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Cap 03
1.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 1. Resolve 90 N force into vector components P and Q where Q = ( 90 N ) sin 40° = 57.851 N Then M B = − rA/BQ = − (0.225 m )(57.851 N ) = −13.0165 N ⋅ m M B = 13.02 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
2.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 2. Fx = ( 90 N ) cos 25° = 81.568 N Fy = ( 90 N ) sin 25° = 38.036 N x = ( 0.225 m ) cos 65° = 0.095089 m y = (0.225 m ) sin 65° = 0.20392 m M B = xFy − yFx = ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N ) = −13.0165 N ⋅ m M B = 13.02 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
3.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 3. Px = ( 3 lb ) sin 30° = 1.5 lb Py = ( 3 lb ) cos 30° = 2.5981 lb M A = xB/ A Py + yB/ A Px = ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb ) = 16.0335 lb ⋅ in. M A = 16.03 lb ⋅ in. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
4.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 4. For P to be a minimum, it must be perpendicular to the line joining points A and B with rAB = ( 3.4 in.)2 + ( 4.8 in.)2 = 5.8822 in. y α = θ = tan −1 x 4.8 in. = tan −1 3.4 in. = 54.689° Then M A = rAB Pmin M A 19.5 lb ⋅ in. or Pmin = = rAB 5.8822 in. = 3.3151 lb ∴ Pmin = 3.32 lb 54.7° or Pmin = 3.32 lb 35.3° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
5.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 5. By definition M A = rB/ A P sin θ where θ = φ + ( 90° − α ) 4.8 in. and φ = tan −1 3.4 in. = 54.689° Also rB/ A = ( 3.4 in.)2 + ( 4.8 in.)2 = 5.8822 in. Then (17 lb ⋅ in.) = ( 5.8822 in.)( 2.9 lb ) sin ( 54.689° + 90° − α ) or sin (144.689° − α ) = 0.99658 or 144.689° − α = 85.260°; 94.740° ∴ α = 49.9°, 59.4° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
6.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 6. (a) (a) M A = rB/ A × TBF M A = xTBFy + yTBFx = ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m or M A = 386 N ⋅ m (b) (b) For FC to be a minimum, it must be perpendicular to the line joining A and C. ∴ M A = d ( FC )min with d = ( 2 m )2 + (1.35 m )2 = 2.4130 m Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min ( FC )min = 160.137 N 1.35 m and φ = tan −1 = 34.019° 2m θ = 90 − φ = 90° − 34.019° = 55.981° ∴ ( FC )min = 160.1 N 56.0° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
7.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 7. (a) M A = xTBFy + yTBFx = ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m or M A = 386 N ⋅ m (b) Have M A = xFC MA 386.41 N ⋅ m or FC = = x 2m = 193.205 N ∴ FC = 193.2 N (c) For FB to be minimum, it must be perpendicular to the line joining A and B ∴ M A = d ( FB )min with d = ( 2 m )2 + ( 0.40 m )2 = 2.0396 m Then 386.41 N ⋅ m = ( 2.0396 m ) ( FC )min ( FC )min = 189.454 N 2m and θ = tan −1 = 78.690° 0.4 m or ( FC )min = 189.5 N 78.7° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
8.
COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 8. ( ) (a) M B = rA/B cos15° W = (14 in.)( cos15° )( 5 lb ) = 67.615 lb ⋅ in. or M B = 67.6 lb ⋅ in. (b) M B = rD/B P sin 85° 67.615 lb ⋅ in. = ( 3.2 in.) P sin 85° or P = 21.2 lb (c) For ( F )min, F must be perpendicular to BC. Then, M B = rC/B F 67.615 lb ⋅ in. = (18 in.) F or F = 3.76 lb 75.0° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 9. 35 in. 5 (a) Slope of line EC = = 76 in. + 8 in. 12 12 Then TABx = (TAB ) 13 12 = ( 260 lb ) = 240 lb 13 5 and TABy = ( 260 lb ) = 100 lb 13 Then M D = TABx ( 35 in.) − TABy ( 8 in.) = ( 240 lb )( 35 in.) − (100 lb )( 8 in.) = 7600 lb ⋅ in. or M D = 7600 lb ⋅ in. (b) Have M D = TABx ( y ) + TABy ( x ) = ( 240 lb )( 0 ) + (100 lb )( 76 in.) = 7600 lb ⋅ in. or M D = 7600 lb ⋅ in. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 10. 35 in. 7 Slope of line EC = = 112 in. + 8 in. 24 24 Then TABx = TAB 25 7 and TABy = TAB 25 Have M D = TABx ( y ) + TABy ( x ) 24 7 ∴ 7840 lb ⋅ in. = TAB ( 0 ) + TAB (112 in.) 25 25 TAB = 250 lb or TAB = 250 lb Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 11. The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y ( d ) where M D = 1152 N ⋅ m (TAB max ) y = TAB max sin θ = ( 2880 N ) sin θ 1.05 m Now sin θ = (d + 0.24 ) + (1.05 ) m 2 2 1.05 ∴ 1152 N ⋅ m = 2880 N (d ) (d + 0.24 ) + (1.05 ) 2 2 or ( d + 0.24 )2 + (1.05)2 = 2.625d or (d + 0.24 ) + (1.05 ) = 6.8906d 2 2 2 or 5.8906d 2 − 0.48d − 1.1601 = 0 Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. Since only the positive value applies here, d = 0.48639 m or d = 486 mm Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 12. with d AB = ( 42 mm )2 + (144 mm )2 = 150 mm 42 mm sin θ = 150 mm 144 mm cosθ = 150 mm and FAB = − FAB sin θ i − FAB cosθ j 2.5 kN = ( − 42 mm ) i − (144 mm ) j 150 mm = − ( 700 N ) i − ( 2400 N ) j Also rB/C = − ( 0.042 m ) i + ( 0.056 m ) j Now M C = rB/C × FAB = ( − 0.042 i + 0.056 j) × ( − 700 i − 2400 j) N ⋅ m = (140.0 N ⋅ m ) k or M C = 140.0 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 13. with d AB = ( 42 mm )2 + (144 mm )2 = 150 mm 42 mm sin θ = 150 mm 144 mm cosθ = 150 mm FAB = − FAB sin θ i − FAB cosθ j 2.5 kN = ( − 42 mm ) i − (144 mm ) j 150 mm = − ( 700 N ) i − ( 2400 N ) j Also rB/C = − ( 0.042 m ) i − ( 0.056 m ) j Now M C = rB/C × FAB = ( − 0.042 i − 0.056 j) × ( − 700i − 2400 j) N ⋅ m = ( 61.6 N ⋅ m ) k or M C = 61.6 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 14. 88 105 ΣM D : M D = ( 0.090 m ) × 80 N − ( 0.280 m ) × 80 N 137 137 = −12.5431 N ⋅ m or M D = 12.54 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 15. Note: B = B ( cos β i + sin β j) B′ = B ( cos β i − sin β j) C = C ( cos α i + sin α j) By definition: B × C = BC sin (α − β ) (1) B′ × C = BC sin (α + β ) (2) Now ... B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α − sin β cos α ) k (3) and B′ × C = B ( cos β i − sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α + sin β cos α ) k (4) Equating the magnitudes of B × C from equations (1) and (3) yields: BC sin (α − β ) = BC ( cos β sin α − sin β cos α ) (5) Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields: BC sin (α + β ) = BC ( cos β sin α + sin β cos α ) (6) Adding equations (5) and (6) gives: sin (α − β ) + sin (α + β ) = 2cos β sin α 1 1 or sin α cos β = sin (α + β ) + sin (α − β ) 2 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 16. Have d = λ AB × rO/ A rB/ A where λ AB = rB/ A and rB/ A = ( −210 mm − 630 mm ) i + ( 270 mm − ( −225 mm ) ) j = − ( 840 mm ) i + ( 495 mm ) j rB/ A = ( −840 mm )2 + ( 495 mm )2 = 975 mm − ( 840 mm ) i + ( 495 mm ) j Then λ AB = 975 mm 1 = ( −56i + 33j) 65 Also rO/ A = ( 0 − 630 ) i + ( 0 − (−225) ) j = − ( 630 mm ) i + ( 225 mm ) j 1 ∴d = ( −56i + 33j) × − ( 630 mm ) i + ( 225 mm ) j 65 = 126.0 mm d = 126.0 mm Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 17. A×B (a) λ = A×B where A = 12i − 6 j + 9k B = − 3i + 9 j − 7.5k Then i j k A × B = 12 − 6 9 − 3 9 − 7.5 = ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 ) k = 9 ( − 4i + 7 j + 10k ) And A × B = 9 (− 4) 2 + (7)2 + (10)2 = 9 165 9 ( − 4i + 7 j + 10k ) ∴λ = 9 165 1 or λ = ( − 4i + 7 j + 10k ) 165 A×B (b) λ = A×B where A = −14i − 2 j + 8k B = 3i + 1.5j − k Then i j k A × B = −14 − 2 8 3 1.5 −1 = ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k = 5 ( −2i + 2 j − 3k ) and A × B = 5 (−2)2 + (2)2 + (−3)2 = 5 17 5 ( −2i + 2 j − 3k ) ∴λ = 5 17 1 or λ = ( − 2i + 2 j − 3k ) 17 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 18. (a) Have A = P × Q i j k P × Q = 3 7 −2 in.2 −5 1 3 = [ (21 + 2)i + (10 − 9) j + (3 + 35)k ] in.2 ( ) ( ) ( = 23 in.2 i + 1 in.2 j + 38 in.2 k ) ∴A= (23)2 + (1)2 + (38) 2 = 44.430 in.2 or A = 44.4 in.2 (b) A = P×Q i j k P × Q = 2 − 4 3 in.2 6 −1 5 = [ (−20 − 3)i + (−18 − 10) j + (−2 + 24)k ] in.2 ( ) ( ) ( = − 23 in.2 i − 28 in.2 j + 22 in.2 k ) ∴A= (− 23)2 + (−28)2 + (22) 2 = 42.391 in.2 or A = 42.4 in.2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 19. (a) Have MO = r × F i j k = − 6 3 1.5 N ⋅ m 7.5 3 − 4.5 = [ (−13.5 − 4.5)i + (11.25 − 27) j + (−18 − 22.5)k ] N ⋅ m = ( −18.00i − 15.75 j − 40.5k ) N ⋅ m or M O = − (18.00 N ⋅ m ) i − (15.75 Ν ⋅ m ) j − ( 40.5 N ⋅ m ) k (b) Have MO = r × F i j k = 2 − 0.75 −1 N ⋅ m 7.5 3 − 4.5 = [ (3.375 + 3)i + (−7.5 + 9) j + (6 + 5.625)k ] N ⋅ m = ( 6.375i + 1.500 j + 11.625k ) N ⋅ m or M O = ( 6.38 N ⋅ m ) i + (1.500 Ν ⋅ m ) j + (11.63 Ν ⋅ m ) k (c) Have MO = r × F i j k = − 2.5 −1 1.5 N ⋅ m 7.5 3 4.5 = [ (4.5 − 4.5)i + (11.25 − 11.25) j + (−7.5 + 7.5)k ] N ⋅ m or M O = 0 This answer is expected since r and F are proportional ( F = −3r ) . Therefore, vector F has a line of action passing through the origin at O. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 20. (a) Have MO = r × F i j k = − 7.5 3 − 6 lb ⋅ ft 3 −6 4 = [ (12 − 36)i + (−18 + 30) j + (45 − 9)k ] lb ⋅ ft or M O = − ( 24.0 lb ⋅ ft ) i + (12.00 lb ⋅ ft ) j + ( 36.0 lb ⋅ ft ) k (b) Have MO = r × F i j k = − 7.5 1.5 −1 lb ⋅ ft 3 −6 4 = [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5)k ] lb ⋅ ft or M O = 0 (c) Have MO = r × F i j k = − 8 2 −14 lb ⋅ ft 3 −6 4 = [ (8 − 84)i + (−42 + 32) j + (48 − 6)k ] lb ⋅ ft or M O = − ( 76.0 lb ⋅ ft ) i − (10.00 lb ⋅ ft ) j + ( 42.0 lb ⋅ ft ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 21. With TAB = − ( 369 N ) j TAB = TAD AD = ( 369 N ) ( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k AD ( 2.4 m )2 + ( −3.1 m )2 + ( −1.2 m )2 TAD = ( 216 N ) i − ( 279 N ) j − (108 N ) k Then R A = 2 TAB + TAD = ( 216 N ) i − (1017 N ) j − (108 N ) k Also rA/C = ( 3.1 m ) i + (1.2 m ) k Have M C = rA/C × R A i j k = 0 3.1 1.2 N ⋅ m 216 −1017 −108 = ( 885.6 N ⋅ m ) i + ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k M C = ( 886 N ⋅ m ) i + ( 259 N ⋅ m ) j − ( 670 N ⋅ m ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 22. Have M A = rC/ A × F where rC/ A = ( 215 mm ) i − ( 50 mm ) j + (140 mm ) k Fx = − ( 36 N ) cos 45° sin12° Fy = − ( 36 N ) sin 45° Fz = − ( 36 N ) cos 45° cos12° ∴ F = − ( 5.2926 N ) i − ( 25.456 N ) j − ( 24.900 N ) k i j k and M A = 0.215 − 0.050 0.140 N ⋅ m − 5.2926 − 25.456 − 24.900 = ( 4.8088 N ⋅ m ) i + ( 4.6125 N ⋅ m ) j − ( 5.7377 N ⋅ m ) k M A = ( 4.81 N ⋅ m ) i + ( 4.61 N ⋅ m ) j − ( 5.74 N ⋅ m ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 23. Have M O = rA/O × R where rA/D = ( 30 ft ) j + ( 3 ft ) k T1 = − ( 62 lb ) cos10° i − ( 62 lb ) sin10° j = − ( 61.058 lb ) i − (10.766 lb ) j AB T2 = T2 AB = ( 62 lb ) ( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k ( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2 = (10 lb ) i − ( 60 lb ) j + (12 lb ) k ∴ R = − ( 51.058 lb ) i − ( 70.766 lb ) j + (12 lb ) k i j k MO = 0 30 3 lb ⋅ ft − 51.058 −70.766 12 = ( 572.30 lb ⋅ ft ) i − (153.17 lb ⋅ ft ) j + (1531.74 lb ⋅ ft ) k M O = ( 572 lb ⋅ ft ) i − (153.2 lb ⋅ ft ) j + (1532 lb ⋅ ft ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 24. (a) Have M O = rB/O × TBD where rB/O = ( 2.5 m ) i + ( 2 m ) j BD TBD = TBD BD − (1 m ) i − ( 2 m ) j + ( 2 m ) k = ( 900 N ) ( −1 m ) + ( − 2 m ) + ( 2 m ) 2 2 2 = − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k Then i j k MO = 2.5 2 0 N⋅m − 300 − 600 600 M O = (1200 N ⋅ m ) i − (1500 N ⋅ m ) j − ( 900 N ⋅ m ) k continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System (b) Have M O = rB/O × TBE where rB/O = ( 2.5 m ) i + ( 2 m ) j BE TBE = TBE BE − ( 0.5 m ) i − ( 2 m ) j − ( 4 m ) k = ( 675 N ) ( 0.5 m ) + ( −2 m ) + ( − 4 m )2 2 2 = − ( 75 N ) i − ( 300 N ) j − ( 600 N ) k Then i j k MO = 2.5 2 0 N⋅m − 75 − 300 − 600 M O = − (1200 N ⋅ m ) i + (1500 N ⋅ m ) j − ( 600 N ⋅ m ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 25. Have M C = rA/C × P where rA/C = rB/C + rA/B = (16 in.)( − cos80° cos15°i − sin 80° j − cos80° sin15°k ) + (15.2 in.)( − sin 20° cos15°i + cos 20° j − sin 20° sin15°k ) = − ( 7.7053 in.) i − (1.47360 in.) j − ( 2.0646 in.) k and P = (150 lb )( cos 5° cos 70°i + sin 5° j − cos 5° sin 70°k ) = ( 51.108 lb ) i + (13.0734 lb ) j − (140.418 lb ) k Then i j k M C = −7.7053 −1.47360 −2.0646 lb ⋅ in. 51.108 13.0734 −140.418 = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k or M C = (19.49 lb ⋅ ft ) i − ( 99.0 lb ⋅ ft ) j − ( 2.12 lb ⋅ ft ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 26. Have M C = rA/C × FBA where rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k and FBA = λ BA FBA = − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2 m = − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i j k ∴ M C = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 27. Have M C = TAD d where d = Perpendicular distance from C to line AD with M C = rA/C TAD and rA/C = ( 3.1 m ) j + (1.2 m ) k AD TAD = TAD AD ( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k TAD = ( 369 N ) ( 2.4 m ) + ( − 3.1 m ) + ( −1.2 m )2 2 2 = ( 216 N ) i − ( 279 N ) j − (108 N ) k Then i j k MC = 0 3.1 1.2 N ⋅ m 216 − 279 −108 = ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k and MC = ( 259.2 N ⋅ m )2 + ( −669.6 N ⋅ m )2 = 718.02 N ⋅ m ∴ 718.02 N ⋅ m = ( 369 N ) d or d = 1.946 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 28. Have M O = TAC d where d = Perpendicular distance from O to rope AC with M O = rA/O × TAC and rA/O = ( 30 ft ) j + ( 3 ft ) k TAC = − ( 62 lb ) cos10° i − ( 62 lb ) sin10° j = − ( 61.058 lb ) i − (10.766 lb ) j Then i j k MO = 0 30 3 lb ⋅ ft − 61.058 −10.766 0 = ( 32.298 lb ⋅ ft ) i − (183.174 lb ⋅ ft ) j + (1831.74 lb ⋅ ft ) k and MO = ( 32.298 lb ⋅ ft )2 + ( −183.174 lb ⋅ ft )2 + (1831.74 lb ⋅ ft )2 = 1841.16 lb ⋅ ft ∴ 1841.16 lb ⋅ ft = ( 62 lb ) d or d = 29.7 ft Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 29. Have M O = TAB d where d = Perpendicular distance from O to rope AB with M O = rA/O × TAB and rA/O = ( 30 ft ) j + ( 3 ft ) k AB TAB = TAB AB ( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k = ( 62 lb ) ( 5 ft ) + ( − 30 ft ) + ( 6 ft )2 2 2 = (10 lb ) i − ( 60 lb ) j + (12 lb ) k Then i j k MO = 0 30 3 lb ⋅ ft 10 − 60 12 = ( 540 lb ⋅ ft ) i + ( 30 lb ⋅ ft ) j − ( 300 lb ⋅ ft ) k and MO = ( 540 lb ⋅ ft )2 + ( 30 lb ⋅ ft )2 + ( −300 lb ⋅ ft )2 = 618.47 lb ⋅ ft ∴ 618.47 lb ⋅ ft = ( 62 lb ) d or d = 9.98 ft Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 30. Have M C = TBD d where d = Perpendicular distance from C to cable BD with M C = rB/C × TB/D and rB/C = ( 2 m ) j BD TBD = TBD BD − (1 m ) i − ( 2 m ) j + ( 2 m ) k = ( 900 N ) ( −1 m ) + ( − 2 m ) + ( 2 m )2 2 2 = − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k Then i j k MC = 0 2 0 N⋅m −300 − 600 600 = (1200 N ⋅ m ) i + ( 600 N ⋅ m ) k and MC = (1200 N ⋅ m )2 + ( 600 N ⋅ m )2 = 1341.64 N ⋅ m ∴ 1341.64 = ( 900 N ) d or d = 1.491 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 31. Have M C = Pd From the solution of problem 3.25 M C = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k Then MC = ( 233.91)2 + ( −1187.48)2 + ( −25.422 )2 = 1210.57 lb ⋅ in. MC 1210.57 lb.in. and d = = P 150 lb or d = 8.07 in. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 32. Have | M D | = FBAd where d = perpendicular distance from D to line AB. M D = rA/D × FBA rA/D = − ( 0.12 m ) j + ( 0.72 m ) k FBA = λ BA FBA = ( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2 m = − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i j k ∴ MD = 0 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k and |MD | = (146.88)2 + (8.64 )2 + (1.44 )2 = 147.141 N ⋅ m ∴ 147.141 N ⋅ m = ( 228 N ) d d = 0.64536 m or d = 0.645 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 33. Have | M C | = FBAd where d = perpendicular distance from C to line AB. M C = rA/C × FBA rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k FBA = λ BA FBA = ( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2 m = − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i j k ∴ M C = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k and | MC | = (146.88)2 + ( 60.48)2 + ( 205.92 )2 = 260.07 N ⋅ m ∴ 260.07 N ⋅ m = ( 228 N ) d d = 1.14064 m or d = 1.141 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 34. (a) Have d = rC/ A sin θ = λ AB × rC/ A where d = Perpendicular distance from C to pipe AB AB 7i + 4 j − 32k with λ AB = = AB ( 7 )2 + ( 4 )2 + ( −32 )2 1 = ( 7i + 4 j − 32k ) 33 and rC/ A = − (14 ft ) i + ( 5 ft ) j + ( L − 22 ) ft k i j k 1 Then λ AB × rC/ A = 7 4 − 32 ft 33 −14 5 L − 22 = 1 33 { } 4 ( L − 22 ) + 32 ( 5 ) i + 32 (14 ) − 7 ( L − 22 ) j + 7 ( 5 ) + 4 (14 ) k ft 1 = ( 4L + 72 ) i + ( −7 L + 602 ) j + 91k ft 33 1 and d = ( 4L + 72 )2 + ( −7 L + 602 )2 + ( 91)2 33 dd 2 1 For (d ) min , = 2 2 ( 4 )( 4L + 72 ) + 2 ( −7 )( −7 L + 602 ) = 0 dL 33 or 65L − 3926 = 0 or L = 60.400 ft But L > Lgreenhouse so L = 30.0 ft 1 (b) with L = 30 ft, d = ( 4 × 30 + 72 )2 + ( −7 × 30 + 602 )2 + ( 91)2 33 or d = 13.51 ft Note: with L = 60.4 ft, 1 d = ( 4 × 60.4 + 72 )2 + ( −7 × 60.4 + 602 )2 + ( 91)2 = 11.29 ft 33 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 35. P ⋅ Q = ( − 4i + 8j − 3k ) ⋅ ( 9i − j − 7k ) = ( − 4 )( 9 ) + ( 8 )( −1) + ( − 3)( − 7 ) = − 23 or P ⋅ Q = −23 P ⋅ S = ( − 4i + 8 j − 3k ) ⋅ ( 5i − 6 j + 2k ) = ( − 4 )( 5 ) + ( 8 )( − 6 ) + ( − 3)( 2 ) = − 74 or P ⋅ S = −74 Q ⋅ S = ( 9i − j − 7k ) ⋅ ( 5i − 6 j + 2k ) = ( 9 )( 5 ) + ( −1)( − 6 ) + ( − 7 )( 2 ) = 37 or Q ⋅ S = 37 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 36. By definition B ⋅ C = BC cos (α − β ) where B = B ( cos β ) i + ( sin β ) j C = C ( cos α ) i + ( sin α ) j ∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β ) or cos β cos α + sin β sin α = cos (α − β ) (1) By definition B′⋅ C = BC cos (α + β ) where B′ = ( cos β ) i − ( sin β ) j ∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β ) or cos β cos α − sin β sin α = cos (α + β ) (2) Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β ) 1 1 or cos α cos β = cos (α + β ) + cos (α − β ) 2 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 37. First note: rB/ A = ( 0.56 m ) i + ( 0.9 m ) j rC/ A = ( 0.9 m ) j − ( 0.48 m ) k rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k rB/ A = ( 0.56 m )2 + ( 0.9 m )2 = 1.06 m rC/ A = ( 0.9 m )2 + ( − 0.48 m )2 = 1.02 m rD/ A = ( − 0.52 m )2 + ( 0.9 m )2 + ( 0.36 m )2 = 1.10 m By definition rB/ A ⋅ rD/ A = rB/ A rD/ A cosθ or ( 0.56i + 0.9 j) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.06 )(1.10 ) cosθ ( 0.56 )( − 0.52 ) + ( 0.9 )( 0.9 ) + ( 0 )( 0.36 ) = 1.166 cosθ cosθ = 0.44494 θ = 63.6° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 38. From the solution to problem 3.37 rC/ A = 1.02 m with rC/ A = ( 0.9 m ) i − ( 0.48 m ) j rD/ A = 1.10 m with rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k Now by definition rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ or ( 0.9 j − 0.48k ) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.02 )(1.10 ) cosθ 0 ( − 0.52 ) + ( 0.9 )( 0.9 ) + ( − 0.48)( 0.36 ) = 1.122cosθ cosθ = 0.56791 or θ = 55.4° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 39. (a) By definition λ BC + λ EF = (1) (1) cosθ where λ BC = ( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k ( 32 )2 + ( − 9 )2 + ( − 24 )2 ft 1 = ( 32i − 9 j − 24k ) 41 − (14 ft ) i − (12 ft ) j + (12 ft ) k λ EF = ( −14 )2 + ( −12 )2 + ( 12 )2 ft 1 = ( −7i − 6 j + 6k ) 11 Therefore ( 32i − 9 j − 24k ) ⋅ ( −7i − 6 j + 6k ) = cosθ 41 11 ( 32 )( −7 ) + ( −9 )( −6 ) + ( −24 )( 6 ) = ( 41)(11) cosθ cosθ = − 0.69623 or θ = 134.1° (b) By definition (TEG ) BC = (TEF ) cosθ = (110 lb )( −0.69623) = −76.585 lb or (TEF ) BC = −76.6 lb Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 40. (a) By definition λ BC ⋅ λ EG = (1) (1) cosθ where λ BC = ( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k ( 32 )2 + ( − 9 )2 + ( − 24 )2 ft 1 = ( 32i − 9 j − 24k ) 41 λ EG = (16 ft ) i − (12 ft ) j + ( 9.75) k (16 )2 + ( −12 )2 + ( 9.75)2 ft 1 = (16i − 12 j + 9.75k ) 22.25 Therefore ( 32i − 9 j − 24k ) ⋅ (16i − 12 j + 9.75k ) = cosθ 41 22.25 ( 32 )(16 ) + ( −9 )( −12 ) + ( −24 )( 9.75) = ( 41)( 22.25) cosθ cosθ = 0.42313 or θ = 65.0° (b) By definition (TEG )BC = (TEG ) cosθ = (178 lb )( 0.42313) = 75.317 lb or (TEG ) BC = 75.3 lb Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 41. First locate point B: d 3.5 = 22 14 or d = 5.5 m (a) d BA = ( 5.5 + 0.5)2 + ( −22 )2 + ( −3)2 = 23 m Locate point D: ( −3.5 − 7.5sin 45° cos15° ) , (14 + 7.5cos 45° ) , ( 0 + 7.5sin 45° sin15° ) m or ( −8.6226 m, 19.3033 m, 1.37260 m ) Then d BD = ( −8.6226 + 5.5)2 + (19.3033 − 22 )2 + (1.37260 − 0 )2 m = 4.3482 m and cosθ ABD = d BA ⋅ d BD = ( 6i − 22 j − 3k ) ⋅ ( −3.1226i − 2.6967 j + 1.37260k ) d BA d BD ( 23)( 4.3482 ) = 0.36471 or θ ABD = 68.6° (b) (TBA )BD = TBA cosθ ABD = ( 230 N )( 0.36471) or (TBA ) BD = 83.9 N Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 42. First locate point B: d 3.5 = 22 14 or d = 5.5 m (a) Locate point D: ( −3.5 − 7.5sin 45° cos10° ) , (14 + 7.5cos 45° ) , ( 0 + 7.5sin 45° sin10° ) m or ( −8.7227 m, 19.3033 m, 0.92091 m ) Then d DC = ( 5.2227 m ) i − ( 5.3033 m ) j − ( 0.92091 m ) k and d DB = ( −5.5 + 8.7227 )2 + ( 22 − 19.3033)2 + ( 0 − 0.92091)2 m = 4.3019 m and cos θ BDC = d DB ⋅ d DC = ( 3.2227i + 2.6967 j − 0.92091k ) ⋅ ( 5.2227i − 5.3033j − 0.92091k ) d DB d DC ( 4.3019 )( 7.5) = 0.104694 or θ BDC = 84.0° (b) (TBD )DC = TBD cosθ BDC = ( 250 N )( 0.104694 ) or (TBD )DC = 26.2 N Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 43. Volume of parallelopiped is found using the mixed triple product (a) Vol = P ⋅ ( Q × S ) 3 −4 1 = − 7 6 − 8 in.3 9 −2 −3 = ( −54 + 288 + 14 − 48 + 84 − 54 ) in.3 = 230 in.3 or Volume = 230 in.3 (b) Vol = P ⋅ ( Q × S ) −5 −7 4 = 6 − 2 5 in.3 −4 8 −9 = ( −90 + 140 + 192 + 200 − 378 − 32 ) in.3 = 32 in.3 or Volume = 32 in.3 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 44. For the vectors to all be in the same plane, the mixed triple product is zero. P ⋅(Q × S ) = 0 −3 −7 5 ∴ 0 = −2 1 −4 8 Sy −6 0 = 18 + 224 − 10S y − 12S y + 84 − 40 So that 22 S y = 286 S y = 13 or S y = 13.00 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 45. Have rC = ( 2.25 m ) k CE TCE = TCE CE ( 0.90 m ) i + (1.50 m ) j − ( 2.25 m ) k TCE = (1349 N ) ( 0.90 ) + (1.50 ) + ( −2.25 ) m 2 2 2 = ( 426 N ) i + ( 710 N ) j − (1065 N ) k Now M O = rC × TCE i j k = 0 0 2.25 N ⋅ m 426 710 −1065 = − (1597.5 N ⋅ m ) i + ( 958.5 N ⋅ m ) j ∴ M x = −1598 N ⋅ m, M y = 959 N ⋅ m, M z = 0 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 46. Have rE = ( 0.90 m ) i + (1.50 m ) j DE TDE = TDE DE − ( 2.30 m ) i + (1.50 m ) j − ( 2.25 m ) k = (1349 N ) ( − 2.30 ) + (1.50 ) + ( − 2.25) m 2 2 2 = − ( 874 N ) i + ( 570 N ) j − ( 855 N ) k Now M O = rE × TDE i j k = 0.90 1.50 0 N ⋅ m − 874 570 − 855 = − (1282.5 N ⋅ m ) i + ( 769.5 N ⋅ m ) j + (1824 N ⋅ m ) k ∴ M x = −1283 N ⋅ m, M y = 770 N ⋅ m, M z = 1824 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 47. Have M z = k ⋅ ( rB ) y × TBA + k ( rC ) y × TCD where M z = − ( 48 lb ⋅ ft ) k ( rB ) y = ( rC ) y = ( 3 ft ) j BA ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k TBA = TBA = (14 lb ) BA ( 4.5) + ( − 3) + ( 9 ) ft 2 2 2 = ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k CD ( 6 ft ) i − ( 3 ft ) j − ( 6 ft ) k TCD = TCD = TCD CD ( 6 ) + ( − 3) + ( − 6 ) ft 2 2 2 TCD = (2i − j − 2k ) 3 Then { } − ( 48 lb ⋅ ft ) = k ⋅ ( 3 ft ) j × ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k T + k ⋅ ( 3 ft ) j × CD ( 2 i − j − 2 k ) 3 or − 48 = −18 − 2TCD TCD = 15.00 lb Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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COSMOS: Complete Online
Solutions Manual Organization System Chapter 3, Solution 48. Have M y = j ⋅ ( rB ) z × TBA × j⋅ ( rC ) z × TCD where M y = 156 lb ⋅ ft ( rB ) z = ( 24 ft ) k; ( rC ) z = ( 6 ft ) k BA ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k TBA = TBA = TBA BA ( 4.5) + ( − 3) + ( 9 ) ft 2 2 2 TBA = ( 4.5i − 3j + 9k ) 10.5 CD ( 6 ft ) i − ( 3 ft ) j + ( 9 ft ) k TCD = TCD = ( 7.5 lb ) CD ( 6 ) + ( − 3) + ( 9 ) ft 2 2 2 = ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k T Then (156 lb ⋅ ft ) = j ⋅ ( 24 ft ) k × BA ( 4.5i − 3j + 9k ) 10.5 { } + j ⋅ ( 6 ft ) k × ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k 108 or 156 = TBA + 30 10.5 TBA = 12.25 lb Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 49. Based on M x = ( P cos φ ) ( 0.225 m ) sin θ − ( P sin φ ) ( 0.225 m ) cosθ (1) M y = − ( P cos φ )( 0.125 m ) (2) M z = − ( P sin φ )( 0.125 m ) (3) Equation ( 3) M z − ( P sin φ )( 0.125 ) By : = Equation ( 2 ) M y − ( P cos φ )( 0.125 ) −4 or = tan φ ∴ φ = 9.8658° − 23 or φ = 9.87° From Equation (2) − 23 N ⋅ m = − ( P cos 9.8658° )( 0.125 m ) P = 186.762 N or P = 186.8 N From Equation (1) 26 N ⋅ m = (186.726 N ) cos 9.8658° ( 0.225 m ) sin θ − (186.726 N ) sin 9.8658° ( 0.225 m ) cosθ or 0.98521sin θ − 0.171341cosθ = 0.61885 Solving numerically, θ = 48.1° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Solutions Manual Organization System Chapter 3, Solution 50. Based on M x = ( P cos φ ) ( 0.225 m ) sin θ − ( P sin φ ) ( 0.225 m ) cosθ (1) M y = − ( P cos φ )( 0.125 m ) (2) M z = − ( P sin φ )( 0.125 m ) Equation ( 3) M z − ( P sin φ )( 0.125 ) By : = Equation ( 2 ) M y − ( P cos φ )( 0.125 ) − 3.5 or = tan φ ; φ = 9.9262° − 20 From Equation (3): − 3.5 N ⋅ m = − ( P sin 9.9262° )( 0.125 m ) P = 162.432 N From Equation (1): M x = (162.432 N )( 0.225 m )( cos 9.9262° sin 60° − sin 9.9262° cos 60° ) = 28.027 N ⋅ m or M x = 28.0 N ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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