:
rE = ( 3 ft ) i + ( 3 ft ) j
rB = ( 3 ft ) sin 30° j + ( 3 ft ) cos 30°k = (1.5 ft ) j + ( 2.598 ft ) k
rD = − ( 3 ft ) i + ( 3 ft ) j
or
or
rA = (10 ft ) sin 30° j − (10 ft ) cos 30°k = ( 5 ft ) j + (8.66 ft ) k
uuur
BE = rE − rB = ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k
uuur
BE = ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BE = 4.243 ft
uuur
BD = rD − rB = − ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k
uuur
BD = − ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BD = 4.243 ft
Then
uuur
ur
BD
T BD = TBD
= TBD ( − 0.707i + 0.3535j − 0.6123k )
BD
uuur
ur
BE
T BE = TBE
= TBE ( 0.707i + 0.3535j − 0.6123k )
BE
continued
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
rB × TBD + rB × TBE + ( 5 ft ) j + ( 8.66 ft ) k  × ( − 75 lb ) j
ΣM C = 0:
i
j
k
i
j
k
0
1.5
2.598 TBD + 0
1.5
2.598 + 649.5 i = 0
or
− 0.707 0.3535 − 0.6123
0.707 0.3535 − 0.6123
Equating the coefficients of the unit vectors to zero:
j:
−1.837 TBD + 1.837 TBE = 0
i:
−1.837 TBD + 1.837 TBE + 649.5 lb = 0
TBD = 176.8 lb TBE = 176.8 lb Force equations:
C x + (176.8 )( − 0.707 ) + (176.8 )( 0.707 ) = 0,
or
Cx = 0
C y + (176.8 )( 0.3535 ) + (176.8 )( 0.3535 ) − 75 lb = 0,
or
C z + (176.8 )( − 0.6123) + (176.8 )( − 0.6123) = 0,
C z = 216.5 lb
or
C y = − 50 lb
C = − ( 50 lb ) j + ( 216.5 lb ) k 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 4, Solution 113.
Free-Body Diagram:
Express the forces in terms of their rectangular components:
uuur
FB
12i − 18j + 36k
2
3
6
TFB = TFB
= TFB
= TFB i − TFB j + TFBk
2
2
2
FB
7
7
(12 ) + ( −18) + ( 36 ) 7
uuur
FC
TFC = TFC
= TFC
FC
−12i − 18j + 36k
2
2
( −12 ) + ( −18) + ( 36 )
2
2
3
6
= − TFBi − TFB j + TFBk
7
7
7
From the free-body diagram, note that
zE
12
=
27 18
z E = 18.00 ft
continued
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
Then, using TED = 2720 lb
TED = 2720
− 27 j + 57.75k
( − 27 )2 + (18 + 39.75)2
TED = ( 32 lb )( − 36j + 77k )
(a)
ΣM A = 0:
rAD × TED + rAF × TFB + rAE × W = 0
i
j
k
i j k
i
j k
i j k
TFB
TFC
or 32 0 0 39.75 +
0 18 −12 +
0 18 −12 + 2720 0 27 −18 = 0
7
7
0 − 36 77
2 −3 6
0 −1 0
− 2 −3 6
Equating the coefficients of the unit vectors to zero
i:
32 ( 39.75 )( 36 ) +
TFB
T
( 72 ) + FC ( 72 ) + 2720 ( −18) = 0
7
7
72
72
TFB +
TFC − 3168 = 0
7
7
j:
(1)
TFB
T
( − 24 ) + FC ( 24 ) = 0
7
7
TFB = TFC
(2)
Substituting Eq. (2) in (1) gives
 72 
2   TFB − 3168 = 0
 7 
or TFB = 154.0 lb TFC = 154.0 lb (b) ΣFx = 0:
Ax +
2
2
(154.0) − (154.0 ) = 0
7
7
Ax = 0
ΣFy = 0:
Ay −
3
3
(154.0) − (154.0 ) − ( 32 )( 36 ) − 2720 = 0
7
7
Ay = 4004 lb
ΣFz = 0:
Az +
6
6
(154.0) + (154.0 ) + ( 32 )( 77 ) = 0
7
7
Az = − 2728 lb
Therefore:
A = ( 4.00 kips ) j − ( 2.73 kips ) k 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 4, Solution 114.
Free-Body Diagram
First express tensions in terms of rectangular components:
uuur
BE
− 8i − 8j + 4k
2
2
1
TBE = TBE
= TBE
= − TBE i − TBE j + TBE k
2
2
2
BE
3
3
3
( − 8) + ( − 8) + ( 4 )
TBF = TBF
uuur
BF
= TBF
BF
8i − 8j + 14k
2
2
(8) + ( − 8) + (14 )
2
=
4
4
7
TBF i − TBF j + TBF k
9
9
9
TCD = TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k )
ΣM A = 0:
or
rAB × TBE + rAB × TBF + rAC × TCD = 0
8j ×
TBE
T
( − 2i − 2 j + k ) + 8j × BF ( 4i − 4 j + 7k )
3
9
+ 10 j × TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) = 0
continued
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
Equating the coefficients of the unit vectors to zero:
(a)
i:
8
56
TBE +
TBF − 10TCD cos φ cosθ = 0
3
9
(1)
k:
16
32
TBE −
TBF − 10TCD cos φ sin θ = 0
3
9
(2)
− 2 ×  Eq. (1)  + Eq. ( 2 ) gives:
  56  32 
− 2   −
 TBF − 10 ( 600 ) cos10° ( − 2 cos 30° + sin 30° ) = 0
  9  9
TBF = 455.00 N
Using this in Eq. (1),
8
56
TBE +
( 455.00 ) − 10 ( 600 ) cos10° cos 30° = 0
3
9
TBE = 857.29 N
Therefore:
TBE = 857 N TBF = 455 N (b)
ΣFx = 0:
ΣFy = 0:
Ax −
2
4
(857.29 N ) + ( 455.00 N ) + ( 600 N ) cos10° sin 30° = 0
3
9
Ax = 73.9 N
2
4
Ay − ( 857.29 N ) − ( 455.00 N ) + ( 600 N ) sin10° = 0
3
9
Ay = 878 N
ΣFz = 0:
Az +
1
7
(857.29 N ) + ( 455.00 N ) − ( 600 N ) cos10° cos 30° = 0
3
9
Az = −127.9 N
Therefore:
A = ( 73.9 N ) i + ( 878 N ) j − (127.9 N ) k 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 4, Solution 115.
Free-Body Diagram:
First express tensions in terms of rectangular components:
uuur
BE
− 8i − 8j + 4k
2
2
1
= TBE
= − TBE i − TBE j + TBE k
TBE = TBE
2
2
2
BE
3
3
3
( − 8) + ( − 8) + ( 4 )
uuur
BF
TBF = TBF
= TBF
BF
8i − 8j + 14k
2
2
(8) + ( − 8) + (14 )
2
=
4
4
7
TBF i − TBF j + TBF k
9
9
9
TCD = TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k )
ΣM A = 0:
or
rAB × TBE + rAB × TBF + rAC × TCD = 0
8j ×
TBE
T
(− 2i − 2 j + k ) + 8j × BF (4i − 4 j + 7k )
3
9
+ 10 j × TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) = 0
continued
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
Equating the coefficients of the unit vectors to zero:
8
56
TBE +
TBF − 10TCD cos φ cosθ = 0
3
9
i:
8
56
(840 N ) + ( 450 N ) = 10TCD cos φ cosθ
3
9
(1)
16
32
TBE −
TBF − 10TCD cos φ sin θ = 0
3
9
k:
16
32
(840 N ) − ( 450 N ) = 10TCD cos φ sin θ
3
9
(a)
(2)
 Eq. ( 2) 
gives:
 Eq. (1) 
16
(840) − 32
( 450)
10TCD cos φ sin θ
3
9
=
8
10TCD cos φ cos θ
(840) + 56
( 450)
3
9
tan θ =
1
1.75
θ = 29.7° θ = 29.745°
(b) Substituting into (1) gives:
8
56
(840 N ) + ( 450 N ) − 10TCD cos8° cos 29.745° = 0,
3
9
or
TCD = 586.19 N
or TCD = 586 N (c)
ΣFx = 0:
Ax −
2
4
(840 N ) + ( 450 N ) + ( 586.19 N ) cos8° sin 29.745° = 0
3
9
Ax = 72.0 N
ΣFy = 0:
Ay −
2
4
(840 N ) − ( 450 N ) − ( 586.19 N ) sin 8° = 0
3
9
Ay = 842 N
ΣFz = 0:
Az +
1
3
(840 N ) + ( 450 N ) − ( 586.19 N ) cos8° cos 29.745° = 0
3
9
Az = −126.0 N
Therefore:
A = ( 72.0 N ) i + ( 842 N ) j − (126.0 N ) k 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 4, Solution 116.
Free-Body Diagram:
Express all forces in terms of rectangular components:
rA = ( 2.4 m ) i
rB = (1.8 m ) j
uuur
AD = − ( 2.4 m ) i + ( 0.3 m ) j + (1.2 m ) k
uuur
BE = − (1.8 m ) i + ( 0.6 m ) j − ( 0.9 m ) k
W = − ( 880 N ) j
Then
uuur
ur
AD
= TAD
T AD = TAD
AD
uuur
ur
BE
T BE = TBE
= TBE
BE
− 2.4i + 0.3j + 1.2k
2
2
( − 2.4 ) + ( 0.3) + (1.2 )
2
8
1
4
= − TADi + TAD j + TADk
9
9
9
2
6
2
3
= − TADi + TAD j − TADk
7
7
7
−1.8i + 0.6 j − 0.9k
2
2
( −1.8) + ( 0.6 ) + ( − 0.9 )
continued
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
ΣM C = 0:
or
rA × TAD + rB × TBE + rA × W = 0
i
2.4
8
−
9
j
0
1
9
k
i j k
0 TAD + 1.8 0 0 TBE + ( 2.4 ) i × ( − 880 ) j = 0
4
6 2
3
−
−
9
7 7
7
Equating the coefficients of the unit vectors to zero:
9.6
5.4
TAD +
TBE = 0
9
7
j:
−
k:
2.4
3.6
TAD +
TBE − 2112 = 0
9
7
or TAD = 2160 N TBE = 2990 N Force equations:
Cx −
8
6
( 2160.0 N ) − ( 2986.7 N ) = 0,
9
7
Cy +
1
2
( 2160.0 N ) + ( 2986.7 N ) − 880 N = 0, or
9
7
Cz +
4
3
( 2160.0 N ) − ( 2986.7 N ) = 0,
9
7
or
or
C x = 4480.0 N
C y = − 213.34 N
C z = 320.01 N
C = ( 4480 N ) i − ( 213 N ) j + ( 320 N ) k 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 4, Solution 117.
Free-Body Diagram:
Express all forces in terms of rectangular components:
rA = ( 2.4 m ) i
rB = (1.8 m ) j
uuur
AD = − ( 2.4 m ) i + ( 0.3 m ) j + (1.2 m ) k
uuur
BE = − (1.8 m ) i + ( 0.6 m ) j − ( 0.9 m ) k
WA = − ( 440 N ) j
WB = − ( 440 N ) j
Then
uuur
ur
AD
T AD = TAD
= TAD
AD
uuur
ur
BE
= TBE
T BE = TBE
BE
− 2.4i + 0.3j + 1.2k
2
2
( − 2.4 ) + ( 0.3) + (1.2 )
2
8
1
4
= − TADi + TAD j + TADk
9
9
9
2
6
2
3
= − TADi + TAD j − TADk
7
7
7
−1.8i + 0.6 j − 0.9k
2
2
( −1.8) + ( 0.6 ) + ( − 0.9 )
continued
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
ΣM C = 0:
or
rA × TAD + rB × TBE + rA × WA + rB × WB = 0
i
2.4
8
−
9
j
0
1
9
k
i j k
0 TAD + 1.8 0 0 TBE + ( 2.4 ) i × (− 440) j + (1.8 ) i × (− 440) j = 0
4
6 2
3
−
−
9
7 7
7
Equating the coefficients of the unit vectors to zero:
9.6
5.4
TAD +
TBE = 0
9
7
j:
−
k:
2.4
3.6
TAD +
TBE − 1848 = 0
9
7
or TAD = 1890 N TBE = 2610 N Force equations:
Cx −
8
6
(1890.00 N ) − ( 2613.3 N ) = 0,
9
7
Cy +
1
2
(1890.00 N ) + ( 2613.3 N ) − 440 N − 440 N = 0,
9
7
Cz +
4
3
(1890.00 N ) − ( 2613.3 N ) = 0,
9
7
or
or
C x = 3920.0 N
or
C y = −76.657 N
C z = 279.99 N
C = ( 3920 N ) i − ( 76.7 N ) j + ( 280 N ) k 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 4, Solution 118.
Free-Body Diagram:
Express all forces in terms of rectangular components:
rA = (140 in.) i
rC = ( 72 in.) i
uuur
EG = − (120 in.) i + (126 in.) k
uuuur
FH = − (120 in.) i − ( 90 in.) k
uur
CI = − ( 72 in.) i + ( 44.8 in.) j
W = − (140 lb ) j
Then
uuur
ur
EG
= TEG
T EG = TEG
EG
−120i + 126k
2
( −120 ) + (126 )
uuuur
ur
FH
T FH = TFH
= TFH
FH
uur
ur
CI
T CI = TCI
= TCI
CI
2
=−
−120i − 90k
( −120 )2 + ( − 90 )2
− 72i + 44.8j
2
( − 72 ) + ( 44.8)
2
=−
20
21
TEG i +
TEGk
29
29
= −0.8TFH i − 0.6TFH k
45
28
TCI i +
TCI j
53
53
continued
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
ΣM D = 0:
or
rE × TEG + rF × TFH + rC × TCI + rA × W = 0
i
j k
i j k
i
j k
TEG
TFH
T
120 0 10
+ 120 0 −10
+ 72 0 0 CI − 19600 lb ⋅ in. = 0
29
5
53
− 20 0 21
−4 0 −3
− 45 28 0
Noting that TCI = TFH and equating the coefficients of the unit vectors to zero:
j:
− 93.793TEG + 80TCI = 0
k:
38.038TCI − 19600 lb ⋅ in. = 0
or TCI = TFH = 515 lb or TEG = 440 lb Force equations:
ΣFx = 0:
Dx −
20
45
4
( 439.50 lb ) − ( 515.28 lb ) − ( 515.28 lb ) = 0,
29
53
5
ΣFy = 0:
Dy +
28
( 515.28 lb ) − 140 lb = 0,
53
ΣFz = 0:
Dz +
21
3
( 439.50 lb ) − ( 515.28) lb = 0,
29
5
or
or
Dx = 1152.83 lb
Dy = −132.223 lb
or
Dz = − 9.0906 lb
D = (1153 lb ) i − (132.2 lb ) j − ( 9.09 lb ) k 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 4, Solution 119.
Free-Body Diagram:
Express all forces in terms of rectangular components:
rB = (120 in.) i
rC = ( 72 in.) i
uuur
EG = − (120 in.) i + (126 in.) k
uuuur
FH = − (120 in.) i − ( 90 in.) k
uur
CI = − ( 72 in.) i + ( 44.8 in.) j
W = − (140 lb) j
Then
uuur
ur
EG
T EG = TEG
= TEG
EG
−120i + 126k
2
uuuur
ur
FH
T FH = TFH
= TFH
FH
uur
ur
CI
T CI = TCI
= TCI
CI
( −120 ) + (126 )
2
=−
−120i − 90k
( −120 )2 + ( − 90 )2
− 72i + 44.8j
2
( − 72 ) + ( 44.8)
2
=−
20
21
TEG i +
TEGk
29
29
= − 0.8TFH i − 0.6TFH k
45
28
TCI i +
TCI j
53
53
continued
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
ΣM D = 0:
or
rE × TEG + rF × TFH + rC × TCI + rA × W = 0
i
j k
i j k
i
j k
TEG
TFH
T
120 0 10
+ 120 0 −10
+ 72 0 0 CI − 19600 lb ⋅ in. = 0
29
5
53
− 20 0 21
−4 0 −3
− 45 28 0
Noting that TCI = TFH and equating the coefficients of the unit vectors to zero:
j:
− 93.793TEG + 80TCI = 0
k:
38.038TCI − 16800 lb ⋅ in. = 0
or TCI = TFH = 442 lb TEG = 377 lb Force equations:
ΣFx = 0:
Dx −
20
45
4
( 376.72 lb ) − ( 441.67 lb ) − ( 441.67 lb ) = 0,
29
53
5
ΣFy = 0:
Dy +
28
( 441.67 lb ) − 140 lb = 0,
53
ΣFz = 0:
Dz +
21
3
( 376.72 lb ) − ( 441.67 lb ) = 0,
29
5
or
or
Dx = 988.15 lb
Dy = − 93.335 lb
or
Dz = − 7.7952 lb
D = ( 998 lb ) i − ( 93.3 lb ) j − ( 7.80 lb ) k 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 4, Solution 120.
Free-Body Diagram:
Geometry:
Using triangle ACD and the law of sines
sin α
sin 50°
or α = 20.946°
=
15 in.
7 in.
β = 50° + 20.946° = 70.946°
Expressing FCD in terms of its rectangular coordinates:
FCD = FCD sin β j + FCD cos β k
= FCD sin 70.946° j + FCD cos 70.946°k
FCD = 0.94521FCD j + 0.32646 FCDk
ΣM B = 0:
( − 26 in.) i × A + ( −13 in.) i + (16 in.) sin 50°j + (16 in.) cos 50°k  × ( − 75 lb ) j
+ ( − 26 in.) i + ( 7 in.) k  × FCD = 0
or
− ( 26 in.) Ayk + ( 26 in.) Az j + (13 in.)( 75 lb ) k + (16 in.)( 75 lb ) cos 50°i
− ( 26 in.) ( 0.94521FCD ) k + ( 26 in.) ( 0.32646 FCD ) j − ( 7 in.) ( 0.94521FCD ) i = 0
continued
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
(a) Setting the coefficients of the unit vectors to zero:
i:
( 75 lb ) (16 in.) cos 50° − ( 0.94521FCD ) ( 7 in.) = 0
FCD = 116.6 lb (b) ΣFx = 0:
k:
Ax = 0
− 0.94521(116.580 lb )  ( 26 in.) + ( 75 lb )(13 in.) − Ay ( 26 in.) = 0
Ay = − 72.693 lb
j:
 0.32646 (116.580 lb )  ( 26 in.) + Az ( 26 in.) = 0
Az = − 38.059 lb
ΣFy = 0:
− 72.693 lb + 0.94521(116.580 lb ) − 75 lb + By = 0
By = 37.500 lb
ΣFz = 0:
− 38.059 lb + 0.32646 (116.580 lb ) + Bz = 0
Bz = 0
Therefore:
A = − ( 72.7 lb ) j − ( 38.1 lb ) k B = ( 37.5 lb ) j 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 4, Solution 121.
Free-Body Diagram:
Express tension in terms of rectangular components:
uuuur
DG = − ( 4.8 in.) j − ( 9 in.) k
uuuur
− 4.7 j − 9k
DG
8
15
TDG = TDG
= TDG
= − TDG j − TDG k
2
2
DG
17
17
( − 4.8) + ( − 9 )
Equilibrium:
ΣM F = 0:
( 6.4 in.) i + ( − 2.4 in.) j × TDG + ( 5.2 in.) j × E
+ ( 7.6 in.) j + ( 9.6 in.) k  × ( − 55 lb ) i = 0
or
15 
 8 
 15 
 TDG i − ( 6.4 in.)   TDG k + ( 6.4 in.)   TDG j
 17 
 17 
 17 
( 2.4 in.) 
− ( 5.2 in.) Ex k + ( 5.2 in.) Ez i + ( 7.6 in.) (55 lb)k − ( 9.6 in.) (55 lb)j = 0
Setting the coefficients of the unit vectors equal to zero:
(a)
j:
15
TDG ( 6.4 in.) − ( 55 lb )( 9.6 in.) = 0
17
TDG = 93.5 lb continued
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
(b)
k:
8

(93.500 lb)  ( 6.4 in.) = 0
17


( 55 lb )( 7.6 in.) − Ex ( 5.2 in.) − 
E x = 26.231 lb
i:
 15

Ez ( 5.2 in.) +  (93.500 lb)  ( 2.4 in.) = 0
17

E z = − 38.077 lb
ΣFx = 0:
− 55 lb + 26.231 lb + Fx = 0
Fx = 28.796 lb
ΣFy = 0:
Fy −
8
( 93.500 lb ) = 0
17
Fy = 44.000 lb
ΣFz = 0:
− 38.077 lb + Fz −
15
( 93.500 lb ) = 0
17
Fz = 120.577 lb
Therefore:
E = ( 26.2 lb ) i − ( 38.1 lb ) k F = ( 28.8 lb ) i + ( 44.0 lb ) j + (120.6 lb ) k 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 4, Solution 122.
Free-Body Diagram:
(
)
W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N
First note
(a)
 ( 0.08 m ) i + ( 0.25 m ) j − ( 0.2 m ) k 
T
TEF = λ EF TEF = 
TEF = EF ( 0.08i + 0.25j − 0.2k )

0.33
( 0.08 )2 + ( 0.25 )2 + ( 0.2 )2 m 

From free-body diagram of rectangular plate
ΣM x = 0:
(147.15 N )( 0.1 m ) − (TEF ) y ( 0.2 m ) = 0
 0.25 

14.715 N ⋅ m − 
 TEF  ( 0.2 m ) = 0
 0.33 

or
TEF = 97.119 N
or
or TEF = 97.1 N (b)
ΣFx = 0:
Ax + (TEF ) x = 0
 0.08 
Ax + 
 ( 97.119 N ) = 0
 0.33 
∴ Ax = −23.544 N
continued
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
ΣM B( z -axis ) = 0:
− Ay ( 0.3 m ) − (TEF ) y ( 0.04 m ) + W ( 0.15 m ) = 0
 0.25 

− Ay ( 0.3 m ) − 
 97.119 N  ( 0.04 m ) + 147.15 N ( 0.15 m ) = 0
 0.33 

or
∴ Ay = 63.765 N
ΣM B( y -axis ) = 0:
Az ( 0.3 m ) + (TEF ) x ( 0.2 m ) + (TEF ) z ( 0.04 m ) = 0
 0.08 

 0.2 

Az ( 0.3 m ) + 
 TEF  ( 0.2 m ) − 
 TEF  ( 0.04 m ) = 0
0.33
0.33






∴ Az = −7.848 N
and A = − ( 23.5 N ) i + ( 63.8 N ) j − ( 7.85 N ) k ΣFy = 0:
Ay − W + (TEF ) y + By = 0
 0.25 
63.765 N − 147.15 N + 
 ( 97.119 N ) + By = 0
 0.33 
∴ By = 9.81 N
ΣFz = 0:
Az − (TEF ) z + Bz = 0
 0.2 
−7.848 N − 
 ( 97.119 N ) + Bz = 0
 0.33 
∴ Bz = 66.708 N
and B = ( 9.81 N ) j + ( 66.7 N ) k 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 4, Solution 123.
Free-Body Diagram:
(
(a)
)
W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N
First note


− ( 0.3 m ) i + ( 0.12 m ) j − ( 0.2 m ) k 
T

T = EH  − ( 0.3) i + ( 0.12 ) j − ( 0.2 ) k 
TEH = λ EH TEH =

 EH
2
2
2
0.38
( 0.3) + ( 0.12 ) + ( 0.2 ) m 

From free-body diagram of rectangular plate
ΣM x = 0:
(147.15 N )( 0.1 m ) − (TEH ) y ( 0.2 m ) = 0
 0.12 

 TEH  ( 0.2 m ) = 0
 0.38 

or
(147.15 N )( 0.1 m ) − 
or
TEH = 232.99 N
or TEH = 233 N (b)
ΣFx = 0:
Ax + (TEH ) x = 0
 0.3 
Ax − 
 ( 232.99 N ) = 0
 0.38 
∴ Ax = 183.938 N
continued
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
ΣM B( z -axis ) = 0:
− Ay ( 0.3 m ) − (TEH ) y ( 0.04 m ) + W ( 0.15 m ) = 0
 0.12
− Ay ( 0.3 m ) − 
( 232.99 N ) ( 0.04 m ) + (147.15 N )( 0.15 m ) = 0
 0.38

or
∴ Ay = 63.765 N
ΣM B( y -axis ) = 0:
Az ( 0.3 m ) + (TEH ) x ( 0.2 m ) + (TEH ) z ( 0.04 m ) = 0
 0.3 

 0.2 

Az ( 0.3 m ) − 
 ( 232.99 N )  ( 0.2 m ) − 
 ( 232.99 )  ( 0.04 m ) = 0
 0.38 

 0.38 

or
∴ Az = 138.976 N
and A = (183.9 N ) i + ( 63.8 N ) j + (139.0 N ) k ΣFy = 0:
Ay + By − W + (TEH ) y = 0
 0.12 
63.765 N + By − 147.15 N + 
 ( 232.99 N ) = 0
 0.38 
∴ By = 9.8092 N
ΣFz = 0:
Az + Bz − (TEH ) z = 0
 0.2 
138.976 N + Bz − 
 ( 232.99 N ) = 0
 0.38 
∴ Bz = −16.3497 N
and B = ( 9.81 N ) j − (16.35 N ) k 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 4, Solution 124.
Free-Body Diagram:
Express tension, weight in terms of rectangular components:
uuur
EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k
uuur
EF
300 i + 1350 j − 700 k
T=T
=T
EF
( 300 )2 + (1350 )2 + ( − 700 )2
=
6
27
14
Ti +
T j− Tk
31
31
31
(
)
W = − (mg ) j = − ( 7 kg ) 9.81 m s 2 j = − (68.67 N)j
ΣM B = 0:
− ( 750 mm ) i × A +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j
+ ( −100 mm ) i + ( 700 mm ) k  × T = 0
or
− ( 750 mm ) Ayk + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i
i
j k
T
+ −100 0 700 ( mm ) = 0
31
6 27 −14
continued
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
Setting the coefficients of the unit vectors equal to zero:
(a)
i:
−
27
T ( 700 mm ) + ( 68.67 N )( 350 mm ) = 0
31
or T = 39.4N T = 39.422 N
(b)
k:
 27

− Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 39.422 N )  (100 mm ) = 0
31


Ay = 29.757 N
j:
14

6

Az ( 750 mm ) −  ( 39.422 N )  (100 mm ) +  ( 39.422 N )  ( 700 mm ) = 0
 31

 31

Az = − 4.7476 N
ΣFx = 0:
Bx +
6
( 39.422 N ) = 0
31
Bx = − 7.6301 N
ΣFy = 0:
29.757 N + B y − 68.67 N +
27
( 39.422 N ) = 0
31
By = 4.5777 N
ΣFz = 0:
− 4.7476 N + Bz −
14
( 39.422 N ) = 0
31
Bz = 22.551 N
Therefore:
A = ( 29.8 N ) j − ( 4.75 N ) k B = − ( 7.63 N ) i + ( 4.58 N ) j + ( 22.6 N ) k 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 4, Solution 125.
Free-Body Diagram:
Express tension, weight in terms of rectangular components:
uur
IF = ( 75 mm ) i + (1350 mm ) j − ( 250 mm ) k
uur
IF
T=T
=T
IF
=
75i + 1350 j − 250 k
( 75)2 + (1350 )2 + ( − 250 )2
3
54
10
Ti +
Tj−
Tk
55
55
55
(
)
W = − (mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − (68.67 N)j
ΣM B = 0:
− ( 750 mm ) i × A +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j
+ (125 mm ) i + ( 250 mm ) k  × T = 0
or
− ( 750 mm ) Ay k + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i
i
j k
T
+ 125 0 250
( mm ) = 0
55
3 54 −10
continued
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
Setting the coefficients of the unit vectors equal to zero:
(a)
i:
−
54
T ( 250 mm ) + ( 68.67 N )( 350 mm ) = 0
55
T = 97.918 N
(b)
k:
or T = 97.9 N  54

− Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 97.918 N )  (125 mm ) = 0
 55

Ay = 50.358 N
j:
 10

3

Az ( 750 mm ) −  ( 97.918 N )  (125 mm ) +  ( 97.918 N )  ( 250 mm ) = 0
55
55




Az = − 4.7475 N
ΣFx = 0:
Bx +
3
( 97.918 N ) = 0
55
Bx = − 5.3410 N
ΣFy = 0:
50.358 N + B y − 68.67 N +
54
( 97.918 N ) = 0
55
By = − 77.826 N
ΣFz = 0:
− 4.7475 N + Bz −
10
( 97.918 N ) = 0
55
Bz = 22.551 N
Therefore:
A = ( 50.4 N ) j − ( 4.75 N ) k B = − ( 5.34 N ) i − ( 77.8 N ) j + ( 22.6 N ) k 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 4, Solution 126.
Free-Body Diagram:
Express forces, weight in terms of rectangular components:
uuur
CE = ( 3 ft ) i + ( 4 ft ) j − ( 2 ft ) k
uuur
CE
FCE = FCE
= FCE
CE
3i + 4 j + 2k
( 3) 2 + ( 4 )2 + ( 2 ) 2
= 0.55709 FCE i + 0.74278 FCE j + 0.37139 FCE k
W = − (mg ) j = − (300 lb)j
ΣM B = 0:
or
( 4 ft ) k × A + (1.5 ft ) i + ( 2 ft ) k  × ( − 300 lb ) j
+ ( 3 ft ) i + ( 4 ft ) k  × FCE = 0
− ( 4 ft ) Ay i + ( 4 ft ) Az j − (1.5 ft )( 300 lb ) k + ( 2 ft )( 300 lb ) i
i
j
k
+
3
0
4
FCE ( ft ) = 0
0.55709 0.74278 0.37139
continued
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
Setting the coefficients of the unit vectors equal to zero:
k:
( 0.74278 FCE ) ( 3 ft ) − ( 300 lb )(1.5 ft ) = 0
FCE = 201.94 lb
or FCE = 202 lb j:
Ax ( 4 ft ) + 0.55709 ( 201.94 lb )  ( 4 ft ) − 0.37139 ( 201.94 lb )  ( 3 ft ) = 0
Ax = − 56.250 lb
i:
− Ay ( 4 ft ) −  0.74278 ( 201.94 lb )  ( 4 ft ) + ( 300 lb )( 2 ft ) = 0
Ay = 0
ΣFx = 0:
− 56.250 lb + Bx + 0.55709 ( 201.94 lb ) = 0
Bx = − 56.249 lb
ΣFy = 0:
0 + By − 300 lb + 0.74278 ( 201.94 lb ) = 0
By = 150.003 lb
ΣFz = 0:
Bz + 0.371391( 201.94 lb ) = 0
Bz = − 74.999 lb
Therefore:
A = − ( 56.3 lb ) i B = − ( 56.2 lb ) i + (150.0 lb ) j − ( 75.0 lb ) k 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 4, Solution 127.
Free-Body Diagram:
Express forces, weight in terms of rectangular components:
uuur
CA = − (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k
uuur
CB = (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k
By symmetry FCA = FCB , and at the load corresponding to buckling
FCA = FCB = 1.8 kN
uuur
CA
FCA = FCA
= (1.8 kN )
CA
−1.2 i + 1.2 j − 0.6 k
( −1.2 )2 + (1.2 )2 + ( − 0.6 )2
FCA = − (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k
uuur
CB
FCB = FCB
= (1.8 kN )
CB
1.2 i + 1.2 j − 0.6 k
(1.2 )2 + (1.2 )2 + ( − 0.6 )2
FCB = (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
ΣM D = 0:
( 2.4 m ) i × E + ( 2.4 m ) i + (1.2 m ) j × FCB + (1.2 m ) j × FCA
+ (1.2 m ) i + ( 0.6 m ) j × Pk = 0
or
i
j
Ex
0 0 + 2.4 1.2 0 kN ⋅ m
E y Ez
1.2 1.2 − 0.6
( 2.4 m )
k
i
j
k
i
j
k
i
j
k
+ 0 1.2 0 kN ⋅ m + (1.2 m ) ( 0.6 m ) 0 = 0
−1.2 1.2 − 0.6
0
0
P
Setting the coefficient of the unit vector i equal to zero:
(a)
i : P(0.6 m) − ( 0.6 )(1.2 ) kN ⋅ m − ( 0.6 )(1.2 ) kN ⋅ m = 0
P = 2.4000 kN
or P = 2.40 kN (b)
By symmetry, Dz = Ez
ΣFz = 0:
Dz + Dz + 2.4 kN − 0.6 kN − 0.6 kN = 0
Dz = Ez = − 0.60000 kN
Therefore:
E z = − ( 0.600 kN ) k 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 128.
Free-Body Diagram:
Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net
moment about this point is zero.
(a) ΣFx = 0:
3 lb + ( 3 lb ) cos 30° − T = 0
T = 5.5981 lb
or T = 5.60 lb (b) ΣFy = 0:
ΣFz = 0:
Dy = 0
Dz − ( 3 lb ) sin 30° = 0
D = (1.500 lb ) k ΣM D = 0:
M D + ( 0.72 in.) j − (1.2 in.) k  × ( −T ) i
+ ( 0.88 in.) j − ( 3 in.) k  × ( 3 lb )(1 + cos 30° ) i − ( 3 lb ) sin 30°k  = 0
i
or M Dx i + M Dy j + M Dz k + 0
−T
(
)
j
k
i
j
k
0
( 0.72 in.) ( −1.2 in.) +
( 0.88 in.) ( − 3 in.) ( 3 lb ) = 0
− sin 30°
0
0
1 + cos 30°
0
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Setting the coefficients of the unit vectors equal to zero:
i:
M Dx − ( 3 lb ) sin 30° ( 0.88 in.) = 0
M Dx = 1.3200 lb ⋅ in.
j:
M Dy + ( 5.5981 lb )(1.2 in.) − ( 3 lb )( 3 in.)(1 + cos 30° ) = 0
M Dy = 10.0765 lb ⋅ in.
k:
M Dz + ( 5.5981 lb )( 0.72 in.) − ( 3 lb )( 0.88 in.)(1 + cos 30° ) = 0
M Dz = 0.89568 lb ⋅ in.
or M D = (1.320 lb ⋅ in.) i + (10.08 lb ⋅ in.) j + ( 0.896 lb ⋅ in.) k 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 129.
Free-Body Diagram:
Express the tension in terms of its rectangular components:
uuuur
DG = − ( 4.8 in.) j − ( 9 in.) k
uuuur
DG
TDC = TDG
= TDG
DG
ΣM E = 0:
− 4.8 j − 9 k
2
( − 4.8) + ( − 9 )
2
=−
8
15
TDG j − TDG k
17
17
M E + ( 6.4 in.) i + ( − 7.6 in.) j × TDG
+ ( 2.4 in.) j + ( 9.6 in.) k  × ( − 44 lb ) i = 0
 15 
 8 
 15 
 TDG i − ( 6.4 in.)   TDGk + ( 6.4 in.)   TDG j
 17 
 17 
 17 
or ( M Ex i + M Ey j + M Ez k ) + ( 7.6 in.) 
+ ( 2.4 in.)( 44 lb ) k − ( 9.6 in.)( 44 lb ) j = 0
Setting the coefficient of the unit vector j equal to zero:
(a) j:
 15

 TDG  ( 6.4 in.) − ( 44 lb )( 9.6 in.) = 0
17


TDG = 74.800 lb
or TDG = 74.8 lb continued
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
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(b)
ΣFx = 0:
E x − 44 lb = 0, or Ex = 44.000 lb
ΣFy = 0:
Ey −
8
( 74.8 lb ) = 0, or E y = 35.200 lb
17
ΣFz = 0:
Ez −
15
( 74.8 lb ) = 0, or Ez = 66.000 lb
17
or E = ( 44.0 lb ) i + ( 35.2 lb ) j + ( 66.0 lb ) k Using the moment equation again and setting the coefficients of the unit vectors i and k equal to zero:
i:
 15 
M Ex + ( 7.6 in.)   ( 74.800 lb ) = 0
 17 
M Ex = −501.60 lb ⋅ in.
k:
8

M Ez + ( 44 lb )( 2.4 in.) −  ( 74.8 lb )  ( 6.4 in.) = 0
17

M Ex = 119.680 lb ⋅ in.
or M E = − ( 502 lb ⋅ in.) i + (119.7 lb ⋅ in.) k 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 4, Solution 130.
Free-Body Diagram:
Express forces and moments in terms of rectangular components:
FDE = FDE
− 40 i − 70 j + 40k
2
2
( − 40 ) + ( − 70 ) + ( 40 )
2
=
FDE
(− 4i − 7 j + 4k )
9
FA = ( 24 N )( sin 20° i − cos 20° j)
B = By j + Bzk ,
(a) ΣFx = 0:
M B = M By j + M Bzk
4
− FDE + 24sin 20° = 0
9
FDE = 18.4691 N
ΣM B = 0:
or
or FDE = 18.47 N rBC × FCF + rBD × FDE + rBA × FA + M B = 0
i
j k
i
j k
i
j
k
18.4691
FCF 0 − 48 36 +
−1 +  M B y j + M B zk 
0 − 80 60 + ( 80 )( 24 ) 0
0
9
−4 −7 4
0 −1 0
sin 20° − cos 20° 0
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Equating the coefficients of the unit vectors to zero:
i:
36FCF +
18.4691
(100 ) + (80 )( 24 )( − cos 20° ) = 0
9
FCF = 44.417 N
or FCF = 44.4 N (b)
j:
18.4691
( − 240 ) + (80 )( 24 )( − sin 20° ) + M By = 0
9
M By = 1149.19 N ⋅ mm
k:
18.4691
( − 320 ) + M Bz = 0
9
M Bz = 656.68 N ⋅ mm
ΣFy = 0:
By − 44.417 −
7
(18.4691) − 24 cos 20° = 0
9
By = 81.3 N
ΣFz = 0:
Bz +
4
(18.4691) = 0
9
Bz = −8.21 N
Therefore:
B = ( 81.3 N ) j − ( 8.21 N ) k M B = (1.149 N ⋅ m ) j + ( 0.657 N ⋅ m ) k 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 131.
Free-Body Diagram:
Express tension, weight in terms of rectangular components:
uuur
EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k
uuur
EF
=T
T=T
EF
=
300 i + 1350 j − 700 k
( 300 )2 + (1350 )2 + ( − 700 )2
6
27
14
Ti +
T j− Tk
31
31
31
(
)
W = − ( mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − ( 68.67 N ) j
ΣM B = 0:
M B +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j
+ ( −100 mm ) i + ( 700 mm ) k  × T = 0
or
( M By j + M Bzk ) + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i
i
j k
T
+ −100 0 700 ( mm ) = 0
31
6 27 −14
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Setting the coefficients of the unit vector i equal to zero:
(a)
i:
−
27
T ( 700 mm ) − ( 68.67 N )( 350 mm ) = 0
31
or T = 39.4 N T = 39.422 N
(b)
ΣFx = 0:
Bx +
6
( 39.422 N ) = 0
31
Bx = − 7.6301 N
ΣFy = 0:
By − 68.67 N +
27
( 39.422 N ) = 0
31
By = 34.335 N
ΣFz = 0:
Bz −
14
( 39.422 N ) = 0
31
Bz = 17.8035 N
B = − ( 7.63 N ) i + ( 34.3 N ) j + (17.80 N ) k Using the moment equation again and setting the coefficients of the unit vectors j and k to zero:
ΣM B ( y − axis) = 0:
14

6

M By −  ( 39.422 N )  (100 mm ) +  ( 39.422 N )  ( 700 mm ) = 0
31
31




M By = − 3.5607 N ⋅ m
ΣM B ( z − axis) = 0:
 27

M Bz + ( 68.67 N )( 375 mm ) −  ( 39.422 N )  (100 mm ) = 0
 31

M Bz = − 22.318 N ⋅ m,
Therefore:
M B = − ( 3.56 N ⋅ m ) j − ( 22.3 N ⋅ m ) k 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 132.
Free-Body Diagram:
Express tensions, load in terms of rectangular components:
uuur
BD = − ( 60 in.) i + ( 25 in.) k
uuur
BE = − ( 60 in.) i + ( 25 in.) j
uuur
CF = − ( 60 in.) i + ( 25 in.) j
BD = BE = CF =
( −60 )2 + ( 25)2 = 65 in.
uuur
BD
12
5
TBD = TBD
= − TBD i + TBD k
BD
13
13
uuur
BE
12
5
TBE = TBE
= − TBE i + TBE j
BE
13
13
uuur
CF
12
5
TCF = TCF
= − TCF i + TCF j
CF
13
13
ΣM A = 0:
or
rB × TBD + rB × TBE + rC × TCF + rG × W = 0
i j k
i j k
i j k
i
j
k
TBD
TBE
TCF
60 0 0
in. + 60 0 0
in. + 60 0 − 30
in. + 60
0
−15 lb ⋅ in. = 0
13
13
13
−12 0 5
−12 5 0
−12 5 0
0 − 500 0
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Equating the coefficients of the unit vectors to zero:
i:
150
TCF − 7500 = 0
13
TCF = 650.00 lb
j:
−
300
360
TBD +
( 650 lb ) = 0
13
13
TBD = 780.00 lb
k:
or TCF = 650 lb or TBD = 780 lb 300
300
TBE − 30000 +
( 650.00 lb ) = 0
13
13
TBE = 650.00 lb
ΣFx = 0:
Ax −
or TBE = 650 lb 12
12
12
( 780 lb ) − ( 650 lb ) − ( 650 lb ) = 0
13
13
13
Ax = 1920.00 lb
ΣFy = 0:
Ay +
5
5
( 780 lb ) + ( 650 lb ) − 500 lb = 0
13
13
Ay = 0
ΣFz = 0:
Az +
5
( 780 lb ) = 0
13
Az = −300.00 lb
Therefore,
A = (1920 lb ) i − ( 300 lb ) k 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 4, Solution 133.
Free-Body Diagram:
Express tensions, load in terms of rectangular components:
uuur
BD = − ( 60 in.) i + ( 25 in.) k
uuur
BE = − ( 60 in.) i + ( 25 in.) j
uuur
CF = − ( 60 in.) i + ( 25 in.) j
BD = BE = CF =
( −60 )2 + ( 25)2 = 65 in.
uuur
BD
12
5
= − TBD i + TBD k
BD
13
13
uuur
BE
12
5
TBE = TBE
= − TBE i + TBE j
BE
13
13
uuur
CF
12
5
TCF = TCF
= − TCF i + TCF j
CF
13
13
TBD = TBD
WG = − ( 500 lb ) j
WC = − ( 800 lb ) j
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
ΣM A = 0:
rB × TBD + rB × TBE + rC × TCF + rG × WG + rC × WC = 0
i
j k
i
j k
i
j k
i
j
k
TBD
TBE
TCF
60 0 0
in. + 60 0 0
in. + 60 0 − 30
in. + 60
0
−15 lb ⋅ in.
13
13
13
−12 0 5
−12 5 0
−12 5 0
0 − 500 0
or
i
j
k
+ 60 0 −30 lb ⋅ in. = 0
0 −800 0
Equating the coefficients of the unit vectors to zero:
i:
150
TCF − 7500 − 24000 = 0
13
TCF = 2730 lb
j:
−
or TCF = 2.73 kips 300
360
TBD +
( 2730 lb ) = 0
13
13
TBD = 3276 lb
k:
or TBD = 3.28 kips 300
300
TBE − 30000 +
( 2730 lb ) − ( 60 )(800 lb ) = 0
13
13
TBE = 650.00 lb
ΣFx = 0:
Ax −
or TBE = 650 lb 12
12
12
( 3276 lb ) − ( 650 lb ) − ( 2730 lb ) = 0
13
13
13
Ax = 6144 lb
ΣFy = 0:
Ay +
5
5
( 2730 lb ) + ( 650 lb ) − 500 lb − 800 lb = 0
13
13
Ay = 0
ΣFz = 0:
Az +
5
( 3276 lb ) = 0
13
Az = −1260.00 lb
Therefore,
A = ( 6.14 kips ) i − (1.260 kips ) k 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 4, Solution 134.
First note
Free-Body Diagram:
TDI = λ DI TDI =
=
TEH = λ EH TEH =
=
TFG = λ FGTFG =
=
− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k
( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m
TDI
TDI
( − 0.65i + 0.2j − 0.44k )
0.81
− ( 0.45 m ) i + ( 0.24 m ) j
( 0.45)2 + ( 0.24 )2 m
TEH
TEH
( − 0.45 i ) + ( 0.24 j)
0.51
− ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k
2
2
2
( 0.45 ) + ( 0.2 ) + ( 0.36 ) m
TFG
TFG
( −0.45i + 0.2j + 0.36k )
0.61
From free-body diagram of frame
ΣM A = 0: rD/ A × TDI + rC/ A × ( −280 N ) j + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j = 0
or
i
j
k
i
j k
i
j k
i
j
k
 TDI 
 TEH 
 TFG 
0.65 0.2
0 
0.32 0 
 + 0.65 0 0 ( 280 N ) + 0
 + 0.45 0 0.06 

 0.81 
 0.51 
 0.61 
−0.65 0.2 −0.44
−0.45 0.24 0
−0.45 0.2 0.36
0 −1 0
i
j
k
+ 0.45 0 0.06 ( 360 N ) = 0
0 −1 0
or
T
0.81
T
0.51
( − 0.088 i + 0.286 j + 0.26 k ) DI + ( − 0.65 k ) 280 N + ( 0.144 k ) EH
+ ( − 0.012 i − 0.189 j + 0.09 k )
TFG
+ ( 0.06 i − 0.45 k )( 360 N ) = 0
0.61
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
From i-coefficient
T 
T 
−0.088  DI  − 0.012  FG  + 0.06 ( 360 N ) = 0
 0.81 
 0.61 
∴ 0.108642TDI + 0.0196721TFG = 21.6
From j-coefficient
(1)
T 
T 
0.286  DI  − 0.189  FG  = 0
0.81


 0.61 
∴ TFG = 1.13959TDI
(2)
From k-coefficient
T 
T 
T 
0.26  DI  − 0.65 ( 280 N ) + 0.144  EH  + 0.09  FG 
0.81
0.51




 0.61 
− 0.45 ( 360 N ) = 0
∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 N
(3)
Substitution of Equation (2) into Equation (1)
0.108642TDI + 0.0196721(1.13959TDI ) = 21.6
∴ TDI = 164.810 N
TDI = 164.8 N or
Then from Equation (2)
TFG = 1.13959 (164.810 N ) = 187.816 N
TFG = 187.8 N or
And from Equation (3)
0.32099 (164.810 N ) + 0.28235TEH + 0.147541(187.816 N ) = 344 N
∴ TEH = 932.84 N
TEH = 933 N or
The vector forms of the cable forces are:
TDI =
164.810 N
( −0.65i + 0.2j − 0.44k )
0.81
= − (132.25 N ) i + ( 40.694 N ) j − ( 89.526 N ) k
TEH =
932.84 N
( − 0.45i + 0.24 j) = − (823.09 N ) i + ( 438.98 N ) j
0.51
187.816 N
( − 0.45i + 0.2 j + 0.36k )
0.61
= − (138.553 N ) i + ( 61.579 N ) j + (110.842 N ) k
TFG =
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Then, from free-body diagram of frame
ΣFx = 0: Ax − 132.25 − 823.09 − 138.553 = 0
∴ Ax = 1093.89 N
ΣFy = 0: Ay + 40.694 + 438.98 + 61.579 − 360 − 280 = 0
∴ Ay = 98.747 N
ΣFz = 0: Az − 89.526 + 110.842 = 0
∴ Az = −21.316 N
or
A = (1094 N ) i + ( 98.7 N ) j − ( 21.3 N ) k 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 4, Solution 135.
Free-Body Diagram:
First note
TDI = λ DI TDI =
=
TEH = λ EH TEH =
=
TFG = λ FGTFG =
− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k
( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m
TDI
( −65i + 20 j − 44k )
81
− ( 0.45 m ) i + ( 0.24 m ) j
( 0.45)2 + ( 0.24 )2 m
TDI
TEH
TEH
( −15i + 8j)
17
− ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k
( 0.45)2 + ( 0.2 )2 + ( 0.36 )2 m
TFG
TFG
( −45i + 20j + 36k )
61
From free-body diagram of frame
=
ΣM A = 0: rD/ A × TDI + rC/ A ×  − ( 280 N ) j + ( 50 N ) k 
+ rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j
or
i
j
k
i
j
k
i
j k
 TDI 
T 
0.65 0.2 0 
0 + 0 0.32 0  EH 
 + 0.65 0
 81 
 17 
−65 20 −44
−15 8 0
0 −280 50
i
j
k
i
j
k
 TFG 
+ 0.45 0 0.06 
 + 0.45 0 0.06 ( 360 N ) = 0
 61 
−45 20 36
0 −1 0
and
T
 81 
T
 17 
( −8.8i + 28.6 j + 26k )  DI  + ( −32.5j − 182k ) + ( 4.8k )  EH 
T 
+ ( −1.2i − 18.9 j + 9.0k )  FG  + ( 0.06i − 0.45k ) ( 360 ) = 0
 61 
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
T 
T 
−8.8  DI  − 1.2  FG  + 0.06 ( 360 ) = 0
 81 
 61 
From i-coefficient
∴ 0.108642TDI + 0.0196721TFG = 21.6
(1)
T 
T 
From j-coefficient 28.6  DI  − 32.5 − 18.9  FG  = 0
81


 61 
∴ 0.35309TDI − 0.30984TFG = 32.5
(2)
From k-coefficient
T 
T 
T 
26  DI  − 182 + 4.8  EH  + 9.0  FG  − 0.45 ( 360 ) = 0
 81 
 17 
 61 
∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344
−3.25 × Equation (1)
−0.35309TDI − 0.063935TFG = −70.201
Add Equation (2)
0.35309TDI − 0.30984TFG =
−0.37378TFG
(3)
32.5
= −37.701
∴ TFG = 100.864 N
TFG = 100.9 N or
Then from Equation (1)
0.108642TDI + 0.0196721(100.864 ) = 21.6
∴ TDI = 180.554 N
TDI = 180.6 N or
and from Equation (3)
0.32099 (180.554 ) + 0.28235TEH + 0.147541(100.864 ) = 344
∴ TEH = 960.38 N
TEH = 960 N or
The vector forms of the cable forces are:
TDI =
180.554 N
( −65i + 20j − 44k )
81
= − (144.889 N ) i + ( 44.581 N ) j − ( 98.079 N ) k
TEH =
960.38 N
( −15i + 8j) = − (847.39 N ) i + ( 451.94 N ) j
17
100.864 N
( −45i + 20j + 36k )
61
= − ( 74.409 N ) i + ( 33.070 N ) j + ( 59.527 N ) k
TFG =
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
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COSMOS: Complete Online Solutions Manual Organization System
Then from free-body diagram of frame
ΣFx = 0: Ax − 144.889 − 847.39 − 74.409 = 0
∴ Ax = 1066.69 N
ΣFy = 0: Ay + 44.581 + 451.94 + 33.070 − 360 − 280 = 0
∴ Ay = 110.409 N
ΣFz = 0: Az − 98.079 + 59.527 + 50 = 0
∴ Az = −11.448 N
Therefore,
A = (1067 N ) i + (110.4 N ) j − (11.45 N ) k 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 4, Solution 136.
Free-Body Diagram:
ΣFx = 0:
Bx + C x = 0, or Bx = − C x
(1)
ΣFy = 0:
Ay + By + C y = 0
(2)
ΣFy = 0:
Az − P = 0, or Az = P = 40.0 lb
(3)
ΣM O = 0:
rOA × A + rOB × B + rOC × C + M Ai − M C k = 0
or
(
)
(
)
(
)
ai × Ay j + Azk + bj × Bxi + By j + ck × C x i + C y j + M Ai − M C k = 0
or
a Ayk − a Az j − bBx j + cCx j − cC y i + M Ai − M C k = 0
continued
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
Equating the coefficients of the unit vectors to zero:
i : − cC y + M A = 0
Cy =
j:
(4)
− a Az + cC x = 0, or using (3)
Cx =
k:
MA
36 lb ⋅ ft
=
= 36.0 lb
c
1 ft
a
9 in.
P=
( 40 lb ) = 30.0 lb
c
12 in.
(5)
a Ay − b Bx − M C = 0, or, using (1) and (5)
b
M
6 in.
0
Ay = − P + C = −
= − 20.0 lb
( 40 lb ) +
c
a
12 in.
9 in.
(6)
Finally substituting into (1) and (2) gives:
Βx = −30.0 lb
By = − Ay − C y = 20.0 lb − 36.0 lb = −16.00 lb
Therefore:
A = − ( 20.0 lb ) j + ( 40.0 lb ) k B = − ( 30.0 lb ) i − (16.00 lb ) j C = ( 30.0 lb ) i + ( 36.0 lb ) j 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 4, Solution 137.
Free-Body Diagram:
ΣFx = 0:
Bx + C x = 0, or Bx = − C x
(1)
ΣFy = 0:
Ay + By + C y = 0
(2)
ΣFz = 0:
Az − P = 0, or Az = P = 60.0 N
(3)
ΣM O = 0:
rOA × A + rOB × B + rOC × C + M Ai − M C k = 0
(
)
(
)
(
)
or
ai × Ay j + Azk + bj × Bxi + By j + ck × C x i + C y j + M Ai − M C k = 0
or
a Ayk − a Az j − bBx j + cCx j − cC y i + M Ai − M C k = 0
continued
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
Equating the coefficients of the unit vectors to zero:
i:
− cC y + M A = 0
Cy =
j:
(4)
− a Az + cC x = 0, or using (3)
Cx =
k:
MA
6.3 N ⋅ m
=
= 35.0 N
c
0.180 m
a
0.240 m
P=
( 60 N ) = 80.0 N
c
0.180 m
(5)
a Ay − b Bx − M C = 0, or, using (1) and (5)
b
M
0.200 m
13 N ⋅ m
Ay = − P + C = −
= − 12.50 N
( 60 N ) +
c
a
0.180 m
0.240 m.
(6)
Finally substituting into (1) and (2) gives:
Βx = −80.0 N
By = − Ay − C y = 12.50 N − 35.0 N = −22.5 N
Therefore:
A = − (12.50 N ) j + ( 60.0 N ) k B = − ( 80.0 N ) i − ( 22.5 N ) j C = ( 80.0 N ) i + ( 35.0 N ) j 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 4, Solution 138.
Free-Body Diagram:
ΣFx = 0:
Bx = 0
ΣM D( x - axis) = 0:
(80 N )( 2.6 m ) − Bz ( 2 m ) = 0
or B = (104.0 N ) k Bz = 104.000 N
ΣM D ( z - axis) = 0:
C y ( 4 m ) − 144 N ⋅ m = 0
C y = 36.000 N
ΣM D( y - axis) = 0:
− C z ( 4 m ) − (104 N )( 6 m ) + ( 80 N )( 6 m ) = 0
C z = − 36.000 N
and C = ( 36.0 N ) j − ( 36.0 N ) k ΣFy = 0:
Dy + 36 = 0, or Dy = −36.000 N
ΣFz = 0:
Dz − 36 + 104 − 80 = 0, or Dz = 12.000 N
Therefore:
D = − ( 36.0 N ) j + (12.00 N ) k 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 4, Solution 139.
Free-Body Diagram:
ΣFx = 0:
Bx = 0
ΣM D ( x - axis) = 0:
(80 N )( 2.6 m ) − Bz ( 2 m ) = 0
or B = (104.0 N ) k Bz = 104.000 N
ΣM D ( z - axis) = 0:
Cy (4 m ) = 0
Cy = 0
ΣM D ( y - axis) = 0:
− C z ( 4 m ) − (104 N )( 6 m ) + ( 80 N )( 6 m ) = 0
C z = − 36.000 N
and C = − ( 36.0 N ) k ΣFy = 0:
ΣFz = 0:
Dy = 0
Dz − 36 + 104 − 80 = 0, or Dz = 12.000 N
Therefore:
D = (12.00 N ) k 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 4, Solution 140.
Free-Body Diagram:
Express the forces in terms of rectangular components:
(
)
W = − ( mg ) j = − ( 3 kg ) 9.81 m/s 2 j = − ( 29.43 N ) j
N B = N B ( 0.8j + 0.6k )
( xB )2 + ( 325 + 75)2 + (100 )2
LAB = 525 mm =
xB = 325 mm
Then,
uuur
BC
TBC = TBC
= TBC
BC
325i + 400 j − 100k
2
2
( 325) + ( 400 ) + ( −100 )
2
=
13
16
4
TBC i + TBC j − TBC k
21
21
21
continued
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
Equilibrium:
ΣM A = 0:
rG/ A × W + rB/ A × N B + rC/ A × TBC = 0
i
j
k
i
j
k
i
j k
T
or 162.5 − 200 50 + 325 − 400 100 N B + 650 0 0 BC = 0
21
− 29.43 0
0
0
0.8 0.6
13 16 − 4
Equating the coefficients of the unit vectors to zero:
i:
1471.5 − 320 N B = 0
N B = 4.5984 N
or
N B = ( 3.6787 N ) j + ( 2.7590 N ) k
j:
− 195 N B +
2600
TBC = 0
21
TBC = 7.2425 N
13
( 7.2425 N ) = 0
21
Ax = −4.4835 N
ΣFx = 0:
Ax +
ΣFy = 0:
Ay − 29.43 N + 3.6787 N +
16
( 7.2425 N ) = 0
21
Ay = 20.233 N
ΣFy = 0:
Az + 2.7590 N −
4
( 7.2425 N ) = 0
21
Az = −1.37948 N
Therefore:
(a)
TBC = 7.24 N (b)
A = − ( 4.48 N ) i + ( 20.2 N ) j − (1.379 Ν ) k N B = ( 3.68 N ) j + ( 2.76 N ) k 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 4, Solution 141.
Free-Body Diagram:
(a) The force acting at E on the free-body diagram of rod AB is
perpendicular to AB and CD. Letting λ E = direction cosines for force E,
λE =
rB/ A × k
rB/ A × k
 − ( 32 in.) i + ( 24 in.) j − ( 40 in.) k  × k
=
2
2
( 32 ) + ( 24 ) in.
= 0.6i + 0.8 j
Also,
W = − (10 lb ) j
B = Bk
E = E ( 0.6i + 0.8 j)
From free-body diagram of rod AB
ΣM A = 0: rG/ A × W + rE/ A × E + rB/ A × B = 0
i
j
k
i
j
k
i
j
k
∴ −16 12 −20 (10 lb ) + −24 18 −30 E + −32 24 −40 B = 0
0 −1 0
0.6 0.8 0
0
0
1
( −20i + 16k )(10 lb ) + ( 24i − 18 j − 30k ) E + ( 24i + 32 j) B = 0
From k-coefficient
160 − 30 E = 0
∴ E = 5.3333 lb
and
E = 5.3333 lb ( 0.6i + 0.8 j)
E = ( 3.20 lb ) i + ( 4.27 lb ) j or
(b) From j-coefficient
−18 ( 5.3333 lb ) + 32 B = 0
∴ B = 3.00 lb
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.
B = ( 3.00 lb ) k COSMOS: Complete Online Solutions Manual Organization System
From free-body diagram of rod AB
ΣF = 0: A + W + E + B = 0
Ax i + Ay j + Az k − (10 lb ) j + ( 3.20 lb ) i + ( 4.27 lb ) j + ( 3.00 lb ) k = 0
From i-coefficient
Ax + 3.20 lb = 0
∴ Ax = −3.20 lb
j-coefficient
Ay − 10 lb + 4.27 lb = 0
∴ Ay = 5.73 lb
k-coefficient
Az + 3.00 lb = 0
∴ Az = −3.00 lb
Therefore
A = − ( 3.20 lb ) i + ( 5.73 lb ) j − ( 3.00 lb ) k 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 4, Solution 142.
Free-Body Diagram:
There is only one unknown of interest and, therefore only one equation is needed:
ΣM AB = 0
Geometry:
 1.05 m 
 = 16.2602°
 3.5 m 
θ = tan −1 
xG = (1.25 m ) cos16.2602° = 1.2 m
yG = 1.95 m − (1.25 m ) sin16.2602° = 1.6 m
uuur
BA
=
λ BA =
BA
−3.6i + 1.05j
( −3.6 )2 + (1.05 )2
=−
24
7
i+
j
25
25
continued
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
rK/A = (1.8 m ) i − ( 0.525 m ) j + ( 0.225 m ) k
rG/A = (1.2 m ) i − (1.95 m − 1.6 m ) j + ( 0.45 m ) k
= (1.2 m ) i − ( 0.35 m ) j + ( 0.45 m ) k
(
)
W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j
uuuur
HG
PHG = PHG
= PHG
HG
−0.05i + 1.6 j − 0.4 k
2
2
( −0.05) + (1.6 ) + ( −0.4 )
2
32
8 
 1
= PHG  − i +
j−
k
33
33 
 33
Now,
ΣM BA = 0:
(
)
(
)
λ BA ⋅ rK / A × W + λ BA ⋅ rG/ A × PHG = 0
− 24
7
0
− 24
7
0
1
PHG
+ 1.2 − 0.35 0.45
=0
1.8 − 0.525 0.225
25
25 )( 33)
(
−1
−8
0 − 245.25
0
32
−
1324.35
342.45
+
P =0
25
25
( )( 35) HG
Therefore: PHG = 127.620 N
or PHG = 127.6 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 4, Solution 143.
Free-Body Diagram:
There is only one unknown of interest and, therefore only one equation is needed:
ΣM AB = 0
Geometry:
 1.05 m 
 = 16.2602°
 3.5 m 
θ = tan −1 
xI = ( 2.50 m ) cos16.2602° = 2.4 m
yI = 1.95 m − ( 2.50 m ) sin16.2602° = 1.25 m
λ BA =
uuur
BA
=
BA
−3.6i + 1.05j
2
( −3.6 ) + (1.05)
2
=−
24
7
i+
j
25
25
continued
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
rK/A = (1.8 m ) i − ( 0.525 m ) j + ( 0.225 m ) k
rI/A = ( 2.4 m ) i − ( 0.7 m ) j + ( 0.45 m ) k
(
)
W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j
uur
JI
= PJI
PJI = PJI
JI
−0.025i + 1.25j − 0.25k
50
10 
 1
= PJI  − i +
j + k
51
51 
 51
( −0.025)2 + (1.25)2 + ( −0.25)2
Now,
ΣM BA = 0:
(
)
(
)
λ BA ⋅ rK / A × W + λ BA ⋅ rI / A × PJI = 0
− 24
7
0
− 24 7
0
1
PJI
+ 12.4 − 0.7 0.45
=0
1.8 − 0.525 0.225
25
25
51)
(
)(
−1 50 −10
0 − 245.25
0
−
1324.35
536.85
+
P =0
25
( 25)( 51) JI
Therefore: PJI = 125.811 N
or PJI = 125.8 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 4, Solution 144.
Free-Body Diagram:
Express forces in terms of their rectangular components:
uuur
BG
− 40 i + 74 j − 32 k
37
16 
 20
TBG = TBG
j−
k
= TBG
= TBG  − i +
2
2
2
BG
45
45
45


−
+
+
−
40
74
32
( ) ( ) ( )
uuuur
BH
TBH = TBH
= TBH
BH
30 i + 60 j − 60 k
2
2 
1
= TBH  i + j − k 
3
3 
3
( 30 )2 + ( 60 )2 + ( − 60 )2
P = − ( 75 lb ) j
uuur
AD
λ AD =
=
AD
80 i − 60 j
2
(80 ) + ( − 60 )
2
= 0.8 i − 0.6 j
continued
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
rB/A = ( 40 in.) i
rC/A = ( 80 in.) i
Now,
ΣM AD = 0:
(
)
(
)
(
)
λ AD ⋅ rB/ A × TBG + λ AD ⋅ rB/ A × TBH + λ AD ⋅ rC/ A × P = 0
0.8 0 − 0.6
0.8 0 − 0.6
0.8 0 − 0.6
TBG
TBH
+ 40 0 0
+ 80 0
40 0
0
0 =0
45
3
− 20 37 −16
1 2 −2
0 − 75 0
−
888
48
TBG −
TBH + 3600 = 0
45
3
Noting that TBG = TBH = T and solving:
T = 100.746 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.
or T = 100.7 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 145.
Free-Body Diagram:
Express forces in terms of their rectangular components:
uuur
BG
− 40 i + 74 j − 32 k
37
16 
 20
= TBG
= TBG  − i +
TBG = TBG
j−
k
2
2
2
BG
45
45
45


−
+
+
−
40
74
32
( ) ( ) ( )
P = − ( 75 lb ) j
λ AD =
uuur
AD
=
AD
80 i − 60 j
2
(80 ) + ( − 60 )
2
= 0.8 i − 0.6 j
rB/ A = ( 40 in.) i
rC/ A = ( 80 in.) i
Now,
ΣM AD = 0:
(
)
(
)
λ AD ⋅ rB/ A × TBG + λ AD ⋅ rC/ A × P = 0
0.8 0 − 0.6
0.8 0 − 0.6
TBG
+ 80 0
40 0
0
0 =0
45
− 20 37 −16
0 − 75 0
−
888
TBG + 3600 = 0
45
Solving for TBG :
TBG = 182.432 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.
or TBG = 182.4 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 146.
Free-Body Diagram:
Express forces in terms of their rectangular components:
W1 = W2 = − ( 30 lb ) j
uuuur
( x − 6) i + y j − 6 k
BH
T=T
=T
BH
( x − 6 )2 + y 2 + ( − 6 )2
λ AF =
uuur
AF
=
AF
6i − 3 j − 6k
2
2
( 6 ) + ( − 3) + ( − 6 )
2
=
2
1
2
i− j− k
3
3
3
rG/ A = ( 3 ft ) i
rB/ A = ( 6 ft ) i
rI / A = ( 6 ft ) i − ( 3 ft ) k
continued
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
Now,
(
ΣM AF = 0:
)
(
)
(
)
λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0
−1 − 2
2
−1 − 2
1
0
0  +
6
0 0
 3
0 − 30 0
( x − 6) y − 6
2
3
60 + ( −36 − 12 y )
T
3
( x − 6 )2 + y 2 + 36
T
3
2
( x − 6 ) + y + 36
2
−1 − 2
1
0 −3   = 0
3
0 − 30 0
2
+ 6
+ 60 = 0
Solving for T :
T =
30
30 + y
( x − 6 )2 + y 2 + 36
It is thus clear that for a given y, T will have its minimum value when x = 6 ft. Denoting this minimum by Tm:
Tm =
30
3+ y
y 2 + 36
Now to find the minimum of Tm, differentiate Tm with respect to y and equate the derivative to zero.
1
1
 1 

−
  ( 3 + y ) 36 + y 2 2 ( 2 y ) − 36 + y 2 2 (1)  30
 2 

dTm
= 
=0
2
dy
(3 + y )
(
)
(
)
Setting the numerator equal to zero and simplifying:
( 3 + y ) y − y 2 − 36 = 0
y = 12 ft
(a) Minimum occurs at:
x = 6 ft, y = 12 ft (b) Using the expression for T:
Tmin =
30
3 + 12
( 6 − 6 )2 + (12 )2 + 36 = 26.833 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.
or Tmin = 26.8 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 147.
Free-Body Diagram:
Express forces in terms of their rectangular components:
W1 = W2 = − ( 30 lb ) j
uuuur
BH
− 6i + y j − 6k
=T
T=T
BH
( −6 )2 + y 2 + ( − 6 )2
uuur
AF
6i − 3 j − 6k
2
1
2
=
= i− j− k
λ AF =
AF
( 6 ) 2 + ( − 3)2 + ( − 6 ) 2 3 3 3
rG/ A = ( 3 ft ) i
rB/ A = ( 6 ft ) i
rI / A = ( 6 ft ) i − ( 3 ft ) k
continued
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
Now,
(
ΣM AF = 0:
)
(
)
(
)
λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0
2 −1 − 2
2 −1 − 2
1
3 0
0  + 6 0 0
3
−6 y −6
0 − 30 0
60 + ( − 36 − 12 y )
2 −1 − 2
1
+ 6 0 −3   = 0
2
( −6 ) + y 2 + 36 0 − 30 0  3 
T
3
( − 6 )2 + y 2 + 36
T
3
+ 60 = 0
Solving for T :
T =
30
30 + y
( −6 )2 + y 2 + 36
T =
30
3+ y
y 2 + 72
Now to find the minimum of T, differentiate T with respect to y and equate the derivative to zero.
1
1
 1 

−
2 2
2 2
  ( 3 + y ) 72 + y
( 2 y ) − 72 + y (1) 30
 2 

dTm
= 
=0
2
dy
(3 + y )
(
)
(
)
Setting the numerator equal to zero and simplifying:
( 3 + y ) y − y 2 − 72 = 0, or y = 24 ft
x = 0, y = 24 ft (a) Minimum occurs at:
(b) Using the expression for T:
Tmin =
30
3 + 24
( 24 )2 + 72 = 28.284 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.
or
Tmin = 28.3 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 148.
Free-Body Diagram:
Express forces in terms of their rectangular components:
(
)
(
)
(
)
WAB = − (1.25 kg/m ) 9.81 m/s 2 ( 0.9 m ) j = − (11.0363 Ν ) j
WBC = − (1.25 kg/m ) 9.81 m/s 2 ( 0.3 m ) j = − ( 3.6788 Ν ) j
WCD = − (1.25 kg/m ) 9.81 m/s 2 (1.35 m ) j = − (16.5544 Ν ) j
uuur
FE
T=T
=T
FE
−0.6i + 0.9 j − 1.35k
2
2
( − 0.6 ) + ( 0.9 ) + ( −1.35)
2
=
T
( −2i + 3j − 4.5k )
33.25
continued
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
uuur
AD
λ AD =
=
AD
0.9i − 0.3j − 1.35k
2
2
( 0.9 ) + ( −0.3) + ( −1.35)
2
=
6
2
9
i− j− k
11
11
11
rG/ A = ( 0.45 m ) i
rF / A = ( 0.6 m ) i
rB/ A = ( 0.9 m ) i
rH/A = ( 0.9 m ) i − ( 0.3 m ) j − ( 0.675 m ) k
Now,
ΣM AD = 0:
(
)
(
)
(
)
(
)
λ AD ⋅ rG/ A × WAB + λ AD ⋅ rF / A × T + λ AD ⋅ rB/ A × WBC + λ AD ⋅ rH / A × WCD = 0
6
−2
−9
6 − 2 −9
6
−2
−9
T
1
1
0.45
0
0   + 0.6 0
0
0
0  
+ 0.9
11
11 33.25
 
 11 
0 −11.0363 0
0 − 3.6788 0
− 2 3 − 4.5
6
−2
−9
1
+ 0.9
− 0.3
− 0.675   = 0
 11 
0 −16.5544
0
4.0634 − 0.34054T + 2.7089 + 6.0950 = 0
Solving for T:
TBG = 37.785 N
or TBG = 37.8 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 4, Solution 149.
Free-Body Diagram:
Express forces in terms of their rectangular components:
(
)
(
)
(
)
WAB = − (1.25 kg/m ) 9.81 m/s 2 ( 0.9 m ) j = − (11.0363 Ν ) j
WBC = − (1.25 kg/m ) 9.81 m/s 2 ( 0.3 m ) j = − ( 3.6788 Ν ) j
WCD = − (1.25 kg/m ) 9.81 m/s 2 (1.35 m ) j = − (16.5544 Ν ) j
uuur
CE
T=T
=T
CE
−0.9 i + 1.2 j − 1.35 k
2
2
( − 0.9 ) + (1.2 ) + ( −1.35)
2
=
T
( −3 i + 4 j − 4.5 k )
45.25
continued
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
uuur
AD
λ AD =
=
AD
0.9 i − 0.3 j − 1.35 k
2
2
( 0.9 ) + ( − 0.3) + ( −1.35)
2
=
6
2
9
i− j− k
11
11
11
rG/ A = ( 0.45 m ) i
rC/ A = ( 0.9 m ) i − ( 0.3 m ) j
rB/ A = ( 0.9 m ) i
rH/A = ( 0.9 m ) i − ( 0.3 m ) j − ( 0.675 m ) k
Now,
ΣM AD = 0:
(
)
(
)
(
)
(
)
λ AD ⋅ rG/ A × WAB + λ AD ⋅ rC/ A × T + λ AD ⋅ rB/ A × WBC + λ AD ⋅ rH / A × WCD = 0
6
−2
−9
6 − 2 −9
6
−2
−9
T
1
1
0.45
0
0   + 0.9 − 0.3 0
0
0  
+ 0.9
11
11 45.25
 
 11 
0 −11.0363 0
0 − 3.6788 0
− 3 4 − 4.5
6
−2
−9
1
+ 0.9
− 0.3
− 0.675   = 0
 11 
0 −16.5544
0
4.0634 − 0.32840T + 2.7089 + 6.0950 = 0
Solving for T:
T = 39.182 N
or T = 39.2 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 4, Solution 150.
Free-Body Diagram:
Express forces in terms of their rectangular components:
uuur
FG
18 i − 6 j − 9 k
T
TFG = TFG
= TFG
= FG ( 6 i − 2 j − 3 k )
2
2
2
FG
7
(18) + ( − 6 ) + ( − 9 )
uuur
AB
λ AB =
=
AB
ΣM AB = 0:
13.5 i + 9 j − 27 k
(13.5)2 + ( 9 )2 + ( − 27 )2
=
3
2
6
i+ j− k
7
7
7
λ AB ⋅ ( rAE × FE ) + λ AB ⋅ ( rBG × TFG ) = 0
3 2 −6
3
2 −6
T 1
1
−1.5 3 − 9  18 + 13.5 −13 0 FG   = 0
7 7
7
0 1 0
6 − 2 −3
18 ( 9 + 27 ) +
TFG
(117 + 162 − 468 + 81) = 0
7
Solving for TFG:
TFG = 42.000 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.
or
TFG = 42.0 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 151.
Free-Body Diagram:
(a) The location of D follows from the geometry of the problem. Since the steel plate is rectangular rD/ A is
perpendicular to rB/ A and therefore:
rD/ A ⋅ rB/ A = 0
Denoting the coordinates of D by (0, y, z):
rD/ A = − ( 0.1 m ) i + yj + ( z − 0.7 m ) k
rB/ A = ( 0.3 m ) i − ( 0.4 m ) k
and
Thus, rD/ A ⋅ rB/ A = − 0.03 − 0.4 z + 0.28 = 0
or
z = 0.625 m.
rD/ A =
( − 0.1 m )2 + y 2 + ( 0.625 m − 0.7 m )2 = 0.75 m
continued
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
Solving for y:
y = 0.73951 m
x = 0, y = 0.740 m, z = 0.625 m Location of D is therefore:
(b) Consider moment equilibrium about axis AB:
uuur
AB
=
λ AB =
AB
0.3i − 0.4k
( 0.3)2 + ( − 0.4 )2
= 0.6i − 0.8k
rD/ A = − ( 0.1 m ) i + ( 0.73951 m ) j − ( 0.075 m ) k
rD/B = − ( 0.4 m ) i + ( 0.73951 m ) j + ( 0.625 m − 0.3 m ) k
N D = N Di
(
)
W = − ( mg ) j = − ( 40 kg ) 9.81 m/s 2 j = − ( 392.4 N ) j
Then,
ΣM AB = 0:
(
)
(
)
λ AB ⋅ rD/ A × N D + λ AB ⋅ rG/B × W = 0
0.6
0
− 0.8
0.6
0
−0.8
− 0.1 0.73951 − 0.075 + − 0.2 0.36976 0.1625 = 0
ND
− 392.4
0
0
0
0
0.59161 N D − 24.525 = 0
N D = 41.455 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.
or N D = ( 41.455 N ) i COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 152.
Free-Body Diagram:
From free-body diagram of beam
ΣFx = 0: Bx = 0
so that
B = By
ΣFy = 0: A + B − (100 + 200 + 300 ) N = 0
A + B = 600 N
or
Therefore, if either A or B has a magnitude of the maximum of 360 N,
the other support reaction will be < 360 N ( 600 N − 360 N = 240 N ) .
(100 N )( d ) − ( 200 N )( 0.9 − d ) − ( 300 N )(1.8 − d )
ΣM A = 0:
+ B (1.8 − d ) = 0
d =
or
720 − 1.8B
600 − B
Since B ≤ 360 N,
d =
720 − 1.8 ( 360 )
600 − 360
ΣM B = 0:
= 0.300 m
or
d ≥ 300 mm
(100 N )(1.8) − A (1.8 − d ) + ( 200 N )( 0.9 ) = 0
d =
or
1.8 A − 360
A
Since A ≤ 360 N,
d =
1.8 ( 360 ) − 360
360
= 0.800 m
or
d ≤ 800 mm
or 300 mm ≤ d ≤ 800 mm 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 4, Solution 153.
Free-Body Diagram:
Have
Cmax = 1000 N
Now
C 2 = C x2 + C y2
∴ Cy =
(1000 )2 − Cx2
(1)
From free-body diagram of pedal
ΣFx = 0: C x − Tmax = 0
∴ C x = Tmax
(2)
ΣM D = 0: C y ( 0.4 m ) − Tmax ( 0.18 m ) sin 60° = 0
∴ C y = 0.38971Tmax
(3)
Equating the expressions for C y in Equations (1) and (3), with C x = Tmax
from Equation (2)
2
= 0.389711Tmax
(1000 )2 − Tmax
2
∴ Tmax
= 868,150
and
Tmax = 931.75 N
or Tmax = 932 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 4, Solution 154.
Free-Body Diagram:
From free-body diagram of inverted T-member
ΣM C = 0: T ( 25 in.) − T (10 in.) − ( 30 lb )(10 in.) = 0
∴ T = 20 lb
or T = 20.0 lb ΣFx = 0: C x − 20 lb = 0
∴ C x = 20 lb
C x = 20.0 lb
or
ΣFy = 0: C y + 20 lb − 30 lb = 0
∴ C y = 10 lb
C y = 10.00 lb
or
Then
and
C =
C x2 + C y2 =
( 20 )2 + (10 )2 = 22.361 lb
 Cy 
−1  10 
 = tan   = 26.565°
 20 
 Cx 
θ = tan −1 
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.
C = 22.4 lb
26.6° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 155.
Free-Body Diagram:
From free-body diagram of frame with T = 300 lb
 5
ΣFx = 0: C x − 100 lb +   300 lb = 0
 13 
∴ C x = −15.3846 lb
or
C x = 15.3846 lb
 12 
ΣFy = 0: C y − 180 lb −   300 lb = 0
 13 
∴ C y = 456.92 lb
Then
and
C =
or
C y = 456.92 lb
(15.3846 )2 + ( 456.92 )2 = 457.18 lb
C x2 + C y2 =
 Cy 
−1  456.92 
 = tan 
 = −88.072°
 −15.3846 
 Cx 
θ = tan −1 
or C = 457 lb
88.1°  12 

ΣM C = 0: M C + (180 lb )( 20 in.) + (100 lb )(16 in.) −   300 lb  (16 in.) = 0
 13 

∴ M C = −769.23 lb ⋅ in.
or M C = 769 lb ⋅ in.
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 4, Solution 156.
Free-Body Diagram:
(a) From free-body diagram of rod AB
ΣM C = 0: P ( l cosθ ) + P ( l sin θ ) − M = 0
or sin θ + cosθ =
(b) For
M
Pl
M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.
sin θ + cosθ =
150 lb ⋅ in.
5
= = 1.25
20
lb
6
in.
(
)( ) 4
sin 2 θ + cos 2 θ = 1
Using identity
(
)
1
sin θ + 1 − sin 2 θ 2 = 1.25
1
2
(1 − sin θ ) = 1.25 − sinθ
2
1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ
2sin 2 θ − 2.5sin θ + 0.5625 = 0
Using quadratic formula
sin θ =
=
or
− ( −2.5) ±
( 6.25) − 4 ( 2 )( 0.5625)
2 ( 2)
2.5 ± 1.75
4
sin θ = 0.95572
∴ θ = 72.886°
and
sin θ = 0.29428
θ = 17.1144°
and
or θ = 17.11° and θ = 72.9° 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 4, Solution 157.
From geometry of forces
 yBE 

 1.5 ft 
β = tan −1 
where
yBE = 2.0 − yDE
= 2.0 − 1.5 tan 35°
= 0.94969 ft
 0.94969 
∴ β = tan −1 
 = 32.339°
 1.5 
and
α = 90° − β = 90° − 32.339° = 57.661°
θ = β + 35° = 32.339° + 35° = 67.339°
Applying the law of sines to the force triangle,
200 lb
T
B
=
=
sin θ
sin α
sin 55°
or
(a)
( 200 lb )
sin 67.339°
T =
=
T
B
=
sin 57.661° sin 55°
( 200 lb )( sin 57.661° ) = 183.116 lb
sin 67.339°
or T = 183.1 lb (b)
B=
( 200 lb )( sin 55° ) = 177.536 lb
sin 67.339°
or B = 177.5 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.
32.3° COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 158.
Free-Body Diagram:
yED = xED = a,
Since
Slope of ED is
45°
∴ slope of HC is
45°
DE =
Also
and
2a
a
1
DH = HE =   DE =
2
2
For triangles DHC and EHC
sin β =
a = 25 mm and
sin β =
R = 125 mm
25 mm
= 0.141421
2 (125 mm )
∴ β = 8.1301°
and
a
2R
c = R sin ( 45° − β )
Now
For
a/ 2
=
R
or β = 8.13° c = (125 in.) sin ( 45° − 8.1301° ) = 75.00 in.
or c = 75.0 in. 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 4, Solution 159.
Free-Body Diagram:
First note
(
)
W = mg = (17 kg ) 9.81 m/s 2 = 166.77 N
h=
(1.2 )2 − (1.125)2 = 0.41758 m
From free-body diagram of plywood sheet
 (1.125 m ) 
ΣM z = 0: C ( h ) − W 
 =0
2


C ( 0.41758 m ) − (166.77 N ) ( 0.5625 m ) = 0
∴ C = 224.65 N
or
C = − ( 225 N ) i
ΣM B( y -axis ) = 0: − ( 224.65 N ) ( 0.6 m ) + Ax (1.2 m ) = 0
or
∴ Ax = 112.324 N
A x = (112.3 N ) i
ΣM B( x-axis ) = 0: (166.77 N ) ( 0.3 m ) − Ay (1.2 m ) = 0
∴ Ay = 41.693 N
or
A y = ( 41.7 N ) j
ΣM A( y -axis ) = 0: ( 224.65 N ) ( 0.6 m ) − Bx (1.2 m ) = 0
∴ Bx = 112.325 N
or
B x = (112.3 N ) i
ΣM A( x-axis ) = 0: B y (1.2 m ) − (166.77 N ) ( 0.9 m ) = 0
∴ By = 125.078 N
or
B y = (125.1 N ) j
∴ A = (112.3 N ) i + ( 41.7 N ) j B = (112.3 N ) i + (125.1 N ) j C = − ( 225 N ) i 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 4, Solution 160.
Free-Body Diagram:
First note
(
)
W = mg = ( 30 kg ) 9.81 m/s 2 = 294.3 N
FEC = λ EC FEC = ( sin15° ) i + ( cos15° ) j FEC
From free-body diagram of cover
(a)
ΣM z = 0:
( FEC cos15° ) (1.0 m ) − W ( 0.5 m ) = 0
or FEC cos15° (1.0 m ) − ( 294.3 N )( 0.5 m ) = 0
∴ FEC = 152.341 N
(b)
or FEC = 152.3 N ΣM x = 0: W ( 0.4 m ) − Ay ( 0.8 m ) − ( FEC cos15° ) ( 0.8 m ) = 0
or ( 294.3 N )( 0.4 m ) − Ay ( 0.8 m ) − (152.341 N ) cos15° ( 0.8 m ) = 0
∴ Ay = 0
ΣM y = 0: Ax ( 0.8 m ) + ( FEC sin15° ) ( 0.8 m ) = 0
or Ax ( 0.8 m ) + (152.341 N ) sin15° ( 0.8 m ) = 0
∴ Ax = −39.429 N
ΣFx = 0: Ax + Bx + FEC sin15° = 0
−39.429 N + Bx + (152.341 N ) sin15° = 0
∴ Bx = 0
ΣFy = 0: FEC cos15° − W + By = 0
or (152.341 N ) cos15° − 294.3 N + B y = 0
∴ By = 147.180 N
or A = − ( 39.4 N ) i B = (147.2 N ) j 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 4, Solution 161.
Free-Body Diagram:
λCD =
First note
=
λCE =
=
− ( 23 in.) i + ( 22.5 in.) j − (15 in.) k
35.5 in.
1
( −23i + 22.5j − 15k )
35.5
( 9 in.) i + ( 22.5 in.) j − (15 in.) k
28.5 in.
1
( 9i + 22.5 j − 15k )
28.5
W = − ( 285 lb ) j
From free-body diagram of plate
(a)
ΣM x = 0:
 22.5  
 22.5  
 T  (15 in.) − 
 T  (15 in.) = 0
 35.5  
 28.5  
( 285 lb )( 7.5 in.) − 
∴ T = 100.121 lb
or T = 100.1 lb continued
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
 23 
 9 
ΣFx = 0: Ax − T 
 +T
=0
35.5


 28.5 
(b)
 23  (
 9 
Ax − (100.121 lb ) 
 + 100.121 lb ) 
=0
35.5


 28.5 
∴ Ax = 33.250 lb
  22.5  
  22.5  
ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T 
  ( 6 in.) − T 
  ( 6 in.) = 0
  35.5  
  28.5  
or

 22.5  
− Ay ( 26 in.) + ( 285 lb )(13 in.) − (100.121 lb ) 
  ( 6 in.)
 35.5  


 22.5  
− (100.121 lb ) 
  ( 6 in.) = 0
 28.5  

∴ Ay = 109.615 lb
  15  
  23  
ΣM B( y -axis ) = 0: Az ( 26 in.) − T 
  ( 6 in.) − T 
  (15 in.)
  35.5  
  35.5  
  15  
  9 
− T 
  ( 6 in.) + T 
  (15 in.) = 0
  28.5  
  28.5  
or
1
 −1
Az ( 26 in.) + 
( 90 + 345) −
( 90 − 135) (100.121 lb ) = 0
28.5
 35.5

∴ Az = 41.106 lb
or A = ( 33.3 lb ) i + (109.6 lb ) j + ( 41.1 lb ) k  22.5 
 22.5 
ΣFy = 0: By − W + T 
 +T
 + Ay = 0
 35.5 
 28.5 
 22.5 22.5 
B y − 285 lb + (100.121 lb ) 
+
 + 109.615 lb = 0
 35.5 28.5 
∴ By = 32.885 lb
 15 
 15 
ΣFz = 0: Bz + Az − T 
 −T
=0
 35.5 
 28.5 
15 
 15
Bz + 41.106 lb − (100.121 lb ) 
+
=0
35.5
28.5


∴ Bz = 53.894 lb
or B = ( 32.9 lb ) j + ( 53.9 lb ) k 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 4, Solution 162.
First note
Free-Body Diagram:
TBG = λBGTBG =
− (18 in.) i + (13.5 in.) k
2
2
(18) + (13.5) in.
TBG
= TBG ( −0.8i + 0.6k )
TDH = λDH TDH =
− (18 in.) i + ( 24 in.) j
(18)2 + ( 24 )2 in.
TDH
= TDH ( −0.6i + 0.8j)
Since λFJ = λDH ,
TFJ = TFJ ( −0.6i + 0.8j)
From free-body diagram of member ABF
ΣM A( x-axis ) = 0:
( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) − (120 lb )(12 in.) = 0
∴ 3.2TFJ + 1.6TDH = 480
ΣM A( z -axis ) = 0:
(1)
( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) − (120 lb )(18 in.) = 0
∴
− 3.2TFJ − 3.2TDH = −960
(2)
Equation (1) + Equation (2)
Substituting in Equation (1)
ΣM A( y -axis ) = 0:
TDH = 300 lb TFJ = 0 ( 0.6TFJ ) ( 48 in.) + 0.6 ( 300 lb ) ( 24 in.) − ( 0.6TBG ) (18 in.) = 0
∴ TBG = 400 lb continued
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
ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0
−0.6 ( 300 lb ) − 0.8 ( 400 lb ) + Ax = 0
∴ Ax = 500 lb
ΣFy = 0: 0.8TFJ + 0.8TDH − 240 lb + Ay = 0
0.8 ( 300 lb ) − 240 + Ay = 0
∴ Ay = 0
ΣFz = 0: 0.6TBG + Az = 0
0.6 ( 400 lb ) + Az = 0
∴ Az = −240 lb
Therefore,
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 = ( 500 lb ) i − ( 240 lb ) k COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 163.
Free-Body Diagram:
First note
λ AE =
− ( 70 mm ) i + ( 240 mm ) k
2
=
2
( 70 ) + ( 240 ) mm
1
( −7i + 24k )
25
rC/ A = ( 90 mm ) i + (100 mm ) k
FC = − ( 600 N ) j
rD/ A = ( 90 mm ) i + ( 240 mm ) k
T = λDF T =
=
− (160 mm ) i + (110 mm ) j − ( 80 mm ) k
(160 )2 + (110 )2 + (80 )2 mm
T
T
( −16 i + 11j − 8k )
21
From the free-body diagram of the bend rod
(
)
(
)
ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rD/ A × T = 0
∴
−7 0 24
−7 0 24
 600 
 T 
90 0 100 
=0
 + 90 0 240 
25 ( 21) 
 25 

0 −1 0
−16 11 −8
600 
 T 
 + (18 480 + 23 760 ) 
 =0
 25 
 25 ( 21) 
( −700 − 2160 ) 
∴ T = 853.13 N
or T = 853 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 5, Solution 1.
\
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
200 × 150 = 30000
−100
250
− 30 000000
6 750 000
2
400 × 300 = 120000
200
150
24 000 000
18000000
Σ
150 000
21000 000
24 750000
X =
ΣxA 21 000000
=
mm
ΣA
150000
or X = 140.0 mm Y =
ΣyA 24 750000
=
mm
ΣA
150 000
or Y = 165.0 mm 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 5, Solution 2.
A,in 2
x ,in.
y ,in.
xA,in 3
yA,in 3
1
10 × 8 = 80
5
4
400
320
2
1
× 9 × 12 = 54
2
13
4
702
216
Σ
134
1102
536
Then
X =
ΣxA 1102
=
ΣA
134
or
and
Y =
ΣyA 1102
=
ΣA
134
or Y = 4.00 in. X = 8.22 in. 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 5, Solution 3.
Then
A, mm 2
x , mm
xA, mm3
1
1
× 90 × 270 = 12 150
2
2
( 90 ) = 60
3
729 000
2
1
× 135 × 270 = 18 225
2
1
(135) = 135
3
2 460 375
Σ
30375
X =
90 +
ΣxA 3189375
mm
=
ΣA
30375
3 189 375
or X = 105.0 mm For the whole triangular area by observation:
Y =
1
( 270 mm )
3
or Y = 90.0 mm 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 5, Solution 4.
A,in 2
x ,in.
1
1
( 21)( 24 ) = 252
2
2
(13)( 40 ) = 520
Σ
2
( 21) = 14
3
21 +
1
(13) = 27.5
2
40 −
y ,in.
xA,in 3
yA,in 3
1
( 24 ) = 32
3
3528
8064
14 300
10 400
17 828
18 464
20
772
Then
X =
ΣxA 17828
=
in.
ΣA
772
or
Y =
ΣyA 18464
=
in.
ΣA
772
or Y = 23.9 in. X = 23.1 in. 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 5, Solution 5.
A, mm 2
x , mm
2
1
π ( 225 )
2
1
( 375)( 225) = 42 188
2
4
−
= 39 761
4 ( 225 )
3π
= − 95.493
125
y , mm
xA, mm3
yA, mm3
95.493
− 3 796 900
3 796 900
75
5 273 500
3 164 100
1 476 600
6 961 000
81 949
Σ
Then
X =
ΣxA 1476600
mm
=
ΣA
81 949
or X = 18.02 mm Y =
ΣyA 6961 000
mm
=
ΣA
81 949
or
Y = 84.9 mm 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 5, Solution 6.
1
2
3
−
π
4
−
A,in 2
x ,in.
y ,in.
xA,in 3
yA,in 3
17 × 9 = 153
8.5
4.5
1300.5
688.5
2
× ( 4.5 ) = −15.9043 8 −
π
4
( 6 )2 = − 28.274
Σ
4 × 4.5
4 × 4.5
= 6.0901 9 −
= 7.0901 − 96.857
3π
3π
−112.761
− 298.19
−182.466
905.45
393.27
10.5465
6.4535
108.822
Then
X =
ΣxA
905.45
=
ΣA 108.822
or
and
Y =
ΣyA 393.27
=
ΣA 108.22
or Y = 3.61 in. X = 8.32 in. 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 5, Solution 7.
A,in 2
2
1
π (16 )
2
− ( 8 )( 8 ) = − 64
Σ
137.06
4
= 201.06
ΣxA 1109.32
=
in.
ΣA
137.06
Then
X =
and
Y = X by symmetry
x ,in.
4 (16 )
3π
xA,in 3
= 6.7906
1365.32
4
− 256
1109.32
or
X = 8.09 in. or Y = 8.09 in. 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 5, Solution 8.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
35 343
63.662
0
2 250 006
0
2
− 4417.9
31.831
− 31.831
−140 626
140 626.2
Σ
30925.1
2 109 380
140 626.2
Then
X =
ΣxA 2109 380
=
ΣA
30 925.1
or
and
Y =
ΣyA 140 625
=
ΣA
30 925.1
or Y = 4.55 mm X = 68.2 mm 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 5, Solution 9.
x
xA
r12
4r1
3π
r13
 π 2  4r1 
 − r1 
=−
3
 4  3π 
r22
4r2
3π
 π 2  4r2  2r23
 r2 
=
3
 2  3π 
A
−
1
Therefore, for X =
4
π
2
Σ
π2
2
π
1
2r23 − r13
3
( 2r − r )
4
2
2
(
2
1
)
ΣxA 4r1
=
:
ΣΑ
3π
(
(
)
4 2r23 − r13
4r1
=
3π
3π 2r22 − r12
)
  r 3

r13  2  2  − 1
  r1 

4


=
2
 r 

3π
r12  2  2  − 1
  r1 



r
2ρ 3 − 1
, where ρ = 2
2
r1
2ρ − 1
or
π =
or
2 ρ 3 − 2πρ 2 + (π − 1) = 0.
Solving numerically for ρ and noting that ρ > 1:
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.
r2
= 3.02 r1
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 10.
First, determine the location of the centroid.
y2 =
From Fig. 5.8A:
Similarly
Then
(
=
2
cos α
r2 π
3
−α
2
y1 =
2
cos α
r1 π
3
−α
2
Σ yA =
)
(
)
(
)
(
)
)
A2 =
A1 =
(
(
)
( π2 − α ) r12
(
)
2 3
r2 − r13 cosα
3
π

π

Σ A =  − α  r22 −  − α  r12
2

2

π

=  − α  r22 − r12
2


Y Σ A = Σ yA
(
Now
( π2 − α ) r22
2
cosα  π
2
cosα  π
r2 π
− α r22  − r1 π
− α r12 
2


 2

3
3
−
−
α
α
2
2
=
and
(
π
2 sin 2 − α
r2 π
3
−α
2
)
 π
 2 3

Y  − α  r22 − r12  =
r2 − r13 cos α

 2
 3
(
)
(
)
Y =
2 r23 − r13 cos α
3 r22 − r12 π2 − α
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
Using Figure 5.8B, Y of an arc of radius
1
( r1 + r2 ) is
2
Y =
=
(
sin π − α
1
( r1 + r2 ) π 2
2
−α
2
(
)
)
1
cos α
(r1 + r2 ) π
2
−α
2
(
(
( r2 − r1 ) r22 + r1 r2 + r12
r23 − r13
=
r22 − r12
( r2 − r1 )( r2 + r1 )
Now
=
(1)
)
)
r22 + r1 r2 + r12
r2 + r1
r2 = r + ∆
Let
r1 = r − ∆
r =
Then
1
( r1 + r2 )
2
2
and
( r + ∆ ) + ( r + ∆ )( r − ∆ ) + ( r − ∆ )
r23 − r13
=
2
2
r2 − r1
(r + ∆) + (r − ∆)
=
2
3r 2 + ∆ 2
2r
In the limit as ∆ → 0 (i.e., r1 = r2 ), then
r23 − r13
3
= r
2
2
2
r2 − r1
=
so that
Y =
3 1
× (r1 + r2 )
2 2
2 3
cos α
× ( r1 + r2 ) π
3 4
−α
2
Which agrees with Eq. (1).
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.
or Y =
1
cos α
!
( r1 + r2 ) π
2
−α
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 11.
Then
X =
A,in 2
x ,in.
xA,in 3
1
27
8.1962
221.30
2
15.5885
3.4641
54.000
3
−18.8495
3.8197
−71.999
Σ
23.739
ΣxA 203.30
=
ΣA
23.739
203.30
or
X = 8.56 in. and by symmetry
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.
Y =0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 12.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
1
( 240 )(150 ) = 18 000
2
160
50
2 880 000
900 000
2
−
1
( 240 )(150 ) = 12 000
3
3
( 240 ) = 180
4
3
(150 ) = 45
10
−2160000
−540 000
720 000
360 000
6000
Σ
Then
X =
ΣxA 720000
=
mm
ΣA
6000
or X = 120.0 mm Y =
ΣyA 360000
=
mm
ΣA
6000
or
Y = 60.0 mm 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 5, Solution 13.
A,in 2
x ,in.
y ,in.
xA,in 3
yA,in 3
1
(18)(8) = 144
−3
4
− 432
576
2
1
( 6 )( 9 ) = 27
2
2
−3
54
−81
(12 )( 9 ) = 84.823
−5.0930
−3.8197
− 432.00
− 324.00
−810.00
171.00
3
Σ
Then
π
4
255.82
X =
ΣxA −810.00
=
in.
255.82
ΣA
Y =
ΣyA 171.00
=
in.
ΣA
255.82
or
X = − 3.17 in. or Y = 0.668 in. 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 5, Solution 14.
X = 90 mm First, by symmetry
1
2
−
3
−
A, mm 2
y , mm
yA, mm3
(180 )(120 ) = 21 600
60
1 296 000
π
4
π
4
( 90 )(120 ) = − 8482.3
120 −
4 × 120
= 69.070
3π
−585 870
( 90 )(120 ) = − 8482.3
120 −
4 × 120
= 69.070
3π
−585 870
4635.4
Σ
Y =
ΣyA 124 260
=
4635.4
ΣA
124 260
or Y = 26.8 mm 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 5, Solution 15.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
18 240
−4
12
72 960
218 880
2
−1920
− 56
54
107520
−103 680
3
− 4071.5
− 41.441
− 41.441
168 731
186 731
Σ
12 248.5
−134171
−53 531.1
Then
and
X =
ΣxA −134171
=
ΣA
12 248.5
Y =
ΣyA −53 531
=
ΣA 12 248.5
or
X = −10.95 mm or Y = − 43.7 mm 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 5, Solution 16.
\
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
2
( 200 )( 200 ) = 26 667
3
75
70
2 000 000
1866 690
2
−
2
(100 )( 50 ) = − 3333.3
3
37.5
− 20
−125 000
66 666
1875 000
1 933 360
23 334
Σ
Then X =
ΣxA 1875 000
=
mm
ΣA
23 334
or X = 80.4 mm Y =
ΣyA 1 933 360
=
mm
ΣA
23 334
or Y = 82.9 mm 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 5, Solution 17.
Locate first Y :
Note that the origin of the X axis is at the bottom of the whole area.
Then
Y =
A, in 2
y , in.
yA, in 3
1
8 × 15 = 120
7.5
900
2
− 4 × 10 = − 40
8
− 320
Σ
80
580
ΣyA 580
=
= 7.2500 in.
ΣA
80
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
Now, to find the first moment of each area about the x-axis:
Area I:
QI = ΣyA =
7.75
5.75
 − ( 4 × 5.75 )  ,
(8 × 7.75) +
2
2 
or QI = 174.125 in 3 !
Area II:
QII = ΣyA = −
7.75
4.25
 − ( 4 × 4.25 )  ,
(8 × 7.25) −
2
2 
or QII = −174.125 in 3 !
Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.
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 5, Solution 18.
A, mm 2
y , mm
yA, mm3
1
(80 )( 20 ) = 1600
90
144 000
2
( 20 )(80 ) = 1600
40
64 000
Σ
Then
Y =
3200
208 000
ΣyA 208 000
=
= 65.000 mm
ΣA
3200
Now, for the first moments about the x-axis:
Area I
QI = ΣyA = 25 ( 80 × 20 ) + 7.5 ( 20 × 15 ) = 42 250 mm3 ,
or QI = 42.3 × 103 mm3 !
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
Area II
QII = ΣyA = − 32.5 ( 20 × 65 ) = 42 250 mm3 ,
or QII = 42.3 × 103 mm3 !
Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.
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 5, Solution 19.
(a) With Qx = Σ yA and using Fig. 5.8 A,
(
)
 2 r sin π − θ 
  r 2 π2 − θ  −
Qx =  3 π 2

−
θ


2


2
= r 3 cos θ − cos θ sin 2 θ
3
(
) ( 32 r sin θ )  12 × 2r cos θ × r sin θ 
(
)
or Qx =
(b) By observation, Qx is maximum when
and then
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.
2 3
r cos3 θ 3
θ =0
Qx =
2 3
r 3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 20.
From the problem statement: F is proportional to Qx . Therefore:
FA
FB
=
, or
( Qx ) A ( Qx )B
FB =
( Qx )B
F
( Qx ) A A
For the first moments:
( Qx ) A =  225 +

12 
3
 ( 300 × 12 ) = 831 600 mm
2
( Qx )B = ( Qx ) A + 2  225 −

Then
FB =
12 
3
 ( 48 × 12 ) + 2 ( 225 − 30 )(12 × 60 ) = 1 364 688 mm
2
1364688
( 280 N ) ,
831600
or FB = 459 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 5, Solution 21.
Because the wire is homogeneous, its center of gravity will coincide with the centroid for the
corresponding line.
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
400
200
0
80 000
0
2
300
400
150
120 000
45 000
3
600
100
300
60 000
180 000
4
150
− 200
225
− 30 000
33 750
5
200
−100
150
− 20 000
30 000
6
150
0
75
0
11 250
Σ
1800
210 000
300 000
Then
X =
ΣxL
210 000
=
= 116.667 mm
ΣL
1800
or X = 116.7 mm and
Y =
ΣyL 300 000
=
= 166.667 mm
ΣL
1800
or Y = 166.7 mm 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 5, Solution 22.
L, in.
x , in.
y , in.
xL, in 2
y , in 2
1
19
9.5
0
180.5
0
2
15
14.5
6
217.5
90
3
4
10
10
40
40
4
10
5
8
50
80
5
8
0
4
0
32
Σ
56
488
242
Then
X =
ΣxL
488
=
ΣL
56
or X = 8.71 in. and
Y =
ΣyA 242
=
56
ΣA
or Y = 4.32 in. 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 5, Solution 23.
Because the wire is homogeneous, its center of gravity will coincide with the centroid for the
corresponding line.
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
600
75
0
45 000
0
2
3752 + 2252 = 437.32
187.5
112.5
81 998
49 199
2
2
− 50 625
50 625
76 373
99 824
3
π
2
Σ
( 225)
−
π
( 225)
1390.75
π
( 225)
Then
X =
ΣxL
76 373
=
ΣL
1390.75
or X = 54.9 mm and
Y =
ΣyL
99 824
=
ΣL
1390.75
or Y = 71.8 mm 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 5, Solution 24.
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
75
37.5
0
2812.5
0
2
150
0
75
0
11 250
3
(150 )π = 471.24
95.492
0
45 000
0
4
75
0
−112.5
0
− 8437.5
47.746
− 47.746
5625.0
− 5625.0
53 437
− 2812.5
5
Σ
( 75)
π
2
= 117.81
889.05
Then
X =
ΣxL 53 437
=
,
ΣL
889.05
or X = 60.1 mm and
Y =
ΣyA − 2812.5
=
ΣA
889.05
or Y = − 3.16 mm 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 5, Solution 25.
O
From Figure 5.8 b:
r =
( 20 in.) sin 30° = 60 in.
π
π
6
Note also that triangle ABO is equilateral, where O is the origin of the coordinate system in the
figure.
For equilibrium:
(a) ΣM A = 0:


 60 
 20 in. −  in.  cos 30° (1.75 lb ) − ( 20 in.) sin 60° TBC = 0
π



Solving for TBC :
TBC = 0.34960 lb
(b) ΣFx = 0:
or
TBC = 0.350 lb Ax + ( 0.34960 lb ) cos 60° = 0
Ax = − 0.174800 lb
ΣFx = 0:
Ay − 1.75 lb + ( 0.34960 lb ) sin 60° = 0
Ay = 1.44724 lb
Therefore:
A = 1.458 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.
83.1° COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 26.
The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity
of the wire must lie directly under B. Also, because the wire is homogeneous the center of gravity
will coincide with the centroid. In other words, x = 0, or ΣxL = 0.
ΣxL = −
2 (150 mm )
π
150 mm
 200 mm 


π (150 mm )  + 
cosθ  (150 mm )
 ( 200 mm ) +  200 mm −
2
2




or
cosθ =
5000
11250
or
θ = 63.6° 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 5, Solution 27.
The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity
of the wire must lie directly under B. Also, because the wire is homogeneous the center of gravity
will coincide with the centroid. In other words, x = 0, or ΣxL = 0.
ΣxL = −
2 (150 mm )
π
150 mm
 200 mm 


π (150 mm )  + 
cosθ  (150 mm )
 ( 200 mm ) +  200 mm −
2
2




or
l 2 + 300l − 197602 = 0.
Solving for l :
l = 319.15, and l = − 619.15, and discarding the negative root
l = 319 mm 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 5, Solution 28.
The centroid coincides with the center of gravity because the wire is homogeneous.
L
1
r
2
2θ r
−
r
2
−
−
r sin θ
− 2r 2 sin θ
θ
l
2
X =
Then
xL
−
l
3
x
r2
2
l2
2
ΣxL
= 0 ⇒ ΣxL = 0 and
ΣL
r2
l2
− 2r 2 sin θ +
= 0, or l = r 1 + 4sin θ
2
2
(a) θ = 15° :
l = r 1 + 4sin15°
or l = 1.427 r (b) θ = 60° :
l = r 1 + 4sin 60°
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.
or l = 2.11 r COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 29.
y =
Then
y =
or
ΣyA
ΣA
 (a + h) 
a
( ab ) − 
  kb ( a − h ) 
2
 2 
ba − kb ( a − h )
1 a (1 − k ) + kh
2 a(1 − k ) + kh
2
=
Let
c =1− k
Then
y =
and
2
ζ =
h
a
a c + kζ 2
2 c + kζ
(1)
Now find a value of ζ (or h) for which y is minimum:
(
dy
a 2kζ ( c + kζ ) − k c + kζ
=
dζ
2
( c + kζ ) 2
2
)=0
or
(
)
2ζ ( c + kζ ) − c + kζ 2 = 0
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.
(2)
COSMOS: Complete Online Solutions Manual Organization System
2cζ + 2ζ 2 − c − kζ 2 = 0
Expanding (2)
ζ =
Then
− 2c ±
or
kζ 2 + 2cζ − c = 0
( 2c )2 − 4 ( k ) ( c )
2k
Taking the positive root, since h > 0 (hence ζ > 0 )
2
h=a
− 2 (1 − k ) + 4 (1 − k ) + 4k (1 − k )
2k
2
(a) k = 0.2:
h=a
(b) k = 0.6:
h=a
− 2 (1 − 0.2 ) + 4 (1 − 0.2 ) + 4 ( 0.2 )(1 − 0.2 )
2 ( 0.2 )
or h = 0.472a !
2
− 2 (1 − 0.6 ) + 4 (1 − 0.6 ) + 4 ( 0.6 )(1 − 0.6 )
2 ( 0.6 )
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.
or h = 0.387a !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 30.
From Problem 5.29, note that Eq. (2) yields the value of ζ that minimizes h.
Then from Eq. (2)
We see
2ζ =
c + kζ 2
c + kζ
(3)
Then, replacing the right-hand side of (1) by 2ζ , from Eq. (3)
We obtain
y =
a
( 2ζ)
2
But
ζ=
h
a
So
y =h
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.
Q.E.D. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 31.
\
Note that y1 = −
=
h
x+h
a
h
(a − x)
a
Choose the area element (EL) as
dA = ( h − y1 ) dx =
h
xdx
a
a
Then
A=
h a
h 1 
1
xdx =  x 2  = ah
∫
0
a
a  2 0
2
Now, noting that xEL = x, and yEL =
1
( h + y1 )
2
a
1
2 a h
2 1 
2

x = ∫ xdA =
x
xdx  = 2  x3  = a
∫
0 
A
ah
3
a
 a  3 0
y =
1 1
2 h 1
2 1 a 2

2
∫ ( h + y1 ) dA = ah ∫ 0  2 ( h + y1 )  ( h − y1 ) dx  = ah  2  ∫ 0 h − y1 dx
A 2


 
(
)
A
1 a  2 h2
h
h
1  2
2
3
 1  1 
=
h − 2 ( a − x )  dx =  x +   2  ( a − x )  =  a − a  = h
∫
0 
ah 
a
a
3  3
a
 3  a 
0

Therefore:
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.
x =
2
a!
3
y =
2
h!
3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 32.
First determine k:
For x = a, y = 0 and therefore
(
)
0 = h 1 − ka3 or k = a −3 , and therefore

x3 
y = h 1 − 3 
a 

Choosing an area element as in the figure:
xEL = x,
yEL =
A = ∫ dA = ∫
∫ xEL dA = ∫
y
,
2
a
ydx =
0
a
xydx =
0
∫
∫
and dA = ydx

x3 
a 
h
1
−

3
0 
dx = h  x −

a 

 x2
x4 
a 
h
x
−
dx
=
b


 −
3
0 

a 
2
a
x4 
3
= ah
3
4
4a  0
a
x5 
3 2
=
ab
3
5a  0 10
2
a
1 a 2
x3 
b2 a 
2 x3 x 6 
b2 
x4
x7 
9
a y 
1
y
dA
ydx
h
x
dx
dx
x
ab 2
=
=
−
=
−
+
=
−
+
=

∫ EL
∫0  2 
∫ 0 
∫ 0 
3
3
6
3
6


2
2
2
28
2
7
a
a
a
a
a
 





0
Now
x =
1
4 3a 2b 2
x
dA
=
= a
∫ EL
A
3ab 10
5
y =
1
4 9ab 2
3
yEL dA =
= b
∫
A
3ab 28
7
and
Therefore:
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.
2
a
5
3
y = b
7
x =
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 33.
For the element (EL) shown
x = a, y = h: h = k1a3
At
k1 =
or
h
a3
a = k 2 h3
k2 =
or
a
h3
Hence, on line 1
y =
h 3
x
a3
y =
h 1/3
x
a1/3
and on line 2
Then
h 
 h
dA =  1/3 x1/3 − 3 x3  dx
a

a
and
yEL =
1  h 1/3
h 3
 1/3 x + 3 x 
2 a
a
a
h 1/3
h 
1
1
 3

∴ A = ∫ dA = ∫
x − 3 x3  dx = h  1/3 x 4/3 − 3 x 4  = ah
1/3
2
a
4a
a

 4a
0
a
0
∫ xEL dA = ∫
a
h 1/3
h 
1
8 2
 3

x − 3 x3  dx = h  1/3 x 7/3 − 3 x5  =
a h
1/3
a
5a
a

 7a
 0 35
a 
x
0 
1
a 
1/3
3 
1/3
3
∫ yEL dA = ∫ 0 2  a1/3 x + a3 x  a1/3 x − a3 x  dx



h
h
h
h
h 2 a  x 2/3 x 6 
h 2  3 x5/3 1 x 6 
=
−
dx
=
−



∫
2/3
6
0
2  a
2  5 a5/3 7 a 6 
a 
a
=
0
8 2
ah
35
From
8 2
 ah 
xA = ∫ xEL dA: x   =
a h
 2  35
or x =
16
a 35
and
8 2
 ah 
yA = ∫ yEL dA: y   =
ah
 2  35
or y =
16
h 35
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 5, Solution 34.
Choose as an area element (EL) the shaded area shown:
π 
dA =  r  dr
2 
xEL =
2r
π
and
r
π 1  2 π 2
r π 
A = ∫r 2  r  dr =  r 2  =
r2 − r12
1  2 
2  2 r
4
(
1
)
Then
r
x =
1
4
4
r2  2r   π

1 3 2
x
dA
rdr
=
=
∫ EL
∫  

 r 
A
π r22 − r12 r1  π   2
π r22 − r12  3  r1
(
)
(
)
or x =
4 r23 − r13
3π r22 − r12
y =
4 r23 − r13
3π r22 − r12
and by symmetry
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 5, Solution 35.
Note that y1 = −
y2 =
b
b
x + b = ( a − x ) , and
a
a
b 2
a − x2
a
Then for the shaded area element:
dA = ( y2 − y1 ) dx =
b 2
a − x 2 − ( a − x )  dx and

a 
a
b  1 
2

 x  1
a 2 − x 2 − ( a − x )  dx =   x a 2 − x 2 + a 2 sin −1    + ( a − x ) 




a
a  2 
 a  2
 0
ab
A = ∫ dA = ∫ 0
=
b  1 π 1 2  ab
(π − 2 )
 × − a =
4
a2 2 2 
Noting that xEL = x, and that yEL =
x =
1
( y1 + y2 ):
2
1
4
ab
2
2
∫ xELdA = ab π − 2 ∫ 0 a  x x − a − x ( a − x )dx
A
(
)
a
3
3
 21
 1 
1 
4
1 
 1
 1
=
 −   a 2 − x 2 2 +  − ax 2 + x3   = 2
  a 2 2 +  − a 3 + a 3  
3  
3  
ab (π − 2 )  3  2 
a (π − 2 )  3 
 2
 2
0
4
(
)
( )
or x =
2a
!
3 (π − 2 )
continued
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
y =
=
1
4
a 1

yEL dA =
y + y1 )  ( y2 − y1 ) dx 
∫
∫
0  ( 2
A
ab (π − 2 )  2

2
ab (π − 2 )
(
) a (π − 2 ) ∫
2
2
∫ y2 − y1 =
2
2
2
ab
0  2
a
(
)
a2 − x2 −
b2
2
a − x )  dx
2(
a

a
2b
4b
4b
a
a
1 2 1 3
= 3
2 ax − x 2 dx = 3
ax − x 2 dx = 3
ax − x 
∫
∫
0
0
3 0
a (π − 2 )
a (π − 2 )
a (π − 2 )  2
(
)
(
)
or y =
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.
2b
!
3 (π − 2 )
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 36.
x =0 First note that symmetry implies
For the element (EL) shown
y = R cos θ, x = R sin θ
dx = R cos θ d θ
dA = ydx = R 2 cos 2θ dθ
Hence
α
1 2
α
 θ sin 2θ 
A = ∫ dA = 2∫ 0 R 2 cos 2 θ dθ = 2R 2  +
 = R ( 2α sin 2α )
4 0
2
2
=
(
)
R3
cos 2 α sin α + 2sin α
3
(
)
R3
cos 2 α sin α + 2sin α
3
y =
R2
( 2α + sin 2α )
2
(
But yA = ∫ yEL dA so
or
α
2
1

cosθ R 2 cos 2 θ dθ = R3  cos 2 θ sin θ + sin θ 
2
3
3

0
α R
∫ yEL dA = 2∫ 0
(
)
)
cos 2 α + 2
2
y = R sin α
3
( 2α + sin 2α )
Alternatively,
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.
y =
2
3 − sin 2 α
R sin α
3
2α + sin 2α
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 37.
x = 0, y = b
At
b = k (0 − a)
2
y=
Now
xEL = x, yEL =
and
dA = ydx =
and
b
a2
b
2
x − a)
2(
a
Then
Then
k =
or
a b
2
A = ∫ dA = ∫ 0
a
y
b
2
=
x − a)
2(
2
2a
b
( x − a )2 dx
a2
a
b 
1
3
x
−
a
= ab
(
)
2 



3
0
3a
( x − a )2 dx =
∫ xEL dA = ∫ 0 x  a 2 ( x − a ) dx  = a 2 ∫ 0 ( x − 2ax + a x )dx
a
=
2


b
a
3
2
2
b  x4 2 3 a2 2 
1 2
x  =
ab
− ax +

3
2
12
a 2  4

a
∫ yEL dA = ∫ 0
=
b

a
b
b2  1
2 b
2
5

x − a )  2 ( x − a ) dx  =
x − a) 
2(
4  (
2a
a
 2a  5
0
1 2
ab
10
1 2
1 
ab
Hence xA = ∫ xEL dA: x  ab  =
 3  12
x =
1 2
1 
yA = ∫ yEL dA: y  ab  =
ab
3
10


y =
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.
1
a 4
3
b 10
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 38.
For the element (EL) shown on line 1 at
x = a, b = k 2 a 2
or
∴ y =
b 2
x
a2
x = a, −2b = k1a3
On line 2 at
∴ y =
k2 =
b
a2
or
k2 =
−2b
a3
−2b 3
x
a3
2b 
 b
dA =  2 x 2 + 3 x3  dx
a

a
Then
A = ∫ dA = ∫
b  x3 2 x 4 
+
 dx = 2 

x 
4a 
a  3
2x
a b  2
0 2
x +
a 
3
a
0
1 1 5
= ab  +  = ab
3 2 6
and ∫ xEL dA = ∫
2b 
a  b 2
x
x + 3 x3  dx =
0  2
a
a

b  x 4 2 x5 
+


5a 
a 2  4
a
1 2
= a 2b  + 
4 5
0
13 2
ab
20
2b 3   b 2 2b 3  
a1 b 2
∫ yEL dA = ∫ 0 2  a 2 x − a3 x   a 2 x + a3 x  dx 

 
 
=
=∫
2

b 2  x5
2
 2b  
−  3 x3   dx =
− 2 x 7 
4

2a  5
7a

a
 

a 1  b 2 
 2 x 
0
2  a
2
a
0
2
13
 1
= b 2a5 
−  = − ab 2
10
7
70


Then
xA = ∫ xEL dA:
yA = ∫ yEL dA:
 5  13 2
x  ab  =
ab
 6  20
 5  13 2
y  ab  −
ab
 6  70
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.
x =
39
a 50
or y = −
39
b 175
or
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 39.
Using the area element shown:
y
,
2
xEL = x,
yEL =
A = ∫ dA = ∫
L 
h 1 +
0 

and dA = ydx
L

x
x2 
x 2 2 x3 
5
− 2 2  dx = h  x +
−
= hL
2
L
2L 3 L  0
6
L 

L

 x 2 1 x3 2 x 4 
x
x2 
x2
x3 
1 2
L
L
∫ xEL dA = ∫ 0 xh 1 + L − 2 L2  dx = h∫ 0  x + L − 2 L2  dx = h  2 + 3 L − 4 L2  = 3 hL





0
2
1 2
h2 L 
x
x2 
h2 L 
x2
x4
x
x2
x3 
∫ yEL dA = 2 ∫ y dx = 2 ∫ 0 1 + L − 2 L2  dx = 2 ∫ 0 1 + L2 + 4 L4 + 2 L − 4 L2 − 4 L3  dx




L
h2 
x3
4 x5 x 2 4 x3 x 4 
4 2
h L
=
− 2 − 3 =
x + 2 + 4 +
L
2 
L  0 10
3L
5L
3L
Now
x =
1
6 1 2 2
∫ xEL dA = 5hL  3 hL  = 5 L and
A


y =
1
6  4 2  12
∫ yEL dA = 5hL  10 h L  = 25 h
A


Therefore:
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.
x =
2
L
5
y =
12
h
25
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 40.
Note that y1 = 0 at x = a, or
(
)
1
a2
0 = 2b 1 − ka 2 , i.e. k =
Also, note that the slope of y2 is
− 3b
and y2 = 0 at x = 2a. Therefore
a
3b
( 2a − x ) .
a
y2 =
Pick the area element dA ( EL ) such that:
for 0 ≤ x ≤ a
dA = ( 3b − y1 ) dx,
and xEL = x,
yEL =
1
( 3b + y1 )
2
and for a ≤ x ≤ 2a
dA = y2 dx,
and xEL = x,
yEL =
1
y2
2
Then:
2b
a
2a
a
2 a 3b

A = ∫ dA = ∫ 0 ( 3b − y1 ) dx + ∫a y2dx = ∫ 0 3b − 2 a 2 − x 2  dx + ∫a
( 2a − x ) dx =
a
a


(
∫
2b 
2 a 3b
a
b + 2 x 2 dx + a
0

a

∫
)
a
2
3b 1
( 2a − x ) dx = b  x + 2 x3  +  −  ( 2a − x )2
a
a  2
3a

0
2a
=
a
continued
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
2  3b 
19
2

ab 1 +  −
ab
− ( 2a − a )  =



3
2
a
6


Now for the centroid:
x =
=

1
6  a 
2 2
3b 2a
∫ xdA = 19ab b∫ 0 x 1 + a 2 x  dx + a ∫a x ( 2a − x ) dx  =
A




a
2a 
6  1 2
1
3
1
6  1 1 
8
1 

 x + 2 x 4  +  ax 2 − x3   =
 +  + 3  4 − − 1 + 

19a  2
3  a  19  2 2 
3
3 
2a
0 a



or x =
y =
18
a!
19
1
6  a1
1
yEL dA =
( 3b + y1 )( 3b − y1 ) dx + ∫ a2a y2 y2dx 
∫
∫

0
A
19ab  2
2

=
1 6  a1
2a 1 2 
y2 dx  =
9b 2 − y12 dx + ∫ a
∫
2 19ab  0 2
2

=
2

2
1 6  a 1  2 4b 2 2
2
2a 9b
2 
 ∫ 0 9b − 4 a − x  dx + ∫a 2 ( 2a − x ) dx 
2 19ab  2 
a
a


=

3b  a 
8
4
2
2a 9

5 + 2 x 2 − 4 x 4  dx + ∫ a 2 ( 2a − x ) dx 
∫

0
19a  
a
a
a


=
a
2a 
3b 
8
4
9
1
3
 5 x + 2 x3 − 4 x5  + 2  −  ( 2a − x ) 
19a 
a 
3a
5a
0 a  3


=
3b 
8 4
3
 5 + −  + 3 (1) 

19 
3 5

(
)
(
)
y =
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.
148
b!
95
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 41.
For y2
x = a, y = b : a = kb2
at
or
a
b2
k =
b 1/2
x
a
Then
y2 =
Now
xEL = x
a
y
b x1/2
x1/2
: yEL = 2 =
, dA = y2dx = b
dx
2
2
2 a
a
and for
0≤ x≤
For
a
1
b  x 1 x1/2 
≤ x ≤ a : yEL = ( y1 + y2 ) =  − +

2
2
2a 2
a 
 x1/2 x 1 
dA = ( y2 − y1 ) dx = b 
− +  dx
 a a 2
Then
a/2
A = ∫ dA = ∫ 0 b
 x1/2 x 1 
x1/2
a
dx + ∫a/2 b 
− +  dx
a
 a a 2
a
=
a/2
 2 x3/2
b  2 3/2 
x2 1 
x
+ b
−
+ x


2a 2  a/2
a 3
0
3 a
3/2
3/2
2 b  a 
3/2
a 
=
  + ( a ) −   
3 a  2 
 2  
2

 1 
a  1
 a  
+ b  −  a 2 −    + ( a ) −    
 2   2 
 2  
 2a 

( )
13
ab
24
1/2
  x1/2 x 1  

a/2  x
a
x
dA
x
b
dx
x
b
=
+
− +   dx



∫ EL
∫0 
∫
a  a/2   a a 2  

=
and
a
a/2
 2 x5/2 x3 x 4 
b  2 5/2 
=
x
+
b
−
+



3a
4  a/2
a  5
0
5 a
continued
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
=
5/2
5/2
2 b  a 
5/2
a 
  + ( a ) −   
5 a  2 
 2  
3
2 
 1  3  a   1  2  a   
+ b  − ( a ) −    + ( a ) −    
 2   4 
 2   
 3a 

=
71 2
ab
240
1/2 
a/2 b x
x1/2

dx 
b
∫ yEL dA = ∫ 0 2
a a 
1 x1/2    x1/2 x 1  
a b x
+ ∫ a/2  − +
− +  dx 
 b 
2a 2
a    a a 2  
a
a/2
3
b2  1 2 
b 2  x 2
1  x 1  


=
+
−
−
x

 
2a  2  0
2  2a 3a  a 2   

 a/2
=
2
2
3
b  a 
2
 a   b2  a 1 
−
  + ( a ) −    −


4a  2 
 2   6a  2 2 
=
11 2
ab
48
Hence xA = ∫ xEL dA:
71 2
 13 
x  ab  =
ab
 24  240
x =
yA = ∫ yEL dA:
 13  11 2
y  ab  =
ab
 24  48
y =
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.
17
a = 0.546a !
130
11
b = 0.423b !
26
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 42.
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Now
Then
and
xEL = r cos θ
and
dL = rd θ
7π /4
7π /4
L = ∫ dL = ∫π /4 rdθ = r [θ ]π /4 =
3
πr
2
7π /4
∫ xEL dL = ∫π /4 r cosθ ( rdθ )
1 
 1
7π /4
2
= r 2 [sin θ ]π /4 = r 2  −
−
 = −r 2
2
2

Thus
3 
xL = ∫ xdL : x  π r  = −r 2 2
2 
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.
x =−
2 2
r 3π
COSMOS: Complete Online Solutions Manual Organization System
SOLUTION 5.43 CONTINUED
(
) (
Then
1/2
dy
= a 2/3 − x 2/3
− x −1/3
dx
Now
xEL = x
and
Then
and
Hence
 dy 
dL = 1 +  
 dx 
2
)
1/2
2

1/2

 
− x −1/3   dx
dx = 1 +  a 2/3 − x 2/3

 

(
) (
)
a
1/3
3
a a
1/3  3 2/3 
dx
=
a
x
= a


1/ 3
2
x
2
0
L = ∫ dL = ∫ 0
a
1/3

3 2
a a
1/3  3 5/3 
x
dL
x
dx
a
x
=
=


∫ EL
∫ 0  x1/3 
5
 = 5a

0


3  3
xL = ∫ xEL dL : x  a  = a 2
2  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.
x =
2
a 5
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 43.
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Now
xEL = a cos3 θ
dL =
and
dx 2 + dy 2
x = a cos3 θ : dx = −3a cos2 θ sin θ dθ
Where
y = a sin 3 θ : dy = 3a sin 2 θ cosθ dθ
Then
(
) (
1/2
)
2
2

dL =  −3a cos 2 θ sin θ dθ + 3a sin 2 θ cosθ dθ 


(
= 3a cosθ sin θ cos 2 θ + sin 2 θ
) dθ
1/2
= 3a cosθ sin θ dθ
π /2
∴ L = ∫ dL = ∫0
=
and
π /2
1

3a cosθ sin θ dθ = 3a  sin 2 θ 
2
0
3
a
2
π /2
3
∫ xELdL = ∫0 a cos θ ( 3a cosθ sin θ dθ )
π /2
2
1

= 3a  − cos5 θ 
 5
0
=
3 2
a
5
3  3
xL = ∫ xEL dL : x  a  = a 2
2  5
Hence
x =
Alternative solution
 x
x = a cos3 θ ⇒ cos 2 θ =  
 a
 y
y = a sin 3 θ ⇒ sin 2 θ =  
 a
 x
∴  
a
2/3
 y
+ 
a
2/3
=1
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.
2/3
2/3
(
y = a 2/3 − x 2/3
)
3/2
2
a !
5
COSMOS: Complete Online Solutions Manual Organization System
(
) (
Then
1/2
dy
= a 2/3 − x 2/3
− x −1/3
dx
Now
xEL = x
and
Then
and
Hence
 dy 
dL = 1 +  
 dx 
2
)
1/2
2

1/2

 
dx = 1 +  a 2/3 − x 2/3
− x −1/3   dx

 

(
) (
)
a
1/3
3
a a
1/3  3 2/3 
dx
=
a
x
= a


1/ 3
2
x
2
0
L = ∫ dL = ∫ 0
a
1/3

3 2
a a
1/3  3 5/3 
x
dL
=
x
dx
=
a
x


∫ EL
∫ 0  x1/3 
5
 = 5a

0


3  3
xL = ∫ xEL dL : x  a  = a 2
2  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.
x =
2
a !
5
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 44.
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Have at
x = a, y = a : a = ka 2
Thus
y =
Then
 dy 
2 
dL = 1 +   dx = 1 +  x  dx
 dx 
a 
1 2
x
a
dy =
and
k =
or
1
a
2
xdx
a
2
2
a
∴ L = ∫ dL = ∫
=
a
0
(
)
a
a
5 + ln 2 + 5 = 1.4789a
2
4
∫ xEL dL = ∫
a

3/2
4
4x2   2  a2  
 
1 + 2 dx  =    1 + 2 x 2  
  3  8 
a
a
 

 
0
a
x
0 
a 2 3/2
5 − 1 = 0.8484a 2
12
xL = ∫ xEL dL: x (1.4789a ) = 0.8484a 2
=
Then
x
4
4x2
a 2
4x2 
1 + 2 x 2 dx =  1 + 2 + ln  x + 1 + 2  
4 a
2
a
a
a 

0

(
)
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.
x = 0.574a COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 45.
xEL = x,
Have
1
πx
x sin
L
2
dA = ydx
and
L/2
A = ∫ dA = ∫0 x sin
and
yEL =
L/2
 L2
πx L
πx
− x cos 
dx =  2 sin
π
L
L
L 0
π
πx
=
L2
π2
πx 
L/2 
x = ∫ xEL dA = ∫0 x  x sin
dx 
L


L/2
 2 L2
 π x  2 L3
 π x  L 2  π x 
=  2 x sin 
 + 3 cos 
 − x sin 

 L  π
 L  π
 L  0
π
Also
L/2 1
y = ∫ yEL dA = ∫0
2
x sin
Hence
L3
L3
π
π3
−2
2
πx
πx 
dx 
 x sin
L 
L

1  2 L2
πx  L
2 L3 
πx
=  2 x sin
−  x − 3  cos 
2  π
L π
L 
π 
=
=
L/2
0

L2  L 
L3
1  1  L3 
1
6 + π2
−
=
(
)
   −



2
2
2  6  8  4π  2 
 96π
(
)
 L2 
z 
 1
xA = ∫ xEL dA: x  2  = L3  2 − 3 
π 
π
π 
or
 L2 
L3  1
2 
− 3
yA = ∫ yEL dA: y  2  =
2 2
π 
 π  96π  π
or y = 0.1653L !
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.
x = 0.363L !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 46.
First note that by symmetry y = 0.
Using the area element shown in the figure,
xEL =
2
2
r cosθ = R cos 2θ cosθ
3
3
dA =
1 2
1
r dθ = R 2 cos 2 2θ dθ
2
2
π
π
π
1
1
A = ∫ dA = R 2 ∫ 4π cos 2 θ dθ = R 2 ∫ 04 cos 2 θ dθ = R 2 ∫ 04 (1 + cos 4θ ) dθ
−
2
2
4
π
1 
1
4 1
= R 2 θ + sin 4θ  = π R 2
2 
4
0 8
π
2

ρ
 1

2


2
2
2
π
3
∫ xEL dA = ∫−4π  3 R cos 2θ cosθ  2 R cos 2θ dθ  = 3 R ∫ 04 cos 2θ cosθ dθ
π
=
π
3
2 3 4
2
R ∫ 0 1 − 2sin 2 θ cosθ dθ = R3 ∫ 04 1 − 6sin 2 θ + 12sin 4 θ − 8sin 6 θ cosθ dθ
3
3
(
)
(
)
π
2 
12
8
4
= R3 sin θ − 2sin 3 θ + sin 5 θ − sin 7 θ 
3 
5
7
0
=
2 3 2
2 12 1 8 1  16 2 3
1− +
R
−
=
R

3
2 
2
5 4 7 8 
105
Now:
x=
1
8  16 2 3 
128 2
xEL dA =
R  =
R
∫
2

A
105π
π R  105

or x = 0.549 R y =0
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 5, Solution 47.
From the solution to problem 5.2:
A = 134 in 2 ,
ΣxA = 1102 in 3,
ΣyA = 536 in 3
and from the solution of problem 5.22
L = 56 in., and ΣxL = 488 in 2
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x-axis:
(
)
Volume = 2π yarea A = 2πΣyA = 2π 536 in 3 = 3367.8 in 3
or
V = 1.949 ft 3 Area = 2π ylength L = 2πΣyL = 2π 6 (15 ) + 10 ( 4 ) + 8 (10 ) + 4 (18 )  = 1520.53 in 2
or
A = 10.56 ft 2 (b) Rotation about x = 19 in.:
(
)
Volume = 2π (19 − xarea ) A = 2π (19 A − ΣxA ) = 2π (19 in ) 134 in 2 − 1102 in 3 


= 9072.9 in 3
or
V = 5.25 ft 3 Area = 2π (19 − xline ) L = 2π (19L − ΣxL ) = 2π (19 in.)( 56 in.) − 488 in 2 
= 3619.1 in 2
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.
A = 25.1 ft 2 COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 48.
From the solution to problem 5.4:
A = 772 in 2 ,
ΣxA = 17828 in 3 ,
( Area )
ΣyA = 18464 in 3
and for the line
(a)
L
x
y
xL
yL
1
13
27.5
0
357.5
0
2
40
34
20
1360
800
3
34
17
40
578
1360
4
212 + 242 = 31.890
10.5
28
334.85
892.92
5
16
21
8
336
128
Σ
134.89
2966.4
3180.9
V = 2π xarea A = 2πΣxA = 2π 17828 in 3 = 112 017 in 3
(
)
or V = 64.8 ft 3 !
(
)
or A = 129.4 ft 2 !
A = 2π xline L = 2πΣxL = 2π 2966.4 in 2 = 18 638.1 in 2
continued
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
(b)
V = 2π ( 40 − yarea ) A = 2π ( 40 A − ΣyA )
(
)
= 2π ( 40 in.) 772 in 2 − 18 464 in 3  = 78 012 in 3


or V = 45.1 ft 3 !
A = 2π ( yline ) y = 40 L = − 2π ΣL ( y − 40 )  = − 2π ( ΣLy − 40ΣL )
= − 2π ( 3180.9 − 40 × 134.89 ) = 13 915.3 in 2
or A = 96.6 ft 2 !
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 5, Solution 49.
From the solution of Problem 5.1:
A = 150000 mm 2 ,
x A = 140 mm,
y A = 165 mm
From the solution of Problem 5.21:
L = 1800 mm,
xL = 116.667 mm,
yL = 166.667 mm
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x-axis:
Ax = 2π yL L = 2π (166.667 mm )(1800 mm ) = 1884 960 mm 2
(
or
A = 1.885 × 106 mm 2 or
V = 155.5 × 106 mm3 )
Vx = 2π y A = 2π (165 mm ) 150 000 mm 2 = 155 509 000 mm 3
(b) Rotation about x = 400 mm:
Ax = 400 mm = 2π ( 400 mm − xL ) L = 2π ( 400 − 116.667 ) mm  (1800 mm ) = 3 204 420 mm 2
or
(
A = 3.20 × 106 mm 2 )
Vx = 2π ( 400 mm − x A ) A = 2π ( 400 − 140 ) mm  150 000 mm 2 = 245 040 000 mm 3
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.
V = 245 × 106 mm3 COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 50.
Applying the second theorem of Pappus-Guldinus, we have
(a) Rotation about axis AA′:
 π ab 
2 2
Volume = 2π yA = 2π ( a ) 
=π a b
 2 
V = π 2a 2b (b) Rotation about axis BB′:
 π ab 
2 2
Volume = 2π yA = 2π ( 2a ) 
 = 2π a b
 2 
V = 2π 2a 2b (c) Rotation about y-axis:
 4a  π ab  2 2
Volume = 2π yA = 2π  
 = πa b
 3π  2  3
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.
V =
2 2
πa b
3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 51.
The area A and circumference C of the cross section of the bar are
A=
π
4
d 2 and C = π d .
Also, the semicircular ends of the link can be obtained by rotating the cross section through a
horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have
for the volume V:
V = 2 (Vside ) + 2 (V end ) = 2 ( AL ) + 2 (π RA) = 2 ( L + π R ) A
2
π
or V = 2 3 in. + π ( 0.75 in.)   ( 0.5 in.)  = 2.1034 in 3
4


or V = 2.10 in 3 For the area A:
A = 2 ( Aside ) + 2 ( Aend ) = 2 ( CL ) + 2 (π RC ) = 2 ( L + π R ) C
or A = 2 3 in. + π ( 0.75 in.)  π ( 0.5 in.)  = 16.8270 in 2
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.
or A = 16.83 in 2 COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 52.
Following the second theorem of Pappus-Guldinus, in each case a
specific generating area A will be rotated about the x axis to produce the
given shape. Values of y are from Fig. 5.8A.
(1) Hemisphere: the generating area is a quarter circle
Have
 4a  π 
V = 2π yA = 2π   a 2 
 3π  4 
or V =
2 3
πa !
3
(2) Semiellipsoid of revolution: the generating area is a quarter ellipse
Have
 4a  π 
V = 2π yA = 2π   ha 
 3π  4 
or V =
2 2
πa h!
3
(3) Paraboloid of revolution: the generating area is a quarter parabola
Have
 3  2 
V = 2π yA = 2π  a  ah 
 8  3 
or V =
1 2
πa h!
2
or V =
1 2
πa h!
3
(4) Cone: the generating area is a triangle
Have
 a  1 
V = 2π yA = 2π   ha 
 3  2 
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 5, Solution 53.
The required volume can be generated by rotating the area shown about the y axis. Applying the second
theorem of Pappus-Guldinus, we have
 5
 1


V = 2π xA = 2π  + 7.5  mm  ×  × 5 mm × 5 mm 
3
2



 
or V = 720 mm3 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 5, Solution 54.
Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by:
AC = π yL = π ΣyL
where the individual lengths are the lengths of the belt cross section that are in contact with the
pulley.
(a)

0.125    0.125 in. 
 
AC = π  2 ( y1L1 ) + y2 L2  = π  2  3 −
+ ( 3 − 0.125 ) in. ( 0.625 in.) 
 in. 

2    cos 20° 
 

or AC = 8.10 in 2 (b)

0.375    0.375 in. 
AC = π  2 ( y1L1 )  = 2π  3 − 0.08 −
 in. 

2    cos 20° 

or AC = 6.85 in 2 (c)

2 ( 0.25 )  
AC = π  2 ( y1L1 )  = π  3 −
 in. π ( 0.25 in.) 
π   

or AC = 7.01 in 2 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
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Chapter 5, Solution 55.
Volume:
The volume can be obtained by rotating the triangular area shown through π radians about the y
axis.
The area of the triangle is:
A=
1
( 52 )( 60 ) = 1560 mm2
2
Applying the theorems of Pappus-Guldinus, we have
(
V = π xA = π ( 52 mm ) 1560 mm 2
)
or V = 255 × 103 mm3 !
The surface area can be obtained by rotating the triangle shown through an angle of π radians about
the y axis.
Considering each line BD, DE, and BE separately:
22
= 31 mm
2
Line BD : L1 = 222 + 602 = 63.906 mm
x1 = 20 +
Line DE : L2 = 52 mm
x2 = 20 + 22 + 26 = 68 mm
Line BE : L3 = 742 + 602 = 95.268 mm
x1 = 20 +
74
= 57 mm
2
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the
lines:
AL = πΣxA = π ( 31)( 63.906 ) + ( 68 )( 52 ) + ( 57 )( 95.268 )  = π [10947.6] = 34.392 × 103 mm 2
The area of the “end triangles”:
1

AE = 2  ( 52 )( 60 )  = 3.12 × 103 mm 2
2

Total surface area is therefore:
A = AL + AE = ( 34.392 + 3.12 ) × 103 mm 2
or A = 37.5 × 103 mm 2 !
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 5, Solution 56.
The mass of the escutcheon is given by m = ( density )V , where V is the volume. V can be
generated by rotating the area A about the x-axis.
From the figure:
L1 = 752 − 12.52 = 73.9510 m
L2 =
37.5
= 76.8864 mm
tan 26°
a = L2 − L1 = 2.9354 mm
φ = sin −1
α=
12.5
= 9.5941°
75
26° − 9.5941°
= 8.2030° = 0.143168 rad
2
Area A can be obtained by combining the following four areas:
Applying the second theorem of Pappus-Guldinus and using Figure 5.8 a, we have
V = 2π yA = 2π ΣyA
continued
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
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Seg
.
A, mm 2
1
1
( 76.886 )( 37.5) = 1441.61
2
2
− α ( 75 ) = − 805.32
2
3
−
1
( 73.951)(12.5) = − 462.19
2
− ( 2.9354 )(12.5 ) = − 36.693
4
y A , mm3
y , mm
1
( 37.5) = 12.5
3
18 020.1
2 ( 75) sin α
sin (α + φ ) = 15.2303
3α
−12 265.3
1
(12.5) = 4.1667
3
−1925.81
1
(12.5) = 6.25
2
− 229.33
Σ
3599.7
Then
(
)
V = 2π ΣyA = 2π 3599.7 mm3 = 22618 mm3
m = ( density )V
(
)(
= 8470 kg/m3 22.618 × 10−6 m3
)
= 0.191574 kg
or m = 191.6 g !
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 57.
The volume of the waste wood is:
Vwaste = Vblank − Vtop , where
2
Vblank = π ( 22 in.) (1.25 in.) = 1900.664 in 3
Vtop = V1 + V2 + V3 + V4
The volumes Vi can be obtained through the use of the theorem of Pappus-Guldinus:
2
2
Vtop = π ( 21.15 in.) ( 0.75 in.)  + π ( 21.4 in.) ( 0.5 in.) 

 

 
( 4 )( 0.5)  in. × π 0.5 in. 2  + 2π   21.15 + ( 4 )( 0.75)  in. × π 0.75 in. 2 
+ 2π   21.4 +
(
)
(
)
 
 
 
3π   4
3π
  4
 

 

= (1053.979 + 719.362 + 26.663 + 59.592 ) in 3 = 1859.596 in 3
Therefore
Vwaste = 1900.664 in 3 − 1859.596 in 3
= 41.068 in 3
Then
Wwaste = γ woodVwaste N tops
(
)(
)
= 0.025 lb/in 3 41.068 in 3 ( 5000 tops )
= 5133.5 lb,or
Vwaste = 5.13 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 58.
The total surface area can be divided up into the top circle, bottom circle, and the edge.
ATotal = ATop circle + ABottom circle + AEdge,or
2
2
ATotal = π ( 21.4 in.)  + π ( 21.15 in.) 

 

 
  

2 ( 0.5 )   π
2 ( 0.75 )   π
+  2π  21.4 +
 in. × ( 0.5 in.)  +  2π  21.15 +
 in. × ( 0.75 in.) 
π   2
π   2
 
  

= (1438.72 + 1405.31 + 107.176 + 160.091)
= 3111.3 in 2
Now, knowing that 1 gallon of lacquer covers 500 ft2, the number of gallons needed, NGallons is
N Gallons = ASurface × coverage × ( number of tops ) × ( number of coats )
N Gallons = 3111.3 in 2 ×
1 Gallon
( 500 ) (144 in 2 )
× 5000 × 3
= 648.19 gal
or N Gallons = 648
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
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Chapter 5, Solution 59.
The mass of the lamp shade is given by
m = ρV = ρ At
where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x axis. Applying the first theorem of Pappus-Guldinus we have
A = 2π yL = 2πΣyL = 2π ( y1L1 + y2 L2 + y3L3 + y4 L4 )
 13 mm
13 + 16 
or A = 2π 
(13 mm ) + 
 mm ×
2
 2 

 16 + 28 
+
 mm ×
2 

( 32 mm )2 + ( 3 mm )2
28 + 33 
 mm ×
2 

(8 mm )2 + (12 mm )2 + 

( 28 mm )2 + ( 5 mm )2 

= 2π ( 84.5 + 466.03 + 317.29 + 867.51) = 10903.4 mm 2
Then
(
)(
)
m = ρ At = 2800 kg/m 3 10.9034 × 10−3 m 2 ( 0.001 m )
or m = 30.5 g 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
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Chapter 5, Solution 60.
Free-Body Diagram:
First note that the required surface area A can be generated by rotating the
parabolic cross section through 2π radians about the x axis. Applying the
first theorem of Pappus-Guldinus, we have
A = 2π yL
2
Now, since
x = ky ,
x = a : a = k ( 7.5 )
at
2
or
a = 56.25 k
(1)
x = ( a + 15 ) mm: a + 15 = k (12.5 )
At
2
or
(2)
a + 15 = 156.25k
Then
Eq. (2) a + 15 156.25k
:
=
Eq. (1)
a
56.25k
or a = 8.4375 mm
Eq. (1) ⇒ k = 0.15
∴ x = 0.15 y 2
and
1
mm
dx
= 0.3 y
dy
2
Now
 dx 
dL = 1 +   dy = 1 + 0.09 y 2 dy
 dy 
So
A = 2π yL
yL = ∫ ydL
and
12.5
∴ A = 2π ∫7.5 y 1 + 0.09 y 2 dy
12.5
3/2 
2 1 
2
= 2π  
 1 + 0.09 y

 3  0.18 
 7.5
(
= 1013 mm 2
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.
)
or A = 1013 mm 2 COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 61.
(a) Note that in the free-body diagram:
R1 =
1
( 4.2 m )( 600 N/m ) = 1260 N,
2
and R2 =
1
( 4.2 m )( 240 N/m ) = 504 N
2
Then for the equivalence of the systems of forces:
ΣFy :
R = R1 + R2 = 1260 + 504 = 1764 N
ΣM A :


1
2
 
 
− x (1764 N ) =  2 + 4.2  m  (1260 N ) +  2 + 4.2  m  ( 504 N ) = 3.8000 m
3
3
 
 


R = 1764 N or x = 3.80 m (b) Equilibrium:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 1764 = 0
A = 1764 N ΣΜ Α = 0:
M A − ( 3.80 m )(1764 N ) = 0
M A = 6.70 kN ⋅ m
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 5, Solution 62.
With R1 = ( 20 lb/ft )(18 ft ) = 360 lb, and R2 =
ΣFy :
or
1
( 60 lb/ft )(18 ft ) = 360 lb:
3
− R = − R1 − R2
R = 360 lb + 360 lb = 720 lb
R = 720 lb + ΣM A :
− x ( 720 lb ) = − ( 9 ft )( 360 lb ) − (13.5 ft )( 360 ft )
x = 11.25 ft 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 5, Solution 63.
R = (1800 N/m )( 3.2 m ) = 5.76 kN
+ ΣM A = 0:
− ( 5.76 kN )(1.2 m + 1.6 m ) + By ( 3.6 m ) = 0, or
By = 4.48 kN
B = 4.48 kN ΣFy = 0:
Ay + 4.48 − 5.76 = 0, or
Ay = 1.28 kN
+ ΣFx = 0:
Ax = 0
Therefore:
A = 1.28 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.
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Chapter 5, Solution 64.
 kN 
R1 =  1.5
 (1.6 m ) = 2.4 kN
m 

R2 =
1  kN 
3
 ( 2.4 m ) = 3.6 kN
2 m 
 kN 
R3 =  3
 (1.6 m ) = 4.8 kN
 m 
Equilibrium:
+ ΣFx = 0:
Ax = 0
+ ΣM B = 0:
( 4.8 m )( 2.4 kN ) + ( 2.4 m )( 3.6 kN ) + ( 0.8 m )( 4.8 kN ) − ( 4.0 m ) Ay = 0
Ay = 6.0000 kN
+ ΣFy = 0:
6 kN − 2.4 kN − 3.6 kN − 4.8 kN + By = 0
By = 4.8000 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.
A = 6.00 kN B = 4.80 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 65.
lb 

R1 =  240  ( 4.8 ft ) = 1152 lb
ft 

R2 =
1
lb 
180  ( 3.6 ft ) = 324 lb
2
ft 
Equilibrium:
+ ΣFx = 0:
Ax = 0
+ ΣFy = 0:
Ay − 1152 lb + 324 lb = 0
Ay = 828.00 lb
+ ΣM A = 0:
A = 828 lb M A − ( 2.4 ft )(1152 lb ) + ( 6 ft )( 324 lb ) = 0
M A = 820.80 lb ⋅ ft
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.
M A = 821 lb ⋅ ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 66.
The distributed load given can be simplified as in the diagram below with the resultants R1 and R2.
The resultants are:
R1 = ( 6 ft )( 30 lb/ft ) = 180 lb, and R2 =
1
( 4.5 ft )(120 lb/ft ) = 270 lb
2
Now, for equilibrium:
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay + 180 − 270 = 0
Ay = 90.0 lb
Therefore:
ΣM A = 0:
A = 90.0 lb
or M = 675 lb ⋅ ft
2


M A + ( 3 ft )(180 lb ) − 1.5 + × 4.5  ft × ( 270 lb ) = 0
3


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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 67.
 kN 
R1 = 1.5
 ( 2.4 m ) = 3.6 kN
m 

R2 =
2  kN 
9
 ( 2.4 m ) = 14.4 kN
3 m 
Equilibrium:
+ ΣFx = 0:
Ax = 0
+ ΣM B = 0:
− ( 3.3 m ) Ay − (1.8 m )( 3.6 kN ) + ( 2.1 m )(14.4 kN ) = 0
Ay = 7.2000 kN
+ ΣFy = 0:
A = 7.20 kN
B = 3.60 kN
7.2 kN + 3.6 kN − 14.4 kN + By = 0
By = 3.6000 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 5, Solution 68.
The resultants:
R1 =
2
( 3.2 ft )(120 lb/ft ) = 256 lb
3
R2 =
1
( 2.4 ft )(120 lb/ft ) = 96 lb
3
R3 =
1
(1.6 ft )( 45 lb/ft ) = 24 lb
3
Then for equilibrium:
ΣFx = 0:
Ax = 0
ΣM B = 0:
3


− ( 7.2 ft ) Ay +  4 + × 3.2  ft × ( 256 lb )
8


3


1

+  1.6 + × 2.4  ft × ( 96 lb ) +  × 1.6 ft  ( 24 lb = 0 )
4
4




Ay = 231.56 lb
ΣFy = 0:
A = 232 lb
B = 144.4 lb
23.56 − 256 − 96 − 24 + By = 0
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 69.
Have
RI =
1
( 9 m )( 2 kN/m ) = 9 kN
2
RII = ( 9 m )(1.5 kN/m ) = 13.5 kN
Then
ΣFx = 0: C x = 0
ΣM B = 0: − 50 kN ⋅ m − (1 m )( 9 kN ) − ( 2.5 m )(13.5 kN ) + ( 6 m ) C y = 0
or
C y = 15.4583 kN
C = 15.46 kN
B = 7.04 kN
ΣFy = 0: By − 9 kN − 13.5 kN + 15.4583 = 0
or
By = 7.0417 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 5, Solution70.
Have
RI =
1
( 9 m ) ( 3.5 − w0 ) kN/m  = 4.5 ( 3.5 − w0 ) kN
2
RII = ( 9 m ) ( w0 kN/m ) = 9w0 kN
(a) Then
or
ΣM C = 0: − 50 kN ⋅ m + ( 5 m )  4.5 ( 3.5 − w0 ) kN  + ( 3.5 m ) ( 9w0 kN ) = 0
9w0 + 28.75 = 0
so
w0 = − 3.1944 kN/m
w0 = 3.19 kN/m
C = 1.375 kN
Note: the negative sign means that the distributed force w0 is upward.
(b)
ΣFx = 0: C x = 0
ΣFy = 0: − 4.5 ( 3.5 + 3.19 ) kN + 9 ( 3.19 ) kN + C y = 0
or
C y = 1.375 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 5, Solution 71.
The distributed load can be represented in terms of resultants:
R1 = ( 8 m )( 300 N/m ) = 2400 N
R2 =
1
( 8 − a ) m  ( 2400 N/m ) = 1200 ( 8 − a ) N
2
For equilibrium:
ΣM B = 0:
1

− 8 Ay + 4 ( 2400 ) +  ( 8 − a )  1200 ( 8 − a )  = 0
3

Ay = 1200 + 50 ( 8 − a )
ΣM A = 0:
2
(1)
2


8By − 4 ( 2400 ) −  a + ( 8 − a )  1200 ( 8 − a )  = 0
3


By = 1200 + 50 (16 + a )( 8 − a )
(a) ΣFy = 0:
(2)
Ay + By − 2400 − 1200 ( 8 − a ) = 0
(3)
Using the requirement By = 2 Ay and (1)
2
3 1200 + 50 ( 8 − a )  − 2400 − 1200 ( 8 − a ) = 0


or
(8 − a )2 − 8 (8 − a ) + 8 = 0, which gives
continued
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
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COSMOS: Complete Online Solutions Manual Organization System
(8 − a ) =
8±
( − 8 )2 − 4 ( 8 )
2
= 6.82843 m or 1.17157 m
a = 1.17157 m or a = 6.82843 m, and therefore
amin = 1.17157 m
or amin = 1.172 m !
(b) ΣFx = 0:
Ax = 0
Equation (1) gives:
Ay = 1200 + 50 ( 6.82843)
2
= 3531.4 N
or A = 3.53 kN
!
B = 7.06 kN
!
By = 2 Ay gives By = 2 ( 3531.4 N ) , and
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 5, Solution 72.
The distributed load can be represented in terms of resultants:
R1 = ( 8 m )( 300 N/m ) = 2400 N
R2 =
1
( 8 − a ) m  ( 2400 N/m ) = 1200 ( 8 − a ) N
2
For equilibrium:
ΣM B = 0:
1

− 8 Ay + 4 ( 2400 ) +  ( 8 − a )  1200 ( 8 − a )  = 0
3

Ay = 1200 + 50 ( 8 − a )
ΣM A = 0:
2
(1)
2


8By − 4 ( 2400 ) −  a + ( 8 − a )  1200 ( 8 − a )  = 0
3


By = 1200 + 50 (16 + a )( 8 − a )
(2)
(a) Dividing Equation (1) by Equation (2):
By
Ay
=
1200 + 50 (16 + a )( 8 − a )
1200 + 50 ( 8 − a )
2
=
(
)
24 + ( 64 − 16a + a )
=
152 − 8a − a 2
88 − 16a + a 2
24 + 128 − 8a − a 2
2
continued
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
Differentiating
By
Ay
:
(
) (
)
2
2
d  By  ( − 8 − 2a ) 88 − 16a + a − 152 − 8a − a ( −16 + 2a )
=0

=
2
da  Ay 
88 − 16a + a 2
(
)
2
a − 20a + 72 = 0
or
a=
or
20 ±
( − 20 )2 − 4 ( 72 )
2
Knowing that a ≤ 8 m: a = 4.7085 m
or
a = 4.71 m !
(b) For equilibrium:
ΣFx = 0:
Ax = 0
and from (1):
Ay = 1200 + 50 ( 8 − 4.7085 )
2
= 1741.70 N
A = 1.742 kN
!
B = 4.61 kN
!
Also,
ΣFy = 0:
1741.70 − 2400 − 1200 ( 8 − 4.7085 ) + By = 0
By = 4608.1 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.
or
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 73.
R1 = ( 3.6 ft )( wA kips/ft ) = 3.6wA kips
R2 =
1
1
( 5.4 ft )  wA kips/ft  = 1.35wA kips
2
2

1

R3 = ( 5.4 ft )  wA kips/ft  = 2.7wA kips
2


Equilibrium:
− (1.8 ft ) ( 3.6wA ) kips  + (1.8 ft ) (1.35wA ) kips 
+ ΣM C = 0:
+ ( 2.7 ft ) ( 2.7 wA ) kips  + ( 2.1 ft )( 6 kips )
− ( 2.4 ft )( 4.5 kips ) − ( 3.6 ft )(1 kip ) = 0
wA = 0.55556 kips/ft
+ ΣFyA = 0:
or
wA = 556 lb/ft RR − ( 3.6 )( 0.55556 ) kips + 1.35 ( 0.55556 ) kips
+ 2.7 ( 0.55556 ) kips − 6 kips − 4.5 kips − 1 kip = 0
Solving for RR :
RR = 7.2500 kips
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.
RR = 7.25 kips COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 74.
R1 = ( 3.6 ft )( wA kips/ft ) = 3.6wA kips
R2 =
1
( 5.4 ft )( 0.6wA kips/ft ) = 1.62wA kips
2
R3 = ( 5.4 ft )( 0.4wA kips/ft ) = 2.16wA kips
Equilibrium:
+ ΣM A = 0:
− (1.8 ft ) ( 3.6wA ) kips  + ( 3.6 ft ) RR
+ ( 5.4 ft ) (1.62wA ) kips  + ( 6.3 ft ) ( 2.16 wA ) kips  − (1.5 ft )( 6 kips )
− ( 6 ft )( 4.5 kips ) − ( 7.2 ft ) P = 0
or
+ ΣFy y = 0:
or
28.836wA + 3.6RR − 7.2 P − 36 = 0
(1)
RR + 3.6wA + 1.62wA + 2.16wA − 6 − 4.5 − P = 0
7.38wA + RR − P − 10.5 = 0
( 28.836 ) Eq. ( 2 ) − ( 7.38) Eq. (1) = 0 gives
2.268RR − 37.098 + 24.3P = 0
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.
(2)
COSMOS: Complete Online Solutions Manual Organization System
Since RR ≥ 0, the maximum acceptable value of P is that for which RR = 0, and
P = 1.52667 kips
or
P = 1.527 kips !
(b) Now, from (2):
7.38wA − 1.52667 − 10.5 = 0
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.
wA = 1.630 kips/ft !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 75.
Noting that the weight of a section of the dam is Wi = γ Vi (Vi being the volume of that section ) :
lb 

W1 = 150 3  (10.5 ft )( 9 ft )(1 ft )  = 14175 lb
ft 

lb   1


W2 = 150 3   (10.5 ft )( 21 ft )(1 ft )  = 16537.5 lb
ft   2


lb 

W3 = 150 3  (18 ft )( 30 ft )(1 ft )  = 81000 lb
ft 

lb   1


W4 = 150 3   ( 3 ft )( 30 ft )(1 ft )  = 6750 lb
2
ft



From the free-body diagram:
x1 = 5.25 ft, x2 =
2
(10.5 ft ) = 7 ft, x3 = 19.5 ft, and x4 = 29.5 ft
3
For the distance a:
a
3
=
, or a = 2.4 ft
24 30
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
Therefore:
lb   1


Ws =  62.4 3   ( 2.4 ft )( 24 ft )(1 ft )  = 1797.12 lb, and
ft   2


xs = 31.5 −
1
( 2.4 ) = 30.7 ft
3
Now, for the pressure force P:
P=
=
1
1
PB A = (γ W hB ) A
2
2
1
lb 
 62.4 3  ( 24 ft ) ( 24 ft )(1 ft ) 
2
ft 
= 17971.2 lb
Then, for equilibrium:
(a) ΣFx = 0:
H −P=0
H = 17971.2 lb
or H = 17.97 kips
ΣFy = 0:
!
V − 14175 − 16537.5 − 81000 − 6750 − 1797.12 = 0
V = 120259.62 lb
or V = 120.3 kips
!
(b) From moment equilibrium:
ΣM A = 0:
1

x (120259.62 lb ) +  × 24 ft  (17971.2 lb ) − ( 5.25 ft )(14175 lb ) − ( 7 ft )(16537.5 lb )
3


(19.5 ft )(81000 lb ) − ( 29.5 ft )( 6750 lb ) − ( 30.7 ft )(1797.12 lb ) = 0
or x = 15.63 ft !
(c) free-body diagram for section of water:
continued
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
For equilibrium:
Ws + P + ( − R ) = 0
ΣF = 0:
where R is the force of the water on the face BD of the dam, and
P = 17971.2 lb, and Ws = 1797.12 lb
Then from the force triangle:
R=
(17971.2 lb )2 + (1797.12 lb )2 = 18.06 kips
 1797.12 
 = 5.71°
 17971.2 
θ = tan −1 
Therefore: R = 18.06 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.
5.71° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 76.
Free-Body Diagram:
Locations of centers of gravity:
5
( 5 m ) = 3.125 m
8
1


x2 = 5 + ( 2 )  m = 6 m
2


1  25

x3 = 7 + ( 4 )  =
m
3 
3

5 

x4 = 7 + ( 4 )  = 9.5 m
8 

x1 =
Weights: Wi = ρi gVi
2

W1 = 2400 kg/m3 9.81 m/s 2  ( 5 m )( 8 m )(1 m )  = 627 840 N
3


(
)(
)
(
)(
)
W2 = 2400 kg/m3 9.81 m/s 2 ( 2 m )( 8 m )(1 m )  = 376 700 N
1

W3 = 2400 kg/m3 9.81 m/s 2  ( 4 m )( 6 m )(1 m )  = 188 352 N
3


(
)(
)
2

W4 = 2400 kg/m3 9.81 m/s 2  ( 4 m )( 6 m )(1 m )  = 156 960 N
3

The pressure force P is:
1
1
P = Aρ gh = ( 6 m )(1 m )   1000 kg/m3 9.81 m/s 2 ( 6 m )  = 176 580 N


2
2
(
)(
)
(
)(
)
Equilibrium:
(a) + ΣFx = 0:
H − 176.580 kN = 0
H = 176.580 kN
+ ΣFy = 0:
H = 176.6 kN
!
V − 627.84 kN − 376.70 kN − 188.352 kN − 156.960 kN = 0
V = 1349.85 kN
(b) + ΣM A = 0:
or
or
V = 1350 kN
x (1349.85 kN ) − ( 3.125 m )( 627.84 kN ) − ( 6 m )( 376.70 kN )
 25 
−
m  (188.352 kN ) − ( 9.5 m )(156.960 kN ) + ( 2 m )(176.580 kN ) = 0
 3

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
x = 5.1337 m
Thus the point of application of the resultant is:
5.13 m to the right of A. !
(c) Free-body diagram and force triangle for the water section BCD
From the force triangle:
R=
(176.580 )2 + (156.960 )2 = 236.26 kN
 156.960 
 = 41.634°
 176.580 
θ = tan −1 
or on the face BD of the dam
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.
R = 236 kN
41.6° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 77.
Free-Body Diagram:
Note that valve opens when B = 0.
Pressures p1 and p2 at top and bottom of valve:
(
)(
)
p = (10 kg/m )( 9.81 m/s ) ( d ) = ( 9810d ) N/m
p1 = 103 kg/m 3 9.81 m/s 2 ( d − 0.225 m ) = ( 9810d − 2207.3) N/m 2
2
3
3
2
2
Force P1 and P2:
P1 =
1
1
p1 A = ( 9810d − 2207.3) N/m 2  ( 0.225 m )( 0.225 m ) 
2
2
= ( 248.32d − 55.872 ) N
P2 =
1
1
p2 A = ( 9810d ) N/m 2  ( 0.225 m )( 0.225 m ) 
2
2
= ( 248.32d ) N
+ ΣM A = 0:
− ( 0.15 − 0.09 ) m  ( 248.32d − 55.872 ) N  + ( 0.09 − 0.075 ) m  ( 248.32d ) N  = 0
Thus d = 0.30000 m, 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.
d = 300 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 78.
Free-Body Diagram:
Note that valve opens when B = 0.
Pressures p1 and p2 at top and bottom of valve:
(
)(
)
p = (10 kg/m )( 9.81 m/s ) ( 0.450 m ) = 4414.5 N/m
p1 = 103 kg/m 3 9.81 m/s 2 ( 0.225 m ) = 2207.3 N/m 2
2
3
3
2
2
Force P1 and P2:
P1 =
1
1
p1 A =
2207.3 N/m 2 ( 0.225 m )( 0.225 m ) 
2
2
(
)
= 55.872 N
P2 =
1
1
p2 A =
4414.5 N/m 2 ( 0.225 m )( 0.225 m ) 
2
2
(
)
= 111.742 N
+ ΣM A = 0:
− ( 0.15 − h ) m  ( 55.872 N ) + ( h − 0.075 ) m  (111.742 N ) = 0
Solving for h: h = 0.100 000 m, or
h = 100.0 mm 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 5, Solution 79.
Since gate is 4 ft wide:
p1 = 4γ ( h − 3)
p2 = 4γ h
p1′ = 4γ ′ ( d − 3)
p2′ = 4γ ′d
)
)
(
(
1
1
( 3 ft ) p1′ − p1 = ( 3 ft ) 4γ ′ ( d − 3) − 4γ ( h − 3) = 6γ ′ ( d − 3) − 6γ ( h − 3)
2
2
1
1
P2′ − P2 = ( 3 ft ) p2′ − p2 = ( 3 ft ) [ 4γ ′d − 4γ h ] = 6γ ′d − 6γ h
2
2
P1′ − P1 =
This gives the free-body diagram:
+ ΣM A = 0:
or
( 3 ft ) B − (1 ft ) ( P1′ − P1 ) − ( 2 ft ) ( P2′ − P2 ) = 0
B=
=
) (
(
1 ′
2 ′
P1 − P1 −
P2 − P2
3
3
)
1
2
6γ ′ ( d − 3) − 6γ ( h − 3)  − [ 6γ ′d − 6γ h ]
3
3
= 2γ ′ ( d − 3) − 2γ ( h − 3) + 4γ ′d − 4γ h
or
+ ΣFx = 0:
B = 6γ ′ ( d − 1) − 6γ ( h − 1)
(
) (
(1)
)
A + B − P1′ − P1 − P2′ − P2 = 0, or using (1)
A + 6γ ′ ( d − 1) − 6γ ( h − 1)  − 6γ ′ ( d − 3) − 6γ ( h − 3)  − [ 6γ ′d − 6γ h ] = 0, or
A = 6γ ′ ( d − 2 ) − 6γ ( h − 2 )
Using the given data in (1) and (2):
h = 6 ft, d = 9 ft, γ = 62.4 lb/ft 3 , γ ′ = 64 lb/ft 3
A = 6 ( 64 )( 9 − 2 ) − 6 ( 62.4 )( 6 − 2 )
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.
(2)
COSMOS: Complete Online Solutions Manual Organization System
= 2688 lb − 1497.6 lb = 1190.4 lb
B = 6 ( 64 )( 9 − 1) − 6 ( 62.4 )( 6 − 1)
= 3072 lb − 1872 lb = 1200 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.
A = 1190 lb
!
B = 1200 lb
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 80.
First, determine the force on the dam face without the silt.
Pw =
Have
=
1
1
Apw = A ( ρ gh )
2
2
1
( 6 m )(1 m )   103 kg/m 3 9.81 m/s 2 ( 6 m ) 


2
(
)(
)
= 176.58 kN
Next, determine the force on the dam face with silt.
Pw′ =
Have
1
( 4.5 m )(1m )   103 kg/m 3 9.81 m/s 2 ( 4.5 m ) 


2
(
)(
)
= 99.326 kN
( Ps )I = (1.5 m )(1 m ) (103 kg/m3 )( 9.81 m/s2 ) ( 4.5 m )
= 66.218 kN
( Ps )II =
1
(1.5 m )(1 m )   1.76 × 103 kg/m3 9.81 m/s 2 (1.5 m ) 


2
(
)(
)
= 19.424 kN
Then
P′ = Pw′ + ( Ps )I + ( Ps )II = 184.97 kN
The percentage increase, % inc., is then given by
% inc. =
(184.97 − 176.58) × 100% = 4.7503%
P′ − Pw
× 100% =
176.58
Pw
% inc. = 4.75% 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 5, Solution 81.
From Problem 5.80, the force on the dam face before the silt is deposited, is Pw = 176.58 kN. The maximum
allowable force Pallow on the dam is then:
Pallow = 1.5Pw = (1.5 )(176.58 kN ) = 264.87 kN
Next determine the force P′ on the dam face after a depth d of silt has settled.
Have
Pw′ =
(
)(
)
1
( 6 − d ) m × (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m 


2
2
= 4.905 ( 6 − d ) kN
( Ps )I =  d (1 m ) (103 kg/m3 )( 9.81 m/s2 ) ( 6 − d ) m 
(
)
= 9.81 6d − d 2 kN
( Ps )II =
(
)(
)
1
 d (1 m )   1.76 × 103 kg/m3 9.81 m/s 2 ( d ) m 


2
= 8.6328d 2 kN
(
)
(
)
P′ = Pw′ + ( Ps )I + ( Ps )II =  4.905 36 − 12d + d 2 + 9.81 6d − d 2 + 8.6328d 2  kN


= 3.7278d 2 + 176.58 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
Now required that P′ = Pallow to determine the maximum value of d.
∴
or
Finally
(3.7278d + 176.58) kN = 264.87 kN
2
d = 4.8667 m
4.8667 m = 20 × 10−3
m
×N
year
or N = 243 years !
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 5, Solution 82.
Pressure force from the water on board AB:
1
Api where p1 and p2 are the pressures at the top and bottom of the board:
2


1
kg 
m
P1 = ( 0.5 m )(1.5 m )  103 3  9.81 2  ( 0.6 m )  = 2207.3 N
2
m 
s 


Pi =
P2 =


1
kg 
m
( 0.5 m )(1.5 m )  103 3  9.81 2  (1 m )  = 3678.8 N

2
m 
s 


Free-Body Diagram:
Ax denotes the force from one piling and is therefore multiplied by two in the free-body diagram.
1

2

+ ΣM A = 0:
− ( 0.3 m ) B +  ( 0.5 ) m  ( 2207.3 N ) +  ( 0.5) m  ( 3678.8 N ) = 0, or
3

3

B = 5313.8 N
4
4
+ ΣFx = 0:
2 Ax + ( 2207.3 N ) + ( 3678.8 N ) = 0, or
5
5
Ax = − 2354.4 N
3
3
5318.8 N − ( 2207.3 N ) − ( 3678.8 N ) + Ay = 0, or
+ ΣFy = 0:
5
5
Ay = −1782.14 N
Therefore: (a) A x = 2.35 kN
(b)
(c)
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 y = 1.782 kN !
B = 5.31 kN !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 83.
Pressure force from the water on board AB:
Pi =
1
Api where p1 and p2 are the pressures at the top and bottom of the board:
2
P1 =


1
kg 
m
( 0.5 m )(1.5 m )  103 3  9.81 2  ( 0.6 m )  = 2207.3 N
2
m 
s 


P2 =


1
kg 
m
( 0.5 m )(1.5 m )  103 3  9.81 2  (1 m )  = 3678.8 N
2
m 
s 


Note that the board can move in two ways: by rotating about A if the rope is pulled upward, and by sliding
down at A if the rope is pulled sideways to the left.
Case 1 (rotation about A):
For minimum tension the rope will be perpendicular to the board.
Free-Body Diagram:
+ ΣM A = 0:
1

2

− ( 0.5 m ) TBC +  ( 0.5 ) m  ( 2207.3 N ) +  ( 0.5 ) m  ( 3678.8 N ) = 0, or
3

3

TBC = 3188.3 N
Case 2 (sliding down at A):
When the board is just about to slide down at A, A y = 0.
continued
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
Free-Body Diagram:
+ ΣM B = 0:
1

2

− ( 0.4 m ) ( 2 Ax ) −  ( 0.5 ) m  ( 3678.8 N ) −  ( 0.5 ) m  ( 2207.3 N ) = 0, or
3
3




2 Ax = − 3372.3 N
+ ΣFx = 0:
− TBC − 3372.3 +
4
4
( 2207.3 N ) + ( 3678.8 N ) = 0, or
5
5
TBC = 1336.58 N
Thus:
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 )min = 1.337 kN
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 84.
Free-Body Diagram:
Force from water pressure:
1
ApB where A is the rectangular cross sectional area through line BD, and pB is the pressure at
2
point B. Thus
1
1
P = A (γ h ) = (16 ft )(10 ft )   62.4 lb/ft 3 (10 ft )  = 18720.0 lb = 18.72 kips


2
2
P−
(
)
1

W = γ V = 62.4 lb/ft 3  ( 3 ft )( 6 ft )( 6 ft )  = 3369.61 lb = 3.3696 kips
2

Equilibrium:
20
+ ΣM A = 0:
(18.72 kips )  ft  − B ( 3 ft ) + ( 3.3696 kips )( 2 ft ) = 0.
 3 
(
)
Solving for B:
B = 43.846 kips, or
+ ΣFx = 0:
B = 43.8 kips 18.72 kips + Ax = 0, or
Ax = −18.7200 kips
+ ΣFy = 0:
Ay − 3.3693 kips + 43.846 kips = 0, or
Ay = − 40.476 kips
A=
( −18.7200 )2 + ( −40.476 )2 = 44.595 kips
 40.476 
 = 65.180°
 18.7200 
θ = tan −1 
Therefore: A = 44.6 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.
65.2° COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 85.
Consider a 1-in. thick section of the gate and a triangular section BDE of water above the gate
Free-Body Diagram:
Pressure force P:
1
1
1
ApB = ( d × 1 in.)(γ d ) = γ d 2 lb
2
2
2
Weight of water section above gate:
P=
4
1 8

WW = γ VW = γ  × d × d × 1 in.  = γ d 2 lb
 2 15
 15
For impending motion of gate: B y = 0, and for equilibrium:
+ ΣM a = 0:
2
1  8   4
d
 1 2 
2
 (16 ) −  d    γ d  −  − 6   γ d  = 0, and
3  15    15
 3
 2

3
d = 27.301 in., 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.
d = 27.3 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 86.
Consider a 1-in. thick section of the gate and a triangular section BDE of water above the gate
Free-Body Diagram:
Pressure force P:
1
1
1
ApB = ( d × 1 in.)(γ d ) = γ d 2 lb
2
2
2
Weight of water section above gate:
P=
4
1 8

WW = γ VW = γ  × d × d × 1 in.  = γ d 2 lb
 2 15
 15
For impending motion of gate: B y = 0, and for equilibrium:
+ ΣM a = 0:
2
1  8   4
d
 1 2 
2
 (16 ) −  d    γ d  −  − (10 − h )   γ d  = 0, and
3
3
15
15
3

 
 
 2


with d = 30 in.
2
1  8   4
d
 1
2
2
 (16 ) −  30    γ 30  −  − (10 − h )   γ 30  = 0, and
3
3
15
15
3
2

 
 



h = 2.8444 in., 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.
h = 2.84 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 87.
Free-Body Diagram:
(
)
W = (125 kg ) 9.81 m/s 2 = 1226.25 N
Denoting the water pressure at a depth h by ph, the forces due to the water pressure P1, P2, P3, P4 can be
obtained as follows:
1
P1 = ADC p( 0.15 m ) , or with
2
(
)(
)
1
P = ( 0.15 m )(1 m )  (1471.50 N/m ) = 110.363 N
2
p( 0.15 m ) = 1000 kg/m3 9.81 m/s 2 ( 0.15 m ) = 1471.50 N/m 2
2
1
P2 = ACB p( 0.15 m) , or
(
)
P2 = ( 0.6 m )(1 m )  1471.50 N/m 2 = 882.90 N
1
ABA p( 0.15 m ) , or
2
1
P3 = ( 0.6 m )(1 m )  1471.50 N/m 2 = 441.45 N
2
1
P4 = ABA p( 0.75 m ) , or with
2
P3 =
(
(
)
)(
)
p( 0.75 m ) = 1000 kg/m3 9.81 m/s 2 ( 0.75 m ) = 7357.5 N/m 2
P4 =
(
)
1
( 0.16 m )(1 m )  7357.5 N/m 2 = 2207.3 N
2
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
Now from the free-body diagram:
− (1.2 m ) D + ( 0.6 m )(1226.25 N ) + ( 0.3 m )(1226.25 N )
+ ΣM A = 0:
− ( 0.6 + 0.05 m )  (110.363 N ) − ( 0.3 m )( 882.90 N ) − ( 0.4 m )( 441.45 N )
− ( 0.2 m )( 2207.3 N ) = 0, or
D = 124.149 N, and D = 124.1 N
+ ΣFx = 0:
!
Ax + 110.363 N + 441.45 N + 2207.3 N = 0, or
Ax = −2759.1 N
+ ΣFy = 0:
Ay − 3 (1226.25 N ) + 882.90 N = 0, or
Ay = 2795.9 N
Then,
A=
( −2759.1)2 + ( 2795.9 )2 = 3930 N, and
θ = tan −1
2795.9
= 45.4°
2759.1
Therefore: A = 3930 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.
45.4° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 88.
Free-Body Diagram:
(
)
W = (125 kg ) 9.81 m/s 2 = 1226.25 N
Denoting the water pressure at a depth h by ph, the forces due to the water pressure P1, P2, P3, P4 can be
obtained as follows:
1
P1 = ADC p( d − 0.6 m ) , or w
2
1
P1 = ( d − 0.6 ) m × (1 m )  γ N/m3 ( d − 0.6 ) m
2
1
2
= γ ( d − 0.6 ) N
2
where γ denotes the specific weight of water. In the same way
1
P2 = ACB p( d − 0.6 m) , or
2
(
(
)
)
P2 = ( 0.6 m ) × (1 m )  γ N/m3 ( d − 0.6 ) m
= 0.6γ ( d − 0.6 ) N
1
ABA p( d − 0.6 m ) , or
2
1
P3 = ( 0.6 m ) × (1 m )  γ N/m3 ( d − 0.6 ) m
2
= 0.3γ ( d − 0.6 ) N
P3 =
(
)
1
ABA p( d m ) , or
2
1
P4 = ( 0.6 m ) × (1 m )  γ N/m3 ( d m )
2
= 0.3γ d N
P4 =
(
)
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
Now from the free-body diagram:
+ ΣM A = 0:
( 0.6 m )(1226.25 N ) + ( 0.3 m )(1226.25 N )

1
1

2
−  0.6 m + ( d − 0.6) m   γ ( d − 0.6) N  − ( 0.3 m ) 0.6γ ( d − 0.6) N 
2
3


2

1

−  ( 0.6 m )  0.3γ ( d − 0.6 ) N  −  ( 0.6 m )  0.3γ ( d − 0.6 ) N + 0.18γ N  = 0, or
3
3




1
1103.63
− 0.036
( d − 0.6 )3 + 0.3 ( d − 0.6 )2 + 0.36 ( d − 0.6 ) =
γ
6
(
)(
)
With γ = 1000 kg/m3 9.81 m/s 2 = 9810 N/m3 , this gives
1
1103.63
− 0.036 = 0.076501
( d − 0.6 )3 + 0.3 ( d − 0.6 )2 + 0.36 ( d − 0.6 ) =
6
9810 N/m3
Solving for d numerically:
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.
d = 0.782 m !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 89.
(a)
Free-body diagram for a 24-in. long parabolic section of water:
In the free body diagram force P is:
P=
1
1
1  3  24  
 3 
AP = A (γ h ) = 
ft 
ft   62.4 lb/ft 3  ft   = 3.9000 lb
2
2
2  12  12  
 12  
(
)
Ww = γ V
 2  4.5  3  24  
= 62.4 lb/ft 3  
ft  ft 
ft  
 3  12  12  12  
= 7.8000 lb
(
)
From the force triangle:
R=
P 2 + Ww 2 =
θ = tan −1
( 3.9 )2 + ( 7.8)2 = 8.7207 lb
Ww
7.8
= tan −1
= 63.435°, or
P
3.9
R = 8.72 lb
(b)
63.4° Free-body diagram for a 24-in. long section of the water:
From (a) WW = 7.8000 lb
From the free-body diagram:
By = 7.8000 lb
+ΣM B = 0:
M B + ( 2.25 − 1.8 ) in. ( 7.8000 lb ) = 0, or
M B = −3.5100 lb ⋅ in.
Therefore, the force-couple system on the gutter is:
R = 7.8 lb ; M = 3.51 lb ⋅ in. 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 5, Solution 90.
Note, for the axes shown
y
yV
−R
−2π R 4
3
− r
8
1 4
πr
4
V
1
(π R ) ( 2R ) = 2π R
2
2
− π r3
3
Σ

r3 
2π  R 3 − 
3

2
3

r4 
−2π  R 4 − 
8 

1
R4 − r 4
Σ yV
8
Y =
=−
1
ΣV
3
R − r3
3
Then
1−
1 r 
 
8 R 
1−
1 r 
 
3 R 
=
(a )
r =
4
3
3
R: y = −
4
1−
1 3
 
3 4
1 3
1−  
3 4
4
3
R
or y = −1.118R 1−
(b)
y = −1.2R : − 1.2R = −
4
or
1 r 
 
8 R 
1 r 
1−  
3 R 
4
3
R
3
r
r
  − 3.2   + 1.6 = 0
R
 
R
Solving numerically
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.
r
= 0.884 R
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 91.
Labeling the two parts of the body as follows:
ΣyV
Then Y =
=
ΣV
( πa h )
( π a h)
7
24
2 2
2
3
2
V
y
yV
1
1 2
πa h
2
h
2
1 2 2
πa h
4
2
1 2
πa h
6
h
4
1
π a 2h 2
24
Σ
2 2
πa h
3
7
π a 2h 2
24
or Y =
7
h 16
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 5, Solution 92.
Labeling the two parts of the body as follows:
V
1
2
Σ
Then Z =
(
(
− 12 a3h
ΣzV
=
2 π a 2h
ΣV
3
1 2
πa h
2
1 2
πa h
6
2 2
πa h
3
−
z
zV
4a
3π
a
2
− a 3h
3
1 3
ah
6
1
− a 3h
2
π
)
)
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.
or Z = −
3a
4π
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 93.
V
x
xV
Rectangular prism
Lab
1
L
2
1 2
L ab
2
Pyramid
1 b
a h
3 2
1 

ΣV = ab  L + h 
6 

Then
Now
L+
ΣxV =
X ΣV = ΣxV
1
h
4
1
1 

abh  L + h 
6
4 

1  2
1 

ab 3L + h  L + h  
6 
4 

so that
 
1  1 
1 
X  ab  L + h   = ab  3L2 + hL + h 2 
6  6 
4 
 
1 h 1 
h 1 h2 

X 1 +
L
3
=
+
+



6 L  6 
L 4 L2 

or
(a) X = ? when h =
Substituting
(1)
1
L
2
h
1
=
into Eq. (1)
L
2
2

1  1  1 
1 11 
X 1 +    = L 3 +   +   
6  2   6 
 2  4  2  

or X =
(b)
57
L
104
X = 0.548L h
= ? when X = L
L
Substituting into Eq. (1)
or
or
1 h 1 
h 1 h2 

L 1 +

 = L  3 + +
6 L 6 
L 4 L2 

1+
1h
1 1h
1 h2
= +
+
6L
2 6 L 24 L2
h2
= 12
L2
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.
∴
h
= 2 3
L
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 94.
Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of
the volume.
V , mm3
y , mm
z , mm
yV , mm 4
zV , mm 4
1
( 60 )(105)(10 ) = 63000
−5
52.5
− 315 000
3 307 500
2
1
2
π ( 30 ) (10 ) = 14 137.2
2
−5
4 ( 30 )
− 70 686
1 664 400
3
(15)( 30 )( 60 ) = 27 000
15
30
405 000
810 000
4
− π (19 ) (10 ) = −11 341.1
−5
105
56 706
−1 190 820
5
1
2
− π (19 ) (15 ) = − 8505.9
2
4 (19 )
30
−186 585
−255 180
−110 565
4 335 900
2
Σ
Then Y =
30 −
3π
= 21.936
105 +
3π
=117.732
84 290
ΣyV −110 565
mm
=
84 290
ΣV
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.
or Y = −1.312 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 95.
Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of
the volume.
Then Z =
V , mm3
z , mm
zV , mm 4
1
( 60 )(105)(10 ) = 63000
52.5
3 307 500
2
1
2
π ( 30 ) (10 ) = 14 137.2
2
4 ( 30 )
1 664 400
3
(15)( 30 )( 60 ) = 27 000
4
105 +
3π
= 117.732
30
810 000
− π (19 ) (10 ) = −11 341.1
105
−1 190 820
5
1
2
− π (19 ) (15 ) = − 8505.9
2
30
− 255 180
Σ
84 290
2
ΣzV 4 335 900
=
mm
ΣV
84 290
4 335 900
or Z = 51.4 mm 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 5, Solution 96.
Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the volume.
V , mm3
x , mm
xV , mm 4
1
(100 )(88)(12 ) =105600
50
5 280 000
2
(100 )(12 )(88) = 105600
50
5 280 000
3
1
( 62 )( 51)(10 ) = 15 810
2
39
616 590
2
( 66 ) = 78
3
−1 389 960
4
−
1
( 66 )( 45)(12 ) = −17 820
2
Σ
Then X =
34 +
209 190
ΣxV 9 786 600
=
mm
ΣV
209190
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.
9 786 600
or X = 46.8 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 97.
Assume that the bracket is homogeneous so that it center of gravity coincides with the centroid of the volume.
V , mm
1
(100 )(88)(12 ) = 105600
2
(100 )(12 )(88) = 105600
3
4
−
z , mm
zV , mm 4
6
633 600
12 +
1
(88) = 56
2
5 913 600
1
( 62 )( 51)(10 ) = 15 810
2
12 +
1
( 51) = 29
3
458 490
1
( 66 )( 45)(12 ) = −17 820
2
55 +
2
( 45) = 85
3
−1 514 700
Σ
Then Z =
3
209 190
ΣzV 5 491 000
=
mm
ΣV
209 190
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.
5 491 000
or Z = 26.2 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 98.
Assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the volume.
Then X =
V , in 3
x , in.
xV , in 4
1
( 8)( 0.9)( 2.7) = 19.44
4
77.76
2
1
( 2.1)( 6)( 2.7) = 17.01
2
2
34.02
3
1
2
π (1.35) ( 0.9) = 2.5765
2
1.8
22.088
2
4
− π ( 0.8 ) ( 0.9 ) = −1.80956
Σ
37.217
ΣxV
119.392
=
in.
ΣV
37.217
8+
8
π
−14.4765
119.392
or X = 3.21 in.
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 5, Solution 99.
Assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the volume.
Then Y =
V , in 3
y , in.
yV , in 4
1
( 8)( 0.9)( 2.7) = 19.44
0.45
8.748
2
1
( 2.1)( 6)( 2.7) = 17.01
2
1.6
27.216
3
1
2
π (1.35) ( 0.9) = 2.5765
2
0.45
1.15943
4
− π ( 0.8 ) ( 0.9 ) = −1.80956
0.45
− 0.81430
Σ
37.217
ΣyV
36.309
=
in.
ΣV
37.217
2
36.309
or Y = 0.976 in. 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 5, Solution 100.
Labeling the five parts of the body as follows, and noting that the center of gravity coincides with
the centroid of the area due to the uniform thickness.
4 × 150 

z5 = −  300 −
 = − 236.34,
3π 

A5 = −
π
2
(150 )2 = −11 250π = − 35 343
xA,106 mm3 yA, 106 mm3 zA, 106 mm3
A mm 2
x , mm
y , mm
z , mm
1
( 600 )( 400 ) = 240000
300
200
0
72
48
0
2
( 300 )( 400 ) = 120000
600
200
−150
72
24
−18
3
− (120 )( 280 ) = − 3360
600
140
− 240
− 20.160
− 4.7040
8.0640
4
( 600 )( 300 ) = 180000
300
400
−150
54
72
− 27
5
− 35 343
240
400
− 236.34
− 2.7000
− 4.5
2.6588
Σ
471 057
169.358
125.159
− 28.583
continued
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
Therefore:
X =
ΣxA 169 358 000
=
= 359.53 mm
471 057
ΣA
or X = 360 mm !
Y =
ΣyA 125159 000
=
= 265.70 mm
471 057
ΣA
or Y = 266 mm !
Z =
ΣzA − 28 583 000
=
= − 60.678 mm
471 057
ΣA
or Z = − 60.7 mm !
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 5, Solution 101.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the area.
A, in 2
x , in.
y , in.
z , in.
xA, in 3
yA, in 3
zA, in 3
1
1
( 4.5)( 3) = 6.75
2
1.5
7
0
10.125
47.25
0
2
( 4.5)(10 ) = 45
2.25
3
4
101.25
135
180
3
− ( 2.25 )( 5 ) = −11.25 − 2.25
1.125
1.5
6
−12.6563
−16.875
− 67.5
2.25
0
4 ( 2.25 )
17.8925
0
71.211
116.611
165.375
183.71
1
4
π
2
( 2.25)2 = 7.9522
Σ
48.452
8+
3π
Then X =
ΣxA 116.611
=
in.
ΣA
48.452
or X = 2.41 in. Y =
ΣyA 165.375
=
in.
ΣA
48.452
or Y = 3.41 in. Z =
ΣzA 183.711
=
in.
48.452
ΣA
or Z = 3.79 in. 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 5, Solution 102.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the area.
X = 150 mm !
First note that by symmetry:
For 1: y = 180 + 96 +
4 (150 )
= 339.7 mm
3π
z =0
For 2: y = 180 +
z =
2 ( 96 )
π
2 ( 96 )
π
= 241.1 mm
= 61.11 mm
For 3: Length DE =
(180 )2 + ( 96 )2 = 204 mm
y = 90 mm,
z = 48 mm
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
A, mm 2
y , mm
z , mm
yA, mm3
zA, mm3
(150 )2 = 35.34 × 103
339.7
0
12.005 × 106
0
( 96 )( 300 ) = 45.24 × 103
244.1
61.11
10.907 × 106
2.765 × 106
3
( 204 )( 300 ) = 61.2 × 103 − 2.25
90
48
5.508 × 106
2.938 × 106
Σ
141.78 × 103
28.420 × 106
5.702 × 106
π
1
2
2
π
2
Then
Y =
ΣyA 28.420 × 106
=
mm
ΣA
141.78 × 103
Z =
ΣzA
5.702 × 106
=
mm
ΣA
141.78 × 103
or Y = 200 mm !
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.
or Y = 40.2 mm !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 103.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the area.
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
( 360 )( 270 ) = 97 200
0
135
0
13 122 000
2
( 339 )( 270 ) = 91 530
168
135
15 377 000
12 356 600
3
1
( 339 )( 72 ) = 12 204
2
224
294
2 733 700
3 588 000
4
( 360 )( 343.63) = 123 707
168
306
20 783 000
37 854 000
5
1
( 343.63)( 45) = 7731.73
2
224
318
1 731 900
2 458 700
6
7731.7
224
318
1 731 900
2 458 700
7
12 204
224
294
2 733 700
3 588 000
8
91 530
168
135
15 377 000
12 356 600
Σ
443 838
60 468 200
87 782 600
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
Then X =
ΣxA 60 468 200
=
mm
ΣA
443 838
or X = 136.2 mm !
Y =
ΣyA 87 782 600
=
mm
ΣA
443 838
or Y = 197.8 mm !
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 5, Solution 104.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the
area.
Note that by symmetry X = 9 in.
A, in 2
y , in.
z , in.
yA, in 3
zA, in 3
1
16.2
1.8
2
29.16
32.4
2
16.2
1.8
2
29.16
32.4
3
97.2
2.7
0
262.44
0
4
1017.876
−15.2789
20.72113
−15552
21091.54
5
1017.876
−15.2789
20.72113
−15552
21091.54
6
− 706.858
−12.7324
23.2676
9000
−16446.9
7
− 706.858
−12.7324
23.2676
9000
−16446.9
8
1017.876
− 22.9183
13.08169
− 23328
13315.54
Σ
1769.511
− 36111.24
22669.6
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
Therefore
X = 9 in. !
Y =
ΣyA − 36111.24
=
ΣA
1769.511
or Y = − 20.4 in. !
Z =
ΣzA 22669.6
=
ΣA 1769.511
or Z = 12.81 in. !
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 5, Solution 105.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the area.
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
π ( 8 )(12 ) = 96π
0
6
0
576π
2
−
10
−128
−160π
12
− 42.667
96π
π
2
π
3
2
Then
2 ( 4)
(8)( 4 ) = −16π
( 4 )2 = 8π
=
π
−
4 ( 4)
3π
8
π
=−
16
3π
4
(8)(12 ) = 96
6
12
576
1152
5
(8)(12 ) = 96
6
8
576
768
6
−
8
− 42.667
− 64π
π
2
( 4 )2 = − 8π
4 ( 4)
3π
=
16
3π
7
( 4 )(12 ) = 48
6
10
288
480
8
( 4 )(12 ) = 48
6
10
288
480
Σ
539.33
1512.6
4287.4
X =
ΣxA 1514.67
=
in. or X = 2.81 in. ΣA
539.33
Y =
ΣyA 4287.4
=
in. or Y = 7.95 in. ΣA
539.33
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 5, Solution 106.
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning
coincides with the centroid of the corresponding area.
yII = yVI = 80 +
( 4 )( 500 ) = 292.2 mm
3π
4
500
( )( ) = 212.2 mm
zII = zVI =
3π
( 2 )( 500 ) = 398.3 mm
yIV = 80 +
π
zIV =
( 2 )( 500 ) = 318.3 mm
π
AII = AVI =
AIV =
π
2
π
4
( 500 )2 = 196 350 mm 2
( 500 )( 680 ) = 534 071 mm 2
A, mm 2
y , mm
z , mm
yA, mm3
zA, mm3
I
(80)(500) = 40 000
40
250
1.6 × 106
10 × 106
II
196 350
292.2
212.2
57.4 × 106
41.67 × 106
III
(80)(680) = 54 400
40
500
0.2176 × 106
27.2 × 106
IV
534 071
398.3
318.3
212.7 × 106
170 × 106
V
(80)(500) = 40 000
40
250
1.6 × 106
10 × 106
VI
196 350
292.2
212.2
57.4 × 106
41.67 × 106
332.9 × 106
300.5 × 106
Σ
1.061 × 106
X = 340 mm Now, symmetry implies
and
(
)
Y ΣA = Σ yA: Y 1.061 × 106 mm 2 = 332.9 × 106 mm 3
(
6
Z ΣA = Σ zA: Z 1.061 × 10 mm
2
) = 300.5 × 10 mm
6
or Y = 314 mm 3
or Z = 283 mm 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 5, Solution 107.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the area.
1
A,in 2
y ,in.
z , in.
yA, in 3
zA, in 3
(15)(14 ) = 120
0
7
0
1470
1.25
7
43.75
245
1.25
0
46.875
0
1.25
7
43.75
245
6.5
0
− 510.51
0
−21.848
4
(14)( 2.5) = 35.0
(15)( 2.5) = 37.5
(14)( 2.5) = 35.0
5
− π ( 5 ) = − 78.540
0
6
1
2
− π (1.5 ) = −1.76715
4
0
7
( 4 )(12 ) = 48
6
10
288
480
8
( 4 )(12 ) = 48
6
10
288
480
Σ
235.43
134.375
1405.79
2
3
2
13 −
4 (1.5 )
3π
= 12.36348
Then Y =
ΣyA 134.375
=
in.
ΣA
235.43
or Y = 0.571 in. Z =
ΣzA 1405.79
=
in.
ΣA
235.43
or Z = 5.97 in. and by symmetry
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.
X = 7.50 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 108.
2
2
2
AB 2 = ( 500 mm ) + ( 750 mm ) + ( 300 mm ) , or
AB = 950 mm
L, mm
x , mm
y , mm
z , mm
xL, mm 2
yL, mm 2
zL, mm 2
AB
950
250
375
150
237.5 × 103
356.25 × 103
142.5 × 103
BD
300
500
0
150
150 × 103
0
45 × 103
DO
500
250
0
0
125 × 103
0
0
OA
750
0
375
0
0
281.25 × 103
0
Σ
2500
512.5 × 103
637.5 × 103
187.5 × 103
Then
X =
ΣxL 512.5 × 103
=
2500
ΣL
or X = 205 mm Y =
ΣyL 637.5 × 103
=
2500
ΣL
or Y = 255 mm Z =
ΣzL 187.5 × 103
=
ΣL
2500
or Z = 75 mm 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 5, Solution 109.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the line
1
2
3
4
Σ
π
2
L, mm
x , mm
y , mm
z , mm
xL, mm 2
yL, mm 2
zL, mm 2
300
280
260
0
140
230
150
0
0
0
0
120
0
39 200
59 800
45 000
0
0
0
0
31 200
( 300 ) = 150π
3  2 × 300  360

=
5 π
π

600
480
π
π
54 000
90 000
72 000
153 000
135 000
103 200
1311.24
Then
X =
ΣxL 153 000
=
ΣL 1311.24
or X = 116.7 mm Y =
ΣyL 135 000
=
ΣL 1311.24
or Y = 103.0 mm Z =
ΣzL 103 200
=
ΣL 1311.24
or Z = 78.7 mm 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 5, Solution 110.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the line.
L,ft
x , ft
y ,ft
xL, ft 2
yL,ft 2
1
10
4cos 45° = 2.8284
5
28.284
50
2
10
4cos 45° = 2.8284
5
28.284
50
3
4π
0
12.5465
0
157.664
4
4π
2 ( 4)
10
32
125.664
5
2π
2 ( 4)
12.5465
16
78.832
Σ
51.416
104.568
462.16
π
π
=
=
8
π
8
π
Then
X =
ΣxL 104.568
=
ΣL
51.416
or
Y =
ΣyL 462.16
=
ΣL
51.416
or Y = 8.99 ft and by symmetry:
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.
X = 2.03 ft COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 111.
First note by symmetry:
Z = 3.00 ft To simplify the calculations replace:
(a) The two rectangular sides with an element of length
L(a) = 2  2 ( 7 ft ) + 2 ( 5 ft )  = 48 ft
and center of gravity at (3.5 ft, 2.5 ft, 3 ft)
(b) The two semicircular members with an element of length
Lb = 2 π ( 3 ft )  = 6π ft
2×3


ft, 3 ft  = ( 2 ft, 6.9099 ft, 3 ft )
and with center of gravity at  2 ft, 5 +
π


(c) The cross members 1 and 2 with an element of length
Lc = 2 ( 6 ft ) = 12 ft
and with center of gravity at ( 2 ft, 5 ft, 3 ft )
(d) This leaves a single straight piece of pipe, labeled (d) in the figure.
Now for the centroid of the frame:
L,ft
x , ft
y ,ft
xL, ft 2
yL,ft 2
(a)
48
3.5
2.5
(b)
6π = 18.8496
2
6.9099
168
37.699
120
130.249
(c)
(d)
12
6
85.850
2
7
5
5
24
42
271.70
60
30
340.25
Σ
Then
X =
ΣxL 271.70
=
ΣL
84.850
or
Y =
ΣyL 340.25
=
ΣL
84.850
or Y = 4.01 ft 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.
X = 3.20 ft COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 112.
Y = Z = 0
First, note that symmetry implies
xI =
5
2π
( 0.5 in.) = 0.3125 in., WI = 0.0374 lb/in 3   ( 0.5 in )3 = 0.009791 lb
8
 3 
(
)
(
)
2
xII = 1.6 in. + 0.5 in. = 2.1 in. WII = 0.0374 lb/in 3 (π )( 0.5 in ) ( 3.2 in.) = 0.093996 lb
2
π 
xIII = 3.7 in. − 1 in. = 2.7 in., WIII = − 0.0374 lb/in 3   ( 0.12 in ) ( 2 in.) = −0.000846 lb
4
 
(
)
2
2
π 
xIV = 7.3 in. − 2.8 in. = 4.5 in., WIV = 0.284 lb/in 3   ( 0.12 in ) ( 5.6 in ) = 0.017987 lb
4
 
(
xV = 7.3 in. +
π
1
( 0.4 in.) = 7.4 in., WV = 0.284 lb/in 3   ( 0.06 in )2 ( 0.4 in.) = 0.000428 lb
4
3
(
Σ
Have
)
)
W , lb
x , in.
xW , in ⋅ lb
I
0.009791
0.3125
0.003060
II
0.093996
2.1
0.197393
III
−0.000846
2.7
−0.002284
IV
0.017987
4.5
0.080942
V
0.000428
7.4
0.003169
0.12136
0.28228
X ΣW = ΣxW : X ( 0.12136 lb ) = 0.28228 in. ⋅ 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.
or X = 2.33 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 113.
Determine first the masses of the component pieces:
π

m1 = 8800 kg/m3  0.0162 − 0.0122 m 2 × ( 0.014 m )  = 0.0108372 kg
4

(
) (
)
π

m2 = 1250 kg/m 3  0.0362 − 0.0162 m 2 × ( 0.014 m )  = 0.0142942 kg
4


(
) (
)
π

m3 = 1250 kg/m 3  0.0602 − 0.0362 m 2 × ( 0.006 m )  = 0.0135717 kg
4

(
) (
)
π

m4 = 1250 kg/m3  0.0802 − 0.0602 m 2 × ( 0.010 m )  = 0.027489 kg
4

(
) (
)
Now, for the center of mass:
Then X =
m, kg
x , mm
xm, kg ⋅ mm
1
0.0108372
7
0.075860
2
0.0142942
7
0.100059
3
0.0135717
3
0.040715
4
0.027489
5
0.137445
Σ
0.066192
0.35408
Σxm
0.35408
or
=
0.066192
Σm
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.
X = 5.35 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 114.
Assume that the stone is homogeneous so that its center of gravity coincides with the centroid of the volume
and locate the center of gravity.
To determine the centroid of the truncated pyramid note that:
3
1
y1 = (1.4 m ) = 1.05 m, and
V1 = ( 0.3 m )( 0.3 m )(1.4 m ) = 0.042 m3
4
3
y2 =
3
( 0.7 m ) = 0.525 m, and
4
V2 = −
1
( 0.15 m )( 0.15 m )( 0.7 m ) = − 0.00525 m3
3
Then Vstone = V1 + V2 = 0.042 m3 − 0.00525 m3 = 0.03675 m3, and
(
)
(
3
3
ΣyV (1.05 m ) 0.042 m + ( 0.525 m ) − 0.00525 m
y=
=
ΣV
0.03675 m3
)
= 1.12500 m
The center of gravity of the stone is therefore 0.425 m (i.e. 1.125 m – 0.7m) above the base.
Now to determine the center of gravity of the marker:
(
)(
)(
)
= ( 7860 kg/m )( 9.81 m/s ) ( 0.3 m )( 0.3 m ) h  = ( 6939.6 h ) N
Wstone = ( ρ gV ) stone = 2570 kg/m3 9.81 m/s 2 0.03675 m3 = 926.53 N
Wsteel = ( ρ gV ) steel
3
2
continued
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
Then
ymar ker =
ΣyW
, or
ΣW
0.3 m =
( 0.425 m )( 926.53 N ) + ( − h2 m ) ( 6939.6 h ) N
, or
( 926.53 + 6939.6 h ) N
h 2 + 0.6 h − 0.033378 = 0.
Solving for h and discarding the negative root, this gives h = 0.051252 m, or
h = 50 mm !
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 5, Solution 115.
Since the brass plates are equally spaced and by the symmetry of the cylinder:
X =Y =0!
For the pipe:
Specific weight of steel: γ s = 0.284 lb/in 3
y1 = 4 in.
outside diameter: 2.5 in.
Inside diameter: 2.5 in. − 2 ( 0.25 in.) = 2.00 in.
π
( 2.5 − 2.0 ) 8 = 14.137 in
2
2
3
Volume:
V1 =
Weight:
W1 = γ sV1 = 0.284 lb/in 3 14.137 in 3 = 4.015 lb
4
(
)(
)
For each brass plate:
Specific weight for brass: γ B = 0.306 lb/in 3
8
2.667 in.
3
1
Volume: V2 = ( 8 )( 4 )( 0.2 ) = 3.2 in 3
2
y2 =
(
)(
)
Weight: W1 = γ sV1 = 0.306 lb/in 3 3.2 in 3 = 0.979 lb
continued
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
For flagpole base:
ΣW = ( 4.015 lb ) + 3 ( 0.979 lb ) = 6.952 lb
ΣyW = ( 4 in.)( 4.015 lb ) + 3 ( 2.667 in.)( 0.979 lb )  = 23.892 in.⋅ lb, or
Y =
ΣyW 23.892 in.⋅ lb
=
= 3.437 in.
ΣW
6.952 lb
Y = 3.437 in. !
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 5, Solution 116.
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is
r 2 = a 2 − x 2 and then
(
x 2 + y 2 = a 2 so that
)
dV = π a 2 − x 2 dx
Component 1
a/2

x3 
a/2
V1 = ∫0 π a 2 − x 2 dx = π  a 2 x − 
3 0

(
=
and
)
11 3
πa
24
∫1 xEL dV = ∫0 x π ( a − x ) dx 
a/2
2
2
a/2
 x2 x4 
= π a 2
− 
4 0
 2
=
Now
7
π a4
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a3  =
π a4
 24
 64
or x1 =
Component 2
a

x3 
a
V2 = ∫a /2 π a 2 − x 2 dx = π  a 2 x − 
3  a/2

(
)
⎧

a3   2  a 
⎪ 2
= π ⎨a ( a ) −  − a   −
3   2
⎪

⎩
( a2 )  ⎫⎪
3
⎬
3 ⎪
 ⎭
continued
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.
21
a!
88
COSMOS: Complete Online Solutions Manual Organization System
=
5
π a3
24
and
a
x π
a/2 
∫2 xELdV = ∫
(
a
 x2 x4 
a − x dx  = π  a 2
−


4  a/2
 2
2
2
)
2
⎧
2
4

a )   2 a2
(
⎪ 2 (a)
− a
= π ⎨a
−
−
2
4  
2
⎪ 
 
⎩
9
=
π a4
64
( )
Now
( a2 )  ⎫⎪
4
⎬
4 ⎪
 ⎭
9
 5

x2V2 = ∫2 xELdV : x2  π a3  =
π a4
 24
 64
or x2 =
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.
27
a!
40
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 117.
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
x2
y2
+
= 1 so that
h2 a 2
The equation of the generating curve is
r2 =
a2 2
h − x 2 and then
2
h
(
)
dV = π
a2 2
h − x 2 dx
2
h
(
)
Component 1
h/2
V1 = ∫0 π
=
and
h/2
a2 2
a2  2
x3 
2
−
=
−
h
x
dx
h
x
π


3 0
h2
h2 
(
)
11 2
πa h
24
2

h/2  a
2
2
x
dV
x
π
=
∫1 EL
∫0  h 2 h − x dx 


(
)
h/2
a2  x2 x4 
= π 2 h2
− 
4 0
h  2
7
π a 2h 2
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a 2h  =
π a 2h 2
 24
 64
=
Now
or x1 =
continued
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.
21
h!
88
COSMOS: Complete Online Solutions Manual Organization System
Component 2
h
V2 = ∫h/2 π
h
a2 2
a2  2
x3 
2
−
=
−
π
h
x
dx
h
x


3 h/2
h2
h2 
(
)
3
⎧
3

h ⎫
h)   2  h 
a 2 ⎪ 2
(
2 ⎪
− h   −
= π 2 ⎨h ( h ) −
⎬
3    2
3 ⎪
h ⎪




⎭
⎩
5
πa 2 h
=
24
()
and
 a2

∫2 xELdV = ∫h/2 x π h2 ( h − x ) dx 
h
2
2


h
=π
a 2  2 x2 x4 
−
h

4  h/2
h2  2
2
⎧
2
4

h )   2 h2
(
a2 ⎪ 2 ( h )
− h
= π 2 ⎨h
−
−
2
4  
2
h ⎪




⎩
9
=
π a 2h 2
64
( )
Now
( h2 )  ⎫⎪
4
⎬
4 ⎪
 ⎭
9
 5

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h 2
24
64


or x2 =
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.
27
h!
40
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 118.
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
x=h−
The equation of the generating curve is
r2 =
h 2
y so that
a2
a2
( h − x ) and then
h
dV = π
a2
( h − x ) dx
h
Component 1
h/2
V1 = ∫0 π
a2
( h − x ) dx
h
h/2
a2 
x2 
= π  hx −

h 
2 0
=
and
3 2
πa h
8
 a2

∫1 xELdV = ∫0 x π h ( h − x ) dx 
h/2


h/2
=π
=
Now
a 2  x 2 x3 
− 
h
h  2
3 0
1
π a 2h 2
12
1
3

x1V1 = ∫1 xEL dV : x1  π a 2h  = π a 2h 2
8
12


or x1 =
continued
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.
2
h!
9
COSMOS: Complete Online Solutions Manual Organization System
Component 2
h
h
V2 = ∫h/2 π
a2
a2 
x2 
( h − x ) dx = π hx − 
2  h/2
h
h 
⎧
2

h)    h 
(
a2 ⎪


= π ⎨ h (h) −
− h  −
2   2
h ⎪



⎩
1
= πa 2 h
8
( h2 )  ⎫⎪
2
⎬
2 ⎪
 ⎭
h
and
 a2

a 2  x 2 x3 
h
∫2 xEL dV = ∫h/2 x π h ( h − x ) dx  = π h  h 2 − 3 



 h/2
2
⎧
2
3

h )   h2
(
a2 ⎪ ( h )
− h
=π
−
−
⎨h
3   2
h ⎪ 2




⎩
1
=
π a 2h 2
12
( )
Now
( h2 )  ⎫⎪
3
⎬
3 ⎪
 ⎭
1
1

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h2
8
 12
or x2 =
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.
2
h!
3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 119.
y = 0!
First note that symmetry implies
z = 0!
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
⎛
x2 ⎞
Now r = b⎜⎜1 − 2 ⎟⎟ so that
a ⎠
⎝
2
2⎛
x2 ⎞
dV = πb ⎜⎜1 − 2 ⎟⎟ dx
a ⎠
⎝
Then
x2 ⎞
a
2⎛
−
π
1
b
⎜
2⎟
0
⎜
⎟
V =∫
⎝
2
⎛
2x2 x4 ⎞
a
dx = ∫0 π b 2 ⎜⎜1 − 2 + 4 ⎟⎟ dx
a ⎠
a
a ⎠
⎝
2⎛
2 x3
x5 ⎞
= π b ⎜⎜ x − 2 + 4 ⎟⎟
3a
5a ⎠
⎝
a
0
2 1⎞
⎛
= π ab 2 ⎜1 − + ⎟
3 5⎠
⎝
8
= π ab 2
15
and
2x2 x4 ⎞
a
2 ⎛
1
x
dV
π
b
x
=
−
+ 4 ⎟⎟ dx
⎜
∫ EL
∫0
⎜
a2
a ⎠
⎝
2⎛ x
2x4
x6 ⎞
= π b ⎜⎜
− 2 + 4 ⎟⎟
4a
6a ⎠
⎝ 2
2
a
0
continued
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
⎛1 1 1⎞
= π a 2b 2 ⎜ − + ⎟
⎝2 2 6⎠
=
Then
1 2 2
πa b
6
1
⎛ 8
⎞
xV = ∫ xEL dV : x ⎜ π ab 2 ⎟ = π a 2b 2
⎝ 15
⎠ 16
or x =
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.
15
a!
6
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 120.
y = 0
First, note that symmetry implies
z = 0
Choose as the element of volume a disk of radius r and
thickness dx.
Then
dV = πr 2dx, xEL = x
Now r = 1 −
1
so that
x
2
1

dV = π 1 −  dx
x

2
1 

= π 1 − + 2  dx
x
x


3
Then
2
1 
1
3 

V = ∫1 π 1 − + 2  dx = π  x − 2 ln x − 
x
x 1
x 



1 
1 
= π  3 − 2 ln3 −  − 1 − 2 ln 1 −  
3 
1 

= ( 0.46944π ) m 3
and
2
3  
x π 1− +
1  
∫ x EL dV = ∫
 
x
3
 x2

1  
dx
=
π
 − 2 x + ln x 


2
x  
2
1
  32
 13
 
= π   − 2 ( 3) + ln 3 −  − 2 (1) + ln1 
 2
 
  2
= (1.09861π ) m
Now
(
)
xV = ∫ x EL dV : X 0.46944π m 3 = 1.09861π m 4
or x = 2.34 m 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 5, Solution 121.
First, by symmetry:
x =a!
y =0!
Next determine the constants k in y = kx1/3 :
x = a, b = ka1/3 or k =
At
b
a1/3
b 1/3
a
x , or x = 3 y 3
1/3
a
b
Choosing horizontal disks of thickness dy for volume elements ( dV in the figure above)
Therefore, y =
2
b
V = ∫0 π  a 2 − ( a − x ) 


b
(
)
= π ∫0 2ax − x 2 dy
a
a2 
b
= π ∫0  2a × 3 y 3 − 6 y 6  dy
b
b


b
a2 
1
1 1 
5
= π 3  2 × y 4 − 3 × y 7  = π a 2b
4
7  0 14
b 
b
1
∫ yELdV , or
V
14 b   a 2 3 a 2 6  
y=
∫ y π  2 y − b6 y  dy 
5π a 2b 0   b3
 
Now y =
14 
y5
1 y8 
= 4  2 ×
− 3

5
5b 
b 8 
b
0
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.
y=
77
b!
100
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 122.
First note that by symmetry:
y =0!
z = 0!
Choose as a volume element a disk of radius y and thickness dx. Then:
xEL = x, and
dV = π y 2dx, or
dV = π h 2 cos 2
πx
2a
dx
Using the identity:
cos 2 x =
1
(1 + cos 2 x ) , this gives
2
dV =
1 2
πx
π h 1 + cos  dx.
2
a 

Then:
V = ∫ dV =
π h2 a 
πx
π h2 
a
πx
a
1
2
∫ 1 + cos a  dx = 2  x + π sin a  = 2 π h a.
2 0


0
continued
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
Also,
∫ xELdV =
=
=
=
π h2 a 
πx


∫  x + x cos a  dx. Integrating by parts,
2 0
π h 2   x 2 
a
a
a
πx
π x  
sin
sin
+
−
x
 

∫
2   2  0 π 
a
a  0 


π h 2  a 2
a
a
πx a
π x  
sin
+
cos
x

π 
a
π
a  0 

2  2
+
π h 2  a 2
a
2a   1 2 2 
4 
+ 0 − 0 −

  = π a h 1 − 2 
2  2
π
π   4
π 

Now,
x=
1
2 1 2 2 
4 
xEL dV =
π a h 1 − 2   , or
∫
2 
V
πh a 4
π 

x=
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.
1 
4 
a 1 − 2  !
2 
π 
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 123.
First note that by symmetry:
x =0!
y =0!
Choosing the volume element shown in the figure, i.e. a cylindrical shell of radius r, height h and
thickness dr:
yEL =
1
y,
2
and
dV = 2π ry dr = 2π r cos
πr
dr , and
2a
πr
a
V = ∫ dV = 2π h ∫0 r cos
dr , or, integrating by parts
2a
a
V = 2π h
2a 
πr
πr 
r sin
dr
− ∫ sin
π 
2a
2a  0
a
π r 2a
πr 

cos 
= 4ah  r sin
+
π
2
a
2a  0

2a 
2

2 
= 4ah  a −
 = 4a h 1 − 
π 
π


Also,
a
2
2
∫ yEL dV = π h ∫0 r cos
π h 2   1 2 
πr
2a
dr =
π h2 a 
πr 


∫ r 1 + cos a  dr
2 0
a
a
πr
π r  
=
+
− ∫ sin
r sin
dr 
 r
2   2  0 π 
a
a  0 
a
continued
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
=
=
π h 2  1
a
a
πr a
π r  
2
+
+
a
r
sin
cos


π 
π
2  2
a
a  0 
π h2  1
2a   1 2 2 
4 
a
2
 a + 0 + 0 −
  = π a h 1 − 2 
π
π  4
2 2
π 

Now,
y=
1
∫ yEL dV =
V
1 2 2 
1
4 
 π a h 1 − 2   , or
2


4
π 

4a 2 h  1 −  
π

y=
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.
(π + 2 ) h !
16
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 124.
Choose as the element of a horizontal slice of thickness dy. For any
number N of sides, the area of the base of the pyramid is given by
Abase = kb 2
where k = k ( N ) ; see note below. Using similar triangles, have
s
h− y
=
b
h
or
s=
b
(h − y)
h
Then
dV = Aslicedy = ks 2dy = k
and
V = ∫0 k
h
=
Also
b2
2
h − y ) dy
2(
h
h
b2
b2  1
2
3
h
y
dy
k
−
=
− (h − y) 
(
)
h2
h 2  3
0
1 2
kb h
3
yEL = y
2

b2 h
h  b
2
so then ∫ y EL dV = ∫0 y k 2 ( h − y ) dy  = k 2 ∫0 h 2 y − 2hy 2 + y 3 dy
h
 h

(
)
h
2
1 
1 2 2
b2  1
kb h
= k 2  h 2 y 2 − hy 3 + y 4  =
3
4 0 12
h 2
Now
1 2 2
1

yV = ∫ y EL dV : y  kb 2h  =
kb h
3
12


or y =
Note:
b 
1
Abase = N  × b × 2 π 
tan N 
2


N
=
b2
π
4 tan N
= k ( N ) b2
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.
1
h Q.E.D. 4
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 125.
Since the spherical cup is uniform, the center of gravity will coincide with the centroid. Also, because the cup is
thin, it can be treated like an area in finding the centroid.
An element of area is obtained by rotating arc ds about the y axis. With the y axis pointing downwards,
dA = 2π rds = 2π ( R sin θ ) Rdθ
= 2π R 2 sin θ dθ
yEL = y = R cosθ
φ
φ
A = ∫ dA = 2π R 2 ∫0 sin θ dθ = 2π R 2 [ − cosθ ]0 = 2π R 2 (1 − cos φ )
∫ yEL dA = ∫0 ( R cosθ ) ( 2π R sin θ dθ ) = 2π R ∫0 cosθ sin θ dθ
φ
3 φ
2
φ
 1

= 2π R3  − cos 2 θ  = π R3 1 − cos 2 φ
2

0
(
)
Then,
y=
1
1
3
2
∫ yEL dA = 2π R 2 1 − cos φ π R 1 − cos φ , or
A
(
)
y=
R
(1 + cos φ )
2
(
)
Using
cos φ =
y=
R−h
h
=1− :
R
R
R 
h 
h
1 + 1 −   = R −
R 
2 
2
The center of gravity is therefore located at a distance of
h h

R − y = R −  R −  = , above the base.(Q.E.D)
2 2

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 5, Solution 126.
(a) Bowl
First note that symmetry implies
x = 0!
z = 0!
for the coordinate axes shown below. Now assume that the bowl may be
treated as a shell; the center of gravity of the bowl will coincide with the
centroid of the shell. For the walls of the bowl, an element of area is
obtained by rotating the arc ds about the y axis. Then
dAwall = ( 2π R sin θ )( Rdθ )
and
Then
and
( yEL ) wall = −R cos θ
π /2
π /2
Awall = ∫π /6 2π R 2 sin θ dθ = 2π R 2 [ − cosθ ]π /6
= π 3R 2
ywall Awall = ∫ ( yEL )wall dA
(
π /2
= ∫π /6 ( − R cosθ ) 2π R 2 sin θ dθ
= π R3 cos 2 θ 
)
π /2
π /6
3
= − π R3
4
π
R2,
By observation
Abase =
Now
y ΣA = ΣyA
4
ybase = −
3
R
2
or
π
3
π  3 


y  π 3R 2 + R 2  = − π R3 + R 2  −
R
4 
4
4  2 

or
y = −0.48763R
R = 350 mm
∴ y = −170.7 mm !
(b) Punch
First note that symmetry implies
x = 0!
z = 0!
and that because the punch is homogeneous, its center of gravity will
coincide with the centroid of the corresponding volume. Choose as the
element of volume a disk of radius x and thickness dy. Then
dV = π x 2dy, yEL = y
continued
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
x2 + y 2 = R2
Now
(
)
dV = π R 2 − y 2 dy
so that
0
1 
0

V = ∫− 3/2 R π R 2 − y 2 dy = π  R 2 y − y 3 
3  − 3/2 R

(
Then
)
3
 
3  1
3   3
= −π  R 2  −
R − −
R  = π 3R3
  2  3  2   8


and
0
0
1 2 2 1 4
2
2
∫ yELdV = ∫− 3/2R ( y ) π R − y dy  = π  2 R y − 4 y 

 − 3/2 R
(
)
2
4




 
1
3
1
3
15
2
= −π  R  −
R − −
R   = − π R4




2  2 
4 2  
64


Now
or
15
3

yV = ∫ yEL dV : y  π 3 R3  = − π R 4
64
8

y =−
5
8 3
R
R = 350 mm
∴ y = −126.3 mm !
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 5, Solution 127.
The centroid can be found by integration. The equation for the bottom of the gravel is:
y = a + bx + cz, where the constants a, b, and c can be determined as follows:
For x = 0, and z = 0: y = − 3 in., and therefore
−
3
1
ft = a, or a = − ft
12
4
For x = 30 ft, and z = 0: y = − 5 in., and therefore
−
5
1
1
ft = − ft + b ( 30 ft ) , or b = −
12
4
180
For x = 0, and z = 50 ft: y = − 6 in., and therefore
−
6
1
1
ft = − ft + c ( 50 ft ) , or c = −
12
4
200
Therefore:
1
1
1
y = − ft −
x−
z
4
180
200
Now
x dV
x = ∫ EL
V
A volume element can be chosen as:
dV = y dxdz,
or
dV =
1
1
1 
x+
z  dx dz, and
1 +
4
45
50 
xEL = x
continued
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
Then
50 30 x 
1


1
∫ xEL dV = ∫0 ∫0 4 1 + 45 x + 50 z  dx dz

30
1 50  x 2
1 3
z 2
= ∫0 
+
x +
x  dz
4  2 135
100  0
=
1 50
∫ ( 650 + 9 z ) dz
4 0
=
1
9 
650 z + z 2 

4
2 0
50
= 10937.5 ft 4
The volume is:
50 30 1 
V ∫ dV = ∫0 ∫0
1
1 
x+
z  dx dz
1 +
4
45
50 
30
1 50 
1 2
z 
= ∫0  x +
x +
x dz
4 
90
50  0
=
1 50 
3 
40 + z  dz
∫
0 
4 
5 
50
1
3 2
=  40 z +
z
4
10  0
= 687.50 ft 3
Then
x dV 10937.5ft 4
x = ∫ EL
=
= 15.9091 ft
V
687.5 ft 3
Therefore:
V = 688 ft 3
x = 15.91 ft
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 5, Solution 128.
Choosing the element of volume shown, i.e. a filament of sides, y, dx, and dz:
dV = y dx dy, and
z EL = z
x
z
b a
V = ∫ dV = ∫ 0 ∫ 0  y0 − y1 − y2  dx dz
a
b

=∫
x
b
y x − y1
0 0
a
2
zx 
1
a 
b
− y2  dz = ∫ 0  y0a − y1a − y2 z  dz
b 0
b 
2a
2


b

1
a z2 
1
1 

=  y0az − y1az − y2
 =  y0 − y1 − y2  ab
b 2 0 
2
2
2 

b
a
xz
z2 
∫ zEL dV = ∫ 0 ∫ 0  y0 z − y1 a − y2 b  dx dz

=∫

a
x2 z
z2x 
b
y zx − y1
− y2
 dz =
0 0
b 0
2a

∫
za
z 2a 
b
y
za
−
y
−
y

 dz
1
2
0 0

2
b 
b
 z 2a
z 2a
z 3a 
1
1  2
1
=  y0
− y1
− y2
 =  y0 − y1 − y2  ab
2
4
3b  0  2
4
3 

continued
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
Now,
z =
1
1
1
1 
1
z EL dV =
y0 − y1 − y2  ab 2 , or
∫

1
1  2

4
3 
V
 y0 − y1 − y2  ab
2
2


1
1
1
y0 − y1 − y2
2
4
3 b
z =
1
1
y0 − y1 − y2
2
2
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 5, Solution 129.
x =0
First note that symmetry implies
Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then
1
dV = 2 xy dz, yEL = y, zEL = z
2
h
h
h
z
Now
and
y = −
z = 1 − 
x = a2 − z2
2 2a
2
a
z

dV = h a 2 − z 2 1 −  dz

a
So
Then
V =∫
=
=
Then
a
h
0
3/2 

z
1 2
 1 
 z 
a − z 1 −  dz = h   z a 2 − z 2 + a 2 sin −1    +
a − z2 
a

 a   3a
 2 

2
(
2
)
a
−a
1 2  −1
a h sin (1) − sin −1 ( −1) 
2
π
2
a 2h
a
1
h
z  


 
2
z




∫ yELdV = ∫ − a  2 × 2 1 − a   h a − z 1 − a  dz 
=
2
h2 a
z
z2 
2
2
a
z
−
−
+
1
2

 dz
∫

a a 2 
4 −a

continued
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
=
3
2 2

h 2  1 
2
2
2
−1  z  
a − z2 2 
  z a − z + a sin    + 
a
a
4  2 
3
  

(
)
3
1  z
a2 z 2
a4
 z   
+ 2 − a2 − z 2 2 +
a − z 2 + sin −1    
8
8
a  4
 a   
(
=
)
a
−a
5h 2a 2  −1
sin (1) − sin −1 ( −1) 

32
 π a 2  5h 2a 2
yV = ∫ yEL dV : y 
h  =
(π )
32
 2 
Then
or y =
a
and

2
z




5
h
16
∫ zELdV = ∫ − a z  h a − z 1 − a  dz 
2

a
3
3
1 z
a2z 2
a 4 −1  z   
 1
= h − a 2 − z 2 2 −  − a 2 − z 2 2 +
a − z2 +
sin    
a 4
8
8
 a   
 3
−a
(
=−
)
(
)
a3h  −1
sin (1) − sin −1 ( −1) 

8
 π a 2h 
π a 3h
zV = ∫ z EL dV : z 
 = −
8
 2 
or z = −
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
4
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 130.
1
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
21 × 22 = 462
1.5
11
693
5082
2
−
1
( 6 )( 9 ) = −27
2
−6
2
162
−54
3
−
1
( 6 )(12 ) = −36
2
8
2
−288
−72
567
4956
Σ
399
Then
X =
Σ xA 567 mm 3
=
ΣA
399 mm 2
or X = 1.421 mm and
Y =
Σ yA 4956 mm 3
=
ΣA
399 mm 2
or Y = 12.42 mm 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 5, Solution 131.
1
2
Σ
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
(10)(15) = 50
3
4.5
7.5
225
375
6.366
16.366
1125
2892
1350
3267
π
4
(15)2 = 176.71
226.71
X Σ A = Σx A
Then
(
)
X 226.71 in 2 = 1350 in 3
or X = 5.95 in. Y ΣA = Σy A
and
(
)
Y 226.71 in 2 = 3267 in 3
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.
or Y = 14.41 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 132.
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Then
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
122 + 62 = 13.416
6
3
80.50
40.25
2
16
12
14
192
224
3
21
1.5
22
31.50
462
4
16
−9
14
−144
224
5
62 + 92 = 10.817
− 4.5
3
− 48.67
32.45
Σ
77.233
111.32
982.7
X ΣL = Σx L
X (77.233 mm) = 111.32 mm 2
and
or X = 1.441 mm Y ΣL = Σ y L
Y (77.233 mm) = 982.7 mm 2
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.
or Y = 12.72 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 133.
First note that for equilibrium, the center of gravity of the wire must lie
on a vertical line through C. Further, because the wire is homogeneous,
its center of gravity will coincide with the centroid of the corresponding
line.
Thus ΣM C = 0, which implies that x = 0
or
Σ xi Li = 0
Hence
L
( L ) + ( − 4 in.)(8 in.) + ( − 4 in.)(10 in.) = 0
2
or
L2 = 144 in 2
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.
or L = 12.00 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 134.
For the element (EL) shown
At
x = a, y = h : h = ka3
Then
x=
h
a3
k =
or
a 1/3
y
h1/3
dA = xdy
Now
=
xEL =
h
A = ∫ dA =∫ 0
Then
a 1/3
y dy
h1/3
1
1 a 1/3
x=
y , yEL = y
2
2 h1/ 3
h
a 1/3
3 a
3
y dy =
y 4/3 = ah
4 h1/3
4
h1/3
0
( )
h
a 1/3  a 1/3  1 a  3 5/3 
3 2
and ∫ xEL dA = ∫
y  1/3 y dy  =
y  =
a h
1/3
2/3 
2h
h
 2 h 5
 0 10
h1
0
∫ yEL dA = ∫
Hence
h  a

y
y1/3dy  =
0  1/3
h

h
a  3 7/3 
3
y  = ah 2
1/3 
h 7
0 7
3 2
3 
xA = ∫ xEL dA : x  ah  =
a h
4
10


x =
2
a 5
3  3
yA = ∫ yEL dA: y  ah  = ah 2
4  7
y =
4
h 7
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 5, Solution 135.
2b = ka 2
For
y1 at x = a, y = 2b
Then
y1 =
By observation
y2 = −
or k =
2b
a2
2b 2
x
a2
b
x
( x + 2b) = b  2 − 
a
a
xEL = x
Now
and for 0 ≤ x ≤ a :
1
b
y1 = 2 x 2
2
a
and
dA = y1dx =
1
b
x
y2 =  2 − 
2
2
a
and
x

dA = y2dx = b  2 −  dx
a


yEL =
2b 2
x dx
a2
For a ≤ x ≤ 2a :
yEL =
Then
a 2b 2
x dx +
2
A = ∫ dA = ∫ 0
=
and
a
2a


a
2
 a
2b  x3 
x 
b
2
+
−
−

 

 
a  
a2  3 0
 2 
a
 2b

2


x
∫ a b  2 − a  dx
2a

2a
=
7
ab
6
 
x

 


0
∫ xEL dA = ∫ 0 x  a 2 x dx  + ∫ a x b  2 − a  dx 
a
2a
=
 2 x3 
2b  x 4 
 + b x −

2 
3a  0
a  4 0

=
1 2
1  2
2
2
3 
2a − ( a )  
a b + b ( 2a ) − ( a )  +

 3a 

2
=
7 2
ab
6
{
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
a b
2  2b
2


2a b 
x  
x

 


∫ yEL dA = ∫ 0 a 2 x  a 2 x dx  + ∫ 0 2  2 − a  b  2 − a  dx 


a
3
2b 2  x5 
b2  a 
x 
= 4   +
−  2 −  
2  3 
a  
a  5 0
=
Hence
2a
a
17 2
ab
30
7  7
xA = ∫ xEL dA: x  ab  = a 2b
6  6
 7  17 2
yA = ∫ yEL dA: y  ab  =
ab
 6  30
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.
x =a y =
17
b 35
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 136.
The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the
second theorem of Pappus-Guldinus and using Fig. 5.8A, we have
V = 2π xA = 2πΣxA = 2π ( x1 A1 + x2 A2 )




 1 1   1 1


3   2R sin 30o
o π
2
= 2π  × R   × R ×
R + 
cos 30   R  

2   3 × π
6

 3 2   2 2

 
6




 R3
R3  3 3
= 2π 
+
π R3
 =
8
 16 3 2 3 
=
Since
3 3
3
π (12 in.) = 3526.03 in 3
8
1 gal = 231 in 3
V =
3526.03 in 3
= 15.26 gal
231 in 3/gal
V = 15.26 gal 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 5, Solution 137.
Have
RI = ( 9 ft )( 200 lb/ft ) = 1800 lb
RII =
Then
1
( 3 ft )( 200 lb/ft ) = 300 lb
2
ΣFx = 0: Ax = 0
ΣM A = 0: − ( 4.5 ft )(1800 lb ) − (10 ft )( 300 lb ) + ( 9 ft ) B y = 0
or
By = 1233.3 lb
B = 1233 lb
A = 867 lb
ΣFy = 0: Ay − 1800 lb − 300 lb + 1233.3 lb = 0
or
Ay = 866.7 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 138.
Have
Then
RI =
1
( 4 m )( 2000 kN/m ) = 2667 N
3
RII =
1
( 2 m )(1000 kN/m ) = 666.7 N
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2667 N − 666.7 N = 0
or
Ay = 3334 N
A = 3.33 kN
M A = 6.33 kN ⋅ m
ΣM A = 0: M A − (1 m )( 2667 N ) − ( 5.5 m )( 666.7 N )
or
M A = 6334 N ⋅ m
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 5, Solution 139.
Consider the free-body diagram of the side.
1
1
Ap = A (γ d )
2
2
Have
P=
Now
ΣM A = 0:
( 9 ft ) T −
d
P=0
3
Then, for d max:
( 9 ft ) ( 0.2 ) ( 40 × 103 lb ) −
d max  1

3
 (12 ft ) ( d max )  62.4 lb/ft d max  = 0
3 2

or
3
216 × 103 ft 3 = 374.4 d max
or
3
d max
= 576.92 ft 3
(
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.
)
d max = 8.32 ft COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 140.
First, assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the corresponding volume.
x , in.
y , in.
I
II
V , in 3
(4)(3.6)(0.75) = 10.8
(2.4)(2.0)(0.6) = 2.88
2.0
3.7
0.375
1.95
21.6
10.656
4.05
5.616
III
π(0.45)2 (0.4) = 0.2545
4.2
2.15
1.0688
0.54711
1.2
0.375
− 0.7068
− 0.22089
32.618
9.9922
IV
Σ
2
− π (0.5) (0.75) = − 0.5890
13.3454
xV , in 4
yV , in 4
X ΣV = Σ x V
Have
(
)
X 13.3454 in 3 = 32.618 in 4
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.
or X = 2.44 in. COSMOS: Complete Online Solutions Manual Organization System
Chapter 5, Solution 141.
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with
the centroid of the corresponding area. Then (see diagram)
zV = 22.5 −
4 ( 6.25 )
3π
= 19.85 mm
AV = −
π
2
( 6.25)2
= − 61.36 mm 2
A, mm 2
x , mm
y , mm
z , mm
xA, mm3
yA, mm3
zA, mm3
I
( 25)( 60) = 1500
12.5
0
30
18 750
0
45 000
II
(12.5)( 60 ) = 750
25
− 6.25
30
18 750
− 4687.5
22 500
III
( 7.5)( 60 ) = 450
28.75
−12.5
30
12 937.5
− 5625
13 500
IV
− (12.5 )( 30 ) = − 375
10
0
37.5
− 3750
0
−14 062.5
V
− 61.36
10
0
19.85
− 613.6
0
−1218.0
Σ
2263.64
46 074
−10 313
65 720
Have
X ΣA = ΣxA
(
)
X 2263.64 mm 2 = 46 074 mm 3
Y ΣA = Σ yA
(
)
Y 2263.64 mm 2 = −10 313 mm 3
or X = 20.4 mm or Y = − 4.55 mm Z ΣA = Σ zA
(
)
Z 2263.64 mm 2 = 65 720 mm 3
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.
or Z = 29.0 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 1.
\
Joint FBDs:
Joint B:
FAB 800 lb FBC
=
=
15
8
17
so
FAB = 1500 lb T
FBC = 1700 lb C
Joint C:
FAC Cx 1700 lb
=
=
8
15
17
FAC = 800 lb T
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 6, Solution 2.
Joint FBDs:
Joint B:
ΣFx = 0:
1
4
FAB − FBC = 0
5
2
ΣFy = 0:
1
3
FAB + FBC − 4.2 kN = 0
5
2
so
Joint C:
7
FBC = 4.2 kN
5
FBC = 3.00 kN C !
12 2
kN
5
FAB = 3.39 kN C !
FAB =
ΣFx = 0:
4
12
(3.00 kN) −
FAC = 0
5
13
FAC =
13
kN
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.
FAC = 2.60 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 3.
Joint FBDs:
Joint B:
FAB FBC 450 lb
=
=
12
13
5
FAB = 1080 lb T
so
FBC = 1170 lb C
Joint C:
ΣFx = 0:
3
12
FAC − (1170 lb ) = 0
5
13
FAC = 1800 lb C
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 6, Solution 4.
Joint FBDs:
Joint D:
FCD FAD 500 lb
=
=
8.4 11.6
8
FAD = 725 lb T
FCD = 525 lb C
Joint C:
ΣFx = 0:
FBC − 525 lb = 0
FBC = 525 lb C
This is apparent by inspection, as is FAC = C y
ΣFx = 0:
8.4
3
(725 lb) − FAB − 375 lb = 0
11.6
5
Joint A:
FAB = 250lb T
ΣFy = 0:
FAC −
4
8
(250 lb) −
(725 lb) = 0
5
11.6
FAC = 700 lb C
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 6, Solution 5.
FBD Truss:
ΣFx = 0 :
Cx = 0
By symmetry: C y = D y = 6 kN
Joint FBDs:
Joint B:
ΣFy = 0:
− 3 kN +
1
FAB = 0
5
FAB = 3 5 = 6.71 kN T
Joint C:
ΣFx = 0:
ΣFy = 0:
Joint A:
ΣFx = 0:
ΣFy = 0:
2
FAB − FBC = 0
5
FBC = 6.00 kN C
3
FAC = 0
5
FAC = 10.00 kN C
6 kN −
6 kN −
4
FAC + FCD = 0
5
FCD = 2.00 kN T
 1

3

− 2
3 5 kN  + 2  10 kN  − 6 kN = 0 check
5

 5

By symmetry:
FAE = FAB = 6.71 kN T
FAD = FAC = 10.00 kN C
FDE = FBC = 6.00 kN C
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 6, Solution 6.
FBD Truss:
ΣM A = 0:
(10.2 m ) C y + ( 2.4 m )(15 kN ) − ( 3.2 m )( 49.5 kN ) = 0
C y = 12.0 kN
Joint FBDs:
Joint FBDs:
Joint C:
FBC
F
12 kN
= CD =
7.4
7.4
8
FBC = 18.50 kN C
FCD = 18.50 kN T
Joint B:
ΣFX = 0:
4
7
(18.5 kN) = 0
FAB −
5
7.4
FAB = 21.875 kN;
ΣFy = 0:
FAB = 21.9 kN C
3
2.4
(21.875 kN) − 49.5 kN +
(18.5 kN) + FBD = 0
5
7.4
FBD = 30.375 kN;
FBD = 30.4 kN C
Joint D:
ΣFx = 0: −
4
7
FAD +
(18.5 kN ) + 15 kN = 0
5
7.4
FAD = 40.625 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.
FAD = 40.6 kN T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 7.
Joint FBDs:
Joint E:
FBE FDE 3 kN
=
=
5
4
3
FBE = 5.00 kN T
FDE = 4.00 kN C
Joint B:
ΣFx = 0:
− FAB +
4
(5 kN) = 0
5
FAB = 4.00 kN T
ΣFy = 0:
FBD − 6 kN −
3
(5 kN) = 0
5
FBD = 9.00 kN C
Joint D:
ΣFy = 0:
3
FAD − 9 kN = 0
5
FAD = 15.00 kN T
ΣFx = 0: FCD −
4
(15 kN) − 4 kN = 0
5
FCD = 16.00 kN C
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 6, Solution 8.
Joint FBDs:
Joint B:
FAB = 12.00 kips C
By inspection:
FBD = 0
Joint A:
FAC FAD 12 kips
=
=
5
13
12
FAC = 5.00 kips C
FAD = 13.00 kips T
Joint D:
ΣFx = 0: FCD −
12
(13 kips) − 18 kips = 0
13
FCD = 30.0 kips C
ΣFy = 0:
5
(13 kips) − FDF = 0
13
FDF = 5.00 kips T
Joint C:
ΣFx = 0: 30 kips −
12
FCF = 0
13
FCF = 32.5 kips T
ΣFy = 0: FCE − 5 kips −
5
(32.5 kips)
13
FCE = 17.50 kips C
Joint E:
by inspection:
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.
FCF = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 9.
FCH = 0 !
First note that, by inspection of joint H:
and
FCG = 0 !
then, by inspection of joint C:
and
Joint D:
FBC = FCD
FBG = 0 !
then, by inspection of joint G:
and
Joint FBDs:
FDH = FGH
FFG = FGH
FBF = 0 !
then, by inspection of joint B:
and
FAB = FBC
FCD FDH 10 kips
=
=
12
13
5
Joint A:
FCD = 24.0 kips T !
so
FDH = 26.0 kips C !
FAB = FBC = 24.0 kips T !
and, from above:
FGH = FFG = 26.0 kips C !
FAF
F
24 kips
= AE =
5
4
41
FAF = 30.0 kips C !
Joint F:
FAE = 25.6 kips T !
ΣFx = 0: FEF −
12
(26 kips) = 0
13
FEF = 24.0 kips C !
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 6, Solution 10.
FBD Truss:
ΣFx = 0: H x = 0
By symmetry: A y = H y = 4 kips
FAC = FCE and FBC = 0
by inspection of joints C and G :
FEG = FGH and FFG = 0
also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF
Joint FBDs:
Joint A:
FAB FAC 3 kips
=
=
5
4
3
FAB = 5.00 kips C
so
FAC = 4.00 kips T
FFH = 5.00 kips C
and, from above,
and
Joint B:
FCE = FEG = FGH = 4.00 kips T
4
4
10
(5 kips) − FBE −
FBD = 0
5
5
109
ΣFx = 0:
ΣFy = 0:
3
3
3
FBD + FBE = 0
( 5 kips ) − 2 −
5
5
109
so
FBD = 3.9772 kips, FBE = 0.23810 kips
FBD = 3.98 kips C
or
Joint E:
FBE = 0.238 kips C
FDF = 3.98 kips C
and, from above,
FEF = 0.238 kips C
ΣFy = 0 :
FDE − 2
3
(0.23810 kips) = 0
5
FDE = 0.286 kips T
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 6, Solution 11.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM G = 0:
3a Ay − 2a (3 kN) − a (6 kN) = 0
A y = 4 kN
by inspection of joint C,
FAC = FCE and FBC = 0
by inspection of joint D,
FBD = FDF and FDE = 6.00 kN C
Joint FBDs:
Joint A:
FAC FAB 4 kN
=
=
21
29
20
FAB = 5.80 kN C
FAC = 4.20 kN C
from above,
Joint B:
ΣFy = 0:
20
20
( 5.80 kN ) − 3 kN − FBE = 0
29
29
FBE =
ΣFx = 0:
FCE = 4.20 kN C
29
20
FBE = 1.450 kN T
21 
29

kN  − FBD = 0
 5.80 kN +
29 
20

FBD = 5.25 kN C
FDF = 5.25 kN C
from above,
Joint F:
ΣFx = 0:
5.25 kN −
21
FEF = 0
29
FEF = 7.25 kN T
ΣFy = 0:
FFG −
20
(7.25 kN) − 1 kN = 0
29
FFG = 6.00 kN C
by inspection of joint G,
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.
FEG = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 12.
FBD Truss:
ΣFx = 0:
Ax = 0
By symmetry: A y = B y = 4.90 kN
FAB = FEG , FAC = FFG , FBC = FEF
and
FBD = FDE , FCD = FDF
5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.9 kN = 0
5
29
ΣFx = 0:
Joint FBDs:
Joint A:
ΣFy = 0:
FAC = 2.8 29 kN
FAC = 15.08 kN T FAD = 17.50 kN C Joint B:
ΣFx = 0:
ΣFy = 0:
4
1
FBC = 0
(17.5 kN − FBD ) −
5
2
3
1
FBC − 2.8 kN = 0
(17.5 kN − FBD ) +
5
2
FBD = 15.50 kN C FBC = 1.6 2 kN;
Joint C:
FBC = 2.26 kN C 2
(2.8 29 kN) = 0
29
FCD = 9.00 kN T 1
3
5
1.6 2 kN + (9 kN) −
(2.8 29 kN) = 0
ΣFx = 0: FCF +
5
2
29
FCF = 7.00 kN T ΣFy = 0:
4
1
1.6 2 kN −
FCD −
5
2
(
(
)
)
from symmetry,
FEG = 17.50 kN C FFG = 15.08 kN T FEF = 2.26 kN C FDE = 15.50 kN C FDF = 9.00 kN T 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 6, Solution 13.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0: (8 m) Gy − (4 m)(4.2 kN) − (2m)(2.8 kN) = 0
G y = 2.80 kN
ΣFy = 0:
Ay − 2.8 kN − 4.2 kN + 2.8 kN = 0
A y = 4.2 kN
Joint FBDs:
5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.2 kN = 0
5
29
ΣFx = 0:
Joint A:
ΣFy = 0:
FAB = 15.00 kN C !
FAC = 12.92 kN T !
FAC = 2.4 29
Joint B:
ΣFx = 0:
ΣFy = 0:
4
1
FBC = 0
(15.00 kN − FBD ) −
5
2
3
1
FBC − 2.8 kN = 0
(15.00 kN − FBD ) +
5
2
FBD = 13.00 kN C !
FBC = 1.6 2 kN,
Joint C:
ΣFy = 0:
(
4
FCD −
5
FBC = 2.26 kN C !
)
2
1
2.4 29 kN −
(1.6 2 kN) = 0
29
2
FCD = 8.00 kN T !
ΣFx = 0:
FCF +
3
(8.00 kN ) −
5
+
5
(2.4 29 kN)
29
1
(1.6 2 kN) = 0
2
FCF = 5.60 kN T !
continued
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
By inspection of joint E,
FDE = FEG
and
FEF = 0 !
Joint F:
ΣFy = 0:
ΣFx = 0:
4
FDF −
5
2
FFG = 0
29
3
5
− 5.6 kN − FDF +
FFG = 0
5
29
FDF = 4.00 kN T !
FFG = 1.6 29 kN
Joint G:
ΣFx = 0:
4
FEG −
5
FFG = 8.62 kN T !
5
(1.6 29 kN) = 0
29
FEG = 10.00 kN C !
from above (joint E)
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.
FDE = 10.00 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 14.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0:
4a H y − 3a (1.5 kN) − 2a (2 kN)
− a (2 kN) = 0
ΣFy = 0:
Ay − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN
+ 3.625 kN = 0
Joint FBDs:
Joint A:
A y = 3.875 kN
FAB FAC
2.625 kN
=
=
1
29
26
FAB = 15.4823 kN,
FAB = 15.48 kN C !
FAC = 14.6597 kN,
FAC = 14.66 kN T !
By inspection of joint C:
Joint B:
H y = 3.625 kN
ΣFy = 0:
FCE = FAC = 14.66 kN T,
2
(15.4823 kN − FBD ) − 2 kN = 0
29
FBD = 10.0971 kN,
ΣFx = 0:
FBC = 0 !
FBD = 10.10 kN C !
5
(15.4823 kN − 10.0971 kN ) − FBE = 0
29
FBE = 5.0000 kN,
FBE = 5.00 kN C !
continued
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
Joint D:
FDF = 10.0971 kN,
By symmetry:
ΣFy = 0:
FDF = 10.10 kN C !
 2

10.0971 kN  − 2 kN = 0
− FDE + 2 
 29

FDE = 5.50 kN T !
FGH FFH
2.625 kN
=
=
1
26
29
Joint H:
By inspection of joint G:
ΣFx = 0 :
Joint F:
FFH = 14.1361 kN
FFH = 14.14 kN C !
FGH = 13.3849 kN
FGH = 13.38 kN T !
FEG = FGH = 13.38 kN T
FEF +
and
FFG = 0 !
2
(10.0971 kN − 14.1361 kN ) = 0
29
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.
FEF = 3.75 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 15.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0:
8a ( J y − 1 kN) − 7a (1 kN) − 6a (2.8 kN)
− 4a (4.5 kN) − 2a (4 kN) − a(1 kN) = 0
J y = 6.7 kN
ΣFy = 0:
− 1 kN − 1 kN + 6.7 kN = 0
Joint FBDs:
Joint A:
Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN
ΣFy = 0:
5 kN −
A y = 6.0 kN
3
5 13
FAB = 0, FAB =
kN
3
13
FAB = 6.01 kN C !
Joint B:

2  5 13
kN  = 0,

13  3

ΣFx = 0:
FAC −
ΣFx = 0:

2  5 13
kN − FBC − FBD  = 0,

13  3

FAC = 3.33 kN T !
5 13
kN
3

3  5 13
kN + FBC − FBD  − 1 kN = 0,

13  3

4 13
FBD − FBC =
kN
3
3
13 kN
FBD =
FBD = 5.41 kN C !
2
1
13 kN
FBC =
FBC = 0.601 kN C !
6
FBC + FBD =
ΣFy = 0:
continued
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
FEF = 0 !
By inspection of joint F:
Joint E:
FDE = FEG
By symmetry:
ΣFy = 0:
2
1
FDE − 4.5 kN = 0,
17
FDE =
9
17 kN
4
FDE = 9.28 kN C !
ΣFx = 0:
2 3
4 9


13 kN  −
17 kN  = 0


13  2
17  4


FDF +
Joint D:
FDF = 6.00 kN T !
ΣFy = 0:
3 3
1 9


13 kN  − 1.4 kN − FCD −
17 kN  = 0


13  2
17  4


FCD = 0.850 kN T !
Joint C:
ΣFy = 0:
0.850 kN −
FCG =
ΣFx = 0:
FCI −
 3
3  13
kN  − FCG = 0

 5
13  6

1.75
kN
3
FCG = 0.583 kN C !
 10
4  1.75
2  13

kN  −
kN  −
kN = 0


5 3
13  6

 3
FCI = 3.47 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 16.
\
FBD Truss:
ΣFx = 0:
ΣM A = 0:
Ax = 0
8a( J y − 1 kN) − 7a(1 kN) − 6a(2.8 kN)
− 4a(4.5 kN) − 2a(4 kN) − a(1 kN) = 0
J y = 6.7 kN
ΣFy = 0:
Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN
−1 kN − 1 kN + 6.7 kN = 0
Joint FBDs:
Joint J:
ΣFy = 0:
(6.7 − 1) kN −
3
FHJ = 0,
13
A = 6.0 kN
FHJ = 1.9 13 kN
FHJ = 6.85 kN C !
Joint H:
ΣFx = 0:
2
(1.9 13 kN) − FIJ = 0,
13
ΣFx = 0:
2
( FGH + FHI − 1.9 13 kN) = 0
13
ΣFy = 0:
3
( FHI − FGH + 1.9 13 kN) − 1 kN = 0
13
26
13 kN,
FGH =
FGH = 6.25 kN C !
15
FHI =
Joint I:
ΣFx = 0:
3.80 kN −
FGI −
13
kN,
6
FHI = 0.601 kN C !

2  13
kN  − FCI = 0

13  6

FCI =
ΣFy = 0:
FIJ = 3.80 kN T !
10.4
kN,
3

3  13
kN  = 0,


13  6

FCI = 3.47 kN T !
FGI = 0.500 kN T !
continued
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
FEF = 0 !
By inspection of joint F,
By symmetry FDE = FEG
Joint E:
ΣFy = 0:
 1

FEG  − 4.5 kN = 0,
2
17


FEG =
9
17 kN
4
FEG = 9.28 kN C !
Joint G:
ΣFy = 0:
3  26
3
1 9
 1

13 kN  − kN − FCG −
17 kN 


5
13  15
17  4
 2

− 2.8 kN = 0
FCG = −
ΣFx = 0:
1.75
kN
3
FCG = 0.583 kN C !
4 9
4  1.75 

17 kN  − FFG −  −


5 3 
17  4

−
2  26

13 kN  = 0

13  15

FFG = 6.00 kN T !
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 6, Solution 17.
FBD Truss:
By load symmetry,
ΣFx = 0:
θ = tan −1
Joint FBDs:
A y = H y = 1600 lb
Ax = 0
6.72 ft
= 16.2602°
23.04 ft
ΣFy′ = 0:
(1600 lb − 400 lb) cosθ − FAC sin θ = 0
FAC = 4114.3 lb
Joint A:
ΣFx = 0:
FAC cos 2θ − FAB cosθ = 0
FAB = 3613.5 lb
ΣFx′ = 0:
Joint B:
Joint C:
FBC = 0.768 kips C !
FCD sin 2θ − (768 lb) cosθ = 0
FCD = 1371.4 lb
ΣFx′′ = 0:
FBD = 3.84 kips C !
FBC − (800 lb) cosθ = 0
FBC = 768.00 lb
ΣFy′′ = 0:
FAB = 3.61 kips C !
3613.5 lb − FBD + (800 lb)sin θ = 0
FBD = 3837.5 lb
ΣFy′ = 0:
FAC = 4.11 kips T !
FCD = 1.371 kips T !
FCE + (1371.4 lb) cos2θ − (768 lb)sin θ − 4114.3 lb = 0
FCE = 2742.9 lb
FCE = 2.74 kips T !
continued
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
Joint E:
ΣFy = 0:
(2742.9 lb)sin 2θ − FDE cosθ = 0
FDE = 1536.01 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.
FDE = 1.536 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 18.
FBD Truss:
A y = H y = 1600 lb
By load symmetry,
ΣFx = 0:
Ax = 0
θ = tan −1
6.72 ft
= 16.2602°
23.04 ft
Joint FBDs:
Joint H:
ΣFy = 0:
1600 lb − 400 lb − FFH sin θ = 0
FFH = 4285.7 lb,
( 4285.7 lb ) cosθ − FGH = 0
ΣFx = 0:
Joint F:
FFH = 4.29 kips C
FGH = 4114.3 lb
ΣFy′ = 0:
FFG − ( 800 lb ) cosθ = 0
FFG = 768.0 lb
ΣFx′ = 0:
FGH = 4.11 kips T
FFG = 0.768 kips C
FDF + ( 800 lb ) sin θ − 4285.7 lb = 0
FDF = 4061.7 lb
FDF = 4.06 kips C
Joint G:
ΣFy = 0:
FDG sin 2θ − (768 lb) cosθ = 0
FDG = 1371.4 lb
ΣFx = 0:
FDG = 1.371 kips T
4114.3 lb − (1371.4 lb ) cosθ − FEG − (768 lb) sin θ = 0
FEG = 2742.9 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.
FEG = 2.74 kips T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 19.
FBD Truss:
ΣFx = 0:
ΣM L = 0:
Ax = 0
(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN)
+ (5.5 m)(6 kN) − (11 m) Ay = 0
A y = 4.5 kN
Joint FBDs:
Joint A:
ΣFy = 0:
4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0
L = 10.3 kN
Joint C:
4.5 kN FAC
F
=
= AB ,
1
2
5
FAC = 9.00 kN T !
FAB = 4.5 5,
FAB = 10.06 kN C !
9 kN FBC
F
=
= CE ,
16
5
281
FBC =
45
kN
16
FBC = 2.81 kN C !
FCE =
Joint B:
ΣFx = 0:
ΣFy = 0:
Solving:
9
281,
16
FCE = 9.43 kN T !
2 
16
FBD + (4.5 5) kN  +
FBE = 0


5
265
1 
FBD + (4.5 5) kN  −

5
3
45
kN = 0
FBE +
16
265
72
5 kN,
11
45
265 kN,
FBE =
176
FBD = −
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.
FBD = 14.64 kN C !
FBE = 4.16 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Joint E:
ΣFx = 0:
16 
 9
 FEG −  16
281 

FEG =
ΣFy = 0:

16  45


281  kN  −
265  kN = 0

265  176



9
281 kN,
11
FEG = 13.72 kN T !
5 9
9

281 kN −
281 kN  +

16
281  11

3  45

265 kN 

265  176

+ FDE = 0,
FDE = −
Joint D:
ΣFx = 0:
ΣFy = 0:
Solving:
45
kN,
22
FDE = 2.05 kN C !
2 
72
10

FDG = 0
5 kN  +
 FDF +
11
5
101

1 
72
1
45

FDG +
5 kN  −
kN = 0
 FDF +
11
22
5
101

7.5
101 kN,
22
FDG = 3.43 kN T !
FDF = − 8.25 5 kN,
FDF = 18.45 kN C !
FDG =
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 6, Solution 20.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM L = 0:
(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN)
− (11 m) Ay = 0
A y = 4.5 kN
Joint FBDs:
ΣFy = 0:
4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0
Joint L:
L = 10.3 kN
10.3 kN FKL FJL
=
=
,
1
2
5
FKL = 20.6 kN T !
FJL = 10.3 5 kN,
FJL = 23.0 kN C !
20.6 kN FJK
F
=
= IK ,
16
5
281
Joint K:
FJK =
51.5
kN
8
FJK = 6.44 kN C !
FIK =
Joint J:
ΣFx = 0:
−
10.3
281 kN,
8
(
FIK = 21.6 kN T!
)
2
16
FHJ + 10.3 5 kN −
FIJ = 0
5
265
(
)
ΣFy = 0:
1
FHJ + 10.3 5 kN −
5
Solving:
FHJ = −
112
5 kN,
11
FHJ = 22.8 kN C !
FIJ = −
1.3
265 kN,
88
FIJ = 0.240 kN C !
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.
3
51.5
kN = 0
FIJ − 6.6 kN +
8
265
COSMOS: Complete Online Solutions Manual Organization System
Joint I:
16 
10.3
16  1.3


281 kN  −
265 kN  = 0
 − FGI +

8
88
281 
265 


ΣFx = 0:
FGI =
ΣFy = 0:
ΣFx = 0:
ΣFy = 0:
−
10.4
kN,
88
FHI = 0.1182 kN T !
2 
112
10

FGH = 0
5 kN  −
 FFH +
11
5
101

1 
112
1
10.4

FGH − 2.2 kN −
5 kN  −
kN = 0
 FFH +
11
88
5
101

Solving:
FFH = − 8.25 5 kN,
FFH = 18.45 kN C !
34
101 kN,
88
FGH = 3.88 kN C !
FGH = −
Joint F:
FGI = 21.3 kN T !
5  112
10.3

281 kN −
281 kN  + FHI

8
281  88

3  1.3

265 kN  = 0
−

265  88

FHI =
Joint H:
112
281 kN,
88
By symmetry:
FDF = 8.25 5 kN
ΣFy = 0:
2
8.25 5 kN − 6 kN − FFG = 0
5
(
)
FFG = 10.50 kN T !
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 6, Solution 21.
Joint FBDs:
FDF
F
500 lb
= EF =
916 916
480
Joint F:
FDF = FEF = 954.17 lb
FDF = 954 lb T !
or
FEF = 954 lb C !
Joint D:
FBD FDE
954.17 lb
=
=
884 240
916
FBD = 920.84 lb
FDE = 250.00 lb
FBD = 921 lb T !
or
FDE = 250 lb C !
By symmetry of joint A vs. joint F
FAB = 954 lb T !
FAC = 954 lb C !
Joint B:
ΣFx = 0:
920.84 lb −
884
4
( 954.17 lb ) + FBE = 0
916
5
FBE = 0 !
ΣFy = 0:
FBC −
240
( 954.17 lb ) = 0
916
FBC = 250.00 lb
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.
FBC = 250 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Joint C:
ΣFx = 0:
884
( 954.17 lb ) − FCE = 0
916
FCE = 920.84 lb
ΣFy = 0:
FCH −
240
(954.17 lb) − 250.00 lb = 0
916
FCH = 500.00 lb
ΣFx = 0:
FCE = 921 lb C !
or
920.84 lb −
or
FCH = 500 lb C !
884
4
( 954.17 lb ) − FHE = 0
916
5
FHE = 0 !
Joint E:
ΣFy = 0:
FEJ −
240
(954.17 lb) − 250.00 lb = 0
916
FEJ = 500.00 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.
or
FEJ = 500 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 22.
Joint FBDs:
FJL
F
500 lb
= KL =
1212 1212
480
Joint L:
FJL = FKL = 1262.50 lb
FJL = 1263 lb T !
FKL = 1263 lb C !
Comparing joint G to joint L:
FGH = 1263 lb T !
FGI = 1263 lb C !
Joint J:
ΣFx = 0:
1188
(1262.50 lb ) − FHJ = 0
1212
FHJ = 1237.50 lb,
ΣFy = 0:
FJK −
240
(1262.50 lb ) − 500 lb = 0
1212
FJK = 750.00 lb,
Joint H:
ΣFx = 0:
FHJ = 1238 lb T !
1237.50 lb −
FJK = 750 lb C !
1188
4
(1262.50 lb ) + FHK = 0
1212
5
FHK = 0 !
ΣFy = 0:
FHI −
240
3
(1262.50 lb) − 500 lb − (0) = 0
1212
5
FHI = 750 lb C !
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
Joint I:
ΣFx = 0:
1188
(1262.50 lb ) − FIK = 0
1212
FIK = 1237.50 lb,
ΣFy = 0:
FIN −
240
(1262.50 lb ) − 750.00 lb = 0
1212
FIN = 1000.00 lb,
Joint K:
ΣFx = 0:
FIK = 1238 lb C !
1237.50 lb −
FIN = 1000 lb C !
1188
4
(1262.50 lb ) − FKN = 0
1212
5
FKN = 0 !
ΣFy = 0:
FKO −
240
(1262.50 lb) − 750.00 lb = 0
1212
FKO = 1000 lb C !
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 6, Solution 23.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM A = 0:
(12 m ) (M y − 1 kN)
− (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0
M y = 5.05 kN
ΣFy = 0:
Ay − 2(1 kN) − 5(1.5 kN) + M y = 0
A y = 4.45 kN
Joint FBDs:
5
FAB = 0,
13
FAB = 8.97 kN C !
ΣFy = 0:
4.45 kN − 1 kN −
ΣFx = 0:
FAC −
12
(8.97 kN ) = 0,
13
FAC = 8.28 kN T !
ΣFx = 0:
12
FCE − 8.28 kN = 0,
13
FCE = 8.97 kN T !
ΣFy = 0:
5
(8.97 kN) − FBC = 0,
13
FBC = 3.45 kN C !
ΣFy = 0:
5
5
(8.97 kN ) − 1.5 kN + 3.45 kN − FBD = 0
13
13
Joint A:
Joint C:
Joint B:
FBD = 14.04 kN
ΣFx = 0:
FBD = 14.04 kN C !
12
12
(8.97 kN) − (14.04 kN) + FBE = 0
13
13
FBE = 4.68 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.
FBE = 4.68 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Joint E:
ΣFx = 0:
6
12
FEH − 4.68 kN − (8.97 kN) = 0
13
37
FEH = 13.1388 kN
ΣFy = 0:
FDE −
5
(8.97 kN ) −
13
or
1
(13.1388) = 0
37
FDE = 5.6100 kN
Joint D:
ΣFx = 0:
ΣFy = 0:
FEH = 13.14 kN T !
or
FDE = 5.61 kN T !
12
12
1
(14.04 kN ) − FDG − FDH = 0
13
13
2
5
5
(14.04 kN ) − FDG − 1.5 kN − 5.61 kN
13
13
1
+
FDH = 0
2
Solving:
FDG = 8.60 kN C !
FDH = 7.10 kN C !
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 6, Solution 24.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM A = 0:
(12 m ) (M y − 1 kN)
− (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0
M y = 5.05 kN
Joint FBDs:
ΣFy = 0:
Ay − 2(1 kN) − 5(1.5 kN) + M y = 0
A y = 4.45 kN
Joint M:
ΣFy = 0:
5.05 kN − 1 kN −
5
FKM = 0,
13
ΣFx = 0:
12
(10.53 kN) − FLM = 0,
13
ΣFx = 0:
9.72 kN −
FKM = 10.53 kN C !
FLM = 9.72 kN T !
6
FJL = 0,
37
FJL = 1.62 37
Joint L:
FJL = 9.85 kN T !
ΣFy = 0:
ΣFx = 0:
Joint K:
ΣFy = 0:
1
(1.62 37 kN) − FKL = 0,
37
12
FIK −
13
FKE = 1.620 kN C !
24
12
FJK − (10.53 kN) = 0
13
577
5
5
(10.53 kN) −
FIK −
13
13
1
FJK
577
− 1.5 kN + 1.62 kN = 0
Solving:
FIK = 10.8136 kN,
FIK = 10.81 kN C !
FJK = 0.26205 kN,
FJK = 0.262 kN T !
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
Joint J:
ΣFx = 0:
24
(0.26205 kN) −
577
6
( FHJ − 9.85 kN) = 0
37
FHJ = 10.1154 kN,
ΣFy = 0:
1
(10.1154 kN ⋅ 9.85 kN) −
37
FHJ = 10.12 kN T !
1
(0.26205 kN) − FIJ = 0
577
FIJ = 0.054541 kN,
Joint I:
ΣFx = 0:
ΣFy = 0:
Solving:
Joint G:
12
24
( FGI − 10.8136 kN ) + FHI = 0
13
25
5
7
(10.8136 kN − FGI ) + FHI − 1.5 kN + 0.05454 kN = 0
13
25
FGI = 8.6029 kN,
FGI = 8.60 kN C !
FHI = 2.1257 kN,
FHI = 2.13 kN C !
By symmetry:
ΣFy = 0:
FIJ = 54.5 N C !
FOG = 8.60 kN C !
 5
2   (8.6029 kN) − 1.5 kN − FGH = 0
 13 
FGH = 5.12 kN T !
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 6, Solution 25.
FBD Truss:
ΣFx = 0: 180 lb − Ax = 0
A x = 180 lb
ΣM A = (12 ft ) G y − ( 6 ft )( 480 lb ) − ( 2 ft )(180 lb )
− (8 ft)(120 lb) = 0,
ΣFy = 0:
G y = 350 lb
Ay − 480 lb − 120 lb + 350 lb = 0
A y = 250 lb
Joint FBDs:
Joint A:
ΣFx = 0:
1
FAC − 180 lb = 0, FAC = 180 2 lb
2
FAC = 255 lb T !
ΣFy = 0:
Joint B:
ΣFx = 0:
ΣFy = 0:
Solving:
1
(180 2 lb) + 250 lb − FAB = 0,
2
FAB = 430 lb C !
4
2
FBD +
FBC = 0
5
5
3
1
430 lb − FBD +
FBC = 0
5
5
180 lb −
FBC = 590 5 lb,
FBC = 1319 lb T !
FBD = 1700 lb C !
Joint C:
ΣFy = 0:
5
1
1
(590 5 lb) −
(180 2 lb) = 0
FCD −
41
5
2
FCD = 154 41 lb,
ΣFx = 0:
FCD = 986 lb T !
4
2
(154 41 lb) −
(590 5 lb)
41
5
1
−
(180 2 lb) = 0
12
FCE = 744 lb T !
FCE +
By inspection of joint G:
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.
FFG = 350 lb C and FEG = 0 !
COSMOS: Complete Online Solutions Manual Organization System
Joint F:
FDF
F
350
= EF =
;
5
1
20
FDF = 1750 lb C !
FEF = 700 5 lb,
FEF = 1565 lb T !
Joint E:
ΣFy = 0:
5
1
(700 5 lb) = 0
FDE − 120 lb −
41
5
FDE = 164 41 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.
FDE = 1050 lb T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 26.
Joint FBDs:
Joint A:
ΣFy = 0:
9
FAC − 1.8 kips = 0,
41
40
FAB − 8.20 kips = 0,
41
ΣFx = 0:
By inspection of joint B,
and then
Joint C:
Note:
DE =
ΣFy = 0:
FAC = 8.20 kips T !
FAB = 8.00 kips C !
FBD = FAB
FBD = 8.00 kips C !
FBC = 0.6 kips,
FBC = 0.600 kips C !
9.2
(4.5 ft) = 2.07 ft,
20
CE = 4.14 ft = 2 DE
1
FCD − 0.6 kips = 0,
5
FCD = 0.6 5 kips
FCD = 1.342 kips T !
ΣFx = 0:
8.2 kips +
(
)
2
0.6 5 kips − FCE = 0
5
FCE = 9.20 kips C !
Joint D:
ΣFy = 0:
(
)
40
2
0.6 5 kips = 0
( FDG − 8.2 kips ) −
41
5
FDG = 9.43 kips T !
ΣFx = 0:
FDE −
(
)
9
1
0.6 5 kips = 0
( 9.43 kips − 8.2 kips ) −
41
5
FDE = 0.330 kips C !
Joint E:
ΣFy = 0:
ΣFx = 0:
5
FEG − 0.33 kips = 0,
13
FEG = 0.858 kips T !
12
( 0.858 kips ) + 9.2 kips − FEF = 0
13
FEF = 9.992 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.
FEF = 9.99 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Joint F:
By vertical symmetry FFG = FFH
ΣFx = 0:
4

9.992 kips − 2  FFG  = 0,
5


FFG = 5.995 kips
FFG = FFH = 6.00 kips C !
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 6, Solution 27.
P6.14
Starting with
ABC, add, in order, joints E, D, F, G, H
∴simple truss
P6.15
Starting with
DEF, add, in order, G, C, B, A, I, H, J
∴simple truss
P6.23
Starting with
ABC, add, in order, E, F, D, H, G, I, J, K, L, M
∴simple truss
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 6, Solution 28.
P6.21
Starting with ABC, add, in order, joints E, D, F, H, J, K, L, I, G, N,
P, Q, R, O, M, S, T
∴simple truss
P6.25
Starting with
ABC, add, in order, joints D, E, F, G
∴simple truss
P6.29
Starting with
ABD, add, in order, joints H, G, F, E, I, G, J
∴simple truss
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 6, Solution 29.
ΣFx = 0:
Fx = 0
Then, by inspection of joint F,
FFG = 0
Then, by inspection of joint G,
FGH = 0
By inspection of joint J,
FIJ = 0
Then, by inspection of joint I,
FHI = 0
FEI = 0
Then, by inspection of joint E,
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.
FBE = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 30.
By inspection of joint D,
FDI = 0
By inspection of joint E,
FEI = 0
Then, by inspection of joint I,
FAI = 0
By inspection of joint F,
FFK = 0
By inspection of joint G,
FGK = 0
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 6, Solution 31.
\
By inspection of joint C:
FBC = 0
Then, by inspection of joint B:
FBE = 0
Then, by inspection of joint E:
FDE = 0
By inspection of joint H :
FFH = 0
and FHI = 0
By inspection of joint Q :
FOQ = 0
and FQR = 0
By inspection of joint J :
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.
FLJ = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 32.
By inspection of joint C :
FBC = 0
By inspection of joint G :
FFG = 0
Then, by inspection of joint F :
FFE = 0
By inspection of joint I :
FIJ = 0
By inspection of joint M :
FMN = 0
Then, by inspection of joint N :
FKN = 0
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 6, Solution 33.
By inspection of joint C:
FBC = 0 !
By inspection of joint M :
FLM = 0 !
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 6, Solution 34.
By inspection of joint A:
FAF = 0
By inspection of joint C :
FCH = 0
By inspection of joint E :
FDE = FEI = 0
By inspection of joint L :
FGL = 0
By inspection of joint N :
FIN = 0
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 6, Solution 35.
(a)
By inspection of joint H :
FCH = 0
Then, by inspection of joint C :
FCG = 0
Then, by inspection of joint G :
FBG = 0
Then, by inspection of joint B :
FBF = 0
By inspection of joint J :
FIJ = 0
Then, by inspection of joint I :
FEI = 0
(b)
FHI = 0
Then, by inspection of joint E :
also,
ΣFx = 0;
FBE = 0
Fx = 0
So, by inspection of joint F :
FFG = 0
And by inspection of joint G :
FGH = 0
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 6, Solution 36.
FBD Truss:
ΣFz = 0:
Bz = 0
ΣM x = 0:
− ( 0.47 m ) C y + ( 0.08 m )( 940 N ) = 0
C y = 160 Nj
Joint FBDs:
Joint D:
where FAD = FAD
FBD = FBD
= FBD
FCD = FCD
ΣFy = 0:
ΣFx = 0:
ΣFz = 0:
− 0.80 m i − 0.08 m k
2
( 0.8) + ( 0.08) m
−10i − k
101
− 0.8 m i − 0.16 m j − 0.2 m k
( 0.8 m )2 + ( 0.16 m )2 + ( 0.2 m )2
− 20i + 4 j − 5k
21
− 0.8 m i + 0.39 m k
2
( 0.8 m ) + ( 0.39 m )
4
FBD − 940 N = 0,
FBD = 4935 N,
21
10
80
20
−
FAD −
FCD −
( 4935 N ) = 0
89
21
101
1
39
5
−
FAD +
FCD −
( 4935 N ) = 0
89
21
101
Solving:
= FAD
2
2
= FCD
− 80i + 39k
89
FBD = 4.94 kN T !
FAD = − 590 101 N,
FAD = 5.93 kN C !
FCD = 1335 N,
FCD = 1.335 kN T !
continued
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
Joint C:
FBC = FBC
0.16 m j − 0.59 m k
( 0.16 m )2 + ( 0.59 m )2
= FBC
16 j − 59k
3737
16
FBC + 160 N = 0
3737
ΣFy = 0:
FBC = −10 3737 N
ΣFz = 0:
− FAC −
(
FBC = 611 N C !
)
59
39
−10 3737 N −
(1335 N ) = 0
89
3737
FAC = 5.00 N T !
Joint B:
FAB = FAB
ΣFy = 0:
− 0.16 m j + 0.12 m k
2
( 0.16 m ) + ( 0.12 m )
−
4
FAB +
5
2
(
− 4 j + 3k
5
= FAB
)
16
4
10 3737 N −
( 4935 N ) = 0
21
3737
FAB = 975 N C !
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 6, Solution 37.
FBD Truss:
ΣFz = 0:
Bz + 987 N = 0
B z = − 987 N k
ΣM x = 0:
− ( 0.47 m ) C y − ( 0.16 m )( 987 N )
+ ( 0.08 m )( 940 N ) = 0
C y = −176.0 N j
Joint FBDs:
Joint D:
FAD = FAD
FBD = FBD
FCD = FCD
ΣFy = 0:
ΣFx = 0:
ΣFz = 0:
− 0.8 m i − 0.08 m k
( 0.8 m )2 + ( 0.08 m )2
FAD
( −10i − k )
101
=
− 0.8 m i − 0.16 m j − 0.2 m k
2
2
( 0.8 m ) + ( 0.16 m ) + ( 0.2 m )
− 0.8 m i + 0.39 m k
( 0.8 m )2 + ( 0.39 m )2
=
2
=
FCD
( − 80i + 39k )
89
4
FBD − 940 N = 0, FBD = 4935 N,
21
10
80
20
−
FAD −
FCD −
( 4935 N ) = 0
89
21
101
1
39
5
−
FAD +
FCD −
( 4935 N ) + 987 N = 0
89
21
101
Solving:
FBD
( − 20i + 4 j − 5k )
21
FCD = − 534 N,
FAD = − 422 101 N,
FBD = 4.94 kN T !
FCD = 534 N C !
FAD = 4.24 kN C !
continued
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
Joint C:
FBC = FBC
0.16 m j − 0.59 m k
2
( 0.16 m ) + ( 0.59 m )
2
=
FBC
(16 j − 59k )
3737
16
FBC − 176 N = 0, FBC = 11 3737 N
3737
ΣFy = 0:
FBC = 672 N T !
ΣFz = 0:
− FAC −
(
)
59
39
11 3737 N +
( 534 N ) = 0
89
3737
FAC = − 415 N,
FAC = 415 N C !
Joint B:
FAB = FAB
ΣFy = 0:
− 0.16 m j + 0.12 m k
( 0.16 m )2 + ( 0.12 m )2
−
4
FAB −
5
(
=
FAB
( − 4 j + 3k )
5
)
16
4
11 3737 N −
( 4935 N ) = 0
21
3737
FAB = −1395 N, FAB = 1.395 kN C !
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 6, Solution 38.
FBD Truss:
ΣM BD = 0:
ΣM x = 0:
ΣFz = 0:
( 4 ft )( 50 lb ) + ( 4 ft )( Ez ) = 0
E z = − ( 50 lb ) k
( 4 ft )( 300 lb ) + ( 4 ft )( − 50 lb ) + ( 4 ft )( Dz ) = 0
D z = − ( 250 lb ) k
Bz − 50 lb − 250 lb = 0
B z = ( 300 lb ) k
ΣM Bz = 0:
( 2 ft )( 300 lb ) − ( 4 ft ) ( C y ) = 0
C y = (150 lb ) j
AB = AC = 20 ft
AD = AE = 6 ft
CD = 4 2 ft
ΣFx = 0:
ΣFy = 0:
B x = − ( 50 lb ) i
Bx + 50 lb = 0
By + 150 lb − 300 lb = 0
B y = (150 lb ) j
FAE = FAE
=
2 ft i − 4 ft j + 4 ft k
( 2 ft )2 + ( 4 ft )2 + ( 4 ft )2
FAE
( 2i − 4 j + 4k )
3
Joint FBDs:
Joint E:
ΣFz = 0:
ΣFx = 0:
2
FAE − 50 lb = 0,
3
1
FDE + ( 75 lb ) = 0,
3
FAE = 75.0 lb T !
FDE = − 25 lb
FDE = 25.0 lb C !
ΣFy = 0:
− FCE −
2
( 75 lb ) = 0,
3
FCE = − 50 lb
FCE = 50.0 lb C !
continued
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
− 2 ft i − 4 ft j + 4 ft k
FAD = FAD
2
( 2 ft ) + ( 4 ft ) + ( 4 ft )
Joint D:
− 4 ft i − 4 ft j
FCD = FCD
ΣFz = 0:
ΣFx = 0:
2
2
( 4 ft ) + ( 4 ft )
2
=
2 ft i + 4 ft k
FAC = FAC
2
( 2 ft ) + ( 4 ft )
)
Joint B:
FAB = FAB
2
=
FAC = 0 !
FBC − 100 lb = 0,
− 2 ft i + 4 ft k
2
( 2 ft ) + ( 4 ft )
ΣFz = 0:
2
=
FBD = 150.0 lb C !
FAC
( i + 2k )
5
2
FAC = 0,
5
ΣFz = 0:
ΣFx = 0:
FAD
( − i − 2 j + 2k )
3
2
FAD − 250 lb = 0, FAD = 375 lb,
FAD = 375 lb T !
3
1
1
FCD − ( 375 lb ) = 0, FCD = −100 2 lb
25 lb −
3
2
FCD = 141.4 lb C !
1
2
− FBD −
−100 2 lb − ( 375 lb ) = 0
3
2
FBD = − 150 lb
Joint C:
=
FCD
( − i − j)
2
(
ΣFy = 0:
2
FBC = 100.0 lb T !
FAB
( − i + 2k )
5
2
FAB + 300 lb = 0
5
FAB = −150 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.
FAB = 335 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 39.
(a)
FBD Truss:
ΣM y = 0:
Bx = 0
Bx = 0
ΣM x = 0:
ΣM z = 0:
Az = − 600 N
(1.7 m ) Az + ( 0.6 m )(1700 N ) = 0,
− (1.7 m ) Ax − (1.125 m )(1700 N ) = 0 Ax = −1125 N
ΣFx = 0:
Cx − 1125 N = 0
Cx = 1125 N
ΣFy = 0:
C y − 1700 N = 0
C y = 1700 N
ΣFz = 0:
Bz − 600 N = 0
Bz = 600 N
A = − (1125 Ν ) i − ( 600 N ) k
so
B = ( 600 Ν ) k
C = (1125 Ν ) i + (1700 N ) j
(b)
Find zero force members:
by inspection of joint B,
FAB = 0 !
by inspection of joint D,
FAD = 0 !
by inspection of joint C ,
FBC = 0 !
Joint FBDs:
Joint E:
FAE = FAE
ΣFy = 0:
−1.125 m i + 1.7 m j − 0.6 m k
2
2
(1.125 m ) + (1.7 m ) + ( 0.6 m )
1.7
FAE − 1700 N = 0,
2.125
2
=
FAE
( −1.125i + 1.7 j − 0.6k )
2.125
FAE = 2125 N
FAE = 2.13 kN T !
ΣFx = 0:
− FBE −
1.125
( 2125 N ) , FBE = −1125 N
2.125
FBE = 1.125 kN C !
ΣFz = 0:
− FDE −
0.6
( 2125 N ) = 0,
2.125
FDE = − 600 N
FDE = 600 N C !
continued
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
Joint D:
FBD = FBD
ΣFz = 0:
−1.125 m i + 0.6 m k
(1.125 m )2 + ( 0.6 m )2
600 N −
0.6
FBD = 0,
1.275
=
FBD
( −1.125i + 0.6k )
1.275
FBD = 1275 N
FBD = 1.275 kN T !
ΣFx = 0:
− FCD −
1.125
(1275 N ) = 0,
1.275
FCD = −1125 N
FCD = 1.125 kN C !
Joint C:
By inspection:
FAC = −1700 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.
FAC = 1.700 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 40.
(a)
(b)
To check for simple truss, start with ABDE and add three members at a time which meet at a single new
joint, successively adding joints G, F, H and C.
∴ This is a simple truss !
There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all
can be determined with the six equilibrium equations. Motion is prevented by the constraints.
∴ Truss is completely constrained, and statically determinate. !
FBD Truss:
ΣM AB = 0:
( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0
G y = − 252 lb
ΣM z = 0:
G y = − 252 lb j
(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0
B y = 504 lb j
ΣFx = 0:
− Ax + 275 lb = 0,
ΣM y = 0:
− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0
Bz = − 240 lb,
A x = − 275 lb i
B z = − 240 lb k
ΣFy = 0:
− Ay + 504 lb − 252 lb = 0,
A y = − 252 lb j
ΣFz = 0:
Az − 240 lb + 240 lb = 0
Az = 0
Determine zero force members:
By inspection of joint C :
FBC = FCD = FGC = 0
By inspection of joint F :
FBF = FEF = FFG = 0
By inspection of joint A:
with Az = 0,
By inspection of joint H :
FDH = 0
FAD = 0
And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three
remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.
The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH.
continued
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
Joint FBDs
FAB = 275 lb T,
By inspection:
from above:
Joint A:
FAE = 252 lb T !
FDE = 0 !
FEF = 0 !
FEH = 240 lb C !
By inspection of joint H
ΣFy = 0:
− 252 lb −
10.08
FBE = 0
1492
FBE = − 373 lb,
Joint E:
ΣFz = 0:
240 lb −
FBE = 373 lb C !
9.6
FEG = 0
14.6
FEG = 365 lb T !
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 6, Solution 41.
(a)
(b)
To check for simple truss, start with ABDE and add three members at a time which meet at a single new
joint, successively adding joints G, F, H and C.
∴ This is a simple truss !
There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all
can be determined with the six equilibrium equations. Motion is prevented by the constraints.
∴ Truss is completely constrained, and statically determinate. !
FBD Truss:
ΣM AB = 0:
( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0
G y = − 252 lb,
ΣM z = 0:
G y = − 252 lb j
(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0
B y = 504 lb j
ΣFx = 0:
− Ax + 275 lb = 0,
ΣM y = 0:
− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0
Bz = − 240 lb,
Ax = − 275 lb i
B z = − 240 lb k
ΣFy = 0:
− Ay + 504 lb − 252 lb = 0,
A y = − 252 lb j
ΣFz = 0:
Az − 240 lb + 240 lb = 0
Az = 0
Determine zero force members:
By inspection of joint C :
FBC = FCD = FGC = 0
By inspection of joint F :
FBF = FEF = FFG = 0
By inspection of joint A:
with Az = 0,
By inspection of joint H :
FDH = 0
FAD = 0
And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three
remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.
The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH.
continued
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
Now, by inspection of joint H,
FGH = 275 lb C !
FCG = 0 !
from above,
FDG = 0 !
FFG = 0 !
FBD Joint G:
FBG = FBG
−10.08j + 9.6k
13.92
FEG = FEG
−11i + 9.6k
14.6
ΣFx = 0:
ΣFy = 0:
275 lb −
−
11
FEG = 0,
14.6
10.08
FBG − 252 lb = 0,
13.92
FEG = 365 lb T !
FBG = − 348 lb
FBG = 348 lb C !
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 6, Solution 42.
FBD Truss:
ΣFx = 0:
Kx = 0
(
)
6a K y − 125 lb − 5a ( 250 lb )
ΣM A = 0:
− 4a ( 250 lb ) − 3a ( 375 lb )
− 2a ( 500 lb ) − a ( 500 lb ) = 0
K y = 937.5 lb
Ay − 3 ( 250 lb ) − 2 ( 500 lb )
ΣFy = 0:
− 375 lb − 125 lb + 937.5 lb = 0
A y = 1312.5 lb
FBD Section ABEC:
( 2 ft ) FCF + ( 4 ft )( 500 lb )
ΣM E = 0:
+ ( 8 ft )( 250 lb − 1312.5 lb ) = 0
FCF = 3250 lb,
ΣFy = 0:
FCF = 3.25 kips T
1312.5 lb − 250 lb − 2 ( 500 lb ) −
FEF = 62.5 5 lb,
ΣFx = 0:
3250 lb +
(
1
FEF = 0
5
FEF = 139.8 lb T
)
2
62.5 5 lb − FEG = 0
5
FEG = 3375 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.
FEG = 3.38 kips C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 43.
FBD Truss:
ΣFx = 0:
Kx = 0
ΣM A = 0:
6a K y − 125 lb − 5a ( 250 lb )
(
)
− 4a ( 250 lb ) − 3a ( 375 lb )
− 2a ( 500 lb ) − a ( 500 lb ) = 0
K y = 937.5 lb
ΣFy = 0:
Ay − 3 ( 250 lb ) − 2 ( 500 lb )
− 375 lb − 125 lb + 937.5 lb = 0
A y = 1312.5 lb
FBD Section IJLK:
ΣM z = 0:
( 4 ft )( 937.5 lb −125 lb ) − ( 2 ft ) FHJ = 0
FHJ = 1625 lb,
ΣFy = 0:
937.5 lb − 125 lb − 250 lb −
FHI = 562.5 5 lbs,
ΣFx = 0:
1.625 kips +
(
FHJ = 1.625 kips C
1
FHI = 0
5
FHI = 1.258 kips C
)
2
0.5625 5 kips − FFI = 0
5
FFI = 2.75 kips T
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 6, Solution 44.
FBD Truss:
ΣFx = 0:
Ax = 0
By load symmetry:
A y = L y = 840 N
ΣM E = 0:
2
24
(1.75 m ) FBD + ( 2 m )( 240 N )
3
25
− ( 4 m )( 840 N − 120 N ) = 0
FBD =
FBD Section ABC:
ΣM B = 0:
FBD = 2.14 kN C
 1.75 
m  FCE − ( 2 m )( 840 N − 120 N ) = 0

 3

FCE =
ΣM A = 0:
15000
N,
7
17280
N,
7
FCE = 2.47 kN T
( 4 m ) 
7

FBE  − ( 2 m )( 240 N ) = 0
 25

FBE =
3000
N,
7
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.
FBE = 429 N C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 45.
FBD Truss:
ΣFx = 0:
Ax = 0
By load symmetry
A y = L y = 840 N
0.4
2
=
3
15
3.5
7
IK = 3 =
4
24
2.3 23
IJ =
=
3
30
slope: GI =
FBD Section JKL:
Resolve FIK and FGJ at L
ΣM J = 0:
( 3 m ) 
7

 24

FIK  − ( 0.4 m ) 
FIK 
25
25




+ ( 3 m )( 840 N − 120 N ) − (1 m )( 240 N ) = 0
FIK = 4210.53 N,
ΣM I = 0:
FIK = 4.21 kN C

  3.5   15

2
FGJ  − 
FGJ 
m
  229
 229
  3

(4 m) 
+ ( 4 m )( 840 N − 120 N ) − ( 2 m )( 240 N ) = 0
FGJ = 3823.0 N,
ΣFx = 0:
FGJ = 3.82 kN T
7
15
30
FIJ = 0
( 4210.5 N ) −
( 3823.0 N ) −
25
229
1429
FIJ = 318 N T
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 6, Solution 46.
\
FBD Truss:
ΣFx = 0:
By symmetry,
Ax = 0
A y = M y = 8.5 kN
FBD Section ABDC:
ΣM D = 0:
(1 m )(1.5 kN − 8.5 kN ) + ( 0.5 m )
40
FCE = 0
41
FCE = 14.35 kN T
ΣM E = 0:
(1 m )(1.5 kN ) + ( 2 m )(1.5 kN − 8.5 kN )
+ ( 0.5 m )
40
FDF = 0,
41
FDF = 25.625 kN
FDF = 25.6 kN C
ΣFx = 0:
40
40
FDE = 0
(14.35 kN − 25.625 kN ) +
41
1721
FDE = 11.4084 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.
FDE = 11.41 kN T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 47.
FBD Truss:
ΣFx = 0:
By symmetry,
Ax = 0
A y = M y = 8.5 kN
FBD Section IMNJ:
ΣM J = 0:
( 2 m )(8.5 kN − 1.5 kN ) − (1 m )(1.5 kN )
− ( 0.5 m )
40
FGI = 0
41
FGI = 25.625 kN,
ΣM G = 0:
FGI = 25.6 kN T
( 3 m )(8.5 kN − 1.5 kN ) − ( 2 m )(1.5 kN )
40
FHJ = 0
41
FHJ = 30.8 kN C
− (1 m )( 3 kN ) − ( 0.5 m )
FHJ = 30.75 kN,
ΣFx = 0:
40
( 30.75 kN − 25.625 kN ) −
41
40
FGJ = 0
2441
FGJ = 6.18 kN T
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 6, Solution 48.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
15
m
6
2 5 5
yD = ⋅ = m
3 2 3
2
2
yE = ⋅ 1 = m
3
3
5
yF − yD = m
6
yG = 1 m
yF =
yD − yG =
2
m
3
continued
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
ΣM D = 0: (1 m )
6
FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0
37
FEG =
8
37 kN
3
FEG = 16.22 kN T !
 1

 3

ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m ) 
FDG  − (1 m ) 
FDG  = 0
 10

 10

FDG =
ΣFx = 0:
4
10 kN
3
6
3
12
FEG −
FDG −
FDF = 0
13
37
10
FDG = 4.22 kN C !
16 − 4 −
FDF = 13 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.
12
FDF = 0
13
FDF = 13.00 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 49.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
so
2
m
3
2 5 5
yH = ⋅ = m
3 2 3
yI =
yH − yI = 1 m
continued
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
ΣM I = 0:
( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
12

FHJ  = 0
 13

FHJ =
ΣM H = 0:
FHJ = 17.33 kN C !
 6

FGI  = 0
 37

( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
FGI =
ΣM L = 0:
52
kN
3
8
37 kN
3
( 2 m )( 2 kN ) − ( 4 m ) FHI = 0
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.
FGI = 16.22 kN T !
FHI = 1.000 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 50.
FBD Truss:
Distance between loads = 1.5 m
ΣFx = 0:
By symmetry,
Ax = 0
A y = K y = 18 kN
FBD Section ABC:
FBD Section ABC:
ΣM D = 0:
(1.5 m ) FCE + (1.5 m )( 6 kN ) − ( 3 m )(18 kN − 3 kN ) = 0
FCE = 22.5 kN T !
ΣM A = 0:
(1.8 m )
4
FCD − (1.5 m )( 6 kN ) = 0
5
FCD = 6.25 kN T !
ΣFy = 0:
18 kN − 3 kN − 6 kN −
8
4
FBD + ( 6.25 kN ) = 0
17
5
FBD = 29.8 kN C !
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 6, Solution 51.
FBD Truss:
Distance between loads = 1.5 m
ΣFx = 0:
By symmetry,
Ax = 0
A y = K y = 18 kN
FBD Section ABC:
FBD Section GHK:
FBD section GHK:
ΣM F = 0:
( 4.5 m )(18 kN − 3 kN ) − ( 3 m )( 6 kN )
− (1.5 m )( 6 kN ) − ( 2.4 m ) FEG = 0
FEG = 16.875 kN,
ΣM K = 0:
 8

FFG  = 0
 73

(1.5 m )( 6 kN ) + ( 3 m )( 6 kN ) − ( 3.6 m ) 
FFG = 8.0100 kN,
ΣFx = 0:
FEG = 16.88 kN T !
15
FFH −
17
FFG = 8.01 kN T !
3
(8.0100 kN ) − 16.875 kN = 0
73
FFH = 22.3 kN C !
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 6, Solution 52.
FBD Truss:
ΣFX = 0:
A x + 4 kips − 4 kips = 0
Ax = 0
By symmetry,
Ay = Ny = 0
FBD Section ABDC:
ΣM D = 0:
(18 ft )( 4 kips ) − ( 9 ft )
4
FCE = 0
5
FCE = 10.00 kips C !
ΣM E = 0:
(18 ft )( 4 kips ) − (12 ft )
3
FDF = 0
5
FDF = 10.00 kips C
ΣFx = 0:
4 kips +
4
(10 kips − 10 kips ) − FDE = 0
5
FDE = 4.00 kips C !
FBD Joint E:
ΣFy = 0:
4
3
FEF − ( 4 kips ) = 0
5
5
FEF = 3.00 kips T !
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 6, Solution 53.
FBD Truss:
ΣFx = 0:
A x + 4 kips − 4 kips = 0
Ax = 0
Ay = Ny = 0
By symmetry,
FBD Section IJNM:
ΣM G = 0:
4
− ( 27 ft )( 4 kips ) + ( 9 ft ) FHI = 0
5
FHI = 15.00 kips T !
ΣM N = 0:
4
5
( 9 ft ) (15 kips ) − ( 27 ft ) FGI = 0
FGI = 4.00 kips C !
FBD Joint I:
ΣFx = 0:
3
4
( 4 kips ) − FIJ = 0
5
5
FIJ = 3.00 kips T !
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 6, Solution 54.
FBD Truss:
ΣFx = 0:
Ax = 0
A y = M y = 7.5 kips
By symmetry,
FBD Section IJNM:
( 9 ft )( 7.5 kips − 0.5 kips ) − ( 4.5 ft )(1 kip )
ΣM J = 0:
 4

− ( 5.25 ft ) 
FGI  = 0,
 17

FGI = 11.4858 kips.
FGI = 11.49 kips T !
(12 ft )( 7.5 kip − 0.5 kips ) − ( 7.5 ft )(1 kip )
ΣM G = 0:
− ( 3 ft )( 3 kips ) + ( 4.5 ft ) FHJ = 0
FHJ = 15 kips C
By inspection of joint H,
ΣFy = 0:
FFH = 15.00 kips C !
1
(11.4858 kips ) + 7.5 kips − 4.5 kips
17
−
3
FGJ = 0
13
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.
FGJ = 6.95 kips T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 55.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry:
A y = v y = 4.5 kips
FBD Joint K:
ΣFx = 0:
ΣFy = 0:
1
( FIK − FKN ) = 0
2
 1

FIK  − 1 kip = 0
2
 2

so
FIK = FKN
FIK =
2
kip
2
FIK = 0.707 kip C !
FBD Section ABEIJ:
ΣFy = 0:
ΣM I = 0:
4.5 kips − 1 kip − 2 (1.5 kips ) −
1 2
3
FJL = 0,
kips +
2
2 2
( 3 ft )(1.5 kips ) + ( 6 ft )(1 kip − 4.5 kips )
+
3
( 3 ft ) ( FJM ) = 0
2
FJM =
FJL = 0 !
11
kips
3
FJM = 6.35 kips T !
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 6, Solution 56.
FBD Truss:
ΣFx = 0:
ΣM N = 0:
Ax = 0
( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb )
+ (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb )
+ (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb )
(
)
+ ( 33.6 ft ) 150 lb − Ay = 0
Ay = 2925 lb
ΣFy = 0:
2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb
− 900 lb − 1950 lb − 300 lb + N y = 0
FBD Section ABC:
N y = 3075 lb
ΣM B = 0:
( 3.6 ft ) FCE − ( 7.2 ft )( 2925 lb − 150 lb ) = 0
FCE = 5550 lb,
ΣM A = 0:
( 7.2 ft )
FCE = 5.55 kips T !
1
2
FBE + ( 3.6 ft )
FBE − ( 7.2 ft )(1050 lb ) = 0
5
5
FBE = 525 5 lb,
ΣFx = 0:
(
FBE = 1.174 kips C !
)
2
FBD − 525 5 lb + 5550 lb = 0
5
FBD = −5031.2 lb
By inspection of joint D, FDF = FBD , so
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.
FDF = 5.03 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 57.
FBD Truss:
ΣFx = 0:
Ax = 0
( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb )
+ (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb )
+ (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb )
+ ( 33.6 ft ) (150 lb − Ay ) = 0
ΣM N = 0:
Ay = 2925 lb
FBD Section KJMN:
2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb
ΣFy = 0:
− 900 lb − 1950 lb − 300 lb + N y = 0
N y = 3075 lb
( 7.2 ft )( 3075 lb − 150 lb ) − ( 2.4 ft )( 300 lb )
ΣM J = 0:
− (1.8 ft ) FIK = 0
FIK = 11300 lb,
ΣM I = 0:
FIK = 11.30 kips T
(11.52 ft )( 3075 lb − 150 lb ) − ( 6.72 ft )( 300 lb )
 2

− ( 4.32 ft )( 750 lb ) + ( 3.96 ft ) 
FHJ  = 0
 5

ΣFx = 0:
FBD Joint I:
FHJ = −3590.5 5 lb,
FHJ = 8.03 kips C !
2
12
−
−3590.9 5 lb −
FIJ − 11300 lb = 0
13
5
(
)
FIJ = − 4461.4 lb,
ΣFx = 0:
11300 lb −
FIJ = 4.46 kip C !
12
( 4461.4 lb ) − FGI = 0
13
FGI = 7181.8 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.
FGI = 7.18 kips T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 58.
FBD Truss:
Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60°
outer members AC, CE, etc. are each 1.0 m
radial members AB, CD, etc. are each 0.4 m
By symmetry , Ay = By = 2.6 kN
ΣFx = 0: A x = 0
FBD Section ABDC:
ΣM F = 0:
( resolve FCE at E )
 1

 1

 2 ( 0.4 m )  FCE sin 40° +  2 ( 0.4 m )  ( FCE cos 40° )




1


+ (1 m ) sin 40° +
( 0.4 m ) ( 0.4 kN )
2


1


− (1 m ) sin 20° + (1 m ) sin 40° +
( 0.4 m ) ( 2.6 kN ) = 0,
2


FCE = 7.3420 kN,
FCE = 7.34 kN C !
(1 m ) cos 40° − ( 0.4 m ) sin 45° + ( 0.4 m ) sin 22.5° = 48.848°
(1 m ) sin 40° + ( 0.4 m ) cos 45° − ( 0.4 m ) cos 22.5°
(1 m ) cos 40° − ( 0.4 m ) sin 45° = 27.566°
and ε = tan −1
(1 m ) sin 40° + ( 0.4 m ) cos 45°
note σ = tan −1
Then,
ΣM C = 0:
( 0.4 m )  FDF cos ( 90° − 22.5° − 48.848°)
− (1 m ) sin 20° ( 2.6 kN ) = 0,
FDF = 2.3464 kN,
and,
ΣFx = 0:
FDF = 2.35 kN T !
FCF ( cos 27.566° ) + ( 2.3464 kN )( cos 48.848° )
− ( 7.3420 kN ) sin 40° = 0
FCF = 3.5819 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.
FCF = 3.58 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 59.
FBD Truss:
Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60°
outer members AC, CE, etc. are each 1.0 m
radial members AB, CD, etc. are each 0.4 m
By symmetry, Ay = By = 2.6 kN
ΣFx = 0: A x = 0
FBD Section ABDFHGEC:
( resolve FGI at I )
( 0.4 m ) FGI sin 80° + (1 m )( sin 80°)(1 kN )
ΣM J = 0:
+ (1 m ) sin 80° + (1 m ) sin 60°  ( 0.6 kN )
+ (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° ( 0.4 kN )
− (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° + (1 m ) sin 20° ( 2.6 kN ) = 0
FGI = 10.8648 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.
FGI = 10.86 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Note:
so,
σ = tan −1
ΣM H = 0:
0.4 m − (1 m ) cos80°
= 12.9443°
(1 m ) sin 80°
( 0.4 m ) cos 60° ( FGJ sin12.9443° )
− ( 0.4 m ) sin 60° ( FGJ cos12.9443° )
− ( 0.4 m ) sin 60° (10.8648 kN ) sin 80°
+ ( 0.4 m ) cos 60° (10.8648 kN ) cos80°
+ ( 0.4 m ) cos 60° (1 kN ) + (1 m ) sin 60° + ( 0.4 m ) cos 60° ( 0.6 kN )
+ (1 m ) sin 60° + (1 m ) sin 40° + ( 0.4 m ) cos 60° ( 0.4 kN )
− (1 m )( sin 60° + sin 40° + sin 20° ) + ( 0.4 m ) cos 60° ( 2.6 kN ) = 0
FGJ = 0.93851 kN,
FGJ = 939 N T !
FBD Joint I:
By symmetry:
ΣFy = 0:
FIK = 10.8648 kN
− FIJ − 1.2 kN + 2 (10.8648 kN ) cos80° = 0
FIJ = 2.5733 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.
FIJ = 2.57 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 60.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry, Ay = Ny = 1.6 kips
FBD Section using cut a–a:
ΣM D = 0:
( 9 ft ) FFH + ( 6 ft )( 0.5 kip ) − (12 ft )(1.6 kips ) = 0
FFH = 1.800 kips T !
ΣFx = 0:
1.800 kips − FDG = 0
FDG = 1.800 kips C !
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 6, Solution 61.
\
FBD Truss:
ΣFx = 0: A x = 0
By symmetry Ay = Ny = 1.6 kips
FBD Joint J:
ΣFx = 0:
− FGJ − FHJ = 0, FHJ = − FGJ
By inspection of joint I, FIJ = 0.6 kip C
FBD Section using cut b-b:
3
 FGJ − ( − FGJ )  − 0.6 kip − 0.5 kip + 1.6 kips = 0
5
ΣFy = 0:
FGJ = −
2.5
kip,
6
ΣM J = 0:
FGJ = 0.417 kip C !
(12 ft )(1.6 kips ) − ( 6 ft )( 0.5 kip ) − ( 4.5 ft )( FHK )
− ( 4.5 ft )( FIL ) = 0
ΣFx = 0:
FIL − FHK = 0 so FIL = FHK
and
FIL = 1.800 kips C !
FHK = 1.800 kips T !
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 6, Solution 62.
FBD Section above a-a:
ΣM G = 0:
 27

FIK  − ( 2.7 m )( 40 kN ) − ( 5.4 m )( 40 kN ) = 0
 793

( 5.9 m ) 
FIK = 57.275 kN,
ΣFy = 0:
FIK = 57.3 kN C
27
( 57.275 kN − FGJ ) = 0
793
FGJ = 57.275 kN T
FBD Section ACIG:
27
( 57.275 kN − 57.275 kN )
793
ΣFy = 0:
+
ΣFx = 0:
18
( FHK − FHJ ) = 0,
949
3 ( 40 kN ) − 2
FHJ = FHK
25
FHK = 0
949
FHK = 53.884 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.
FHK = 53.9 kN C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 63.
FBD Section ACFBD:
ΣM D = 0:
( 4.3 m )
27
FFI − ( 2.7 m )( 40 kN ) = 0
793
FFI = 26.196 kN,
ΣFy = 0:
FFI = 26.2 kN C !
27
( 26.196 kN − FDG ) = 0
793
FDG = 26.196 kN T
FBD Section ACFD:
ΣFy = 0:
27
( 26.196 kN − 26.196 kN )
793
+
ΣFx = 0:
54
( FEI − FEG ) = 0,
6397
FEI = FEG
−2
59
( FEG ) + 2 ( 40 kN )
6397
−
8
( 26.196 kN + 26.196 kN ) = 0
793
FEG = 44.136 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.
FEG = 44.1 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 64.
FBD Truss:
ΣFx = 0:
ΣM F = 0:
Ax = 0
( 2.4 m )(8 kN ) + ( 5.1 m )(12 kN )
− ( 7.5 m ) Ay = 0,
A y = 10.72 kN
Since only CD can provide an upward force necessary for
equilibrium, it must be in tension, and FBE = 0
FBD Section ABC:
ΣFy = 0:
10.72 kN − 12 kN +
FCD = 2.4229 kN,
ΣM C = 0:
1.68
FCD = 0
3.18
FCD = 2.42 kN T (1.68 m ) FBD − ( 2.4 m )(10.72 kN ) = 0
FBD = 15.3143 kN,
ΣFx = 0: FCE +
FBD = 15.31 kN C 2.7
( 2.4229 kN ) − 15.3143 kN = 0
3.18
FCE = 13.26 kN T 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 6, Solution 65.
FBD Truss:
ΣFx = 0:
ΣM F = 0:
Ax = 0
( 2.4 m )(8 kN ) + ( 5.1 m )( 6 kN ) − ( 7.5 m ) Ay = 0
A y = 6.64 kN
FBD Section ABC:
Since only BE can provide the downward force necessary for
equilibrium,
it must be in tension, so FCD = 0
ΣFy = 0:
6.64 kN − 6 kN −
FBE = 1.21143 kN,
ΣM = 0:
FBE = 1.211 kN T !
(1.68 m ) FCE − ( 2.4 m )( 6.64 kN ) = 0
FCE = 9.4857 kN,
ΣFx = 0:
1.68
FBE = 0
3.18
9.4857 kN +
FCE = 9.49 kN T !
2.7
(1.21143 kN ) − FBD = 0
3.18
FBD = 10.51 kN C !
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 6, Solution 66.
FBD Truss:
ΣFx = 0:
ΣM C = 0:
Cx = 0
(8 ft )( 6 kips ) − (8 ft )( 9 kips ) + (16 ft ) G y
− ( 24 ft )(12 kips ) = 0
G y = 19.5 kips
FBD Section ABC:
ΣFy = 0: 19.5 kips + C y − 6 kips − 9 kips − 12 kips = 0
C y = 7.5 kips
Since only BE can provide the downward force necessary for
equilibrium, it must be in tension, so CD is slack, FCD = 0
ΣFy = 0: − 6 kips + 7.5 kips −
3
FBE = 0
5
FBE = 2.50 kips T !
FBD Section FGH:
Since only EF can provide the downward force necessary for equilibrium,
it must be in tension, so DG is slack, FDG = 0
ΣFy = 0: 19.5 kips − 12 kips −
3
FEF = 0
5
FEF = 12.00 kips T !
Knowing that FCD = FDG = 0, inspection of joint D gives
FDE = 0 !
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 6, Solution 67
FBD Truss:
( 32 ft ) H y − ( 24 ft )(12 kips ) − (16 ft )( 9 kips )
ΣM A = 0:
− ( 8 ft )( 6 kips ) = 0,
ΣFy = 0:
H y = 15 kips
Ay − 6 kips − 9 kips − 12 kips + 15 kips = 0
A y = 12 kips
ΣFx = 0:
FBD Section ABC:
Ax = 0
Since only BE can provide the downward force necessary for
equilibrium, it must be in tension, so CD is slack, FCD = 0
ΣFy = 0: 12 kips − 6 kips −
3
FBE = 0
5
FBE = 10.00 kips T !
Since only EF can provide the downward force necessary for equilibrium,
it must be in tension, so DG is slack, FDG = 0
FBD Section FGH:
ΣFy = 0: 15 kips − 12 kips −
3
FEF = 0
5
FEF = 5.00 kips T !
Knowing that FCD = FDF = 0, inspection of joint D gives
FDE = 0 !
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 6, Solution 68
FBD Section ABDC:
Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be
slack, FCF = 0
4
FDE = 0,
5
ΣFx = 0:
20 kN −
ΣM D = 0:
( 3.2 m ) FCE − ( 2.4 m )( 20 kN ) = 0
FDE = 25.0 kN C !
FCE = 15.00 kN T !
ΣFy = 0: FDF − 15.00 kN −
3
( 25 kN ) = 0
5
FDE = 30.0 kN C !
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 6, Solution 69.
FBD Section ABFE:
It is not apparent which counter is active, so we guess that FEH = 0;
ΣM F = 0:
 24

FEG + 40 kN  − ( 4.8 m )( 20 kN ) = 0
 745

( 3.2 m ) 
FEG = −11.3728 kN
FEG = 11.37 kN C !
ΣM G = 0:
 24

FFH  − ( 7.2 m )( 20 kN ) − (1.3 m )( 40 kN ) = 0
 745

( 5.8 m ) 
FFH = 38.432 kN,
ΣFx = 0:
20 kN −
FFH = 38.4 kN C !
13
15
( 38.432 kN − 11.3728 kN ) − FFG = 0
17
745
FFG = 8.0604 kN,
FFG = 8.06 kN T !
Since FEG is in tension, our guess was correct. A negative answer would be impossible, indicating
an incorrect guess.
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 6, Solution 70.
Structure (a):
Non-simple truss with r = 3, m = 16, N = 10,
so 2N > m + r
∴ partially constrained !
Structure (b):
Non-simple truss with r = 3, m = 15, N = 10,
so 2N > m + r
∴ partially constrained !
Structure (c):
Non-simple truss with r = 4, m = 16, N = 10,
so N = 2m + r
Examine Forces:
FHJ = 0, FIJ = P C
By inspection of joint J,
Then, by inspection of joint H,
By inspection of joint I,
FFH = 0,
FGI = 0,
FGH = P C
Iy = 0
Although several members carry no load with the given truss loading, they do
constrain the motion of GH and IJ. Truss ABFG is a simple truss with r = 3, m = 11,
N = 7 (2N = r + m)
so the structure is completely constrained, and determinate. !
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 6, Solution 71.
Structure (a):
Structure (b):
Structure (c):
Simple truss with r = 4, m = 16, n = 10
So m + r = 20 = 2n so
completely constrained and determinate !
Compound truss with r = 3, m = 16, n = 10
so
So m + r = 19 < 2n = 20
partially constrained !
Non-simple truss with r = 4, m = 12, n = 8
So m + r = 16 = 2n but must examine further, note that reaction forces
A x and H x are aligned, so no equilibrium equation will resolve them.
Also
consider
∴ Statically indeterminate !
For ΣFy = 0: H y = 0, but then
ΣM A ≠ 0 in FBD Truss,
∴ Improperly constrained !
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 6, Solution 72.
Structure (a):
Simple truss (start with ABC and add joints alphabetical to complete
truss), with
r = 4, m = 13, n = 8
r + m = 17 > 2n = 16
so
Structure is completely constrained but indeterminate. !
Structure (b):
From FBD II:
FBD I:
FBD II:
ΣM G = 0 ⇒
Jy
ΣFx = 0 ⇒
Fx
ΣM A = 0 ⇒
Fy
ΣFy = 0 ⇒
Ay
ΣFx = 0 ⇒
Ax
ΣFy = 0 ⇒
Gy
Thus have two simple trusses with all reactions known,
so structure is completely constrained and determinate. !
Structure (c):
Structure has r = 4, m = 13, n = 9
so
r + m = 17 < 2n = 18,
structure is partially constrained !
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 6, Solution 73.
Non-Simple truss with r = 4, m = 12, N = 8, so 2N = r + m,
but we must examine further:
Structure (a):
ΣFx = 0: E x = 0
By symmetry, FBD = FDF and FCD = FDG
FBD joint D:
Vertical equilibrium cannot be satisfied with FCD = FDG,
Improperly constrained !
Structure (b):
Non-simple truss with r = 3, m = 10, N = 7,
so 2N = m + r
∴ Partially constrained !
Structure (c):
Non-simple truss with r = 3, m = 13, N = 8,
but we must examine further
so 2N = r + m,
ΣM A = 0: 3a G y − ( a + 3a + 4a ) P = 0
Gy =
Joint H:
ΣFy = 0:
Joint F:
1
FFH − P = 0
2
FFH = P 2 T
ΣFy = 0: FFG − P −
8
P
3
Joint G:
(
)
1
P 2 =0
2
FFG = 2P C
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.
ΣFy ≠ 0
Improperly constrained !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 74.
Structure (a):
No. of members
m = 12
No. of joints
n=8
m + r = 16 = 2n
No. of react. comps.
r =4
unks = eqns
FBD of EH:
ΣM H = 0 → FDE ; ΣFx = 0 → FGH ; ΣFy = 0 → H y
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. !
Structure (b):
No. of members
m = 12
No. of joints
n=8
m + r = 15 < 2n = 16
No. of react. comps.
r =3
unks < eqns
partially constrained !
Note: Quadrilateral DEHG can collapse with joint D moving downward:
in (a) the roller at F prevents this action.
Structure (c):
No. of members
m = 13
No. of joints
n=8
m + r = 17 > 2n = 16
No. of react. comps.
r =4
unks > eqns
completely constrained but indeterminate !
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 6, Solution 75.
Structure (a):
Rigid truss with r = 3, m = 14, n = 8
so
r + m = 17 > 2n = 16
so completely constrained but indeterminate !
Structure (b):
Simple truss (start with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8
so
r + m = 16 = 2n
so completely constrained and determinate !
Structure (c):
Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but
horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by
any equilibrium equation.
∴ structure is improperly constrained !
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 6, Solution 76.
\
FBD member AE:
Ax = 120 lb
ΣFx = 0: − Ax + 120 lb = 0,
on ABCD
ΣM A = 0:
A x = 120.0 lb
!
(8 in.)( C ) − ( 2 in.)(120 lb ) = 0
C = 30 lb
ΣFy = 0: 30 lb − Ay = 0,
Ay = 30 lb
A y = 30.0 lb !
on ABCD
FBD member ABCD:
ΣFx = 0: 120 lb − Bx = 0,
ΣM B = 0:
B x = 120.0 lb
!
( 6 in.) D − ( 4 in.)( 30 lb ) − ( 4 in.)( 30 lb )
− ( 2 in.)(120 lb ) = 0
D = 80.0 lb !
ΣFy = 0: 30 lb − By − 30 lb + 80 lb = 0
B y = 80.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 77.
FBD member ABC:
Note that BD is a two-force member
(a)
ΣΜ C = 0:
(16 in.) 
4

FBD  − ( 24 in.)( 80 lb ) = 0
5

FBD = 150 lb,
(b)
FBD = 150.0 lb
36.9°
ΣFy = 0:
Cy −
3
(150 lb ) = 0,
5
C y = 90 lb
ΣFx = 0:
Cx −
4
(150 lb ) + 80 lb = 0,
5
Cx = 40 lb
C = 98.5 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.
66.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 78.
FBD Frame:
ΣM A = 0:
( 0.25 m ) Dx − ( 0.95 m )( 480 N ) = 0
D x = 1824 N
FBD member DF:
Note that BE is a two-force member, Ex = Ey
ΣFx = 0:
−1824 N + Ex = 0,
E x = 1824 N
so
ΣM D = 0:
E y = 1824 N
( 0.50 m )(1824 N ) − ( 0.75 m ) C + ( 0.95 m )( 480 N ) = 0
C = 1824 N
ΣFy = 0:
− D y + 1824 N − 1824 N + 480 N = 0
D y = 480 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 6, Solution 79.
FBD Frame:
ΣM A = 0:
( 0.25 m ) Dx − 400 N ⋅ m = 0
D x = 1600 N
E x = 1600 N
∴ E y = 1600 N
FBD member DF:
Note BE is a two-force member, so Ex = Ey
ΣFx = 0: 1600 N − E x = 0,
ΣM D = 0: − ( 0.50 m )(1600 N ) + ( 0.75 m ) C − 400 N ⋅ m = 0
C = 1600 N
ΣFy = 0: Dy − 1600 N + 1600 N = 0
Dy = 0 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 80.
FBD Ring:
(a)
ΣM A = 0:
(8 in.) ( FBC cos 35° ) − (8 in.)( 6 lb ) = 0
FBC = 7.3246 lb,
(b)
ΣFx = 0:
FBC = 7.32 lb C
Ax − ( 7.3246 lb ) cos 35° = 0
Ax = 6 lb
ΣFy = 0:
Ay + ( 7.3246 lb ) sin 35° − 6 lb = 0
Ay = 1.79876 lb
A = 6.26 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.
16.69°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 81.
FBD Ring:
(a)
FBC =
(b)
(8 in.) ( FBC cos 20° − 6 lb ) = 0
ΣM A = 0:
6 lb
,
cos 20°
ΣFx = 0:
ΣFy = 0:
FBC = 6.39 lb C
Ax −
Ay −
6 lb
cos 20° = 0,
cos 20°
Ax = 6 lb
6
sin 20° − 6 lb = 0,
cos 20°
Ay = 8.1838 lb
A = 10.15 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.
53.8°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 82.
FBD Frame:
ΣM A = 0:
 2

180 N  = 0
 2

( 0.2 m ) Bx − ( 0.36 m ) 
Bx = 162 2 N,
ΣFx = 0: −162 2 N + Ax −
B x = 229 N
2
(180 N ) = 0
2
Ax = 252 2 N,
ΣFy = 0: By −
2
(180 N ) = 0
2
B y = 90 2 N,
FBD member ABD:
ΣM C = 0:
A x = 356 N
B y = 127.3 N
( 0.20 m ) ( 252 2 N ) − ( 0.08 m ) ( 90 2 N )
− ( 0.20 m ) D = 0,
D = 216 2 N
D = 305 N
ΣFx = 0: 252 2 N − 162 2 N − C x = 0
Cx = 90 2 N,
C x = 127.3 N
ΣFy = 0: 90 2 N + C y − 216 2 N = 0
C y = 126 2 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.
C y = 178.2 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 83.
FBD Frame:
ΣM B = 0:
( 0.2 m ) Ax − 60 N ⋅ m = 0
A x = 300 N
ΣFx = 0: 300 N − Bx = 0,
ΣFy = 0: By = 0,
B x = 300 N
By = 0
FBD member ABD:
ΣM C = 0:
( 0.20 m )( 300 N ) − ( 0.20 m ) Dy = 0
D y = 300 N
ΣFy = 0:
C y − 300 N = 0
ΣFx = 0:
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.
C y = 300 N
Cx = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 84.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
F

F

2 


2 
(8 in.)  CE  + ( 2 in.)  CE  − ( 6 in.)( 24 lb ) = 0
FCE = 14.4 2 lb,
so
E x = 14.40 lb
E y = 14.40 lb
ΣFx = 0:
− Ax + 14.4 lb = 0
A x = 14.40 lb
ΣFy = 0:
Ay − 24 lb + 14.40 lb = 0
A y = 9.60 lb
(b) FBD CE:
!
!
!
!
Note AC is a two-force member
ΣM E = 0:
1
 4

FAC +
FAC  − (1 in.)( 24 lb ) = 0
17
 17

( 3 in.) 
FAC = 1.6 17 lb,
(
)
(
)
ΣFx = 0: E x −
4
1.6 17 lb = 0,
17
ΣFy = 0:
1
1.6 17 lb − 24 lb = 0
17
Ey +
A x = 6.40 lb
!
A y = 1.600 lb
!
E x = 6.40 lb
!
E y = 22.4 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.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 85.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
1
 1

FCE +
FCE  − ( 0.3 m )( 320 lb )
2
 2

( 0.4 m ) 
E x = 120.0 N
FCE = 120 2 N,
E y = 120.0 N
(b) FBD CE:
(
)
1
120 2 N = 0,
2
ΣFx = 0:
− Ax +
ΣFy = 0:
Ay − 320 N +
(
A x = 120.0 N
)
1
120 2 N = 0
2
A y = 200 N
!
!
!
!
Note: AC is a two-force member
ΣM E = 0:
1
 1

FAC +
FAC  − ( 0.15 m )( 320 N ) = 0
2
 2

( 0.25 m ) 
FAC = 96 2 N,
A x = 96.0 N
A y = 96.0 N
ΣFx = 0: Ex − 96.0 N = 0
ΣFy = 0:
E y − 320 N + 96.0 N = 0
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.
E x = 96.0 N
E y = 224 N
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 86.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
 1

 1

− ( 8 in.) 
FCE  − ( 2 in.) 
FCE  + 192 lb ⋅ in. = 0
 2

 2

E x = 19.20 lb
FCE = 19.2 2 lb,
E y = 19.20 lb
(b) FBD CE:
ΣFx = 0:
Ax − 19.2 lb = 0,
ΣFy = 0:
Ay − 19.2 lb = 0,
A x = 19.20 lb
A y = 19.20 lb
!
!
!
!
Note: AC is a two-force member
ΣM E = 0:
4
 1

FAE +
FAE  + 192 lb ⋅ in. = 0
17
 17

( 3 in.) 
Ax =
4
FAE ,
17
A x = 51.2 lb
!
Ay =
1
FAE ,
17
A y = 12.80 lb
!
ΣFx = 0: Ex − 51.2 lb = 0,
E x = 51.2 lb
!
ΣFy = 0:
E y = 12.80 lb
!
FAE = −12.8 17 lb,
E y − 12.80 lb = 0,
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 6, Solution 87.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
 2

FCE  = 0,
120 N ⋅ m − ( 0.4 m ) 
2


FCE = 150 2 N
E x = 150.0 N
E y = 150.0 N
(b) FBD CE:
ΣFx = 0:
Ax − 150.0 N = 0,
ΣFy = 0:
Ay − 150.0 N = 0,
A x = 150.0 N
!
!
!
A y = 150.0 N
!
A x = 240 N
!
Note: AC is a two-force member
 2

ΣM E = 0: − ( 0.25 m ) 
FAC  + 120 N ⋅ m = 0
 2

FAC = 240 2 N,
A y = 240 N
ΣFx = 0: 240 N − Ex = 0,
ΣFy = 0:
E y − 240 N = 0,
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.
E x = 240 N
E y = 240 N
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 88.
FBD Frame:
Note: In analysis of entire frame, location of M is immaterial. Note also
that AB, BC, and FG are two force members.
ΣM H = 0:
( 6 ft ) I y − 180 lb ⋅ ft = 0,
ΣFy = 0:
30 lb −
ΣM C = 0:
( 2.5 ft ) 
I y = 30.0 lb
!
(a) FBD AI:
5
FAB = 0,
13
12

78 lb − FFG  = 0,
 13

ΣFx = 0: FBC −
12
78 lb − 72 lb = 0,
13
FAB = 78.0 lb
22.6° !
FGH = 72.0 lb
!
FBC = 144.0 lb
!
(b) FBD AI:
FAB = 78.0 lb
As above, ΣFy = 0 yields
Then:
 12

ΣM C = 0: ( 2.5 ft )  78.0 lb − FFG  − 180 lb ⋅ ft = 0
 13

22.6° !
FFG = 0 !
ΣFx = 0:
FBC −
12
( 78.0 lb ) = 0,
13
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.
FBC = 72.0 lb
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 89.
(a) FBD ACF:
Note: BC is a two-force member
ΣM F = 0:
 1

FBC  = 0,
 26

( 0.1 m )(120 N ) − ( 0.3 m ) 
B x = 200 N
FBC = 40 26 N,
ΣFx = 0:
(b) FBD BCE:
ΣFy = 0:
(
)
5
40 26 N − Fx = 0,
26
(
B y = 40.0 N
!
Fx = 200 N
!
Fy = 160.0 N
!
Fx = 320 N
!
Fy = 160.0 N
!
B x = 320 N
!
)
1
40 26 N = 0,
26
Fy − 120 N −
!
Note ACF is a two-force member
ΣM B = 0:
 1

FCF  = 0,
 5

( 0.4 m )(120 N ) − ( 0.3 m ) 
FCF = 160 5 N,
ΣFx = 0: Bx −
ΣFy = 0:
(
(
)
2
160 5 N = 0,
5
)
1
160 5 N − 120 N − B y = 0
5
B y = 40.0 N
!
(c) Moving the 120 N force from D to E does not affect the reactions. The answers are the same as in part (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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 90.
(a) FBD AF:
Note: BC is a two-force member
ΣM F = 0:
 1

FBC  − 48 N ⋅ m = 0,
 26

( 0.3 m ) 
FBC = 160 26 N
so
(b & c) FBD BCE:
B x = 800 N
!
B y = 160.0 N
!
ΣFx = 0:
Fx − 800 N = 0,
Fx = 800 N
!
ΣFy = 0:
− Fy + 160 N = 0,
Fy = 160.0 N
!
Note: ACF is a two-force member, and that the application point of the
couple is immaterial.
ΣM B = 0:
 1

FCF  − 48 N ⋅ m = 0
 5

( 0.3 m ) 
FCF = 160 5 N,
Fx = 320 N
!
Fy = 160.0 N
!
ΣFx = 0: 320 N − Bx = 0,
B x = 320 N
!
ΣFy = 0:
B y = 160.0 N
!
−160 N + By = 0,
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 6, Solution 91.
\
First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged.
ΣM C = 0:
rT1 − rT2 = 0
(a) Replace each force with an equivalent force-couple.
(b) Cut cable and replace forces on pulley with equivalent pair of forces at A as above.
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.
T1 = T2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 92.
FBD Pipe 2:
ΣFx′ = 0:
NG −
3
( 220 lb ) = 0,
5
NG = 132 lb
ΣFy′ = 0:
ND −
4
( 220 lb ) = 0,
5
N D = 176 lb
FBD Pipe 1:
θ = 90° − 2 tan −1
ΣFx′ = 0:
3
= 16.2602°
4
N F cos (16.2602° ) − 132 lb −
3
( 220 lb ) = 0
5
N F = 275.00 lb
ΣFy′ = 0:
N C + ( 275 lb ) cos (16.2602° ) −
4
( 220 lb ) = 0
5
NC = 99.00 lb
FBD Frame & Pipes:
ΣM A = 0:


4
5
( 48 in.) E y − 24 in. + 24 in. + (10 in.)  (220 lb) = 0


E y = 256
ΣFy = 0:
ΣFx = 0:
2
lb,
3
Ay − 2 ( 220 lb ) + 256
Ay = 183

1
lb
3
Ax − E x = 0,
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.
E y = 257 lb
A y = 183.3 lb
2
lb = 0
3
Ax = E x
COSMOS: Complete Online Solutions Manual Organization System
FBD BE:
ΣM B = 0:
( 24 in.)  256

2 
 15 
lb  − (18 in.) E x +  in.  ( 275 lb ) = 0
3 
 4

E x = 399.5 lb,
E x = 400 lb
From above,
A x = 400 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 93.
FBD Pipe 2:
FBD Pipe 1:
ΣFx' = 0:
NG −
3
( 220 lb ) = 0,
5
NG = 132 lb
ΣFy′ = 0:
ND −
4
( 220 lb ) = 0,
5
N D = 176 lb
ΣFx′ = 0:
ΣFy′ = 0:
ΣM B = 0:
FBD BE:
NF −
NC −
3
( 220 lb ) − 132 lb = 0,
5
4
( 220 lb ) = 0,
5
N F = 264 lb
NC = 176 lb
( 24 in.) E y − ( 32 in.) Ex + ( 5 in.) ( 264 lb ) = 0
4Ex − 3E y = 165
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.
(1)
COSMOS: Complete Online Solutions Manual Organization System
FBD Frame with Pipes:
ΣM A = 0: ( 48 in.) E y − (14 in.) Ex + ( 5 in.) ( 264 lb )
− ( 35 in. + 45 in.) (176 lb ) = 0
(2)
solving with (1) above:
E y = 355
ΣFx = 0:
2
lb,
3
Ax − 308 lb −
E y = 356 lb
4
3
( 264 lb ) + ⋅ 2 (176 lb ) = 0
5
5
A x = 308 lb
ΣFy = 0:
Ay + 355
!
!
2
3
4
lb − ( 264 lb ) − ⋅ 2 (176 lb ) = 0
3
5
5
A y = 84.3 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.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 94.
FBD AC:
ΣM c = 0:
( 0.6 m ) Ax − ( 0.06 m )(170 N ) = 0,
A x = 17.00 N
FBD CE:
ΣFx = 0:
−17 N +
ΣFx = 0:
133 N −
15
(170 N ) − Cx = 0,
17
Cx = 133 N
15
(170 N ) + Ex = 0,
17
E x = 17.00 N
ΣM C = 0:
( 0.48 m )(17.00 N ) + ( 0.90 m ) E y
− ( 0.45 m ) (170 N ) = 0,
E y = 75.9 N
FBD Frame:
ΣFy = 0:
Ay − 170 N + 75.9 N = 0,
A y = 94.1 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 6, Solution 95.
FBD Frame & pulley:
ΣM B = 0:
( 0.6 m ) Ay − ( 0.075 m )( 240 N ) = 0,
A y = 30.0 N
ΣFy = 0:
− 30.0 N + By − 240 N = 0,
B y = 270 N
FBD AD:
ΣFx = 0:
− Ax + Bx = 0,
Ax = Bx
ΣM D = 0:
( 0.200 m ) Ax − ( 0.075 m ) ( 240 N )
+ ( 0.30 m ) ( 30 N ) = 0
A x = 45.0 N
From above
Ax = Bx ,
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.
B x = 45.0 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 96.
(a)
FBD Entire machine:
ΣM A = 0:
( 95 in.) (16 kips) − ( 35 in.)(18 kips ) + (145 in.) ( 2B )
− (170 in.) ( 50 kips ) = 0,
B = 26.241 kips
ΣFy = 0:
B = 26.2 kips
−16 kips + 2 A − 18 kips + 2 ( 26.241 kips ) − 50 kips = 0,
A = 15.76 kips
(b)
FBD Motor unit:
ΣM D = 0:
( 30 in.) Cx + (85 in.)( 52.482 kips ) − (110 in.)( 50 kips ) = 0
Cx = 34.634 kips,
C x = 34.6 kips
ΣFx = 0:
Dx − 34.634 kips = 0,
D x = 34.6 kips
ΣFy = 0:
52.482 kips − 50 kips − Dy = 0,
D y = 2.48 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 6, Solution 97.
(a)
FBD Entire Machine:
(145 in.) (2B) − ( 35 in.)(18 kips )
ΣM A = 0:
− (170 in.)( 50 kips ) = 0
2B = 62.966 kips,
ΣFy = 0:
B = 31.5 kips
!
2 A + 62.966 kips − 18 kips − 50 kips = 0
2 A = 5.034 kips,
A = 2.52 kips
!
where A and B are single
wheel forces
(b)
ΣM D = 0:
FBD Motor unit:
(85 in.)( 62.966 kips ) − (110 in.)( 50 kips )
+ (30 in.) Cx = 0,
Cx = 4.9297 kips
C x = 4.93 kips
ΣFy = 0:
− Dy − 50 kips + 62.966 kips = 0
D y = 12.97 kips
ΣFx = 0:
!
!
Dx − C x = 0, Dx = Cx ,
D x = 4.93 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.
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 98.
FBD Frame:
( 0.625 m ) F − ( 0.75 m )( 4 kN ) − (1.25 m )( 3 kN ) = 0
ΣM A = 0:
Fx = 10.8 kN
ΣFx = 0:
Ax − 10.8 kN = 0,
ΣFy = 0:
Ay − 4 kN − 3 kN = 0,
FBD BF: (I)
A x = 10.80 kN
A y = 7.00 kN
FBD ABD: (II)
I:
ΣM C = 0:
( 0.375 m )(10.8 kN ) − ( 0.25 m ) Bx = 0, Bx = 16.2 kN,
II:
ΣM D = 0:
( 0.25 m )(10.8 kN + 16.2 kN ) + ( 0.5 m ) By − (1.00 m )( 7.0 kN ) = 0,
By = 0.5 kN,
ΣFx = 0:
−10.8 kN − 16.20 kN + Dx = 0,
Dx = 27 kN,
ΣFy = 0:
7.0 kN − 0.5 kN − Dy = 0,
Dy = 6.5 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.
B x = 16.20 kN
B y = 500 N
D x = 27.0 kN
D y = 6.50 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 99.
FBD Frame:
( 0.3 m ) Dx = − ( 0.4 m ) ( 900 N ) = 0,
ΣM A = 0:
Dx = 1200 N
D x = 1.200 kN
ΣFx = 0: 1.200 kN − Ax = 0,
ΣFy = 0:
Ax = 1.200 kN,
A x = 1.200 kN
Ay − 900 N = 0
A y = 900 N
Note: AG is a two-force member.
ΣM A = 0:
( 0.2 m )  By − ( 0.4 m ) ( 900 N ) = 0,
ΣFy = 0:
900 N + 1800 N − 900 N −
FBD AC:
−
4
( 3000 N ) − 1200 N + Bx = 0,
5
On GBEH:
ΣM E = 0:
ΣFx = 0:
FBD DF:
ΣFy = 0:
3
FAG = 0
5
( FAGx = 2400 N, FAGy = 1800 N )
FAG = 3000 N
ΣFx = 0:
By = 1800 N
Bx = 3600 N
G x = 2.40 kN
,
G y = 1.800 kN !
B x = 3.60 kN
,
B y = 1.800 kN
3
FDH = 0,
5
1200 N − Ex = 0
− ( 0.2 m )
!
FDH = 0
Ex = 1200 N
Ey = 0
On GBEH:
E x = 1.200 kN
H x = 0,
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.
,
Ey = 0 !
Hy = 0 !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 100.
FBD Frame:
( 6 in.) Dx − (13.5 in.) ( 48 lb ) − (16.5 in.) ( 20 lb ) = 0
ΣM A = 0:
Dx = 163 lb
ΣFx = 0:
− A x + 20 lb + 48 lb + 163 lb = 0
Ax = 231 lb
ΣFy = 0:
Ay = 0
FBD DE:
ΣM E = 0:
(19.5 in.)(163 lb ) − ( 6 in.) Bx = 0
Bx = 529.75 lb
ΣM C = 0:
( 4.5 in.) By − ( 7.5 in.)( 231 lb ) = 0,
B x = 530 lb
On ABC:
ΣFx = 0:
FBD ABC:
ΣFy = 0:
On ABC:
By = 385 lb
,
− 231 lb + 529.75 lb − Cx = 0,
C y − 385 lb = 0,
B y = 385 lb
Cx = 298.75 lb
C y = 385 lb
C x = 299 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.
,
C y = 385 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 101.
FBD Frame:
ΣM E = 0:
( 6 in.)(84 lb ) − ( 42 in.) Ay = 0
Ay = 12.00 lb
FBD CF: (I)
FBD ABC: (II)
I:
ΣM D = 0:
( 3.5 in.) Cx + ( 7 in.) C y − ( 5 in.) (84 lb ) = 0,
II:
ΣM B = 0:
(12 in.) C y − ( 3.5 in.) Cx − (12 in.) (12 lb ) = 0,
II:
ΣFx = 0:
ΣFy = 0:
Cx + 2C y = 120 lb
3.5 Cx − 12 C y = 144 lb
Solving:
Cx = 60.632 lb,
C y = 29.684 lb
On ABC:
C x = 60.6 lb,
C y = 29.7 lb
− Bx + 60.632 lb = 0,
Bx = 60.632 lb,
B x = 60.6 lb
12 lb + By + 29.684 lb = 0,
By = − 41.684 lb,
B y = 41.7 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 102.
FBD Stool:

w 1
N
= ( 56 kg )  9.81
 = 274.68 N
2 2
kg 

ΣM A = 0:
( 0.350 m ) By − ( 0.150 m )( 274.68 N ) = 0,
By = 117.72 N
FBD BE:
ΣM E = 0:
 1

FFG  = 0
 2

(1.25 m )(117.72 N ) − ( 0.15 m ) 
FFG = 98.1 2 N
ΣFx = 0:
ΣFy = 0:
Ex −
1
(98.1 2) N = 0,
2
− Ey +
E x = 98.1 N
1
(98.1 2 N) + 117.72 N = 0
2
E y = 215.82 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.
E y = 216 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 103.
FBD BC:
Note that only one of the forces B y , B′y exists.
(
)
M C B y or B′y − ( 0.15 m )
ΣM C = 0:

2 
FFG = 
M B or B′y
 0.15 m  C y


(
(
So we seek a maximum of M C By or B′y
1
FFG = 0
2
)
(1)
)
( )
M C ( B′y ) = ( 0.175 m ) B′y
M C By = ( 0.125 m ) By
where
FBD stool:
w = ( 56 kg ) ( 9.81 N/kg ) = 549.36 N
consider contact at A′, A y = 0, either
ΣM A′ = ( 0.425 m ) By − ( 0.225 m )
w
=0
2
( )
M C By = ( 0.03309 m ) w
By = 0.26471 w,
or ( M A′ ) = ( 0.475 m ) B′y − ( 0.225 m )
(i)
w
=0
2
( )
B′y = ( 0.23684 ) w, M C B′y = ( 0.04145 m ) w
(ii)
consider contact at A, A′y = 0, either
ΣM A = 0:
( 0.35 m ) By − ( 0.15 m )
w
=0
2
( )
By = 0.21429 w, M C By = ( 0.02678 m ) w
or
ΣM A = 0:
B′y = 0.1875 w,
( 0.40 m ) B′y − ( 0.15 m )
(iii)
w
=0
2
( )
M C B′y = ( 0.03281 m ) w
(iv)
(a) The maximum is (ii), with contact at A′ and B′, and
!

2 
(b) FFG max = 
 ( 0.04145 m ) ( 549.36 N ) = 214.69 N,
 0.15 m 
FFG max = 215 N T !
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 6, Solution 104.
Members FBDs:
I:
ΣM A = 0:
(12.8 ft ) Bx − ( 32 ft ) By − ( 20 ft )(14 kips ) = 0
II:
ΣM C = 0:
( 7.2 ft ) Bx + ( 24 ft ) By − (12 ft )( 21 kips ) = 0
Solving: Bx = 27.5 kips, By = 2.25 kips,
I:
ΣFx = 0:
A x = 27.5 kips
ΣFy = 0:
Ax − 27.5 kips = 0,
Ax = 27.5 kips,
(a)
!
Ay − 14 kips − 2.25 kips = 0, Ay = 16.25 kips,
(b)
A y = 16.25 kips
!
B x = 27.5 kips
!
B y = 2.25 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 6, Solution 105.
Member FBDs:
II:
ΣM C = 0:
( 7.2 ft ) Bx − ( 24 ft ) By − (12 ft )(14 kips ) = 0
I:
ΣM A = 0:
(12.8 ft ) Bx + ( 32 ft ) By − ( 20 ft )( 21 kips ) = 0
Solving: Bx = 28.75 kips, By = 1.675 kips,
I:
ΣFx = 0:
A x = 28.8 kips
ΣFy = 0:
Ax − 28.75 kips = 0,
Ax = 28.75 kips,
(a)
!
Ay − 21 kips + 1.625 kips, Ay = 19.375 kips ,
A y = 19.38 kips
(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.
!
B x = 28.8 kips
!
B y = 1.625 kips
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 106.
\
FBD Frame:
ΣM A = 0:
(12 in.) Dx − ( 28 in.)( 411 lb ) = 0
D x = 959 lb
FBD DF:
Note that BF and CE are two-force members.
4
15
FCE −
FBF = 0
5
17
ΣFx = 0:
959 lb +
ΣM D = 0:
(12 in.) 
Solving:
FBF = 357 lb,
ΣM E = 01:
( 22.5 in.)  ( 357 lb ) − (12 in.) Dy = 0
3

 8

FCE  + ( 34.5 in.) 
FBF  = 0
5

 17

D y = 315 lb ,
8
17
so

(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.
D = 1009 lb
18.18°
FBF = 357 lb T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 107.
FBD Frame:
(12 in.) Dx − ( 34.5 in.)( 274 lb ) = 0
ΣM A = 0:
Dx = 787.75 lb
FBD DF:
ΣM D = 0:
(12 in.) 
ΣFx = 0:
787.75 lb +
Solving:
ΣM E = 0:
3

 8

FCE  + ( 34.5 in.) 
FBF − 274 lb  = 0
5
17




4
15
FCE −
FBF = 0
5
17
FBF = 684.25 lb,
( 22.5 in.)  ( 684.25 lb ) − 274 lb  − (12 in.) Dy = 0
Dy = 90 lb,
8
17

so
(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.
D = 793 lb
6.52°
FBF = 684 lb T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 108.
FBD AC:
Note that BE is a two force member
ΣFy = 0:
1
FBE − P = 0
5
FBE = 5 P
ΣΜ Α = 0:
 1
 5
( 0.4 m ) 
( 5 P ) + ( 0.1 m )  25 ( 5 P )
− a P + ( 0.3 m ) C = 0
so
( a − 0.6 m ) P = ( 0.3 m ) C
since P > 0 and C ≥ 0, a – 0.6 m ≥ 0
a ≥ 0.6 m
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 6, Solution 109.
Member FBDs:
ΣΜ Α = 0:
( 0.6 m )
4
4

FCF − ( 0.9 m )  FDG  = 0
5
5


4
4
ΣΜ Ε = 0: − ( 0.3 m ) FCF + ( 0.6 m ) FDG − ( 0.9 m )( 800 N ) = 0
5
5
Solving:
FCF = 9000 N,
FDG = 6000 N
FCF = 9.00 kN C
FDG = 6.00 kN T
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 6, Solution 110.
Member FBDs:
4
FDG = 0
5
ΣM A = 0:
( 0.3 m ) FBF − ( 0.9 m )
ΣM E = 0:
4
− ( 0.3 m ) FBF + ( 0.6 m ) FDG − ( 0.9 m )( 0.8 kN ) = 0
5
Solving:
FBE = − 3 kN,
FBF = − 7.2 kN
so
FBF = 7.20 kN T
FDG = 3.00 kN C
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 6, Solution 111.
Member FBDs:
4
4
FCH − ( 0.3 m ) FBG = 0
5
5
ΣM A = 0:
( 0.6 m )
ΣM E = 0:
4
4
− ( 0.9 m ) FCH + ( 0.6 m ) FBG − ( 0.9 m )( 0.8 kN ) = 0
5
5
Solving:
FBG = 6 kN, FCH = 3 kN
so
FBG = 6.00 kN T
FCH = 3.00 kN C
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 6, Solution 112.
Member FBDs:
I:
ΣM J = 0:
2a C y + a Cx − a P = 0
Solving:
I:
II:
Cx =
II:
ΣM K = 0: 2a C y − a C x = 0
P
P
, Cy =
2
4
ΣFx = 0:
1
FBG − Cx = 0,
2
FBG = 2C x =
2
P,
2
ΣFy = 0:
FAF +
1  2 
P
P  − P + = 0,

4
2  2 
FAF =
ΣFx = 0:
Cx −
1
FDG = 0,
2
ΣFy = 0:
−
FDG = 2C x =
P
1  2 
P  + FEH = 0,
+

4
2  2 
FBG = 0.707 P T
P
,
4
FAF = 0.250 P T
2
P,
2
FDG = 0.707 P T
P
4
FEH = 0.250 P C
FEH = −
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 6, Solution 113.
Note that, if we assume P is applied to BF, each individual member FBD
looks like:
Fshort = 2Flong =
so
2
Fload
3
(by moment equations about S and L).
Labeling each interaction force with the letter corresponding to the joint
of application, we have:
F =
2(P + E)
3
= 2B
C
D
=
3
2
B
A
C =
=
3
2
E =
so
2(P + E)
= 2 B = 6C = 18E
3
A = 2C = 3E
F =
2
P
P +

3
26 
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.
E=
P
26
so D = 2E =
P
13
A=
3P
13
P + E = 27 E
F=
9P
13
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 114.
Note that, if we assume P is applied to EG, each individual member FBD
looks like
so
2Fleft = 2 Fright = Fmiddle
Labeling each interaction force with the letter corresponding to the joint
of its application, we see that
B = 2 A = 2F
C = 2B = 2D
G = 2C = 2H
P + F = 2G ( = 4C = 8B = 16 F ) = 2E
From
P + F = 16 F , F =
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.
P
15
so A =
P
15
D=
2P
15
H=
4P
15
E=
8P
15
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 115.
(a) Member FBDs:
ΣM B = 0: aF2 = 0
FBD II:
ΣFy = 0:
By = 0
FBD I:
ΣM A = 0:
aF2 − 2aP = 0
F2 = 0
but F2 = 0
so P = 0
not rigid for P ≠ 0 !
(b) Member FBDs:
(
)
Note: 7 unknowns Ax , Ay , Bx , By , F1, F2 , C but only 6 independent equations.
System is statically indeterminate !
System is, however, rigid !
(c) FBD whole:
FBD right:
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
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COSMOS: Complete Online Solutions Manual Organization System
FBD I:
ΣM A = 0: 5aBy − 2aP = 0
ΣFy = 0: Ay − P +
2
P=0
5
By =
Ay =
a
5a
Bx −
By = 0
2
2
FBD II:
ΣM c = 0:
FBD I:
ΣFx = 0: Ax + Bx = 0
2
P
5
3
P
5
Bx = 5By
Ax = − Bx
B x = 2P
A x = 2P
A = 2.09P
16.70° !
B = 2.04P
11.31° !
System is rigid !
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 6, Solution 116.
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the
third force, at B, must be parallel to the link forces.
(a) FBD whole:
a 4
1
B + 5a
B=0
4 17
17
ΣM A = 0:
− 2aP −
ΣFx = 0:
Ax −
ΣFy = 0:
Ay − P +
4
B=0
17
1
B=0
17
B = 2.06 P
B = 2.06P
14.04° !
A = 2.06P
14.04° !
A x = 2P
Ay =
P
2
rigid !
(b) FBD whole:
Since B passes through A,
ΣM A = 2aP = 0 only if P = 0
∴ no equilibrium if P ≠ 0
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.
not rigid !
COSMOS: Complete Online Solutions Manual Organization System
(c) FBD whole:
ΣM A = 0: 5a
1
3a 4
B+
B − 2aP = 0
4 17
17
ΣFx = 0: Ax +
4
B=0
17
ΣFy = 0: Ay − P +
1
B=0
17
B=
17
P
4
B = 1.031P
14.04° !
A = 1.250P
36.9° !
Ax = − P
Ay = P −
P 3P
=
4
4
System is rigid !
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 6, Solution 117.
(a) Member FBDs:
FBD I:
ΣM A = 0: aF1 − 2aP = 0
F1 = 2P;
FBD II:
ΣM B = 0: − aF2 = 0
F2 = 0
ΣFy = 0: Ay − P = 0
ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2P
B x = 2P
ΣFy = 0: By = 0
FBD I:
ΣFx = 0: − Ax − F1 + F2 = 0
Ay = P
Ax = F2 − F1 = 0 − 2P
so B = 2P
!
A x = 2P
so A = 2.24P
26.6° !
Frame is rigid !
(b) FBD left:
FBD whole:
a
a
5a
P + Ax −
Ay = 0
2
2
2
FBD I:
ΣM E = 0:
FBD II:
ΣM B = 0: 3aP + aAx − 5aAy = 0
Ax − 5 Ay = − P
Ax − 5 Ay = −3P
This is impossible unless P = 0 ∴ not rigid !
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
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COSMOS: Complete Online Solutions Manual Organization System
(c) Member FBDs:
FBD I: ΣFy = 0: A − P = 0
FBD II:
ΣM D = 0: aF1 − 2aA = 0
F1 = 2 P
ΣFx = 0: F2 − F1 = 0
F2 = 2 P
ΣM B = 0: 2aC − aF1 = 0
C =
ΣFx = 0: F1 − F2 + Bx = 0
ΣFx = 0: By + C = 0
F1
= P
2
A= P
!
C= P
!
Bx = P − P = 0
By = −C = − P
B= P !
Frame is rigid !
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 6, Solution 118.
FBD ABC:
ΣM C = 0: 0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30° 
+ 0.030 m + ( 0.30 m ) cos 30°  ( 400 N ) cos 30°
 12

 5

− ( 0.03 m )  FBD  − ( 0.045 m )  FBD  = 0
13
13




FBD = 3097.64 N
ΣFx = 0:
5
( 3097.64 N ) − ( 400 N ) sin 30° − Cx = 0
13
C x = 991.39 N
ΣFy = 0:
12
( 3097.64 N ) − ( 400 N ) cos 30° − C y = 0
13
C y = 2512.9 N
FBD Blade:
(a)
Vertical component at D =
12
( 3097.64 N )
13
= 2.86 kN
(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.
C = 2.70 kN
68.5°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 119.
FBD Blade:
ΣFy = 0:
FBD ABC:
ΣM C = 0:
3 kN −
12
FBD = 0: FBD = 3.25 kN C
13
0.045 m + ( 0.3 m ) sin 30°  P sin 30°
+ 0.030 m + ( 0.3 m ) cos 30° P cos 30°
− ( 0.045 m )
− ( 0.03 m )
5
( 3.25 kN )
13
12
( 3.25 kN ) = 0
13
P = 0.41968 kN
so
ΣFx = 0:
P = 420 N
(a)
− ( 0.41968 kN ) sin 30° +
5
( 3.25 kN ) − Cx = 0
13
C x = 1.04016 kN
ΣFy = 0:
− ( 0.41968 kN ) cos30° +
12
( 3.25 kN ) − C y = 0
13
C y = 2.6365 kN
so
(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.
C = 2.83 kN
68.5°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 120.
FBD Stamp D:
ΣFy = 0:
E − FBD cos 20° = 0,
ΣM A = 0:
( 0.2 m )( sin 30°)( FBD cos 20°)
FBD ABC:
E = FBD cos 20°
+ ( 0.2 m )( cos 30° )( FBD sin 20° )
− ( 0.2 m ) sin 30° + ( 0.4 m ) cos15°  ( 250 N ) = 0
FBD = 793.64 N C
and, from above,
E = ( 793.64 N ) cos 20°
E = 746 N
(a)
ΣFx = 0:
!
Ax − ( 793.64 N ) sin 20° = 0
A x = 271.44 N
ΣFy = 0:
Ay + ( 793.64 N ) cos 20° − 250 N = 0
A y = 495.78 N
so
(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.
A = 565 N
61.3° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 121.
\
FBD Stamp D:
ΣFy = 0:
900 N − FBD cos 20° = 0,
FBD = 957.76 N C
(a)
FBD ABC:
ΣM A = 0:
( 0.2 m )( sin 30° )  ( 957.76 N ) cos 20°
+ ( 0.2 m )( cos 30° )  ( 957.76 N ) sin 20°
− ( 0.2 m ) sin 30° + ( 0.4 m ) cos15° P = 0
P = 301.70 N,
P = 302 N
(b)
ΣFx = 0:
Ax − ( 957.76 N ) sin 20° = 0
A x = 327.57 N
ΣFy = 0:
− Ay + ( 957.76 N ) cos 20° − 301.70 N = 0
A y = 598.30 N
so
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 = 682 N
61.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 122.
FBD ABC:
( 45 mm ) sin 25° − 10 mm 
θ = sin −1 
ΣM C = 0:
100 mm
= 5.1739°
( 70 mm )(110 N ) − ( 45 mm ) sin 25° FBD cos 5.1739°
+ ( 45 mm ) cos 25° FBD sin 5.1739° = 0
FBD = 504.50 N
ΣFx = 0:
( 504.50 N ) cos 5.1739° − (110 N ) sin 25° − Cx = 0,
Cx = 455.96 N
FBD CE:
ΣFx = 0:
455.96 N − Q = 0
Q = 455.96 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)
Q = 456 N
(b)
FBD = 540 N T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 123.
Member FBDs:
θ = tan −1
= tan −1
( BC + CD ) sin 60°
AF + ( AB + BC − CD ) cos 60°
( 2.5 in. + 1 in.) sin 60°
4.5 in. + (1.5 in. + 2.5 in. − 1 in.) cos 60°
θ = 26.802°
From FBD CDE:
ΣM C = 0:
( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 26.802°) = 0,
FDF = 150.234 lb C
ΣFx = 0:
(150.234 lb ) cos ( 26.802°) − ( 20 lb ) sin 60° − Cx = 0,
Cx = 116.774 lb
ΣFy = 0:
(150.234 lb ) sin 26.802° − ( 20 lb ) cos 60° − C y = 0
C y = 57.742 lb
From FBD ABC:
ΣM A = 0:
(1.5 in.) sin 60° Bx + (1.5 in.) cos 60° By − ( 4 in.) sin 60°  (116.774 lb )
+ ( 4 in.) cos 60° ( 57.742 lb ) = 0,
3 Bx + By = 385.37 lb
continued
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.
(1)
COSMOS: Complete Online Solutions Manual Organization System
From FBD BFG:
ΣM G = 0:
− (1.5 in.) (150.234 lb ) sin 26.802° + (1.5 in.) cos 30°  (150.234 lb ) cos 26.802°
− (1.5 in.) cos 30° Bx − 6 in. + (1.5 in.) sin 30°  B y = 0,
Bx = 243.32 lb,
Solving (1) and (2):
ΣFx = 0:
ΣFy = 0:
3Bx − 9 B y = 96.775 lb
By = − 36.075 lb
243.32 lb − (150.234 lb ) cos 26.802° − Gx = 0,
G x = 109.226 lb
G y = 31.667 lb
G y − (150.234 lb ) sin 26.802° + 36.075 lb = 0,
On G,
G = 113.7 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.
16.17° !
(2)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 124.
Member FBDs:
θ = tan −1
= tan −1
( BC + CD ) sin 60°
AF + AB cos 75° + ( BC − CD ) cos 60°
( 2.5 in. + 1 in.) sin 60°
4.5 in. + (1.5 in.) cos 75° + ( 2.5 in. − 1 in.)  cos 60°
θ = 28.262°
From FBD CDE:
ΣM C = 0:
( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 28.262°) = 0
FDF = 150.069 lb
ΣFx = 0:
(150.069 lb ) cos 28.262° − ( 20 lb ) sin 60° − Cx = 0
Cx = 114.859 lb
ΣFy = 0:
(150.069 lb ) sin 28.262° − ( 20 lb ) cos 60° − C y = 0
C y = 61.058 lb
From FBD ABC:
ΣM A = 0:
(1.5 in.) sin 75° Bx − (1.5 in.) cos 75° By − (1.5 in.) sin 75° + ( 2.5 in.) sin 60° (114.859 lb )
+ (1.5 in.) cos 75° + ( 2.5 in.) cos 60° ( 61.058 lb ) = 0
or 1.44889 Bx − 0.38823 B y = 315.07 lb
(1)
continued
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
From FBD BFG:
ΣM G = 0:
− (1.5 in.) (150.069 lb ) sin 28.262° + (1.5 in.) cos15° (150.069 lb ) cos 28.262° 
− (1.5 in.) cos15° Bx + ( 6 in.) + (1.5 in.) sin15°  B y = 0
or 1.44889 Bx − 6.3882 B y = 84.926 lb
Solving (1) and (2):
Bx = 227.74 lb,
(2)
By = 38.358 lb
ΣFx = 0:
227.74 lb − (150.069 lb ) cos 28.262° − Gx = 0,
G x = 95.560 lb
ΣFy = 0:
G y − (150.069 lb ) sin 28.262° + 38.358 lb = 0,
G y = 32.700 lb
So force on G is
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.
G = 101.8 lb
18.89° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 125.
FBD Piston:
ΣFy = 0:
180 lb − FCE cos 20° = 0,
FCE = 191.552 lb C
FBD Joint C:
FBC
191.552°
=
sin 40°
sin 40°
FBC = 191.552 lb
FBD AB:
ΣM A = 0:
( 6 in.)(191.552 lb ) cos 30° − M = 0
M = 995 lb ⋅ in.
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 6, Solution 126.
FBD AB:
ΣM A = 0:
( 6 in.) FBC cos 30° − 660 lb ⋅ in. = 0,
FBC = 127.017 lb T
FBD Piston:
180 lb
C
cos φ
ΣFy = 0:
180 lb − FCE cos φ = 0,
FCE =
ΣFy = 0:
180 lb
cos φ + (127.017 lb ) sin 30°
cos φ
− FCD cos φ = 0,
FBD Joint C:
FCD cos φ = 243.51 lb
ΣFx = 0:
(1)
(127.017 lb ) cos30° −
180 lb
sin φ − FCD sin φ = 0
cos φ
FCD sin φ = 110.000 lb − (180 lb ) tan φ
divide (2) by (1)
(2)
tan φ = 0.45173 − 0.73919 tan φ
tan φ = 0.25974,
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.
φ = 14.56°
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Chapter 6, Solution 127.
(a) FBD E:
ΣFx = 0:
FDE cos α −160 N = 0,
where α = tan −1
FBD Joint D:
β = tan −1
FDE =
160 N
cosα
38.8 mm
= 22.823°, so FDE = 173.59 N C
92.2 mm
37.5 mm − 18.8 mm
= 25.732°
38.8 mm
FCD cos ( 45° − 25.732° )
ΣFx = 0:
− (173.59 N ) cos ( 25.732° + 22.823° ) = 0
FCD = 121.717 N C
FBD BC:
ΣM B = 0:
( 0.045 m )(121.717 N ) sin 45° − M = 0,
M = 3.873 N ⋅ m,
M = 3.87 N ⋅ m
(b) FBD E:
ΣFx = 0:
FDE cos α −160 N = 0,
where α = tan −1
FDE =
160 N
cos α
20 mm + 35.4 mm − 26.4 mm
= 16.8584°
92.2 mm + 27.8 mm − 24.3 mm
so FDE = 167.185 N C
continued
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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FBD Joint D:
β = tan −1
27.8 mm − 24.3 mm
= 7.5520°
26.4 mm
γ = tan −1
20 mm + 35.4 mm − 26.4 mm
= 42.363°
24.3 mm + 7.5 mm
ΣFy′ = 0:
(167.185 N ) sin (16.8584° + 42.363°)
− FCD sin ( 90° − 7.5520° − 42.363° ) = 0,
FCD = 223.07 N C
FBD BC:
δ = tan −1
35.4 mm
= 51.857°
27.8 mm
ΣM B = 0: M − ( 0.045 m )( 223.07 N ) sin ( 90° − 51.857° − 7.5520° ) = 0
M = 5.12 N ⋅ m !
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 6, Solution 128.
(a) FBD BC:
( 0.045 m ) FCD sin 45° − 6.00 N ⋅ m = 0
ΣM B = 0:
FCD = 188.562 N C
FBD Joint D:
α = tan −1
38.8 mm
= 22.823°
92.2 mm
β = tan −1
37.5 mm − 18.8 mm
= 25.732°
38.8 mm
ΣFx′ = 0:
(188.562 N ) cos ( 45° − 25.732°)
− FDE cos ( 22.823° + 25.732° ) = 0, FDE = 268.92 N C
FBD E:
( 268.92 N ) cos ( 22.823°) − P = 0,
ΣFx = 0:
P = 248 N
(b) FBD BC:
δ = tan −1
35.4 mm
= 51.857°
27.8 mm
β = tan −1
27.8 mm − 24.3 mm
= 7.5520°
26.4 mm
ΣM B = 0:
!
( 0.045 m ) FCD sin ( 90° − 51.857° − 7.5510°) − 6.00 N ⋅ m = 0
FCD = 262.00 N C
continued
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
FBD Joint D:
β = 7.5520°,
α = tan −1
γ = tan −1
20 mm + 35.4 mm − 26.4 mm
= 42.363°
24.3 mm + 7.5 mm
20 mm + 35.4 mm − 26.4 mm
= 16.8584°
92.2 mm + 27.8 mm − 24.3 mm
ΣFy′ = 0:
FDE sin (16.8584° + 42.363° )
− ( 262.00 N ) sin ( 90° − 7.5520° − 42.363° ) = 0,
FDE = 196.366 N C
FBD E:
ΣFx = 0:
(196.366 N )( cos16.8584°) − P = 0,
P = 187.9 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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 129.
Member FBDs:
BCD:
AB:
ΣFy = 0:
By = 0
ΣM C = 0:
( 0.128 m ) Bx − ( 0.10 m )(80 N ) = 0,
ΣM A = 0:
(0.15 m) ( 62.5 N ) − M = 0,
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.
Bx = 62.5 N
M = 9.38 N ⋅ m
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 130.
FBD BCD:
d = 0.24 m
ΣFy = 0:
0.100 m
= 0.1875 m
0.128 m
(80 N ) sin 45° − By = 0
By = 40 2 N
ΣM C = 0:
( 0.100 m )(80 N ) cos 45°
+ ( 0.1875 m )( 80 N ) sin 45°
(
+ ( 0.24 m ) 40 2 N
)
− ( 0.128 m ) Bx = 0
Bx = 233.13 N
FBD AB:
ΣM A = 0:
( 0.2 m ) ( 40 2 N ) − ( 0.15 m )( 233.13 N ) = 0
M = 46.3 N ⋅ m.
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 131.
FBD BD:
θ = 90° − 25° = 65°
ΣM B = 0:
 2 ( 0.2 m ) cos 25° D sin 65° −15 N ⋅ m = 0
D = 45.654 N
FBD AC:
ΣM A = 0:
M A = ( 0.2 m )( 45.654 N ) = 0
M A = 9.13 N ⋅ m
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 132.
FBD BD:
ΣM B = 0:
 2 ( 0.2 m ) cos 25° C −15 N ⋅ m = 0
C = 41.377 N
FBD AC:
ΣM A = 0:
M A − ( 0.2 m )( 41.377 N ) sin 65° = 0
M A = 7.50 N ⋅ m
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 133.
FBD Member CD:
Noting that B is Perpendicular to CD,
ΣFx′ = 0:
Dy sin 30° − ( 80 lb ) cos 30° = 0,
D y = 138.564 lb
FBD Machine:
ΣM A = 0:
10 in.
(138.564 lb ) − M = 0,
sin 30°
M = 2771.3 lb ⋅ in.
M = 2.77 kip ⋅ in.
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 134.
FBD CD:
Since B is ⊥ to CD
ΣFx′ = 0:
Dy sin 30° − ( 80 lb ) cos 30° = 0
D y = 138.564 lb
FBD Whole:
a = 10 in. − ( 8 in.) cos 30° = 3.07180 in.
b=
a
= 5.32051 in.
tan 30°
d = b − 4 in. = 1.32051 in.
ΣM A = 0:
(10 in.)(80 lb ) + (1.32051 in.)(138.564 lb ) − M = 0
M = 983 lb ⋅ in.
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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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 135.
FBD ABD:
First note, by inspection, that P = W = 250 lb.
Note: that BC is a two-force member
ΣM D = 0:
( 5 in.)
3
1
FBC + ( 3 in.)
FBC − ( 9 in.)( 250 lb ) = 0
10
10
FBC = 125 10 lb T
ΣFx = 0:
Dx −
(
)
3
125 10 lb = 0
10
D x = 375 lb
ΣFy = 0:
250 lb −
(
)
1
125 10 lb − Dy = 0
10
D y = 125 lb
(note, that Dy = 125 lb is evident by symmetry of the barrel)
so D = 395 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
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18.43°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 136.
\
FBD whole:
By symmetry A = B = 22.5 kN
FBD ADF:
ΣM F = 0:
( 75 mm ) CD − (100 mm )( 22.5 kN ) = 0
CD = 30 kN
ΣFx = 0:
ΣFy = 0:
Fx − CD = 0
Fx = CD = 30 kN
22.5 kN − Fy = 0
Fy = 22.5 kN
so F = 37.5 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.
36.9°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 137.
FBD Joint A:
TB TC 2.1 kN
=
=
17 17
30
TB = TC = 1.19 kN
FBD BF and CF w/cables to A:
By symmetry
Ex = Fx = 1.05 kN
and E y = Fy
FBD Machine:
By symmetry
H x = I x = 1.05 kN
and H y = I y
ΣM D = 0:
FBD BF:
( 0.075 m ) Fy − ( 0.08 m ) (1.05 kN )
15
(1.19 kN )
17
8
− ( 0.16 m )
(1.19 kN ) = 0
16
Fy = 4.48 kN
− ( 0.15 m )
ΣM G = 0:
( 0.075 m ) ( 4.48 kN ) + ( 0.08 m ) (1.05 kN )
+ ( 0.02 m ) (1.05 kN ) − ( 0.15 m ) H y = 0
FBD FGH:
H y = 2.94 kN
ΣFy = 0:
2.94 kN − G y + 4.48 kN = 0 ,
G y = 7.42 kN
ΣFx = 0:
−1.05 kN + Gx + 1.05 kN = 0 ,
Gx = 0
so
H = 3.12 kN
70.3° !
G = 7.42 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 6, Solution 138.
FBD Machine:
By inspection, P = W = 110 lb,
By symmetry:
ΣFy = 0:
H x = Ix,
110 lb − 2H y = 0 ,
ΣFy = 0:
H y = I y = 55 lb,
FCE = FBD
By symmetry:
FBD ABC:
Hy = Iy
4

110 lb − 2  FCE  = 0 ,
5


FCE = FBD =
550
lb C
8
 4  550  
 3  550  
lb   + (17 in.)  
lb  



5  8
5  8
( 2.5 in.) 
ΣM H = 0:
− (4 in.) FFG = 0
FBD DFH:
FFG = 209.69 lb,
ΣFx = 0:
209.69 lb −
FFG = 210 lb
3  550 
lb  − H x = 0

5 8

H x = 168.440 lb
so
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.
H = 177.2 lb
18.08°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 139.
FBD top handle:
Note CD and DE are two-force members
ΣM A = 0:
( 4 in.)
6
5
FCD − (1.5 in.)
FCD − (13.2 in.) ( 90 lb ) = 0
61
61
FCD = 72 61 lb
FDE = FCD = 72 61 lb
By symmetry:
FBD Joint D:
ΣFx = 0:
D−2
(
)
5
72 61 lb = 0,
61
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.
D = 720 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 140.
FBD top handle:
By symmetry the horizontal force at F must be zero
ΣFx = 0: Dx = 0
ΣM F = 0:
( 0.015 m ) Fy − ( 0.185 m )(135 N ) = 0
Dy = 1665 N
FBD ABD:
ΣM B = 0:
( 40 mm ) (1665 N ) − ( 30 mm ) A = 0
A= 2220 N
A= 2.22 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 6, Solution 141.
FBD cutter AC:
ΣM C = 0:
( 32 mm )1.5 KN − ( 28 mm ) Ay − (10 mm ) Ax = 0
 11 
10 Ax + 28  Ax  = 48 kN
 13 
Ax = 1.42466 kN
Ay = 1.20548 kN
FBD handle AD:
ΣM D = 0:
(15 mm )(1.20548 kN ) − ( 5 mm )(1.42466 kN )
− (70 mm) P = 0
P = 0.1566 kN = 156.6 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 6, Solution 142.
FBD CD:
θ = tan − 1
ΣM D = 0:
0.85 in. − 0.3 in.
= 22.932°
2.6 in. − 1.3 in.
( 3.4 in.)( 30 lb ) − ( 0.8 in.) FAC sin 22.932°
+ ( 0.3 in.) FAC cos 22.932° = 0
FAC = 2879.6 lb
ΣFx = 0: Dx − ( 2879.6 lb ) cos 22.932° = 0,
Dx = 2652.0 lb
ΣFy = 0: Dy − ( 2879.6 lb ) sin 22.932° + 30 lb = 0,
Dy = 1092.00 lb
FBD BD:
ΣM B = 0:
( 2.6 in.) F + ( 0.2 in.)(1092.00 lb )
− ( 0.85 in.)( 2652.0 lb ) = 0
F = 783 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 143.
FBD Joint B:
α = tan − 1
1
= 5.7106°
10
ΣFx′ = 0: FBC cos ( 30° + 5.7106° ) − ( 54 lb ) cos 30° = 0
FBC = 57.595 lb T
FCD = 57.595 lb
By symmetry:
FBD Joint C:
ΣFx′′ = 0: 2 ( 57.595 lb ) sin ( 5.7106° ) − P = 0
P = 11.4618 lb
so
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.
P = 11.46 lb
30.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 144.
FBD BC:
ΣM B = 0:
(1 in.) Cx − ( 2.65 in.) ( 270 lb ) = 0
Cx = 715.5 lb
FBD CF:
( 3.5 in.) P − ( 0.75 in.) FDE cos 40°
+ ( 0.65 in.) FDE sin 40° = 0
ΣM C = 0:
FDE = 22.333 P
ΣFx = 0: C x − ( 22.333 P ) cos 40° + P sin 30° = 0
P = 43. 081 lb
P = 43.1 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.
60.0˚
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 145.
FBD Handles:
ΣM C = 0: − aBx + b (100 N − 100 N ) = 0
Bx = 0
FBD top handle:
ΣM A = 0:
( 28 mm ) By − (110 mm ) (100 N ) = 0
By = 392.86 N
FBD top blade:
ΣM D = 0:
( 40 mm ) ( 392.86 N ) − ( 30 mm ) E = 0
E = 523.81 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.
E = 524 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 146.
FBD BE:
ΣM B = 0:
( 746 mm ) C − ( 640 mm ) ( 0.25 kN ) = 0
C=
160
kN
746
FBD Blade:
ΣM A = 0:
 2  160

kN  


 5  746
( 38 mm ) F − ( 47 mm ) 
 1  160

kN   = 0
− ( 23 mm ) 


 5  746
F = 8.82 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 6, Solution 147.
FBD Boom with bucket and man:
θ = tan − 1
3 ft + ( 6 ft ) sin 35°
= 62.070°
( 6 ft ) cos 35° − 1.5 ft
ΣM A = 0: ( 6 ft ) cos 35° ( FBD sin 62.070° )
− ( 6 ft ) sin 35° ( FBD cos 62.070° )
− ( 9ft ) cos 35°  (1400 lb ) − ( 20 ft ) cos 35°  ( 450 lb ) = 0
FBD = 17693 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.
FBD = 6.48 kips
62.1°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 148.
FBD AED:
ΣM A = 0:
(a)
( 36 in.) (1500 lb ) − ( 25 in.) F = 0,
F = 2160 lb
ΣFy = 0: Ay − 1500 lb = 0,
ΣFy = 0: Ax − 2160 lb = 0,
FBD Entire machine:
ΣM B = 0:
Ay = 1500 lb
Ax = 2160 lb
(84 in.) (1500 lb ) − ( 25 in.) Gx = 0,
Gx = 5040 lb
ΣM C = 0:
FBD AG:
( 24 in.) ( Gy + 1500 lb )
− (12.5 in.) ( 5040 lb − 2160 lb ) = 0,
Gy = 0
(b)
ΣFy = 0: C y − 1500 lb + 0 = 0,
C y = 1500 lb
ΣFx = 0:
Cx − 2160 lb − 5040 lb = 0,
Cx = 7200 lb
so on BCF,
C x = 7.20 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.
,
C y = 1.500 kips
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 149.
FBD boom:
θ = tan −1
Note:
( 3.2sin 24° − 1) m
( 3.2cos 24° − 0.6 ) m
θ = 44.73°
(a)
ΣM A = 0:
( 6.4 m ) cos 24° ( 2.3544 kN )
− ( 3.2 m ) cos 24° B sin 44.73°
+ ( 3.2 m ) sin 24° B cos 44.73° = 0
B = 12.153 kN
B = 12.15 kN
44.7°
ΣFx = 0: Ax − (12.153 kN ) cos 44.73° = 0
A x = 8.633 kN
(b)
ΣFy = 0:
− 2.3544 kN + (12.153 kN ) sin 44.73° − Ay = 0
A y = 6.198 kN
On boom:
A = 10.63 kN
35.7°
On carriage:
A = 10.63 kN
35.7°
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
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 150.
FBD Bucket (one side):
(a)
ΣM D = 0:
( 0.8 m ) ( 2.4525 kN ) − ( 0.5 m ) FAB = 0
FAB = 3.924 kN
w = ( 250 kg ) ( 9.81 N/kg ) = 2452.5 N = 2.4525 kN
FBD link BE:
ΣM E = 0:
( 0.68 m )( 3.924 kN ) − ( 0.54 m ) 
80

FCD 
 89

 39

+ ( 0.35 m )  FCD  = 0,
 89

FCD = 7.682 kN
FCD = 7.68 kN C
FBD top blad
(b)
FBD Entire linkage:
ΣM G = 0:
( 2.5 m ) ( 2.4525 kN )
1
1
FGH − ( 0.6 m )
FGH = 0,
2
2
FFH = 21.677 kN,
FFH = 21.7 kN C
+ ( 0.2 m )
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 6, Solution 151.
\
FBD Central Gear:
By symmetry: F1 = F2 = F3 = F
ΣM A = 0:
3 ( rA F ) − 15 N ⋅ m = 0,
ΣM C = 0:
rB ( F − F4 ) = 0,
ΣFx′ = 0:
C x′ = 0
ΣFy′ = 0:
C y′ − 2 F = 0,
F =
5
N⋅m
rA
FBD Gear C:
F4 = F
C y′ = 2 F
Gears B and D are analogous, each having a central force of 2F
FBD Spider:
ΣM A = 0:
75 N ⋅ m − 3 ( rA + rB ) 2 F = 0
25 N ⋅ m − ( rA + rB )
10
N⋅m = 0
rA
rA + rB
r
= 2.5 = 1 + B ,
rA
rA
FBD Outer gear:
Since
rA = 24 mm,
ΣM A = 0:
rB = 1.5 rA
rB = 36.0 mm !
(a)
3 ( rA + 2rB ) F − M E = 0
3 ( 24 mm + 72 mm )
(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
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5 N⋅m
−ME = 0
24 mm
M E = 60.0 N ⋅ m
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 152.
FBD Gear A: looking from C
(a)
M A = 15 lb ⋅ ft
ΣM A = 0:
rA = 4 in.
M A − P rA = 0
P=
M A 180 lb ⋅ in.
=
rA
4 in.
P = 45 lb
FBD Gear B: looking from F
ΣM B = 0:
M 0 − rB P = 0
M 0 = rB P = ( 2.5 in.)( 45 lb ) = 112.5 lb ⋅ in.
M 0 = 112.5 lb ⋅ in. i !
FBD ABC: looking down
(b)
( 2 in.)( 45 lb ) − ( 5 in.) C = 0
ΣM B = 0:
ΣFz = 0:
FBD BEG:
45 lb − B + 18 lb = 0
By analogy, using FBD DEF
ΣFz = 0:
E = 63 lb k
C = 18 lb k
B = −63 lb k
F = 18 lb k
Gz + 63 lb − 63 lb = 0
Gz = 0
ΣFy = 0
Gy = 0
ΣM G = 0
M G − ( 6.5 in.)( 63 lb ) = 0
M G = ( 410 lb ⋅ in.) i !
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
FBD CFH:
ΣF = 0:
Hz = H y = 0
ΣM H = 0
M H = − ( 6.5 in.)(18 lb )
= −117 lb ⋅ in.
M G = − (117.0 lb ⋅ in.) i !
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 6, Solution 153.
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke
can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be
normal to the plane of the crosspiece.
If the crosspiece arm attached to shaft CF is horizontal, the plane of the crosspiece is normal to shaft AC, so
couple M C is along AC.
FBDs shafts with yokes:
(a)
ΣM x = 0:
FBD CDE:
M C cos30° − 50 N ⋅ m = 0
M C = 57.735 N ⋅ m
FBD BC: ΣM x′ = 0: M A − M C = 0
(b)
M A = 57.7 N ⋅ m
ΣM C = 0: M A i′ + ( 0.5 m ) Bz j′ − ( 0.5 m ) By′ k = 0
ΣF = 0: B + C = 0
so
B=0
C=0
FBD CDF : ΣM Dy = 0: − ( 0.6 m ) Ez + ( 57.735 N ⋅ m ) sin 30° = 0
E z = 48.1 N k
ΣFx = 0: Ex = 0
ΣM Dz = 0: ( 0.6 m ) E y = 0
Ey = 0
0
ΣF = 0: C + D + E = 0
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.
so E = ( 48.1 N ) k
D = −E = − ( 48.1 N ) k
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 154.
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke
can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be
normal to the plane of the crosspiece.
If the crosspiece arm attached to CF is vertical, the plane of the crosspiece is normal to CF, so the couple M C
is along CF.
(a)
FBD CDE:
ΣM x = 0: M C − 50 N ⋅ m = 0
FBD BC:
ΣM x′ = 0: M A − M C cos 30° = 0
M C = 50 N ⋅ m
M A = ( 50 N ⋅ m ) cos 30°
M A = 43.3 N ⋅ m
(b)
ΣM Cy′ = 0: M C sin 30° + ( 0.5 m ) Bz = 0
ΣM Cz = 0: − ( 0.5 m ) By = 0
ΣF = 0: B + C = 0
FBD CDE:
Bz = −
( 50 N ⋅ m )( 0.5) = −50 N
0.5 m
so B = − ( 50.0 N ) k
By = 0
C = −B
so
C = ( 50 N ) k on BC
ΣM Dy = 0: − ( 0.4 m ) C z − ( 0.6 m ) Ez = 0
4
Ez = − ( 50 N )   = −33.3 N
6
ΣM Dz = 0: E y = 0
ΣFx = 0: Ex = 0
ΣF = 0: C + D + E = 0
so E = − ( 33.3 N ) k
− ( 50 N ) k + D − ( 33.3 N ) k = 0
D = ( 83.3 N ) k
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 6, Solution 155.
FBD Joint A:
By inspection of FBD of entire device, FA = w = 1500 lb
By symmetry, FAB = FAC
FBD CDF:
ΣFy = 0:
 1

FAC  = 0,
1500 lb − 2 
 5

ΣFx = 0:
Dx − Fx −
(
FAC = 750 5 lb
)
2
750 5 lb = 0,
5
Dx − Fx = 1500 lb
ΣFy = 0:
(
(1)
)
1
750 5 lb = 0,
5
Fy − D y +
Dy − Fy = 750 lb
ΣM D = 0:
FBD EF:
(2)
 1

 2

(4.6 ft) 
750 5 lb  + ( 0.6 ft ) 
750 5 lb 
 5

 5

(
)
(
)
− (1 ft ) Fy − ( 3.6 ft ) Fx = 0,
3.6 Fx + Fy = 4350 lb
By symmetry:
ΣM F = 0:
Ex = Dx ,
E y = Dy
Hy = Jy =
w
= 750 lb,
2
(4), (5)
H y = 750 lb
!
( 3 ft ) E y − ( 3.6 ft ) Ex − (1 ft ) H y + ( 0.6 ft )( 750 lb ) = 0
3.6Ex − 3E y + H y = 450 lb
ΣFx = 0:
(3)
Dx + Fx − H x = 0,
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.
(6)
(7)
COSMOS: Complete Online Solutions Manual Organization System
Solving Eqns, (1)–(7), on EFH
Fx = 540 lb
!
Fy = 2410 lb
!
E x = 2040 lb
E y = 3160 lb
H x = 2580 lb
H y = 750 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.
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 156.
FBD Truss:
ΣM E = 0: ( 9 ft ) Fy − ( 6.75 ft )( 4 kips ) − (13.5 ft )( 4 kips ) = 0
Fy = 9 kips
ΣFy = 0: − E y + 9 kips = 0
E y = 9 kips
ΣFx = 0: − Ex + 4 kips + 4 kips = 0
E x = 8 kips
By inspection of joint E:
FEF = 8.00 kips T !
Joint FBDs:
Joint F:
FEC = 9.00 kips T !
By inspection of joint B:
FAB = 0 !
FBD = 0 !
ΣFx = 0:
Joint C:
4
FCF − 8 kips = 0
5
ΣFy = 0: FDF −
3
(10 kips) = 0
5
ΣFx = 0: 4 kips −
FCF = 10.00 kips C !
FDF = 6.00 kips T !
4
(10 kips ) + FCD = 0
5
FCD = 4.00 kips T !
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
ΣFy = 0: FAC − 9 kips +
3
(10 kips ) = 0
5
FAC = 3.00 kips T !
Joint A:
ΣFx = 0: 4 kips −
4
FAD = 0
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.
FAD = 5.00 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 157.
FBD Truss:
ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0
L y = 0.75 kip
FJK = 0 !
Inspection of joints K , J , and I , in order, shows that
FIJ = 0 !
FHI = 0 !
Joint FBDs:
FIK = FKL ; FHJ = FJL and FGI = FIK
and that
0.75
F
F
= JL = KL
1
8
5
FJL = 2.1213 kips
FKL = 1.6771 kips
FJL = 2.12 kips C !
FKL = 1.677 kips T !
FHJ = 2.12 kips C !
and, from above:
FGI = FIK = 1.677 kips T !
ΣFx = 0:
2
1
( FFH + FGH ) − ( 2.1213 kips ) = 0
5
2
ΣFy = 0:
1
1
( FGH + FFH ) + ( 2.1213 kips ) = 0
5
2
Solving:
FFH = 2.516 kips
FGH = − 0.8383 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.
FFH = 2.52 kips C !
FGH = 0.838 kips T !
COSMOS: Complete Online Solutions Manual Organization System
ΣFx = 0:
2
( FDF − 2.516 kips ) = 0
5
ΣFy = 0: FFG − 0.5 kip +
FDF = 2.52 kips C
1
( 2 )( 2.516 kips ) = 0
5
FFG = 1.750 kips T !
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 6, Solution 158.
FBD Truss:
ΣM F = 0: (10 m ) G y − ( 7.5 m )( 80 kN ) − ( 8 m )( 30 kN ) = 0
G y = 84 kN
ΣFx = 0: − Fx + 30 kN = 0
Fx = 30 kN
ΣFy = 0: Fy + 84 kN − 80 kN = 0
Fy = 4 kN
FEG = 0 !
By inspection of joint G:
FCG = 84 kN C !
84 kN
F
F
= CE = AC = 10.5 kN
8
61
29
Joint FBDs:
FCE = 82.0 kN T !
FAC = 56.5 kN C !
ΣFy = 0:
2
6
FAE +
(82.0 kN ) − 80 kN = 0
5
61
FAE = 19.01312
ΣFx = 0: − FDE −
1
(19.013 kN ) +
5
FDE = 43.99 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.
FAE = 19.01 kN T !
5
(82.0 kN ) = 0
61
FDE = 44.0 kN T !
COSMOS: Complete Online Solutions Manual Organization System
ΣFx = 0:
1
FDF − 30 kN = 0
5
FDF = 67.082 kN
ΣFy = 0:
FDF = 67.1 kN T !
2
( 67.082 kN ) − FBF − 4 kN = 0
5
FBF = 56.00 kN
ΣFx = 0: 30 kN +
5
FBD −
61
FBF = 56.0 kN C !
5
FAB = 0
29
ΣFy = 0: 56 kN −
6
FBD −
61
Solving:
FBD = 42.956 kN
FBD = 43.0 kN T !
FAB = 61.929 kN
FAB = 61.9 kN C !
ΣFy = 0:
3
FAB = 0
29
6
2
( 42.956 N ) + ( FAD − 67.082 N ) = 0
61
5
FAD = 30.157 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.
FAD = 30.2 N T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 159.
FBD Truss:
ΣFx = 0: A x = 0
By load symmetry, A y = I y = 8 kips
FBD Section:
ΣM D = 0:
( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0
FCE = 14 kips
ΣM E = 0:
FCE = 14.00 kips T
( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips )
( 4.5 ft )
FDF =
7
FDF = 0
51.25
8 51.25
kips
4.5
ΣFy = 0: 8 kips − 2 kips − 4 kips +
FDF = 12.73 kips C
1.5 8 51.25
kips −
51.25 4.5
FDE = −1.692 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.
3
FDE = 0
58
FDE = 1.692 kips C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 160.
FBD Section:
p =
Note: BG
ΣM A = 0:
8.4
( 2.7 m ) = 1.89 m
12.0
( 2.7 m ) FFG − ( 3.6 m )(8 kN ) − ( 7.8 m )(8 kN ) − (12 m )( 4 kN ) = 0
FFG = 51.56 kN
ΣM G = 0:
FFG = 51.6 kN C
(1.89 m ) 
12

FAB  − ( 4.2 m )( 8 kN ) − ( 8.4 m )( 4 kN ) = 0
 12.3

FAB = 36.44 kN
ΣM D = 0:
FAB = 36.4 kN T
( 4.2 m )(8 kN ) + (8.4 m )(8 kN ) − ( 8.4 m )  FAG  = 0
3
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.

FAG = 20.0 kN T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 161.
FBD Truss:
ΣM A = 0:
( 24 ft ) K x − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) = 0
K x = 4.5 kips
FBD Section:
ΣM F = 0:
(12 ft )( 4.5 kips ) − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) + ( 36 ft ) FEJ = 0
FEJ = 1.500 kips T
ΣM J = 0:
(18 ft )( 2 kips ) + (12 ft )( 4.5 kips ) − ( 36 ft ) FAF = 0
FAF = 2.50 kips T
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 6, Solution 162.
FBD Frame:
ΣM F = 0:
(10.8 ft ) Ay − (12 ft )( 4.5 kips ) = 0
A y = 5.00 kips
ΣFx = 0:
− Ax + 4.5 kips = 0
A x = 4.50 kips
FBD member ABC:
Note: BE is a two-force member
ΣM C = 0:
(12 ft ) FBE + (10.8 ft )( 5 kips ) − (18 ft )( 4.5 kips ) = 0
FBE = 2.25 kips
ΣFx = 0:
Cx + 2.25 kips − 4.5 kips = 0
C x = 2.25 kips
ΣFy = 0:
C y − 5 kips = 0
C y = 5.00 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 6, Solution 163.
FBD Frame:
ΣM E = 0:
( 3.75 in.) Bx + (8.75 in.)( 75 lb ) = 0
Bx = 175 lb
FBD member ACE:
B x = 175.0 lb
ΣFx = 0:
Ex − Bx = 0
ΣFy = 0:
E y + By − 75 lb = 0
E x = 175.0 lb
By = 75 lb − E y
ΣM C = 0:
− (1.25 in.)( 75 lb ) + ( 3.75 in.)(175 lb ) − ( 4.5 in.) E y = 0
E y = 125.0 lb
Thus
By = 75 lb − 125 lb = −50 lb
B y = 50.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 164.
FBD Frame:
Q = 65 N
P = 15 N
ΣM D = 0:
( 0.25 m )( P + Q ) − (.15 m )( P + Q ) − ( 0.08 m ) Ex = 0
Ex = 1.2 ( P + Q ) = 100 N
ΣFx = 0:
(b)
Dx − Ex = 0 = Dx − 100 N
(a)
ΣFy = 0:
E x = 100.0 N
Dx = 100 N
D x = 100.0 N
E y − Dy − 2P − 2Q = 0
E y = D y + 2 ( P + Q ) = Dy + 160 N
FBD member BF:
ΣM C = 0:
( 0.15 m )( 65 N ) − ( 0.1 m ) Dy − ( 0.04 m )(100 N )
− ( 0.25 m )(15 N ) = 0
Dy = 20 N
From above
(a)
D y = 20.0 N
E y = 20 N + 160 N = 180 N (b) E y = 180.0 N
ΣFx = 0:
− C x + 100 N = 0
(a)
ΣFy = 0:
C x = 100.0 N
− 65 N + C y − 20 N − 15 N = 0
(a)
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.
C y = 100.0 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 165.
Member FBDs:
FBD I:
ΣM C = 0:
R ( FBD sin 30° )
−  R (1 − cos 30° )  ( 200 N ) − R ( 100 N ) = 0
FBD = 253.6 N
ΣFx = 0:
(b)
FBD = 254 N T
− C x + ( 253.6 N ) cos 30° = 0
C x = 219.6 N
ΣFy = 0:
C y + ( 253.6 N ) sin 30° − 200 N − 100 N = 0
C y = 173.2 N
(c) so C = 280 N
ΣM A = 0:
aP − a ( 253.6 N ) cos 30°  = 0
FBD II:
(a)
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.
38.3°
P = 220 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 166.
FBD jaw AB:
ΣFx = 0: Bx = 0
ΣM B = 0:
( 0.01 m ) G − ( 0.03 m ) A = 0
A=
ΣFy = 0:
G
3
A + G − By = 0
By = A + G =
4G
3
FBD handle ACE:
By symmetry and FBD jaw DE: D = A =
E y = By =
ΣM C = 0:
G
, Ex = Bx = 0,
3
4G
3
( 0.105 m )( 240 N ) + ( 0.015 m )
G
4G
− ( 0.015 m )
=0
3
3
G = 1680 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 6, Solution 167.
Note: By symmetry the vertical components of pin forces C and D are
zero.
FBD handle ACF: (not to scale)
ΣFy = 0:
Ay = 0
ΣM C = 0:
(13.5 in.)( 20 lb ) − (1.5 in.) Ax = 0
ΣFx = 0:
C − Ax − 20 lb = 0 C = (180 + 20 ) lb = 200 lb
Ax = 180 lb
FBD blade DE:
ΣM D = 0:
( 9 in.) E − ( 3 in.)(180 lb ) = 0
E = 60.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 1.
FBD FRAME:
ΣM A = 0:
( 0.3 m ) Dx − ( 0.4 m )( 900 N ) = 0
∴ D x = 1200 N
FBD JEHDF:
ΣFx = 0:
1200 N + V = 0
V = −1200 N
ΣFy = 0: F = 0
ΣM J = 0:
( 0.15 m )(1200 N ) − M = 0
M = + 180 N ⋅ m
Thus,
( on JE )
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.
F =0
V = 1200 N
M = 180.0 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 2.
FBD FRAME:
ΣM A = 0:
( 0.95 m )( 480 N ) − ( 0.25 m ) Dx = 0,
D x = 1824 N
ΣFy = 0:
FBD ABC:
Ay + Dy + 480 N = 0
(1)
Note: BE is a two-force member
ΣM B = 0:
( 0.75 m ) Ay = 0
Then, from (1) above,
Ay = 0
Dy = − 480 N
D y = 480 N
FBD sect. DJ:
ΣFx = 0:
F − 1824 N = 0
ΣFy = 0:
− 480 N – V = 0
ΣM J = 0:
F = 1.824 kN
V = − 480 N
V = 480 N
M = 120.0 N ⋅ m
M + ( 0.25 m )( 480 N ) = 0
M = −120 N ⋅ m
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 7, Solution 3.
FBD CEF:
ΣFx = 0: C x −
ΣΜ E = 0:
1
(180 N )
2
C x = 90 2 N
( 0.2 m ) C y − ( 0.05 m ) ( 90 2 N )
− ( 0.15 m + 0.08 m )(180 N ) = 0
C y = 126 2 N
FBD sect. CJ:
ΣFx = 0: 90 2 N − F = 0
ΣFy = 0: −126 2 N + V = 0
ΣM J = 0:
F = 127.3 N
V = 178.2 N
( 0.05 m ) ( 90 2 N ) − ( 0.10 m ) (126 2 N ) − M = 0
M = 11.46 N ⋅ m
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 7, Solution 4.
FBD Frame:
Note: AC is a two-force member, resolve FAC at C:
ΣM E = 0: ( 0.25 m + 0.25 m )
1
FAC − ( 0.15 m )( 320 N ) = 0
2
FAC = 96 2 N
FBD sect. AB:
ΣFx = 0:
− 96 N +
ΣFy = 0: 96 N +
3
7
V +
F =0
4
4
7
3
V − F =0
4
4
Solving:
V = 8.50 N
41.4° F = 135.5 N
46.8° 4− 7 
ΣM B = 0: M − ( 0.3 m )( 96 N ) + 
 10 m  ( 96 N ) = 0


M = 15.799 N ⋅ m,
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.
M = 15.80 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 5.
FBD Frame:
Note: AB is a two-force member, so
Ay
Ax
=
12
5
ΣM C = 0:
(1)
(15 in.) Ax − ( 24 in.)( 78 lb ) = 0
A x = 124.8 lb
From (1) above,
A y = 52.0 lb
FBD AJ:
ΣFx = 0:
−124.8 lb + F = 0
ΣFy = 0:
52 lb − V = 0
ΣM J = 0:
M − (10 in.)( 52 lb ) = 0
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.
F = 124.8 lb
V = 52.0 lb
M = 520 lb ⋅ in.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 6.
FBD CD:
Note: AB is a two-force member
ΣΜ C = 0:
5

 12

FAB  + ( 7.5 in.)  FAB 
 13

 13

(18 in.) 
− ( 24 in.)( 78 lb ) = 0
FAB = 135.2 lb
ΣFx = 0:
Cx −
12
(135.2 lb ) = 0
13
C x = 124.8 lb
ΣFy = 0:
Cy +
5
(135.2 lb ) − ( 78 lb ) = 0
13
C y = 26 lb
FBD CK:
ΣFx′ = 0:
−F +
12
5
(124.8 lb ) + ( 26 lb ) = 0
13
13
F = 125.2 lb
ΣFy′ = 0:
V −
5
12
(124.8 lb ) + ( 26 lb ) = 0
13
13
V = 24.0 lb
ΣM K = 0:
22.6° 67.4° ( 5 in.)(124.8 lb ) − (12 in.)( 26 lb ) − M = 0
M = 312 lb ⋅ in.
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 7, Solution 7.
FBD half-section:
By symmetry,
Ay = B y =
W
2
Where W = ( 9 kg )( 9.81 N/kg ) = 88.29 N
FBD AJ:
ΣFx = 0: F = 0
ΣFy = 0:
F =0
W W
−
−V = 0
2
2
ΣM J = 0: M − ( r − x )
but
x=
V =0
W
=0
2
2 W

= 2.406 N ⋅ m
, so M = 1 −  r
π 2
π

2r
M = 2.41 N ⋅ m
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 7, Solution 8.
FBD AJ:
Note: Cut is just left of contact with ground.
W = ( 9 kg )( 9.81 N/kg ) = 88.29 N
Also
x=
ΣFx = 0:
ΣFy = 0:
2r
π
and
r = 0.15 m
F =0
−
ΣM J = 0:
M =
F=0 W
+V =0
2
x
W
−M =0
2
2
( 0.15 m ) 
π
88.29 N 

2


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.
V = 44.1 N
M = 4.22 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 9.
SOLUTION
FBD AB:
FBD sect. BJ:
(
)
B y = 12 lb
ΣM A = 0:
r By − 12 lb = 0
ΣFy′ = 0:
(12 lb ) cos 30° − (12 lb ) sin 30° − F = 0
F = 4.39 lb
ΣFx′ = 0:
(12 lb ) cos 30° + (12 lb ) sin 30° − V = 0
V = 16.39 lb
ΣM J = 0:
60° 30° ( 4 in.) sin 30° (12 lb ) + ( 4 in.)(1 − cos 30° )(12 lb ) − M = 0
M = 30.4 lb ⋅ in.
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 7, Solution 10.
FBD AB:
ΣM A = 0:
ΣΜ J = 0:
FBD BJ:
(
)
r By − 12 lb = 0
B y = 12 lb
( 4 in.) sin θ (12 lb ) + ( 4 in.)(1 − cosθ )(12 lb ) − M = 0
M = ( 48 lb ⋅ in.)(1 − cosθ + sin θ )
(a) to maximize M, set
(1)
dM
=0
dθ
dM
= ( 48 lb ⋅ in.)( sin θ + cosθ ) = 0
dθ
so tan θ = −1
θ = − 45°, 135°
(only θ = 135° is on ring)
(b)
ΣFy′ = 0:
F + (12 lb ) cosθ − (12 lb ) sin θ = 0
θ = 135°, so
F = 16.97 lb@135°
ΣFx′ = 0:
θ = 135° (12 lb ) cosθ + (12 lb ) sin θ − V = 0
F = 16.97 lb
45° θ = 135°, V = 0 From (1) above, M = ( 48 lb ⋅ in.)(1 − cos135° + sin135° ) = 115.88 lb ⋅ in.
M = 115.9 lb ⋅ in.
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 7, Solution 11.
FBD Frame:
ΣM A = 0:
(16.4 in.)
24
FEC − (12.6 in.)(120 lb ) = 0
25
FEC = 93.75 lb
FBD CBD:
ΣΜ C = 0:
(16.2 in.)(120 lb ) − (13.5 in.) By = 0
By = 144 lb
ΣFy = 0:
144 lb +
7
( 93.75 lb ) − 120 lb − C y = 0
25
C y = 50.25 lb
ΣFx = 0:
Cx −
24
( 93.75 lb ) = 0
27
C x = 90.0 lb
FBD CJ:
ΣFy′ = 0:
F + ( 50.25 lb ) cos30° − ( 90.0 lb ) sin 30° = 0
F = 1.482 lb
ΣFx′ = 0:
V − ( 90 lb ) cos 30° + ( 50.25 lb ) sin 30° = 0
V = 103.1 lb
ΣM O = 0:
60° 30° (8.4 in.)( 50.25 lb + 1.482 lb ) − M = 0
M = 435 lb ⋅ in.
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 7, Solution 12.
FBD Frame:
ΣM A = 0:
(16.4 in.)
24
FEC − (12.6 in.)(120 lb ) = 0
25
FEC = 93.75 lb
FBD CBD:
ΣM C = 0:
(16.2 in.)(120 lb ) − (13.5 in.) By = 0
B y = 144 lb
FBD DK:
ΣFx′ = 0:
(144 lb − 120 lb ) sin 30° − F = 0
F = 12.00 lb
ΣFy′ = 0:
(144 lb − 120 lb ) cos 30° − V = 0
V = 20.8 lb
ΣΜ O = 0:
30° 60° (8.4 in.)(12 lb − 144 lb ) + (11.1 in.)(120 lb ) − M = 0
M = 223 lb ⋅ in.
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 7, Solution 13.
FBD AB:
( 0.8 m )(1.8 kN ) − ( 0.24 m ) Ax = 0
ΣΜ B = 0:
A x = 6.0 kN
Geometry:
y = kx 2 ; at B, 0.24 m = k ( 0.8 m )
so
k = 0.375
2
1
m
1
2

J , y J =  0.375  ( 0.48 m ) = 0.0864 m
m

at
dy
= 2kx
dx
slope of parabola
at J ,
dy
1

= 2  0.375  ( 0.48 m ) = 0.36 = tan θ J
dx
m

θ J = 19.799°
FBD AJ:
ΣFx′ = 0:
( 6 kN ) cos19.799° − (1.8 kN ) sin19.799° − F = 0
F = 6.26 kN
ΣFy′ = 0:
( 6 kN ) sin19.799° − (1.8 kN ) cos19.799° − V = 0
V = 0.3387 kN
ΣM J = 0:
19.80° V = 339 N
70.2° ( 0.48 m )(1.8 kN ) − ( 0.0864 m )( 6 kN ) − M = 0
M = 0.3456 kN ⋅ m
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.
M = 346 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 14.
FBD AB:
ΣΜ B = 0:
LP − hAx = 0, Ax =
L
P
h
Geometry:
h
hx 2
y
,
=
L2
L2
y = kx 2 , at B: h = kL2 , so k =
at J: y J = ka 2 =
slope =
FBD AJ:
ΣM J = 0:
ha 2
L2
dy
= 2kx,
dx
aP −
at J : slope =
2ha
= tan −1 θ J
2
L
ha 2 LP
−M =0
L2 h

a2 
M = P  a −

L 

dM
=0
da
a

or P 1 − 2  = 0,
L

To maximize set
a=
L
2
2

L 




L
PL
2
Then M max = P  −    =
2
L 
4




PL
4
at a =
L
2
M max =
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 7, Solution 15.
FBD Frame with pulley and cord:
ΣΜ Α = 0:
(1.8 m ) Bx − ( 2.6 m )( 360 N ) − ( 0.2 m )( 360 N ) = 0
B x = 560 N
FBD BE:
Note: Cord forces have been moved to pulley hub as per Problem 6.91.
ΣΜ Ε = 0:
(1.4 m )( 360 N ) + (1.8 m )( 560 N ) − ( 2.4 m ) By = 0
By = 630 N
FBD BJ:
ΣFx′ = 0:
F + 360 N −
3
4
( 630 N − 360 N ) − ( 560 N ) = 0
5
5
F = 250 N
ΣFy′ = 0:
V +
4
3
( 630 N − 360 N ) − ( 560 N ) = 0
5
5
V = 120.0 N
ΣM J = 0:
36.9° 53.1° M + ( 0.6 m )( 360 N ) + (1.2 m )( 560 N )
− (1.6 m )( 630 N ) = 0
M = 120.0 N ⋅ m
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 7, Solution 16.
FBD Frame with pulleys and cord:
ΣM B = 0:
(1.8 m ) Ax − ( 2.0 m )( 360 N ) − ( 2.6 m )( 360 N ) = 0
A x = 920 N
FBD AE:
Note: Cord forces have been moved to pulley hub as per Problem 6.91.
ΣM E = 0:
( 2.4 m ) Ay − (1.8 m )( 360 N ) = 0
A y = 270 N
FBD AK:
ΣFx = 0:
920 N − 360 N − F = 0
F = 560 N
ΣFy = 0:
− 270 N + 360 N − V = 0
V = 90.0 N
ΣM K = 0:
(1.6 m )( 270 N ) − (1.0 m )( 360 N ) − M = 0
M = 72.0 N ⋅ m
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 7, Solution 17.
FBD Frame:
ΣM A = 0:
(1 m ) Dy − ( 2.675 m )( 360 N ) = 0
D y = 963 N
FBD DEF:
ΣM F = 0:
(1 m )( 963 N ) − ( 2.4 m ) Dx − ( 0.125 m )( 360 N ) = 0
D x = 3825 N
FBD DJ:
ΣFy′ = 0:
F−
12
5
( 963 N ) − ( 382.5 N ) = 0
13
13
F = 1036 N
ΣFx′ = 0:
F = 1.036 kN
5
12
( 963 N ) − ( 382.5 N ) − V = 0
13
13
V = 17.31 N
ΣM J = 0:
67.4°
22.6°
( 0.5 m )( 963 N ) − (1.2 m )( 382.5 N ) − M = 0
M = 22.5 N ⋅ m
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 7, Solution 18.
FBD AC:
Note: Cord forces moved to pulley hub as per Problem 6.91.
To determine θ the coordinates of C are
xC = 2.55 m, yC =
4
( 2.55 m ) = 3.40 m, and
3
xG = 0, yG = 3.75 m
∴ θ = tan −1
3.75 m − 3.40 m
= 7.8153°
2.55 m
ΣM A = 0:
( 3.75 m ) ( 360 N ) cos 7.8153°
− ( 2.55 m )( 360 N ) − ( 3 m ) FBG = 0
FBG = 139.820 N
FBD KC:
ΣFy′ = 0:
F−
4
4
( 360 N ) − ( 360 N ) cos  7.8153° + tan −1  = 0
5
3

F = 462.8 N
ΣFx′ = 0:
− V + 139.820 N +
F = 463 N
53.1°
3
( 360 N )
5
4

− ( 360 N ) sin  7.8153° + tan −1  = 0
3

V = 41.1 N
ΣM K = 0:
− M − (1.5 m )(139.820 N ) −
36.9°
4
( 2.75 m )( 360 N )
5
4

+ ( 2.75 m )( 360 N ) sin  7.8153° + tan −1  = 0
3

M = 61.7 N ⋅ m
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 7, Solution 19.
FBD Frame and pipe:
W = (10 ft )(18.5 lb/ft ) = 185 lb
ΣM A = ( 24 in.) Cx − (12.6 in.)185 lb = 0
C x = 97.125 lb
FBD pipe:
By symmetry N E = N D = N
ΣFy = 0:
 21 
2  N  − 185 N = 0, N = 127.738 lb
 29 
Also note
20 
20 8

= in.
a = r tan  tan −1  = ( 2.8 in.)
21
21 3


FBD BC:
ΣM B = 0:
8 
 in.  (127.735 lb ) + (12 in.)( 97.125 lb )
3 
− (12.6 in.) C y = 0
C y = 119.534 lb
FBD CJ:
ΣFx′ = 0:
F −
21
20
( 97.125 lb ) − (119.534 lb ) = 0
29
29
F = 152.8 lb
ΣFy′ = 0:
−V −
20
21
( 97.125 lb ) + (119.534 lb ) = 0
29
29
V = 19.58 lb
ΣM C = 0:
46.4° 44.6° M − ( 8.7 in.)(19.58 lb ) = 0
M = 170.3 lb ⋅ in.
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 7, Solution 20.
FBD Frame:
W = (10 ft )(18.5 lb/ft ) = 185 lb
ΣM C = 0:
( 24 in.) Ax − (12.6 in.)(185 lb ) = 0
A x = 97.125 lb
FBD Pipe
By symmetry N E = N D
ΣFy = 0:
Also note
FBD AD:
ΣM B = 0:
 21

2  N D  − 185 lb = 0,
 29

N D = 127.738 lb
20 

 20  8
a = r tan  tan −1
 = ( 2.8 in.)   = in.
21


 21  3
(12.6 in.)( 97.125 lb ) − (12.6 in.) Ay
8 
−  in.  (127.738 lb ) = 0
3 
A y = 65.465 lb
FBD AK:
ΣFx′ = 0:
ΣFy′ = 0:
ΣM A = 0:
21
20
( 97.125 lb ) + ( 65.465 lb ) − F = 0
29
29
F = 115.5 lb
44.6° 20
21
( 97.125 lb ) − ( 65.465 lb ) + V = 0
29
29
V = 19.58 lb
46.4° M − ( 8.7 in.)(19.58 lb ) = 0
M = 170.3 lb ⋅ in.
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 7, Solution 21.
(a)
FBD Rod:
ΣFx = 0:
ΣM D = 0:
Ax = 0
aP − 2aAy = 0
Ay =
P
2
ΣFx = 0: V = 0
FBD AJ:
ΣFy = 0:
P
−F =0
2
F=
P
2
ΣM J = 0: M = 0
(b)
FBD Rod:
ΣM A = 0
4 
3 
2a  D  + 2a  D  − aP = 0
5 
5 
ΣFx = 0:
ΣFy = 0:
Ax −
D=
4 5
P=0
5 14
Ay − P +
3 5
P=0
5 14
5P
14
Ax =
Ay =
2P
7
11P
14
continued
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
2
P −V = 0
7
ΣFx = 0:
FBD AJ:
V =
11P
−F =0
14
ΣFy = 0:
ΣM J = 0:
(c)
FBD Rod:
FBD AJ:
2P
7
a
F=
11P
14
M =
2
aP
7
2P
−M =0
7
ΣM A = 0:
a  4D 

 − aP = 0
2 5 
D=
ΣFx = 0:
Ax −
4 5P
=0
5 2
Ax = 2 P
ΣFy = 0:
Ay − P −
ΣFx = 0:
3 5P
=0
5 2
Ay =
5P
2
5P
2
2P − V = 0
V = 2P
ΣFy = 0:
5P
−F =0
2
F=
ΣM J = 0:
5P
2
a ( 2P ) − M = 0
M = 2aP
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 7, Solution 22.
(a)
FBD Rod:
ΣM D = 0:
aP − 2aA = 0
P
2
A=
ΣFx = 0:
V −
P
=0
2
V =
FBD AJ:
ΣFy = 0:
ΣM J = 0:
F =0
M −a
P
=0
2
M =
(b)
FBD Rod:
ΣM D = 0:
A=
P
2
aP −
aP
2
a4 
 A = 0
25 
5P
2
continued
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
FBD AJ:
ΣFx = 0:
3 5P
−V = 0
5 2
V =
ΣFy = 0:
3P
2
4 5P
−F =0
5 2
F = 2P
M =
(c)
FBD Rod:
ΣM D = 0:
3 
4 
aP − 2a  A  − 2a  A  = 0
5


5 
A=
ΣFx = 0:
5P
14
 3 5P 
V −
=0
 5 14 
V =
ΣFy = 0:
3
aP
2
3P
14
4 5P
−F =0
5 14
F=
ΣM J = 0:
2P
7
 3 5P 
M − a
=0
 5 14 
M =
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.
3
aP
14
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 23.
FBD Rod:
FBD JB:
ΣM A = 0:
Note: r =
2r 

rB −  r −
W =0
π 

r sin
φ
φ
2 =
2
x=
ΣM J = 0:
r sin
30°
2

B = 1 − W
π

60°
2 = 3r
π
180°
π
3r 1 3r r 
3
−
=  3− 
2
2π
2
π
M +
r
2
r
3  2 
1 −  W −  3 −  W  = 0
π
π  3 
2
2
M = 0.774Wr
on BK
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 7, Solution 24.
FBD Rod:
2r 
 1  
2r 
B − r −
W =0
π 
 2  
ΣM A = 0:
1 1
B = 2  − W
2 π 
FBD JB:
 60° 
r sin 

 2  = 3r
r =
30°
π
180°
π
x = r sin 60° − r sin 30°
x=
1
3
 3 − r
π
2
21 1
 r sin 60° + r (1 − cos 60° )  2  2 − π W


ΣM J = 0:
−
1
3  2W
−M =0
 3 − r
π 3
2
M = − 0.01085Wr
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.
M = 0.01085Wr
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 25.
FBD Rod:
ΣFy = 0:
Ay − W = 0
Ay = W
ΣM B = 0:
r ( Ax − W ) +
2r
π
W =0
2

A x = 1 − W
π

FBD AJ:
W′ =
θ
2θ
W =
W
π
π
2
r =r
ΣM J = 0:
sin
θ
2 = 2r sin θ
θ
θ
2
2
2

r sin θ 1 − W − r (1 − cosθ )W
π

 2r
 2θ
θ θ
W −M =0
+  sin  cos − r cosθ 
θ
2
2


 π
2θ


M = Wr  sin θ − 1 + cosθ −
cosθ 
π


For Mmax,
dM
2
2θ


sin θ  = 0
= Wr  cosθ − sin θ − cosθ +
dθ
π
π


or tan θ =
π −2
π − 2θ
Solving numerically:
θ = 0.48338 rod = 27.7°
M max = 0.0777Wr
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 7, Solution 26.
FBD Rod:
2

1 1
2 rB − r 1 − W = 0
B = 2  − W
π
2 π 

1
1 1
1 1
2  − W = 0 A x =  − W
ΣFx = 0:
Ax −
2
π
2
2 π 


1
1 1
1 1
2  − W = 0 A y =  + W
ΣFy = 0:
Ay − W +
2
2 π 
2 π 
FBD AJ:
W′ =
ΣM A = 0:
θ
2θ
W =
W
π
π
2
2r
θ
r =
sin
θ
2
ΣM J = 0:
1 1
1 1
M + r sin θ  − W − r (1 − cosθ )  +  W
π
2


2 π 
 2r
  2θ 
θ
θ
+  sin cos  − r cosθ   W  = 0
θ
2
2



 π
 1 1 

2θ
M = Wr  +  (1 − sin θ − cosθ ) +
cosθ 
π
 2 π 

For M max ,
 1 1 

dM
2
2θ
sin θ  = 0
= Wr  +  ( − cosθ + sin θ ) + cosθ −
dθ
π
π
 2 π 

or tan θ =
π −2
π + 2 − 4θ
Solving numerically θ = 0.27539 rad = θ1
or θ = 1.16164 rad = θ 2
M (θ1 ) = − 0.0230Wr , M (θ 2 ) = 0.0362Wr
so M max = 0.0362Wr
at θ = θ 2 = 66.6°
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 7, Solution 27.
Note: α =
180° − 60°
π
= 60° =
2
3
r =
r
α
sin α =
3r
3
3 3
r
=
π 2
2π
Weight of section = W
ΣFy′ = 0:
F−
ΣM O = 0:
120
4
= W
270 9
4
W cos30° = 0
9
rF − ( r sin 60° )
F =
2 3
W
9
4W
−M =0
9
2 3 3 3 3 4
2 3 1 
−
−  Wr
M = r
W = 
π 
2π 2 9 
 9
 9
M = 0.0666Wr
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 7, Solution 28.
ΣFx = 0:
FBD Rod:
ΣM B = 0:
FBD AJ:
rAy +
2r W 2r 2W
−
=0
π 3
π 3
Ay =
2W
3π
α =
Note:
Ax = 0
60°
π
= 30° =
2
6
Weight of segment = W
F =
ΣM J = 0:
M =
r
α
sin α =
( r cosα − r sin 30° )
60
2W
=
270
9
r
3r
sin 30° =
π /6
π
2W
2W
+ ( r − r sin 30° )
−M =0
9
3π
 3 1
2W  3r 3 r
3r 
1 
− +
− +

 = Wr 

9 π 2
2 2π 
9 3π 
 3π
M = 0.1788Wr
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 7, Solution 29.
(a) FBD Beam:
ΣM C = 0:
ΣFy = 0:
LAy − M 0 = 0
Ay =
M0
L
C=
M0
L
− Ay + C = 0
Along AB:
ΣFy = 0:
−
ΣM J = 0:
M0
−V = 0
L
x
V =−
M0
+M =0
L
M0
L
M =−
straight with M = −
M0
at B
2
M0
−V = 0
L
V =−
M0
x
L
Along BC:
ΣFy = 0:
ΣM K = 0:
−
M +x
M0
− M0 = 0
L
straight with M =
(b) From diagrams:
M0
L
x

M = M 0 1 − 
L


M0
at B
2
M = 0 at C
V max =
M0
everywhere
L
M max =
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.
M0
at B
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 30.
FBD Beam:
(a)
ΣFx = 0:
Ax = 0, by symmetry Ay = By = P
ΣFy = 0:
Along AB:
P −V = 0
V = P
ΣM J = 0:
M − xP = 0
M = Px
Along BC:
ΣFy = 0:
P − P −V = 0
V =0
ΣM K = 0:
M − xP + ( x − a ) P = 0
M = Pa
Along CD:
ΣFy = 0:
P − P − P −V = 0
V = −P
ΣM L = 0:
M − xP + ( x − a ) P
+ ( x − L + a) P = 0
M = P ( L − x)
Note: Symmetry in M diag. follows symmetry of FBD
(b)
V max = P along AB and CD M max = Pa along BC 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 7, Solution 31.
FBD Section:
ΣFy = 0: − V −
1  x
x  w0  = 0
2  L
V =−
(a)
1 w0 2
x
2 L
1
V ( L ) = − w0 L
2
ΣM J = 0: M +
1 1 
x 
x  x  w0   = 0
3  2  L 
M =−
1 w0 3
x
6 L
M ( L) = −
1
w0 L2
6
(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.
V max =
1
w0 L at B 2
M max =
1
w0 L2 at B 6
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 32.
(a) FBD Beam:
ΣFx = 0:
ΣM B = 0:
Bx = 0
aP + 2aC − 3.5aP = 0
C = 1.25P
ΣFy = 0:
− P + By + 1.25P − P = 0
By = 0.75P
Along AB:
ΣFy = 0:
−P − V = 0
V = −P
ΣM J = 0:
M + xP = 0
M = − Px
Along BC:
ΣFy = 0:
− P + 0.75P − V = 0
V = − 0.25P
ΣM K = 0:
M + xP − ( x − a )( 0.75P ) = 0
M ( 3a ) = −1.5Pa
M = − 0.75Pa − 0.25Px
Along CD:
ΣFy = 0:
V −P=0
ΣM L = 0:
− M − x1 P = 0
V =P
M = − Px1
M (1.5a ) = −1.5Pa
V max = P along AB and CD
(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.
M max = 1.5Pa at C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 33.
(a) FBD Beam:
ΣFx = 0:
Cx = 0
ΣM C = 0:
( 3.6 ft )(1 kip ) − ( 3 ft )( 4 kips ) + ( 6 ft )( 2 kips )
− ( 9.6 ft ) B = 0
B = 0.375 kip
ΣFy = 0: −1 kip + C y − 4 kips + 2 kips − 0.375 kip = 0
C y = 3.375 kips
Along AC:
ΣFy = 0:
−1 kip − V = 0
V = −1 kip
ΣM J = 0:
M + x (1 kip ) = 0
M = − (1 kip ) x
M ( 3.6 ft ) = 3.6 kip ⋅ ft
Along CD:
ΣFy = 0:
−1 kip + 3.375 kips − V = 0
V = 2.375 kips
ΣM K = 0:
M + x (1 kip ) − ( x − 3.6 ft )( 3.375 kips ) = 0
M = −12.15 kip ⋅ ft + ( 2.375 kips ) x
M ( 6.6 ft ) = 3.525 kip ⋅ ft
continued
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
Along EB:
ΣFy = 0:
V − 0.375 kips = 0
V = 0.375 kips
ΣM L = 0:
− M − x1 ( 0.375 kips ) = 0
M = − ( 0.375 kips ) x1
M ( 3.6 ft ) = −1.35 kip ⋅ ft
Along DE:
ΣFy = 0:
V + 2 kips − 0.375 kips = 0
V = −1.625 kips
Also M is linear
(b)
V max = 2.38 kips along CD
M max = 3.60 kips at C
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 7, Solution 34.
(a)
FBD Beam:
ΣM B = 0:
(.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips
ΣFy = 0: 12.625 kips − 10 kips − 8 kips − 4 kips + B = 0
B = 9.375 kips
Along AC:
ΣFy = 0:
12.625 kips − V = 0
V = 12.625 kips
ΣM J = 0:
M − x (12.625 kips ) = 0
M = (12.625 kips ) x
M = 22.725 kip ⋅ ft at C
Along CD:
ΣFy = 0: 12.625 kips − 10 kips − V = 0
V = 2.625 kips
ΣM K = 0:
M + ( x − 1.8 ft )(10 kips ) − x (12.625 kips ) = 0
M = 18 kip ⋅ ft + ( 2.625 kips ) x
M = 29.8125 kip ⋅ ft at D ( x = 4.5 ft )
continued
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
Along DE:
ΣFy = 0:
(12.625 − 10 − 8) kips − V = 0
V = −5.375 kips
ΣM L = 0: M + x1 ( 8 kips ) + ( 2.7 ft + x1 )(10 kips )
− ( 4.5 ft + x1 )(12.625 kips ) = 0
M = 29.8125 kip ⋅ ft − ( 5.375 kips ) x1
M = 5.625 kip ⋅ ft at E
( x1 = 4.5 ft )
Along EB:
ΣFy = 0:
V + 9.375 kips = 0
ΣM N = 0:
V = 9.375 kips
x2 ( 9.375 kip ) − M = 0
M = ( 9.375 kips ) x2
M = 5.625 kip ⋅ ft at E
From diagrams:
(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.
V max = 12.63 kips on AC
M max = 29.8 kip ⋅ ft at D
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 35.
(a)
ΣM E = 0:
FBD Beam:
(1.1 m )( 540 N ) − ( 0.9 m ) C y + ( 0.4 m )(1350 N ) − ( 0.3 m )( 540 N ) = 0
C y = 1080 N
ΣFy = 0:
− 540 N + 1080 N − 1350 N
−540 N + E y = 0
E y = 1350 N
Along AC:
ΣFy = 0:
− 540 N − V = 0
V = −540 N
ΣM J = 0:
x ( 540 N ) + M = 0
M = − ( 540 N ) x
Along CD:
ΣFy = 0:
− 540 N + 1080 N − V = 0
V = 540 N
ΣM K = 0:
M + ( 0.2 m + x1 )( 540 N ) − x1 (1080 N ) = 0
M = −108 N ⋅ m + ( 540 N ) x1
M = 162 N ⋅ m at D ( x1 = 0.5 m )
continued
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
Along DE:
ΣFy = 0:
ΣM N = 0:
V + 1350 N − 540 N = 0
V = −810 N
M + ( x3 + 0.3 m ) ( 540 N ) − x3 (1350 N ) = 0
M = −162 N ⋅ m + ( 810 N ) x3
M = 162 N ⋅ m at D ( x3 = 0.4 )
Along EB:
(b)
ΣFy = 0:
V − 540 N = 0
V = 540 N
ΣM L = 0:
M + x2 ( 540 N ) = 0
M = −540 N x2
( x2 = 0.3 m )
M = −162 N ⋅ m at E
V max = 810 N on DE
From diagrams
M max = 162.0 N ⋅ m at D and E
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 7, Solution 36.
(a) FBD Beam:
a =1m
ΣFx = 0:
ΣFy = 0:
Bx = 0
− 1.5 kN + 2 kN − 4 kN + 5 kN − By = 0
By = 1.5 kN
ΣM B = 0:
a  4 (1.5 kN ) − 3 ( 2 kN ) + 2 ( 4 kN ) − 1( 5 kN )  − M B = 0
M B = ( 3 kN ) a = 3 kN ⋅ m
Along AC:
ΣFy = 0:
ΣM J = 0:
Along CD:
ΣFy = 0:
ΣM K = 0:
−1.5 kN − V = 0
V = −1.5 kN
M − x (1.5 kN ) = 0
M = − (1.5 kN ) x
M (1 m ) = −1.5 kN ⋅ m
−1.5 kN + 2 kN − V = 0
V = 0.5 kN
M + x (1.5 kN ) − ( x − 1 m )( 2 kN ) = 0
M = − 2 kN ⋅ m + ( 0.5 kN ) x
M ( 2 m ) = −1 kN ⋅ m
continued
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
Along EB:
ΣFy = 0:
ΣM L = 0:
V − 1.5 kN = 0
V = 1.5 kN
− M − x1 (1.5 kN ) − 3 kN ⋅ m = 0
M = − 3 kN ⋅ m − (1.5 kN ) x1,
M (1 m ) = − 4.5 kN ⋅ m
Along DE:
ΣFy = 0:
V + 5 kN − 1.5 kN = 0
V = − 3.5 kN
Also M is linear here
(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.
V max = 3.50 kN along DE
M max = 4.50 kN ⋅ m at E
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 37.
(a) FBD Beam:
ΣM A = 0: − (1.3 m ) (1.8 kN/m )( 2.6 m )  − (1.6 m )( 4 kN ) + ( 4 m ) B = 0
B = 3.121 kN
ΣFy = 0:
Ay − (1.8 kN/m )( 2.6 m ) − 4 kN + 3.121 kN = 0
A y = 5.559 kN
Along AC:
ΣFy = 0:
5.559 kN − (1.8 kN/m ) x − V = 0
V = 5.559 kN − (1.8 kN/m ) x
ΣM J = 0:
M +
x
(1.8 km ) x  − x ( 5.559 kN ) = 0
2
M = ( 5.559 kN ) x − ( 0.9 kN/m ) x 2
Along CD:
ΣFy = 0:
5.559 kN − x (1.8 kN/m ) − 4 kN − V = 0
V = (1.559 kN ) − (1.8 kN/m ) x
ΣM K = 0: M + ( x − 1.6 m )( 4 kN ) +
x
(1.8 kN/m ) x  − x ( 5.559 kN ) = 0
2
M = 6.4 kN ⋅ m + (1.559 kN ) x − ( 0.9 kN/m ) x 2
continued
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
Along DB:
ΣFy = 0:
V + 3.121 kN = 0
V = − 3.121 kN
(b)
ΣM L = 0:
− M + x1 ( 3.121 kN ) = 0
M = ( 3.121 kN ) x1
V max = 5.56 kN at A
M max = 6.59 kN ⋅ m at C
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 7, Solution 38.
(a) FBD Beam:
ΣFx = 0:
ΣFy = 0:
Ax = 0
Ay + ( 2 m )( 24 kN/m ) − 48 kN − 8 kN = 0
A y = 8 kN
ΣM A = 0:
M A + (1 m )( 2 m )( 24 kN/m ) − ( 3.5 m )( 48 kN )
− ( 2 m )( 8 kN ) = 0,
M A = 152 kN ⋅ m
Along AC:
ΣFy = 0:
8 kN + x ( 24 kN ⋅ m ) − V = 0
V = 8 kN + ( 24 kN/m ) x
ΣM J = 0:
M + 152 kN ⋅ m − x ( 8 kN ) −
x
( 24 kN/m ) x = 0
2
M = (12 kN/m ) x 2 + ( 8 kN ) x − 152 kN ⋅ m
Along DB:
ΣFy = 0:
V − 8 kN = 0
V = 8 kN
ΣM K = 0:
M + x1 ( 8 kN ) = 0,
M = − ( 8 kN ) x1
continued
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
Along CD:
ΣFy = 0:
ΣM L = 0:
V − 48 kN − 8 kN = 0,
V = 56 kN
M + ( x1 − 0.5 m )( 48 kN ) + x1 ( 8 kN ) = 0
M = 24 kN ⋅ m − ( 56 kN ) x1
(b)
V max = 56.0 kN along CD
M max = 152.0 kN ⋅ m at A
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 7, Solution 39.
(a) FBD Beam:
by symmetry,
C y = Gy =
Cx = 0, and
1
 2 (12 lb/in )(10 in ) + 2 (100 lb ) + (150 lb ) 
2
C y = G y = 295 lb
Along AC:
ΣFy = 0:
− (12 lb/in.) x − V = 0
 lb 
V = − 12
x
 in. 
ΣM J = 0:
M +
x
(12 lb/in.) x = 0,
2
 lb 
M = −  6  x2
 in. 
Along CD:
ΣFy = 0:
− (12 lb/in.)(10 in.) + 295 lb − V = 0
V = 175 lb
ΣM K = 0:
M + ( x − 5 in.)(12 lb/in.)(10 in.) − ( x − 10 in.)( 295 lb ) = 0
M = − 2350 lb ⋅ in. + (175 lb ) x
continued
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
Along DE:
ΣFy = 0:
ΣM N = 0:
− (12 lb/in.)(10 in.) + 295lb − 100 lb − V = 0,
V = 75 lb
M + ( x − 16 in.)(100 lb ) − ( x − 10 in.)( 295 lb )
+ ( x − 5 in.)(12 lb/in.)(10 in.) = 0
M = − 750 lb ⋅ in + ( 75 lb ) x
Complete diagrams using symmetry.
(b)
V max = 175.0 lb along CD and FG
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.
M max = 900 lb ⋅ in. at E
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 40.
(a) FBD Beam:
ΣM D = 0:
( 6 ft )(1 kip/ft )( 6 ft ) − ( 7.5 ft )( 2 kips/ft )( 9 ft )
− (12 ft ) Fy + (15 ft )( 33 kips ) = 0
Fy = 33 kips
ΣFy = 0:
− (1 kip/ft )( 6 ft ) + Dy − ( 2 kips/ft )( 9 ft ) − 33 kips + 33 kips = 0
D y = 24 kips
Along AC:
ΣFy = 0:
− (1 kip/ft ) x − V = 0,
ΣM J = 0:
M +
V = − (1 kip/ft ) x
x
(1 kip ⋅ ft ) x = 0,
2
1

M = −  kip/ft  x 2
2

Along CD:
ΣFy = 0:
ΣM K = 0:
− (1 kip/ft )( 6 ft ) − V = 0,
V = − 6 kips
M + ( x − 3 ft )(1 kip/ft )( 6 ft ) = 0
M = 18 kip ⋅ ft − ( 6 kips ) x
continued
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
Along DE:
ΣFy = 0:
− 6 kips + 24 kips − V = 0,
V = 18 kips
ΣM L = 0:
( x − 3 ft )( 6 kips ) − ( x − 9 ft )( 24 kips ) + M = 0
M = −198 kip ⋅ ft + (18 kips ) x
Along FB:
ΣFy = 0:
ΣM N = 0:
V + 33 kips = 0,
V = − 33 kips
x1 ( 33 kips ) − M = 0,
M = ( 33 kips ) x1
Along EF:
ΣFy = 0: V − ( 2 kips/ft ) x2 − 33 kips + 33 kips = 0
V = ( 2 kips/ft ) x2
ΣM O = 0:
M +
x2
( 2 kips/ft ) x2 − ( 3 ft ) ( 33 kips ) = 0
2
M = 99 kip ⋅ ft − (1 kip/ft ) x22
V max = 33.0 kips along FB
(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.
M max = 99.0 kip ⋅ ft at F
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 41.
(a) FBD Beam:
( 4 m )( w) − ( 2 m )(12 kN/m ) = 0
ΣFy = 0:
w = 6 kN/m
Along AC:
− x ( 6 kN/m ) − V = 0,
ΣFy = 0:
V = − ( 6 kN/m ) x
V = −6 kN at C ( x = 1 m )
x
( 6 kN/m )( x ) = 0
2
M = − ( 3 kN/m ) x 2
M = −3 kN ⋅ m at C
ΣM J = 0:
M +
Along CD:
ΣFy = 0:
− (1 m )( 6 kN/m ) + x1 ( 6 kN/m ) − V = 0
V = ( 6 kN/m )(1 m − x1 ) ,
ΣM K = 0:
V = 0 at x1 = 1 m
x1
( 6 kN/m ) x1 = 0
2
M = −3 kN ⋅ m − ( 6 kN ) x1 + ( 3 kN/m ) x12
M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) −
M = −6 kN ⋅ m at center ( x1 = 1 m )
Finish by symmetry.
(b) From diagrams:
V max = 6.00 kN at C and D
M max = 6.00 kN at center
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 7, Solution 42.
(a) FBD Beam:
ΣFy = 0:
(12 m ) w − ( 6 m )( 3 kN/m ) = 0
w = 1.5 kN/m
Along AC:
x (1.5 kN/m ) − V = 0,
ΣFy = 0:
V = (1.5 kN/m ) x
V = 4.5 kN at C
ΣM J = 0:
M −
x
(1.5 kN/m )( x ) = 0
2
M = ( 0.75 kN/m ) x 2 ,
M = 6.75 N ⋅ m at C
Along CD:
x (1.5 kN/m ) − ( x − 3 m )( 3 kN/m ) − V = 0
ΣFy = 0:
V = 9 kN − (1.5 kN/m ) x,
ΣM K = 0:
V = 0 at x = 6 m
x
 x − 3m
M +
 ( 3 kN/m )( x − 3 m ) − (1.5 kN/m ) x = 0
2
2


M = −13.5 kN ⋅ m + ( 9 kN ) x − ( 0.75 kN/m ) x 2
M = 13.5 kN ⋅ m at center ( x = 6 m )
Finish by symmetry.
(b) From diagrams:
V max = 4.50 kN at C and D
M max = 13.50 kN ⋅ m at center
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 7, Solution 43.
(a) FBD Beam:
ΣFy = 0: ( 8 m ) w − 2 ( 6 kN ) − ( 4 m )( 5 kN/m ) = 0
w = 4 kN/m
Along AC:
ΣFy = 0:
( 4 kN/m ) x − V = 0
V = ( 4 kN/m ) x
ΣM J = 0: M −
x
( 4 kN/m ) x = 0,
2
M = ( 2 kN/m ) x 2
Along CD:
( 4 kN/m ) x − 6 kN − V = 0
ΣFy = 0:
V = ( 4 kN/m ) x − 6 kN
ΣM K = 0:
M + ( x − 1 m )( 6 kN ) −
x
( 4 kN/m ) x = 0
2
M = ( 2 kN/m ) x 2 − ( 6 kN ) x + 6 kN ⋅ m
Note: V = 0 at x = 1.5 m where M = 1.5 kN/m
Along DE:
ΣFy = 0:
( 4 kN/m )( 2 m ) − 6 kN − (1 kN/m )( x − 2 m ) − V = 0
V = 4 kN − (1 kN/m ) x
 x − 2 m
ΣM L = 0: M + 
 (1 kN/m )( x − 2 m ) + ( x − 1 m ) 6 kN
2


− ( x − 1 m )( 4 kN/m )( 2 m ) = 0
continued
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
1

M = −  kN/m  x 2 + ( 4 kN ) x − 4 kN ⋅ m
2


Note: V = 0 at x = 4 m, where M = 4 kN ⋅ m
Complete diagrams using symmetry.
(b)
V max = 4 kN at C and F
M max = 4 kN ⋅ m at center
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 7, Solution 44.
(a) FBD Beam:
ΣFy = 0:
w (1.5 m ) − 2 ( 3.6 kN ) = 0
w = 4.8 kN/m
Along AC:
ΣFy = 0:
( 4.8 kN/m ) x − V = 0,
ΣM J = 0:
M −
V = ( 4.8 kN/m ) x
x
( 4.8 kN/m ) x = 0,
2
M = ( 2.4 kN/m ) x 2
Along CD:
ΣFy = 0:
( 4.8 kN/m ) x − 3.6 kN − V = 0
V = ( 4.8 kN/m ) x − 3.6 kN
ΣM K = 0:
M + ( x − 0.3 m )( 3.6 kN ) −
x
( 4.8 kN/m ) x = 0
2
M = 1.08 kN ⋅ m − ( 3.6 kN ) x + ( 2.4 kN/m ) x 2
Note: V = 0 at x = 0.75 m, where M = − 0.27 kN ⋅ m
Complete diagrams using symmetry.
V max = 2.16 kN at C and D
(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.
M max = 270 N ⋅ m at center
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 45.
FBD CE:
ΣFx = 0:
Cy = 0
ΣFy = 0:
C y − 4 kN = 0
C y = 4 kN
ΣM C = 0:
M C − ( 0.5 m )( 4 kN ) = 0
M C = 2 kN ⋅ m
ΣFx = 0:
Beam AB:
ΣFy = 0:
ΣM A = 0:
Ax = 0
Ay − 4 kN − 2 kN − 1 kΝ = 0
M A − 2 kN ⋅ m − ( 0.5 m )( 4 kN ) − (1 m )( 2 kN )
− (1.5 m )(1 kN ) = 0,
Along AC:
ΣFy = 0:
7 kN − V = 0
V = 7 kN
ΣM J = 0:
Ay = 7 kN
M + 7.5 kN ⋅ m − x ( 7 kN ) = 0
M = ( 7 kN ) x − 7.5 kN ⋅ m
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.
M A = 7.5 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Along DB:
ΣFy = 0:
V − 1 kN = 0
V = 1 kN
ΣM K = 0:
− M + x1 (1 kN ) = 0
M = (1 kN ) x1
Along CD:
ΣFy = 0:
ΣM M = 0:
V − 2 kN − 1 kN = 0
V = 3 kN
M + ( x1 − 0.5 m )( 2 kN ) + x1 (1 kN ) = 0
M = 1 kN ⋅ m − ( 3 kN ) x1
Note: M exhibits a discontinuity at C, equal to 2 kN ⋅ m, the value of MC.
From the diagrams,
V max = 7.00 kN along AC
M max = 7.50 kN at A
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 7, Solution 46.
FBD CE or DF:
ΣFx = 0: Cx , Dx = 0
ΣFy = 0: C y − 750 N = 0,
C y = 750 N
Dy = 750 N
ΣM C = 0: M C − ( 0.3 m )( 750 N ) = 0
M C = 225 N ⋅ m = M D
Beam AB:
ΣM A = 0:
( 0.9 m ) Dy − 2 ( 225 N ⋅ m ) − ( 0.3 m )( 750 N )
− ( 0.9 m )( 750 N ) − (1.2 m )( 540 N ) = 0
Dy = 2220 N
ΣFy = 0: Ay − 2 ( 750 N ) − 540 N + 2220 N = 0
Ay = −180 N
A y = 180 N
ΣFy = 0: −180 N − V = 0
Along AC:
V = −180 N
ΣM J = 0: M + x (180 N ) = 0
M = (180 N ) x
Along CD:
ΣFy = 0: −180 N − 750 N − V = 0,
ΣM K = 0:
V = − 930 N
M − 225 N ⋅ m + ( x − 0.3 m )( 750 N ) + x (180 N ) = 0
M = 450 N ⋅ m − ( 930 N ) x
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
Along DB:
ΣFy = 0: V − 540 N = 0
V = 540 N
ΣM N = 0: M + x1 ( 540 N ) = 0
M = − ( 540 N ) x1
Note: The discontinuities in M, at C and D, equal 225 N ⋅ m, M C and M D
From the diagrams
V max = 930 N along CD
M max = 387 N ⋅ m at D
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 7, Solution 47.
FBD Angle:
ΣFy = 0:
T − Cy = 0 Cy = T
( 0.3 ft ) T − M C = 0, M C = ( 0.3 ft ) T
By symmetry, Dy = T , M D = ( 0.3 ft ) T
ΣM C = 0:
ΣFy = 0:
2T − 810 lb − (100 lb/ft )( 9 ft ) = 0
T = 855 lb.
(a)
Beam AB:
From above M C = M D = ( 0.3 ft )( 855 lb ) = 256.5 lb ⋅ ft
ΣFy = 0: − x (100 lb/ft ) − V = 0
Along AC:
V = − (100 lb/ft ) x
ΣM J = 0: M +
x
(100 lb/ft ) x = 0,
2
M = − ( 50 lb/ft ) x 2
M C = 256.5 lb ⋅ ft
Along CI:
ΣFy = 0:
− (100 lb/ft ) x + 855 lb − V = 0
V = 855 lb − (100 lb/ft ) x
ΣM K = 0:
M − 256.5 lb ⋅ ft +
x
(100 lb/ft ) x
2
− ( x − 3.6 ft )( 855 lb ) = 0
M = − ( 50 lb/ft ) x 2 + ( 855 lb ) x − 2821.5 lb ⋅ ft
Complete diagrams using symmetry
Note: Discontinuities in M, at C and D, equal M C and M D
(b)
V max = 495 lb at C and D
M max = 648 lb ⋅ ft at C and D
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 7, Solution 48.
FBD Angle:
ΣFy = 0: T − C y = 0
Cy = T
ΣM C = 0: M C − ( 0.3 ft ) T = 0,
M C = ( 0.3 ft ) T
By symmetry, Dy = T and M D = ( 0.3 ft ) T
(a)
Beam AB:
ΣFy = 0: 2T − (100 lb/ft )( 9 ft ) − 810 lb = 0
T = 855 lb
From above, M C = M D = ( 0.3 ft )( 855 lb ) = 256.5 lb ⋅ ft
ΣFy = 0: − (100 lb/ft ) x − V = 0
Along AC:
V = − (100 lb/ft ) x
ΣM J = 0:
M +
x
(100 lb/ft ) x = 0
2
M C = 256.5 lb ⋅ ft
Along CI:
ΣFy = 0:
M = ( 50 lb/ft ) x 2
855 lb − (100 lb/ft ) x − V = 0
V = 855 lb − (100 lb/ft ) x
ΣM K = 0: M − 256.5 lb ⋅ ft +
x
(100 lb/ft ) x
2
− ( x − 2.7 ft )( 855 lb ) = 0
M = − ( 50 lb/ft ) x 2 + ( 855 lb ) x − 2052 lb ⋅ ft
Complete diagrams using symmetry
Note: The discontinuities in M, at C and D, equal M C and M D
(b)
V max = 585 lb at C and D M max = 783 lb ⋅ ft at I 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 7, Solution 49.
Note: D passes through C, so
FBD Whole:
Dy
Dx
0.35 m
= 0.7
0.5 m
=
ΣM H = 0: ( 0.65 m )( 200 N ) − ( 0.5 m ) ( 0.7 Dx ) + ( 0.1 m ) ( Dx )
+ ( 0.25 m )( 400 N ) − ( 0.15 m )( 200 N ) = 0
D x = 800 N
D y = 560 N
ΣFx = 0: 800 N − H x = 0
ΣFy = 0:
H x = 800 N
560 N − 400 N − 2 ( 200 N ) + H y = 0
H y = 240 N
Beam AB with forces at D & H replaced by forces and couples at E and G.
Horizontal forces not shown to avoid clutter.
ΣFy = 0:
Along AE:
− 200 N − V = 0
V = − 200 N
ΣM J = 0:
x ( 200 N ) + M = 0
M = − ( 200 N ) x
Along EF:
ΣFy = 0:
− 200 N + 560 N − V = 0
V = 360 N
ΣM K = 0: M − 80 N ⋅ m + x ( 200 N ) − ( x − 0.15 m )( 560 N ) = 0
M = ( 360 N ) x − 4 N ⋅ m
Along GB:
ΣFy = 0: V − 400 N = 0
V = 200 N
ΣM L = 0:
M + x1 ( 200 N ) = 0
M = ( 200 N ) x1
continued
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
ΣFy = 0: V + 240 N − 200 N = 0
Along FG:
V = − 40 N
(
)
M − 160 N ⋅ m + x1 − 0.15 m ( 240 N ) − x1 ( 200 ) N = 0
ΣM N = 0:
M = 124 N ⋅ m + ( 40 N ) x1
From diagrams,
V max = 360 N along EF M max = 140.0 N ⋅ m at F 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 7, Solution 50.
FBD AB + Pulley & Cord:
 5 
 12 
ΣM A = 0: ( 48 in.)  D  + ( 20 in.)  D  − (100 in.)(120 lb ) = 0
13


 13 
D = 325 lb
so D x = 300 lb
ΣFy = 0: − Ay + 125 lb − 120 lb = 0
, D y = 125 lb
A y = 5 lb
Neglecting the diameter of pulley G, the cord EG has slope 3/4,
and tension 120 lb, E x = 96 lb
(a)
E y = 72 lb
,
Beam AB with forces at D and G replaced by forces and couples at E
and F. Horizontal forces are omitted to avoid clutter.
Along AE:
ΣFy = 0:
− 5 lb − V = 0,
V = − 5 lb
ΣM J = 0: x ( 5 lb ) + M = 0,
M = ( − 5 lb ) x
Along EF:
ΣFy = 0:
ΣM K = 0:
− 5 lb + 197 lb − V = 0,
V = 192 lb
M + 6000 lb ⋅ in. + x ( 5 lb ) − ( x − 48 in.)(197 lb ) = 0
M = (192 lb ) x − 15456 lb ⋅ in.
Along FB:
ΣFy = 0: V − 120 lb = 0, V = 120 lb
ΣM L = 0: − M − x1 (120 lb ) = 0
M = − (120 lb ) x1
(b)
V max = 192.0 lb along EF
From diagrams,
M max = 6240 lb ⋅ in. = 520 lb ⋅ ft at E
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 7, Solution 51.
ΣFy = 0: Lw − 2P = 0
w=
2P
L
Along AC:
ΣM J = 0: M −
x  2P 

x = 0
2 L 
M =
P 2
x
L
Along CD:
ΣM K = 0: M + ( x − a ) P −
x  2P 

x = 0
2 L 
M =
P 2
x − Px + Pa
L
Complete diagram using symmetry
Note: M min = Pa −
1
L
(center)
PL at x =
2
4
Setting M max = − M min :
Solving a = −
P 2
PL
L2
a = − Pa +
or a 2 + La −
=0
4
L
4
L
2
±
L, positive root a =
2
2
(a)
Then, with L = 1.5 m,
a = 0.31066 m
(b)
and, with P = 3.6 kN,
M max =
Pa 2
L
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.
2 −1
L
2
a = 0.311 m M max = 232 N ⋅ m COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 52.
ΣFy = 0: T − C y = 0
FBD Angle CE:
Cy = T
ΣM C = 0: M C − ( 0.3 ft ) T = 0,
by symmetry D y = T
M C = ( 0.3 ft ) T
and M D = ( 0.3 ft ) T
ΣFy = 0: 2T − 810 lb − ( 9 ft )(100 lb/ft ) = 0
Beam AB:
T = 855 lb
From above M C = M D = 256.5 lb ⋅ ft
ΣM J = 0: M −
Along AC:
x
(100 lb/ft ) x = 0
2
M = − ( 50 lb/ft ) x 2
Along CI:
M C = 256.5 lb ⋅ ft
ΣM K = 0:
x
(100 lb/ft ) x
2
− ( x − 4.5 ft + a )( 855 lb ) = 0
M − 256.5 lb ⋅ ft +
M =  − 50 x 2 + 855 ( x + a ) − 3591 lb ⋅ ft with a in ft
Complete M diagram using symmetry
At x = ( 4.5 − a ) ft, M min = − 50 ( 4.5 − a ) lb ⋅ ft
2
At x = 4.5 ft, M max = ( 855 a − 756 ) lb ⋅ ft
Setting M max = − M min : a 2 − 26.1 a + 35.37 = 0
(a)
Solving: a = 13.05 ± 11.6160,
a < 4.5 so
(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.
a = 1.434 ft
giving M max = 470 lb ⋅ ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 53.
Replacing the 1500 N force with equivalent force and couple at D,
M C = 0:
(1.25 m ) P − (1.2 m )(1500 N ) − 1800 N ⋅ m
+ ( 2.4 m ) E y − ( 3.65 m ) 2P = 0
assume P in N: E y = (1500 + 2.5208 P ) N
ΣFy = 0:
C y − P − 2P − 1500 + (1500 + 2.5208 P ) = 0
C y = 0.47917 P
ΣM J = 0: M + x P = 0
Along AC:
M = − Px
ΣM K = 0:
Along CD:
M + x P − ( x − 1.25 m )( 0.47917 P ) = 0
M = − 0.5208 Px − 0.5990 P
at x = 2.45− (left of D), M = −1.875 P
at x = 2.45+ (right of D), M = 1800 − 1.875 P
ΣM E = 0: − M − (1.25 m )( 2 P ) = 0
At E:
M = − 2.5 P
(a)
Setting M max = − M min : 1800 − 1.875 P = 2.5 P
P = 411.43
P = 411 N (b)
M max = 2.5 P
M max = 1029 N ⋅ m
M max = 1.029 kN ⋅ m 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 7, Solution 54.
Since there are no distributed loads, M is piecewise linear, and only pts A,
C, and D need be considered.
At A:
ΣM A = 0: M + ( 0.75 m )( 4 kN ) + (1.75 m )(16 kN )
− (1.75 m + a )( 8 kN ) = 0
(With a in m) M = ( 8 a − 17 ) kN ⋅ m
At C:
ΣM C = M + (1 m )(16 kN ) − (1 m + a m )( 8 kN ) = 0
M = ( 8 a − 8 ) kN ⋅ m
ΣM D = 0: M − ( a m )( 8 kN ) = 0
At D:
M = 8 a kN ⋅ m
(a)
Apparently M max = 8 a kN ⋅ m at D,
and M min = ( 8 a − 17 ) kN ⋅ m at A
Setting M max = − M min : 8 a = 17 − 8 a
a=
17
m
16
a = 1.063 m
(b)
and M max = 8 a =
17
kN ⋅ m
2
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.
M max = 8.50 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 55.
ΣM A = 0: a Dy − ( 5 ft )( 500 lb ) − (10 ft )( 500 lb ) = 0
Dy =
7500 lb ⋅ ft
a
Since there are no distributed loads, M is piecewise linear, so only points
C and D need be considered. Assume a in ft.
ΣM D = 0:
At D:
M + (10 − a ) ft  ( 500 lb ) = 0
M D = − 500 (10 − a ) lb ⋅ ft
At C:
ΣM C = 0: M + ( 5ft )( 500 lb ) − 5 − (10 − a )  ft
7500
lb = 0
a
37500 

M C =  5000 −
 lb ⋅ ft
a 

(a)
Apparently M max = M C and M min = M D (recall 5 < a < 10 )
Setting M C = − M D : 5000 −
37500
= 5000 − 500 a
a
37500 = 500 a 2
a = 75
(b)
(
)
M max = 500 10 − 75 lb ⋅ ft
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 = 8.66 ft
M max = 670 lb ⋅ ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 56.
ΣM A = 0: aDy − ( 5 ft )( 250 lb ) − (10 ft )( 500 lb ) = 0
Dy =
With a in ft, Dy =
6250 lb ⋅ ft
a
6250
lb
a
Since there are no distributed loads, M is piecewise linear, so only points
C and D need be considered.
ΣM D = 0:
At D:
(10 − a ) ft  ( 500 lb ) + M D = 0
M D = − 500 (10 − a ) lb ⋅ ft
At C:
ΣM C = 0:
 6250 
M C + ( 5 ft )( 500 lb ) − 5 − (10 − a )  ft 
lb  = 0
 a

31250 

M C =  3750 −
 lb ⋅ ft
a 

(a)
Apparently M max is M C and M min is M D ( 5 < a < 10 )
Setting M C = − M D : 3750 −
31250
= − 500 a + 5000
a
500 a 2 − 1250 − 31250 = 0 or a 2 − 2.5 a − 62.5 = 0
a = 1.25 ± 8.004, positive root a = 9.254 ft
a = 9.25 ft (b)
M max = 500 (10 − a ) lb ⋅ ft
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.
M max = 373 lb ⋅ ft COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 57.
ΣM J = 0: − M −
M due to distributed load:
x
wx = 0
2
1
M = − wx 2
2
ΣM J = 0: − M + xw = 0
M due to counter weight:
(a) Both applied:
M = wx
M = Wx −
And here M =
w 2
x
2
dM
W
= W − wx = 0 at x =
dx
w
W2
> 0 so M max ; M min must be at x = L
2w
So M min = WL −
1 2
wL . For minimum M max set M max = −M min , so
2
W2
1
= −WL + wL2 or W 2 + 2wLW − w2 L2 = 0
2w
2
W = −wL ± 2w2 L2 (need +)
W =
(b) w may be removed
( 2 − 1) wL = 0.414wL ( 2 − 1) wL
2
W2
M max =
=
2w
2
2
M = Wx, M max = WL at A
Without w,
M = Wx −
With w (see part a)
M min = WL −
For minimum M max , set M max ( no w ) = −M min ( with w )
WL = − WL +
M max = 0.858wL2 w 2
x ,
2
M max =
W2
W
at x =
2w
w
1 2
wL at x = L
2
1 2
1
wL → W = wL →
2
4
With
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.
M max =
1 2
wL 4
W =
1
wL 4
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 58.
(a) FBD Beam:
ΣM C = 0:
ΣFy = 0:
LAy − M 0 = 0
− Ay + C = 0
Ay =
M0
L
C=
M0
L
M0
 dV

= w = 0 .
at A, and remains constant 
L
 dx

Moment Diag: M starts at zero at A and decreases linearly
M
M
M 
 dM
= V = − 0  to − 0 at B, where M jumps by M 0 to + 0 .

2
2
dx
L


M
M continues to decrease with slope − 0 to zero at C.
L
Shear Diag: V = −
(b) From diagrams:
V max =
M0
everywhere
L
M max =
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.
M0
at B
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 59.
(a) and (b)
By symmetry Ay = By
ΣFy = 0: 2 Ay − 2 P = 0
Ay = P
By = P
Shear Diag:
V is piecewise constant with discontinuities equal to P at A, B, C and D in
the direction of the loads.
Moment Diag:
M is piecewise linear with slope equal to + P on AB, 0 on BC, –P on CD.
M B = Pa
V max = P along AB and CD
M max = Pa along BC
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 7, Solution 60.
(a) and (b)
Shear Diag:
 dV

= − w  starting at 0 at A, and
Since w is linear, V is quadratic 
 dx

1
decreasing to − w0 L at B.
2
Moment Diag:
 dm

= V  to
M is zero at A and decreases cubically 
 dx

1 1
1

2
 − w0 L  L = − w0 L at B.
3 2
6

V max =
M max =
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.
1
w0 L at B
2
1
w0 L2 at B
6
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 61.
(a) and (b)
ΣM B = 0:
aP + 2aC y − 3.5aP = 0
C y = 1.25P
ΣFy = 0:
By − 2 P + 1.25P = 0
By = 0.75P
Shear Diag:
V is piecewise constant, equal to –P from A to B, jumping up 0.75P, at B,
to − 0.25P, and jumping up 1.25 P, at C, to + P.
Moment Diag:
 dM

M is zero at A, decreasing linearly 
= V = − P  to − Pa at B, and
 dx

 dM

further, 
= V = − 0.25P  to − Pa − ( 0.25P )( 2a ) = −1.5Pa at C. M
 dx

 dM

then increases linearly 
= V = P  to −1.5Pa + P (1.5a ) = 0 at D,
 dx

as it must.
V max = P along AB and CD
M max = 1.5Pa at C
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 7, Solution 62.
(a) and (b)
ΣM B = 0:
( 0.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips )
+ ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips
Shear Diag:
 dV

V is piecewise constant, 
= 0  with discontinuities at each
 dx

concentrated force. (equal to force)
V max = 12.63 kips
Moment Diag:
 dM

= V  throughout.
M is zero at A, and piecewise linear 
 dx

M C = (1.8 ft )(12.625 kips ) = 22.725 kip ⋅ ft
M D = 22.725 kip ⋅ ft + ( 2.7 ft )( 2.625 kips )
= 29.8125 kip ⋅ ft
M E = 29.8125 kip ⋅ ft − ( 4.5 ft )( 5.375 kips )
= 5.625 kip ⋅ ft
M B = 5.625 kip ⋅ ft − ( 0.6 ft )( 9.375 kips ) = 0
M max = 29.8 kip ⋅ ft
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 7, Solution 63.
(a) and (b)
FBD Beam:
ΣM E = 0:
(1.1 m )( 0.54 kN ) − ( 0.9 m ) C y
+ ( 0.4 m )(1.35 kN ) − ( 0.3 m )( 0.54 kN ) = 0
C y = 1.08 kN
ΣFy = 0:
− 0.54 kN + 1.08 kN − 1.35 kN + E − 0.54 kN = 0
E = 1.35 kN
Shear Diag:
 dV

V is piecewise constant, 
= 0 everywhere  with discontinuities at
 dx

each concentrated force. (equal to the force)
V max = 810 N
Moment Diag:
M is piecewise linear starting with M A = 0
M C = 0 − 0.2 m ( 0.54 kN ) = 0.108 kN ⋅ m
M D = 0.108 kN ⋅ m + ( 0.5 m )( 0.54 kN ) = 0.162 kN ⋅ m
M E = 0.162 kN ⋅ m − ( 0.4 m )( 0.81 kN ) = − 0.162 kN ⋅ m
M B = 0.162 kN ⋅ m + ( 0.3 m )( 0.54 kN ) = 0
M max = 0.162 kN ⋅ m = 162.0 N ⋅ m
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 7, Solution 64.
(1.5 m ) E y −  m  ( 360 N/m )(1 m ) − (1.3 m )( 600 N )
1
2
ΣM A = 0:

− ( 2 m )( 420 N ) = 0
ΣFy = 0:
(a)
E y = 1200 N
Ay + 1200 N − ( 360 N/m )(1 m ) − 600 N − 420 N = 0
A y = 180.0 N
Shear Diag:
V jumps to Ay = 180 N at A, then decreases linearly
 dV

= − w = − 360 N/m  to 180 N − ( 360 N/m )(1 m ) = −180 N at C.

dx


From C, V is piecewise constant ( w = 0 ) with jumps of − 600 N at D,
+ 1200 N at E, − 420 N at B.
Moment Diag:
M starts at zero at A with slope
dM
= V = 180 N/m, decreasing to zero
dx
1
(180 N )( 0.5 m ) = 45 N ⋅ m. M is zero again
2
at C, decreasing to − (180 N )( 0.3 m ) = − 54 N ⋅ m at D. M then
at x = 0.5 m . There M =
decreases by ( 780 N )( 0.2 m ) = 156 N ⋅ m to − 210 N ⋅ m at E, and
increases by ( 420 N )( 0.5 m ) = 210 N ⋅ m to zero at B.
(b)
From the diagrams,
V max = 780 N along EB
M max = 210 N ⋅ m at E
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 7, Solution 65.
(a)
Shear Diag:
 dV

V is zero at A with constant slope 
= − w = −1 kip/ft  decreasing to
 dx

− 3.6 kips at C. V then jumps 9 kips to 5.4 kips and is constant to D.
Then V increases with constant slope 1.5 kips/ft for 3 ft, to 9.9 kips at B.
This is also equal to By .
Moment Diag:
 dM

= V  decreasing to − 3.6 kips at C,
M is zero at A, with zero slope 
 dx

1
where M = ( − 3.6 kips )( 3.6 ft ) , M C = − 6.48 kip ⋅ ft. M then increases
2
linearly with slope 5.4 kips to − 6.48 kip ⋅ ft + ( 5.4 kips )(1.8 ft )
= 3.24 kip ⋅ ft at D. Finally, M increases, with increasing slope, to
 5.4 kips + 9.9 kips 
M B = 3.24 kip ⋅ ft + 
 ( 3 ft ) ,
2


(b)
M B = 26.19 kip ⋅ ft.
From the diagrams,
V max = 9.90 kips at B
M max = 26.2 kip ⋅ ft at 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 66.
(a)
ΣM B = 0: ( 4 m ) Ay − ( 2.7 m )(1.8 kN/m )( 2.6 m )
− ( 2.4 m )( 4 kN ) = 0
A y = 5.559 kN
Shear Diag:
At A, V jumps up 5.559 kN, then decreases with uniform slope of
−1.8 kN/m to 2.679 kN at C. V then jumps down 4 kN to −1.321 kN,
and continues with uniform slope −1.8 kN/m to − 3.121 kN at D. V is
then constant to B. Note: By = 3.121 kN
Moment Diag:
dM
= V = 5.559 kN. The slope decreases to
dx
 5.559 + 2.679

kN  (1.6 m ) ,
2.679 kN at C, where M = 
2


M C = 6.59 kN ⋅ m. At C the slope drops to −1.321 kN and continues to
M is zero at A, with slope
 1.321 + 3.121 
decrease, M D = 6.59 kN ⋅ m − 
 kN (1 m ) = 4.37 kN ⋅ m.
2


M then decreases with uniform slope − 3.121 kN, to zero at B.
From the diagrams,
(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.
V max = 5.56 kN at A
M max = 6.59 kN ⋅ m at C
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 67.
(a)
ΣFy = 0:
Ay + ( 24 kN/m )( 2 m ) − 48 kN − 8 kN = 0
A y = 8 kN
ΣM A = 0:
M A − (1 m )( 24 kN/m )( 2 m ) + ( 3.5 m )( 48 kN )
+ ( 4 m )( 8 kN ) = 0
M A = −152 kN ⋅ m
Shear Diag:
V jumps to 8 kN at A, and increases with uniform slope
dV
= − w = 24 kN/m to 56 kN at C. V is constant at 56 kN to D, then
dx
drops by 8 kN to 8 kN at D, is then constant at 8 kN to B.
Moment Diag:
M starts at M A = −152 kN ⋅ m, with slope 8 kN, which increases to 56 kN
 8 + 56

at C, where M = −152 kN ⋅ m + 
kN  ( 2 m ) = − 88 kN ⋅ m. Then
 2

M increases with uniform slope 56 kN to − 88 kN ⋅ m + ( 56 kN )(1.5 m )
= − 4 kN ⋅ m at D, and finally, with slope 8 kN, to zero at B.
From the diagrams,
(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.
V max = 56 kN along CD
M max = 152.0 kN ⋅ m at A
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 68.
Note: Cx is omitted to avoid clutter.
(a)
By symmetry C y = G y
ΣFy = 0:
2C y − 2 (12 lb/in.)(10 in.) − 2 (100 lb ) − 150 lb = 0
C y = 295 lb
G y = 295 lb
Shear Diag:
dV
= − w = −12 lb/in. which is uniform to
dx
C, where V = − (12 lb/in.)(10 in.) = −120 lb ⋅ in. V jumps 295 lb to
At A, V = 0 and has slope
+ 175 lb at C, is constant to D where it drops 100 lb to 75 lb, is constant
to E where it drops 150 lb to − 75 lb. The diagram can be completed
using symmetry.
Moment Diag:
M is zero at A, with zero slope, which decreases linearly to −120 lb at C,
1
where M = − (120 lb )(10 in.) = − 600 lb ⋅ in. M then increases, with
2
uniform slope 175 lb, to − 600 lb ⋅ in. + (175 lb )( 6 in.) = 450 lb ⋅ in. at D.
M then increases, at uniform slope 75 lb, to 450 lb + (75 lb)(6 in.)
= 900 lb ⋅ in. at E. The diagram can be completed using symmetry.
From the diagrams,
(b)
V max = 175.0 lb along CD and FG
M max = 900 lb ⋅ in. at center E
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 7, Solution 69.
(a)
(18 ft )(1 kip/ft )( 6 ft ) − (12 ft ) Dy
ΣM F = 0:
+ ( 4.5 ft )( 2 kips/ft )( 9 ft ) + ( 3 ft )( 33 kips ) = 0
D y = 24 kips
ΣFy = 0:
24 kips + Fy + 33 kips − (1 kip/ft )( 6 ft )
− ( 2 kips/ft )( 9 ft ) = 0
Fy = − 33 kips
Fy = 33 kips
Shear Diag:
dV
= −1 kip/ft is uniform to C, where V = − 6 kips .
dx
Then V is constant to D where it jumps up 24 kips to + 18 kips, and
dV
= − 2 kips/ft and V decreases by
remains constant to E. From E to F,
dx
18 kips to zero at F, where it drops 33 kips, is constant to B, and jumps
V = 0 at A and
33 kips to zero.
Moment Diag:
At A, M = 0 and
dM
starts at zero, decreasing to − 6 kips at C, where
dx
1
( 6 kips )( 6 ft ) = −18 kip ⋅ ft. M then decreases linearly by
2
( 6 kips )( 3 ft ) to − 36 kip ⋅ ft at D, and increases linearly by
M =
(18 kip )( 3 ft ) to +18 kip ⋅ ft at E. From E to F,
dM
decreases from
dx
1
(18 kips )( 9 ft ) to 99 kip ⋅ ft, at F.
2
M then decreases linearly to zero at B.
18 kips to zero as M increases by
From the diagrams,
(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.
V max = 33.0 kips along FB
M max = 99.0 kips at F
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 70.
(a)
ΣFy = 0:
Ay + ( 200 N/m )( 0.8 m ) − (120 N/m )( 0.3 m ) = 0
Ay = −124 N
ΣM A = 0:
A y = 124 N
M A − 60 N ⋅ m − ( 0.4 m )( 200 N/m )( 0.8 m )
+ (1.35 m )(120 N/m )( 0.3 m ) = 0
M A = 75.4 N ⋅ m
Shear Diag:
dV
= 200 N/m from A to C and V increases by
dx
( 200 N/m )( 0.8 m ) = 160 N to + 36 N at C. It remains at 36 N to D,
V drops to 124 N at A.
then decreases linearly to zero at B. Note V = 0 where
x
124 N
=
,
0.8 m 160 N
or x = 0.62 m.
Moment Diag:
dM
= −124 N. The slope increases to zero
dx
1
at x = 0.62 m, where M = 75.4 N ⋅ m − (124 N )( 0.62 m )
2
= 36.96 N ⋅ m. The slope then increases as M increases by
1
( 36 N )( 0.18 m ) = 3.24 N ⋅ m to 40.2 N ⋅ m at C, where it drops
2
60 N ⋅ m to −19.8 N ⋅ m. M increases linearly by
M jumps to 75.4 at A where
( 36 N )( 0.4 m ) = 14.4 N ⋅ m to − 5.4 N ⋅ m, and finally M increases
quadratically by
1
dM
is
( 36 N )( 0.3 m ) = 5.4 N ⋅ m to zero at B where
2
dx
also zero.
From the diagrams,
(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.
V max = 124.0 N at A
M max = 75.4 N at A
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 71.
(a)
− ( 3 m )( 9 kN ) − 27 kN ⋅ m − ( 9 m )(12 kN ) + (12 m ) F
ΣM A = 0:
+ (16.5 m )( 3 kN ) + 22.5 kN ⋅ m = 0
F = 7.5 kN
ΣFy = 0:
Ay − 9 kN − 12 kN + 7.5 kN + 3 kN = 0
A y = 10.5 kN
Shear Diag:
V is piecewise constant, with jumps at A, C, E, F, and B, equal to the
forces there.
Moment Diag:
M is piecewise linear with jumps at D and B equal to the couples there.
M C = (10.5 kN )( 3 m ) = 31.5 kN ⋅ m
M D − = 31.5 kN ⋅ m + (1.5 kN )( 3 m ) = 36.0 kN ⋅ m
M D+ = 36 kN ⋅ m + 27 kN ⋅ m = 63 kN ⋅ m
M E = 63 kN ⋅ m + (1.5 kN )( 3 m ) = 67.5 kN ⋅ m
M F = 67.5 kN ⋅ m − (10.5 kN )( 3 m ) = 36 kN ⋅ m
M B − = 36 kN ⋅ m − ( 3 kN )( 4.5 m ) = 22.5 kN ⋅ m
Finally M drops 22.5 kN ⋅ m to zero at B
(b)
From the diagrams,
V max = 10.50 kN along AC and EF
M max = 67.5 kN ⋅ m at E
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 7, Solution 72.
(a)
( 3 m ) By − ( 2.1 m )( 2.5 kN/m )( 4.2 m ) = 0
ΣM A = 0:
B y = 7.35 kN
ΣFy = 0:
Ay − ( 2.5 kN/m )( 4.2 m ) + 7.35 kN = 0
A y = 3.15 kN
Shear Diag:
dV
= − 2.5 kN/m throughout, and jumps at A and B equal to
dx
the forces there.
V has slope
VB − = 3.15 kN − ( 2.5 kN/m )( 3 m ) = − 4.35
VB + = − 4.35 kN + 7.35 kN = 3 kN
VC = 3 kN − ( 2.5 kN/m )(1.2 m ) = 0
Note, V = 0 where 3.15 kN − ( 2.5 kN/m ) x = 0, x = 1.26 m.
Moment Diag:
dM
= 3.15 kN. The slope decreases to zero at
dx
x = 1.26 m and to − 4.35 kN at B, jumps to 3.0 kN and decreases to
0 at C.
1
M D = ( 3.15 kN )(1.26 m ) = 1.9845 kN ⋅ m
2
1
M B = 1.9845 kN ⋅ m − ( 4.35 kN )( 3 m − 1.26 m ) = −1.80 kN ⋅ m
2
1
M C = −1.80 kN ⋅ m + ( 3 kN )(1.2 m ) = 0
2
At A, M = 0 and
V max = 4.35 kN at B
From the diagrams,
(b)
M max = 1.985 kN at D (1.26 m from A )
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 7, Solution 73.
(a)
Note: C x omitted to avoid clutter.
ΣM C = 0: − ( 0.5 m )( 2 kN/m )( 2 m ) + (1.5 m )(1 kN )
− ( 3 m )( 4.5 kN/m )(1 m ) + ( 3.5 m ) By = 0
B y = 4.0 kN
ΣFy = 0:
C y − ( 2 kN/m )( 2 m ) + 1 kN − ( 4.5 kN/m )(1 m )
+ 4.0 kN = 0
C y = 3.5 kN
Shear Diag:
dV
= − 2 kN/m, so V decreases to −1 at C, then
dm
jumps 3.5 kN to + 2.5 kN, and continues to decrease, with the same
slope, to − 0.5 kN at D, jumps 1 kN to + 0.5 kN from D to E. Then V
decreases at rate 4.5 kN/m, to − 4.0 kN at B. Note that V = 0 where
8
( − 4.5 kN/m )( x ) = − 4 kN, x = m, and where ( 2 kN/m )( y ) = 2.5,
9
5
y = m.
4
Moment Diag:
dM
At A, M and
= 0, with the slope decreasing to −1 kN at C, where
dx
1
M = − (1 kN )( 0.5 m ) = − 0.25 kN ⋅ m. The slope jumps to 2.5 kN
2
and decreases to zero at F and to − 0.5 kN at D.
At A, V = 0 and
M F = − 0.25 kN ⋅ m +
1
5
( 2.5 kN )  m  = 1.3125 kN ⋅ m
2
4 
M D = 1.3125 kN ⋅ m −
1
1
( 0.5 kN )  m  = 1.25 kN ⋅ m.
2
4 
From D, M increases by ( 0.5 kN )(1 m ) to 1.75 kN ⋅ m at G. M continues
to increase to 1.75 kN ⋅ m +
then decreases by
(b)
1
1
( 0.5 kN )  m  = 1.7778 kN ⋅ m at G and
2
9 
1
8
( 4 kN )  m  = 1.7778 kN ⋅ m, to zero at B.
2
9 
V max = 4.00 kN at B
From the diagrams,
4

M max = 1.778 kN ⋅ m at G  m from B  9


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 7, Solution 74.
(a)
ΣFy = 0:
Ay − ( 2 kips/ft )( 8 ft ) − 3 kips + 7 kips = 0
A y = 12 kips
M A + ( 4 ft )( 2 kips/ft )( 8 ft ) − (14 ft )( 3 kips )
ΣM A = 0:
− ( 20 ft )( 7 kips ) = 0
M A = 34 kip ⋅ ft
Shear Diag:
V jumps to 12 kips at A, then decreases at 2 kips/ft to − 4 kips at C to D.
V drops 3 kips to − 7 kips from D to B and jumps 7 kips to zero. Note:
V = 0 where 12 kips − ( 2 kips/ft ) x = 0, x = 6 ft.
Moment Diag:
M jumps to 34 kip ⋅ ft at A and then increases with decreasing slope to
1
34 kip ⋅ ft + (12 kips )( 6 ft ) = 70 kip ⋅ ft at E, and decreases by
2
1
( 4 kips )( 2 ft ) = 4 kip ⋅ ft, to 66 kip ⋅ ft at C. M then decreases by
2
( 4 kips )( 6 ft ) to 42 kip ⋅ ft at D, and by ( 7 kips )( 6 ft ) to zero at B.
M max = 70 kip ⋅ ft at E ΣFy = 0:
Ay − ( 2 kips/ft )( 8 ft ) − 3 kips + 10 kips = 0
A y = 9 kips
(b)
ΣM A = 0:
M A + ( 4 ft )( 2 kips/ft )( 8 ft ) + (14 ft )( 3 kips )
− ( 20 ft )(10 kips ) = 0
M A = 94 kip ⋅ ft
Shear Diag:
V jumps to 9 kips at A, then decreases, at 2 kips/ft , to − 7 kips at C to D,
drops 3 kips to −10 kips from D to B and jumps 10 kips to 0.
Note: V = 0 where 9 kips − ( 2 kips/ft ) x = 0, x = 4.5 ft.
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
Moment Diag:
M jumps to 94 kip ⋅ ft at A and increases to
94 kip ⋅ ft +
1
( 9 kips )( 4.5 ft ) = 114.25 kip ⋅ ft, then decreases by
2
1
( 7 kips )( 3.5 ft ) to 102 kip ⋅ ft at C. M decreases linearly by
2
( 7 kips )( 6 ft ) to 60 kip ⋅ ft at D, then by (10 kips )( 6 ft ) to zero at
B.
M max = 114.3 kip ⋅ ft at E ( 4.5 ft from A) 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 7, Solution 75.
(15 ft ) C y + ( 9 ft )( 600 lb ) − ( 4.5 ft )(800 lb/ft )( 9 ft ) = 0
ΣM A = 0:
C y = 1800 lb = 1.8 kips
ΣFy = 0:
Ay + 1800 lb − ( 800 lb )( 9 ft ) + 600 lb = 0
A y = 4800 lb = 4.8 kips
Shear Diag:
V jumps to 4.8 kips at A then decreases linearly, at 0.8 kips/ft, to
− 2.4 kips at B, jumps 0.6 kips to −1.8 kips, is constant to C, and jumps
1.8 kips to zero.
Note: V = 0 at D, where 4.8 kips − ( 800 kip/ft ) x = 0, x = 6.0 ft.
Moment Diag:
M starts at zero and increases with decreasing slope to
1
( 4.8 kips )( 6 ft ) = 14.4 kip ⋅ ft at D, then decreases by
2
1
( 2.4 kips )( 3 ft ) to 10.8 kip ⋅ ft at B. M then decreases with slope
2
−1.8 kips to zero at C.
M max = 14.40 kips at D (6 ft from A)
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 7, Solution 76.
(a)
ΣM A = 0:
( 6 m ) By − (1 m )( 20 kN/m )( 2 m ) − 24 kN ⋅ m
− ( 5 m )( 20 kN/m )( 2 m ) = 0
B y = 44 kN
ΣFy = 0: Ay − 2 ( 20 kN/m )( 2 m ) + 44 kN = 0
A y = 36 kN
Shear Diag:
V jumps to 36 kN at A, then decreases with slope –20 kN/m to − 4 kN
at C, is constant to E, then decreases with slope –20 kN/m to − 44 kN
at B.
Note: V = 0 at F where 36 kN − ( 20 kN/m ) x = 0, x = 1.8 m.
Moment Diag:
Starting at zero M increases with decreasing slope to
1
(36 kN )(1.8 m)
2
1
( 4 kN )( 0.2 m ) to 32 kN ⋅ m at C ,
2
then with slope − 4 kN to 28 kN ⋅ m at D, where it jumps to 52 kN ⋅ m,
M decreases with slope − 4 kN to 48 kN ⋅ m at E, then with increasingly
= 32.4 kN ⋅ m at F , decreases by
(b)
 4 + 44

negative slope by 
kN  ( 2 m ) to zero at B.
2


M max = 52 kN ⋅ m ( at D ) W
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 7, Solution 77.
(a)
ΣM A = 0:
( 6 m ) By − (1 m )( 20 kN/m )( 2 m ) + 24 kN ⋅ m
− ( 5 m )( 20 kN/m )( 2 m ) = 0
ΣFy = 0:
B y = 36 kN
Ay − 2 ( 20 kN/m )( 2 m ) + 36 kN = 0
A y = 44 kN
Shear Diag:
V jumps to 44 kN at A, then decreases with slope − 20 kN/m to 4 kN at
C, is constant to E, then decreases with slope − 20 kN/m to −36 kN at B.
V = 0 at F where − 36 kN + ( 20 kN/m ) x = 0, x = 1.8 m.
Moment Diag:
 44 + 4 
kN ( 2 m )
M starts at zero, increases with decreasing slope to 
 2

= 48 kN ⋅ m at C , increases with slope 4 kN to 52 kN ⋅ m at D, drops
24 kN ⋅ m to 28 kN ⋅ m then increases with slope 4 kN to 32 kN ⋅ m at E.
1
Then M increases with decreasing slope, by ( 4 kN )( 0.2 m ) to
2
32.4 kN ⋅ m at F and decreases with increasingly negative slope to
zero at B.
(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.
M max = 52.0 kN ⋅ m ( at D ) COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 78.
(a)
Note: The 2 kip force at E has been replaced by the equivalent force and
couple at C.
ΣM A = 0: − ( 6 ft )(1 kip/ft )(12 ft ) + 8 kip ⋅ ft − (12 ft )( 2 kips )
+ ( 24 ft ) Dy − ( 32 ft )(1 kip ) = 0
ΣFy = 0:
D y = 5 kips
Ay − (1 kip/ft )(12 ft ) − 2 kips + 5 kips − 1 kip = 0
A y = 10 kips
Shear Diag:
From 10 kips at A, V decreases with slope −1 kip/ft to − 2 kips at C,
drops 2 kips, is constant at − 4 kips to D, jumps 5 kips, and is constant at
1 kip to B.
V = 0 at E, where 10 kips − (1 kip/ft ) x = 0, x = 10 ft.
Moment Diag:
From zero at A, M increases with decreasing slope to
1
(10 kips)(10 ft )
2
1
( 2 kips )( 2 ft ) to 48 kip ⋅ ft at C, drops
2
8 kip ⋅ ft to 40 kip ⋅ ft, then decreases with slope − 4 kips to − 8 kip ⋅ ft at
D. Finally M increases with slope 1 kip to zero at B.
= 50 kip ⋅ ft at F , decreases by
(b)
M max = 50 kip ⋅ ft at F (10.00 ft from A ) W
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 7, Solution 79.
(a)
Note: The 2 kip force at E has been replaced by the equivalent force and
couple at C.
ΣM A = 0: − ( 6 ft )(1 kip/ft )(12 ft ) + (12 ft )( 2 kips ) − 8 kip ⋅ ft
+ ( 24 ft ) Dy − ( 32 ft )(1 kip ) = 0
ΣFy = 0:
Ay − (1 kip/ft )(12 ft ) + 2 kips −
Ay =
Dy =
11
kips
3
11
kips − 1 kip = 0
3
22
kips
3
Shear Diag:
22
14
Starting at
kips at A, V decreases with slope −1 kip/ft to −
kips
3
3
8
11
kips
at C, jumps 2 kips and remains constant at − kips to D, jumps
3
3
and remains constant at 1 kip to B, drops to zero.
V = 0 at F, where
22
kip − (1 kip/ft ) x = 0,
3
x=
22
ft.
3
Moment Diag:
Starting from zero, M increases with decreasing slope to
1  22
 22 
kips 
ft  = 26.889 kip ⋅ ft at F . M then decreases by

2 3
 3 
1  14
 14 
kips  ft  to 16 kip ⋅ ft at C , jumps to 24 kip ⋅ ft, decreases

2 3
 3 
8
with slope − kips to − 8 kip ⋅ ft at D, and finally increases with slope
3
1 kip to zero at B.
(b)
M max = 26.9 kip ⋅ ft at F ( 7.33 ft from A) W
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 7, Solution 80.
(b)
 x

w = w0  4   − 1
 L

(a) Distributed load
dV
= −w, and V ( 0 ) = 0, so
dx
Shear:
x
x/L
0
0
V = ∫ − wdx = − ∫
V =∫
x/L
0
x
Lwd  
L
2
 x

 
x
 x   x 
wo L 1 − 4    d   = w L   − 2  
0
 L 
 L   L 
L 



Notes: At x = L, V = −w0 L
x
x
  = 2 
L
L
And V = 0 at
2
x
1
=
L
2
or
1
x
Also V is max where w = 0  = 
4
L
Vmax =
1
w0 L
8
M ( 0 ) = 0,
Moment:
x
x/L
0
0
M = ∫ vdx = L ∫
2
M = w0 L ∫
x/L
0
dM
=V
dx
x x
V  d  
L L
2
 x 
x  x
  − 2    d  
 L    L 
 L 
2
 x 
x 
V = w0 L   − 2     L  
 L 
 1  x  2 2  x 3 
M = w0 L    −    3  L  
 2  L 
2
M max =
1
L
w0 L2 at x =
24
2
1
M min = − w0 L2 at x = L
6
M max =
(c)
w0 L2
L
at x =
24
2
M max = −M min =
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.
w0 L2
at B 6
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 81.
(b)
ΣFy = 0: Ay −
1
1
( 2w0 ) ( 2a ) + w0a = 0
2
2
Ay =
3
w0a
2
2
1
7 1
( 2a ) ( 2w0 ) ( 2a ) −  a  ( w0 ) ( a ) = 0
3
2
3 2
ΣM A = 0: M A +
3
M A = − w0a 2
2
(a) for 0 ≤ x ≤ 2a:
w=
w0
x
a
V =
3
M = − w0a 2 +
2
∫ a xdx = 2a (3a − x ) W
xw
0
3
w0a −
2
w0
2
2
0
∫ 2a (3a − x ) dx
xw
0
2
2
0
M =
(
)
(
)
w0
− 9a 3 + 9a 2 x − x3 W
6a
Note: V = 0 at x = 3a, where M = 0.232 w0a 2
At x = 2a,
M =
1
w0a 2
6
for 2a ≤ x ≤ 3a :
1
V = − w0a +
2
w = − 3w0 +
w0
x
a
x w
0
∫ a (3a − x ) dx
2a
V =
M =
1
w0a +
6
∫ 2a ( − 9a + 6ax − x ) dx
x w
0
2
2
2a
M =
(c)
w0
− 9a 2 + 6ax − x 2 W
2a
(
)
w0
27a3 − 27a 2 x + 9ax 2 − x3 W
6a
M max =
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.
3
w0a 2 at A ( x = 0 ) W
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 82.
(a)
FBD Beam:
ΣM B = 0:
1
2
( 3a )  w0 ( 3a ) − 5aAy = 0
ΣFy = 0: 0.9w0a −
A y = 0.9w0a

1
w0 ( 3a ) + B = 0
2
B = 0.6w0a
Shear Diag:
V = Ay = 0.9w0a from A to C and V = B = − 0.6w0a from B to D.
x1
. If x1 is measured right to left,
3a
dV
x w
dM
= + w and
= − V . So, from D, V = − 0.6w0a + ∫ 0 1 0 x1dx1,
dx1
dx1
3a
Then from D to C, w = w0
2

1  x1  
V = w0a  − 0.6 +   
6  a  

2
x 
Note: V = 0 at  1  = 3.6, x1 =
a
3.6 a
Moment Diag:
 dM

M = 0 at A, increasing linearly 
= 0.9w0a  to M C = 0.9w0a 2.
 dx1

 dM

Similarly M = 0 at B increasing linearly 
= 0.6w0a  to
dx


M D = 0.6w0a 2. Between C and D
2

1  x1  

M = 0.6w0a + w0a ∫ 0.6 −   dx1,
0 
6 a  


2
x1
3

x  1 x  
M = w0a 2 0.6 + 0.6  1  −  1  
 a  18  a  

(b)
At
x1
=
a
3.6, M = M max = 1.359w0a 2 x1 = 1.897a left of D
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 7, Solution 83.
ΣM A = 0: LBy −
V =−
w0 L
+
20
L1

 w0 L  = 0
54

∫
By =
x1 w
w
0 3
x dx1 = 0 3
3 1
0 L
20L
( 5x − L )
4
1
4
w0 
4
5 ( L − x ) − L4  W
3 

20 L 
V =
or
w0 L
20
Note: V = 0 at x1 = 5−1/ 4 L = 0.6687 L
or
x = 0.3313 L
M =0+
=
∫ 20L (5x − L ) dx
x1
0
w0
4
1
3
(
)
or
M =
w0
x15 − L4 x1
20L3
4
1
w0 
5
L − x ) − L4 ( L − x )  W
(
3 

20L 
M max = M at x1 = 0.6687 L,
x = 0.331 L W
M max = 0.0267 w0 L2 W
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 7, Solution 84.
(a)
ΣM C = 0: ( 395 − 215 ) N ⋅ m + ( 0.1 m )(1000 N/m )( 0.2 m )
FBD section AC:
− ( 0.2 m )VA = 0,
VA = 1000 N
ΣM B = 0: 395 N ⋅ m + ( 0.45 m )(1 kN/m )( 0.4 m )
− ( 0.65 m )(1 kN ) + ( 0.25 m ) P = 0
FBD whole:
(b)
P = 0.3 kN,
P = 300 N W
ΣFy = 0: 1000 N − 400 N − 300 N − Q = 0
Q = 300 N W
Shear Diag:
Starting at 1000 N, V decreases with slope − 1000 N/m to 600 N at D, drops
300 N and is constant to B where it drops 300 N to zero.
Moment Diag:
Starting at − 395 N ⋅ m, M increases with decreasing slope to −395 N ⋅ m
 1000 + 600 
N ( 0.4 m ) = − 75 N ⋅ m at D, then increases with slope 300 N
+


2
to zero at 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 85.
(a)
ΣM C = 0:
( 395 − 215) N ⋅ m − ( 0.2 m ) Ay
FBD section AC:
+ ( 0.1 m )(1000 N/m )( 0.2 m ) = 0
A y = 1000 N
ΣM B = 0: 395 N ⋅ m + ( 0.325 m )(1000 N/m )( 0.65 m )
+ ( 0.25 m ) P = 0 − ( 0.65 m )(1000 N ) = 0
(b)
FBD Whole:
P = 175.0 N W
ΣFy = 0: 1000 N − 175 N − Q − (1000 N/m )( 0.65 m ) = 0
Q = 175.0 N W
Shear Diag:
V starts at 1000 N, and has slope − 1000 N/m throughout, but with drops of
175 N at D and B.
Moment Diag:
M starts at − 395 N ⋅ m and increases with decreasing slope to − 295 N ⋅ m
 1000 + 600 
+
N ( 0.4 m ) = − 75 N ⋅ m at D. There is a discontinuity in slope,


2
 425 + 175 
N  ( 0.25 m ) to zero at B.
and M increases by 
2


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 7, Solution 86.
(a)
FBD AC:
ΣM C = 0: 2.7 kN ⋅ m − ( 0.2 m ) Ay = 0,
A y = 1.35 kN
( 0.2 m ) Dy − 2.5 kN ⋅ m = 0,
D y = 1.25 kN
ΣM D = 0:
ΣM C = 0: − ( 2 m )(1.35 kN ) − ( 2 m )( 0.4 kN/m )( 4 m ) − ( 4 m ) Q
+ ( 6 m )(1.25 kN ) = 0,
FBD AB:
Q = 0.4 kN
Q = 400 N W
ΣFy = 0:
1.35 kN − P − ( 0.4 kN/m )( 4 m ) − 0.4 kN + 1.25 kN = 0
P = 0.6 kN
P = 600 N W
(b)
FBD Whole:
Shear Diag:
V is constant at 1.35 kN from A to C, drops 0.6 kN, then decreases with
slope –0.4 kN/m ( −1.6 kN ) to − 0.85 kN at D, drops 0.4 kN to
−1.25 kN, and is constant to B.
V = 0 where 0.75 kN – (0.4 kN/m)x = 0,
x = 1.875 m.
Moment Diag:
From zero at A, M increases with slope 1.35 kN to 2.70 kN ⋅ m at C, the
slope drops to 0.75 kN and then decreases to zero at E, where
1
M = 2.7 kN ⋅ m + ( 0.75 kN )(1.875 m ) = 3.403 kN ⋅ m. This curve
2
1
continues to D where M = 3.403 kN ⋅ m − ( 0.85 kN )( 2.125 m )
2
= 2.50 kN ⋅ m, then M decreases with slope −1.25 kN to zero at 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 87.
(a)
FBD AC:
ΣM C = 0: 2.7 kN ⋅ m − (1.35 m ) Ay = 0,
A y = 2 kN
ΣM D = 0: 2.5 kN ⋅ m − ( 2 m ) By = 0,
B y = 1.25 kN
ΣM C = 0: ( 6.65 m )(1.25 kN ) − ( 4.65 m ) Q
− ( 2.325 m )( 0.4 kN/m )( 4.65 m ) − (1.35 m )( 2 kN ) = 0
Q = 0.27699 kN
FBD DB:
Q = 277 N
W
ΣFy = 0: 2 kN − P − 0.277 kN − ( 0.4 kN/m )( 4.65 m ) + 1.25 kN = 0
P = 1.113 kN
(b)
FBD Whole:
W
Shear Diag:
V is constant at 2 kN from A to C, drops 1.113 kN to 0.887 kN, then decreases
with slope − 0.4 kN/m to 0.887 kN − ( 0.4 kN/m ) ( 4.65 m )
= − 0.973 kN, drops 0.277 kN to −1.25 kN and is constant to B.
V = 0 where 0.887 kN − ( 0.4 kN/m ) x = 2.2175 m at E .
Moment Diag:
Starting from zero, M increases with slope 2 kN to 2.7 kN ⋅ m at C. The
slope drops to 0.887 kN and decreases to zero at E where M = 2.7 kN
1
+ ( 0.887 kN )( 2.2175 m ) = 3.68 kN ⋅ m. This curve continues to D
2
where M = 2.5 kN ⋅ m, then M decreases linearly to zero at 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 88.
FBD Cable:
ΣM A = 0:
( 5 m ) Dy − ( 4 m )( 2.25 kN ) − ( 2 m )( 3 kN ) = 0
D y = 3 kN
ΣFy = 0:
Ay − 3 kN − 2.25 kN + 3 kN = 0
A y = 2.25 kN
Point D:
ΣFx = 0: − Ax + Dx = 0,
Ax = Dx
(1)
Since Ax = Dx and Dy > Ay , TCD is Tmax
ΣFy = 0: 3 kN −
ΣFx = 0: −
Point A:
3
TCD = 0
5
TCD = 5 kN
4
( 5 kN ) + Dy = 0
5
D x = 4 kN
From(1),
A x = 4 kN
dB
2.25 kN
=
2m
4 kN
(a )
d B = 1.125 m W
(b) A x = 4.00 kN
W
A y = 2.25 kN W
(c)
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.
Tmax = 5.00 kN W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 89.
FBD Cable:
ΣM A = 0:
( 5 m ) Dy − ( 4 m )( 2.25 kN ) − ( 2 m )( 3 kN ) = 0
D y = 3 kN
ΣFy = 0:
Ay − 3 kN − 2.25 kN + 3 kN = 0
A y = 2.25 kN
ΣFx = 0: − Ax + Dx = 0
Point D:
Since Ax = Dx , and Dy > Ay ,
Ax = Dx
Tmax is TCD = 3.6 kN
1 + dC2
dC
1m
=
=
3 kN
Dx
3.6 kN
1.2 dC = 1 + dC2 ,
1.44 dC2 = 1 + dC2
Point A:
dC = 1.50756 m
1 m 
Also Dx = 3 kN 
 = 1.98997 kN = Ax
 dC 
dB
2m
=
2.25 kN 1.9900 kN
(a) d B = 2.26 m W
(b) dC = 1.508 m W
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 7, Solution 90.
ΣM A = 0:
4aE y − 3a (1.2 kips ) − 2a ( 0.8 kips ) − a ( 0.4 kips ) = 0
FBD Cable:
E y = 1.4 kips
ΣFy = 0:
AY − ( 0.4 + 0.8 + 1.2 kips ) + 1.4 kips = 0
A y = 1.0 kips
ΣFx = 0:
− Ax + Ex = 0,
Since Ax = Ex and E y > Ay ,
FBD CDE:
ΣM C = 0:
Ax = Ex
Tmax = TDE
( 30 ft )(1.4 kips ) − (12 ft ) Ex − (15 ft )(1.2 kips ) = 0
(a)
E x = 2.00 kips W
E y = 1.400 kips W
(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.
Tmax = TDE = 2.44 kips W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 91.
FBD Cable:
ΣM A = 0:
4aE y − 3a (1.2 kips ) − 2a ( 0.8 kips ) − a ( 0.4 kips ) = 0
E y = 1.4 kips
ΣFy = 0:
Ay − ( 0.4 + 0.8 + 1.2 kips ) + 1.4 kips = 0
A y = 1.0 kips
ΣFx = 0:
Point E:
− Ax + Ex = 0,
Ax = Ex
Since Ax = Ex , and E y > Ay ,
Tmax = TDE
Ex =
( 5 kips )2 + (1.4 kips )2 = 4.8 kips
FBD CDE:
ΣM C = 0:
( 30 ft )(1.4 kips ) − dC ( 4.8 kips ) − (15 ft )(1.2 kips ) = 0
dC = 5.00 ft
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 7, Solution 92.
(a) FBD Whole:
ΣM A = 0:
(1.2 m ) Ex + ( 4 m ) E y − (1 m )(1.8 kN )
− ( 2 m )( 3.6 kN ) − ( 3 m )(1.2 m ) = 0
1.2Ex + 4E y = 12.6 kN
FBD CDE:
ΣM C = 0:
(1)
( 2 m ) E y − ( 0.6 m ) Ex − (1 m )(1.2 kN ) = 0
− 0.6Ex + 2E y = 1.2 kN
Solving (1) and (2)
(2)
E x = 4.25 kN
E y = 1.875 kN
E = 4.65 kN
(a)
(b) From FBD whole:
AB:
ΣFx = 0
ΣFy = 0
− Ax + 4.25 kN = 0
23.8°
A x = 4.25 kN
Ay − 1.8 kN − 3.6 kN − 1.2 kN + 1.875 kN = 0
A y = 4.725 kN
ΣM B = 0:
d B ( 4.25 kN ) − (1 m )( 4.725 kN ) = 0
d B = 1.112 m
DE:
ΣM D = 0:
(1 m )(1.875 kN ) − ( d D − 1.2 m )( 4.25 kN ) = 0
d D = 1.641 m
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 7, Solution 93.
FBD DE:
FBD Whole:
ΣFy = 0:
E y − 1.2 kN = 0
E y = 1.2 kN
ΣM D = 0:
(1 m )(1.2 kN ) − ( dC − 1.2 m ) Ex = 0
ΣM A = 0:
(1.2 m ) Ex + ( 4 m )(1.2 kN ) − ( 3 m )(1.2 kN )
(1)
− ( 2 m )( 3.6 kN ) − (1 m )(1.8 kN ) = 0
E x = 6.5 kN
dC = 1.385 m
(a) then, from (1)
ΣFx = 0:
ΣFy = 0:
− Ax + 6.5 kN = 0
A x = 6.5 kN
Ay − 1.8 kN − 3.6 kN − 1.2 kN + 1.2 kN = 0
A y = 5.4 kN
So
(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.
A = 8.45 kN
39.7°
E = 6.61 kN
10.46°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 94.
FBD Cable:
Hanger forces at A and F act on the supports, so A y and Fy act on the
cable.
ΣM F = 0:
( 6 ft + 12 ft + 18 ft + 24 ft )( 400 lb )
− ( 30 ft ) Ay − ( 5 ft ) Ax = 0
Ax + 6 Ay = 4800 lb (1)
FBD ABC:
ΣM C = 0:
( 7 ft ) Ax − (12 ft ) Ay + ( 6 ft )( 400 lb ) = 0
(2)
Solving (1) and (2) A x = 800 lb
Ay =
2000
lb
3
ΣFx = 0:
From FBD Cable:
− 800 lb + Fx = 0
FBD DEF:
Fx = 800 lb
ΣFy = 0:
2000
lb − 4 ( 400 lb ) + Fy = 0
3
Fy =
Since Ax = Fx and Fy > Ay ,
Tmax = TEF =
(a) Tmax = 1229.27 lb,
ΣM D = 0:
2800
lb
3
(800 lb )
2
 2800 
lb 
+
 3

2
Tmax = 1.229 kips
(12 ft ) 
2800 
lb  − d D ( 800 lb ) − ( 6 ft )( 400 lb ) = 0
 3

(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.
d D = 11.00 ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 95.
FBD CDEF:
ΣM C = 0:
(18 ft ) Fy − ( 9 ft ) Fy − ( 6 ft + 12 ft )( 400 lb ) = 0
Fx − 2Fy = − 800 lb
FBD Cable:
ΣM A = 0:
(1)
( 30 ft ) Fy − ( 5 ft ) Fx − ( 6 ft )(1 + 2 + 3 + 4 )( 400 lb ) = 0
Fx − 6Fy = − 4800 lb
Solving (1) and (2), Fx = 1200 lb
ΣFx = 0:
Point F:
ΣFy = 0:
(2)
, Fy = 1000 lb
− Ax + 1200 lb = 0 ,
A x = 1200 lb
Ay + 1000 lb − 4 ( 400 lb ) = 0 ,
A y = 600 lb
Since Ax = Ay and Fy > Ay , Tmax = TEF
Tmax =
FBD DEF:
(1 kip )2 + (1.2 kips )2
Tmax = 1.562 kips (a)
ΣM D = 0:
(12 ft )(1000 lb ) − d D (1200 lb ) − ( 6 ft )( 400 lb ) = 0
(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.
d D = 8.00 ft COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 96.
ΣM A = 0:
( 9 ft ) P − ( 6 ft )( 30 lb ) − a ( 20 lb ) = 0
FBD BC:
20a 

P =  20 +
 lb with a in ft.
9 

ΣFx = 0:
ΣFy = 0:
− TAB x + P = 0
(1)
TAB x = P
TAB y − 20 lb − 30 lb = 0
But
TAB x
a
=
TAB y
7
Solving (1) and (2),
a = 4.0645 ft,
so
TAB y = 50 lb
P=
50a
7
(2)
P = 29.032 lb
(a) P = 29.0 lb (b) a = 4.06 ft 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 7, Solution 97.
FBD C:
ΣM B = 0:
( 2 ft )( 25 lb ) − ( 6 ft − a )( 30 lb ) = 0
a=
13
ft
3
a = 4.33 ft
FBD BC:
ΣM A = 0:
( b + 2 ft )( 25 lb ) − ( 6 ft )( 30 lb ) − 
13 
ft  ( 20 lb ) = 0
 3 
b=
32
ft
3
b = 10.67 ft
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 7, Solution 98.
ΣFx = 0: −
FBD B:
ΣFy = 0:
12
3
TAB + TBC + 1.32 kN = 0
13
5
5
4
TAB − TBC = 0
13
5
Solving: TAB = 2.08 kN, TBC = 1 kN
By inspection of A,
A = 2.08 kN
(a)
FBD C:
ΣFx = 0:
ΣFy = 0:
12
3
TCD − (1 kN ) = 0 ,
13
5
22.6°
TCD = 0.65 kN
4
5
(1 kN ) + ( 0.65 kN ) − w = 0
5
13
w = 1.05 kN
(b) m =
(c)
w
1050 N
=
= 107.03 kg
g
9.81 m/s 2
From above
m = 107.0 kg
TAB = 2.08 kN
TBC = 1.000 kN
TCD = 650 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 7, Solution 99.
FBD C:
(
)
W = mg = (150 kg ) 9.81 m/s 2 = 1471.5 N = 1.4715 kN
ΣFx = 0:
12
3
TCD − TBC = 0
13
5
5
4
TCD + TBC − 1.4715 kN = 0
13
5
ΣFy = 0:
Solving: TCD = 0.91093 kN = 911 N, TBC = 1.40143 kN
FBD B:
By inspection of D,
ΣFx = 0:
ΣFy = 0:
P−
(a)
D = 911 N
22.6°
12
3
TAB + (1.40143 kN ) = 0
13
5
5
4
TAB − (1.40143 kN ) = 0,
13
5
(b)
From above
(c)
TAB = 2.91497 kN
P = 1.850 kN
TAB = 2.91 kN
TBC = 1.401 kN
TCD = 911 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 7, Solution 100.
FBD E:
Ey
Ex
T
=
= DE
1
1
2
FBD CDE:
ΣM C = 0:
Ex = E y =
35
wD
23
35

wD  − ( 5.1 m ) wD − (1.6 m ) wC = 0
 23

(8.1 m − 3 m ) 
wC = 1.66304wD ,
mC = 1.66304 mD
FBD B:
(a)
ΣFx = 0:
4
− TBC + Ex = 0,
5
ΣFx = 0:
45 
1
24
TAB −
TBF = 0
 Ex  −
54 
25
2
ΣFy = 0:
Solving:
(1)
( 6.5 m ) E y − ( 4.2 m ) Ex − ( 3.5 m ) wD = 0
2.3Ex = 3.5wD ,
ΣM B = 0:
Ex = E y
TBC =
mC = 56.5 kg 5
Ex
4
1
7
3 5 
TAB −
TBF −  Ex  = 0
25
5 4 
2
Ex
31
=
TBF ,
4
25
TBF =
TBF =
25
Ex
124
25  35 
  wD = 0.30680wD
124  23 
wD = ( 34 kg )( 9.81 N/kg )
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.
(b)
TBF = 102.3 N COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 101.
FBD E:
Ey
Ex
=
1
1
ΣM C = 0:
( 6.5 m ) E y − ( 4.2 m ) Ex − ( 3.5 m ) wD = 0
2.3Ex = 3.5wD ,
FBD CDE:
ΣM B = 0:
Ex = E y
(8.1 m − 3 m )
Ex = E y =
35
wD
23
35
wD − ( 5.1 m ) wD − (1.6 m ) wC = 0
23
wD = 0.60131 wC ,
mD = 0.60131 mC = 33.072 kg
mD = 33.1 kg
(a)
4
ΣFx = 0: − TBC = Ex ,
5
Point B:
ΣFx = 0:
ΣFy = 0:
Solving:
TBC =
5
Ex
4
45 
1
24
TAB −
TBF = 0
 Ex  −
54 
25
2
1
7
3 5 
TAB −
TBF +  Ex  = 0
25
5 4 
2
31
1
TBF = Ex ,
25
4
TBF =
TBF =
25
Ex
124
(
25  35
 25 35
( 33.072 kg ) 9.81 m/s2
 wD  =
124  23
 124 23
(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.
)
TBF = 99.5 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 102.
(a) FBD half wire:
Since h = 1.5 m L = 30 m we can approximate the load as evenly
distributed horizontally, and the length S = L.
kg  
N

W =  0.6
 ( 30 m ) = 176.58 N
  9.81
m 
kg 

ΣM B = 0:
(15 m )(176.58 N ) − (1.5 m ) TC = 0
TC = 1765.8 N
Tmax = TB = TC2 + W 2
Tmax =
(b)
(1765.8 N )2 + (176.58 N )2 ,
Tmax = 177.5 kN
2
4


2  yB 
2  yB 

S B = xB 1 + 
 − 
 + ...
3  xB 
5  xB 




2
4


2  1.5 
2  1.5 
= ( 30 m ) 1 + 
−


 + ... = 30.05 m
3  30 
5  30 


Note: the third term in the brackets is unnecessary
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.
S tot = 60.1 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 103.
Half-cable FBDs:
T1x = T2 x to create zero horizontal force on tower → thus T01 = T02
FBD I:
(15 m )  w ( 30 m ) − h1T0 = 0
ΣM B = 0:
( 450 m ) w
h =
2
1
FBD II:
T0
( 2 m ) T0 − (10 m )  w ( 20 m ) = 0
ΣM B = 0:
T0 = (100 m ) w
( 450 m ) w = 4.50 m h =
2
(a)
ΣFx = 0:
FBD I:
1
T1x − T0 = 0
T1x = (100 m ) w
ΣFy = 0:
T1y − ( 30 m ) w = 0
T1y = ( 30 m ) w
T1 =
(100 m )2 + ( 30 m )2 w
(
= (104.4 m )( 0.4 kg/m ) 9.81 m/s 2
)
= 409.7 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.
(100 m ) w
COSMOS: Complete Online Solutions Manual Organization System
ΣFy = 0:
FBD II:
T2y − ( 20 m ) w = 0
T2y = ( 20 m ) w
T2 x = T1x = (100 m ) w
T2 =
(100 m )2 + ( 20 m )2 w = 400.17 N
(b)
T1 = 410 N T2 = 400 N *Since h
L it is reasonable to approximate the cable weight as being distributed uniformly along the
horizontal. The methods of section 7.10 are more accurate for cables sagging under their own weight.
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 7, Solution 104.
FBD half-span:
(a)
 2075 ft 

 ( 23032.5 kips ) − ( 464 ft ) T0 = 0
 2 
ΣM B = 0:
T0 = 47, 246 kips
Tmax = T02 + W 2 =
( 47, 246 kips )2 + ( 23, 033 kips )2 = 56, 400 kips
2
4


2 y
2 y
s = x 1 +   −   + "
3 x 
5 x


(b)
2
4


2  464 ft 
2  464 ft 
sB = ( 2075 ft ) 1 + 
−


 + "
3  2075 ft 
5  2075 ft 


sB = 2142 ft
l = 2sB
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.
l = 4284 ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 105.
FBD half-span:
W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips
ΣM B = 0: ( 875 ft )(17, 065 kips ) − ( 316 ft ) T0 = 0
T0 = 47, 246 kips
Tmax = T02 + W 2 =
( 47, 246 kips )2 + (17, 063 kips )2
(a)
Tmax = 50, 200 kips
2
4


2 y
2 y
s = x 1 +   −   + "
3 x 
5 x


2
4


2  316 ft 
2  316 ft 
−
sB = (1750 ft ) 1 + 


 + "
3  1750 ft 
5  1750 ft 


= 1787.3 ft
(b)
*
To get 3-digit accuracy, only two terms are needed.
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.
l = 2sB = 3575 ft
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 106.
FBD pipe:
neglecting friction
ΣM B = 0:
FBD half-cord:
r (Tmax − WB ) = 0,
Tmax = WB = 60 lb
Assuming the weight to be evenly distributed horizontally, and
S=L
W = ( 0.02 lb/ft )( 75 ft ) = 1.5 lb
T0 =
( 60 lb )2 − (1.5 lb )2 = 59.981 lb
ΣM B = ( 37.5 ft )(1.5 lb ) − h ( 59.981 lb ) = 0
(a)
(b)
h = 0.93780 ft, h = 11.25 in.
θ B = sin −1
1.5 lb
= 1.43254°,
60 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.
θ B = 1.433°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 107.
(a) FBD ship:
ΣFx = 0:
T0 − 300 N = 0,
T0 = 300 N
FBD half-span:*
Tmax = T02 + W 2 =
(b)
ΣM A = 0: hTx −
L
W = 0,
4
2


2 4
s = x 1 +   + L
3 x


( 2.5 m ) =
h=
( 300 N )2 = ( 54 N )2 = 305 N LW
4Tx
but
yA = h =
LW
4Tx
so
yA
W
=
xA
2Tx
2

L
2  53.955 N 
1 + 
 − L → L = 4.9732 m
2 
3  600 N 

So h =
LW
= 0.2236 m
4Tx
*See note Prob. 7.103
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.
h = 224 mm COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 108.
2
4


2 y
2 y
s = x  1 +   −   + L
3 x
5 x 


Knowing
2
2


2 h 
2 h 
l = 2sTOT = L 1 + 
 − 
 + L
3  L/2 
5  L/2 




Winter:
2
4


2  386 ft 
2  386 ft 
L
lw = ( 4260 ft ) 1 + 
−
+
 = 4351.43 ft



3  2130 ft 
5  2130 ft 


Summer:
2
4


2  394 ft 
2  394 ft 
ls = ( 4260 ft )  1 + 
 − 
 + L = 4355.18 ft
3  2130 ft 
5  2130 ft 


∆l = ls − lw = 3.75 ft 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 7, Solution 109.
FBD whole:
ΣM D = 0:
( 2m ) TAx − ( 5 m )( 98.1 + 147.15) N = 0
TAx = 1103.6 N
ΣFx = 0:
FBD half-cable:
ΣFy = 0:
T0 − TAx = 0,
TAy = 49.05 N = 0,
ΣM A = 0:
T0 = 1103.6 N
TAy = 49.05 N
h (1103.6 N ) − ( 2.5 m )( 49.05 N ) = 0
h = 0.11111 m
h = 111.1 mm
(a)
(b)
T
θ = tan −1 Ay = tan −1
TAx
49.05
= 2.5449°,
1103.6
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.
θ = 2.54°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 110.
FBD AB:
ΣM A = 0:
(1100 ft ) TBy − ( 496 ft ) TBx − ( 550 ft )W = 0
11TBy − 4.96TBx = 5.5W
(1)
FBD CB:
ΣM C = 0:
( 550 ft ) TBy − ( 278 ft ) TBx − ( 275 ft )
11TBy − 5.56TBx = 2.75W
Solving (1) and (2)
TBy = 28,798 kips
Solving (1) and (2)
TBx = 51, 425 kips
tan θ B =
Tmax = TB = TB2x + TB2y ,
So that
W
=0
2
(2)
TB y
TBx
(a)
Tmax = 58,900 kips
(b)
θ B = 29.2°
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 7, Solution 111.
Cable profile:
Eqn:
y=
w 2
x
2T0
xB − x A = 45 in.
at A:
4 in. =
w
( xB − 45 in.)2
2T0
(1)
at B:
1 in. =
w 2
xB
2T0
(2)
∴
( xB − 45 in.)2 = 4
or
xB2
xB2 + 30 xB − 675 = 0
xB = ( −15 ± 30 ) in.
xB = 15 in.
x A = xB − 45 in.
x A = − 30 in.
(a)
lowest point (x = 0) is 30 in. from A
T0 =
From (2),
wxB2 1  0.18 lb 
2
= 
 (15 in.)
2
2  12 in. 
T0 = 1.6875 lb
Tmax occurs at A where slope is maximum
Tmax = T02 + ( wx A ) =
2
(1.6875 lb )2 + 
0.18 lb 
2
 ( − 30 in.) = 1.74647 lb
 12 in. 
(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.
Tmax = 1.747 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 112.
ΣM A = 0:
y AT0 −
a
wa = 0
2
ΣM B = 0: − yBT0 +
yA =
d = ( yB − yB ) =
But
∴
wa 2
2T0
yB =
(
w 2
b − a2
2T0
wb 2
2T0
)
2
2
T0 = TB2 − ( wb ) = Tmax
− ( wb )
2
2
2
− ( wb )  = w2 ( b 2 − a 2 ) = L2 w2 ( 4b 2 − 4 Lb + L2 )
( 2d )2 Tmax

2
or
Using
b
wb = 0
2

T2 
4 L2 + d 2 b 2 − 4 L3b +  L4 − 4d 2 max
=0
w2 

(
)
L = 6 m,
d = 0.9 m,
Tmax = 8 kN,
yields
b = ( 2.934 ± 1.353) m,
(
)
w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m
b = 4.287 m
( since b > 3 m )
(a)
a = 6 m − b = 1.713 m 2
2
T0 = Tmax
− ( wb ) = 7156.9 N
yA
wa
=
= 0.09979,
2T0
xA
yB
wb
=
= 0.24974
2T0
xB
2
2




2  yA 
2  yB 



l = s A + sB = a 1 + 
+
L
+
b
1
+
+
L



3  xA 
3  xB 








2
2
2
2


= (1.713 m ) 1 + ( 0.09979 )  + ( 4.287 m ) 1 + ( 0.24974 )  = 6.19 m
3
3




(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.
l = 6.19 m COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 113.
x
wx − yT0 = 0
2
ΣM P = 0:
Geometry:
wx 2
2T0
so
y
wx
=
x
2T0
and d = yB − y A =
w 2
b − a2
2T0
y =
(
)
Also
2
2


2  yA  
2  yB  
l = s A + sB a 1 +    + b 1 +   
3  a  
3  b  


FBD Segment:
2  y A 
w2 3
y  
a + b3
  +  B   =
 b   6T02
3  a 

2
l−L
2
(
)
(
2
3
3
1
4d 2
2d a +b
3
3
=
a +b =
6 b2 − a 2 2
3 b2 − a 2 2
(
)
(
)
(
)
)
Using l = 6.4 m, L = 6 m, d = 0.9 m, b = 6 m − a, and solving for a,
knowing that a < 3 ft
a = 2.2196 m
a = 2.22 m
(a)
T0 =
Then
(
w 2
b − a2
2d
(
)
)
w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m
And with
And
b = 6 m − a = 3.7804 m
Tmax = TB = T02 + ( wb )
=
T0 = 4338 N
2
( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2
Tmax = 5362 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.
(b)
Tmax = 5.36 kN
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 114.
FBD Cable:
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
(Where ΣM B loads includes all applied loads)
x

ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0
L

FBD AC:
(2)
(Where ΣM C left includes all loads left of C)
x
(1) − ( 2 ):
L
hT0 −
x
ΣM B loads + ΣM C left = 0
L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
FBD Beam:
ΣM C = 0: xABy − ΣM C left − M C = 0
FBD AC:
x
( 4 ) − ( 5):
L
−
(5)
x
ΣM B loads + ΣM C left + M C = 0
L
Comparing (3) and (6)
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.
M C = hT0
Q.E.D.
(6)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 115.
FBD Beam:
ΣM D = 0:
(1 m )( 2.25 kN ) + ( 3 m )( 3 kN ) − ( 5 m ) ABy = 0
A By = 2.25 kN
ΣFy = 0: 2.25 kN − 3 kN − 2.25 kN + DBy = 0
D By = 3 kN
FBD AB:
ΣM B = 0:
M B − ( 2 m )( 2.25 kN ) = 0,
M B = 4.5 kN ⋅ m
Note, since A and D are in line horizontally, ACy = A By and DCy = D By .
Also, since Dy > Ay ,
Tmax = TCD = 3.6 kN
2
2
T0 = TCD
− DCy
=
Cable:
( 3.6 kN )2 − ( 3 kN )2
T0 = 3.96 kN
dB =
M B 4.5 kN ⋅ m
=
= 2.2613 m
T0
3.96 kN
d B = 2.26 m
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 7, Solution 116.
FBD Beam:
ΣM E = 0:
(1 m )(1.2 kN ) + ( 2 m )( 3.6 kN ) + ( 3 m )(1.8 kN )
− ( 4 m ) ABy = 0
A By = 3.45 kN
ΣFx = 0: ABx = 0
FBD AB:
ΣM B = 0: M B − (1 m )( 3.45 kN ) = 0,
M B = 3.45 kN ⋅ m
FBD AC:
ΣM C = 0: M C + (1 m )(1.8 kN ) − ( 2 m )( 3.45 kN ) = 0
M C = 5.1 kN ⋅ m
ΣM D = 0: M D + (1 m )( 3.6 kN ) + ( 2 m )(1.8 kN ) − ( 3 m )( 3.45 kN ) = 0
FBD AD:
M D = 3.15 kN ⋅ m
hC = dC − 0.6 m = 1.8 m − 0.6 m = 1.2 m
Cable:
T0 =
M C 5.1 kN ⋅ m
=
= 4.25 kN
hC
1.2 m
hB =
M B 3.45 kN ⋅ m
=
= 0.81176 m
T0
4.25 kN
d B = hB + 0.3 m = 1.11176 m,
hD =
d B = 1.112 m
M D 3.15 kN ⋅ m
=
= 0.74118 m
T0
4.25 kN
d D = hD + 0.9 m = 1.64118 m,
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.
d D = 1.641 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 117.
FBD Beam:
ΣFx = 0:
ABx = 0
By symmetry A By = FBy = 800 lb
FBD AC:
ΣM C = 0: M C + ( 6 ft )( 400 lb ) − (12 ft )( 800 lb ) = 0
M C = 7200 lb ⋅ ft
By symmetry, M D = M C = 7200 lb ⋅ ft
∴
hC = hD
Cable:
hC = dC − 3 ft = 12 ft − 3 ft = 9 ft
d D = hD + 2 ft = 9 ft + 2 ft = 11 ft
d D = 11.00 ft
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 7, Solution 118.
FBD Beam:
ΣFx = 0:
ABx = 0
By symmetry A By = FBy = 800 lb
FBD AC:
ΣM C = 0:
M C + ( 6 ft )( 400 lb ) − (12 ft )( 800 lb ) = 0
M C = 7200 lb ⋅ ft
By symmetry, M D = M C = 7200 lb ⋅ ft
∴
Cable:
hC = hD
hC = dC − 3 ft = 9 ft − 3 ft = 6 ft
d D = hD + 2 ft = 6 ft + 2 ft = 8 ft
d D = 8.00 ft
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 7, Solution 119.
FBD Elemental segment:
ΣFy = 0:
Ty ( x + ∆x ) − Ty ( x ) − w ( x ) ∆ x = 0
Ty ( x + ∆x )
So
T0
Ty
But
T0
=
−
Ty ( x )
T0
=
dy
dx
dy
dy
−
w( x)
dx x + ∆x dx x
=
T0
∆x
So
In
lim :
∆x → 0
w( x)
d2y
=
2
T0
dx
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.
Q.E.D.
w( x)
∆x
T0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 120.
w ( x ) = w0 cos
πx
L
From Problem 7.119
w ( x ) w0
πx
d2y
=
=
cos
2
T0
T0
L
dx


dy
= 0
 using
dx 0


dy W0 L
πx
=
sin
dx
T0π
L
So
y =
But
And
w L2 
π
L
y   = h = 0 2 1 − cos 
2
T0π 
2
T0 = Tmin
Tmax = TA = TB :
TBy =
so
TBy
T0
=
w0 L2 
πx
1 − cos
  using y ( 0 ) = 0 
2 
L  
T0π 
so
T0 =
Tmin =
w0 L2
π 2h
dy
wL
= 0
dx x = L
T0π
2
w0 L
π
2
TB = TBy
+ T02 =
w0 L
π
 L 
1+ 

πh 
2
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.
w0 L2
π 2h
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 121.
Elemental Segment:
dx = cosθ ds,
But
w0
ds
cos 2 θ
w0
w( x) =
cos3 θ
w ( x ) dx =
Load on segment*
so
From Problem 7.119
d2y
w( x)
w0
=
=
2
T0
dx
T0 cos3 θ
In general
d2y
d  dy 
d
dθ
=
( tan θ ) = sec2 θ
 =
2
dx  dx  dx
dx
dx
So
dθ
w0
w0
=
=
3
2
dx
T0 cosθ
T0 cos θ sec θ
or
T0
cosθ dθ = dx = rdθ cosθ
w0
Giving r =
T0
= constant. So curve is circular arc
w0
*For large sag, it is not appropriate to approximate ds by dx.
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.
Q.E.D.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 122.
(
)
w = ( 0.07 kg/m ) 9.81 m/s 2 = 0.6867 N/m
FBD half-span:
L = 10 m
SCB =
12 m
=6m
2
SCB = c sinh
xB
,
c
6 m = c sinh
5m
c
Solving numerically, c = 4.6954 m
yB = c cosh
 5m 
xB
= ( 4.6954 m ) cosh 
 = 7.6188 m
c
 4.6954 m 
hB = yB − c = 7.6188 m − 4.6954 m = 2.9234 m
(a)
hB = 2.92 m
TB = wyB = ( 0.6867 N/m )( 7.6188 m ) = 5.2318 N
(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.
TB = 5.23 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 123.
sB = 30 ft,
w=
hB = 24 ft,
120 lb
= 2 lb/ft
60 ft
L
2
xB =
yB2 = c 2 + sB2 = ( hB + c )
2
= hB2 + 2chB + c 2
( 30 ft ) − ( 24 ft )
sB2 − hB2
=
2hB
2 ( 24 ft )
2
c=
2
c = 6.75 ft
Then
sB = c sinh
xB
s
→ xB = c sinh −1 B
c
c
 30 ft 
xB = ( 6.75 ft ) sinh −1 
 = 14.83 ft
 6.75 ft 
(a)
L = 2 xB = 29.7 ft
Tmax = TB = wyB = w ( c + hB ) = ( 2 lb/ft )( 6.75 ft + 24 ft ) = 61.5 lb
(b)
Tmax = 61.5 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 124.
S B = 250 ft
FBD half-span:
w = 2.8 lb/ft
hB = 125 ft
yB = hB + c = 125 ft + c
yB2 − sB2 = c 2
(125 ft − c )2 − ( 250 ft )2 = c 2
c = 187.5 ft
xB
,
c
250 ft = (187.5 ft ) sinh
xB
4
= sinh −1 = 1.0986,
c
3
xc = 205.99 ft
sB = c sinh
xB
c
(a)
span L = 2 xB = 411.98 ft
L = 412 ft
(b)
Tmax = wyB = ( 2.8 lb/ft )(125 ft + 187.5 ft ) = 875 lb
Tmax = 875 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 125.
FBD half-span:
sB = 65 m,
(
hB = 30 m
)
w = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m
yB2 = c 2 + sB2
( c + hB )2 = c 2 + sB2
2
c=
( 65 m ) − ( 30 m )
sB2 − hB2
=
2hB
2 ( 30 m )
2
= 55.417 m
Now
sB = c sinh
xB
s
 65 m 
→ xB = c sinh −1 B = ( 55.417 m ) sinh −1 

c
c
 55.417 m 
= 55.335 m
L = 2 xB = 2 ( 55.335 m ) = 110.7 m Tmax = wyB = w ( c + hB ) = ( 33.35 N/m )( 55.417 m + 30 m ) = 2846 N
Tmax = 2.85 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 7, Solution 126.
FBD Cable:
s = 30 m
30 m


= 15 m 
 so sB =
2


(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
hB = 12 m
yB2 = ( c + hB ) = c 2 + s 2B
2
c=
So
sB2 − hB2
2hB
2
2
15 m ) − (12 m )
(
c=
= 3.375 m
2 (12 m )
Now
sB = c sinh
xB
s
 15 m 
→ xB = c sinh −1 B = ( 3.375 m ) sinh −1 

c
c
 3.375 m 
xB = 7.4156 m
P = T0 = wc = ( 2.943 N/m )( 3.375 m )
L = 2 xB = 2 ( 7.4156 m )
(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.
P = 9.93 N
L = 14.83 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 127.
FBD Cable:
sT = 30 m,
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
P = T0 = wc,
c=
c=
P
w
30 N
= 10.1937 m
2.943 N/m
yB2 = ( hB + c ) = c 2 + sB2
2
h 2 + 2ch − sB2 = 0,
sB =
30 m
= 15 m
2
h 2 + 2 (10.1937 m ) h − 225 m 2 = 0
h = 7.9422 m
sB = c sinh
(a)
xA
s
 15 m 
→ xB = c sinh −1 B = (10.1937 m ) sinh −1 

c
c
 10.1937 m 
= 12.017 m
L = 2 xB = 2 (12.017 m ) (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.
L = 24.0 m
h = 7.94 m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 128.
FBD half-span:
xD =
3.6 w
= 1.8 m
2
kg 
m

w =  3.72
 9.81 2  = 36.4932 N/m
m
s



sD =
length 3.8 m
=
= 1.9 m
2
2
sD = c sinh
xD
c
1.9 m = c sinh
1.8 m
c
Solving numerically, c = 3.1433 m
yD2 − sD2 = c 2
yD2 = (1.9 m ) + ( 3.1433 m )
2
2
yD = 3.6729
h = yD − c = 3.6729 m − 3.1433 m = 0.5296 m
h = 0.530 m
(a)
h = 530 mm
(b)
TDx = 114.7 N
N

TDx = wc =  36.4932  ( 3.1433 m ) = 114.71 N
m

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 7, Solution 129.
FBD half-span:
sB =
90 m
= 45 m
2
xB =
L 60 m
=
= 30 m
2
2
TB = Tmax = 300 N
sB = c sinh
xB
,
c
Solving numerically,
yB = c cosh
45 m = c sinh
30 m
c
c = 18.495 m
 30 m 
xB
= (18.495 m ) cosh 

c
 18.495 m 
= 48.651 m
hB = yB − c = 48.651 m − 18.495 m = 30.156 m
Tmax = wyB
(a)
h = 30.2 m
(b)
m = 56.6 kg
300 N = w ( 48.651 m )
w = 6.1664
N
m
N

W = w ( length ) =  6.1664  ( 90 m ) = 554.97 N
m

m=
W 554.97 N
=
= 56.57 kg
g
9.81 m/s 2
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 7, Solution 130.
sB =
45 ft
= 22.5 ft
2
xB =
L = 20 ft
L
= 10 ft
2
sB = c sinh
xB
c
22.5 ft = c sinh
10 ft
c
Solving numerically:
yB = c cosh
c = 4.2023 ft
xB
c
= ( 4.2023 ft ) cosh
10 ft
= 22.889 ft
4.2023 ft
hB = yB − c = 22.889 ft − 4.202 ft
hB = 18.69 ft 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 7, Solution 131.
l = total length
xB
,
c
yB = c cosh
L + c = c cosh
1 = cosh
or
So
xB
c
c=
L
L
−
2c c
L
= 4.933
c
Solving numerically,
sB = c sinh
L/2
c
l
L
= c sinh
2
2c
l
 L
2sinh  
 2c 
=
10 ft
= 0.8550 ft
 4.933 
2sinh 

 2 
L = 4.933 c = ( 4.933)( 0.8550 ft ) = 4.218 ft
(a)
from y = c cosh
x
c
tan θ B =
L = 4.22 ft
dy
x
= sinh
dx
c
dy
L
 4.933 
= sinh
= sinh 
 = 5.848
dx B
2c
 2 
(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.
θ B = 80.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 132.
(
)
w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m
L = 48 m,
Tmax ≤ 1800 N
Tmax = wyB → yB =
yB ≤
Tmax
w
1800 N
= 61.162 m
29.43 N/m
x
yB = c cosh B ,
c
Solving numerically,
48 m
61.162 m = c cosh 2 *
c
c = 55.935 m
h = yB − c = 61.162 m − 55.935 m
h = 5.23 m *Note: There is another value of c which will satisfy this equation. It is much smaller, thus corresponding to a
much larger h.
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 7, Solution 133.
Pulley B:
TB = wa
but
TB = wyB ,
yB = a
so
now
yB = c cosh
xB
=a
c
So
a = c cosh
3 ft
c
sB = c sinh
also
(1)
xB
c
24 ft − a
3 ft
= c sinh
2
c
or
12 ft = c sinh
= c sinh
3 ft a
+
c
2
(2)
3 ft c
3 ft
+ cosh
c
2
c
c = 1.13194 ft
Solving numerically
or 17.7167 ft
a = c cosh
from (1)
3 ft
3 ft
= (1.13194 ft ) cosh
c
1.13194 ft
a = 8.05 ft
or


3 ft
a = (17.7167 ft ) cosh 

 17.7167 ft 
a = 17.97 ft
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 7, Solution 134.
FBD Cable:
s A = length = 10 ft
ΣFx = 0:
− F + T0 = 0,
y A = c + h = c cosh
So
1+
xA
a
h
= c cos h = c cosh
c
c
c
h
h
= cosh
c
c
Solving numerically,
s A = c sinh
F = T0
h
= 1.61614
c
h
= c sinh (1.61614 ) = 2.41748c = 10 ft
c
So c = 4.1365 ft,
h = 6.6852 ft
y A = c + h = 10.8217 ft
 20 lb 
F = T0 = wc = 
 ( 4.1365 ft ) = 8.2730 lb
 10 ft 
(a)
F = 8.27 lb
(b)
h = a = 6.69 ft
(c)
Tmax = 21.6 lb
 lb 
Tmax = wy A =  2  (10.8217 ft ) = 21.643 lb
 ft 
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 7, Solution 135.
FBD Cable:
Pulley
TA = mg = Tmax
m

mg = ( 40 kg )  9.81 2  = 392.4 N
s 

Also L = 15 m,
So x A = −
15 m
= − 7.5 m,
2
y A = c cosh
So
h=5m
c cosh
xB = + 7.5 m
xA
=c+h
c
− 7.5 m
=c+5m
c
Solving numerically,
sB = c sinh
c = 6.3175 m
 7.5 m 
xB
= ( 6.3175 m ) sinh 
 = 9.390 m
c
 6.3175 m 
cable length = 2sB = 18.78 m
(a)
Tmax = wyB = w ( c + h ) = w ( 6.3175 m + 5 m )
w=
(b)
392.4 N
= 34.672 N/m
11.3175 m
mass/length =
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.
w 34.672 N/m
kg
=
= 3.53
g
9.81 N/kg
m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 136.
yD = c cosh
xD
c
h + c = c cosh
a
c
10.8 ft


12 ft = c  cosh
− 1
c


Solving numerically, c = 6.2136 ft
Then
yB = ( 6.2136 ft ) cosh
10.8 ft
= 18.2136 ft
6.2136 ft
F = Tmax = wyB = (1.5 lb/ft )(18.2136 ft )
F = 27.3 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 137.
yD = c cosh
xD
c
c + h = c cosh
a
c
a


h = c  cosh − 1
c


18 ft


12 ft = c  cosh
− 1
c


Solving numerically
c = 15.162 ft
yB = h + c = 12 ft + 15.162 ft = 27.162 ft
F = TD = wyD = (1.5 lb/ft )( 27.162 ft ) = 40.74 lb
F = 40.7 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 138.
(
)
w = 4 kg/m 9.81 m/s 2 = 39.24 N/m
P = T0 = wc,
c=
P
800 N
=
w 39.24 N/m
c = 20.387 m
y = c cosh
x
c
dy
x
= sinh
dx
c
tan θ = −
dy
−a
a
= − sinh
= sinh
dx −a
c
c
a = c sinh −1 ( tan θ ) = ( 20.387 m ) sinh −1 ( tan 60° )
a = 26.849 m
y A = c cosh
a
26.849 m
= ( 20.387 m ) cosh
= 40.774 m
c
20.387 m
b = y A − c = 40.774 m − 20.387 m = 20.387 m
So
s = c sinh
(a)
B is 26.8 m right and 20.4 m down from A a
26.849 m
= ( 20.387 m ) sinh
= 35.31 m
c
20.387 m
(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.
s = 35.3 m COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 139.
(
)
w = ( 4 kg/m ) 9.81 m/s 2 = 39.24 N/m
P = T0 = wc
c=
P
600 N
=
w 39.24 N/m
c = 15.2905 m
x
y = c cosh ,
c
dy
x
= sinh
dx
c
dy
−a
a
= − sinh
= sinh
dx − a
c
c
At A:
tan θ = −
So
a = c sinh −1 ( tan θ ) = (15.2905 m ) sinh −1 ( tan 60° ) = 20.137 m
yB = h + c = c cosh
a
c
a


h = c  cosh − 1
c




20.137 m
= (15.2905 m )  cosh
− 1
15.2905 m


= 15.291 m
So
s = c sinh
(a)
B is 20.1 m right and 15.29 m down from A a
20.137 m
= (15.291 m ) sinh
= 26.49 m
c
15.291 m
(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.
s = 26.5 m COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 140.
Cable:
Since y A = c cosh
xA
,
c
yB = c cosh
xB
c
( xA < 0 )
and xB − x A = 8 m
x
c
( c + 0.5 m ) = c cosh A
x
 c 
( c + 1.2 m ) = c cosh  B 
1.2 m 
0.5 m  8 m

−1 
cosh −1 1 +
 + cosh 1 +
=
c 
c 
c


Solving numerically,
So
c = 9.9987 m


0.5 m  
x A = cosh −1  1 +
  ( 9.9987 m )
9.9987 m  


= 3.15 m
C is 3.15 m from house (a)
Tmax = TB = wyB
kg  
N

=  2.1
 (1.2 m + 9.9987 m )
  9.81
m 
kg 

= 230.7 N
(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.
Tmax = 231 N COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 141.
y A = c cosh
−a
= c + 6 ft
c
6 ft 

a = c cosh −1 1 +

c 

yB = c cosh
b
= c + 11.4 ft
c
11.4 ft 

b = c cosh −1 1 +

c 

So
Solving numerically,

6 ft 
11.4 ft  

−1 
a + b = c cosh −1 1 +
 + cosh 1 +
  = 36 ft
c 
c 



c = 20.446 ft
11.4 ft 

b = ( 20.446 ft ) cosh −1 1 +
 = 20.696 ft
20.446 ft 

(a)
C is 20.7 ft left of and 11.4 ft below B
 20.696 ft 
Tmax = wyB = ( 0.3 lb/ft )( 20.446 ft ) cosh 
 = 9.554 lb
 20.446 ft 
(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.
Tmax = 9.55 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 142.
(a)
tan θ =
dy
x
= sinh
dx
c
s = c sinh
x
= c tan θ, Q.E.D.
c
So
( cosh x = sinh x + 1)
y = c ( tan θ + 1) = c sec θ
And
y = c secθ , Q.E.D.
(b)
Also
y 2 = s2 + c2 ,
2
2
2
2
2
2
2
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 7, Solution 143.
TB = Tmax = wyB
= wc cosh
ξ =
Let
L
2c
so
xB
L  2c 
L
= w   cosh
c
2 L 
2c
Tmax =
wL
cosh ξ
2ξ

dTmax
wL 
1
=
 sinh ξ − cosh ξ 
dξ
2ξ 
ξ

tanh ξ −
min Tmax ,
For
Solving numerically
(Tmax )min =
=0
ξ
ξ = 1.1997
wL
cosh (1.1997 ) = 0.75444wL
2 (1.1997 )
Lmax =
(a)
If
Tmax
T
= 1.3255 max
0.75444w
w
(
)
Tmax = 32 kN and w = ( 0.34 kg/m ) 9.81 m/s 2 = 3.3354 N/m
Lmax = 1.3255
(b)
1
32.000 N
= 12 717 m
3.3354 N/m
Lmax = 12.72 km
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 7, Solution 144.
ymax = c cosh
Tmax = wymax ,
ymax =
L
=h+c
2c
ymax =
80 lb
= 40 ft
2 lb/ft
c cosh
Solving numerically,
Tmax
w
9 ft
= 40 ft
c
c1 = 2.6388 ft
c2 = 38.958 ft
h = ymax − c
h1 = 40 ft − 2.6388 ft
h1 = 37.4 ft
h2 = 40 ft − 38.958 ft
h2 = 1.042 ft
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 7, Solution 145.
Tmax = wyB = 2wsB
y B = 2sB
c cosh
L
L
= 2c sinh
2c
2c
tanh
L
1
=
2c
2
L
1
= tanh −1 = 0.549306
2c
2
hB
y −c
L
= B
= cosh
−1
c
c
2c
= 0.154701
hB
hB /c
=
L
2( L/2c)
=
0.5 ( 0.154701)
= 0.14081
0.549306
hB
= 0.1408
L
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 7, Solution 146.
Tmax = wyB = wc cosh
(a)
L
2c
dTmax
L
L
L

= w  cosh
−
sinh 
dc
2c 2c
2c 

For
min Tmax ,
tanh
L
2c
=
2c
L
dTmax
=0
dc
L
= 1.1997
2c
yB
L
= cosh
= 1.8102
2c
c
h
y
= B − 1 = 0.8102
c
c
h
h  1 h  2c  
0.8102
=
= 0.3375
= 0.338   =
L  2 c  L   2 (1.1997 )
L
T0 = wc,
(b)
Tmax = wc cosh
T0 = Tmax cosθ B ,
But
So
L
,
2c
Tmax
L
y
= cosh
= B
2c
T0
c
Tmax
= secθ B
T0
 yB 
−1
 = sec (1.8102 )
c


θ B = sec−1 
= 56.46°
Tmax = wyB = w
θ B = 56.5° yB  2c  L 
L
   = w (1.8102 )
c  L  2 
2 (1.1997 )
Tmax = 0.755wL 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 7, Solution 147.
FBD AB:
4 
3 
ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0
5


5 
C = 100 lb
ΣFx = 0: − Ax +
4
(100 lb ) = 0
5
A x = 80 lb
ΣFy = 0: Ay +
3
(100 lb ) − 70 lb = 0
5
A y = 10 lb
FBD AJ:
ΣFx′ = 0: F − ( 80 lb ) sin 30° − (10 lb ) cos30° = 0
F = 48.66 lb
F = 48.7 lb
60° ΣFy′ = 0: V − ( 80 lb ) cos30° + (10 lb ) sin 30° = 0
V = 64.28 lb
V = 64.3 lb
30° ΣM 0 = 0: ( 8 in.)( 48.66 lb ) − (8 in.)(10 lb ) − M = 0
M = 309.28 lb ⋅ in.
M = 309 lb ⋅ in.
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 7, Solution 148.
FBD Whole:
ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) = 0
B x = 130 lb
FBD BE with pulleys and
cord:
ΣM E = 0: ( 5.4 ft )(130 lb ) − ( 7.2 ft ) By
+ ( 4.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B y = 150 lb
ΣFx = 0: Ex − 130 lb = 0
E x = 130 lb
ΣFy = 0: E y + 150 lb − 90 lb − 90 lb = 0
E y = 30 lb
ΣFx′ = 0: − F − 90 lb +
FBD JE and pulley:
4
3
(130 lb ) + ( 90 lb − 30 lb ) = 0
5
5
F = 50.0 lb
3
4
(130 lb ) + ( 30 lb − 90 lb ) = 0
5
5
V = −30 lb
V = 30.0 lb
ΣFy′ = 0: V +
ΣM J = 0: − M + (1.8 ft )(130 lb ) + ( 2.4 ft )( 30 lb ) + ( 0.6 ft )( 90 lb )
− ( 3.0 ft )( 90 lb ) = 0
M = 90.0 lb ⋅ ft
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 7, Solution 149.
FBD Rod:
ΣFx = 0:
ΣM B = 0:
Ax = 0
2r
π
W − rAy = 0
α = 15°, weight of segment = W
FBD AJ:
r =
r
α
ΣFy′ = 0:
sin α =
2W
π
F=
Ay =
2W
π
30° W
=
90°
3
r
sin15° = 0.9886r
π /12
cos 30° −
W
cos30° − F = 0
3
W 3  2 1
 − 
2 π 3
2W 
W

ΣM 0 = M + r  F −
=0
 + r cos15°
3
π 

M = 0.0557 Wr
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 7, Solution 150.
(a)
Along AC:
ΣFy = 0:
ΣM J = 0:
− 3 kip − V = 0
V = −3 kips
M + x ( 3 kips ) = 0
M = ( 3 kips ) x
M = −9 kip ⋅ ft at C
Along CD:
ΣFy = 0:
− 3 kips − 5 kips − V = 0
V = −8 kips
ΣM K = 0:
M + ( x − 3 ft )( 5 kips ) + x ( 3 kips ) = 0
M = +15 kip ⋅ ft − ( 8 kips ) x
M = −16.2 kip ⋅ ft at D ( x = 3.9 ft )
Along DE:
ΣFy = 0:
− 3 kips − 5 kips + 6 kips − V = 0
V = −2 kips
ΣM L = 0:
M − x1 ( 6 kips ) + (.9 ft + x1 )( 5 kips )
+ ( 3.9 ft + x1 )( 3 kips ) = 0
M = −16.2 kip ⋅ ft − ( 2 kips ) x1
M = −18.6 kip ⋅ ft at E
( x1 = 1.2 ft )
continued
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
Along EB:
ΣFy = 0: −3 kips − 5 kips + 6 kips − 4 kips − V = 0
ΣM N = 0:
V = −6 kips
M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips )
+ ( 5.1 ft + x2 )( 3 kips ) − (1.2 ft + x2 )( 6 kips ) = 0
M = −18.6 kip ⋅ ft − ( 6 kips ) x2
M = −33 kip ⋅ ft at B ( x2 = 2.4 ft )
(b)
From diagrams:
V max = 8.00 kips on CD M max = 33.0 kip ⋅ ft at 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 151.
(a)
By symmetry:
Ay = B = 8 kN +
1
( 4 kN/m )( 5 m )
2
A y = B = 18 kN
Along AC:
ΣFy = 0:
18 kN − V = 0
ΣM J = 0:
M − x (18 kN )
V = 18 kN
M = (18 kN ) x
M = 36 kN ⋅ m at C ( x = 2 m )
Along CD:
ΣFy = 0:
18 kN − 8 kN − ( 4 kN/m ) x1 − V = 0
V = 10 kN − ( 4 kN/m ) x1
V = 0 at x1 = 2.5 m ( at center )
ΣM K = 0: M +
x1
( 4 kN/m ) x1 + (8 kN ) x1 − ( 2 m + x1 )(18 kN ) = 0
2
M = 36 kN ⋅ m + (10 kN/m ) x1 − ( 2 kN/m ) x12
M = 48.5 kN ⋅ m at x1 = 2.5 m
Complete diagram by symmetry
(b)
From diagrams:
V max = 18.00 kN on AC and DB
M max = 48.5 kN ⋅ m at center
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 7, Solution 152.
By symmetry:
Ay = B = 60 kN − P
Along AC:
ΣM J = 0: M − x ( 60 kN − P ) = 0
M = ( 60 kN − P ) x
M = 120 kN ⋅ m − ( 2 m ) P at x = 2 m
Along CD:
ΣM K = 0:
M + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − Px
M = 120 kN ⋅ m − ( 4 m ) P at x = 4 m
Along DE:
ΣM L = 0:
M − ( x − 4 m ) P + ( x − 2 m )( 60 kN )
− x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − ( 4 m ) P
(const)
Complete diagram by symmetry
For minimum M max , set M max = −M min
120 kN ⋅ m − ( 2 m ) P = − 120 kN ⋅ m − ( 4 m ) P 
(a)
(b)
P = 40.0 kN
M min = 120 kN ⋅ m − ( 4 m) P
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.
M max = 40.0 kN ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 153.
FBD Beam:
ΣM A = 0:
( 4 m ) B − (1 m )( 20 kN/m )( 2 m ) − M = 0
B = 10 kN +
(a)
(a)
B = 10 kN
(b)
B = 13 kN
ΣFy = 0:
M
4m
Ay − ( 20 kN/m )( 2 m ) + B = 0
Ay = 40 kN − B
(a)
A y = 30 kN
(b)
A y = 27 kN
Shear Diags:
 dV

VA = Ay , then V is linear 
= −20 kN/m  to C.
 dx

VC = Ay − ( 20 kN/m )( 2 m ) = Ay − 40 kN
(b)
(a)
VC = −10 kN
(b)
VC = −13 kN
V = 0 = Ay − ( 20 kN/m ) x1 at x1 =
(a)
x1 = 1.5 m
(b)
x1 = 1.35 m
Ay m
20 kN
V is constant from C to B.
continued
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
Moment Diags:
 dM

M A = applied M . Then M is parabolic 
decreases with V 
dx


M is max where V = 0. M max = M +
(a)
M max =
1
Ay x1.
2
1
( 30 kN )(1.5 m ) = 22.5 kN ⋅ m
2
1.500 m from A
(b)
M max = 12 kN ⋅ m +
1
( 27 kN )(1.35 m ) = 30.225 kN ⋅ m
2
M max = 30.2 kN, 1.350 m from A
M C = M max −
1
VC ( 2 m − x1 )
2
(a)
M C = 20 kN ⋅ m
(b)
M C = 26 kN ⋅ m
 dM

Finally, M is linear 
= VC  to zero at B.
 dx

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 7, Solution 154.
(b)
(a)
x

Distributed load w = w0 1 − 
L


ΣM A = 0:
L 1

 w0 L  − LB = 0
3 2

ΣFy = 0:
Ay −
1


 total = w0 L 
2


B=
1
wL
w0 L + 0 = 0
2
6
w0 L
6
Ay =
w0 L
3
Shear:
VA = Ay =
w0 L
,
3
x
dV
x

= −w → V = VA − ∫ w0 1 −  dx
0
dx
L

Then
 1 x 1  x 2 
1 w0 2
w L
V =  0  − w0 x +
x = w0 L  − +   
2 L
 3 
 3 L 2  L  
Note: At x = L, V = −
w0 L
;
6
2
x
x
x 2
V = 0 at   − 2   + = 0 →
=1−
3
L
L
L
 
 
1
3
Moment:
M A = 0,
Then
x
x/L  x   x 
 dM 

 = V → M = ∫0 Vdx = L ∫0 V   d  
 dx 
L L
M = w0 L2 ∫
x / L 1
0
2
x 1 x    x
 − +   d 
 3 L 2  L    L 
 1  x  1  x  2 1  x 3 
M = w0 L2    −   +   
6  L  
 3  L  2  L 
continued
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
 x
=1−
M max  at
 L
1
2
 = 0.06415w0 L
3 
(a)
 1 x 1  x 2 
V = w0 L  − +     3 L 2  L  
 1  x  1  x  2 1  x 3 
M = w0 L2    −   +    6  L  
 3  L  2  L 
(c)
M max = 0.0642 w0 L2 at x = 0.423L 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 7, Solution 155.
(
)
L 4w0
Lx − x 2 dx = 0
ΣFy = 0: wg L − ∫
0 L2
(a)
wg L =
Define ξ =
4w0  1 2 1 3  2
 LL − L  = w0 L
3  3
L2  2
wg =
2w0
3
 x  x 2  2
x
dx
so dξ =
→ net load w = 4w0  −    − w0
L
L
 L  L   3
 1

w = 4w0  − + ξ − ξ 2 
6


or
ξ
 1

V = V ( 0 ) − ∫ 4w0 L  − + ξ − ξ 2  dξ
0
 6

1
6
1
2
1
3


= 0 + 4w0 L  ξ + ξ 2 − ξ 3 
V =
x
ξ
2
M = M 0 + ∫ Vdx = 0 + w0 L2 ∫
0
0
3
=
(b)
(
2
w0 L ξ − 3ξ 2 + 2ξ 3
3
)
(ξ − 3ξ + 2ξ ) dξ
2
3
(
2
1  1
1
w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 ξ 2 − 2ξ 3 + ξ 4
3
2  3
2
Max M occurs where V = 0 → 1 − 3ξ + 2ξ 2 = 0 → ξ =
)
1
2
1 1
1  w L2

1 2
M  ξ =  = w0 L2  − +  = 0
2 3
48

 4 8 16 
M max =
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.
w0 L2
at center of beam
48
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 156.
(a) FBD cable:
ΣM E = 0:
( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN )
− ( 3 m ) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
ΣM C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax − ( 8 m ) Ay = 0
FBD ABC:
Ax − 8 Ay = −4.8 kN
Solving (1) and (2)
Ax = 3.2 kN
Ay = 1 kN
So A = 3.35 kN
(b) cable:
(2)
17.35°
ΣFx = 0: − Ax + Ex = 0
Ex = Ax = 3.2 kN
ΣFy = 0: Ay − (1.2 + 0.8 + 1.2 ) kN + E y = 0
E y = 3.2 kN − Ay = ( 3.2 − 1) kN = 2.2 kN
So E = 3.88 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.
34.5°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 157.
FBD AC:
FBD CB:
ΣM A = 0: (13.5 ft ) T0 −
a
( 57.5 lb/ft ) a = 0
2
(
)
T0 = 2.12963 lb/ft 2 a 2
ΣM B = 0:
60 ft − a
( 57.5 lb/ft )( 60 ft − a ) − ( 6 ft ) T0 = 0
2
(
)
6T0 = 28.75 lb/ft 2 3600 ft 2 − (120 ft ) a + a 2 
Using (1) in (2),
Solving:
0.55a 2 − (120 ft ) a + 3600 ft 2 = 0
a = (108 ± 72 ) ft,
a = 36 ft (180 ft out of range)
So
C is 36 ft from A
(a)
C is 6 ft below and 24 ft left of B
T0 = 2.1296 lb/ft 2 ( 36 ft ) = 2760 lb
2
W1 = ( 57.5 lb/ft )( 36 ft ) = 2070 lb
(b)
Tmax = TA = T02 + W12 =
( 2760 lb )2 + ( 2070 lb )2 = 3450 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.
(2)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 158.
sB = 100 ft,
w=
4 lb
= 0.02 lb/ft
200 ft
Tmax = 16 lb
Tmax = TB = wyB
yB =
TB
16 lb
=
= 800 ft
w
0.02 lb/ft
c 2 = yB2 − sB2
c=
But
(800 ft )2 − (100 ft )2 = 793.73 ft
yB = xB cosh
xB
y
→ xB = c cosh −1 B
c
c
 800 ft 
= ( 793.73 ft ) cosh −1 
 = 99.74 ft
 793.73 ft 
L = 2 xB = 2 ( 99.74 ft ) = 199.5 ft
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 8, Solution 1.
FBD Block B:
Tension in cord is equal to W A = 25 lb from FBD’s of block A and
pulley.
ΣFy = 0:
N − WB cos 30° = 0,
N = WB cos 30°
(a) For smallest WB , slip impends up the incline, and
F = µ s N = 0.35WB cos30°
ΣFx = 0:
F − 25 lb + WB sin 30° = 0
( 0.35cos30° + sin 30° )WB = 25 lb
WB min = 31.1 lb
(b) For largest WB , slip impends down the incline, and
F = − µ s N = − 0.35 WB cos30°
ΣFx = 0:
Fs + WB sin 30° − 25 lb = 0
( sin 30° − 0.35cos30° )WB = 25 lb
W B max = 127.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 2.
FBD Block B:
Tension in cord is equal to WA = 40 lb from FBD’s of block A and
pulley.
(a)
ΣFy = 0:
N − ( 52 lb ) cos 25° = 0,
N = 47.128 lb
Fmax = µ s N = 0.35 ( 47.128 lb ) = 16.495 lb
ΣFx = 0:
Feq − 40 lb + ( 52 lb ) sin 25° = 0
So, for equilibrium, Feq = 18.024 lb
Since Feq > Fmax , the block must slip (up since F > 0)
∴ There is no equilibrium
(b) With slip,
F = µk N = 0.25 ( 47.128 lb )
F = 11.78 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.
35°
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Chapter 8, Solution 3.
FBD Block:
Tension in cord is equal to P = 40 N, from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣF y = 0 :
N − (98.1 N ) cos 20° + (40 N ) sin 20° = 0
N = 78.503 N
Fmax = µ s N = ( 0.30 )( 78.503 N ) = 23.551 N
For equilibrium:
ΣFx = 0:
( 40 N ) cos 20° − ( 98.1 N ) sin 20° − F = 0
Feq = 4.0355 N < Fmax ,
F = Feq
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.
∴ Equilibrium exists
F = 4.04 N
20°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 4.
Tension in cord is equal to P = 62.5 N, from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣFy = 0:
N − ( 98.1 N ) cos 20° + ( 62.5 N ) sin15° = 0
N = 76.008 N
Fmax = µ s N = ( 0.30 )( 76.008 N ) = 22.802 N
For equilibrium:
ΣFx = 0:
( 62.5 N ) cos15° − ( 98.1 N ) sin 20° − F = 0
Feq = 26.818 N > Fmax
so no equilibrium,
and block slides up the incline
Fslip = µ x N = ( 0.25 )( 76.008 N ) = 19.00 N
F = 19.00 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.
20°
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Chapter 8, Solution 5.
Tension in cord is equal to P from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣFy = 0:
N − ( 98.1 N ) cos 20° + P sin 25° = 0
(1)
ΣFx = 0:
P cos 25° − ( 98.1 N ) sin 20° + F = 0
(2)
For impending slip down the incline, F = µ s N = 0.3 N and solving
(1) and (2),
PD = 7.56 N
For impending slip up the incline, F = − µ s N = − 0.3 N and solving
(1) and (2),
PU = 59.2 N
so, for equilibrium
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.
7.56 N ≤ P ≤ 59.2 N
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Chapter 8, Solution 6.
FBD Block:
(
)
W = ( 20 kg ) 9.81 m/s 2 = 196.2 N
For θ min motion will impend up the incline, so F is downward and
F = µs N
ΣFy = 0:
ΣFx = 0:
(1) + ( 2 ):
N − ( 220 N ) sin θ − (196.2 N ) cos 35° = 0
F = µ s N = 0.3 ( 220 sin θ + 196.2 cos 35° ) N
(1)
( 220 N ) cosθ − F − (196.2 N ) sin 35° = 0
(2)
0.3 ( 220 sin θ + 196.2cosθ ) N
= ( 220 cosθ ) N − (196.2sin 35° ) N
or
220cosθ − 66sin θ = 160.751
Solving numerically:
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.
θ = 28.9°
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Chapter 8, Solution 7.
FBD Block:
For Pmin motion will impend down the incline, and the reaction force R
will make the angle
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
with the normal, as shown.
Note, for minimum P, P must be ⊥ to R, i.e. β = φs (angle between
P and x equals angle between R and normal).
β = 19.29°
(b)
then P = (160 N ) cos ( β + 40° )
= (160 N ) cos 59.29° = 81.71 N
(a)
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.
Pmin = 81.7 N
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Chapter 8, Solution 8.
FBD block (impending motion
downward)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
(a) Note: For minimum P,
P⊥R
So
β = α = 90° − ( 30° + 14.036° ) = 45.964°
and
P = ( 30 lb ) sin α = ( 30 lb ) sin ( 45.964° ) = 21.567 lb
P = 21.6 lb
(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.
β = 46.0°
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Chapter 8, Solution 9.
FBD Block:
For impending motion. φ s = tan −1 µ s = tan −1 ( 0.40 )
φ s = 21.801°
Note β1,2 = θ1,2 − φ s
10 lb
15 lb
=
sinφs sinβ1,2
From force triangle:
 15 lb
 33.854°
sin ( 21.801° )  = 
10 lb
 146.146°
β1,2 = sin −1 
55.655°
So θ1,2 = β1,2 + φ s = 
167.947°
So
(a)
equilibrium for
0 ≤ θ ≤ 55.7°
(b)
equilibrium for
167.9° ≤ θ ≤ 180°
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 8, Solution 10.
FBD A with pulley:
Tension in cord is T throughout from pulley FBD’s
ΣFy = 0:
2T − 20 lb = 0,
T = 10 lb
FBD E with pulley:
For θ max , motion impends to right, and
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
From force triangle,
20 lb
10 lb
=
,
sin (θ − φs ) sinφ s
2sin φ s = sin (θ − φ s )
θ = sin −1 ( 2sin19.2900° ) + 19.2900° − 60.64°
θ max = 60.6°
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 8, Solution 11.
FBD top block:
ΣFy = 0:
N1 − 196.2 N = 0
N1 = 196.2 N
(a) With cable in place, impending motion of bottom block requires
impending slip between blocks, so F1 = µ s N1 = 0.4 (196.2 N )
F1 = 78.48 N
FBD bottom block:
ΣFy = 0:
N 2 − 196.2 N − 294.3 N = 0
N 2 = 490.5 N
F2 = µ s N 2 = 0.4 ( 490.5 N ) = 196.2 N
ΣFx = 0:
− P + 78.48 N + 196.2 N = 0
P = 275 N
FBD block:
(b) Without cable AB, top and bottom blocks will move together
ΣFy = 0:
N − 490.5 N = 0,
Impending slip:
ΣFx = 0:
N = 490.5 N
F = µ s N = 0.40 ( 490.5 N ) = 196.2 N
− P + 196.2 N = 0
P = 196.2 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 8, Solution 12.
FBD top block:
Note that, since φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.8° > 15°, no motion
will impend if P = 0, with or without cable AB.
(a) With cable, impending motion of bottom block requires impending
slip between blocks, so F1 = µ s N
ΣFy′ = 0:
N1 − W1 cos15° = 0,
N1 = W1 cos15° = 189.515 N
F1 = µ s N1 = ( 0.40 )W1 cos15° = 0.38637 W1
F1 = 75.806 N
FBD bottom block:
ΣFx′ = 0:
T − F1 − W1 sin15° = 0
T = 75.806 N + 50.780 N = 126.586 N
(
)
W2 = ( 30 kg ) 9.81 m/s 2 = 294.3 N
ΣFy = 0 :
N 2 − (189.515 N ) cos (15° ) − 294.3 N
+ ( 75.806 N ) sin15° = 0
N 2 = 457.74 N
F2 = µ s N 2 = ( 0.40 )( 457.74 N ) = 183.096 N
FBD block:
ΣFx = 0:
− P + (189.515 N ) + ( 75.806 N ) cos15°
+ 126.586 N + 183.096 N = 0
P = 361 N
(b) Without cable, blocks remain together
ΣFy = 0:
N − W1 − W2 = 0
N = 196.2 N + 294.3 N
= 490.5 N
F = µ s N = ( 0.40 )( 490.5 N ) = 196.2 N
ΣFx = 0:
− P + 196.2 N = 0
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.
P = 196.2 N
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Chapter 8, Solution 13.
FBD A:
Note that slip must impend at both surfaces simultaneously.
N1 + T sin θ − 16 lb = 0
ΣFy = 0:
N1 = 16 lb − T sin θ
Impending slip:
F1 = µ s N1 = ( 0.20 )(16 lb − T sin θ )
F1 = 3.2 lb − ( 0.2 ) T sin θ
(1)
F1 − T cosθ = 0
(2)
ΣFx = 0:
FBD B:
ΣFy = 0:
N 2 − N1 − 24 lb = 0,
N 2 = N1 + 24 lb
= 30 lb − T sin θ
Impending slip:
F2 = µ s N 2 = ( 0.20 )( 30 lb − T sin θ )
= 6 lb − 0.2 T sin θ
ΣFx = 0:
10 lb − F1 − F2 = 0
10 lb = µ s ( N1 + N 2 ) = ( 0.2 )  N1 + ( N1 + 24 lb ) 
10 lb = 0.4 N1 + 4.8 lb,
Then
Then
N1 = 13 lb
F1 = µ s N1 = ( 0.2 )(13 lb ) = 2.6 lb
(1):
T sin θ = 3.0 lb
( 2 ):
T cosθ = 2.6 lb
Dividing tan θ =
3
,
2.6
θ = tan −1
3
= 49.1°
2.6
θ = 49.1°
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 8, Solution 14.
FBD’s:
Note: Slip must impend at both surfaces simultaneously.
A:
ΣFy = 0:
N1 − 20 lb = 0,
Impending slip:
N1 = 20 lb
F1 = µ s N1 = ( 0.25 )( 20 lb ) = 5 lb
ΣFx = 0:
− T + 5 lb = 0,
T = 5 lb
ΣFy′ = 0:
N 2 − ( 20 lb + 40 lb ) cosθ − ( 5 lb ) sin θ = 0
N 2 = ( 60 lb ) cosθ − ( 5 lb ) sin θ
B:
Impending slip:
ΣFx′ = 0:
F2 = µ s N 2 = ( 0.25 )( 60cosθ − 5sin θ ) lb
− F2 − 5 lb − ( 5 lb ) cosθ + ( 20 lb + 40 lb ) sin θ = 0
− 20cosθ + 58.75sin θ − 5 = 0
Solving numerically,
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.
θ = 23.4°
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Chapter 8, Solution 15.
FBD:
For impending tip the floor reaction is at C.
(
)
W = ( 40 kg ) 9.81 m/s 2 = 392.4 N
For impending slip φ = φs = tan −1 µ s = tan −1 ( 0.35 )
φ = 19.2900°
tan φ =
0.8 m
,
EG
EG =
0.4 m
= 1.14286 m
0.35
EF = EG − 0.5 m = 0.64286 m
(a)
α s = tan −1
EF
0.64286 m
= tan −1
= 58.109°
0.4 m
0.4 m
α s = 58.1°
(b)
P
W
=
sin19.29° sin128.820
P = ( 392.4 N )( 0.424 ) = 166.379 N
P = 166.4 N
Once slipping begins, φ will reduce to φk = tan −1 µk .
Then α max will increase.
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 8, Solution 16.
First assume slip impends without tipping, so F = µ s N
FBD
ΣFy = 0:
N + P sin 40° − W = 0,
N = W − P sin 40°
F = µ s N = 0.35 (W − P sin 40° )
ΣFx = 0:
F − P cos 40° = 0
0.35W = P ( cos 40° + 0.35sin 40° )
Ps = 0.35317 W
(1)
Next assume tip impends without slipping, R acts at C.
ΣM A = 0:
( 0.8 m ) P sin 40° + ( 0.5 m ) P cos 40° − ( 0.4 m )W = 0
Pt = 0.4458W > Ps from (1)
(
∴ Pmax = Ps = 0.35317 ( 40 kg ) 9.81 m/s 2
)
= 138.584 N
(a)
Pmax = 138.6 N (b) Slip is impending Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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Chapter 8, Solution 17.
FBD Cylinder:
For maximum M, motion impends at both A and B
ΣFx = 0:
FA = µ A N A;
FB = µ B N B
N A − FB = 0
N A = FB = µ B N B
FA = µ A N A = µ Aµ B N B
ΣFy = 0:
N B + FA − W = 0
1
or
NB =
and
FB = µ B N B =
1 + µ Aµ B
W
µB
W
1 + µ Aµ B
FA = µ Aµ B N B =
µ Aµ B
W
1 + µ Aµ B
ΣM C = 0: M − r ( FA + FB ) = 0
(a) For
µA = 0
N B (1 + µ Aµ B ) = W
and
M = Wr µ B
1 + µA
1 + µ Aµ B
µ B = 0.36
M = 0.360Wr
(b) For
µ A = 0.30
and
µ B = 0.36
M = 0.422Wr
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 8, Solution 18.
FBD’s:
FBD Drum:
(a)
ΣM D = 0:
 10 
 ft  F − 50 lb ⋅ ft = 0
 12 
F = 60 lb
Impending slip: N =
F
µs
=
60 lb
= 150 lb
0.40
FBD arm:
ΣM A = 0:
( 6 in.) C + ( 6 in.) F − (18 in.) N = 0
C = − 60 lb + 3 (150 lb ) = 390 lb
Ccw = 390 lb (b) Reversing the 50 lb ⋅ ft couple reverses the direction of F, but the magnitudes of F and N are not changed.
Then, using the FBD arm:
ΣM A = 0:
( 6 in.) C − ( 6 in.) F − (18 in.) N = 0
C = 60 lb + 3 (150 lb ) = 510 lb
Cccw = 510 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
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Chapter 8, Solution 19.
For slipping, F = µ k N = 0.30 N
FBD’s:
(a) For cw rotation of drum, the friction force F is as shown.
From FBD arm:
ΣM A = 0:
( 6 in.)( 600 lb ) + ( 6 in.) F − (18 in.) N = 0
600 lb + F − 3
F =
F
=0
0.30
600
lb
9
Moment about D = (10 in.) F = 666.67 lb ⋅ in.
M cw = 55.6 lb ⋅ ft
(b) For ccw rotation of drum, the friction force F is reversed
ΣM A = 0:
( 6 in.)( 600 lb ) − ( 6 in.) F − (18 in.) N = 0
600 lb − F − 3
F
=0
0.30
F =
600
lb
11
 10  600 
Moment about D =  ft 
lb  = 45.45 lb ⋅ ft
 12  11 
M ccw = 45.5 lb ⋅ ft 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
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Chapter 8, Solution 20.
FBD:
(a)
ΣM C = 0:
r ( F − T ) = 0,
T = F
Impending slip: F = µ s N or N =
ΣFx = 0:
F
µs
=
T
µs
F + T cos ( 25° + θ ) − W sin 25° = 0
T 1 + cos ( 25° + θ )  = W sin 25°
ΣFy = 0:
(1)
N − W cos 25° + T sin ( 25° + θ ) = 0
 1

+ sin ( 25° + θ )  = W cos 25°
T
0.35


Dividing (1) by (2):
(2)
1 + cos ( 25° + θ )
= tan 25°
1
+ sin ( 25° + θ )
0.35
Solving numerically, 25° + θ = 42.53°
θ = 17.53° (b) From (1)
T (1 + cos 42.53° ) = W sin 25°
T = 0.252W 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
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Chapter 8, Solution 21.
FBD ladder:
Note: slope of ladder =
L = 6.5 m, so AC =
4.5 m
12
13
=
, so AC = ( 4.5 m )
= 4.875
1.875 m
5
12
4.875 m
3
= L,
6.5 m
4
and DC = BD =
AD =
1
L
2
1
L
4
For impending slip: FA = µ s N A ,
FC = µ s NC
 12 
Also θ = tan −1   − 15° = 52.380°
 5
FA − W sin15° + FC cosθ − NC sin θ = 0
ΣFx = 0:
FA = W sin15° − µ s
10
10
W cosθ +
W sin θ
39
39
= ( 0.46192 − 0.15652µ s )W
ΣFy = 0:
N A − W cos15° + FC sin θ + NC cosθ = 0
N A = W cos15° − µ s
10
10
W sin θ − W cosθ
39
39
= ( 0.80941 − 0.20310µ s )W
But FA = µ N A :
0.46192 − 0.15652µ s = 0.80941µ s − 0.20310µ s2
µ s2 − 4.7559µ s + 2.2743
µ s = 0.539, 4.2166
µ s min = 0.539 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
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Chapter 8, Solution 22.
FBD ladder:
Slip impends at both A and B, FA = µ s N A , FB = µ s N B
ΣFx = 0:
ΣFy = 0:
FA − N B = 0,
N B = FA = µ s N A
N A − W + FB = 0,
N A + FB = W
N A + µs N B = W
(
)
N A 1 + µ s2 = W
ΣM O = 0:
( 6 m ) N B +  m W −  m  N A = 0
5
4
6µ s N A +
µ s2 +

(
5
2

)
5
5
N A 1 + µ s2 − N A = 0
4
2
24
µs − 1 = 0
5
µ s = − 2.4 ± 2.6
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.
µ s min = 0.200
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Chapter 8, Solution 23.
FBD rod:
(a) Geometry:
BE =
L
cosθ
2
EF = L sin θ
So
or
L

DE =  cosθ  tan β
2

DF =
L cosθ
2 tan φ s
1
 L cosθ
L  cosθ tan β + sin θ  =
2
 2 tan φs
tan β + 2 tan θ =
1
1
1
=
=
= 2.5
tan φ s
µ s 0.4
Also,
L sin θ + L sin β = L
or
sin θ + sin β = 1
Solving Eqs. (1) and (2) numerically
θ1 = 4.62°
(1)
(2)
β1 = 66.85°
θ 2 = 48.20° β 2 = 14.75°
θ = 4.62° and θ = 48.2°
Therefore,
(b) Now
and
or
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
T
W
=
sin φs
sin ( 90 + β − φ s )
T =W
sin φs
sin ( 90 + β − φ s )
For
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.
θ = 4.62°
T = 0.526W
θ = 48.2°
T = 0.374W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 24.
FBD:
Assume the weight of the slender rod is negligible compared to P.
First consider impending slip upward at B. The friction forces will be
directed as shown and FB,C = µ s N B,C
ΣM B = 0:
( L sinθ ) P − 
a 
 NC = 0
 sin θ 
NC = P
ΣFx = 0:
L 2
sin θ
a
NC sin θ + FC cosθ − N B = 0
NC ( sin θ + µ s cosθ ) = N B
so
ΣFy = 0:
NB = P
L 2
sin θ ( sin θ + µ s cosθ )
a
− P + NC cosθ − FC sin θ − FB = 0
P = NC cosθ − µ s NC sin θ − µ s N B
so P = P
L 2
L
sin θ ( cosθ − µ s sin θ ) − µ s P sin 2 θ ( sin θ + µ s cosθ )
a
a
Using θ = 35° and µ s = 0.20, solve for
(1)
L
= 13.63.
a
To consider impending slip downward at B, the friction forces will be
reversed. This can be accomplished by substituting µ s = − 0.20 in
L
= 3.46.
equation (1). Then solve for
a
Thus, equilibrium is maintained for
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.
3.46 ≤
L
≤ 13.63
a
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Chapter 8, Solution 25.
FBD ABC:
ΣM C = 0:
0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30°
+ 0.030 m + ( 0.30 m ) cos 30° ( 400 N ) cos 30°
 12

 5

− ( 0.03 m )  FBD  − ( 0.045 m )  FBD  = 0
 13

 13

FBD = 3097.64 N
FBD Blade:
ΣFx = 0:
N −
25
( 3097.6 N ) = 0
65
N = 1191.4
F = µ s N = 0.20 (1191.4 N ) = 238.3 N
ΣFy = 0:
P+F−
60
( 3097.6 N ) = 0
65
P = 2859.3 − 238.3 = 2621.0 N
Force by blade
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.
P = 2620 N
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Chapter 8, Solution 26.
FBD CD:
Note: The plate is a 3-force member, and for minimum µ s , slip
impends at C and D, so the reactions there are at angle φ s from the
normal.
From the FBD,
OCG = 20° + φ s
and
ODG = 20° − φ s
Then OG = ( 0.5 in.) tan ( 20° + φs )
 1.2 in.

+ 0.5 in.  tan ( 20° − φs )
and OG = 
 sin70°

Equating, tan ( 20° + φs ) = 3.5540 tan ( 20° − φs )
Solving numerically,
φ s = 10.5652°
µ s = tan φs = tan (10.5652° )
µ s = 0.1865 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 8, Solution 27.
FBD pin A:
From FBD Whole the force at A = 750 lb
4
( FAB′ − FAB ) = 0, FAB′ = FAB
5
ΣFx = 0:
3
750 lb − 2 FAB = 0, FAB = 625 lb
5
ΣFy = 0:
FBD Casting:
ΣFx = 0:
N D′ − N D = 0, N D′ = N D = N
FD
Impending slip
FD = FD′ = µ N D , or N D =
ΣFy = 0:
2FD − 750 lb = 0, FD = 375 lb
ND =
µs
375 lb
µs
FBD ABCD:
ΣM C = 0:
(12 in.)
375 lb
µs
4
5
(12 in.) N − ( 6 in.) F − ( 42.75 in.) ( 625 lb ) = 0
= ( 6 in.)( 375 lb ) + ( 42.75 in.)
4
( 625 lb ) = 0
5
µ s = 0.1900 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 8, Solution 28.
From FBD Whole, and neglecting weight of clamp compared to 550 lb plate, P = − W Since AB is a
two-force member, B is vertical and B = W .
FBD BCD:
ΣM C = 0:
(1.85 in.)W − ( 2.3 in.) D cos 40°
− ( 0.3 in.) D sin 40° = 0,
D = 0.94642W
FBD EG:
( 0.9 in.) NG − (1.3 in.) FG − (1.3 in.) N D cos 40° = 0
ΣM E = 0:
Impending slip:
FG = µ s NG
Solving: ( 0.9 − 1.3µ s ) NG = 0.94250W
(1)
FBD Plate:
By symmetry NG = NG′ ,
ΣFy = 0:
FG = FG′ = µ s NG
2 FG − W = 0,
Substitute in (1): ( 0.9 − 1.3µ s )
FG =
W
,
2
NG =
W
2µ s
W
= 0.94250W
2µ s
Solving, µ s = 0.283,
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.
µsm = 0.283
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Chapter 8, Solution 29.
FBD table + child:
(
)
W = 16 kg ( 9.81 m/s ) = 156.96 N
WC = 18 kg 9.81 m/s 2 = 176.58 N
2
T
(a) Impending tipping about E , N F = FF = 0, and
ΣM E = 0:
( 0.05 m )(176.58 N ) − ( 0.4 m )(156.96 N ) + ( 0.5 m ) P cosθ − ( 0.7 m ) P sin θ = 0
33cosθ − 46.2sin θ = 53.955
Solving numerically
θ = −36.3°
and
θ = −72.6°
−72.6° ≤ θ ≤ −36.3° Therefore
Impending tipping about F is not possible
(b) For impending slip:
FE = µ s N E = 0.2 N E
ΣFx = 0: FE + FF − P cosθ = 0
or
FF = µ s N F = 0.2 N F
0.2 ( N E + N F ) = ( 66 N ) cos θ
ΣFy = 0: N E + N F − 176.58 N − 156.96 N − P sin θ = 0
N E + N F = ( 66sin θ + 333.54 ) N
So
Solving numerically,
330 cosθ = 66sin θ + 333.54
θ = −3.66°
and
θ = −18.96°
Therefore,
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.
−18.96° ≤ θ ≤ −3.66° COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 30.
Geometry of four-bar:
Considering the geometry when α = 0,
1/ 2
2
2
LCD = ( 60 mm − 52 mm ) + ( 36 mm + 22 mm ) 


= 58.549 mm
In general, 52 mm − ( 36 mm ) sin α = 60 mm − ( 58.549 mm ) sin β
 36sin α + 8 
so β = sin −1 

 58.549 
(a) FBD ACE:
α =0
β = 7.8533°, note that the links at E and K are prevented from pivoting
downward by the small blocks
FE
FCD sin β − FE = 0, FCD =
ΣFy = 0:
sin 7.8533°
ΣM A = 0:
( 60 mm ) 
FE

 cos 7.8533° − ( 32 mm ) FE − ( 212 mm ) N E = 0
 sin 7.8533° 
Impending slip on pad N E =
FE
µs
, so

212 
 435.00 − 32 −
 FE = 0
µs 

µ s = 0.526
(b) α = 30°, β = 26.364°
ΣFx = 0:
ΣFy = 0:
−
3
FAB + FCD cos 26.364° − N E = 0
2
1
− FAB + FCD sin 26.364° − FE = 0
2
Eliminating FAB ,
FCD ( 0.89599 − 0.76916 ) − N E + FE = 0
Impending slip FE = µ s N E , so 0.126834 FE = (1 − µ s ) N E
( 60 mm ) FCD cos 26.364°
− ( 212 mm ) N E − ( 32 mm ) µ s N E = 0
53.759 FCD = ( 212 − 32µ s ) N E = 0
ΣM A = 0:
212 − 32µ s
53.759
=
1 − µs
0.12634
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.
µ s = 0.277
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Chapter 8, Solution 31.
FBD ABD:
(15 mm ) N A − (110 mm ) FA = 0
ΣΜ D = 0:
Impending slip:
FA = µ SA N A
So
15 − 110µ SA = 0 µ SA = 0.136364
µ SA = 0.1364 FA − Dx = 0, Dx = FA = µ SA N A
ΣFx = 0:
FBD Pipe:
r = 60 mm
ΣFy = 0:
NC − N A = 0,
NC = N A
FBD DF:
ΣM F = 0:
( 550 mm ) FC − (15 mm ) NC − ( 500 mm ) Dx = 0
Impending slip:
FC = µ SC NC = µ SC N A
So,
550µ SC N A − 15N A − 500µ SA N A = 0
550µ SC = 15 + 500 ( 0.136364 )
µ SC = 0.1512 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 8, Solution 32.
FBD Plate:
Assume reactions as shown, at ends of sleeves,
For impending slip
FA = µ s N A ,
FB = µ s N B
P sin θ − µ s N A − µ s N B = 0
ΣFx = 0:
N A + N B = 2.5 P sin θ
ΣFy = 0:
N A − N B − P cosθ = 0,
N A − N B = P cosθ
P
( 2.5sin θ + cosθ ) ,
2
NB =
Solving: N A =
ΣM B = 0:
P
( 2.5sin θ − cosθ ) (1)
2
( 23.5 in.) P sin θ − (16 in.) N A + (1 in.) FA = 0
P
2
( 23.5 in.) P sin θ − 16 in. − 0.4 (1 in.) ( 2.5sin θ + cosθ ) = 0
4sin θ − 7.8cosθ = 0,
(2)
θ = 62.9°
For θ > 62.9°, the panel will be self locking, ∴ motion for θ ≤ 62.9°.
As θ decreases, N B will reverse direction at 2.5sin θ − cosθ = 0,
(see equ. 1) or at θ = 21.8°. So for θ ≤ 21.8°
ΣFx = 0 :
P sin θ − µ s ( N A + N B ) = 0
N A + N B = 2.5 P sin θ
ΣFy = 0:
N A + N B − P cosθ = 0,
N A + N B = P cosθ
∴ 2.5sin θ = cosθ ,
θ = 21.8°
So impending motion for 21.8° ≤ θ ≤ 62.9° W
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 8, Solution 33.
FBD Plate:
Assuming reactions as shown, at ends of sleeves,
For impending slip
ΣFx = 0:
FA = µ s N A ,
FB = µ s FB
P sin θ − µ s ( N A + N B ) = 0
ΣFy = 0:
N A − N B − P cosθ = 0,
Solving:
NA =
N A + N B = 2.5 P sin θ
(1)
N A − N B = P cosθ
(2)
P
( 2.5sin θ + cosθ ) ,
2
NB =
P
( 2.5sin θ − cosθ )
2
Note that, for θ < 21.8°, N B becomes negative, so we must change equ. 2 to
N A + N B = P cosθ ,
( 2′ )
but equ. (1) does not change. Solving (1) and ( 2′ ) gives P cosθ = 2.5P sin θ ,
or θ = 21.8°, so the lower limit for impending slip is θ = 21.8°.
For θ ≥ 21.8°, the forces are as shown, and
ΣM B = 0:
( 23.5 in.) P sin θ + xP cosθ + (1 in.) FA − (16 in.) N A = 0
P
2
( 23.5 in.) P sin θ + x P cosθ + 0.4 (1 in.) − (16 in.) ( 2.5sin θ + cosθ ) = 0
or
4sin θ − ( 7.8 in.) − x  cosθ = 0,
tan θ = 1.950 −
x
4 in.
(a) For x = 4 in., tan θ = 1.950, θ = 43.5°. For θ > 43.5° self locking
∴ impending motion for 21.8° ≤ θ ≤ 43.5° W
(b) As x increases from 4 in., the upper bound for θ decreases, becoming
21.8° ( tan θ = 0.4000 ) when x = ( 4 in.)(1.950 − 0.400 ) = 6.2 in.
Thus xmax = 6.20 in. W
at which θ must equal 21.8°.
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 8, Solution 34.
FBD Collar:
Impending motion down:
a
−a
cosθ
Stretch of spring x = AB − a =
 a

 1

− a  = (1.5 kN/m )( 0.5 m ) 
− 1
Fs = kx = k 
cos
cos
θ
θ




 1

= ( 0.75 kN ) 
− 1
 cos θ

N − Fs cosθ = 0
ΣFx = 0:
N = Fs cosθ = ( 0.75 kN )(1 − cosθ )
Impending motion up:
Impending slip:
F = µ s N = ( 0.4 )( 0.75 kN )(1 − cosθ )
= ( 0.3 kN )(1 − cosθ )
+ down, – up
ΣFy = 0:
Fs sin θ ± F − W = 0
( 0.75 kN )( tan θ − sin θ ) ± ( 0.3 kN )(1 − cosθ ) − W = 0
or
W = ( 0.3 kN ) [ 2.5 ( tan θ − sin θ ) ± (1 − cosθ )]
with θ = 30°:
Wup = 0.01782 kN
( OK )
Wdown = 0.0982 kN
( OK )
Equilibrium if 17.82 N ≤ W ≤ 98.2 N W
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 8, Solution 35.
Geometry:
1 m − ( 0.5 m ) cos α  tan θ = ( 0.5 m ) sin α
tan θ ( 2 − cos α ) = sin α
θ = 30° → α = 60°
then LAB = (1 m ) cos 30° =
FBD B:
3
m
2
1 
 kN   3
m − m
Fs = k ( LAB − L0 ) = 1.5
 
m  2
2 

Fs = 0.75
( 3 − 1) kN = 549.04 N
ΣFx = 0:
F + W sin 60° − 549.04 N = 0
F = 549.04 N −
ΣFy = 0:
W
2
N − W cos 60° = 0,
N =
3
W
2
For impending slip upward, F is as shown and F = µ s N , so
549.04 N −
W
3
W,
= 0.40
2
2
Wmin = 648.61 N
For impending slip downward, F is reversed, or F = − µ s N , so
549.04 N −
m=
W
( 9.81 m/s )
2
W
3
W,
= − 0.40
2
2
so
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.
Wmax = 3575 N
66.1 kg ≤ m ≤ 364 kg W
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Chapter 8, Solution 36.
FBD Collar:
Note: BC is a two-force member, and for M max , slip will impend to the
right.
ΣFy = 0:
FBC cosθ − N = 0,
Impending slip:
F = µ s N = µ s FBC cosθ
ΣFx = 0:
FBC sin θ − F − P = 0
N = FBC cosθ
FBC ( sin θ − µ s cosθ ) = P
FBD AB:
ΣM A = 0:
M − ( 2l ) FAB cosθ = 0
M = 2l cosθ
P
sin θ − µ s cosθ
M max =
2 Pl
W
tan θ − µ s
For µ s = tan θ , M max = ∞ 
 self locking W
For µ s > tan θ , M max < 0 
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 8, Solution 37.
Geometry:
L L L
+ −
θ = cos −1 2 4 2 = 60°
L
2
For min
FBD AB:
a
L
slip will impend to right and reactions will be at
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900° from normal.
Note: AB is a three-force member
CD = a tan ( 60 + φs ) = ( L − a ) tan ( 60° − φ s )
a tan ( 79.29° ) = ( L − a ) tan ( 40.71° )
6.1449 =
L
−1
a
a
= 0.13996
L
min
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
= 0.1400 W
L
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Chapter 8, Solution 38.
FBD A:
Note: Rod is a two force member. For impending slip the reactions are at
angle
φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.801°
FBD B:
Consider first impending slip to right
9 lb
FAB =
= 3.8572 lb
tan 66.801
ΣFy = 0:
N B − ( 3.8522 lb ) sin 30° − ( 6 lb ) cos 30° = 0
N B = 7.1223 lb,
FB = µ s N B = 0.40 ( 7.1223 lb )
FB = 2.8489 lb
ΣFx = 0:
− 2.8489 lb + ( 3.8572 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0
Pmin = − 2.508 lb
FBD A:
Next consider impending slip to left
FAB = ( 9 lb ) tan 66.801° = 21.000 lb
FBD B:
ΣFy = 0:
N B − ( 21 lb ) sin 30° − ( 6 lb ) cos 30° = 0,
N B = 15.6959 lb
FB = µ s N B = 0.4 (15.6959 lb ) = 6.2784 lb
ΣFx = 0:
6.2784 lb + ( 21 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0
Pmax = 21.465 lb
equilibrium for − 2.51 lb ≤ P ≤ 21.5 lb W
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 8, Solution 39.
FBD AB:
ΣM A = 0:
8 in 2 + 4 in 2 ( N ) − M A = 0
N =
(12 lb ⋅ ft )(12 in./ft ) = 16.100 lb
8.9443 in.
F = µ s N = 0.3 (16.100 lb ) = 4.83 lb
Impending motion:
Note: For max MC, need F in direction shown; see FBD BC.
FBD BC + collar:
ΣM C = 0:
or
MC =
M C − (17 in.)
1
2
2
N − ( 8 in.)
N − (13 in.)
F =0
5
5
5
17 in.
16 in.
26 in.
(16.100 lb ) +
(16.100 lb ) +
( 4.830 lb ) = 293.77 lb ⋅ in.
5
5
5
( MC )max = 24.5 lb ⋅ ft
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 8, Solution 40.
FBD yoke:
ΣFx = 0:
P − N = 0,
N = P = 8 lb
F = µ s N = 125 ( 8 lb )
For impending slip,
F = 2 lb
For M max , F on yoke is down as shown
FBD wheel and slider:
For M min , F on yoke is up.
(a) For M max the 2 lb force is up as shown.
ΣM B = 0:
M B − ( 3 in.) sin 65° ( 8 lb ) − ( 3 in.) cos 65° ( 2 lb ) = 0
M B max = 24.3 lb ⋅ in. W
(b) For M min the 2 lb force is reversed, and
ΣM B = 0:
M B − ( 3in .) sin 65°  ( 8 lb ) + ( 3 in.) cos 65°  ( 2 lb ) = 0
M B min = 19.22 lb ⋅ in.
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.
W
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Chapter 8, Solution 41.
FBD Rod:
ΣM A = 0:
( 20 in.) N1 − (12.5 in.)(12 lb ) = 0
N1 = 7.5 lb.
FBD Cylinder:
ΣFy = 0:
N 2 − 7.5 lb − 36 lb = 0,
N 2 = 43.5 lb
since µ1 = µ2 and N1 < N 2 , slip will impend at top of cylinder first, so
F1 = µ s N1 .
F1 = 0.35 ( 7.5 lb ) = 2.625 lb
ΣM D = 0:
( 4.25 in.) P − (12.5 in.)( 2.625 lb ) = 0,
P = 7.7206 lb
Pmax = 7.72 lb W
To check slip analysis above, ΣFy = 0:
N 2 − 36 lb − 7.5 lb = 0
N 2 = 43.5 lb
F2max = µ s N 2 = 0.35 ( 43.5 lb ) = 15.225 lb
ΣFx = 0:
P − F1 − F2 = 0,
7.72 lb − 2.625 lb − F2 = 0
F2 = 5.095 lb < Fmax ,
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.
OK
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 42.
FBD pulley:
Note that φSA = tan −1 µ SA = tan −1 ( 0.5 ) = 26.565° < 30°, Cable is needed
to keep A from sliding downward.
ΣFy = 0:
2T − WB = 0,
T =
WB
,
2
WB = 2T
(1)
FBD block A:
(a) For minimum WB , there will be impending slip of block A
downward, and FA = µ SA N A as shown.
ΣFy′ = 0:
N A − WA cos 30° = 0,
N A = WA cos 30°
= 23.544 N cos 30° = 20.390 N
(
)
WA = ( 2.4 kg ) 9.81 m/s 2 = 23.544 N
FA = ( 0.50 )( 20.390 N ) = 10.195 N
FBD block C:
ΣFx′ = 0:
T − WA sin 30° + FA = 0
T = ( 23.544 N ) sin 30° − 10.195 N = 1.577 N
From (1)
WB = 2T = 3.154 N,
mB =
3.154 N
(
9.81 m/s 2
)
= 0.322 kg,
mB min = 322 g ΣFy = 0:
NC − WC = 0,
NC = 58.86 N
FC max = µ SC NC = 0.30 ( 58.86 N ) = 17.658 N
Since T = 1.577 N < FC max , block B doesn’t slip and above answer for
mB min is correct.
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
(b) For mB max assume impending slip of block C to left, FC = Fmax
ΣFx = 0:
− T + FC = 0,
From (1) WB = 2T = 35.316 N,
T = FC = FC max = 17.658 N
mB =
WB
35.316 N
=
= 3.6 kg
g
9.81 m/s 2
From FBD block A,
ΣFx = 0:
T − WA sin 30° + FA = 0,
FA = WA sin 30° − T
FA = ( 23.544 N ) sin 30° − 17.658 N = − 5.886,
Since FA < FA max , A does not slip
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.
FAmax = 10.195 N
M B max = 3.6 kg COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 43.
FBD A:
For impending motion A must start up and C down the incline. Since the
normal force between A and B is less than that between B and C, and the
friction coefficients are the same, Fmax will be reached first between A
and B, and B and C will stay together.
ΣFy = 0:
N1 − ( 4 lb ) cos 30° = 0,
Impending slip:
F1 = µ s N1 = 2 3µ s lb
ΣFx = 0:
T − ( 4 lb ) sin 30° − 2 3µ s lb = 0
(
FBD B and C:
N1 = 2 3 lb
)
T = 2 1 + 3µ s lb
ΣFy = 0:
N 2 − 2 3 lb − ( 3 lb + 8 lb ) cos 30° = 0
N2 =
Impending slip:
ΣFx = 0:
(1)
15
3 lb
2
15
3µ s lb
2

15  
3  µs  lb − ( 3 + 8) lbsin 30° = 0
T +  2 3 +
 
2

F2 = µ s N 2 =
 11 19

3µ s  lb
T = −
2
2

FBD B:
Equating (1) and (2):
(
(2)
)
4 1 + 3µ s lb = 11 − 19 3µ s
23 3 µ s = 7,
µ s min = 0.1757 W
To check slip reasoning above:
ΣFy = 0:
N3 − 2 3 lb − ( 3 lb ) cos 30° = 0,
F3max = µ s N3 =
ΣFx = 0:
N3 =
7
3 lb
2
7
3µ s
2
− ( 3 lb ) sin 30° + 2 3µ s lb − F3 = 0
F3 = 2 3 ( 0.1757 ) lb −
3
lb = − 0.891 lb
2
F3 < F3max , OK
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 8, Solution 44.
FBD rod:
3 in.
N B − ( 4.5 in.) cosθ  W = 0
cosθ
ΣM A = 0:
or
N B = (1.5cos 2 θ )W
Impending motion:
FB = µ s N B = (1.5µ s cos 2 θ )W
= ( 0.3cos 2 θ )W
ΣFx = 0:
N A − N B sin θ + FB cosθ = 0
or
N A = (1.5cos 2 θ )W ( sin θ − 0.2 cosθ )
Impending motion:
FA = µ s N A
= ( 0.3cos 2 θ )W ( sin θ − 0.2 cosθ )
ΣFy = 0: FA + N B cosθ + FB sinθ − W = 0
(
FA = W 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ
or
)
Equating FA’s
0.3cos 2 θ ( sin θ − 0.2cosθ ) = 1 − 1.5cos3 θ − 0.3cos 2 θ sinθ
0.6cos 2 θ sin θ + 1.44cos3 θ = 1
Solving numerically
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.
θ = 35.8° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 45.
FBD pin A:
ΣFx = 0:
12
3
FAB − FAC = 0
13
5
ΣFy = 0:
5
4
FAB + FAC − P = 0
13
5
Solving:
FAB =
13
P,
21
NB −
12 13
⋅
P = 0,
13 21
FAC =
20
P
21
FBD B:
ΣFx = 0:
NB =
For Pmin slip of B impends down, so FB = µ s N B =
ΣFy = 0:
FBD C:
11 12
5 13
⋅
P−
⋅
P − 18 lb = 0,
20 21
13 21
12
P
21
11
NB
20
Pmin = 236.25 lb
(For P < 236.25 lb, A will slip down)
ΣFy = 0:
NC − 80 lb −
4 20
16
⋅
P = 0, NC = 80 lb +
P
5 21
21
For Pmax slip of C impends to right, FC = µ s NC
FC =
or
ΣFx = 0:
11 
16 
44
P  = 44 lb +
P
 80 lb +
20 
21 
105
3 20
⋅
P − FC = 0,
5 21
12
44
P = 44 lb +
P
21
105
Pmax = 288.75 lb
∴ equilibrium 236 ≤ P ≤ 289 W
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 8, Solution 46.
φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801°, slip impends at wedge/block wedge/wedge and block/incline
FBD Block:
R2
530 lb
=
sin 41.801° sin 46.398°
R2 = 487.84 lb
FBD Wedge:
P
487.84 lb
=
sin 51.602° sin 60.199°
P = 441 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 47.
φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.801°, and slip impends at wedge/lower block, wedge/wedge, and upper
block/incline interfaces.
FBD Upper block and wedge:
R2
530 lb
=
sin 41.801° sin 38.398°
R2 = 568.76 lb
FBD Lower wedge:
P
568.76 lb
=
sin 51.602° sin 68.199°
P = 480 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 48.
(
)
WD = (18 kg ) 9.81 m/s 2 = 176.58 N
Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.0362°
FBD Lever:
ΣM C = 0:
( 0.3 m )( 350 N ) − ( 0.4 m )(176.58 N )
− ( 0.525 m ) RA cos 4.0362°
+ ( 0.05 m ) RA sin 4.0362° = 0
RA = 66.070 N
ΣFx = 0:
ΣFy = 0:
( 66.07 N ) sin 4.0362° + Cx = 0,
Cx = − 4.65 N
( 66.07 N ) cos 4.0362° − 350 N − 176.58 N = 0
FBD Wedge:
P
66.070 N
=
sin18.072° sin 75.964°
P = 21.1 lb
(a)
(b)
P = 21.1 lb
C x = 4.65 N
C y = 461 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 8, Solution 49.
(
)
WD = (18 kg ) 9.81 m/s 2 = 176.58 N
Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.0362°
FBD Lever:
ΣM C = 0:
( 0.3 m )( 350 N ) − ( 0.4 m )(176.58 N )
− ( 0.525 m ) RA cos 24.036°
− ( 0.05 m ) RA sin 24.036° = 0
RA = 68.758 N
ΣFx = 0:
ΣFy = 0:
C x − ( 68.758 N ) sin 24.036° = 0,
Cx = 28.0 N
C y − 350 N − 176.58 N + ( 68.758 N ) cos 24.036° = 0
C y = 464 N
FBD Wedge:
P
68.758 N
=
sin 38.072° sin 75.964°
(a)
(b)
P = 43.7 N
C x = 28.0 N
C y = 464 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 8, Solution 50.
For steel/steel contact,
φ s1 = tan −1 µ s1 = tan −1 ( 0.3) = 16.6992°
For steel/concrete interface, φ s2 = tan −1 µ s2 = tan −1 ( 0.6 ) = 30.964°
FBD Plate CD:
ΣFy = 0:
N − 90 kN = 0,
F = µ s1 N = 0.3 ( 90 kN ) = 27 kN
Impending slip:
ΣFx = 0:
F = 90 kN
F − Q = 0,
Q = F = 27 kN
FBD Top wedge assuming impending slip between wedges:
ΣFy = 0:
Rw = 100.74 kN
Rw cos 26.699° − 90 kN = 0,
ΣFx = 0:
P − 27 kN − (100.74 kN ) sin 26.699° = 0
P = 72.265 kN,
(a)
P = 72.3 kN
(b)
Q = 27.0 kN
To check above assumption; note that bottom wedge is a two-force member so the reaction of the floor on that
wedge is Rw, at 26.699° from the vertical. This is less than φ s2 = 30.964°, so the bottom wedge doesn’t slip
on the concrete.
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 8, Solution 51.
For steel/steel contact, φ s1 = tan −1 µ s1 = tan −1 ( 0.30 ) = 16.6992°
For steel/concrete contact, φ s2 = tan −1 µ s2 = tan −1 ( 0.60 ) = 30.964°
FBD Plate CD and top wedge:
Q = 90 kN tan 26.6992° = 45.264 kN
Rw =
90 kN
= 100.741 kN
cos 26.6992°
FBD Bottom wedge: slip impends at both surfaces
P
100.714 kN
=
sin 57.663° sin 59.036°
(a)
P = 99.3 kN
(b)
Q = 45.3 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 8, Solution 52.
FBD Wedge:
φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801°
By symmetry RB = RC
ΣFy = 0:
2RC sin ( 29.801° ) − P = 0,
P = 0.9940 RC
RC
175 lb
,
=
sin 41.801° sin18.397 lb
P = 367.3 lb
FBD Block C:
(a)
P = 367 lb
b) Note: That increasing friction between B and the incline will mean that block B will not slip, but the above
calculations will not change.
(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.
P = 367 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 53.
FBD Block C:
φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.8014°
ΣFx = 0:
RACx − RCFx = 0
ΣFy = 0:
so
RCFy − RACy − 175 lb = 0
RCFy
RCFx
−
RACy
RACx
=
175 lb
RACx
cot ( 20° + φ ) − cot ( 32.2° ) > 0
φ < 12.2° < φ s = 21.8°
so block C does not slip (or impend)
FBD Block B:
(a) φ B = tan −1 µ B = tan −1 ( 0.4 ) = 21.8014°
RB
175 lb
,
=
sin 41.8014° sin 46.3972°
RB = 161.083 lb
(b) φ B = tan −1 µ B = tan −1 ( 0.6 ) = 30.9638°
RB
175 lb
,
=
sin 50.9638° sin 37.2330°
RB = 224.65 lb
FBD Wedge:
P
RB
=
,
sin 59.6028° sin 52.1986°
P = 1.09163 RB
(a)
RB = 161.083 lb,
P = 175.8 lb
(b)
RB = 224.65 lb,
P = 245 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 54.
Since vertical forces are equal and µ s ground > µ s wood, assume no impending motion of board. Then there
will be impending slip at all wood/wood contacts, φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
FBD Top wedge:
R1 =
8 kN
= 8.4758 kN
cos19.29°
R1
P
=
sin 52.710° cos 56.580°
P = 8.892 kN
To check assumption, consider
FBD wedges + board:
F1 = µ1 8 kN = 0.35 ( 8 kN ) = 2.8 kN
ΣFy = 0:
NG − 8 kN = 0,
NG = 8 kN
FG max = µG NG = ( 0.6 )( 8 kN ) = 4.8 kN
ΣFx = 0:
FG − F1 = 0,
FG = F1 = 2.8 kN
FG < FG max ,
OK
∴ P = 8.89 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 8, Solution 55.
Assume no impending motion of board on ground. Then there will be impending slip at all wood/wood
interfaces.
FBD Top wedge:
Wedge is a two-force member so R 2 = − R1
and θ = 2φs = 2 tan −1 µ s = 2 tan −1 ( 0.35 )
θ = 38.6°
To check assumption, consider
FBD wedges + board:
F1 = µ1 8 kN = 0.35 ( 8 kN ) = 2.8 kN
ΣFy = 0:
NG − 8 kN = 0,
NG = 8 kN
FG max = µG NG = ( 0.6 )( 8 kN ) = 4.8 kN
ΣFx = 0:
FG − F1 = 0,
FG = F1 = 2.8 kN
FG < FG max ,
OK
∴ P = 8.89 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 8, Solution 56.
FBD Cylinder:
Slip impends at B
φ SC = tan −1 ( 0.35 ) = 19.2900°
ΣM A = 0:
r RC cos (12° + 19.29° ) −
4r
W =0
3π
RC = 0.49665, W = 124.163 lb
FBD Wedge:
φ SF = tan −1 µ SF = tan −1 ( 0.50 ) = 26.565°
P
124.163 lb
=
sin 58.855° sin 63.435°
P = 117.5 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 57.
FBD tip of screwdriver:
φs = tan −1 µ s = tan −1 ( 0.12 ) = 6.8428°
by symmetry R1 = R2
ΣFy = 0:
2R1 sin ( 6.8428° + 8° ) − 3.5 N = 0
R1 = R2 = 6.8315 N
If P is removed quickly, the vertical components of R1 and R2 vanish, leaving the horizontal components
H1 = H 2 = ( 6.8315 N ) cos14.8428°
= 6.6035 N
Side forces = 6.60 N This is only instantaneous, since 8° > φ s , so the screwdriver will be forced out.
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 8, Solution 58.
As the plates are moved, the angle θ will decrease.
(a)
φ s = tan −1 µ s = tan −1 0.2 = 11.31°. As θ decreases, the minimum angle at the contact approaches
12.5° > φs = 11.31°, so the wedge will slide up and out from the slot.
(b)
φ s = tan −1 µ s = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this
time the angle at the other contact is 25° − 16.7° = 8.3° < φ s ) The wedge binds in the slot.
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 8, Solution 59.
FBD Wedge:
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
by symmetry R1 = R2
ΣFy = 0:
2 R1 sin 22.29° − 60 lb = 0
R2 = 79.094 lb
When P is removed, the vertical component of R1 and R2 will vanish, leaving the horizontal components
H1 = H 2 = ( 79.094 lb ) cos 22.29°
= 73.184 lb
Final forces H1 = H 2 = 73.2 lb
Since these are at 3° ( < φs ) from the normal, the wedge is self-locking and will remain in place.
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 8, Solution 60.
FBD Cylinder:
(
)
W = ( 80 kg ) 9.81 m/s 2 = 784.8 N
ΣM G = 0:
FA − FB = 0,
ΣM D = 0:
dN B − dN A + rW = 0,
so
N A > NB,
FA = FB
(1)
N A = NB +
W
3
(2)
FA max > FB max
∴ slip impends first at B. FB = µ s N B = 0.25 N B
ΣM A = 0: ( r cos 30° ) N B − ( r sin 30° )W − r (1 + sin 30° )( 0.25 N B ) = 0
r
=
note d =
tan 30
N B = 1.01828W = 799.15 N
3r
FB = 0.25 N B = 199.786 N
From (2) above, N A = 799.15 N +
784.8 N
= 1252.25 N
3
From (1), FA = FB = 199.786 N
ΣFy = 0:
FBD Wedge:
NC − (1252.25 N ) cos10° + 199.786 N sin10° = 0
NC = 1198.53 N
Impending slip FC = µ s NC = 0.25 (1198.53 N ) = 299.63 N
ΣFx = 0:
P − 299.63 N − (199.786 N ) cos10°
− (1252.25 N ) sin10° = 0
P = 714 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.
20.0°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 61.
FBD Cylinder:
(
)
W = ( 80 kg ) 9.81 m/s 2 = 784.8 N
For impending slip at B,
ΣM A = 0:
FB = µ sB N B = 0.30 N B
( r cos 30° ) N B − r (1 + sin 30° )( 0.30 N B )
− r sin 30°W = 0
N B = 1.20185W = 943.21 N
FB = 0.30 N B = 0.36055W
ΣM G = 0:
ΣFx = 0:
r ( FA − FB ) = 0,
FA = FB = 0.36055W
N A sin 30° + FA cos30° − N B = 0
NA =
− ( 0.36055W ) cos30° + 1.20185W
sin 30°
N A = 1.77920W
For minimum µ A , slip impends at A, so
µ A min =
FA
0.36055W
=
= 0.2026
N A 1.77920W
µ A min = 0.203
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 8, Solution 62.
FBD plank + wedge:
(8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft )
ΣM A = 0:
− ( 2 ft )
1
( 48 lb/ft )( 3 ft )
2

5  1
−  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0
3  2

N B = 185 lb
ΣFy = 0:
 48 + 96

NW + 185 lb − 
lb/ft  ( 3ft )
2


+
1
( 96 lb/ft )( 5 ft ) = 0
2
NW = 271 lb
Since NW > N B , and all µ s are equal, assume slip impends at B and between wedge and floor, and not at A.
Then FW = µ s NW = 0.45 ( 271 lb ) = 121.95 lb
FB = µ s N B = 0.45 (185 lb ) = 83.25 lb
ΣFx = 0:
P − 121.95 lb − 83.25 lb = 0,
P = 205.20 lb
Check Wedge for assumption
271 lb − RA cosθ = 0
ΣFy = 0:
ΣFx = 0:
so tan θ =
205.2 lb − 121.95 lb − RA sin θ = 0
83.25
= 0.3072 < µ s + tan 9°
271
so no slip here
∴ (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.
P = 205 lb
impending slip at B
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 63.
FBD plank + wedge:
ΣM A = 0:
(8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft )
1
( 48 lb/ft )( 3 ft )
2

5 
−  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0
3 

− ( 2 ft )
NW = 185 lb
ΣFy = 0:
 48 + 96

N A + 185 lb − 
lb/ft  ( 3ft )
2


1
− ( 96 lb/ft )( 5 ft ) = 0
2
N A = 271 lb
Since N A > NW , and all µ s are equal, assume impending slip at top and bottom of wedge and not at A. Then
FW = µ s NW = 0.45 (185 N )
FW = 83.25 lb
FBD Wedge:
φ s = tan −1 µ s = tan −1 ( 0.45 ) = 24.228°
ΣFy = 0:
185 lb − RB cos ( 24.228° + 9° ) = 0
RB = 221.16 lb
ΣFx = 0:
( 221.16 lb ) sin 33.228° + 83.25 lb − P = 0
P = 204.44 lb
Check assumption using plank/wedge FBD
ΣFx = 0:
FA + FW − P = 0,
FA = 204.44 lb − 83.25 lb = 121.19 lb
FA max = µ s N A = 0.45 ( 271 lb ) = 121.95 lb
FA < FA max , OK
∴ (a)
P = 204 lb
(b)
no impending slip at A
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 8, Solution 64.
(
)
(
)
WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N
Slip must impend at all surfaces simultaneously, F = µ s N
FBD I: A + B
ΣFy = 0:
N B − 150 N − 98.1 N − 490.5 N = 0,
impending slip:
FBD II: A
N B = 738.6 N
FB = µ s N B = ( 738.6 N ) µ s
N A = ( 738.6 N ) µ s
ΣFx = 0:
N A − FB = 0,
ΣFy′ = 0:
FAB + ( 738.6µ s ) N  sin 20° − (150 N + 98.1 N ) cos 20° = 0
FAB =  233.14 − ( 252.62 ) µ s  N
ΣFx′ = 0:
( 738.6µ s ) N  cos 20° − (150 N + 98.1 N ) sin 20° − N AB = 0
N AB = 84.855 + ( 694.06 ) µ s  N
µs =
FAB
233.14 − 252.62µ s
=
N AB
84.855 + 694.06µ s
µ s2 = 0.48623µ s − 0.33591 = 0
µ s = − 0.24312 ± 0.62850
Positive root
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.
µ s = 0.385
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 65.
(
)
(
)
WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N
Slip impends at all surfaces simultaneously
FBD I: A + B
ΣFx = 0:
ΣFy = 0:
N A − FB = 0,
N A = FB = µ s N B
(1)
FA − (150 N + 98.1 N + 490.5 N ) + N B = 0
µ s N A + N B = 738.6 N
Solving (1) and (2) N B =
FBD II: B
ΣFx′ = 0:
FB =
738.6 µ s
N
1 + µ s2
N AB + ( 490.5 N ) cos 70° − N B cos 70° − FB sin 70° = 0
N AB =
ΣFy′ = 0:
738.6 N
,
1 + µ s2
(2)
738.6 N
( cos 70° + µ s sin 70° ) − ( 490.5 N ) cos 70° (1)
1 + µ s2
−FAB − ( 490.5 N ) sin 70° + N B sin 70° − FB cos 70° = 0
FAB =
738.6 N
( sin 70° − µs cos 70° ) − ( 490.5 N ) sin 70° = 0
1 + µ s2
Setting FAB = µ s N AB ,
µ s3 − 6.8847µ s2 − 2.0116µ s + 1.38970 = 0
Solving numerically, µ s = − 0.586, 0.332, 7.14
Physically meaningful solution:
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.
µ s = 0.332
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 66.
FBD jack handle:
See Section 8.6
ΣM C = 0:
aP − rQ = 0 or P =
r
Q
a
FBD block on incline:
(a)
Raising load
Q = W tan (θ + φ s )
P=
r
W tan (θ + φs )
a
continued
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
PROBLEM 8.66 CONTINUED
(b)
Lowering load if screw is self-locking ( i.e.: if φs > θ )
Q = W tan (φs − θ )
P=
(c)
r
W tan (φs − θ )
a
Holding load is screw is not self-locking ( i.e: if φs < θ )
Q = W tan (θ − φs )
P=
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.
r
W tan (θ − φ s )
a
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 67.
FBD large gear:
ΣM C = 0:
(12 in.)W − 7.2 kip ⋅ in. = 0,
W = 0.600 kips
= 600 lb
Block on incline:
θ = tan −1
0.375 in.
= 2.2785°
2π (1.5 in.)
φ s = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = W tan (θ + φ s )
= ( 600 lb ) tan 9.1213° = 96.333 lb
FBD worm gear:
r = 1.5 in.
ΣM B = 0:
(1.5 in.)( 96.333 lb ) − M = 0
M = 144.5 lb ⋅ in.
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 8, Solution 68.
FBD large gear:
ΣM C = 0:
(12 in.)W − 7.2 kip ⋅ in. = 0
W = 0.600 kips = 600 lb
Block on incline:
θ = tan −1
0.375 in.
= 2.2785°
2π (1.5 in.)
φ s = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = W tan (φ s − θ )
= ( 600 lb ) tan 4.5643° = 47.898 lb
FBD worm gear:
r = 1.5 in.
ΣM B = 0:
M − (1.5 in.)( 47.898 lb ) = 0
M = 71.8 lb ⋅ in.
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 8, Solution 69.
Block/incline analysis:
θ = tan −1
0.125 in.
= 2.4238°
2.9531 in.
φ s = tan −1 ( 0.35 ) = 19.2900°
Q = 47250 tan ( 21.714° ) = 18.816 lb
Couple =
d
 0.94 
in.  (18.516 lb ) = 8844 lb ⋅ in.
Q=
2
 2

Couple = 7.37 lb ⋅ ft
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 8, Solution 70.
FBD joint D:
FAD = FCD
By symmetry:
ΣFy = 0: 2FAD sin 25° − 4 kN = 0
FAD = FCD = 4.7324 kN
FBD joint A:
FAE = FAD
By symmetry:
ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0
FAC = 8.5780 kN
Block and incline A:
θ = tan −1
2 mm
= 4.8518°
π ( 7.5 mm )
φ s = tan −1 µ s = tan −1 0.15 = 8.5308°
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
PROBLEM 8.70 CONTINUED
Q = ( 8.578 kN ) tan (13.3826° )
= 2.0408 kN
Couple at A:
M A = rQ
 7.5

=
mm  ( 2.0408 kN )
2


= 7.653 N ⋅ m
By symmetry: Couple at C:
M C = 7.653 N ⋅ m
Total couple M = 2 ( 7.653 N ⋅ m )
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.
M = 15.31 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 71.
FBD joint D:
By symmetry:
FAD = FCD
ΣFy = 0: 2FAD sin 25° − 4 kN = 0
FAD = FCD = 4.7324 kN
FBD joint A:
By symmetry:
FAE = FAD
ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0
FAC = 8.5780 kN
Block and incline at A:
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
PROBLEM 8.71 CONTINUED
θ = tan −1
2 mm
= 4.8518°
π ( 7.5 mm )
φ s = tan −1 µ s = tan −1 0.15
φ s = 8.5308°
φ s − θ = 3.679°
Q = ( 8.5780 kN ) tan 3.679°
Q = 0.55156 kN
Couple at A: M A = Qr
 7.5 mm 
= ( 0.55156 kN ) 

 2 
= 2.0683 N ⋅ m
By symmetry:
Couple at C : M C = 2.0683 N ⋅ m
Total couple M = 2 ( 2.0683 N ⋅ m )
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.
M = 4.14 N ⋅ m
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 72.
FBD lower jaw:
By symmetry B = 540 N
ΣFy = 0:
− 540 N + A − 540 N = 0,
A = 1080 N
(a) since A > B when finished, adjust A first when there will be no force
Block/incline at B:
(b) θ = tan −1
4 mm
= 6.0566°
12π mm
φ s = tan −1 µ s = tan −1 ( 0.35) = 19.2900°
Q = ( 540 N ) tan 25.3466° = 255.80 N
Couple = rQ = ( 6 mm )( 255.80 N ) = 1535 N ⋅ mm
M = 1.535 N ⋅ m
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 8, Solution 73.
FBD lower jaw:
By symmetry B = 540 N
ΣFy = 0:
− 540 N + A − 540 N = 0,
A = 1080 N
since A > B, A should be adjusted first when no force is required.
If instead, B is adjusted first,
Block/incline at A:
θ = tan −1
4 mm
= 6.0566°
12π mm
φ s = tan −1 µ s = tan −1 ( 0.35) = 19.2900°
Q = (1080 N ) tan 25.3466° = 511.59 N
Couple = rQ = ( 6 mm )( 511.59 N ) = 3069.5 N ⋅ mm
M = 3.07 N ⋅ m
Note that this is twice that required if A is adjusted first.
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 8, Solution 74.
Block/incline:
θ = tan −1
0.25 in.
= 2.4302°
1.875π in.
φ s = tan −1 µ s = tan −1 ( 0.10 ) = 5.7106°
Q = (1000 lb ) tan ( 8.1408° ) = 143.048 lb
Couple = rQ = ( 0.9375 in.)(143.048 lb ) = 134.108 lb ⋅ in.
M = 134.1 lb ⋅ in.
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 8, Solution 75.
FBD Bucket:
(
)
= ( 0.18 m ) sin ( tan 0.30 ) = 0.05172 m
rf = r sin φ s = r sin tan −1 µ s
−1
ΣM A = 0:
(1.6 m + 0.05172 m ) T − ( 0.05172 m )W = 0
T = 0.031314W

kN 
= 0.031314 ( 50 Mg )  9.81

Mg 

= 15.360 kN
T = 15.36 kN !
NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 76.
FBD Windlass:
(
rf = rb sin φs = rb sin tan −1 µ s
(
)
)
= (1.5 in.) sin tan −1 0.5 = 0.67082 in.
ΣM A = 0:
( 8 − 0.67082 ) in. P − ( 5 + 0.67082 ) in. 160 lb = 0
P = 123.797 lb
P = 123.8 lb NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 77.
FBD Windlass:
(
rf = r sin φs = r sin tan −1 µ s
(
)
)
= (1.5 in.) sin tan −1 0.5 = 0.67082 in.
ΣM A = 0:
( 8 + 0.67082 ) in. P − ( 5 + 0.67082 ) in. (160 lb ) = 0
P = 104.6 lb NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 78.
FBD Windlass:
(
rf = r sin φs = r sin tan −1 µ s
(
)
)
= (1.5 in.) sin tan −1 0.50 = 0.67082 in.
ΣM A = 0:
( 8 + 0.67082 ) in. P − ( 5 − 0.67082 ) in. (160 lb ) = 0
P = 79.9 lb NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 79.
FBD Windlass:
(
rf = r sin φs = r sin tan −1 µ s
(
)
)
= (1.5 in.) sin tan −1 0.50 = 0.67082 in.
ΣM A = 0:
( 8 − 0.67082 ) in. P − ( 5 − 0.67082 ) in. (160 lb ) = 0
P = 94.5 lb NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 80.
(a) FBD lever (Impending CW
rotation):
ΣM C = 0:
( 0.2 m + rf ) ( 75 N ) − ( 0.12 m − rf ) (130 N ) = 0
rf = 0.0029268 m = 2.9268 mm
sin φs =
rf
rs
*
 −1 rf 
 −1 2.9268 mm 
µ s = tan φs = tan  sin
 = tan  sin

18 mm 
rs 


= 0.34389
µ s = 0.344 (b) FBD lever (Impending CCW
rotation):
( 0.20 m − 0.0029268 m )( 75 N )
ΣM D = 0:
− ( 0.12 m + 0.0029268 m ) P = 0
P = 120.2 N NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 81.
Pulley FBD’s:
rp = 30 mm
Left:
(
rf = raxle sin φk = raxle sin tan −1 µ k
)
*
(
= ( 5 mm ) sin tan −1 0.2
)
= 0.98058 mm
Left:
ΣM C = 0:
Right:
( rp − rf ) ( 600 lb ) − 2rpTAB = 0
TAB =
or
30 mm − 0.98058 mm
( 600 N ) = 290.19 N
2 ( 30 mm )
TAB = 290 N ΣFy = 0:
290.19 N − 600 N + TCD = 0
TCD = 309.81 N
or
TCD = 310 N Right:
( rp + rf ) TCD − ( rp − rf ) TEF = 0
ΣM G = 0:
or
TEF =
30 mm + 0.98058 mm
( 309.81 N ) = 330.75 N
30 mm − 0.98058 mm
TEF = 331 N NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 82.
Pulley FBDs:
Left:
rp = 30 mm
(
rf = raxle sin φk = raxle sin tan −1 µ k
)
*
(
= ( 5 mm ) sin tan −1 0.2
)
= 0.98058 mm
( rp + rf ) ( 600 N ) − 2rpTAB = 0
ΣM C = 0:
30 mm + 0.98058 mm
( 600 N ) = 309.81 N
2 ( 30 mm )
TAB =
or
TAB = 310 N ΣFy = 0:
Right:
TAB − 600 N + TCD = 0
TCD = 600 N − 309.81 N = 290.19 N
or
TCD = 290 N ( rp − rf ) TCD − ( rp + rf ) TEF = 0
ΣM H = 0:
TEF =
or
30 mm − 0.98058 mm
( 290.19 N )
30 mm + 0.98058 mm
TEF = 272 N NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there is
little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79, µ s = 0.50,
and the error made by using the approximation is about 11.8%.
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 8, Solution 83.
FBD link AB:
Note: That AB is a two-force member. For impending motion, the pin
forces are tangent to the friction circles.
r
θ = sin −1 f
25 in.
where
(
rf = rp sin φs = rp sin tan −1 µ s
)
*
(
)
= (1.5 in.) sin tan −1 0.2 = 0.29417 in.
Then
θ = sin −1
0.29417 in.
= 1.3485°
12.5 in.
(b)
Rvert = R cosθ
θ = 1.349° Rhoriz = R sin θ
Rhoriz = Rvert tan θ = ( 50 kips ) tan1.3485° = 1.177 kips
(a) Rhoriz = 1.177 kips NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 84.
FBD gate:
(
)
(
)
W1 = 66 kg 9.81 m/s 2 = 647.46 N
W2 = 24 kg 9.81 m/s 2 = 235.44 N
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
)
= ( 0.012 m ) sin tan −1 0.2 = 0.0023534 m
ΣM C = 0:
( 0.6 m − rf )W1 + ( 0.15 m − rf ) P − (1.8 m + rf )W2 = 0
P=
(1.80235 m )( 235.44 N ) − ( 0.59765 m )( 647.46 N )
( 0.14765 m )
= 253.2 N
P = 253 N !
NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 85.
It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single combined weight of
( 90 kg ) ( 9.81 m/s2 ) = 882.9 N, located at a distance x =
right of B.
(
rf = rs sin φs = rs sin tan −1 µ s
(1.8 m )( 24 kg ) − ( 0.6 m )( 66 kg ) = 0.04 m to the
90 kg
) = ( 0.012 m ) sin ( tan 0.2)
*
−1
= 0.0023534 m
FBD pulley + gate:
β = sin −1
α = tan −1
0.04 m
= 14.931°
0.15 m
rf
0.0023534 m
= 0.8686°
0.15524 m
OB
= sin −1
OB =
0.15
= 0.15524 m
cos α
then
θ = α + β = 15.800°
P = W tan θ = 249.8 N
P = 250 N !
NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 86.
FBD gate:
(
)
W = 24 kg ( 9.81 m/s ) = 235.44 N
W1 = 66 kg 9.81 m/s 2 = 647.46 N
2
2
(
rf = rs sin φs = rs sin tan −1 µ s
(
)
*
)
= ( 0.012 m ) sin tan −1 0.2 = 0.0023534 m
ΣM C = 0:
( 0.6 m + rf )W1 + ( 0.15 m + rf ) P − (1.8 m − rf )W2 = 0
P=
(1.79765 m )( 235.44 N ) − ( 0.60235 m )( 647.46 N )
0.15235 m
= 218.19 N
P = 218 N !
NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 87.
It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single weight of
(1.8 m )( 24 kg ) − ( 0.15 m )( 66 kg ) = 0.04 m to the
( 90 kg )( 9.81 N/kg ) = 882.9 N, located at a distance x =
90 kg
right of B.
FBD pulley + gate:
(
rf = rs sin φs = rs sin tan −1 µ s
) = ( 0.012 m ) sin ( tan 0.2)
*
−1
rf = 0.0023534 m
β = sin −1
α = tan −1
0.04 m
= 14.931°
0.15 m
rf
0.0023534 m
= 0.8686°
0.15524 m
OB
= sin −1
OB =
0.15 m
= 0.15524 m
cos α
then
θ = α − β = 14.062°
P = W tan θ = 221.1 N
P = 221 N !
NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 88.
FBD Each wheel:
(
rf = raxle sin φ = raxle sin tan −1 µ
ΣFx = 0:
P
− R sin θ = 0
4
R cosθ −
ΣFy = 0:
)
W
=0
4
∴ tan θ =
sin θ =
but
P
W
rf
rw
P = W tanθ
or
r
= axle sin tan −1 µ
rw
(
)
(a) For impending motion, use µ s = 0.12
sin θ =
0.5 in.
sin tan −1 0.12
5 in.
(
)
θ = 0.68267°
P = W tan θ = ( 500 lb ) tan ( 0.68267° )
P = 5.96 lb (b) For constant speed, use µ k = 0.08
sin θ =
1
sin tan − 1 0.08
10
(
)
θ = 0.45691°
P = ( 500 lb ) tan ( 0.45691° )
P = 3.99 lb NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 89.
FBD Each wheel:
For equilibrium (constant speed) the two forces R and
W
must be equal
2
and opposite, tangent to the friction circle, so
rf
sin θ =
where θ = tan −1 ( slope )
rw
(
)
sin tan −1 0.03 =
rw = (12.5 mm )
(
rB sin tan −1 µ k
rw
)
(
) = 49.666 mm
sin ( tan 0.03)
sin tan −1 0.12
−1
d w = 99.3 mm NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal
bearing and axle friction problems. While this approximation may be valid for very small values of µ , there
is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79,
µ s = 0.50, and the error made by using the approximation is about 11.8%.
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 8, Solution 90.
FBD
(8 in.) Q − M = 0,
ΣM O = 0:
Q=
M
8 in.
but, from equ. 8.9,
M =
2
2
 7 in. 
µk WR = ( 0.60 )(10.1 lb ) 

3
3
 2 
= 14.14 lb
so,
Q=
14.14
,
8
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.
Q = 1.768 lb
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 91.
2
R3 − R13
1
D23 − D13
µ s P 22
µ
=
P
s
3
3
R2 − R12
D22 − D12
Eqn. 8.8 gives
M =
so
( 0.030 m ) − ( 0.024 m )
1
M = ( 0.15 )( 80 kg ) 9.81 m/s 2
3
( 0.030 m )2 − ( 0.024 m )2
(
)
3
M = 1.596 N ⋅ m
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.
3
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 92.
Let the normal force on ∆A be ∆N , and
∆N
k
=
r
∆A
∆F = µ∆N , ∆M = r ∆F
As in the text
The total normal force
2π  R k

P = lim Σ∆N = ∫ 0  ∫ 0 rdr  dθ
∆A → 0
 r

(
R
)
P = 2π ∫ 0 kdr = 2π kR
The total couple
or
P
2π R
k
2π  R

M worn = lim Σ∆M = ∫ 0  ∫ 0 r µ rdr  dθ
∆A → 0
r


R
M worn = 2πµ k ∫ 0 rdr = 2πµ k
R2
P R2
= 2πµ
2
2π R 2
or
M worn =
1
µ PR
2
Now
M new =
2
µ PR
3
Thus
k =
[Eq. (8.9)]
1 µ PR
M worn
3
= 22
= = 75%
µ PR 4
M new
3
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 8, Solution 93.
Let normal force on ∆A be ∆N , and
∆N
k
=
∆A
r
∆F = µ∆N , ∆M = r ∆F
As in the text
The total normal force P is
2π  R k

P = lim Σ∆N = ∫ 0  ∫ R 2 rdr  dθ
1 r
∆A → 0


P = 2π ∫R 2 kdr = 2π k ( R2 − R1 )
R
The total couple is
k =
or
1
P
2π ( R2 − R1 )
k
2π  R

M worn = lim Σ∆M = ∫ 0  ∫R 2 r µ rdr  dθ
∆A → 0
r
 1

∫ ( rdr ) = πµ k (
R
M worn = 2πµ k R 2
1
R22 − R12
(
)
) = 2π ( R − R )
πµ P R22 − R12
2
M worn =
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.
1
1
µ P ( R2 + R1 )
2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 94.
Let normal force on ∆A be ∆N , and
∆A = r ∆s∆φ
∆N = k ∆A
so
∆N
= k,
∆A
∆s =
∆r
sin θ
where φ is the azimuthal angle around the symmetry axis of rotation
∆Fy = ∆N sin θ = kr ∆r ∆φ
P = lim Σ∆Fy
Total vertical force
∆A → 0
2π
P = ∫0
(
( ∫ krdr ) dφ = 2π k ∫ rdr
R2
R1
P = π k R22 − R12
Friction force
)
R2
R1
or
k =
Total couple
M = 2π
(
)
∆F = µ∆N = µ k ∆A
∆M = r ∆F = r µ kr
Moment
π
P
R22 − R12
∆r
∆φ
sin θ
2π  R µ k

M = lim Σ∆M = ∫ 0  ∫R 2
r 2dr  dφ
∆A → 0
 1 sin θ

µ k R2 2
2 πµ
P
r dr =
∫
R1
2
sin θ
3 sin θ π R2 − R32
(
)
M =
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.
(R − R )
3
2
3
3
2 µ P R23 − R13
3 sin θ R22 − R12
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 95.
If normal force per unit area (pressure) of the center is PO , then as a function
r

of r, P = PO 1 − 
R


r
2π R

ΣFN = W = ∫ PdA = ∫ 0 ∫ 0 PO 1 −  rdrdθ
R

2
R3 
R2
2π  R
W = PO ∫ 0 
d
P
−
=
θ
2
π

O
3R 
6
 2
so PO =
3W
π R2
For slipping, dF = µk ( PdA)
r
2π R 
Moment = ∫ rdF = µk PO ∫ 0 ∫ 0 r  1 −  rdrdθ
R

3
R4 
R3
2π  R
d
P
= µ k PO ∫ 0 
−
=
θ
2
πµ

k O
4 R 
12
 3
so M = 2πµ k
3W R3
1
= µ k WR
π R 2 12 2
ΣM O = 0:
(8 in.) Q − M = 0
1
 7 in. 
0.6 )(10.1 lb ) 
(

M
2
 2 
Q=
=
8 in.
(8 in.)
Q = 1.326 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 96.
FBD pipe:
θ = sin −1
0.025 in. + 0.0625 in.
= 1.00257°
5 in.
P = W tan θ for each pipe, so also for total
P = ( 2000 lb ) tan (1.00257° )
P = 35.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 97.
FBD disk:
tan θ = slope = 0.02
b = r tan θ = ( 60 mm )( 0.02 )
b = 1.200 mm
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 8, Solution 98.
FBD wheel:
r = 230 mm
b = 1 mm
θ = sin −1
b
r
b

P = W tan θ = W tan  sin −1  for each wheel, so for total
r

(
)
1 

P = (1000 kg ) 9.81 m/s 2 tan  sin −1

230 

P = 42.7 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 8, Solution 99.
FBD wheel:
(
)
rf = raxle sin φ = raxle sin tan −1 µ ,
rw =
rf
sin θ
+
b
tan θ
For small θ , sin θ
tan θ , so tan θ
R cosθ −
ΣFy = 0:
rf + b
rw
W
=0
4
− R sin θ +
ΣFx = 0:
P
=0
4
P
W
Solving: tan θ =
so P = W tan θ = W
(a) For impending slip, use µ s ,
µs or µk
rf + b
rw
 0.5 in. 
−1
rf = 
 sin tan 0.12 = 0.029786 in.
 2 
(
so P = ( 500 lb )
)
0.02986 in. + 0.25 in.
= 55.96 lb
2.5 in.
P = 56.0 lb (b) For constant speed, use µ k ,
 0.5 in. 
−1
rf = 
 sin tan 0.08 = 0.019936 in.
 2 
so P = ( 500 lb )
(
)
( 0.019936 + 0.25) in. = 53.99 lb
2.5 in.
P = 54.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 100.
FBD wheel:
For equilibrium (constant speed), R and
W
are equal and opposite
2
and tangent to the friction circle as shown
(
)
(
rf = raxle sin tan −1 µk = (12.5 mm ) sin tan −1 0.12
)
rf = 1.48932 mm
From diagram, rw =
For small θ , sin θ
tan θ = slope
rw =
rf
sin θ
+
b
tan θ
tan θ , so rw
rf + b
tan θ
1.48932 mm + 1.75 mm
= 107.977 mm
0.03
d w = 216 mm 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 8, Solution 101.
β = 4π rad
Two full turns of rope →
(a)
T
T1
µ s β = ln 2
µs =
or
µs =
1
β
T
ln 2
T1
1
20 000 N
= 0.329066
ln
4π
320 N
µ s = 0.329 (b)
β =
=
1
µs
T
ln 2
T1
1
80 000 N
ln
0.329066
320 N
= 16.799 rad
β = 2.67 turns 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 8, Solution 102.
FBD A:
(
)
WA = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣFx = 0:
TA − WA sin 30° = 0,
TA =
WA
2
ΣFx′ = 0:
WB sin 30° − TB = 0,
TB =
WB
2
FBD B:
(a) Motion of B impends up incline and mB = 8 kg
TA
= eµs β ,
TB
µs =
=
1
β
1
β
ln
ln
TA
1 W
= ln A
TB
β WB
mA
3  10 kg 
= ln 

π  8 kg 
mB
From hint, β is not dependent
on shape of support
µ s = 0.21309
µ s = 0.213 (b) For maximum mB , motion of B impend down incline
TB
= eµs β ,
TA
TB = TAe
0.21309
π
3
= 1.250TA
∴ WB = 1.25WA and mB = 1.25 mA = 1.25 (10 kg )
mB max = 12.50 kg 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 8, Solution 103.
FBD A:
ΣFx = 0:
TA − WA sin 30° = 0,
TA =
WA
2
ΣFx′ = 0:
WB sin 30° − TB = 0,
TB =
WB
2
FBD B:
For mB min , motion of B impends up incline
π
And
0.50
TA
= e 3 = 1.68809
TB
But
m A WA
T
=
= A = 1.68809
mB
WB
TB
so mB min = 5.9238 kg
From hint, β is not dependent
on shape of C
For mB max , motion of B impends down incline
π
so
0.50
mB
W
T
µ
= B = B = e s β = e 3 = 1.68809
mA WA TA
so mB max = 16.881 kg
For equilibrium
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.
5.92 kg ≤ mB ≤ 16.88 kg COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 104.
β = 1.5 turns = 3π rad
For impending motion of W up
P = We µs β = (1177.2 N ) e(
0.15 )3π
= 4839.7 N
For impending motion of W down
P = We− µs β = (1177.2 N ) e (
− 0.15 )3π
= 286.3 N
For equilibrium
286 N ≤ P ≤ 4.84 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 8, Solution 105.
Horizontal pipe:
Contact angles β H =
Vertical pipe
π
Contact angle βV = π
2
µ sH = 0.25
µ sV = 0.2
For P to impend downward,
(
)
(
)
µ π
µ π
µ π
µ π
P =  e sH 2  Q =  e sH 2  e µsV π R =  e sH 2  e µsV π  e sH 2  (100 lb )








π µ +µ
Pmax = e ( sH sV )  (100 lb ) = (100 lb ) e0.45π = 411.12 lb


For 100 lb to impend downward, the ratios are reversed, so
100 lb = Pe0.45π ,
Pmin = 24.324 lb
So, for equilibrium,
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.
24.3 lb ≤ P ≤ 411 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 106.
Horizontal pipe
Vertical pipe
π
Contact angles β H =
Contact angle βV = π
2
µ sH = 0.30
µ sV = ?
For Pmin , the 100 lb force impends downward, and
µ
100 lb =  e

sH
π
2
R =  e µ




sH
π
2
(
)
(
)
 e µ π Q =  e µ π π2  e µ π  e µ








sV
s
sV
sH
π
2
P


π 0.30 + µ sV ) 
100 lb = e (
( 20 lb ) , so eπ ( 0.30 + µsV ) = 5


(a) For Pmax the force P impends downward, and the ratios are reversed,
so
Pmax = 5 (100 lb ) = 500 lb (b) π ( 0.30 + µ sV ) = ln 5
µ sV =
1
π
ln 5 − 0.30 = 0.21230
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.
µ sV = 0.212 COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 107.
FBD motor and mount:
Impending belt slip: cw rotation
T2 = T1e µs β = T1e0.40π = 3.5136 T1
ΣM D = 0:
(12 in.)(175 lb ) − ( 7 in.) T2 − (13 in.) T1 = 0
2100 lb = ( 7 in.)( 3.5136 ) + 13 in. T1
T1 = 55.858 lb,
T2 = 3.5136 T1 = 196.263 lb
FBD drum at B:
ΣM B = 0:
M B − ( 3 in.)(196.263 lb − 55.858 lb ) = 0
M B = 421 lb ⋅ in. r = 3 in.
(Compare to 857 lb ⋅ in. using V-belt, Problem 8.130)
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 8, Solution 108.
FBD motor and mount:
Impending belt slip: ccw rotation
T1 = T2e µs β = T2e0.40π = 3.5136 T2
ΣM D = 0:
(12 in.)(175 lb ) − (13 in.) T1 − ( 7 in.) T2 = 0
2100 lb = (13 in.)( 3.5136 ) + 7 in. T2 = 0
T2 = 39.866 lb,
T1 = 3.5136 T2 = 140.072 lb
FBD drum at B:
ΣM B = 0:
( 3 in.)(140.072 lb − 39.866 lb ) − M B = 0
M B = 301 lb ⋅ in. 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 8, Solution 109.
FBD lower portion of belt:
ΣFy = 0:
48 N − N D = 0,
N D = 48 N
Slip on both platen and wood
FD = µkD N D = 0.10 ( 48 N ) = 4.8 N
FE = µkE N E = ( 48 N ) µ kE
FBD Drum A (assume free to rotate)
ΣFx = 0:
TA − TB − 4.8 N − µ kE ( 48 N ) = 0
TB = TA + 4.8 N + µ kE ( 48 N )
(1)
ΣM A = 0:
rA (TA − TT ) = 0,
(2)
ΣM B = 0:
M B + r (TT − TB ) = 0
TT = TA
FBD Drive drum B
TB = TT +
2.4 N ⋅ m
= TT + 96 N
0.025 N
Impending slip on drum, TB = TT e µs β = TT e0.35π
so TT + 96 N = TT e0.35π , TT = 47.932 N
TB = 143.932 N
From (2) above, TA = TT , so
(a)
Tmin lower = 47.9 N From (1) above, 143.932 N = 47.932 N + 4.8 N + µkE ( 48 N )
So
(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.
µ kE = 1.900 COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 110.
FBD Flywheel:
( 0.225 m )(TB − TA ) − 12.60 N ⋅ m = 0
ΣM C = 0:
TB − TA = 56 N,
TB = TA + 56 N
Also, since the belt doesn’t change length, the additional stretch in
spring B equals the decrease in stretch of spring A. Thus the increase
in TB equals the decrease in TA.
Thus TB + TA = ( 70 N + ∆T ) + ( 70 N − ∆T ) = 140 N
(TA + 56 N ) + TA = 140 N,
TA = 42 N
TB = 42 N + 56 N = 98 N
(a)
TA = 42.0 N TB = 98.0 N For slip TB = TAe µk β , or µ k =
µk =
1
π
ln
1
β
ln
TB
TA
98
= 0.2697
42
(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.
µ k = 0.230 COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 111.
FBD Flywheel:
Slip of belt: TB = TA e µk β = TA e0.20π
Also, since the belt doesn’t change length, the increase in stretch of
spring B equals the decrease in stretch of spring A. Therefore the
increase in TB equals the decrease in TA , and the sum is unchanged,
so TA + TB = 80 N + 80 N = 160 N
(
)
∴ TA 1 + e0.20π = 160 N, so TA = 55.663 N
TB = 104.337 N
(a)
TA = 55.7 N TB = 104.3 N ΣM C = 0:
( 0.225 m )(TB − TA ) − M C = 0
M C = ( 0.225 m )(104.337 N − 55.663 N )
(b) M C = 10.95 N ⋅ m 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 8, Solution 112.
FBD Lever:
ΣM E = 0:
( 60 mm )( 240 N ) − ( 40 mm ) FBD cos 30° = 0
FBD = 415.69 N
FBD Drum:
Belt slip:
T2 = T1 e µk β
= ( 415.69 N ) e
0.25( 5.5851)
= 1679.44 N
ΣM C = 0:
r (T2 − T1 ) − M = 0
( 0.08 m )(1679.44 N − 415.69 N ) − M = 0
M = 101.1 N ⋅ m 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 8, Solution 113.
FBD Drum:
(a) With M E = 125 lb ⋅ ft
ΣM E = 0:
( 7 in.) (TA − TC ) − (125 lb ⋅ ft ) = 0
TA − TC = 214.29 lb
Belt slip: TA = TC e µk β = TC e
( ) = 3.0028T
0.30 76π
so 2.0028 TC = 214.9 lb,
C
TC = 106.995 lb
TA = 321.28 lb
FBD Lever:
ΣM B = 0:
P=
(15 in.) P + ( 2 in.) TC − ( 7.5 in.) TA = 0
(1)
( 7.5 in.)( 321.28 lb ) − ( 2 in.)(106.995 lb )
17 in.
P = 129.2 lb (b) With M E = 125 lb ⋅ ft , the drum analysis will be reversed, and will yield TA = 106.995 lb,
TC = 321.28 lb
Eqn. (1) will remain the same, so
P=
( 7.5 in.)(106.995 lb ) − ( 2 in.)( 321.28 lb )
17 in.
P = 9.41 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 114.
FBD Lever:
If brake is self-locking, no force P is required
( 2 in.) TC − ( 7.5 in.) TA = 0
ΣM B = 0:
TC = 3.75 TA
For impending slip on drum: TC = TA e µs β
∴ e µs β = 3.75, or µ s =
With β =
1
β
ln 3.75
7π
,
6
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.
µ s = 0.361 COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 115.
FBD Lever:
ΣM B = 0:
( 40 mm ) TC − (100 mm ) TA = 0,
TC = 2.5 TA
FBD Drum:
(a) For impending slip ccw: TC = Tmax = 4.5 kN
so
TA =
ΣM D = 0:
TC
= 1.8 kN
2.5
M D + ( 0.16 m )(1.8 kN − 4.5 kN ) = 0
M D = 0.432 kN ⋅ m
M D = 432 N ⋅ m (b) For impending slip ccw, TC = TA e µs β
or µ s =
1
T
3
ln C =
ln 2.5 = 0.21875
β TA 4π
µ s = 0.219 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 8, Solution 116.
(a) For minimum mC with blocks at rest, impending slip of A is down/left.
Note: φs = tan −1 µ s = tan −1 0.30 = 16.7° < 30°, so mC min > 0
FBD A:
(
)
WA = ( 6 kg ) 9.81 m/s 2 = 58.86 N
ΣFy = 0:
N A − WA cos 30° = 0,
N A = WA cos 30°
Impending slip: FA = µ s N A = 0.30WA cos 30°
ΣFx = 0:
TA + FA − WA sin 30° = 0, TA = WA ( sin 30° − 0.30 cos 30° )
= 14.1377 N
FBD Drum:
If blocks don’t move, belt slips on drum, so
0.2 0.87266 )
14.1377 N = TA = TC e µk β = TC e (
= 1.19069 TC
so
TC = 11.8735 N
FBD C:
ΣFy′ = 0:
NC − WC cos 20° = 0,
NC = WC cos 20°
Impending slip: FC = µ s NC = 0.30WC cos 20°
ΣFx′ = 0:
0.30WC cos 20° + WC sin 20° − 11.8735 N = 0
WC = 19.0302 N,
mC =
WC
= 1.93988 kg
9.81 m/s 2
mC = 1.940 kg continued
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
(b) For motion of A to impend up/right
FBD A:
As in part (a) N A = WA cos 30°, FA = 0.30WA cos 30°
ΣFx = 0:
TA − WA ( sin 30° + 0.30cos 30° ) = 0
TA = 44.722 N
Also, as in part (a) TA = TC e µk β = 1.19069 TC , so TC =
44.722 N
1.19069
TC = 37.560 N
FBD C:
As in part (a) FC = 0.30WC cos 20°
ΣFx′ = 0:
WC ( sin 20° − 0.30cos 20° ) − 37.560 N = 0
WC = 624.83 N,
mC =
WC
= 63.69 kg
9.81 m/s 2
mC = 63.7 kg (c) For uniform motion of A up and B down, and minimum mC , there will be impending slip of the rope on
the drum.
FBD A is same as in (b) but FA = µk N A = 0.20WA cos30°
and
ΣFx = 0:
TA − WA ( sin 30° + 0.20cos 30° ) = 0,
TA = 39.625 N
Drum analysis, with impending slip, TA = TC e µs β
39.625 N = TC e
0.30( 0.87266 )
= 1.29926 TC
or TC = 30.498 N
FBD C is same as in (b), but FC = µ k NC = 0.20WC cos 20°
and
ΣFx′ = 0:
WC ( sin 20° − 0.20 cos 20° ) − TC = 0
WC =
30.498 N
= 197.933 N,
0.154082
mC =
197.934 N
9.81 m/s 2
mC = 20.2 kg 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 8, Solution 117.
Geometry and force rotation:
Let
EBC = α = cos −1
50 mm
= 60° =
100 mm
s DBE , FAE , GAE
Then contact angles are β B = 360° − 120° = 240° =
upper cylinder, and β A = 30° =
π
6
4π
rad for cord on
3
rad for each cord
contact on lower cylinder.
Let the force in section FC = TF
Let the force in section DG = TG
With A fixed and the cord moving,
TG = We µk β A = We
( ) = 1.13985W
0.25 π6
For maximum W, slip impends on drum B, so
TB = TF e µs β B or TF = TG e− µs β B
TF = 1.13985We
( ) = 0.32441W
−0.30 43π
For slip at F
W1 = TF e µk β A = 0.32441We
( ) = 0.36978W
0.25 π6
so W = 2.7043W1 and m = 2.7043 in.
= 2.7043 ( 75 kg )
m = 203 kg 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 8, Solution 118.
Geometry and force notation:
Note: θ = sin −1
βC = β D =
π
2
π
r
= 30° = rad, so contact angles are:
2r
6
π
+
=
6
2π
,
3
βE = π
(a) For all pulleys locked, slip impends at all contacts
If WA impends downward, T1 = (16 lb ) e µs β E , T2 = T1e µs β D , WA = T2e µ s βC
so WA = (16 lb ) e s ( C
µ β + βD + βE )
= (16 lb ) e
( ) = 69.315 lb
0.20 73π
If WA impends upward all ratios are inverted, so WA = (16 lb ) e
( )
−0.20 73π
= 3.6933 lb
3.69 lb ≤ WA ≤ 69.3 lb W
For equilibrium,
(b) If pulley D is free to rotate, T1 = T2 while the other ratios remain as in (a)
For WA impending down WA = (16 lb ) e s ( C
µ β + βE )
= (16 lb ) e
( )
0.20 53π
WA = 45.594 lb
For WA impending upward, WA = (16 lb ) e
( ) = 5.6147 lb
−0.2 53π
For equilibrium
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.
5.61 lb ≤ WA ≤ 45.6 lb W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 119.
Geometry and force notation:
θ = sin −1
r
π
= 30° = , so contact angles are:
2r
6
βC = β D =
π
2
+
π
6
=
2π
,
3
βE = π
(a) D and E fixed, so slip on these surfaces. For maximum N A , slip impends on pulley C
WA = T2e µs βC , and T1 = T2e µk β D ,
(16 lb ) = T1e µk β E
WA = (16 lb ) e
so
− µk ( β E + β D ) µ s βC
e
= (16 lb ) e
( )e0.20( 23π ) = 11.09 lb −0.15 53π
(b) C and D fixed, so slip there. For maximum WA , slip impends on E
so T1 = (16 lb ) e µs β E , T1 = T2e µk β D , T2 = WAe µk βC
so WA = (16 lb ) e µs β E e
− µk ( βC + β D )
= (16 lb ) e0.20π e
( ) = 16 lb
−0.15 43π
WA = 16.00 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 120.
Geometry and force notation:
θ = sin −1
5 in.
π
= 30° = rad, so contact angles are:
10 in.
6
βC = π −
π
6
=
5π
π π
2π
=
, βD = +
,
6
2 6
3
βE =
π
2
(a) All pulleys locked with impending slip at all.
If WA impends upward, T1 = WAe µs βC ,
T2 = T1e µs β D ,
WA = (16 lb ) e
(16 lb ) = T2e µs β E , so
− µ s ( βC + β D + β E )
= (16 lb ) e
(
)
−0.20 56 + 46 + 63 π
WA = 4.5538 lb
If WA impends downward all ratios are inverted
so WA = (16 lb ) e
+0.20( 2π )
= 56.217 lb
4.55 lb ≤ WA ≤ 56.2 lb For equilibrium,
(b) Pulley D is free to rotate so T1 = T2 , other ratios are the same
If WA impends upward, WA = (16 lb ) e
− µ s ( βC + β E )
= (16 lb ) e
( )
−0.20 43π
WA = 6.9229 lb
If WA impends downward, ratios are inverted, WA = (16 lb ) e
( )
+0.20 43π
WA = 36.979 lb
For equilibrium
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.
6.92 lb ≤ WA ≤ 37.0 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 121.
Geometry and force notation:
θ = sin −1
βC = π −
5 in.
π
= 30° = rad, so contact angles are:
10 in.
6
π
6
=
5π
π π
2π
=
, βD = +
,
6
2 6
3
βE =
π
2
(a) D and E fixed, so slip at these surfaces,
For maximum WA , slip impends on C.
WA = T1e µs βC , T2 = T1e µk β D , 16 lb = T2e µk β E
so WA = (16 lb ) e
= (16 lb ) e
− µ k ( β D + β E ) µ s βC
e
( ) e0.20( 56π ) = 15.5866 lb
−0.15 76π
WA max = 15.59 lb (b) C and D fixed, so slip at these surfaces—impending slip on E
T1 = WAe µk βC , T2 = T1e µk β D ,
so WA = (16 lb ) e
− µk ( βC + β D ) µ s β E
e
T2 = (16 lb ) e µs β E
= (16 lb ) e
( ) e0.20( π2 )
−0.15 32π
WA = 10.8037 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.
WA max = 10.80 lb COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 122.
FBD drum B:
ΣM B = 0:
( 0.02 m )(TA − T ) − 0.30 N ⋅ m = 0
TA − T =
0.30 N ⋅ m
= 15 N
0.02 m
Impending slip: TA = Te µs β = Te0.40π
(
)
Solving; T e0.40π − 1 = 15 N
T = 5.9676 N
If C is free to rotate P = T
Pmin = 5.97 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 8, Solution 123.
FBD drum B:
ΣM B = 0:
( 0.02 m )(TA − T ) − 0.3 N ⋅ m = 0
TA − T = 15 N
Impending slip:
(
TA = Te µs β B = Te0.40π
)
Solving, T e0.40π − 1 = 15 N
T = 5.9676 N
If C is frozen, tape must slip there, so
P = Te µk βC = ( 5.9676 N ) e
( ) = 9.5599 N
0.30 π2
Pmin = 9.56 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 8, Solution 124.
FBD pin B:
T1 = T2
(a) By symmetry:
 2 
ΣFy = 0: B − 2 
T =0
 2 1 


Drum:
or
B=
2T1 =
2T2
(1)
For impending rotation :
T3 > T1 = T2 > T4 , so T3 = Tmax = 5.6 kN
(
−0.25 π4 + π6
Then
T1 = T3e− µs β L = ( 5.6 kN ) e
or
T1 = 4.03706 kN = T2
and
T4 = T2e− µs β R = ( 4.03706 kN ) e
or
T4 = 2.23998 kN
)
( )
−0.25 34π
ΣM F = 0: M 0 + r (T4 − T3 + T2 − T1 ) = 0
or
M 0 = ( 0.16 m )( 5.6 kN − 2.23998 kN ) = 0.5376 kN ⋅ m
Lever:
M 0 = 538 N ⋅ m
(b) Using Equation (1)
B=
2T1 =
2 ( 4.03706 kN )
= 5.70927 kN
ΣM D = 0:
( 0.05 m )( 5.70927 kN ) − ( 0.25 m ) P = 0
P = 1.142 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 8, Solution 125.
FBD pin B:
T1 = T2
(a) By symmetry:
 2 
ΣFy = 0: B − 2 
T1  = 0
 2 
FBD Drum:
or
B=
(1)
2T1
For impending rotation :
T4 > T2 = T1 > T3 , so T4 = Tmax = 5.6 kN
( )
−0.25 34π
Then
T2 = T4e− µ s β R = ( 5.6 kN ) e
or
T2 = 3.10719 kN = T1
and
T3 = T1e− µs β L = ( 3.10719 kN ) e
or
T3 = 2.23999 kN
(
−0.25 π4 + π6
)
ΣM F = 0: M 0 + r (T2 − T1 + T3 − T4 ) = 0
M 0 = (160 mm )( 5.6 kN − 2.23999 kN ) = 537.6 N ⋅ m
FBD Lever:
M 0 = 538 N ⋅ m
(b) Using Equation (1)
B=
2T1 =
2 ( 3.10719 kN )
B = 4.3942 kN
ΣM D = 0:
( 0.05 m )( 4.3942 kN ) − ( 0.25 m ) P = 0
P = 879 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 8, Solution 126.
FBD wrench:
Note: EC =
( 0.2 m ) , EA = EC − 0.03 m
sin 65°
θ = 65°
so β = 295° = 5.1487 rad
ΣM E = 0:
 0.20 m

 0.20 m

− 0.03 m  F − 
cos 65° − 0.03 m  T = 0

 sin 65°

 sin 65°

T = 3.01408F
ΣFx = 0:
N sin 65° + F cos 65° − T = 0
Impending slip: N =
 sin 65°

, so F = 
+ cos 65°  = T
µs
 µs

F
or
sin 65°
µs
+ cos 65° = 3.01408
µ s = 0.3497
Must still check slip of belt on pipe
FBD small portion of belt at A:
ΣFn = 0:
N1 − N 2 = 0
Impending slip, both sides: F1 = µ s N1,
so
F2 = µ s N 2
F1 = F2 = F
ΣFt = 0:
2 F − TA = 0, TA = 2F
Impending slip of belt on pipe: T = TAe µs β
or µ s =
1
β
ln
T
1
3.01408
=
= 0.0797
ln
2F
5.1487
2
Above controls, so for self-locking, need
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.
µ s = 0.350
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 127.
FBD wrench
Note: EC =
( 0.20 m ) , EA = EC − 0.03 m
sin 75°
θ = 75°
so β = 285° = 4.9742 rad
ΣM E = 0:
 0.20 m

 0.20 m

− 0.03 m  F − 
cos 75° − 0.03 m  T = 0

 sin 75°

 sin 75°

T = 7.5056 F
ΣFx = 0:
N sin 75° + F cos 75° − T = 0
Impending slip: N =
 sin 75°

, so F = 
+ cos 75°  = T = 7.5056 F
µs
 µs

F
sin 75°
µs
+ cos 75° = 7.5056
µ s = 0.1333
Must still check impending slip of belt on pipe
FBD small portion of belt at A
ΣFn = 0:
N1 − N 2 = 0
Impending slip F1 = µ s N1,
F2 = µ s N 2
so
F1 = F2 = F
ΣFt = 0:
2 F − TA = 0, TA = 2 F
Impending slip of belt on pipe T = TAe µ s β
or µ s =
1
β
ln
T
1
7.5056
=
= 0.2659
ln
2F
4.9742
2
This controls, so for self locking,
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.
µ s min = 0.267
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 128.
ΣFn = 0:
∆N − T + (T + ∆T )  sin
∆N = ( 2T + ∆T ) sin
or
ΣFt = 0:
∆F = ∆T cos
Impending slipping:
∆F = µ s ∆N
In limit as
∆T cos
∆θ
2
∆θ
(T + ∆T ) − T  cos
− ∆F = 0
2
or
So
∆θ
=0
2
∆θ
2
∆θ
∆θ
sin ∆θ
= µ s 2T sin
+ µ s ∆T
2
2
2
∆θ → 0: dT = µ sTdθ ,
or
dT
= µ s dθ
T
dT
So
T2
β
∫ T1 T = ∫ 0 µ s dθ ;
and
T
ln 2 = µ s β
T1
or T2 = T1e µs β Note: Nothing above depends on the shape of the surface, except it is
assumed to be a smooth curve.
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 8, Solution 129.
Small belt section:
Side view:
∆N
α
∆θ
sin − T + (T + ∆T )  sin
=0
2
2
2
ΣFy = 0:
2
ΣFx = 0:
∆θ
(T + ∆T ) − T  cos
− ∆F = 0
2
∆F = µ s ∆N ⇒ ∆T cos
Impending slipping:
In limit as ∆θ → 0:
End view:
dT =
µ sTdθ
α
sin
So
or
or
dT
µs
=
dθ
α
T
sin
2
or
2
dT
T2
∫ T1 T =
∆θ
2T + ∆T
∆θ
= µs
sin
α
2
2
sin
2
µs β
dθ
α ∫0
sin
2
T
µβ
ln 2 = s
α
T1
sin
2
µ β /sin α2
T2 = T1e s
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 8, Solution 130.
FBD motor and mount:
Impending belt slip, cw rotation
µs β
T2 = T1e
sin α2
( 0.40π )
T2 = T1e sin18° = 58.356T1
ΣM D = 0:
(12 in.)(175 lb ) − (13 in.) T1 − ( 7 in.) T2 = 0
2100 lb = 13 in. + ( 7 in.)( 58.356 )  T1
T1 = 4.9823 lb,
T2 = 58.356T1 = 290.75 lb
FBD drum at B:
ΣM B = 0:
M B + ( 3 in.)( 4.9823 lb − 290.75 lb ) = 0
M B = 857 lb ⋅ in. (Compare to 421 lb ⋅ in. using flat belt, Problem 8.107)
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 8, Solution 131.
Geometry:
θ = sin −1
2 in.
= 7.1808° = 0.12533 rad
16 in.
β A = π − 2θ = 2.8909 rad
Since β B > β A , impending slip on A will control the
maximum couple transmitted
FBD A:
ΣM A = 0:
60 lb ⋅ in. + ( 2 in.)(T1 − T2 ) = 0
T2 − T1 = 30 lb
µs β
Impending slip: T2 = T1e
sin α2
 ( 0.35)( 2.8909 )

so T1  e sin18°
− 1 = 30 lb




T1 = 1.17995 lb
T2 = 31.180 lb
FBD B:
ΣFx = 0:
P − ( 31.180 lb + 1.17995 lb ) cos 7.1808° = 0
P = 32.1 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 132.
FBD block:
ΣFn = 0:
N − (1000 N ) cos 30° − ( 200 N ) sin 30° = 0
N = 966.03 N
Assume equilibrium:
ΣFt = 0:
F + ( 200 N ) cos 30° − (1000 N ) sin 30° = 0
F = 326.8 N = Feq.
But
Fmax = µ s N = ( 0.3) 966 N = 290 N
Feq. > Fmax
and
impossible ⇒ Block moves F = µk N
= ( 0.2 )( 966.03 N )
Block slides down
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.
F = 193.2 N
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 133.
FBD block (impending
motion to the right)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
P
W
=
sin φs
sin (θ − φ s )
sin (θ − φs ) =
(a)
W
sin φs
P
W = mg
(
)
 ( 30 kg ) 9.81 m/s 2

m = 30 kg: θ − φ s = sin −1 
sin14.036° 


120 N
= 36.499°
∴ θ = 36.499° + 14.036°
(b)
m = 40 kg: θ − φs = sin
or θ = 50.5°
(
 ( 40 kg ) 9.81 m/s 2
−1 

120 N
) sin14.036°

= 52.474°
∴ θ = 52.474° + 14.036°
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.
or θ = 66.5°
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Chapter 8, Solution 134.
FBDs
Top block:
(a) Note: With the cable, motion must impend at both contact surfaces.
ΣFy = 0:
N1 − 40 lb = 0
F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb
Impending slip:
Bottom block:
ΣFx = 0:
T − F1 = 0
ΣFy = 0:
N 2 − 40 lb − 60 lb = 0
Impending slip:
T − 16 lb = 0
T = 16 lb
N 2 = 100 lb
F2 = µ s N 2 = 0.4 (100 lb ) = 40 lb
ΣFx = 0:
FBD blocks:
N1 = 40 lb
− P + 16 lb + 16 lb + 40 lb = 0
P = 72.0 lb
(b) Without the cable, both blocks will stay together and motion will
impend only at the floor.
ΣFy = 0:
Impending slip:
ΣFx = 0:
N − 40 lb − 60 lb = 0
N = 100 lb
F = µ s N = 0.4 (100 lb ) = 40 lb
40 lb − P = 0
P = 40.0 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 135.
FBD ladder:
Motion impends at both A and B, so
FA = µ s N A
ΣM A = 0:
NB =
or
a
W =0
2
lN B −
ΣFx = 0:
FA +
ΣFy = 0:
2.5W
13
5
12
FB −
NB = 0
13
13
NA −
l = 19.5 ft
a
7.5 ft
W =
W
2l
39 ft
2.5
W
13
µs N A +
a = 7.5 ft
NB =
or
FB = µ s N B = µ s
Then
a
5
=
l
13
FB = µ s N B
and
12.5
(13)
µ sW −
30
(13)2
W =0
( 30 − 12.5µs )
W
(13)
NA − W +
2
2
µs
12
5
FB +
NB = 0
13
13
 30 − 12.5µ s
 W
+ 30µ s + 12.5 
=W

2
µs

 (13)
b 12
=
l
13
or
µ s2 − 5.6333µ s + 1 = 0
µ s = 2.8167 ± 2.6332
or
µ s = 0.1835
and
µ s = 5.45
The larger value is very unlikely unless the surface is treated with
some “non-skid” material.
In any event, the smallest value for equilibrium is µ s = 0.1835
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 8, Solution 136.
FBD window:
(
)
W
2
N A − ND = 0
N A = ND
T = ( 2 kg ) 9.81 m/s 2 = 19.62 N =
ΣFx = 0:
FA = µ s N A
Impending motion:
ΣM D = 0:
(
W = ( 4 kg ) 9.81 m/s
2
) = 39.24 N
ΣFy = 0:
FD = µ s N D
( 0.36 m )W − ( 0.54 m ) N A − ( 0.72 m ) FA = 0
W =
3
N A + 2µ s N A
2
NA =
2W
3 + 4µ s
FA − W + T + FD = 0
FA + FD = W − T
=
W
2
Now
FA + FD = µ s ( N A + N D ) = 2µ s N A
Then
W
2W
= 2µ s
2
3 + 4µ s
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.
µ s = 0.750 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 137.
FBD Collar:
Stretch of spring x = AB − a =
a
−a
cosθ
 a

 1

− a  = (1.5 kN/m )( 0.5 m ) 
− 1
Fs = k 
θ
θ
cos
cos




 1

= ( 0.75 kN ) 
− 1 = ( 750 N )( sec θ − 1)
θ
cos


Fs cosθ − W + N = 0
ΣFy = 0:
W = N + ( 750 N ) (1 − cosθ )
or
Impending slip:
F = µ s N (F must be +, but N may be positive or negative)
Fs sin θ − F = 0
ΣFx = 0:
or
F = Fs sin θ = ( 750 N )( tan θ − sin θ )
(a) θ = 20°:
F = ( 750 N )( tan 20° − sin 20° ) = 16.4626 N
Impending motion:
(Note: for
N =
F
µs
16.4626 N
= 41.156 N
0.4
=
N < 41.156 N, motion will occur, equilibrium for
N > 41.156)
W = N + ( 750 N )(1 − cos 20° ) = N + 45.231 N
But
So equilibrium for W ≤ 4.07 N and W ≥ 86.4 N W
(b) θ = 30°:
F = ( 750 N )( tan 30° − sin 30° ) = 58.013 N
Impending motion:
N =
F
µs
=
58.013
= 145.032 N
0.4
W = N + ( 750 N )(1 − cos 30° ) = N ± 145.03 N
= −44.55 N ( impossible ) , 245.51 N
Equilibrium for W ≥ 246 N W
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 8, Solution 138.
FBD pin C:
FAB = P sin10° = 0.173648P
FBC = P cos10° = 0.98481P
ΣFy = 0:
FBD block A:
N A − W − FAB sin 30° = 0
N A = W + 0.173648P sin 30° = W + 0.086824P
or
ΣFx = 0:
FA − FAB cos 30° = 0
FA = 0.173648P cos30° = 0.150384P
or
FA = µ s N A
For impending motion at A:
NA =
Then
FA
µs
: W + 0.086824 P =
0.150384
P
0.3
P = 2.413W
or
ΣFy = 0:
N B − W − FBC cos 30° = 0
N B = W + 0.98481P cos30° = W + 0.85287 P
FBD block B:
ΣFx = 0:
FBC sin 30° − FB = 0
FB = 0.98481P sin 30° = 0.4924 P
FB = µ s N B
For impending motion at B:
Then
or
NB =
FB
µs
: W + 0.85287 P =
0.4924P
0.3
P = 1.268W
Thus, maximum P for equilibrium
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.
Pmax = 1.268W W
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Chapter 8, Solution 139.
φs = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
750 lb
=
sin104.036° sin16.928°
R2 = 2499.0 lb
FBD wedge B:
P
2499.0
=
sin 73.072° sin 75.964°
P = 2464 lb
P = 2.46 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 8, Solution 140.
Block on incline:
θ = tan −1
0.1 in.
= 3.0368°
2π ( 0.3 in.)
φs = tan −1 µ s = tan −1 0.12 = 6.8428°
Q = ( 500 lb ) tan 9.8796° = 87.08 lb
Couple on each side
M = rQ = ( 0.3 in.)( 87.08 lb ) = 26.12 lb ⋅ in.
Couple to turn = 2M = 52.2 lb ⋅ in. 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 8, Solution 141.
FBD pulley:
ΣFy = 0:
R − 103.005 N − 49.05 N − 98.1 N = 0
R = 250.155 N
ΣM O = 0:
( 0.12 m )(103.005 N − 98.1 N ) − rf ( 250.155 N ) = 0
rf = 0.0023529 m = 2.3529 mm
r
φ s = sin −1 f
rs

r 


rs 

µ s = tan φs = tan  sin −1 f  = tan  sin −1
2.3529 mm 

30 mm 
µ s = 0.0787 W
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 8, Solution 142.
FBD wheel:
ΣM E = 0:
M E = ( 7.5 in.)(T2 − T1 )
or
FBD lever:
ΣM C = 0:
Impending slipping:
So
( 4 in.)(T1 + T2 ) − (16 in.)( 25 lb ) = 0
T1 + T2 = 100 lb
or
or
− M E + ( 7.5 in.)(T2 − T1 ) = 0
T2 = T1e µs β
T2 = T1e
( ) = 3.2482T
0.25 32π
1
T1 (1 + 3.2482 ) = 100 lb
T1 = 23.539 lb
and
M E = ( 7.5 in.)( 3.2482 − 1)( 23.539 lb ) = 396.9 lb ⋅ in.
M E = 397 lb ⋅ in. W
Changing the direction of rotation will change the direction of M E and
will switch the magnitudes of T1 and T2 .
The magnitude of the couple applied will not change. W
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 8, Solution 143.
FBD block:
ΣFn = 0:
NC − ( 200 lb ) cos 30° = 0; N = 100 3 lb
ΣFt = 0:
TC − ( 200 lb ) sin 30° ∓ FC = 0
TC = 100 lb ± FC
FBD Drum:
(1)
where the upper signs apply when FC acts
(a) For impending motion of block , FC
, and
(
)
FC = µ s NC = 0.35 100 3 lb = 35 3 lb
(
)
TC = 100 − 35 3 lb
So, from Equation (1):
TC = WAe µk β
But belt slips on drum, so
−0.25( 3 )
WA =  100 − 35 3 lb  e


(
)
2π
WA = 23.3 lb W
and FC = µ s NC = 35 3 lb
(b) For impending motion of block , FC
(
)
From Equation (1):
TC = 100 + 35 3 lb
Belt still slips, so
−0.25( 3 )
WA = TC e− µk β =  100 + 35 3 lb  e


(
)
2π
WA = 95.1 lb W
continued
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
PROBLEM 8.143 CONTINUED
(c) For steady motion of block , FC
, and FC = µk NC = 25 3 lb
(
)
T = 100 + 25 3 lb.
Then, from Equation (1):
Also, belt is not slipping on drum, so
−0.35( 3 )
WA = TC e− µ s β =  100 + 25 3 lb  e


(
)
2π
WA = 68.8 lb W
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 9, Solution 1.
b2 − b1
x + b1
a
First note:
y=
Have
I y = ∫ x 2dA
b2 − b1
x + b1
a
a
x 2d ydx
0 0
=∫ ∫
a
 b − b1

= ∫ 0 x2  2
x + b1  dx
 a

a
 1 b − b1 4 1 3 
= 2
x + b1x 
3
4 a
0
=
1 3
a ( b1 + 3b2 )
12
Iy =
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.
1 3
a ( b1 + 3b2 ) 12
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 2.
At
5
x = a, y = b :
∴ y =
b
a
5
2
5
b = ka 2
x2
or
dI y =
1 3
x dy
3
=
Then
Iy =
x=
or
a
b
b
5
a2
2
2
5
y5
1 a3 65
y dy
3 b 65
1 a3 b 65
∫ y dy
3 b 65 0
1 5 a3 115
=
y
3 11 b 65
=
k =
b
0
5 a3 115
b
33 b 65
or I y =
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.
5 3
a b
33
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 3.
At x = 0:
First note:
At x = a :
a = k ( 2b − b )
k =
∴ x=
Have
2
c = −b
or
or
0 = k (b + c)
2
a
b2
a
2
y − b)
2(
b
I y = ∫ x 2dA
=∫
a
2
y − b)
2b b 2 (
x 2dxdy
0
0
∫
3
=
1 2b  a
2
y − b )  dy
∫
0  2(
3
b

2b
1 a3 1
7
=
× ( y − b)
6
3b
7
b
=
1 3
ab
21
Iy =
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.
1 3
a b
21
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 4.
y = kx 2 + c
Have
At
x = 0, y = b :
or
c=b
b = k ( 0) + c
At
x = 2a, y = 0:
or
k =−
y =−
Then
=
Then
I y = ∫ x 2dA,
2a
I y = ∫ a x 2dA =
2
0 = k ( 2a ) + b
b
4a 2
b 2
x +b
4a 2
b
4a 2 − x 2
4a 2
(
dA = ydx =
)
(
)
b
4a 2 − x 2 dx
4a 2
(
)
b 2a 2
2
2
∫ x 4a − x dx
4a 2 a
2a
b  2 x3 x5 
=
− 
 4a
3
5 a
4a 2 
=
b
b
8a3 − a3 −
32a5 − a5
2
3
20a
=
7a3b 31a3b
−
3
20
(
)
(
)
Iy =
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.
47 3
a b
60
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 5.
b2 − b1
x + b1
a
First note:
y =
Have
I x = ∫ y 2dA
=∫
b2 − b1
x + b1
a
a
y 2d ydx
0 0
∫
3
1 a  b − b1

= ∫0  2
x + b1  dx
3  a

4
1  1 a   b2 − b1

= ×
x + b1 

3  4 b2 − b1   a

a
0
=
1
a
b24 − b14
12 b2 − b1
=
1
a
( b2 + b1 )( b2 − b1 ) b22 + b12
12 b2 − b1
=
1
a ( b1 + b2 ) b12 + b22
12
(
)
(
(
)
)
Ix =
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.
1
a ( b1 + b2 ) b12 + b22 12
(
)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 6.
SOLUTION
5
At x = a, y = b : b = ka 2
or k =
∴y =
b
5
a2
b
a
5
2
I x = ∫ y 2dA
5
x2
dA = xdy
a 2 
b
= ∫ 0 y 2  2 y 5 dy 
 b 5

a
5 175
= 2 ×
y
b 5 17
b
0
17
5a b 5
=
17 b 52
or I x =
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.
5 3
ab 17
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 7.
At x = 0: 0 = k ( b + c )
First note:
or
2
c = −b
At x = a : a = k ( 2b − b )
Have
or
k =
a
b2
∴
x=
a
2
y − b)
2(
b
2
I x = ∫ y 2dA
=∫
a
2
y − b)
2b b 2 (
y 2dxdy
0
b
∫
=
a 2b 2
2
y ( y − b ) dy
2 ∫b
b
=
a 2b 4
y − 2by 3 + b 2 y 2 dy
2 ∫b
b
=
a 1 5 1 4 1 2 3
 y − by + b y 
2
3
b2  5
b
=
a  1
1
1 2
1 2 3 
5
4
3
1 5 1
4
 ( 2b ) − ( 2b ) + b ( 2b )  −  b − b b + b b  
2
3
5
2
3
b 2   5
 

(
)
2b
( )
( )
8 1 1 1
 32
= ab3 
−8+ − + − 
3 5 2 3
 5
=
31 3
ab
30
Ix =
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.
31 3
ab 30
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 8.
Have
y = kx 2 + c
At
x = 0, y = b: b = k (0) + c
or
c=b
At
x = 2a, y = 0: 0 = k (2a) 2 + b
or
k =−
Then
y =
Now
dI x =
=
b
4a 2
b
4a 2 − x 2
4a 2
(
)
1 3
y dx
3
3
1 b3
4a 2 − x 2 dx
6
3 64a
(
)
continued
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
Then
I x = ∫ dI x
=
3
1 b3 2 a
4a 2 − x 2 dx
6 ∫a
3 64a
=
b3
2a
64a 6 − 48a 4 x 2 + 12a 2 x 4 − x 6 dx
6 ∫a
192a
(
)
(
)
2a
b3 
12 2 5 x 7 
6
4 3
64
a
x
16
a
x
a x −
=
−
+


5
7 a
192a 6 
=
b3 
64a 7( 2 − 1) − 16a 7 ( 8 − 1)
192a 6 
+
=
12 7
1

a ( 32 − 1) − (128 − 1) 
5
7

ab3 
372 127 
3
−
 64 − 112 +
 = 0.043006ab
192 
5
7 
I x = 0.0430ab3 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 9, Solution 9.
x2
y2
+
=1
a 2 b2
x = a 1−
y2
b2
dA = xdy
dI x = y 2dA = y 2 xdy
b
b
I x = ∫ dI x = ∫ − b xy 2dy = a ∫ − b y 2 1 −
Set: y = b sin θ
y2
dy
b2
dy = b cosθ dθ
π
I x = a ∫ 2π b 2 sin 2 θ 1 − sin 2 θ b cosθ dθ
−
2
π
π
−
−
= ab3 ∫ 2π sin 2 θ cos 2 θ dθ = ab3 ∫ 2π
2
2
1 2
sin 2θ dθ
4
π
π
1
1
1
1

2
= ab3 ∫ 2π (1 − cos 4θ ) dθ = ab3 θ − sin 4θ 
− 2
4
8
4

−π
2
=
2
1 3  π  π  π 3
ab  −  −   = ab
8
 2  2  8
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.
Ix =
1
π ab3 8
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 10.
At
2a = kb3
x = 2a, y = b:
or
k =
2a
b3
Then
x=
2a 3
y
b3
or
y =
Now
dI x =
b
( 2a )
1
1
3
x3
1 3
1 b3
y dx =
xdx
3
3 2a
2a
Then
1 b3 2 a
1 b3 1 2
I x = ∫ dI x =
xdx
x
=
∫
3 2a a
6 a 2 a
=
b3
4a 2 − a 2
12a
(
)
Ix =
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.
1 3
ab 4
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 11.
−a
At x = a : b = k 1 − e a 


First note:
or
Have
k =
b
1 − e−1
I x = ∫ y 2dA
b
a
1− e
= ∫0 ∫0
−x 

 1− e a 
 2

 y dydx
−1 
3
3
=
−x
1  b  a
1 − e a  dx

−1  ∫ 0 
31 − e 


=
−x
−2 x
−3 x
1  b  a
1 − 3e a + 3e a − e a  dx

−1  ∫ 0 
31 − e 


3
3
a
−x
1 b  
 a  −2 x  a −3x  
x − 3( − a ) e a + 3 −  e a −  − e a 
= 
−1  
31 − e  
 2
 3
 0
3
1  b  
1
1 
 
= 
a + 3ae−1 − 1.5ae−2 + ae −3  −  3a − 1.5a + a  
−1  
3  1 − e  
3
3 
 
=
1
ab3 
11 
1.91723 − 
3 1 − e −1 3 
6
(
)
= 0.1107ab3
I x = 0.1107ab3 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 9, Solution 12.
x2
y2
+
=1
a 2 b2
y = b 1−
x2
a2
dA = 2 ydx
dI y = x 2dA = 2 x 2 ydx
a
a
I y = ∫ dI y = ∫ 0 2 x 2 ydx = 2b ∫ 0 x 2 1 −
x = a sin θ
Set:
x2
dx
a2
dx = a cosθ dθ
π
I y = 2b∫ 02 a 2 sin 2 θ 1 − sin 2 θ a cosθ dθ
3
π
= 2a b 02 sin 2 θ cos 2 θ dθ
∫
π
3
π
= 2a b∫ 02
1 2
sin 2θ dθ
4
π
1
1
1
1

2
= a3b∫ 02 (1 − cos 4θ ) dθ = a3b θ − sin 4θ 
2
2
4
4

0
=
1 3 π
 π
a b  − 0  = a3b
4
2
 8
Iy =
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.
1 3
πa b 8
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 13.
At
x = 2a, y = b : 2a = kb3
2a 3
y
b3
Then
x=
or
y =
Now
I y = ∫ x 2dA
Then
I y = ∫ a x2
b
1
2a 3
( )
1
( 2a )
1
3
1
b
( 2a )
1
3
2a
x 3 dx
7
∫ a x 3 dx
b
3 103
=
x
1
( 2a ) 3 10
=
=
x3
dA = ydx
2a
=
b
3b
10 ( 2a )
1
3
2a
a
 2a 103 − a 103 
( )


10
3ba3  103
2 − 1 3 
1 

10 ( 2 ) 3 
= 2.1619a3b
or I y = 2.16a3b 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 9, Solution 14.
First note:
or
Have
−a
At x = a : b = k 1 − e a 


k =
b
1 − e−1
I y = ∫ x 2dA
b
a
1− e
= ∫0 ∫0
−x 

1− e a 
 2

 x d ydx
−1 
−x 
a
 b 
= ∫ 0 x2 
1 − e a   dx
−1 

1 − e 
a


−x

2
 
b 1 3
e a  1  2
 1
x
x
2
x
2
=
−
−
−
−
+






3 
1 − e−1  3
 a
 
 1   a 
− 


 a

0
=
2


b   1 3
a
3 −1  a
3
+
a
a
e

 2 + 2 + 2   − a × 2 
−1 
a
1 − e   3
a
 

=
a3b  1

+ 5e −1 − 2 
−1 
1− e 3

(
)
= 0.273a3b
I y = 0.273 a3b 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 9, Solution 15.
1
b = k1a 4
x = a, y = b:
At
k1 =
or
b = k2a 4
b
a4
b 4
x
a4
k2 =
b
Then
y1 =
and
x1 =
Now
a x4
y2 − y1 dx = b 0  1

A = ∫(
a
b
y2 =
y4
)
1
a4
1
x4
a
a
x2 = 4 y 4
b
1
1
4
1
4
b
∫

1
a
4
−
x 4 
dx
a4 

a
 4 x 54 1 x5 
 = 3 ab
= b
−
5
 5 a 14 5 a 4 

0
I x = ∫ y 2dA
Then
dA = ( x1 − x2 ) dy
 a 1

a
b
I x = ∫ 0 y 2  1 y 4 − 4 y 4  dy
 4

b
b

b
 4 y 134
1 y 7 
= a
−
7 b4 
13 b 14

0
1
 4
= ab3  − 
 13 7 
or I x =
Now
kx =
Ix
=
A
15 3
ab
91
=
3
ab
5
15 3
ab 91
25 2
b = 0.52414b
91
or k x = 0.524b 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 9, Solution 16.
First note:
At x = a :
2b = k a
or
k=
2b
a
y1 =
b
x
a
Straight line:
Now:
∴ y2 =
2b
a
x
b 
a  2b 12
A = 2∫ 0 
x − x  dx
a 
 a
a
 4 x 23
1 2 
= 2b 
−
x
 3 a 2a 

0
=
Have
5
ab
3
I x = ∫ y 2dA
1
2b 2
x
a
= 2 0 b a y 2dydx
x
a
∫ ∫
=
2 a  8b3 32 b3 3 
x − 3 x  dx
∫ 

3 0  a 23
a


2b3  2
8 5
1
=
 × 3 x 2 − 3 x 4 
3  5 a2
4a

a
0
Ix =
And
kx =
=
59 3
ab 30
Ix
A
59 3
ab
30
5
ab
3
= b 1.18
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.
k x = 1.086 b COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 17.
At
x = a, y = b:
k1 =
or
Then
Now
y1 =
A = ∫(
1
b = k1a 4
b = k2a 4
b
a4
b
b 4
x
a4
k2 =
y2 =
and

1
a
4
a x4
y2 − y1 dx = b 0  1

)
∫
1
a4
−
b
a
1
4
1
x4
x 4 
dx
a4 

a
 4 x 54 1 x5 
 = 3 ab
= b
−
1
4
5
5 a4 5 a 

0
dA = ( y2 − y1 ) dx
Now
I y = ∫ x 2dA
Then
 b 1

b
a
I y = ∫ 0 x 2  1 x 4 − 4 x 4  dx
 4

a
a


9
a
4
a x4
= b 0 1

∫
−
x 6 
dx
a4 

a
 4 x 134
1 x 7 
= b
−
7 a4 
13 a 14

0
1 
 4
= b  a3 − a3 
7 
 13
or I y =
Now
ky =
Iy
A
=
15 3
ab
91
=
3
ab
5
15 3
a b
91
25
a = 0.52414a
91
or k y = 0.524a 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 9, Solution 18.
First note:
Now:
At x = a :
2b = k a
or
k=
2b
a
Straight line:
y1 =
b
x
a
∴ y2 =
2b
a
x
b 
a  2b 12
A = 2∫ 0 
x − x  dx
a 
 a
a
 4 x 23
1 2 
= 2b 
−
x
 3 a 2a 

0
=
Have
5
ab
3
I y = ∫ x 2dA
1
2b 2
x
a
= 2 0 b a x 2dydx
x
a
∫ ∫
 2b 12 b 
a
x − x  dx
= 2∫ 0 x 2 
a 
 a
a
7

2 x2
1 x 4 

= 2b 2 ×
−

7 a 4 a 

 0
=
9 3
ab
14
Iy =
9 3
a b
14
continued
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
And
ky =
Iy
A
=
9 3
ab
14
5
ab
3
=a
27
70
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.
k y = 0.621 a COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 19.
First note:
At x = 0: b = c cos ( 0 )
c=b
or
At x = 2a : b = b sin k ( 2a )
2ka =
or
k=
Then
π
2
π
4a
π
π 
2a 
A = ∫ a  b sin
x − b cos
x  dx
4a
4a 

2a
π
4a
π 
 4a
= b  − cos
x−
sin
x
4a
π
4a  a
 π
=−
=
Have
4ab 
1 
 1
+
(1) − 

π 
2 
 2
4ab
π
( 2 − 1)
I x = ∫ y 2dA
π
2 a b sin 4a x
= ∫a ∫
b cos
=
π
4a
x
y 2dydx
1 2a  3 3 π

3
3 π
∫  b sin 4a x − b cos 4a x  dx
3 a 

2a
b3  4a
π
1 4a
π   4a
π
1 4a 3 π  
x+
cos3
x  −  sin
x−
sin
x 
=
 − cos
3  π
4a
3π
4a   π
4a
3π
4a   a
=
3
3
4ab3  
1  1
1 1 
1
1  1   


−
1
+
−
−
+
−
+

3π  
3   2 3  2 
3  2   
2



=
4ab3  5
2
2− 

3π  6
3
continued
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
Ix =
2
ab3 5 2 − 4
9π
(
)
= 0.217 ab3
And
kx =
=
I x = 0.217 ab3 Ix
A
2
ab3 5 2 − 4
9π
4
ab 2 − 1
π
(
(
)
)
= 0.642b
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.
k x = 0.642 b COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 20.
At x = 0: b = c cos ( 0 )
First note:
c=b
or
At x = 2a : b = b sin k ( 2a )
2ka =
or
k=
π
2
π
4a
π
π 
2a 
A = ∫ a  b sin
x − b cos
x  dx
4a
4a 

Then
2a
π
4a
π 
 4a
= b  − cos
x−
sin
x
4a
π
4a  a
 π
=−
=
Have
4ab 
1 
 1
+
(1) − 

π 
2 
 2
4ab
π
( 2 − 1)
I y = ∫ x 2dA
π
2 a b sin 4a x 2
= ∫a ∫
b cos
π
4a
x
x dydx
π
π 
2a

= ∫ a x 2  b sin
x − b cos
x  dx
4
a
4
a 

continued
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




2
π
2
π
x
π 
 2x
= b 
+
−
sin
x
cos
x
cos
x
2
3

π 
4a
4a
4a 
π 
  π 







   4a 
 4a 

 4a 

2a




2
2x
π
2
π
x
π 
cos
x
sin
x
sin
x
−
−
+

3
  π 2
π 
4a
4a
4a  
π 
 



 

 4a 
 4a 
  4a 
  a
2a
64a3b 
π
π  π  
π
π 
π 2 2 
Iy =
sin
x
cos
x
x
sin
x
cos
x
2
x 
−
+
+
−







4a  2a  
4a
4a  
π 3  4a
16a 2  
a
=

64a3b  
π 2    1
1 
π 2   
1
π
1
2
2
+
−
−
+
−
(
)(
)
(
)






 




4    2
16   
π 3  
2  


I y = 1.482a3b = 1.48228a3b
And
ky =
=
Iy
A
1.48228a3b
4ab
2 −1
π
(
)
= 1.676a
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.
k y = 1.676a COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 21.
(a)
dI x =
1 3
a dx
3
Ix =
1 3 a
a3
2
a ∫ − a dx =
( 2a ) = a 4
3
3
3
dI y = x 2dA = x 2adx
x
a
I y = a − a x 2dx = a 
a
2 4
 = a
 3  −a 3
∫
JO = I x + I y =
J O = kO2 A
3
2 4 2 4
a + a
3
3
4 4
a
JO
2
k =
= 3 2 = a2
A
3
2a
2
JO =
4 4
a 3
kO = a
2
3
(b)
dI x =
1 3
a dx
12
Ix =
a 3 2a
a 3 2a 1 4
dx
=
[ x] = 6 a
∫
12 0
12 0
dI y = x 2dA = x 2 ( adx )
x
2a
I y = a 0 x 2dx = a 
∫
JO = I x + I y =
J O = kO2 A
3
2a
8 4
 = a
3
 3 0
1 4 8 4 17 4
a + a =
a
6
3
6
17 4
a
J
17 2
kO2 = O = 6 2 =
a
A
12
2a
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.
JO =
17 4
a 6
kO = a
17
12
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 22.
Have: A = ( 2a )( 3b ) −
1
( 2a )( b )
2
= 5ab
Have:
JP = Ix + I y
Where:
I x = ∫ y 2dA
a 2b
= 2∫ 0 ∫ b y 2dydx
− x
a
=
3
2 a
3
 b  
−
−
2
b
x

∫ ( )  a   dx
3 0 

 
2 
x4 
= b3  8x + 3 
3 
4a 
=
And
a
0
11 3
ab
2
I y = ∫ x 2dA
a 2b
= 2∫ 0 ∫ b x 2dydx
− x
a

a
 b 
= 2∫ 0 x 2  2b −  − x   dx
 a 

a
1 4
2
= 2b  x3 +
x 
4a  0
3
=
11 3
ab
6
continued
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
Then:
Also:
JP =
kP =
=
11 3 11 3
ab + a b
2
6
JP =
11
ab a 2 + 3b 2 6
kP =
11 2
a + 3b 2 30
(
)
JP
A
11
ab a 2 + 3b 2
6
5ab
(
)
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 9, Solution 23.
y1 :
x = a,
At
y = 2b: 2b = ma
or
m=
2b
a
Then
y1 =
2b
x
a
y2 :
x = 0, y = b: b = k ( 0 ) + c or
At
x = a, y = 2b: 2b = ka 2 + b
At
k =
or
b
a2
b 2
b
x + b = 2 x2 + a2
a2
a
(
y2 =
Then
Now
c=b
)
2b 
a b
A = ∫ ( y2 − y1 ) dx = ∫ 0  2 x 2 + a 2 −
x dx
a 
a
(
)
a
 b 1
 b 
=  2  x3 + a 2 x  − x 2 
 a 0
a  3
=
Now
1
b 1 3
 b
a + a3  − a 2 = ab
2
3
a 3
 a
a
I y = ∫ x 2dA = ∫ 0 x 2 ( y2 − y1 ) dx
2b 
a
b
= ∫ 0 x2  2 x2 + a 2 −
x dx
a 
a
(
)
a
b  1
1
 2b x 4 
= 2  x5 + a 2 x3  −

3
a  5
 a 4 0
=
b  a5 1 5  2b a 4
1 3
+ a  −
=
ab
2

3  a 4
30
a  5
continued
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
And
1 
1
I x = ∫ dI x = ∫  y23 − y13  dx
3
3 

=
3
1 a  b3 2
8b3 3 
2
x
+
a
−
x  dx

∫
3 0  a6
a3 
=
1 b3 a  1 6

x + 3x 4a 2 + 3x 2a 4 + a 6 − 8x3  dx
∫
3 a3 0  a3

(
)
(
)
1 b3  1  x 7 3 2 5 3 4 3
8 4
6 
Ix =
a
x
a
x
a
x
+
+
+
−

3  3
 4x 
3 a  a  7
5
3


=
Finally
0


1 b3  1  a 7 3 7
26 3
+ a + a 7 + a 7  − 2a 4  =
ab
3  3

3 a  a  7
5

 105
26 3
1 3
ab +
ab
105
30
JP = Ix + I y =
or J P =
And
a
kP =
JP
=
A
ab
7a 2 + 52b 2 210
(
ab
7a 2 + 52b 2
210
1
ab
3
or k P =
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.
(
)
)
7a 2 + 52b 2
70
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 24.
x = r 2 − y2
First note:
JP = Ix + I y
r
I x = ∫ y 2dA = 2∫ r ∫ 0
r 2 − y2
2
y 2dxdy
r
= 2∫ r y 2 r 2 − y 2 dy
2
Let
y = r sin θ
Then
dy = r cosθ dθ ;
Thus
y = r:
θ =
π
2
;
y=
r
:
2
θ =
π
6
π
I x = 2∫ π2 r 2 sin 2 θ ( r cosθ )( r cosθ dθ )
6
Now
sin 2θ = 2sin θ cosθ
Ix =
thus
sin 2 θ cos 2 θ =
1 2
sin 2θ
4
1 4 π2 2
r ∫ π sin 2θ dθ
2
6
π
1  θ sin 4θ  2
= r4  −

2 2
8 π
6
=
1 4π
π
1
3
+ ×
r  −

2  4 12 8
2 
=
3
r4  π
 +

4 3
8 
r
I y = ∫ x 2dA = 2∫ r ∫0
r 2 − y2
2
=
x 2dxdy
3
2 r 2
2
∫ r r − y 2 dy
3 2
(
)
continued
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
Let
y = r sin θ
Then
Iy =
Now
cos 4 θ = cos 2 θ 1 − sin 2 θ = cos 2 θ −
Thus
Iy =
2 π2 3
3
∫
π r cos θ ( r cos θ dθ )
3 6
(
)
(
)
1 2
sin 2θ
4
2 4 π2  2
1

r ∫ π  cos θ − sin 2 2θ  dθ
3
4
6 

π
2  θ sin 2θ  1  θ sin 4θ   2
= r 4  +
−  −

3  2
4  4 2
8  π
6
Then
=
1 4  π 1 π   π 1
3 1 π 1 1
3 
r  − ×  −  + ×
− × + × ×

3  2 4 2   6 2
2
4 6 4 4
2  
=
r4  π 3 3 
 −

4  3
8 
JP =
=
r4  π
3  r4  π 3 3 
 +
+
 −

43
8  4  3
8 
r4
8π − 3 3
48
(
)
J P = 0.415r 4 Now
r
A = 2∫ r xdy
2
r
= ∫ r r 2 − y 2 dy
2
continued
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
Let
y = r sin θ
π
A = 2∫π2 ( r cosθ )( r cosθ dθ )
6
π
2 θ
sin 2θ  2
= 2r  +

4 π
2
6
π π 1
3
= r 2  − − ×

2 
2 6 2
=
Have
kP =
=
r2
4π − 3 3
12
(
)
JP
A
r4
8π − 3 3
48
r2
4π − 3 3
12
(
(
)
)
k P = 0.822 r 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 9, Solution 25.
J O ≡ I 0 = ∫ r 2dA
(a) Have
dA =
Where
Then
3π
rdr
2
R
 3π 
J O = ∫R 2 r 2 
r  dr
1
 2 
R
=
3π 4 2
3π 4
r
=
R2 − R14
8
8
R1
(
)
JO =
)
( I x )1 = ( I x )2 = ( I x )3
By inspection
I x = 3 ( I x )1
so that
( )1 + ( I y )2 + ( I y )3 = 3 ( I y )1
Iy = Iy
Symmetry implies
( I x )1 = ( I y )1
Then
Ix = I y
Now
JO = I x + I y
Then
(
I x = ( I x )1 + ( I x )2 + ( I x )3
(b) Now
Similarly,
3π 4
R2 − R14 8
Ix = I y =
JO
3π 4
=
R2 − R14
2
16
(
)
or I x = I y =
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.
3π 4
R2 − R14 16
(
)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 26.
3π 4
R2 − R14
8
(
JO =
(a) From Problem 9.25
R
A = ∫ dA = ∫R 2
And
1
3π
3π 2
rdr =
R2 − R12
2
4
(
)
3π 4
R2 − R14
R22 + R12 R22 − R12
J
1 2
2
O
8
kO =
=
=
=
R2 + R12
2
2
3π 2
A
2
2
R
−
R
2
2
1
R2 − R1
4
(
(
) (
)
Now
Rm =
1
( R1 + R2 )
2
Then
R2 = Rm +
Then
)
And
kO2 =
For
t
t
2
)(
(
)
)
(
t = R2 − R1
and
R1 = Rm −
and
t
2
2
2
1 
t
t 
1

 Rm +  +  Rm −   = Rm2 + t 2
2 
2
2  
4

R1, R2
kO2
Rm2
kO
Or
(b) Have
)
R − kO
% error = m
× 100% =
kO
1− 1+
=
Rm2 +
Rm −
Rm2 +
1 t 


4  Rm 
1 t 
1+ 

4  Rm 
2
Rm 1 2
t
4 × 100%
1 2
t
4
2
× 100%
continued
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
Then
t
= 1: % error =
Rm
1− 1+
1+
1
4
1
4 × 100%
or % error = −10.56% t
1
= : % error =
Rm
4
1− 1+
11
 
44
11
1+  
4 4
2
× 100%
2
or % error = −0.772% 1
t
= : % error =
Rm 16
1 1 
1− 1+  
4  16 
1+
1 1 
 
4  16 
2
2
× 100%
or % error = −0.0488% 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 9, Solution 27.
π
Have:
a cos 2θ
A = 2∫ 04 ∫ 0
rdrdθ
π
= ∫ 04 a 2 cos 2 2θ dθ
π
2 θ
sin 4θ  4
=a  +

8 0
2
=
Have:
π
8
a2
J O = ∫ r 2dA
π
a cos 2θ
= 2∫ 04 ∫ 0
r 2 ( rdrdθ )
π
1
= ∫ 04 a 4 cos 4 2θ dθ
2
Now:
(
cos 4 2θ = cos 2 2θ 1 − sin 2 2θ
= cos 2 2θ −
)
1 2
sin 4θ
4
π
Then:
1
1


J O = a 4 ∫ 04  cos 2 2θ − sin 2 4θ  dθ
2
4


π
1  θ sin 4θ  1  θ sin 8θ   4
= a 4  +
−  −

2  2
8  4 2
16   0
=
1 4π 1 π 
a  − × 
4 4 4 4
JO =
3π 4
a 64
continued
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
And:
kO =
=
JO
A
3π 4
a
64
π
8
a2
kO =
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
6
4
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 28.
y =
By observation
b
y
2h
x=
or
h
x
b
2
 b 
dA = xdy = 
y  dy
 2h 
Now
b 3
y dy
2h
And
dI x = y 2dA =
Then
I x = ∫ dI x = 2∫ 0
h b
2h
y 3dy
h
=
y =
From above
Now
b y4
1
= bh3
h 4 0 4
2h
x
b
2h 

dA = ( h − y ) dx =  h −
x  dx
b 

=
h
( b − 2 x ) dx
b
And
dI y = x 2dA = x 2
Then
I y = ∫ dI y = 2∫ 02
b
h
( b − 2 x ) dx
b
h 2
x ( b − 2 x ) dx
b
b
h 1
1 2
= 2  bx3 − x 4 
b 3
2 0
3
4
h b  b 
1b 
1 3
bh
=2    −    =
b  3  2 
2  2   48
continued
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
Now
JO = I x + I y =
1 3
1 3
bh +
bh
4
48
or J O =
And
bh
12h 2 + b 2
J
1
2
48
O
kO =
12h 2 + b 2
=
=
1
A
24
bh
2
(
)
(
bh
12h 2 + b 2 48
(
)
)
or kO =
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.
12h 2 + b 2
24
COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 29.
First the circular area is divided into an increasing number of identical
circular sectors. The sectors can be approximated by isosceles triangles.
For a large number of sectors the approximate dimensions of one of the
isosceles triangles are as shown.
For an isosceles triangle (see Problem 9.28)
JO =
b = r ∆θ
Then with
=
dJ O sector
dθ
and
)
h=r
1 4 
r ∆θ 12 + ∆θ 2 


48
(
)
 ∆ J O sector 
2 
1 4
= lim 
 = lim  r 12 + ( ∆θ )  
∆θ → 0

 ∆θ
 ∆θ → 0  48
=
Then
(
1
( r ∆θ )( r ) 12r 2 + ( r ∆θ )2 
48
( ∆ J O )sector
Now
bh
12h 2 + b 2
48
1 4
r
4
( J O )circle = ∫ dJ O sector = ∫ 02π
1 4
1
2π
r dθ = r 4 [θ ]0
4
4
or ( J O )circle =
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.
π
2
r4 COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 30.
From the solution to sample Problem 9.2, the centroidal polar moment of
inertia of a circular area is
( J C )cir =
π
2
r4
The area of the circle is
Acir = π r 2
So that
A2
 J C ( A )  =
cir
2π
Two methods of solution will be presented. However, both methods
depend upon the observation that as a given element of area dA is moved
closer to some point C, The value of J C will be decreased ( J C = ∫ r 2dA ;
as r decreases, so must J C ).
Solution 1
Imagine taking the area A and drawing it into a thin strip of negligible
width and of sufficient length so that its area is equal to A. To minimize
the value of ( J C ) A , the area would have to be distributed as closely as
p