mecanica vectorial traabjoi.docx

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° !



(a)

(b)

We measure:

R = 57 lb, α = 86° R = 57 lb


86° !



(a)

Parallelogram law:

(b)

Triangle rule:

We measure: R = 10.5 kN

α = 22.5°

R = 10.5 kN


22.5° !



(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°




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 !




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


Faa′ = 266 N !



Using the triangle rule and the Law of Cosines, Have: β = 180° − 45°

β = 135° Then: R 2 = ( 900 ) + ( 600 ) − 2 ( 900 )( 600 ) cos 135° 2

2

or R = 1390.57 N

Using the Law of Sines, 600 1390.57 = sin γ sin135° or γ = 17.7642° and α = 90° − 17.7642°

α = 72.236°


(a)

α = 72.2° !

(b)

R = 1.391 kN !



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 !




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 !




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


(a)

α = 7.48° !

(b)

R = 138.4 N !



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 !




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 !




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 !




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)


TBC = 4.88 N

6.00° !

R = 69.8 N !



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


76.6° !



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


85.8° !



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


22.5° "



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


72.0° !




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


 20 kN  sin α =   sin110°  41.357 kN  = 0.45443

α = 27.028° Hence:

φ = α + 45° = 72.028° R = 41.4 kN


72.0° !




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


 30 kN  sin α =   sin110°  41.357 kN  = 0.68164

α = 42.972° Finally:

φ = α + 45° = 87.972° R = 41.4 kN


88.0° !



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




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




Determine the following distances:

680 N Force:

dOA =

( −160 mm )2 + ( 300 mm )2

dOB =

( 600 mm )2 + ( 250 mm )2

dOC =

( 600 mm )2 + ( −110 mm )2

Fx = 680 N

= 340 mm

= 650 mm = 610 mm

( −160 mm ) 340 mm Fx = − 320 N !

( 300 mm )

Fy = 680 N

340 mm Fy = 600 N !

390 N Force:

Fx = 390 N

( 600 mm ) 650 mm Fx = 360 N !

Fy = 390 N

( 250 mm ) 650 mm Fy = 150 N !

610 N Force:

Fx = 610 N

( 600 mm ) 610 mm Fx = 600 N !

Fy = 610 N

( −110 mm ) 610 mm Fy = −110 N !




We compute the following distances:

OA =

( 48)2 + ( 90 )2

= 102 in.

OB =

( 56 )2 + ( 90 )2

= 106 in.

OC =

(80 )2 + ( 60 )2

= 100 in.

Then: 204 lb Force:

Fx = − ( 204 lb )

48 , 102

Fy = + ( 204 lb )

90 , 102

Fx = −96.0 lb

Fy = 180.0 lb

212 lb Force:

Fx = + ( 212 lb )

56 , 106

Fx = 112.0 lb

90 , 106

Fy = 180.0 lb

Fx = − ( 400 lb )

80 , 100

Fx = −320 lb

Fy = − ( 400 lb )

60 , 100

Fy = −240 lb

Fy = + ( 212 lb ) 400 lb Force:




(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




(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 !




(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 "




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


Py = 218 lb !



(a)

P=

45 N cos 20° or P = 47.9 N !

(b)

Px = ( 47.9 N ) sin 20° or Px = 16.38 N !




(a)

P=

18 N sin 20° or P = 52.6 N !

(b)

Py =

18 N tan 20° or Py = 49.5 N !




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° !




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


61.4° !



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


6.68° !



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


45.0° !



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:


R = 45.2 lb

62.3° !



(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°




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


26.1° !



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


88.3° !



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 !




(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 !




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 !




(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 !




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°

(b)

TBC =

400 sin 60° = 347.73 lb sin 95°


FAC = 169.7 lb ! TBC = 348 lb !



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 !




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 !




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 !




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°




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:

− TCD =

17 24 TBC + TCD = 0 25 293 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


WC = 97.7 N !



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 !




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




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


FD = 1.678 kN



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




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 "



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 "



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)


TBC = 265 lb !




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


0 ≤ W ≤ 148.2 lb !


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 © 2007 The McGraw-Hill Companies.



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 !




(a)

For TAB to be a minimum TAB must be perpendicular to TAC

∴ α + 10° = 60°

(b)

or

α = 50.0° W

or

TAB = 35.0 lb W

Then TAB = ( 70 lb ) sin 30°




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°




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°





ΣFx = 0: 2Tx − 1200 N = 0 Tx = 600 N

(Tx )2 + (Ty )

2

= T2

( 600 N )2 + (Ty )

2

= ( 870 N )

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 "




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


TBC = 346 lb !



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 !




At Collar A ...

ΣFy = 0:

− 9 lb +

or

h 2

12 + h 2

Fs = 0

hFs = 9 144 + h 2

Fs = k ( L′AB − LAB )

Now

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 + h

2

)

− 12 2  = 9 144 + h 2 

144 + h 2 = 12 2 h

Solving Numerically ...


h = 16.81 in.



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




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


7.50°



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




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




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


Q = 3.53 kN



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




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° !




(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° !




(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° !




(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° !




(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° !




(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° !




(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° !




(a)

Fx = − ( 220 N ) cos 60° cos 35° = − 90.107 N Fx = − 90.1 N W

Fy = ( 220 N ) sin 60°

= 190.526 N Fy = 190.5 N W Fz = − ( 220 N ) cos 60° sin 35° = − 63.093 N Fz = − 63.1 N W

(b)

cosθ x =

−90.107 Ν 220 N

θ x = 114.2° W cosθ y =

190.526 N 220 N

θ y = 30.0° W cosθ z =

−63.093 N 220 N

θ z = 106.7° W




(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° !




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° !




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° !




(a)

(

We have ( cosθ x ) + cosθ y 2

( cosθ y )

2

= 1 − ( cosθ x ) − ( cosθ z )

Since Fy < 0 we must have Thus

2 ) + ( cosθ z )2 = 1 2

2

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 !




(a)

(

We have ( cosθ x ) + cosθ y 2

2 ) + ( cosθ z )2 = 1

(

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 !




(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° !




(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° !




!!!" DB = ( 36 in.) i − ( 42 in.) j − ( 36 in.) k DB =

( 36 in.)2 + ( − 42 in.)2 + ( − 36 in.)2

TDB = TDBλDB = TDB TDB =

= 66 in.

!!!" DB DB

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 !




!!!" EB = ( 36 in.) i − ( 45 in.) j + ( 48 in.) k EB =

( 36 in.)2 + ( − 45 in.)2 + ( 48 in.)2

TEB = TEBλEB = TEB TEB =

= 75 in.

!!!" EB EB

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 (TEB ) z

= − 36.0 lb ! = 38.4 lb !




!!!" BA = ( 4 m ) i + ( 20 m ) j − ( 5 m ) k BA = F = F λ 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 = 21 m  BA F = ( 400 N ) i + ( 2000 N ) j − ( 500 N ) k

Fx = + 400 N, Fy = + 2000 N, Fz = − 500 N !




!!!" DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) k DA = F = F λ 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 = 25.2 m  DA F = ( 200 N ) i + (1000 N ) j + ( 740 N ) k

Fx = + 200 N, Fy = + 1000 N, Fz = + 740 N !




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 TBG = TBG λBG = TBG

TBG =

uuuv BG BG

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




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 TBH = TBH λBH = TBH TBH =

uuuuv BH BH

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



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°




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°




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


R = 1122 N

θ x = 147.7°

θ y = 61.6° θ z = 104.2°



!!!" 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 =




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°



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°

cos θ x =

cos θ y = cos θ z =

−1013.75 lb 1338.35 lb

873.78 lb 1338.35 lb


θ y = 139.2° θ z = 49.2°



!!!" AB = − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k

Cable AB:

AB =

( − 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2

TAB = TAB

!!!" AB 600 N  − ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) k  = 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

TAC = TAC

TAC = −

Load P:

= 750 mm

= 850 mm

!!!" AC TAC  − ( 600 mm ) i + ( 320 mm ) j − ( 510 mm ) k  = AC 850 mm 

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 !




uuur AB = − ( 4 m ) i − ( 20 m ) j + ( 5 m ) k

Cable AB:

AB =

( − 4 m)2 + ( −20 m)2 + (5 m )2

TAB = TAB

= 21 m uuur AB T = 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

TAC = TAC

= 23.6 m uuur AC 1770 N (12 m ) i − ( 20 m ) j + ( 3.6 m ) k  = 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

TAD = TAD =

= 25.2 m uuur AD TAD  − ( 4 m ) i − ( 20 m ) j + (14.8 m ) k  = 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 !



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

TAB =

= 860 mm

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:

ΣF = 0:

i component:

− 0.6 TAB

k component:

−

TAB + TAC + TAD + W = 0 25 + 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 !




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 !


(3)



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

AD = ( − 5.60 m ) + ( − 3.30 m ) = 6.50 m uuur AB TAB = TAB = ( − 4.20i − 5.60j) AB 7.00 m 2

TAB = TAB λ AB

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


(1)


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 !




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


!



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

FBA = FBAλBA =

= 14.4 in.

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



 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 "




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




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

TAC = T λ AC = TAC

= 35.4 m

uuur AC TAC (18 m ) i − ( 30 m ) j + ( 5.4 m ) k  = 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

TAB = T λ AB = TAB

= 31.5 m

uuur AB TAB  − ( 6 m ) i − ( 30 m ) j + ( 7.5 m ) k  = 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

TAD = T λ AD = TAD

= 37.8 m

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



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 "




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 !




!!!" 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 = TAC =

(1 ft )2 + ( −8 ft )2 + ( 4 ft )2

= 9 ft

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 = TAD =

(1.75 ft )2 + ( −8 ft )2 + ( −1 ft )2

= 8.25 ft

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 !




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



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




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 j: TBE + TCF + TDG − W

(1)

65 =0 8

(2)

k : − TBE sin 45° + TCF sin 30° − TDG sin15° = 0

(3)

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 !




d DA =

( 400 mm )2 + ( − 600 mm )2

d DB =

( − 200 mm )2 + ( − 600 mm )2 + (150 mm )2

d DC =

( − 200 mm )2 + ( − 600 mm )2 + ( −150 mm )2

= 721.11 mm = 650 mm = 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 − TDB − TDC + W = 0 13 13 3 3 TDB − TDC = 0 13 13

i component:

0.55470 TDA −

(1)

j component:

−0.83205 TDA

(2)

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 !


(3)



d DA =

( 400 mm )2 + ( − 600 mm )2

d DB =

( − 200 mm )2 + ( − 600 mm )2 + ( 200 mm )2

d DC =

( − 200 mm )2 + ( − 600 mm )2 + ( − 200 mm )2

= 721.11 mm = 663.32 mm = 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

0.55470 TDA − 0.30151TDB − 0.30151TDC = 0

i component:

−0.83205 TDA − 0.90454 TDB − 0.90454 TDC + W = 0

j component:

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 !


(1) (2) (3)



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 !




d BA =

( 6 m )2 + ( 6 m )2 + ( 3 m )2

d AC =

( −10.5 m )2 + ( − 6 m )2 + ( − 8 m )2

d AD =

( − 6 m )2 + ( − 6 m )2 + ( 7 m )2

d AE =

( 6 m )2 + ( − 4.5 m )2

FBA = FBAλBA =

=9m

= 14.5 mm

= 11 mm

= 7.5 m

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



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




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 !




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 !


(5)



d AB =

( − 30 ft )2 + ( 24 ft )2 + ( 32 ft )2

d AC =

( − 30 ft )2 + ( 20 ft )2 + ( −12 ft )2

TAB = TAB λ AB =

= 50 ft = 38 ft

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




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




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 !




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 !




uuur DA = − ( 8 ft ) i + ( 40 ft ) j + (10 ft ) k

( − 8 ft )2 + ( 40 ft )2 + (10 ft )2

DA =

TDA =

= 42 ft

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

( 3 ft )2 + ( 36 ft )2 + ( − 8 ft )2

DB = TDB =

= 37 ft

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

DC =

TDC

TDC = At D

( a − 8)2 + 585

=

( a − 8)2 + 585 ft

( 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 ( a − 8)2 + 585 3

( a − 8) + 585 2

=0

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., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.


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




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 !




d AB =

( − 2.7 m )2 + ( 2.4 m )2 + ( − 3.6 m )2

d AC =

( 2.4 m )2 + (1.8 m )2

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

= 5.1 m

=3m

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



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 9 8 2 TAB + TADE − TADE = 0 17 9 3

i component:

−

j component:

8 16 2 TAB + 0.8 TAC + TADE + TADE − W = 0 17 9 3

k component:

−

12 2 1 TAB + 0.6 TAC − TADE + TADE = 0 17 9 3

(1) (2) (3)

Simplifying (1), (2), (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′)

Setting W = 1400 N and solving (1), (2), (3) simultaneously: TAB = 203 N " TAC = 149.6 N " TADE = 485 N "




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)


W = 2060 N !



Free-Body Diagrams of collars

For both Problems 2.127 and 2.128:

( AB )2 (1 m )2

Here

= x2 + y 2 + z 2

= ( 0.40 m ) + y 2 + z 2 2

y 2 + z 2 = 0.84 m 2

or

Thus, with y given, z is determined. Now λ AB

uuur AB 1 = = ( 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



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


Q = 1.963 kN !



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 !




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




We compute the following distances: OA =

( 70 )2 + ( 240 )2

OB =

( 210 )2 + ( 200 )2

= 290 mm

OC =

(120 )2 + ( 225)2

= 255 mm

= 250 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:




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 !




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 !




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 !




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


32.3°



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° !




Fz = F cosθ z = ( 600 lb ) cos136.8°

(a)

= −437.38 lb

Fz = −437 lb !

Then: F 2 = Fx2 + Fy2 + Fz2 2 ( ) + ( −437.38 lb )2

So: ( 600 lb ) = ( 200 lb ) + Fy 2

2

Hence: Fy = −

(b)

cosθ x =

( 600 lb )2 − ( 200 lb )2 − ( −437.38 lb )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


θ y = 126.7° !



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° !




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




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

TAB = T λ AB = TAB

= 68 in.

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

TAC = T λ AC = TAC

= 75 in.

uuur AC T = 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



TAD = T λ AD = TAD

uuur AD T = 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 "




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

TAB = T λ AB = TAB

= 0.88 m

uuur AB TAB  − ( 0.48 m ) i + ( 0.72 m ) j − ( 0.16 m ) k  = 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

TAC = T λ AC = TAC

= 0.77 m

uuur AC TAC ( 0.24 m ) i + ( 0.72 m ) j − ( 0.13 m ) k  = 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



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 !


mecanica vectorial traabjoi.docx

Recommend Documents