Sol Cap 02 - Edicion 8

COSMOS: Complete Online Solutions Manual Organization System 

Chapter 2, Solution 1. 

(a) 

(b) 

We measure: 

R = 37 lb, α = 76° 

R = 37 lb 76° ! 

Vector Mechanics for Engineers: Statics and Dynamics, 8/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) 

(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° 

R = 5.4 kN 12° ! 

(b) 

Triangle rule: 

We measure: 

R = 5.4 kN α = 12° 

R = 5.4 kN 12° ! 






(a) 

Using the triangle rule and the Law of Sines 

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 

F bb ′ 

= 

sinα 

200 N 

sin 45° 

F ′ = 276 N ! 

bb 






(a) 


sinα sin 45° 

= 

120 N 200 N 

sinα = 0.42426 

α = 25.104° 

or α = 25.1°! 

(b) β + 45° + 25.104° = 180° 

β = 109.896° 


F aa ′ 200 N 

= 

sin β sin 45° 

F aa ′ 200 N 

= 

sin109.896° sin 45° 

or 

F ′ = 266 N ! 

aa 






Using the triangle rule and the Law of Cosines, 

Have: β = 180° − 45° 

Then: 

or 

β = 135° 

R 

2 

2 2 

( ) ( ) ( )( ) 

= 900 + 600 − 2 900 600 cos 135° 

R = 1390.57 N 

Using the Law of Sines, 

600 1390.57 

sinγ = 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 

= = 

sinα 

sin 38° 

sin β 

α = 90° − 28° = 62 ° , β = 180° − 62° − 38° = 80° 

Then: 

F2 R 30 lb 

= = 

sin 62° sin 38° sin80° 

or (a) F 2 = 26.9 lb ! 

(b) 

R = 18.75 lb ! 







F1 R 20 lb 

= = 

sinα 

sin 38° 

sin β 

α = 90° − 10° = 80 ° , β = 180° − 80° − 38° = 62° 

Then: 

F1 R 20 lb 

= = 

sin80° sin 38° sin 62° 

or (a) F 1 = 22.3 lb ! 

(b) 

R = 13.95 lb ! 






Using the Law of Sines: 

60 N 80 N 

sinα = 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 ! 







Have: β = 180° − ( 35° + 25° 

) 

= 120° 

Then: 

P R 80 lb 

= = 

sin 35° sin120° sin 25° 

or (a) 

(b) 

P = 108.6 lb ! 

R = 163.9 lb ! 







(a) Have: 

80 lb 70 lb 

sinα = sin 35° 

sinα = 0.65552 

α = 40.959° 

or α = 41.0°! 

(b) β = 180 − ( 35° + 40.959° 

) 

= 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° 

And: 

(b) R = ( 80 lb) 

cos35° 

or 

P = 45.9 lb ! 

or R = 65.5 lb ! 






For T BC to be a minimum, 

R and T BC must be perpendicular. 

Thus T BC = ( 70 N) 

sin 4° 

= 4.8829 N 

And R = ( 70 N) 

cos 4° 

= 69.829 N 

(a) 

(b) 

T = 4.88 N 6.00° ! 

BC 

R = 69.8 N ! 






Using the force triangle and the Laws of Cosines and Sines 

We have: 

γ = 180° − ( 15° + 30° 

) 

= 135° 

2 

2 2 

Then: ( ) ( ) ( )( ) 

R = 15 lb + 25 lb − 2 15 lb 25 lb cos135° 

or 

2 

= 1380.33 lb 

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, 

( ) ( ) ( )( ) 

2 2 2 

R = 45 lb + 15 lb − 2 45 lb 15 lb cos135° 

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: 

or 

2 

2 2 

( ) ( ) ( )( ) 

R = 5 kN + 8 kN − 2 5 kN 8 kN cos105° 

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° " 







We have: γ = 180° − ( 45° + 25° ) = 110° 

2 

2 2 

Then: ( ) ( ) ( )( ) 

and 

R = 30 kN + 20 kN − 2 30 kN 20 kN cos110° 

2 

= 1710.42 kN 

R = 41.357 kN 

20 kN 41.357 kN 


⎛ 20 kN ⎞ 

sinα 

= ⎜ 

sin110° 

41.357 kN 

⎟ 

⎝ ⎠ 

= 0.45443 

α = 27.028° 

Hence: φ = α + 45° = 72.028° 

R = 41.4 kN 72.0° ! 







We have: γ = 180° − ( 45° + 25° ) = 110° 

2 

2 2 

Then: ( ) ( ) ( )( ) 

and 

R = 30 kN + 20 kN − 2 30 kN 20 kN cos110° 

2 

= 1710.42 kN 

R = 41.357 kN 

20 kN 41.357 kN 


⎛ 20 kN ⎞ 

sinα 

= ⎜ 

sin110° 

41.357 kN 

⎟ 

⎝ ⎠ 

= 0.45443 

α = 27.028° 

Hence: φ = α + 45° = 72.028° 

R = 41.4 kN 72.0° ! 







We have: γ = 180° − ( 45° + 25° ) = 110° 

2 

2 2 

Then: ( ) ( ) ( )( ) 

and 

R = 30 kN + 20 kN − 2 30 kN 20 kN cos110° 

2 

= 1710.42 kN 

R = 41.357 kN 

30 kN 41.357 kN 


⎛ 30 kN ⎞ 

sinα 

= ⎜ 

sin110° 

41.357 kN 

⎟ 

⎝ ⎠ 

= 0.68164 

α = 42.972° 

Finally: φ = α + 45° = 87.972° 

R = 41.4 kN 88.0° ! 






F = 2.4 kN cos50° 

2.4 kN Force: ( ) 

x 

F = 1.85 kN cos 20° 

1.85 kN Force: ( ) 

x 

( ) 

F = 2.4 kN sin 50° 

y 

( ) 

F = 1.85 kN sin 20° 

y 

F x = 1.543 kN ⊳ 

F = 1.839 kN ⊳ 

y 

F x = 1.738 kN ⊳ 

F = 1.40 kN cos35° 

1.40 kN Force: ( ) 

x 

( ) 

F =− 1.40 kN sin 35° 

y 

F = 0.633 kN ⊳ 

y 

F x = 1.147 kN ⊳ 

F =−0.803 kN ⊳ 

y 






F = 5kips cos40° 

5 kips: ( ) 

x 

or 

F x = 3.83 kips ⊳ 

( ) 

F = 5kips sin40° 

y 

or F = 3.21 kips ⊳ 

y 

F =− 7kips cos70° 

7 kips: ( ) 

x 

or F =−2.39 kips ⊳ 

x 

( ) 

F = 7 kips sin 70° 

y 

or F = 6.58 kips ⊳ 

y 

F =− 9kips cos20° 

9 kips: ( ) 

x 

( ) 

F = 9 kips sin 20° 

y 

or F =−8.46 kips ⊳ 

x 

or F = 3.08 kips ⊳ 

y 






Determine the following distances: 

680 N Force: 

d 

d 

d 

OA 

OB 

OC 

F x 

( ) ( ) 

2 2 

= − 160 mm + 300 mm = 340 mm 

( ) ( ) 

2 2 

= 600 mm + 250 mm = 650 mm 

( ) ( ) 

2 2 

= 600 mm + − 110 mm = 610 mm 

( −160 mm) 

= 680 N 340 mm 

390 N Force: 

F y = 

F x = 

F y = 

( 300 mm) 

680 N 340 mm 

( 600 mm) 

390 N 650 mm 

( 250 mm) 

390 N 650 mm 

F =− 320 N ! 

x 

F = 600 N ! 

y 

F = 360 N ! 

x 

610 N Force: 

F x = 

( 600 mm) 

610 N 610 mm 

F = 150 N ! 

y 

F y 

( −110 mm) 

= 610 N 610 mm 

F = 600 N ! 

x 

F =− 110 N ! 

y 






We compute the following distances: 

Then: 

204 lb Force: 

212 lb Force: 

( ) ( ) 

2 2 

OA = 48 + 90 = 102 in. 

( ) ( ) 

2 2 

OB = 56 + 90 = 106 in. 

( ) ( ) 

2 2 

OC = 80 + 60 = 100 in. 

( ) 48 

F x =− 204 lb , 

F x =−96.0 lb ⊳ 

102 

( ) 90 

F y =+ 204 lb , 

F y = 180.0 lb ⊳ 

102 

( ) 56 

F x =+ 212 lb , 

F x = 112.0 lb ⊳ 

106 

400 lb Force: 

( ) 90 

F y =+ 212 lb , 

F y = 180.0 lb ⊳ 

106 

( ) 80 

F x =− 400 lb , 

F x =−320 lb ⊳ 

100 

( ) 60 

F y =− 400 lb , 

F y =−240 lb ⊳ 

100 






(a) 

P = 

= 

P y 

sin 35 

° 

960 N 

sin 35° 

or 

P = 1674 N ⊳ 

(b) 

P = 

x 

= 

P 

y 

tan35° 

960 N 

tan 35° 

or 

P x = 1371 N ⊳ 






(a) 

P = 

P = 

P x 

cos 40 

° 

30 lb 

cos 40° 

or 

P = 39.2 lb ! 

(b) P = P tan 40° 

y 

x 

( ) 

P = 30 lb tan 40° 

y 

or P = 25.2 lb ! 

y 






(a) 

P = 100 N 

y 

P = 

P y 

sin 75 

° 

100 N 

P = 

sin 75° 

or 

P = 103.5 N " 

(b) 

P = 

x 

P 

y 

tan 75° 

P x = 

100 N 

tan 75° 

or P = 26.8 N " 

x 






We note: 

CB exerts force P on B along CB, and the horizontal component of P is 

Then: 

(a) P = Psin 50° 

x 

P x = 

260 lb. 

P = 

= 

P x 

sin50 

° 

260 lb 

sin50° 

(b) P = P tan 50° 

x 

= 339.40 lb 

P = 339 lb ! 

y 

P = 

y 

= 

Px 

tan 50° 

260 lb 

tan 50° 

= 218.16 lb 

P 218 lb ! 

y = 






(a) 

P = 

45 N 

cos 20° 

or 

P = 47.9 N ! 

(b) ( ) 

P = 47.9 N sin 20° 

x 

or P = 16.38 N ! 

x 






(a) 

18 N 

P = 

sin 20° 

or 

P = 52.6 N ! 

(b) 

P y = 

18 N 

tan 20° 

or P = 49.5 N ! 

y 






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 

( 4.428 kN) ( 1.669 kN) 

R =Σ F = i + j 

( 4.428 kN) ( 1.669 kN) 

R = + 

= 4.7321 kN 

1.669 kN 

tanα = 

4.428 kN 

α = 20.652° 

2 2 

R = 4.73 kN 20.6° ! 







( ) ( ) 

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 

( 7.02 kips) ( 12.87) 

R =Σ F =− i + j 

( ) ( ) 

2 2 

R = − 7.02 kips + 12.87 kips = 14.66 kips 

−1 12.87 

α = tan ⎛ 

⎜ 

⎞ 

⎟ = 61.4° 

⎝−7.02 

⎠ 

R = 14.66 kips 61.4° ! 







( 48.0 lb) ( 90.0 lb) 

F OA =− i + j 

( 112.0 lb) ( 180.0 lb) 

F OB = i + j 

OC =− 

( 320 lb) − ( 240 lb) 

F i j 

( 256lb) ( 30lb) 

R =Σ F =− i + j 

( 256lb) ( 30lb) 

R = − + 

= 257.75 lb 

2 2 

tan 

30 lb 

256 lb 

α = − 

α =− 6.6839° 

R = 258 lb 6.68° ! 






From Problem 2.23: 

( 320 N) ( 600 N) 

F OA =− i + j 

( 360 N) ( 150 N) 

F OB = i + j 

OC = 

( 600N) − ( 110N) 

F i j 

( 640 N) ( 640 N) 

R =Σ F = i + j 

( 640 N) ( 640 N) 

R = + 

= 905.097 N 

2 2 

tanα = 

640 N 

640 N 

α = 45.0° 

R = 905 N 45.0° ! 






Cable BC Force: 

100-lb Force: 

156-lb Force: 

and 

Further: 

Thus: 

( ) 84 

F x =− 145 lb =− 105 lb 

116 

( ) 80 

F y = 145 lb = 100 lb 

116 

( ) 3 

F x =− 100 lb =− 60 lb 

5 

( ) 4 

F y =− 100 lb =− 80 lb 

5 

( ) 12 

F x = 156 lb = 144 lb 

13 

( ) 5 

F y =− 156 lb =− 60 lb 

13 

R =Σ F =− 21 lb, R =Σ F =− 40 lb 

x x y y 

( ) ( ) 

2 2 

R = − 21 lb + − 40 lb = 45.177 lb 

tanα = 

40 

21 

−1 40 

α = tan = 62.3° 

21 

R = 45.2 lb 62.3° ! 






(a) Since R is to be horizontal, R y = 0 

Then, R =Σ F = 0 

y 

y 

( ) α ( ) 

( 13) cosα 

= ( 7) 

sinα 

+ 9 

90 lb + 70 lb sin − 130 lb cosα 

= 0 

2 

( ) 

13 1 − sin α = 7 sinα 

+ 9 

2 2 

Squaring both sides: ( ) ( ) ( ) 

169 1 − sin α = 49 sin α + 126 sinα 

+ 81 

2 

( ) α ( ) 

218 sin + 126 sinα 

− 88 = 0 

Solving by quadratic formula: sinα = 0.40899 

or α = 24.1°! 

(b) 

Since R is horizontal, R = R x 

Then, R = Rx = ΣFx 

F x 

( ) ( ) 

Σ = 70 cos 24.142° + 130 sin 24.142° 

or 

R = 117.0 lb ! 






300-N Force: 

400-N Force: 

600-N Force: 

and 

( ) 

F = 300 N cos 20° = 281.91 N 

x 

( ) 

F = 300 N sin 20° = 102.61 N 

y 

( ) 

F = 400 N cos85° = 34.862 N 

x 

( ) 

F = 400 N sin85° = 398.48 N 

y 

( ) 

F = 600 N cos5° = 597.72 N 

x 

( ) 

F =− 600 N sin 5°=− 

52.293 N 

y 

R 

R 

x 

y 

=Σ F = 914.49 N 

x 

=Σ F = 448.80 N 

y 

( ) ( ) 

2 2 

R = 914.49 N + 448.80 N = 1018.68 N 

Further: 

tanα = 

448.80 

914.49 

−1 448.80 

α = tan = 26.1° 

914.49 

R = 1019 N 26.1° ! 






Σ F x : 

R 

x 

=Σ F 

x 

( ) ( ) ( ) 

R = 600 N cos50° + 300 N cos85° − 700 N cos50° 

x 

R =− 38.132 N 

x 

Σ F y : 

R 

y 

=Σ F 

y 

( ) ( ) ( ) 

R = 600 N sin 50°+ 300 N sin85° + 700 N sin50° 

y 

R y = 

1294.72 N 

( 38.132 N) ( 1294.72 N) 

R = − + 

R = 1295 N 

1294.72 N 

tanα = 

38.132 N 

α = 88.3° 

2 2 

R = 1.295 kN 88.3° ! 






We have: 

84 12 3 

Rx =Σ Fx =− TBC 

+ − 

116 13 5 

or R =− 0.72414T 

+ 84 lb 

and 

x 

BC 

( 156 lb) ( 100 lb) 

80 5 4 

Ry =Σ Fy = TBC 

− − 

116 13 5 

( 156 lb) ( 100 lb) 

R 

y 

= 0.68966T 

− 140 lb 

BC 

(a) 

So, for R to be vertical, 

R 

x 

=− 0.72414T 

+ 84 lb = 0 

BC 

T = 116.0 lb ! 

BC 

(b) Using 

T BC = 

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, R x = 0 

Then, R =Σ F = 0 

x 

x 

( ) ( ) ( ) ( ) 

600 N cosα + 300 N cos α + 35° − 700 N cosα 

= 0 

Expanding: ( ) 

3 cosαcos35°− sinαsin 35° − cosα 

= 0 

Then: 

tanα 

= 

⎛1 

⎞ 

cos35°− ⎜ ⎟ 

⎝3 

⎠ 

sin 35° 

α = 40.265° 

α = 40.3°! 

(b) Since R is vertical, R = Ry 

Then: R = Ry = Σ Fy 

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 

(a) Setting R y = 0 in Equation (2): 

( ) ( ) 

R =Σ F = 300 N + 400 N cosα + 600 N sinα 

(1) 

x 

x 

( ) ( ) 

R =Σ F = 400 N sinα − 600 N cosα 

(2) 

y 

y 

Thus 

600 

tanα = = 1.5 

400 

α = 56.3°! 

(b) Substituting for α in Equation (1): 

( ) ( ) 

R = 300 N + 400 N cos56.3° + 600 N sin 56.3° 

x 

R x = 

1021.11 N 

R = R x = 1021 N ! 






(a) Require R =Σ F = 0: 

y 

y 

( ) ( ) 

900 lb cos 25°+ 1200 lb sin 35°− sin 65°= 

0 

or 

T = 1659.45 lb 

AE 

T AE 

T = 1659 lb ! 

AE 

(b) 

R =Σ F 

x 

R =− ( 900 lb) sin 25°− ( 1200 lb) cos35°− ( 1659.45 lb) 

cos65° 

R = 2060 lb ! 






Free-Body Diagram 

Force Triangle 

Law of Sines: 

FAC 

TBC 

400 lb 

= = 

sin 25° sin 60° sin 95° 

(a) 

400 lb 

F AC = sin 25 ° = 169.691 lb 

sin 95° 

F = 169.7 lb ! 

AC 

(b) 

400 

T BC = sin 60 ° = 347.73 lb 

sin 95° 

T = 348 lb ! 

BC 






Free-Body Diagram: 

Σ F x = 

0: 

4 21 

− TCA 

+ TCB 

= 0 

5 29 

or 

T 

CB 

⎛29 ⎞⎛4 

⎞ 

= ⎜ T 

21 ⎟⎜ 

5 ⎟ 

⎝ ⎠⎝ ⎠ 

CA 

3 20 

Σ F y = 0: TCA 

+ TCB 

− ( 3 kN) 

= 0 

5 29 

3 20⎛29 4 ⎞ 

CA + CA 3 kN 0 

5 29 

⎜ × − = 

21 5 

⎟ 

⎝ ⎠ 

Then T 

T ( ) 

or 

T CA = 

2.2028 kN 

(a) 

(b) 

T = 2.20 kN ! 

CA 

T = 2.43 kN ! 

CB 







Σ = 0: − F sin50° + F sin 70° = 0 

F y 

B 

C 

F 

C 

sin 50° 

= 

sin 70° 

( F ) 

B 

Σ = 0: − F cos50° − F cos70° + 940 N = 0 

F x 

B 

C 

sin 50 

FB 

⎢ 

⎡ cos50°+ cos70° ⎛ 

⎜ 

° ⎞ 

⎟⎥ 

⎤ = 940 

⎣ 

⎝sin 70° 

⎠⎦ 

F B = 

1019.96 N 

F C 

sin 50° 

= 

sin 70° 

( 1019.96 N ) 

or 

F = 831 N ! 

C 

F = 1020 N ! 

B 







Σ = 0: − T cos25° − T cos 40° + ( 70 lb) 

cos10° = 0 

(1) 

F x 

AB 

AC 

Σ = 0: T sin 25°− T sin 40°+ ( 70 lb) 

sin10°= 0 

(2) 

F y 

AB 

Solving Equations (1) and (2) simultaneously: 

AC 

(a) T = 38.6 lb ! 

AB 

(b) T = 44.3 lb ! 

AC 







(a) Σ F x = 0: − TAB 

cos30° + Rcos65° = 0 

cos30 

R = 

° T AB 

cos65° 

Σ = 0: − T sin 30° + Rsin 65° − ( 550 N) 

= 0 

F y 

AB 

T AB 

⎛ cos30° 

⎞ 

⎜− sin 30° + sin 65° − 550 = 0 

cos65° 

⎟ 

⎝ 

⎠ 

or 

T = 405 N ! 

AB 

cos30° 

cos65° 

(b) R = ( 450 N ) 

or 

R = 830 N ! 






Free-Body Diagram At B: 

12 17 

Σ Fx = 0: − TBA + TBC 

= 0 

13 293 

or T = 1.07591 T 

BA 

BC 

Σ F y = 

T 

T 

T 

0: 

BC 

BC 

BC 

or 

5 2 

TBA 

+ TBC 

− 300 N = 0 

13 293 

⎛ 5 ⎞ 293 

= ⎜300 

− TBA 

⎟ 

⎝ 13 ⎠ 2 

= 2567.6 − 3.2918T 

BA 

( T ) 

= 2567.6 − 3.2918 1.07591 

T BC = 

565.34 N 

BC 

Free-Body Diagram At C: 

17 24 

Σ F x = 0: − T 0 

293 

BC + T 

25 

CD = 

T CD 

17 ⎛25 

⎞ 

= ( 565.34 N) 

⎜ 293 

24 

⎟ 

⎝ ⎠ 

T CD = 

584.86 N 

2 7 

Σ F y = 0: − T 0 

293 

BC + T 

25 

CD − W C = 

2 7 

W C =− + 

293 

25 

( 565.34 N) ( 584.86 N) 

or 

W = 97.7 N ! 

C 







Σ F x = 

0: 

−8 kips + 15 kips − T D cos 40° = 0 

T D = 

9.1378 kips 

Σ = 0: ( 9.1378 kips) 

sin 40°− = 0 

F y 

T C 

T = 9.14 kips ! 

D 

T = 5.87 kips ! 

C 







Σ F y = 

0: 

− 9 kips + TD 

sin 40° = 0 

T = 14.0015 kips 

D 

T = 14.00 kips ⊳ 

D 

( ) 

Σ F x = 

−6 kips + − 14.0015 kips cos 40° = 0 

T B 

0: 

T = 16.73 kips 

B 

T = 16.73 kips ⊳ 

B 







Σ = 0: F + ( 2.3 kN) sin15° − ( 2.1 kN) 

cos15° = 0 

F x 

C 

Σ = 0: F − ( 2.3 kN) cos15° + ( 2.1 kN) 

sin15° = 0 

F y 

D 

or 

or 

F = 1.433 kN ⊳ 

C 

F = 1.678 kN ⊳ 

D 







Σ = 0: − F B cos15° + 2.4 kN + ( 1.9kN) 

sin15° = 0 

F x 

or 

F B = 

2.9938 kN 

Σ = 0: F − ( 1.9 kN) cos15° + ( 2.9938 kN) 

sin15° = 0 

F y 

D 

F = 2.99 kN ⊳ 

B 

F = 1.060 kN ⊳ 

D 






From Similar Triangles we have: 

L 

2 

( 2.5 m) ( 8 L) ( 5.45 m) 

2 2 2 

− = − − 

− 6.25 = 64 −16 L − 29.7025 

or 

L = 2.5342 m 

And 

5.45 m 

cos β = 

8 m − 2.5342 m 

or β = 4.3576° 

Then 

cosα = 

2.5 m 

2.5342 m 

or α = 9.4237° 


Σ F x = 

0: 

or 

( ) 

−T 

cosα − 35 N cosα + T cos β = 0 

T ABC 

ABC 

( ) 

ABC 

35 cos9.4237° 

= 

cos4.3576°− cos9.4237° 

T ABC = 

3255.9 N 

Σ F y = 

0: 

( ) 

T sinα + 35 N sinα + T sin β − W = 0 

ABC 

ABC 

( ) ( ) 

sin 9.4237° 3255.9 N + 35 N + 3255.9 N sin 4.3576° − W = 0 

or W = 786.22 N 

(a) 

(b) 

W = 786 N " 

T = 3.26 kN " 

ABC 






From Similar Triangles we have: 

L 

2 

( 3 m) ( 8 L) ( 4.95 m) 

2 2 2 

− = − − 

− 9 = 64 −16 L − 24.5025 

or 

L = 3.0311 m 

Then 

4.95 m 

cos β = 

8 m − 3.0311 m 

or β = 4.9989° 

And 

3m 

cosα = 

3.0311 m 

or α = 8.2147° 


(a) Σ F x = 0: 

−T cosα − T cosα + T cos β = 0 

or 

ABC DE ABC 

T 

DE 

cos β − cosα 

= 

T 

cosα 

Σ F y = 

0: 

ABC 

( ) 

T sinα + T sinα + T sin β − 720 N = 0 

ABC DE ABC 

Substituting for α and β gives 

T ABC 

T ABC 

⎡ 

⎛cos 

β − cosα 

⎞ ⎤ 

⎢sinα + sinα⎜ 

⎟ + sin β⎥= 

720 

⎣ 

⎝ cosα 

⎠ ⎦ 

T ABC 

= 

( 720) 

cos 

sin ( α + β ) 

( 720) 

cos8.2147° 

( °+ ° ) 

= 

sin 8.2147 4.9989 

α 

T ABC = 

3117.5 N 

or 

T = 3.12 kN " 

ABC 

cos 4.9989°− cos8.2147° 

3117.5 N 

cos8.2147° 

(b) = 

( ) 

T DE 

T DE = 

20.338 N 



© 2007 The McGraw-Hill Companies. 

or 

T = 20.3 N " 

DE




3 15 15 

Σ F x = 0: − TAC 

+ TBC 

− ( 150 lb ) = 0 

5 17 17 

or 

17 

− TAC + 5TBC 

= 750 

(1) 

5 

4 8 8 

Σ F y = 0: TAC 

+ TBC 

− ( 150 lb ) − 190 lb = 0 

5 17 17 

or 

17 

T 

5 

+ 2T 

= 1107.5 

(2) 

AC BC 

Then adding Equations (1) and (2) 

7T BC = 1857.5 

or 

T BC = 

265.36 lb 

Therefore (a) T AC = 169.6 lb ! 

(b) 

T = 265 lb ! 

BC 







3 15 15 

Σ F x = 0: − TAC 

+ TBC 

− ( 150 lb ) = 0 

5 17 17 

or 

17 

− TAC 5 

+ 5TBC 

= 750 

(1) 

4 8 8 

Σ F y = 0: TAC 

+ TBC 

− ( 150 lb ) − W = 0 

5 17 17 

or 

17 17 

TAC 

+ 2TBC 

= 300 + W 

5 4 

(2) 

Adding Equations (1) and (2) gives 

17 

7TBC 

= 1050 + W 

4 

or 

17 

TBC 

= 150 + W 

28 

Using Equation (1) 

17 ⎛ 17 ⎞ 

− TAC 

+ 5 150 W 750 

5 

⎜ + = 

28 

⎟ 

⎝ ⎠ 

or 

25 

TAC 

= W 

28 

Now for T ≤ 240 lb ⇒ 

25 

TAC 

:240= 

W 

28 

or 

W = 269 lb 

17 

TBC 

: 240 = 150 + W 

28 

or 

W = 148.2 lb 

Therefore 0 ≤ W ≤ 148.2 lb ! 






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: 

or 

and 

Then: 

or 

4 35 12 

Σ Fx 

= 0: − ( 3W) + ( W) 

+ Fs 

= 0 

5 37 37 

Fs 

= 4.4833W 

3 12 35 

Σ Fy 

= 0: ( 3W) + ( W) 

+ Fs 

− 400 N = 0 

5 37 37 

3 ( 3W) + 12 ( W) + 35 ( 4.4833W) 

− 400 N = 0 

5 37 37 

W = 62.841 N 

and 

or 

(a) 

(b) Have spring force 

Where 

and 

F s = 

281.74 N 

( ) 

F = k L − L 

s AB O 

( ) 

F = k L − L 

AB AB AB O 

W = 62.8 N ⊳ 

( ) ( ) 

2 2 

L = 0.360 m + 1.050 m = 1.110 m 

AB 

So: 

L = 758 mm ⊳ 

O 

( ) 

281.74 N = 800 N/m 1.110 − m 

L O 

or 






Free-Body Diagram At A: 

First Note ... 

2 2 

With L = ( 22 in. ) + ( 16.5 in. ) 

AB 

L AB = 

AD 

L AD = 

F = k L − L 

Then AB AB ( AB O ) 

(a) F x 0: 

27.5 in. 

( 30 in. ) ( 16 in. ) 

L = + 

34 in. 

2 2 

= ( 9 lb/in. )( 27.5 in. − 22.5 in. ) 

= 45 lb 

= ( − ) 

= ( 3 lb/in. )( 34 in. − 22.5 in. ) 

F k L L 

AD AD AD O 

= 34.5 lb 

4 7 15 

− 45 lb + T AC + 34.5lb = 0 

5 25 17 

or T = 19.8529 lb 

Σ = ( ) ( ) 

(b) F y 0: 

Σ = ( ) ( ) ( ) 

AC 

3 24 8 

45 lb + 19.8529 lb + 34.5 lb − W = 0 

5 25 17 

T = 19.85 lb ! 

AC 

W = 62.3 lb ! 






(a) For T AB to be a minimum 

TAB 

must be perpendicular to 

∴ α + 10° = 60° 

TAC 

T = 70 lb sin 30° 

(b) Then ( ) 

AB 

or α = 50.0° 

or 

T = 35.0 lb 

AB 






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. 



Force triangle is isoceles with 2β = 180°− 85° 

β = 47.5° 

P = 2( 900 N) 

cos47.5° = 1216 N 

Since P > 0, solution is correct 

(a) 

P = 1216 N ! 

α = 180°− 55° − 47.5° = 77.5° 

(b) α = 77.5°! 






Note: Refer to Note in Problem 2.60 



(a) Law of Cosines 

P 

2 

2 2 

( ) ( ) ( )( ) 

= 1400 N + 700 N − 2 1400 N 700 N cos85° 

or 

P = 1510 N ! 

(b) Law of Sines 

sin β sin85° 

= 

1400 N 1510 N 

sin β = 0.92362 

β = 67.461° 

α = 180° − 55° − 67.461° 

or α = 57.5°! 






Free-Body Diagram At C: Σ F x = 0: 

2T 

− 1200 N = 0 

x 

T x 

= 600 N 

( ) ( ) 2 

2 2 

x + y = 

T T T 

2 2 

2 

y 

( 600 N) + ( T ) = ( 870 N) 

T y 

= 630 N 

By similar triangles: 

AC 1.8 m 

= 

870 N 630 N 

AC = 2.4857 m 

L = 2( AC ) 

L = 2( 2.4857 m) 

L = 4.97 m 

L = 4.97 m " 






T BC must be perpendicular to F AC to be as small as possible. 

Free-Body Diagram: C 

Force Triangle is a Right Triangle 

(a) We observe: α = 55° 

α = 55°! 

(b) ( ) 

T = 400 lb sin 60° 

BC 

or 

T = 346.41 lb T = 346 lb ! 

BC 

BC 






At Collar A ... Have F = k( L′ 

− L ) 

For stretched length 

For unstretched length 

s AB AB 

( 12 in. ) ( 16 in. ) 

L ′ = + 

AB 

L′ 

AB = 

20 in. 

L = 12 2 in. 

AB 

2 2 

Then F s = ( − ) 

4 lb/in. 20 12 2 in. 

F s 

= 12.1177 lb 

For the collar ... 

Σ F y = 

0 

4 

− W + ( 12.1177 lb ) = 0 

5 

W = 9.69 lb ! 






At Collar A ... 

Σ F y = 

0: 

h 

− 9lb+ F 0 

2 2 

s = 

12 + h 

or 

hF = 9 144 + h 

s 

Now F = k( L′ 

− L ) 

Where the stretched length 

2 

s AB AB 

( ) 2 2 

12 in. 

L′ = + h 

AB 

L AB 

= 12 2 in. 

Then 

hF = 9 144 + h 

s 

Becomes 

⎡ 

⎢ ( ) 

2 

h 3 lb/in. 144 + − 12 2 

⎤ 

= 9 144 + 

⎣ 

h ⎥⎦ 

h 

h − 3 144 + h = 12 2 h 

or ( ) 

2 

2 2 

Solving Numerically ... 

h = 16.81 in. ⊳ 






Free-Body Diagram: B (a) Have: T + F + T = 0 

where magnitude and direction of 

of 

F AB is known. 

BD AB BC 

T BD are known, and the direction 

Then, in a force triangle: 

By observation, T BC is minimum when α = 90.0°⊳ 

(b) Have T = ( 310 N) sin ( 180° − 70° − 30° 

) 

BC 

= 305.29 N 

TBC 

= 305 N ⊳ 







Since T 

AB 

= T = 140 lb, Force triangle is isosceles: 

BC 

With 2β + 75° = 180° 

β = 52.5° 

Then α = 90° − 52.5° − 30° 

α = 7.50° 

P 

( 140 lb) 

cos52.5 

2 = ° 

P = 170.453 lb 

P = 170.5 lb 7.50° ⊳ 






Free-Body Diagram of Pulley 

(a) 

(b) 

(c) 

(d) 

2 

( )( ) 

Σ F = 0: 2T 

− 280 kg 9.81 m/s = 0 

y 

T = 

1 

( 2746.8 N ) 

2 

2 

( )( ) 

Σ F = 0: 2T 

− 280 kg 9.81 m/s = 0 

y 

T = 

1 

( 2746.8 N ) 

2 

2 

( )( ) 

Σ F = 0: 3T 

− 280 kg 9.81 m/s = 0 

y 

1 

T = ( 2746.8 N ) 

3 

2 

( )( ) 

Σ F = 0: 3T 

− 280 kg 9.81 m/s = 0 

y 

1 

T = ( 2746.8 N ) 

3 

T = 1373 N ⊳ 

T = 1373 N ⊳ 

T = 916 N ⊳ 

T = 916 N ⊳ 

(e) 

2 

( )( ) 

Σ F = 0: 4T 

− 280 kg 9.81 m/s = 0 

y 

T = 

1 

( 2746.8 N ) 

4 

T = 687 N ⊳ 






Free-Body Diagram of Pulley and 

Crate 

(b) 

2 

( )( ) 

Σ F = 0: 3T 

− 280 kg 9.81 m/s = 0 

y 

1 

T = ( 2746.8 N ) 

3 

T = 916 N ⊳ 

(d) 

2 

( )( ) 

Σ F = 0: 4T 

− 280 kg 9.81 m/s = 0 

y 

T = 

1 

( 2746.8 N ) 

4 

T = 687 N ⊳ 






Free-Body Diagram: Pulley C 

(a) F T ( ) ( ) 

Σ = 0: cos30° − cos50° − 800 N cos50° = 0 

x 

ACB 

Hence 

T ACB = 

2303.5 N 

(b) ( ) ( ) 

Σ F = 0: T sin 30°+ sin 50° + 800 N sin 50°− Q = 0 

y 

ACB 

TACB 

= 2.30 kN ⊳ 

( )( ) ( ) 

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 

( ) 

Σ F = 0: T cos30° − cos50° − Pcos50° = 0 

x 

ACB 

or P = 0.34730T ACB 

(1) 

( ) 

Σ F = 0: T sin30° + sin50° + Psin50° − 2000 N = 0 

y 

ACB 

or 1.26604T + 0.76604P = 2000 N 

(2) 

(a) Substitute Equation (1) into Equation (2): 

ACB 

( ) 

1.26604T + 0.76604 0.34730T 

= 2000 N 

ACB 

ACB 

Hence: 

T ACB 

= 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 = 230lbcos25 

( ) ° 

Q = 54.378 lb 

Equivalent loading at A: 

Law of Cosines: 

2 2 2 

( ) ( ) ( ) ( )( ) ( α) ( α) 

120 lb = 100 lb + 54.378 lb − 2 100 lb 54.378 lb cos 125° − cos 125° − = − 0.132685 

This gives two values: 125°− α = 97.625° 

Thus for R < 120 lb: 

α = 27.4° 

125°− α =− 97.625° 

α = 223° 

27.4°< α < 223°! 






(a) ( ) 

F = 950 lb sin 50° cos 40° 

x 

= 557.48 lb 

F y 

=− ( 950 lb) 

cos50° 

F = 557 lb ! 

x 

=− 610.65 lb 

F =−611 lb ! 

y 

( ) 

F = 950 lb sin50° sin 40° 

z 

= 467.78 lb 

F = 468 lb ! 

z 

(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) F ( ) 

x 

=− 810 lb cos 45° sin 25° 

=− 242.06 lb 

F y 

=− ( 810 lb) 

sin 45° 

F =−242 lb ! 

x 

=− 572.76 lb 

F =−573 lb ! 

y 

F z 

= ( 810 lb) 

cos 45° cos 25° 

= 519.09 lb 

F = 519 lb ! 

z 

(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) F ( ) 

x 

= 900 N cos30° cos 25° 

= 706.40 N 

F y 

= ( 900 N) 

sin 30° 

F = 706 N ! 

x 

= 450.00 N 

F = 450 N ! 

y 

F z 

=− ( 900 N) 

cos30° sin 25° 

=− 329.04 N 

F =−329 N ! 

z 

(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) ( ) 

F =− 1900 N sin 20° sin 70° 

x 

=− 610.65 N 

F =− 611 N ! 

x 

( ) 

F = 1900 N cos 20° 

y 

= 1785.42 N 

F = 1785 N ! 

y 

( ) 

F = 1900 N sin 20° cos70° 

z 

= 222.26 N 

F = 222 N ! 

z 

(b) 

cos 

θ x 

= 

−610.65 N 

1900 N 

or θ x = 108.7°! 

1785.42 N 

cosθ y = 

1900 N 

or θ y = 20.0°! 

cosθ = 

z 

222.26 N 

1900 N 

or θ z = 83.3°! 






(a) ( ) 

F = 180 lb cos35° sin 20° 

x 

= 50.430 lb 

( ) 

F =− 180 lb sin 35° 

y 

F = 50.4 lb ! 

x 

=− 103.244 lb 

F =− 103.2 lb ! 

y 

( ) 

F = 180 lb cos35° cos 20° 

z 

= 138.555 lb 

F = 138.6 lb ! 

z 

(b) 

50.430 lb 

cosθ x = 

180 lb 

or θ x = 73.7°! 

cos 

θ y 

= 

−103.244 lb 

180 lb 

or θ y = 125.0°! 

138.555 lb 

cosθ z = 

180 lb 

or θ z = 39.7°! 






(a) ( ) 

F x = 180 lb cos30° cos 25° 

= 141.279 lb 

(b) 

( ) 

F =− 180 lb sin 30° 

y 

=− 90.000 lb 

( ) 

F z = 180 lb cos30° sin 25° 

= 65.880 lb 

141.279 lb 

cosθ x = 

180 lb 

cos 

θ y 

−90.000 lb 

= 

180 lb 

F = 141.3 lb ! 

x 

F =− 90.0 lb ! 

y 

F = 65.9 lb ! 

z 

or θ x = 38.3°! 

or θ y = 120.0°! 

cosθ = 

z 

65.880 lb 

180 lb 

or θ z = 68.5°! 






(a) ( ) 

F = − 220 N cos60° cos35° 

x 

= − 90.107 N 

F =−90.1 N 

x 

( ) 

F = 220 N sin 60° 

y 

= 190.526 N 

F y = 190.5 N 

( ) 

F = − 220 N cos60° sin 35° 

z 

= − 63.093 N 

F =−63.1 N 

z 

(b) 

cos 

θ x 

= 

− 90. 

107 Ν 

220 N 

θ = 114.2° 

x 

190.526 N 

cosθ y = 

220 N 

θ = 30.0° 

y 

cos 

θ z 

= 

−63.093 N 

220 N 

θ = 106.7° 

z 






(a) 

F = 180 N 

x 

With 

Fx 

= Fcos60° cos35° 

180 N = F cos60° cos35° 

or F = 439.38 N 

F = 439 N ! 

(b) 

cosθ = 

x 

180 N 

439.48 N 

θ = 65.8°! 

x 

( ) 

F = 439.48 N sin 60° 

y 

F y = 

380.60 N 

380.60 N 

cosθ y = 

439.48 N 

θ = 30.0°! 

y 

( ) 

F =− 439.48 N cos60° sin 35° 

z 

F =− 126.038 N 

z 

cos 

θ z 

= 

−126.038 N 

439.48 N 

θ = 106.7°! 

z 






2 2 2 

x y z 

F = F + F + F 

( 65 N) ( 80 N) ( 200 N) 

2 2 2 

F = + − + − 

F = 225 N ! 

Fx 

cosθ x = = 

F 

65 N 

225 N 

θ = 73.2°! 

x 

cosθ 

y 

Fy 

= = 

F 

−80 N 

225 N 

θ = 110.8°! 

y 

Fz 

cosθ 

z = = 

F 

− 200 N 

225 N 

θ = 152.7°! 

z 






2 2 2 

x y z 

F = F + F + F 

( 450 N) ( 600 N) ( 1800 N) 

2 2 2 

F = + + − 

Fx 

450 N 

cosθ x = = 

F 1950 N 

Fy 

600 N 

cosθ y = = 

F 1950 N 

F = 1950 N ! 

θ = 76.7°! 

x 

θ = 72.1°! 

y 

Fz 

cosθ 

z = = 

F 

−1800 N 

1950 N 

θ = 157.4°! 

z 






2 2 

2 

x y z 

(a) We have ( θ ) ( θ ) ( θ ) 

cos + cos + cos = 1 

( cosθy) = 1 − ( cosθx) − ( cosθz) 

2 2 2 

Since F y < 0 we must have cosθ y < 0 

2 2 

Thus cosθ =− 1 − ( cos 43.2° ) − cos( 83.8° 

) 

y 

cosθ y =− 0.67597 

θ = 132.5°! 

y 

(b) Then: 

F 

= 

F 

y 

cosθ 

y 

F 

−50 lb 

= − 0.67597 

73.968 lb 

F = 

And 

F = Fcosθ 

x 

x 

( ) 

F = 73.968 lb cos 43.2° 

x 

F = 53.9 lb ! 

x 

F = Fcosθ 

z 

z 

( ) 

F = 73.968 lb cos83.8° 

z 

F = 7.99 lb ! 

z 

F = 74.0 lb ! 






2 2 

2 

x y z 

(a) We have ( θ ) ( θ ) ( θ ) 

cos + cos + cos = 1 

2 

2 

( θ ) = − ( θ ) − ( θ ) 2 

or cos 1 cos cos 

z x y 

Since F z < 0 we must have cosθ z < 0 

2 2 

Thus cosθ =− 1 − ( cos113.2° ) − cos( 78.4° 

) 

z 

cosθ z =− 0.89687 

θ = 153.7°! 

z 

(b) Then: 

F 

Fz 

−35 lb 

= = 

cosθ 

− 0.89687 

z 

F = 39.025 lb 

And 

F = Fcosθ 

x 

x 

( ) 

F = 39.025 lb cos113.2° 

x 

F =− 15.37 lb ! 

x 

F = Fcosθ 

y 

y 

( ) 

F = 39.025 lb cos78.4° 

y 

F = 7.85 lb ! 

y 

F = 39.0 lb ! 






(a) We have F = Fcosθ 

y 

y 

( ) 

F = 250 N cos 72.4° 

y 

F y = 

75.592 N 

F = 75.6 N ! 

y 

Then 

2 2 2 2 

= x + y + z 

F F F F 

( ) ( ) ( ) 

2 2 2 2 

250 N = 80 N + 75.592 N + Fz 

F z = 

224.47 N 

F = 224 N ! 

z 

Fx 

(b) cosθ x = 

F 

cosθ = 

x 

80 N 

250 N 

θ = 71.3°! 

x 

cosθ = 

z 

cosθ = 

z 

Fz 

F 

224.47 N 

250 N 

θ = 26.1°! 

z 






(a) Have F = Fcosθ 

x 

x 

( ) 

F = 320 N cos104.5° 

x 

F =− 80.122 N 

x 

F =− 80.1 N ! 

x 

Then: 

2 2 2 2 

= x + y + z 

F F F F 

2 

( 320 N) = ( − 80.122 N) + + ( − 120 N) 

2 2 F 2 

y 

F y = 

285.62 N 

F = 286 N ! 

y 

(b) 

cosθ = 

y 

cosθ = 

y 

F 

y 

F 

285.62 N 

320 N 

θ = 26.8°! 

y 

cosθ = 

z 

Fz 

F 

cos 

θ z 

= 

−120 N 

320 N 

θ = 112.0°! 

z 






!!!" 

DB = i − j − k 

( 36 in. ) ( 42 in. ) ( 36 in. ) 

( ) ( ) ( ) 

2 2 2 

DB = 36 in. + − 42 in. + − 36 in. = 66 in. 

T 

= T λ = T 

DB DB DB DB 

!!!" 

DB 

DB 

T 55 lb 

( 36 in. ) ( 42 in. ) ( 36 in. ) 

DB = 

66 in. 

⎡ ⎣ i − j − k ⎤ ⎦ 

( 30 lb) ( 35 lb) ( 30 lb) 

= i − j − k 

T = ! 

∴ ( ) 30.0 lb 

DB 

x 

( ) 35.0 lb 

T =− ! 

DB 

y 

( ) 30.0 lb 

T =− ! 

DB 

z 






!!!" 

EB = i − j + k 

( 36 in. ) ( 45 in. ) ( 48 in. ) 

( ) ( ) ( ) 

2 2 2 

EB = 36 in. + − 45 in. + 48 in. = 75 in. 

T 

= T λ = T 

EB EB EB EB 

!!!" 

EB 

EB 

T 60 lb 

( 36 in. ) ( 45 in. ) ( 48 in. ) 

EB = 

75 in. 

⎡ ⎣ i − j + k ⎤ ⎦ 

( 28.8 lb) ( 36 lb) ( 38.4 lb) 

= i − j + k 

T = ! 

∴ ( ) 28.8 lb 

EB 

x 

( ) 36.0 lb 

T =− ! 

EB 

y 

( ) 38.4 lb 

T = ! 

EB 

z 






!!!" 

BA = i + j − k 

( 4 m) ( 20 m) ( 5 m) 

( ) ( ) ( ) 

2 2 2 

BA = 4 m + 20 m + − 5 m = 21 m 

!!!" 

F BA 2100 N 

= Fλ 

BA = F = ⎡( 4 m ) + ( 20 m ) − ( 5 m ) ⎤ 

BA 21 m 

⎣ i j k ⎦ 

( 400 N) ( 2000 N) ( 500 N) 

F = i + j − k 

F =+ 400 N, F =+ 2000 N, F =− 500 N ! 

x y z 






!!!" 

DA = i + j + k 

( 4 m) ( 20 m) ( 14.8 m) 

( ) ( ) ( ) 

2 2 2 

DA = 4 m + 20 m + 14.8 m = 25.2 m 

!!!" 

F DA 1260 N 

= Fλ 

DA = F = ⎡( 4 m ) + ( 20 m ) + ( 14.8 m ) ⎤ 

DA 25.2 m 

⎣ i j k ⎦ 

( 200 N) ( 1000 N) ( 740 N) 

F = i + j + k 

F =+ 200 N, F =+ 1000 N, F =+ 740 N ! 

x y z 






uuuv 

BG =− i + j − k 

( 1 m) ( 1.85 m) ( 0.8 m) 

( 1 m) ( 1.85 m) ( 0.8 m) 

2 2 2 

BG = − + + − 

BG = 2.25 m 

T 

= T λ = T 

BG BG BG BG 

uuuv 

BG 

BG 

T 450 N 

( 1 m) ( 1.85 m) ( 0.8 m) 

BG = 

2.25 m 

⎡ ⎣− i + j − k ⎤ ⎦ 

( 200 N) ( 370 N) ( 160 N) 

=− i + j − k 

∴ ( T BG ) x 

= −200 N ⊳ 

( ) 370 N 

T = ⊳ 

BG 

y 

( T BG ) z 

=−160.0 N ⊳ 






uuuuv 

BH = i + j − k 

( 0.75 m) ( 1.5 m) ( 1.5 m) 

( 0.75 m) ( 1.5 m) ( 1.5 m) 

2 2 2 

BH = + + − 

T 

= 2.25 m 

= T λ = T 

BH BH BH BH 

uuuuv 

BH 

BH 

T 600 N 

( 0.75 m ) ( 1.5 m ) ( 1.5 m ) 

BH = 

2.25 m 

⎡ ⎣ i + j − k ⎤ ⎦ 

( 200 N) ( 400 N) ( 400 N) 

= i + j − k 

∴ ( T BH ) x 

= 200 N ⊳ 

( ) 400 N 

T = ⊳ 

BH 

y 

( ) 400 N 

T =− ⊳ 

BH 

z 






( 4 kips)[ cos30 sin 20 sin 30 cos30 cos 20 ] 

( 1.18479 kips) i ( 2 kips) j ( 3.2552 kips) 

k 

( 8 kips)[ cos 45 sin15 sin 45 cos 45 cos15 ] 

( 1.46410 kips) i ( 5.6569 kips) j ( 5.4641 kips) 

k 

( 0.27931 kip) ( 3.6569 kips) ( 2.2089 kips) 

P = ° ° i − ° j + ° ° k 

= − + 

Q = − ° ° i + ° j − ° ° k 

=− + − 

R = P + Q = − i + j − 

k 

( 0.27931 kip) ( 3.6569 kips) ( 2.2089 kips) 

2 2 2 

R = − + + − 

R = 4.2814 kips 

or R = 4.28 kips ⊳ 

Rx 

−0.27931 kip 

cosθ 

x = = = − 0.065238 

R 4.2814 kips 

Ry 

3.6569 kips 

cosθ y = = = 0.85414 

R 4.2814 kips 

Rz 

− 2.2089 kips 

cosθz 

= = = − 0.51593 

R 4.2814 kips 

or θ x = 93.7°⊳ 

θ = 31.3°⊳ 

y 

θ = 121.1°⊳ 

z 






( 6 kips)[ cos30 sin 20 sin 30 cos30 cos 20 ] 

P = ° ° i − ° j + ° ° k 

( 1.77719 kips) ( 3 kips) ( 4.8828 kips) 

= i − j + 

k 

( 7 kips)[ cos 45 sin15 sin 45 cos 45 cos15 ] 

Q = − ° ° i + ° j − ° ° k 

( 1.28109 kips) ( 4.94975 kips) ( 4.7811 kips) 

( 0.49610 kip) ( 1.94975 kips) ( 0.101700 kip) 

=− i + j − 

k 

R = P + Q = i + j + 

k 

( 0.49610 kip) ( 1.94975 kips) ( 0.101700 kip) 

2 2 2 

R = + + 

R = 2.0144 kips 

or R = 2.01 kips ⊳ 

Rx 

0.49610 kip 

cosθ x = = = 0.24628 

R 2.0144 kips 

Ry 

1.94975 kips 

cosθ y = = = 0.967906 

R 2.0144 kips 

Rz 

0.101700 kip 

cosθ z = = = 0.050486 

R 2.0144 kips 

or θ x = 75.7°⊳ 

θ = 14.56°⊳ 

y 

θ = 87.1°⊳ 

z 






AB 

uuur 

AB =− i + j + k 

( 600 mm) ( 360 mm) ( 270 mm) 

( 600 mm) ( 360 mm) ( 270 mm) 

2 2 2 

AB = − + + 

AB = 750 mm 

uuuv 

AC =− 600 mm i + 320 mm j − 510 mm k 

( ) ( ) ( ) 

( 600 mm) ( 320 mm) ( 510 mm) 

2 2 2 

AC = − + + − 

AC = 850 mm 

uuur 

T AB 510 N 

AB = TAB 

= ⎡− ( 600 mm) + ( 360 mm) + ( 270 mm) 

⎤ 

AB 750 mm 

⎣ i j k ⎦ 

( 408 N) ( 244.8 N) ( 183.6 N) 

T AB =− i + j + k 

uuur 

T AC 765 N 

AC = TAC 

= ⎡− ( 600 mm) + ( 320 mm) − ( 510 mm) 

⎤ 

AC 850 mm 

⎣ i j k ⎦ 

( 540 N) ( 288 N) ( 459 N) 

T AC =− i + j − k 

AC 

( 948 N) ( 532.8 N) ( 275.4 N) 

R = T + T = − i + j − k 

Then 

and 

cos 

R = 1121.80 N 

R = 1122 N ⊳ 

θ x 

−948 N 

= θ x = 147.7°⊳ 

1121.80 N 

532.8 N 

cosθ y = 

1121.80 N 

θ y = 61.6°⊳ 

− 275.4 N 

cosθ z = 

1121.80 N 

θ z = 104.2°⊳ 






AB 

!!!" 

AB =− i + j + k 

( 600 mm) ( 360 mm) ( 270 mm) 

( ) ( ) ( ) 

2 2 2 

AB = − 600 mm + 360 mm + 270 mm = 750 mm 

AB = 750 mm 

!!!" 

AC =− 600 mm i + 320 mm j − 510 mm k 

( ) ( ) ( ) 

( ) ( ) ( ) 

2 2 2 

AC = − 600 mm + 320 mm + − 510 mm = 850 mm 

AC = 850 mm 

!!!" 

T AB 765 N 

AB = TAB 

= ⎡− ( 600 mm) + ( 360 mm) + ( 270 mm) 

⎤ 

AB 750 mm 

⎣ i j k ⎦ 

( 612 N) ( 367.2 N) ( 275.4 N) 

T AB =− i + j + k 

!!!" 

T AC 510 N 

AC = TAC 

= ⎡− ( 600 mm) + ( 320 mm) − ( 510 mm) 

⎤ 

AC 850 mm 

⎣ i j k ⎦ 

( 360 N) ( 192 N) ( 306 N) 

T AC =− i + j − k 

AC 

( 972 N) ( 559.2 N) ( 30.6 N) 

R = T + T = − i + j − k 

Then 

R = 1121.80 N 

R = 1122 N ! 

−972 N 

cosθ x = θ x = 150.1°! 

1121.80 N 

559.2 N 

cosθ y = θ y = 60.1°! 

1121.80 N 

−30.6 N 

cosθ z = θ z = 91.6°! 

1121.80 N 






Have T ( 760 lb)( sin 50 cos 40 i cos50 j sin 50 sin 40 k ) 

AB = ° ° − ° + ° ° 

AC 

AC 

( cos 45 sin 25 sin 45 cos 45 cos 25 ) 

T = T − ° ° i − ° j + ° ° k 

(a) R = T + T 

( R ) = 0 

A AB AC A x 

( ) 

( R ) F 0: 

∴ = Σ = 

A 

x 

760 lb sin 50° cos 40° − cos 45° sin 25° = 0 

T AC 

x 

or 

T = 1492.41 lb 

AC 

∴ = 1492 lb ⊳ 

(b) ( RA) =Σ Fy 

= ( − 760 lb) cos50°− ( 1492.41 lb) 

sin 45° 

y 

( R A) y 

=− 1543.81 lb 

( RA) Fz 

( 760 lb) sin 50 sin 40 ( 1492.41 lb) 

cos 45 cos 25 

z 

( R A) z 

= 1330.65 lb 

∴ = − ( 1543.81 lb) + ( 1330.65 lb) 

Then 

T AC 

=Σ = ° °+ ° ° 

R j k 

A 

R A = 2038.1 lb 

R A = 2040 lb ⊳ 

0 

cosθ x = 

2038.1 lb 

θ x = 90.0°⊳ 

cos 

θ y 

−1543.81 lb 

= θ y = 139.2°⊳ 

2038.1 lb 

1330.65 lb 

cosθ z = θ z = 49.2°⊳ 

2038.1 lb 






Have T = T ( sin 50° cos 40° i − cos50° j + sin 50° sin 40° 

k ) 

AB 

AB 

( 980 lb)( cos 45 sin 25 sin 45 cos 45 cos 25 ) 

T AC = − ° ° i − ° j + ° ° k 

(a) R = T + T 

( R ) = 0 

A AB AC A x 

AB 

( R ) F 0: 

∴ = Σ = 

A 

x 

( ) 

T sin 50° cos 40° − 980 lb cos 45° sin 25° = 0 

x 

or 

T AB = 

499.06 lb 

∴ = 499 lb ⊳ 

T AB 

(b) ( RA) F 

y y ( ) ( ) 

( R A) y 

=− 1013.75 lb 

( RA) F 

z z ( ) ( ) 

( R A) z 

= 873.78 lb 

∴ R = − ( 1013.75 lb) j + ( 873.78 lb) 

k 

Then 

and 

A 

=Σ =− 499.06 lb cos50°− 980 lb sin 45° 

=Σ = 499.06 lb sin 50° sin 40°+ 980 lb cos 45° cos 25° 

R = 1338.35 lb 

R = 1338 lb ⊳ 

A 

0 

cosθ x = θ x = 90.0°⊳ 

1338.35 lb 

A 

cos 

θ y 

−1013.75 lb 

= θ y = 139.2°⊳ 

1338.35 lb 

873.78 lb 

cosθ z = θ z = 49.2°⊳ 

1338.35 lb 






!!!" 

Cable AB: AB =− ( 600 mm) + ( 360 mm) + ( 270 mm) 

i j k 

( ) ( ) ( ) 

2 2 2 

AB = − 600 mm + 360 mm + 270 mm = 750 mm 

!!!" 

T AB 600 N 

AB = TAB 

= ⎡− ( 600 mm) + ( 360 mm) + ( 270 mm) 

⎤ 

AB 750 mm 

⎣ i j k ⎦ 

( 480 N) ( 288 N) ( 216 N) 

T AB =− i + j + k 

!!!" 

AC =− 600 mm i + 320 mm j − 510 mm k 

Cable AC: ( ) ( ) ( ) 

( ) ( ) ( ) 

2 2 2 

AC = − 600 mm + 320 mm + − 510 mm = 850 mm 

!!!" 

T AC TAC 

AC = TAC 

= ⎡− ( 600 mm) + ( 320 mm) − ( 510 mm) 

⎤ 

AC 850 mm 

⎣ i j k ⎦ 

60 32 51 

TAC =− TACi + TACj − TACk 

85 85 85 

Load P: 

P =−P 

j 

51 

(a) ( RA) =Σ Fz = 0: ( 216 N) 

− TAC 

= 0 

or T 360 N 

z 

AC = ! 

85 

32 

RA =Σ Fy = 0: 288 N + TAC 

− P = 0 or P = 424 N ! 

y 

85 

(b) ( ) ( ) 






uuur 

Cable AB: AB =−( 4m) − ( 20m) + ( 5m) 

i j k 

( ) ( ) ( ) 

2 2 2 

AB = − 4 m + − 20 m + 5 m = 21m 

uuur 

T AB TAB 

AB = TAB 

= ⎡−( 4m) − ( 20m) + ( 5m) 

⎤ 

AB 21 m 

⎣ i j k ⎦ 

uuur 

AC = 12 m i − 20 m j + 3.6 m k 

Cable AC: ( ) ( ) ( ) 

( ) ( ) ( ) 

2 2 2 

AC = 12 m + − 20 m + 3.6 m = 23.6 m 

uuur 

T AC 1770 N ( 12 m ) ( 20 m ) ( 3.6 m 

AC = TAC 

= ⎡ − + ) ⎤ 

AC 23.6 m 

⎣ i j k ⎦ 

( 900 N) ( 1500 N) ( 270 N) 

= i − j + k 

uuur 

i j k 

Cable AD: AD =−( 4 m) − ( 20 m) + ( 14.8 m) 

Now... 

( ) ( ) ( ) 

2 2 2 

AD = − 4 m + − 20 m + 14.8 m = 25.2 m 

uuur 

T AD TAD 

AD = TAD 

= ⎡−( 4 m) − ( 20 m) + ( 14.8 m) 

⎤ 

AD 25.2 m 

⎣ i j k ⎦ 

T AD 

( 10 m) ( 50 m) ( 37 m) 

= ⎡− − − ⎤ 

63 m 

⎣ i j k ⎦ 

R = T + T + T and R = R; R = R = 0 

AB AC AD j x z 

4 10 

Σ Fx = 0: − TAB + 900 − TAD 

= 0 

(1) 

21 63 

5 37 

Σ Fy = 0: TAB + 270 − TAD 

= 0 

(2) 

21 63 

Solving equations (1) and (2) simultaneously yields: 

T = 1.775 kN ! 

AD 

T = 3.25 kN ! 

AB 






2 2 

( ) ( ) 

2 2 

( ) ( ) 

( ) ( ) ( ) 

TAB 

⎡ ( 450 mm) ( 600 mm) 

⎤ 

d = − 450 mm + 600 mm = 750 mm 

AB 

d = 600 mm + − 320 mm = 680 mm 

AC 

2 2 2 

d = 500 mm + 600 mm + 360 mm = 860 mm 

AD 

T AB = − + 

750 mm 

⎣ i j ⎦ 

T = − 0.6i + 0.8 j T 

AB 

( ) AB 

TAC 

⎡( 600 mm) ( 320 mm) 

T AC = 

680 mm 

⎣ j − k ⎤⎦ 

T ⎛15 8 ⎞ 

AC = ⎜ − TAC 

⎝ 

⎟ 

17 17 

⎠ 

T TAD 

AD = ⎡( 500 mm) + ( 600 mm) + ( 360 mm) 

⎤ 

860 mm 

⎣ ⎦ 

T ⎛25 30 18 ⎞ 

AD = ⎜ + + TAD 

⎝ 

⎟ 

43 43 43 

⎠ 

W =−W j 

At point A: Σ F = 0: T + T + T + W = 0 

i component: 

k component: 

j component: 

From Equation (1): 


AB AC AD 

25 

− 0.6TAB 

+ TAD 

= 0 

43 

or 

T 

AB 

= ⎛5⎞⎛25⎞ 

⎜ T 

⎝ 

⎟⎜ ⎟ 

3⎠⎝43⎠ 

18 18 

− TAC 

+ TAD 

= 0 

17 43 

17 18 

or TAC 

= ⎛ ⎞⎛ ⎞ 

⎜ ⎟⎜ ⎟TAD 

(2) 

⎝ 8 ⎠⎝43⎠ 

15 30 

0.8TAB + TAC + TAD 

− W = 0 

17 43 

15 ⎛17 18 ⎞ 30 

0.8TAB + ⎜ ⋅ TAD + TAD 

− W = 0 

17 ⎝ 

⎟ 

8 43 ⎠ 43 

255 

0.8TAB 

+ TAD 

− W = 0 

(3) 

172 

⎛5⎞⎛25⎞ 

6kN= ⎜ TAD 

⎝ 

⎟⎜ ⎟ 3 ⎠⎝ 43 ⎠ 

or T AD = 6.1920 kN 

AD 

255 

0.8( 6 kN) + ( 6.1920 kN) 

− W = 0 

172 

(1) 

∴ W = 13.98 kN ! 





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. 

T 

T 

AB 

AC 

⎛5⎞⎛25⎞ 

= ⎜ T 

⎝ ⎟⎜ ⎟ 

3⎠⎝43⎠ 

AD 

⎛17⎞⎛18⎞ 

= ⎜ T 

⎝ ⎟⎜ ⎟ 

8 ⎠⎝43⎠ 

AD 

(1) 

(2) 

From Equation (1) 


255 

0.8TAB 

+ TAD 

− W = 0 

(3) 

172 

T AB 

⎛5⎞⎛25⎞ 

= ⎜ 

⎝ ⎟⎜ ⎟ 

3⎠⎝43⎠ 

( 4.3 kN) 

or 

T AB = 

4.1667 kN 

255 

0.8( 4.1667 kN) + ( 4.3 kN) 

− W = 0 

172 

∴ W = 9.71 kN ! 






uuur 

AB =− 

( 4.20 m) i −( 5.60 m) 

2 2 

( ) ( ) 

( 2.40 m) ( 5.60 m) ( 4.20 m) 

AB = − 4.20 m + − 5.60 m = 7.00 m 

uuur 

AC = i − j + k 

( ) ( ) ( ) 

( 5.60 m) − ( 3.30 m) 

j 

2 2 2 

AC = 2.40 m + − 5.60 m + 4.20 m = 7.40 m 

uuur 

AD =− j k 

( ) ( ) 

2 2 

AD = − 5.60 m + − 3.30 m = 6.50 m 

uuur 

AB TAB 

TAB = TABλAB = TAB 

= ( −4.20i − 5.60j) 

AB 7.00 m 

T ⎛ 3 4 ⎞ 

AB = ⎜− − TAB 

⎝ 

⎟ 

5 i 5 

j ⎠ 

uuur 

AC TAC 

TAC = TACλAC = T AC = 2.40i − 5.60j+ 

4.20k 

AC 7.40 m 

( ) 

T ⎛12 28 21 ⎞ 

AC = ⎜ − + TAC 

⎝ 

⎟ 

37 i 37 j 37 

k ⎠ 

uuur 

AD TAD 

TAD = TADλAD = TAD 

= −5.60j − 3.30k 

AD 6.50 m 

T ⎛ 56 33 ⎞ 

AD = ⎜− − T 

⎝ 

⎟ 

65 j 65 

k ⎠ 

P = P j 

AD 

For equilibrium at point A: Σ F = 0 

TAB + TAC + TAD 

+ P = 

( ) 

0 

i component: 

3 12 

− TAB 

+ TAC 

= 0 

5 37 

or 

T 

AB 

20 

= TAC 

(1) 

37 





j component: 

4 28 56 

− TAB − TAC − TAD 

+ P = 0 

5 37 65 

4 28 56⎛65 7 ⎞ 

− TAB − TAC − ⎜ ⋅ TAC 

+ P = 0 

5 37 65⎝ 

⎟ 

11 37 ⎠ 

4 700 

− TAB 

− TAC 

+ P = 0 

(2) 

5 407 

k component: 

21 33 

TAC 

− TAD 

= 0 

37 65 

or 

T 

AD 

⎛65⎞⎛ 7 ⎞ 

= ⎜ T 

⎝ ⎟⎜ ⎟ 

11⎠⎝37⎠ 

AC 

(3) 


⎛20⎞ 

259 N = ⎜ TAC 

⎝ 

⎟ 37 ⎠ 

or T AC = 479.15 N 

4 700 

− 259 N − 479.15 N + P = 0 

5 407 

From Equation (2): ( ) ( ) 

∴ P =1031 N ! 






See Problem 2.103 for the analysis leading to the linear algebraic Equations (1), (2), and (3) 

T 

AB 

20 

= TAC 

(1) 

37 

4 700 

− TAB 

− TAC 

+ P = 0 (2) 

5 407 

T 

AD 

⎛65 ⎞⎛ 7 ⎞ 

= ⎜ T 

11 ⎟⎜ 

37 ⎟ 

⎝ ⎠⎝ ⎠ 

AC 

(3) 

Substituting for T = 444 N into Equation (1) 

AC 

20 444 N 

37 

240 N 

Gives T = ( ) 

And from Equation (3) 

or 

AB 

T AB = 

4 700 

− ( 240 N) − ( 444 N) 

+ P = 0 

5 407 

∴ P = 956 N ! 






BA 

2 2 

( 11 in. ) ( 9.6 in. ) 14.6 in. 

2 2 

( 9.6 in. ) ( 7.2 in. ) 12.0 in. 

2 2 2 

( ) ( ) ( ) 

FBA 

( 11 in. ) ( 9.6 in. ) 

d = − + = 

d = + − = 

CA 

d = 9.6 in. + 9.6 in. + 4.8 in. = 14.4 in. 

DA 

F BA = FBAλBA 

= ⎡ − + ⎤ 

14.6 in. ⎣ i j ⎦ 

11 9.6 

F ⎡ ⎛ ⎞ ⎛ ⎞ 

= ⎤ 

BA ⎢− ⎜ ⎟i 

+ ⎜ ⎟j 

⎥ 

⎣ ⎝14.6 ⎠ ⎝14.6 

⎠ ⎦ 

F FCA 

CA = FCAλCA 

= ⎡( 9.6 in. ) − ( 7.2 in. ) ⎤ 

12.0 in. ⎣ j k ⎦ 

4 3 

F ⎡ ⎛ ⎞ ⎛ ⎞ 

= ⎤ 

CA ⎢⎜ ⎟j 

− ⎜ ⎟k 

⎥ 

⎣⎝5⎠ ⎝5⎠ 

⎦ 

F FDA 

DA = FDAλDA 

= ⎡( 9.6 in. ) + ( 9.6 in. ) + ( 4.8 in. ) ⎤ 

14.4 in. ⎣ i j k ⎦ 

2 2 1 

F ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

= ⎤ 

DA ⎢⎜ ⎟i + ⎜ ⎟j + ⎜ ⎟k 

⎥ 

⎣⎝3⎠ ⎝3⎠ ⎝3⎠ 

⎦ 

P =−P j 

At point A: Σ F = 0: F + F + F + P = 0 

BA CA DA 

i 

component: 

11 2 

− ⎛ ⎜ ⎞ FBA 

+ ⎛ ⎞ FDA 

= 0 

14.6 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠ 

(1) 

j 

component: 

⎛ 9.6 ⎞ ⎛4 ⎞ ⎛2 

⎞ 

⎜ FBA FCA FDA 

P 0 

14.6 

⎟ + ⎜ + − = 

5 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

(2) 

k component: 

3 1 

− ⎛ ⎜ ⎞ FCA 

+ ⎛ ⎞ FDA 

= 0 

5 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠ 

(3) 

continued 






⎛14.6 ⎞⎛2 

⎞ 

FBA 

= ⎜ F 

11 ⎟⎜ 

3 ⎟ 

⎝ ⎠⎝ ⎠ 

⎛14.6 ⎞⎛2 

⎞ 

29.2 lb = ⎜ 

11 

⎟⎜ 

3 

⎟ 

⎝ ⎠⎝ ⎠ 

DA 

F DA 

or 

F = 

DA 

33lb 

Solving Eqn. (3) for F CA gives: 

F 

CA 

⎛5 

⎞ 

= ⎜ F 

9 ⎟ 

⎝ ⎠ 

DA 

F CA 

⎛5 

⎞ 

= ⎜ 

9 ⎟ 

⎝ ⎠ 

( 33 lb) 

Substituting into Eqn. (2) for FBA, FDA, 

andF CA in terms of F DA gives: 

⎛ 9.6 ⎞ ⎛4 ⎞⎛5 ⎞ ⎛2 

⎞ 

⎜ ( 29.2 lb) ( 33 lb) ( 33 lb) 

P 0 

14.6 

⎟ + ⎜ + − = 

5 

⎟⎜ 

9 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 

∴ 

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 

(1) 

⎝14.6 ⎠ ⎝3 

⎠ 


⎛ 9.6 ⎞ ⎛4 ⎞ ⎛2 

⎞ 

⎜ ⎟FBA + ⎜ ⎟FCA + ⎜ ⎟FDA 

− P = 0 

⎝14.6 ⎠ ⎝5 ⎠ ⎝3 

⎠ 

3 1 

− ⎛ ⎜ ⎞ ⎟FCA 

+ ⎛ ⎜ ⎞ 

⎟FDA 

= 0 (3) 

⎝5⎠ ⎝3⎠ 

F 

BA 

⎛14.6 ⎞⎛2 

⎞ 

= ⎜ ⎟⎜ ⎟ F 

⎝ 11 ⎠⎝3 

⎠ 

⎛5 

⎞ 

From Equation (3): FCA 

= ⎜ ⎟ FDA 

⎝9 

⎠ 

Substituting into Equation (2) for F and F gives: 

BA 

⎛ 9.6 ⎞⎛14.6 ⎞⎛2 ⎞ ⎛4 ⎞⎛5 ⎞ ⎛2 

⎞ 

⎜ ⎟⎜ ⎟⎜ ⎟FDA + ⎜ ⎟⎜ ⎟FDA + ⎜ ⎟FDA 

− P = 0 

⎝14.6 ⎠⎝ 11 ⎠⎝3 ⎠ ⎝5 ⎠⎝9 ⎠ ⎝3 

⎠ 

Since 

P = 45 lb 

and 

and 

F BA 

F CA 

CA 

⎛838 

⎞ 

or ⎜ ⎟F 

⎝495 

⎠ 

DA 

DA 

= P 

⎛838 

⎞ 

⎜ ⎟F DA = 45 lb 

⎝495 

⎠ 

or F DA = 26.581 lb 

⎛14.6 ⎞⎛2 

⎞ 

= ⎜ ⎟⎜ ⎟ ( 26.581 lb) 

⎝ 11 ⎠⎝3 

⎠ 

⎛5 

⎞ 

= ⎜ ⎟ 

⎝9 

⎠ 

( 26.581 lb) 

or 

or 

(2) 

F = 23.5 lb ⊳ 

BA 

F = 14.77 lb ⊳ 

CA 

and F = 26.6 lb ⊳ 

DA 






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 = i − j + k 

( 18 m) ( 30 m) ( 5.4 m) 

( ) ( ) ( ) 

2 2 2 

AC = 18 m + − 30 m + 5.4 m = 35.4 m 

uuur 

T AC TAC 

AC = Tλ 

AC = TAC 

= ⎡( 18 m) − ( 30 m) + ( 5.4 m) 

⎤ 

AC 35.4 m 

⎣ i j k ⎦ 

AC 

AC 

( 0.50847 0.84746 0.152542 ) 

T = T i − j + k 

and 

uuur 

AB =− i − j + k 

( 6 m) ( 30 m) ( 7.5 m) 

( ) ( ) ( ) 

2 2 2 

AB = − 6 m + − 30 m + 7.5 m = 31.5 m 

uuur 

T AB TAB 

AB = Tλ 

AB = TAB 

= ⎡−( 6 m) − ( 30 m) + ( 7.5 m) 

⎤ 

AB 31.5 m 

⎣ i j k ⎦ 

AB 

AB 

( 0.190476 0.95238 0.23810 ) 

T = T − i − j + k 

uuur 

Finally AD =−( 6 m) −( 30 m) −( 22.2 m) 

i j k 

( ) ( ) ( ) 

2 2 2 

AD = − 6 m + − 30 m + − 22.2 m = 37.8 m 

uuur 

T AD TAD 

AD = Tλ 

AD = TAD 

= ⎡−( 6 m) − ( 30 m) − ( 22.2 m) 

⎤ 

AD 37.8 m 

⎣ i j k ⎦ 

AD 

AD 

( 0.158730 0.79365 0.58730 ) 

T = T − i − j − k 

continued 





With P = Pj 

, at A: 

Σ F = 0: T + T + T + j = 0 

AB AC AD P 

Equating the factors of i, j, and k to zero, we obtain the linear algebraic 

equations: 

i : − 0.190476T + 0.50847T − 0.158730T 

= 0 (1) 

AB AC AD 

j : −0.95238T − 0.84746T − 0.79365T + P = 0 (2) 

AB AC AD 

k : 0.23810T + 0.152542T − 0.58730T 

= 0 (3) 

AB AC AD 

In Equations (1), (2) and (3), set T AB = 3.6 kN, and, using conventional 

methods for solving Linear Algebraic Equations (MATLAB or Maple, 

for example), we obtain: 

T AC = 

T AD = 

1.963 kN 

1.969 kN 

P = 6.66 kN " 






Based on the results of Problem 2.107, particularly Equations (1), (2) and (3), we substitute T AC = 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 

T AB = 

T AD = 

4.77 kN 

2.61 kN 

P = 8.81 kN ! 






At A Σ F = 0 

Substituting for 

!!!" 

AB =− i − j + k 

( 6.5 ft) ( 8 ft) ( 2 ft) 

( ) ( ) ( ) 

2 2 2 

AB = − 6.5 ft + − 8 ft + 2 ft = 10.5 ft 

T TAB 

AB = ⎡−( 6.5 ft) − ( 8 ft) + ( 2 ft) 

⎤ 

10.5 ft ⎣ i j k ⎦ 

T AB 

( 0.61905i 0.76190j 0.190476k 

) 

= − − + 

!!!" 

AC = i − j + k 

( 1ft) ( 8ft) ( 4ft) 

( ) ( ) ( ) 

2 2 2 

AC = 1ft + − 8ft + 4ft = 9ft 

T TAC 

AC = ⎡( 1ft) − ( 8ft) + ( 4ft) 

⎤ 

9ft⎣ 

i j k ⎦ 

( 0.111111 0.88889 0.44444 ) 

= T i − j + k 

AC 

!!!" 

AD = i − j − k 

( 1.75 ft) ( 8 ft) ( 1 ft) 

( ) ( ) ( ) 

2 2 2 

AD = 1.75 ft + − 8 ft + − 1 ft = 8.25 ft 

T TAD 

AD = ⎡( 1.75 ft) − ( 8 ft) − ( 1 ft) 

⎤ 

8.25 ft ⎣ i j k ⎦ 

T AD 

( 0.21212i 0.96970j 0.121212k 

) 

= − − 

Σ F = 0: − 0.61905T + 0.111111T + 0.21212T 

= 0 

(1) 

x AB AC AD 

Σ F = 0: −0.76190T − 0.88889T − 0.96970T + W = 0 

(2) 

y AB AC AD 

Σ F = 0: 0.190476T + 0.44444T − 0.121212T 

= 0 

(3) 

z AB AC AD 

W = 320 lb and Solving Equations (1), (2), (3) simultaneously yields: 

T = 86.2 lb ! 

AB 

T = 27.7 lb ! 

AC 

T = 237 lb ! 

AD 






See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below. 

− 0.61905T + 0.111111T + 0.21212T 

= 0 (1) 

AB AC AD 

−0.76190T − 0.88889T − 0.96970T + W = 0 (2) 

AB AC AD 

0.190476T + 0.44444T − 0.121212T 

= 0 (3) 

AB AC AD 

Now substituting for T = 220 lb Gives: 

AD 

Solving Equations (4) and (6) simultaneously gives 

− 0.61905T 

+ 0.111111T 

+ 46.662 = 0 

(4) 

AB 

AC 

−0.76190T − 0.88889T − 213.33 + W = 0 

(5) 

AB 

AC 

0.190476T 

+ 0.44444T 

− 26.666 = 0 

(6) 

AB 

AC 

T = 79.992 lb and T = 25.716 lb 

AB 

Substituting into Equation (5) yields 

W = 297 lb ⊳ 

AC 






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 = λ = 

AB 

BE 

cos 45° i + 8j − sin 45° 

k 

65 

⎛cos 45° i + 8j − sin 45° 

k ⎞ 

TBE = TBEλ 

BE = TBE 

⎜ ⎟ 

⎝ 65 ⎠ 

⎛cos30° i + 8j + sin 30° 

k ⎞ 

TCF = TCFλ 

CF = TCF 

⎜ ⎟ 

⎝ 65 ⎠ 

At A: Σ F = 0: T + T + T + W + P = 0 

BE CF DG 

⎛− cos15° i + 8j − sin15° 

k ⎞ 

TDG = TDGλ 

DG = TDG 

⎜ ⎟ 

⎝ 

65 ⎠ 

Then, isolating the factors of i, j, and k, we obtain three algebraic equations: 

TBE TCF TDG 

i : cos 45°+ cos30°− cos15°+ P = 0 

65 65 65 

or T cos 45°+ T cos30°− T cos15°+ P 65 = 0 

(1) 

BE CF DG 

8 8 8 

j : TBE + TCF + TDG 

− W = 0 

65 65 65 

or 

65 

TBE + TCF + TDG 

− W = 0 

(2) 

8 

TBE TCF TDG 

k : − sin 45° + sin 30° − sin15° = 0 

65 65 65 

or − T sin 45° + T sin 30° − T sin15° = 0 

(3) 

With P = 0 and the tension in cord BE = 0.2 lb: 

BE CF DG 

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: 

T CF = 

T DG = 

0.669 lb 

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 : T cos 45°+ T cos30°− T cos15°+ 65P 

= 0 

(1) 

BE CF DG 

65 

j : TBE + TCF + TDG 

− W = 0 

(2) 

8 

k : − T sin 45° + T sin 30° − T sin15° = 0 (3) 

BE CF DG 

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 T CF 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 

( P ) 

= − 1.729 + 0.668 lb 

Hence, for T CF > 0 

− 1.729P 

+ 0.668 > 0 

or 

P < 0.386 lb 

∴ φ ≤ P < 0.386 lb ! 






( ) ( ) 

2 2 

d = 400 mm + − 600 mm = 721.11 mm 

DA 

( ) ( ) ( ) 

2 2 2 

d = − 200 mm + − 600 mm + 150 mm = 650 mm 

DB 

( ) ( ) ( ) 

2 2 2 

d = − 200 mm + − 600 mm + − 150 mm = 650 mm 

DC 

T DA = TDAλDA 

T DA 

( 400 mm) i ( 600 mm) 

= ⎡ − 

721.11 mm ⎣ 

= 0.55470i 

− 0.83205j 

T DA 

T DB = TDBλDB 

At point D Σ F =0: T + T + T + W 0 

i component: 

j component: 

k component: 

2 

Setting ( )( ) 

DA DB DC = 

W = 16 kg 9.81 m/s = 156.96 N 

( ) 

T DB 

And Solving Equations (1), (2), and (3) simultaneously: 

j ⎤ 

⎦ 

( 200 mm) ( 600 mm) + ( 150 mm) 

= ⎡− − 

⎤ 

650 mm ⎣ i j k ⎦ 

4 12 3 

= T ⎛ DB ⎜− i − j + k 

⎞ 

13 13 13 

⎟ 

⎝ 

⎠ 

T DC = TDCλDC 

T TDC 

DC = ⎡−( 200 mm) − ( 600 mm) − ( 150 mm) 

⎤ 

650 mm ⎣ i j k ⎦ 

⎛ 4 12 3 ⎞ 

= T DC ⎜− i − j − k 

13 13 13 

⎟ 

⎝ 

⎠ 

W = Wj 

4 4 

0.55470TDA − TDB − TDC 

= 0 

(1) 

13 13 

12 12 

−0.83205TDA − TDB − TDC 

+ W = 0 

(2) 

13 13 

3 3 

TDB 

− TDC 

= 0 

(3) 

13 13 

T = 62.9 N ! 

DA 

T = 56.7 N ! 

DB 

T = 56.7 N ! 

DC 






( ) ( ) 

2 2 

d = 400 mm + − 600 mm = 721.11 mm 

DA 

( ) ( ) ( ) 

2 2 2 

d = − 200 mm + − 600 mm + 200 mm = 663.32 mm 

DB 

( ) ( ) ( ) 

2 2 2 

d = − 200 mm + − 600 mm + − 200 mm = 663.32 mm 

DC 

T DA = TDAλDA 

T DA 

( 400 mm) i ( 600 mm) 

= ⎡ − 

721.11 mm ⎣ 

T DA 

( 0.55470i 

0.83205j 

) 

= − 

T DB = TDBλDB 

T DB 

j ⎤ 

⎦ 

( 200 mm) ( 600 mm) + ( 200 mm) 

= ⎡− − 

⎤ 

663.32 mm ⎣ i j k ⎦ 

T DB 

( 0.30151i 0.90454 j + 0.30151k 

) 

= − − 

T DC = TDCλDC 

T DC 

( 200 mm) ( 600 mm) − ( 200 mm) 

= ⎡− − 

⎤ 

663.32 mm ⎣ i j k ⎦ 

T DC 

At point D Σ F =0: T + T + T + W 0 

DA DB DC = 

( 0.30151i 0.90454j 0.30151k 

) 

= − − − 

i component: 0.55470T − 0.30151T − 0.30151T 

= 0 

(1) 

DA DB DC 

j component: −0.83205T − 0.90454T − 0.90454T + W = 0 

(2) 

DA DB DC 

k component: 0.30151T 

− 0.30151T 

= 0 

(3) 

2 


W = 16 kg 9.81 m/s = 156.96 N 


DB 

T = 62.9 N ! 

DA 

T = 57.8 N ! 

DB 

DC 

T = 57.8 N ! 

DC 






From the solutions of 2.107 and 2.108: 

TAB 

TAC 

TAD 

= 0.5409P 

= 0.295P 

= 0.2959P 

Using 

P = 8 kN: 

T = 4.33 kN ! 

AB 

T = 2.36 kN ! 

AC 

T = 2.37 kN ! 

AD 






( ) ( ) ( ) 

2 2 2 

d = 6m + 6m + 3m = 9m 

BA 

( ) ( ) ( ) 

2 2 2 

d = − 10.5 m + − 6 m + − 8 m = 14.5 mm 

AC 

( ) ( ) ( ) 

2 2 2 

d = − 6m + − 6m + 7m = 11mm 

AD 

( ) ( ) 

2 2 

d = 6m + − 4.5m = 7.5m 

AE 

F 

F BA = FBAλBA 

= 

BA ⎡ ( 6m) + ( 6m) + ( 3m) 

⎤ 

9m 

⎣ i j k ⎦ 

AC = TACλAC 

2 2 1 

= F ⎛ BA ⎜ i + j + k 

⎞ 

⎟ 

⎝3 3 3 ⎠ 

T AC 

T = ⎡−( 10.5 m) − ( 6 m) − ( 8 m) 

AD = TADλAD 

14.5 m 

⎣ 

21 12 16 

= T ⎛ AC ⎜− i − j − k 

⎞ 

⎟ 

⎝ 29 29 29 ⎠ 

T AD 

i j k ⎦ 

⎤ 

T = ⎡−( 6m) − ( 6m) + ( 7m) 

AE = WAEλAE 

11 m 

⎣ 

i j k ⎤ 

⎦ 

⎛ 6 6 7 ⎞ 

= T AD ⎜− i − j + k ⎟ 

⎝ 11 11 11 ⎠ 

W ⎡ ⎤ 

7.5 m 

⎣ 

j ⎦ 

W = ( 6 m) i − ( 4.5 m) 

W 

O 

( 0.8i 

0.6j 

) 

= W − 

=−W 

At point A: Σ F =0: F + T + T + W + W 0 

BA AC AD AE O = 

j 

i component: 

2 21 6 

FBA − TAC − TAD 

+ 0.8W 

= 0 

(1) 

3 29 11 

continued 





j component: 

2 12 6 

FBA − TAC − TAD 

− 1.6W 

= 0 

(2) 

3 29 11 

1 16 7 

k component: FBA − TAC + TAD 

= 0 

(3) 

3 29 11 

2 


W = 20 kg 9.81 m/s = 196.2 N 


F = 1742 N ⊳ 

BA 

T = 1517 N ⊳ 

AC 

T = 403 N ⊳ 

AD 






Σ F x 

= 0: 

( )( ) ( )( ) ( )( ) 

− T sin30° sin 50° + T sin30° cos 40° + T sin 30° cos60° = 0 

AD BD CD 

Dividing through by sin 30° and evaluating: 

− 0.76604T + 0.76604T + 0.5T 

= 0 

(1) 

AD BD CD 

Σ F y 

= 0: 

( ) ( ) ( ) 

− T cos30° − T cos30° − T cos30° + 62 lb = 0 

AD BD CD 

or TAD + TBD + T CD = 71.591 lb 

(2) 

Σ F z 

= 0: 

Solving Equations (1), (2), and (3) simultaneously: 

T sin 30° cos50° + T sin 30° sin 40° − T sin 30° sin 60° = 0 

AD BD CD 

or 0.64279T + 0.64279T − 0.86603T 

= 0 

(3) 

AD BD CD 

T = 30.5 lb ! 

AD 

T = 10.59 lb ! 

BD 

T = 30.5 lb ! 

CD 






From the solutions to Problems 2.111 and 2.112, have 

Applying the method of elimination to obtain a desired result: 

Multiplying (2′) 

by sin 45° and adding the result to (3): 

T 

CF 

TBE + TCF + TDG 

= 0.2 65 

(2′) 

− T sin 45° + T sin 30° − T sin15° = 0 (3) 

BE CF DG 

T cos45°+ T cos30°− T cos15°− P 65 = 0 (1′ 

) 

BE CF DG 

( ) T ( ) 

sin 45°+ sin 30° + sin 45°− sin15° = 0.2 65 sin 45° 

DG 

or T = 0.94455 − 0.37137T 

Multiplying (2′) 

by sin 30° and subtracting (3) from the result: 

T 

BE 

( ) T ( ) 

sin 30°+ sin 45° + sin 30°+ sin15° = 0.2 65 sin 30° 

DG 

or T = 0.66790 − 0.62863T 

(5) 

Substituting (4) and (5) into (1′) 

: 

BE 

1.29028 −1.73205T 

− P 65 = 0 

∴ T DG is taut for 

DG 

P < 

1.29028 lb 

65 

or 0 ≤ P ≤ 0.1600 lb! 

CF 

DG 

DG 






( ) ( ) ( ) 

2 2 2 

d = − 30 ft + 24 ft + 32 ft = 50 ft 

AB 

( ) ( ) ( ) 

2 2 2 

d = − 30 ft + 20 ft + − 12 ft = 38 ft 

AC 

T TAB 

AB = TABλAB 

= ⎡− ( 30 ft) + ( 24 ft) + ( 32 ft) 

⎤ 

50 ft 

⎣ i j k ⎦ 

= − 0.6i + 0.48j + 0.64k 

T AB 

( ) 

T TAC 

AC = TACλAC 

= ⎡− ( 30 ft) + ( 20 ft) − ( 12 ft) 

⎤ 

38 ft 

⎣ i j k ⎦ 

16 30 

N = Ni + Nj 

34 34 

W =−( 175 lb) 

j 

30 20 12 

= T ⎛ AC ⎜− i + j − k 

⎞ 

⎟ 

⎝ 38 38 38 ⎠ 

At point A: Σ F =0: T + T + N + W 0 

i component: 

j component: 

AB AC = 

30 16 

−0.6TAB 

− TAC 

+ N = 0 

(1) 

38 34 

20 30 

0.48TAB 

+ TAC 

+ N − 175 lb = 0 

(2) 

38 34 

12 

k component: 0.64TAB 

− TAC 

= 0 

(3) 

38 

Solving Equations (1), (2), and (3) simultaneously: 

T = 30.9 lb ⊳ 

AB 

T = 62.5 lb ⊳ 

AC 






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: Σ F =0: T + T + N + W + P 0 

AB AC = 

i component: 

j component: 

k component: 

30 16 

−0.6TAB 

− TAC 

+ N = 0 

38 34 

(1) 

20 30 

0.48TAB 

+ TAC 

+ N − 175 lb = 0 

38 34 

(2) 

12 

0.64TAB 

− TAC 

38 

− 45 lb = 0 

(3) 

Solving (1), (2), and (3) simultaneously: 

T = 81.3 lb ⊳ 

AB 

T = 22.2 lb ⊳ 

AC 






Note: BE shares the same unit vector as AB. 

Thus: 

( 25 mm) cos45° i + ( 200 mm) j − ( 25 mm) 

sin 45° 

k 

λBE 

= λAB 

= 

201.56 mm 

T TBE 

BE = TBEλBE 

= ⎡( 25 mm) cos45 ° + ( 200 mm) − ( 25 mm) 

sin 45° 

⎤ 

201.56 mm ⎣ 

i j k ⎦ 

T TCF 

CF = TCFλCF 

= ⎡( 25 mm) cos30 ° + ( 200 mm) + ( 25 mm) 

sin 30° 

⎤ 

201.56 mm ⎣ 

i j k ⎦ 

T TDG 

DG = TDGλDG 

= ⎡− ( 25 mm) cos15 ° + ( 200 mm) − ( 25 mm) 

sin15° 

⎤ 

201.56 mm ⎣ 

i j k ⎦ 

W = −Wj ; P = Pk 

At point A: ΣF =0: TBE + TCE + TDG 

+ W + P =0 

i component: 0.087704T + 0.107415T − 0.119806T 

= 0 

(1) 

BE CF DG 

j component: 0.99226T + 0.99226T + 0.99226T − W = 0 

(2) 

BE CF DG 

k component: − 0.087704T + 0.062016T − 0.032102T + P = 0 

(3) 

BE CF DG 

Setting W = 10.5 N and P = 0, and solving (1), (2), (3) simultaneously: 

T = 1.310 N ! 

BE 

T = 4.38 N ! 

CF 

T = 4.89 N ! 

DG 






See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 

i component: 0.087704T + 0.107415T − 0.119806T 

= 0 

(1) 

BE CF DG 

j component: 0.99226 T + 0.99226 T + 0.99226 T − W = 0 

(2) 

BE CF DG 

k component: − 0.087704T + 0.062016T − 0.032102T + P = 0 

(3) 

BE CF DG 

Setting W = 10.5 N and P = 0.5 N, and solving (1), (2), (3) simultaneously: 

T = 4.84 N ! 

BE 

T = 1.157 N ! 

CF 

T = 4.58 N ! 

DG 






At D ΣF = 0: 

Σ F x = 

Σ F z = 

0: 

0: 

uuur 

DA =− i + j + k 

( 8ft) ( 40ft) ( 10ft) 

( ) ( ) ( ) 

2 2 2 

DA = − 8ft + 40ft + 10ft = 42ft 

T TADB 

DA = ⎡− ( 8ft) + ( 40ft) + ( 10ft) 

⎤ 

42 ft 

⎣ i j k ⎦ 

= T ADB − 0.190476i + 0.95238j + 0.23810k 

uuur 

DB = 3ft i + 36ft j − 8ft k 

( ) 

( ) ( ) ( ) 

( ) ( ) ( ) 

2 2 2 

DB = 3ft + 36ft + − 8ft = 37ft 

T TADB 

DB = ⎡( 3ft) + ( 36ft) − ( 8ft) 

⎤ 

37 ft 

⎣ i j k ⎦ 

= T ADB 0.081081i + 0.97297j − 0.21622k 

uuur 

DC = a − 8ft i − 24ft j − 3ft k 

( ) 

( ) ( ) ( ) 

2 

( a 8ft) ( 24ft) ( 3ft) 

2 2 

DC = − + − + − 

( a ) 

( a ) 2 

= − 8 + 585 ft 

TDC 

TDC 

= ⎡( a 8ft) ( 24ft) ( 3ft) 

⎤ 

2 ⎣ − i − j − k ⎦ 

− 8 + 585 

( a − 8) 

( a − ) 2 + 

− 0.190476TADB + 0.081081TADB + TDC 

= 0 

8 585 

3 

0.23810TADB − 0.21622TADB − TDC 

= 0 

− 8 + 585 

( a ) 2 

(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 T ADB gives: 

Σ F x = 0: − 0.109395TADB 

+ 0.52705TDC 

= 0 

(3) 

continued 





And writing Σ F = 0 gives: 

y 

1.92535T − 0.84327T − W = 0 

(4) 

ADB 

Substituting into equation (3) for T = 17 lb gives: 

DC 

DC 

( ) 

− 0.109395 + 0.52705 17 lb = 0 

T ADB 

or 

T = 81.9 lb ⊳ 

ADB 

Substituting into equation (4) for T = 17 lb and T = 81.9 lb gives: 

DC 

( ) ( ) 

ADB 

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: 

− 0.109395T 

+ 0.52705T 

= 0 

(3) 

ADB 

DC 

1.92535T − 0.84327T − W = 0 

(4) 

ADB 

DC 

Substituting for W = 120 lb and solving equations (3) and (4) simultaneously yields 

T = 68.6 lb ! 

ADB 

T = 14.23 lb ! 

DC 






( ) ( ) ( ) 

2 2 2 

d = − 2.7 m + 2.4 m + − 3.6 m = 5.1 m 

AB 

( ) ( ) 

2 2 

d = 2.4 m + 1.8 m = 3 m 

AC 

( ) ( ) ( ) 

2 2 2 

d = 1.2 m + 2.4 m + − 0.3 m = 2.7 m 

AD 

( ) ( ) ( ) 

2 2 2 

d = − 2.4 m + 2.4 m + 1.2 m = 3.6 m 

AE 

T AB = TABλAB 

T AB 

( 2.7 m) ( 2.4 m) ( 3.6 m) 

= ⎡− + − ⎤ 

5.1 m ⎣ i j k ⎦ 

9 8 12 

= T ⎛ AB ⎜− i + j − k 

⎞ 

17 17 17 

⎟ 

⎝ 

⎠ 

T AC = TACλAC 

T 

T AC 

( 2.4 m) j ( 1.8 m) 

= ⎡ + 

3m 

⎣ 

T AC 

( 0.8j 

0.6k 

) 

= + 

= 2T 

λ 

AD ADE AD 

T ADE 

k ⎤ 

⎦ 

2 

= ⎡( 1.2 m) + ( 2.4 m) − ( 0.3 m) 

⎤ 

2.7 m 

⎣ i j k ⎦ 

⎛ 8 16 2 ⎞ 

= T ADE ⎜ i + j − k 

9 9 9 

⎟ 

⎝ 

⎠ 

continued 





T AE = TAEλAE 

T ADE 

( 2.4 m) ( 2.4 m) ( 1.2 m) 

= ⎡− + + ⎤ 

3.6 m 

⎣ i j k ⎦ 

⎛ 2 2 1 ⎞ 

= T ADE ⎜− i + j + k 

3 3 3 

⎟ 

⎝ 

⎠ 

W =−Wj 

At point A: ΣF = 0: T + T + T + T W = 0 

i component: 

j component: 

k component: 

Simplifying (1), (2), (3): 

AB AC AD AE + 

9 8 2 

− TAB + TADE − TADE 

= 0 

(1) 

17 9 3 

8 16 2 

TAB + 0.8 TAC + TADE + TADE 

− W = 0 

(2) 

17 9 3 

12 2 1 

− TAB + 0.6 TAC − TADE + TADE 

= 0 

(3) 

17 9 3 

− 81T 

+ 34T 

= 0 

(1) ′ 

AB 

ADE 

72T + 122.4T + 374T = 153 W 

(2) ′ 

AB AC ADE 

− 108T + 91.8T + 17T 

= 0 

(3) ′ 

AB AC ADE 

Setting W = 1400 N and solving (1), (2), (3) simultaneously: 

T = 203 N " 

AB 

T = 149.6 N " 

AC 

T = 485 N " 

ADE 






See Problem 2.125 for the analysis leading to the linear algebraic Equations ( 1 ′), ( 2 ′), 

and ( 3′ ) below: 

i component: 81T 

34T 

0 

− + = ( 1′ ) 

AB 

j component: 72 T 122.4 T 37.4 T 153 W 

ADE 

+ + = ( 2′ ) 

AB AC ADE 

k component: 108 T 91.8 T 17 T 0 

− + + = ( 3′ ) 

AB AC ADE 

Setting T AB = 300 N and solving (1), (2), (3) simultaneously: 

(a) 

(b) 

(c) 

T = 221 N ! 

AC 

T = 715 N ! 

ADE 

W = 2060 N ! 






Free-Body Diagrams of collars For both Problems 2.127 and 2.128: 

Here ( ) ( ) 

( ) 2 2 2 2 

AB = x + y + z 

2 2 2 2 

1m = 0.40m + y + z 

or 

y 

2 2 2 

+ z = 0.84 m 

Thus, with y given, z is determined. 

Now 

uuur 

AB 1 

λ AB = = ( 0.40i − yj + zk) 

m = 0.4i − yk + zk 

AB 1m 

Where y and z are in units of meters, m. 

From the F.B. Diagram of collar A: 

Σ F = 0: N i + N k + Pj + T λ = 0 

Setting the jcoefficient to zero gives: 

With P = 680 N, 

x z AB AB 

P − yT AB = 

0 

T AB 

= 

680 N 

y 

Now, from the free body diagram of collar B: 

Σ F = 0: N i + N j + Qk − T λ = 0 

x y AB AB 

continued 





Setting the k coefficient to zero gives: 

Q − TABz 

= 

And using the above result for T AB we have 

680 N 

Q = TABz = z 

y 

0 

Then, from the specifications of the problem, y = 300 mm = 0.3 m 

and 

z 

2 2 

∴ z = 

( ) 2 

= 0.84 m − 0.3 m 

0.866 m 

(a) 

or 

and 

680 N 

T AB = = 2266.7 N 

0.30 

(b) Q = 2266.7( 0.866) 

= 1963.2 N 

or 

T = 2.27 kN ! 

AB 

Q = 1.963 kN ! 






From the analysis of Problem 2.127, particularly the results: 

y 

2 2 2 

+ z = 0.84 m 

T AB 

Q 

= 

= 

680 N 

y 

680 N 

z 

y 

With y = 550 mm = 0.55 m, we obtain: 

and 

2 2 

( ) 2 

z = 0.84 m − 0.55 m 

∴ z = 0.73314 m 

(a) 

or 

and 

680 N 

T AB = = 1236.36 N 

0.55 

T = 1.236 kN ! 

AB 

(b) Q = 1236.36( 0.73314) 

N = 906 N 

or 

Q = 0.906 kN ! 







(a) 

Have: 

20 lb 14 lb 

sinα = sin 30° 

sinα = 0.71428 

(b) β = 180° − ( 30° + 45.6° 

) 

α = 45.6°⊳ 

= 104.4° 

Then: 

R 14 lb 

= 

sin104.4° sin 30° 

R = 

27.1 lb ⊳ 






We compute the following distances: 

500 N Force: 

( ) ( ) 

2 2 

OA = 70 + 240 = 250 mm 

( ) ( ) 

2 2 

OB = 210 + 200 = 290 mm 

( ) ( ) 

2 2 

OC = 120 + 225 = 255 mm 

F x 

F y 

⎛ 70 ⎞ 

=−500 N⎜ 250 

⎟ 

⎝ ⎠ 

⎛240 

⎞ 

=+ 500 N⎜ 250 

⎟ 

⎝ ⎠ 

F =− 140.0 N ! 

x 

F = 480 N ! 

y 

435 N Force: 

510 N Force: 

F x 

F y 

F x 

F y 

⎛210 

⎞ 

=+ 435 N⎜ 290 

⎟ 

⎝ ⎠ 

⎛200 

⎞ 

=+ 435 N⎜ 290 

⎟ 

⎝ ⎠ 

⎛120 

⎞ 

=+ 510 N⎜ 255 

⎟ 

⎝ ⎠ 

⎛225 

⎞ 

=−510 N⎜ 255 

⎟ 

⎝ ⎠ 

F = 315 N ! 

x 

F = 300 N ! 

y 

F = 240 N ! 

x 

F =− 450 N ! 

y 






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: 

(a) 

450 N 

P = = 549.3 N 

cos35° 

P = 549 N ! 

(b) ( ) 

P = 315 N ! 

x 

P = 450 N tan 35° 

x 

= 315.1 N 








Law of Sines: 

TAC 

TBC 

5 kN 

= = 

sin115° sin 5° sin 60° 

(a) 

5 kN 

T AC = sin115 ° = 5.23 kN 

sin 60° 

T = 5.23 kN ! 

AC 

(b) 

5 kN 

T BC = sin 5 ° = 0.503 kN 

sin 60° 

T = 0.503 kN ! 

BC 







First, consider the sum of forces in the x-direction because there is only one unknown force: 

or 

Now 

or 

( ) ( ) 

Σ F = 0: T cos32° − cos42° − 20 kN cos 42° = 0 

x 

ACB 

0.104903T ACB = 14.8629 kN 

T ACB = 141.682 kN 

( ) ( ) 

Σ F = 0: T sin 42°− sin 32° + 20 kN sin 42°− W = 0 

y 

ACB 

( )( ) ( )( ) 

141.682 kN 0.139211 + 20 kN 0.66913 − W = 0 

(b) 

T = 141.7 kN ! 

(a) 

ACB 

W = 33.1 kN ! 






Free-Body Diagram: Pulley A 

Σ F = 0: 2Psin 25° − Pcosα 

= 0 

x 

and 

cosα 

= 0.8452 or α = ± 32.3° 

For α =+ 32.3° 

Σ F = 0: 2Pcos 25°+ Psin 32.3°− 350 lb = 0 

y 

or 

P = 149.1 lb 32.3° ⊳ 

For α =− 32.3° 

Σ F = 0: 2Pcos 25° + Psin − 32.3° − 350 lb = 0 

y 

or 

P = 274 lb 32.3° ⊳ 






(a) F = Fsin 30° sin 50° = 220.6 N (Given) 

x 

F 

220.6 N 

= = 575.95 N 

sin30 ° sin50° 

F = 576 N ! 

(b) 

Fx 

220.6 

cosθ x = = = 0.38302 

F 575.95 

θ = 67.5°! 

x 

Fy 

= Fcos30° = 498.79 N 

Fy 

498.79 

cosθ y = = = 0.86605 

F 575.95 

θ = 30.0°! 

y 

Fz 

=− Fsin 30° cos50° 

=− ( 575.95 N) 

sin 30° cos50° 

=− 185.107 N 

Fz 

−185.107 

cosθ 

z = = = − 0.32139 

F 575.95 

θ = 108.7°! 

z 






(a) F Fcosθ 

( ) 

Then: 

z 

= = 600 lb cos136.8° 

z 

=− 437.38 lb 

F =− 437 lb ! 

2 2 2 2 

= x + y + z 

F F F F 

z 

2 2 2 

2 

F y 

So: ( 600 lb) = ( 200 lb) + ( ) + ( − 437.38 lb) 

2 2 2 

Hence: F =− ( 600 lb) −( 200 lb) −( − 437.38 lb) 

y 

=− 358.75 lb 

F =− 359 lb ! 

y 

(b) 

Fx 

200 

cosθ x = = = 0.33333 

θ x = 70.5°! 

F 600 

Fy 

−358.75 

cosθ 

y = = = − 0.59792 θ y = 126.7°! 

F 600 






( 500 lb)[ cos30 sin15 sin 30 cos30 cos15 ] 

( 500 lb)[ 0.2241i 0.50j 0.8365k 

] 

( 112.05 lb) i ( 250 lb) j ( 418.25 lb) 

k 

( 600 lb)[ cos 40 cos 20 sin 40 cos40 sin 20 ] 

( 600 lb)[ 0.71985i 0.64278j 0.26201k 

] 

( 431.91 lb) i ( 385.67 lb) j ( 157.206 lb) 

k 

( 319.86 lb) ( 635.67 lb) ( 261.04 lb) 

P = − ° ° i + ° j + ° ° k 

= − + + 

=− + + 

Q = ° ° i + ° j − ° ° k 

= + − 

= + − 

R = P + Q = i + j + k 

( ) ( ) ( ) 

2 2 2 

R = 319.86 lb + 635.67 lb + 261.04 lb = 757.98 lb 

R = 758 lb ! 

Rx 

319.86 lb 

cosθ x = = = 0.42199 

R 757.98 lb 

θ = 65.0°! 

Ry 

635.67 lb 

cosθ y = = = 0.83864 

R 757.98 lb 

θ = 33.0°! 

Rz 

261.04 lb 

cosθ z = = = 0.34439 

R 757.98 lb 

θ = 69.9°! 

x 

y 

z 






Equating to zero the coefficients of i, j, k: 

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 

and TAB = TABλ 

AB = T AB 

AB = ( − 0.48i + 0.8j − 0.36k) 

TAB 

AB 

uuur 

TAC = TACλ 

AC = T AC 

AC = ( 0.88235j + 0.47059k) 

TAC 

AC 

uuur 

TAD = TADλ 

AD = T AD 

AD = ( 0.51948i + 0.77922j − 0.35065k) 

TAD 

AD 

Equilibrium Condition with W =−Wj 

Σ F = 0: T + T + T − Wj 

= 0 

AB AC AD 

Substituting the expressions obtained for TAB, TAC, and TADand 

factoring i, j, and k: 

( − 0.48TAB + 0.51948TAD ) i + ( 0.8TAB + 0.88235TAC + 0.77922TAD 

−W) 

+ ( − 0.36T + 0.47059T − 0.35065T 

) k = 0 

AB AC AD 

− 0.48T 

+ 0.51948T 

= 0 

AB 

AD 

0.8T + 0.88235T + 0.77922T − W = 0 

AB AC AD 

− 0.36T + 0.47059T − 0.35065T 

= 0 

AB AC AD 

Substituting T AB = 3 kN in Equations (1), (2) and (3) and solving the resulting set of equations, using 

conventional algorithms for solving linear algebraic equations, gives 

T AC = 

T AD = 

4.3605 kN 

2.7720 kN 

W = 8.41 kN ⊳ 

j 






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 = i − j+ 

k 

( 32 in. ) ( 48 in. ) ( 36 in. ) 

( ) ( ) ( ) 

2 2 2 

AB = − 32 in. + − 48 in. + 36 in. = 68 in. 

uuur 

T AB TAB 

AB = Tλ 

AB = TAB 

= ⎡−( 32 in. ) − ( 48 in. ) + ( 36 in. ) ⎤ 

AB 68 in. 

⎣ i j k ⎦ 

AB 

AB 

( 0.47059 0.70588 0.52941 ) 

T = T − i − j + k 

uuur 

and AC = ( 45 in. ) − ( 48 in. ) + ( 36 in. ) 

i j k 

( ) ( ) ( ) 

2 2 2 

AC = 45 in. + − 48 in. + 36 in. = 75 in. 

uuur 

T AC TAC 

AC = Tλ 

AC = TAC 

= ⎡( 45 in. ) − ( 48 in. ) + ( 36 in. ) ⎤ 

AC 75 in. 

⎣ i j k ⎦ 

AC 

AC 

( 0.60 0.64 0.48 ) 

T = T i − j + k 

uuur 

Finally, AD = ( 25 in. ) − ( 48 in. ) − ( 36 in. ) 

i j k 

( ) ( ) ( ) 

2 2 2 

AD = 25 in. + − 48 in. + − 36 in. = 65 in. 

continued 





uuur 

T AD TAD 

AD = Tλ 

AD = TAD 

= ⎡( 25 in. ) − ( 48 in. ) − ( 36 in. ) ⎤ 

AD 65 in. 

⎣ i j k ⎦ 

AD 

AD 

With W = Wj , at A we have: 

( 0.38461 0.73846 0.55385 ) 

T = T i − j − k 

Σ F = 0: T + T + T + j = 0 

AB AC AD W 

Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: 

i : − 0.47059T + 0.60T − 0.38461T 

= 0 

(1) 

AB AC AD 

j : −0.70588T − 0.64T − 0.73846T + W = 0 

(2) 

AB AC AD 

k : 0.52941T + 0.48T − 0.55385T 

= 0 

(3) 

AB AC AD 

In Equations (1), (2) and (3), set T AD = 120 lb, and, using conventional methods for solving Linear Algebraic 

Equations (MATLAB or Maple, for example), we obtain: 

T AB = 

T AC = 

32.6 lb 

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 =− i + j − k 

( 0.48 m) ( 0.72 m) ( 0.16 m) 

( ) ( ) ( ) 

2 2 2 

AB = − 0.48 m + 0.72 m + − 0.16 m = 0.88 m 

uuur 

T AB TAB 

AB = Tλ 

AB = TAB 

= ⎡− ( 0.48 m) + ( 0.72 m) − ( 0.16 m) 

⎤ 

AB 0.88 m 

⎣ i j k ⎦ 

AB 

AB 

( 0.54545 0.81818 0.181818 ) 

T = T − i + j − k 

and 

uuur 

AC = i + j − k 

( 0.24 m) ( 0.72 m) ( 0.13 m) 

( ) ( ) ( ) 

2 2 2 

AC = 0.24 m + 0.72 m − 0.13 m = 0.77 m 

uuur 

T AC TAC 

AC = Tλ 

AC = TAC 

= ⎡( 0.24 m) + ( 0.72 m) − ( 0.13 m) 

⎤ 

AC 0.77 m 

⎣ i j k ⎦ 

AC 

AC 

( 0.31169 0.93506 0.16883 ) 

T = T i + j − k 

At A: Σ F = 0: T + T + P + Q + W = 0 

AB 

AC 

Noting that T AB = T AC because of the ring A, we equate the factors of 

i, j,andk to zero to obtain the linear algebraic equations: 

( ) 

i : − 0.54545 + 0.31169 T + P = 0 

or P = 0.23376T 

( ) 

j : 0.81818 + 0.93506 T − W = 0 

continued 





With W = 1200 N: 

or W = 1.75324T 

( ) 

k : −0.181818 − 0.16883 T + Q = 0 

or Q = 0.35065T 

1200 N 

T = = 684.45 N 

1.75324 

P = 160.0 N ! 

Q = 240 N !

Sol Cap 02 - Edicion 8

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