Baixe Vector Mechanics for Engineers - Statics 8E Solutions - chapter 3 e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity! COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 1. Resolve 90 N force into vector components P and Q where ( )90 N sin 40Q = ° 57.851 N= Then /B A BM r Q= − ( )( )N851.57m225.0−= 13.0165 N m= − ⋅ 13.02 N mB = ⋅M COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 2. ( )90 N cos 25xF = ° 81.568 N= ( )90 N sin 25yF = ° 38.036 N= ( )0.225 m cos65x = ° 0.095089 m= ( ) °= 65sinm225.0y 0.20392 m= B y xM xF yF= − ( )( ) ( )( )0.095089 m 38.036 N 0.20392 m 81.568 N= − 13.0165 N m= − ⋅ 13.02 N mB = ⋅M COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 5. By definition / sinA B AM r P θ= where ( )90θ φ α= + ° − and 1 4.8 in.tan 3.4 in. φ −   =     54.689= ° Also ( ) ( )2 2/ 3.4 in. 4.8 in.B Ar = + 5.8822 in.= Then ( ) ( )( ) ( )17 lb in. 5.8822 in. 2.9 lb sin 54.689 90 α⋅ = ° + ° − or ( )sin 144.689 0.99658α° − = or 144.689 85.260 ; 94.740α° − = ° ° 49.9 , 59.4α∴ = ° ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 6. (a) (b) (a) /A B A BF= ×M r T y xA BF BFM xT yT= + ( )( ) ( )( )2 m 200 N sin 60 0.4 m 200 N cos60= ° + ° 386.41 N m= ⋅ or 386 N mA = ⋅M (b) For CF to be a minimum, it must be perpendicular to the line joining A and C. ( )minA CM d F∴ = with ( ) ( )2 22 m 1.35 md = + 2.4130 m= Then ( )( )min386.41 N m 2.4130 m CF⋅ = ( )min 160.137 NCF = and 1 1.35 mtan 34.019 2 m φ −   = = °    90 90 34.019 55.981θ φ= − = ° − ° = ° ( )min 160.1 NC∴ =F 56.0° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 7. (a) (b) (c) y xA BF BF M xT yT= + ( )( ) ( )( )2 m 200 N sin 60 0.4 m 200 N cos60= ° + ° 386.41 N m= ⋅ or 386 N m A = ⋅M Have A C M xF= or 386.41 N m 2 m A C M F x ⋅= = 193.205 N= 193.2 N C ∴ =F For B F to be minimum, it must be perpendicular to the line joining A and B ( ) minA B M d F∴ = with ( ) ( )2 22 m 0.40 m 2.0396 md = + = Then ( )( ) min 386.41 N m 2.0396 m C F⋅ = ( ) min 189.454 N C F = and 1 2 m tan 78.690 0.4 m θ −   = = °    or ( ) min 189.5 N C =F 78.7° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 10. Slope of line 35 in. 7 112 in. 8 in. 24 EC = = + Then 24 25 ABx ABT T= and 7 25 ABy ABT T= Have ( ) ( )D ABx AByM T y T x= + ( ) ( )24 77840 lb in. 0 112 in. 25 25 AB AB T T∴ ⋅ = + 250 lb AB T = or 250 lb AB T = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 11. The minimum value of d can be found based on the equation relating the moment of the force ABT about D: ( ) ( )maxD AB yM T d= where 1152 N mDM = ⋅ ( ) ( )max max sin 2880 N sinAB AByT T θ θ= = Now ( ) ( )2 2 1.05 msin 0.24 1.05 md θ = + + ( ) ( ) ( ) 2 2 1.051152 N m 2880 N 0.24 1.05 d d    ∴ ⋅ =   + +   or ( ) ( )2 20.24 1.05 2.625d d+ + = or ( ) ( )2 2 20.24 1.05 6.8906d d+ + = or 25.8906 0.48 1.1601 0d d− − = Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. Since only the positive value applies here, 0.48639 md = or 486 mmd = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 12. with ( ) ( )2 242 mm 144 mmABd = + 150 mm= 42 mmsin 150 mm θ = 144 mmcos 150 mm θ = and sin cosAB AB ABF Fθ θ= − −F i j ( ) ( )2.5 kN 42 mm 144 mm 150 mm  = − − i j ( ) ( )700 N 2400 N= − −i j Also ( ) ( )/ 0.042 m 0.056 mB C = − +r i j Now /C B C AB= ×M r F ( ) ( )0.042 0.056 700 2400 N m= − + × − − ⋅i j i j ( )140.0 N m= ⋅ k or 140.0 N mC = ⋅M COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 15. Note: ( )cos sinB β β= +B i j ( )cos sinB β β′ = −B i j ( )cos sinC α α= +C i j By definition: ( )sinBC α β× = −B C (1) ( )sinC BC α β′ × = +B (2) Now ... ( ) ( )cos sin cos sinB Cβ β α α× = + × +B C i j i j ( )cos sin sin cosBC β α β α= − k (3) and ( ) ( )cos sin cos sinB Cβ β α α′ × = − × +B C i j i j ( )cos sin sin cosBC β α β α= + k (4) Equating the magnitudes of ×B C from equations (1) and (3) yields: ( ) ( )sin cos sin sin cosBC BCα β β α β α− = − (5) Similarly, equating the magnitudes of ′ ×B C from equations (2) and (4) yields: ( ) ( )sin cos sin sin cosBC BCα β β α β α+ = + (6) Adding equations (5) and (6) gives: ( ) ( )sin sin 2cos sinα β α β β α− + + = or ( ) ( )1 1sin cos sin sin 2 2 α β α β α β= + + − COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 16. Have /AB O Ad = × rλ where / / B A AB B Ar = r λ and ( )/ 210 mm 630 mmB A = − −r i ( )( )270 mm 225 mm+ − − j ( ) ( )840 mm 495 mm= − +i j ( ) ( )2 2/ 840 mm 495 mmB Ar = − + 975 mm= Then ( ) ( )840 mm + 495 mm 975 mmAB λ − = i j ( )1 56 33 65 = − +i j Also ( ) ( )/ 0 630 0 ( 225)O A = − + − −r i j ( ) ( )630 mm 225 mm= − +i j ( ) ( ) ( )1 56 33 630 mm 225 mm 65 d  ∴ = − + × − + i j i j 126.0 mm= 126.0 mmd = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 17. (a) ×= × A B A B λ where 12 6 9= − +A i j k 3 9 7.5= − + −B i j k Then 12 6 9 3 9 7.5 × = − − − i j k A B ( ) ( ) ( )45 81 + 108 18= − −27 + 90 + −i j k ( )9 4 7 10= − + +i j k And 2 2 29 ( 4) (7) (10) 9 165× = − + + =A B ( )9 4 7 10 9 165 − + + ∴ = i j k λ or ( )1 4 7 10 165 = − + +i j kλ (b) ×= × A B A B λ where 14 2 8= − − +A i j k 3 1.5= + −B i j k Then 14 2 8 3 1.5 1 × = − − − i j k A B ( ) ( ) ( )2 12 24 14 21 6= − + − + − +i j k ( )5 2 2 3= − + −i j k and 2 2 25 ( 2) (2) ( 3) 5 17× = − + + − =A B ( )5 2 2 3 5 17 − + − ∴ = i j k λ or ( )1 2 2 3 17 = − + −i j k λ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 20. (a) Have O = ×M r F 7.5 3 6 lb ft 3 6 4 = − − ⋅ − i j k [ ](12 36) ( 18 30) (45 9) lb ft= − + − + + − ⋅i j k or ( ) ( ) ( )24.0 lb ft lb ft lb ftO = − ⋅ + 12.00 ⋅ + 36.0 ⋅M i j k (b) Have O = ×M r F 7.5 1.5 1 lb ft 3 6 4 = − − ⋅ − i j k [ ](6 6) ( 3 3) (4.5 4.5) lb ft= − + − + + − ⋅i j k or 0O =M (c) Have O = ×M r F 8 2 14 lb ft 3 6 4 = − − ⋅ − i j k [ ](8 84) ( 42 32) (48 6) lb ft= − + − + + − ⋅i j k or ( ) ( ) ( )76.0 lb ft lb ft lb ftO = − ⋅ − 10.00 ⋅ + 42.0 ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 21. With ( )369 NAB = −T j ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2.4 m 3.1 m 1.2 m 369 N 2.4 m 3.1 m 1.2 m AB AD ADT AD − − = = + − + − i j k T ( ) ( ) ( )216 N 279 N 108 NAD = − −T i j k Then 2A AB AD= +R T T ( ) ( ) ( )216 N 1017 N 108 N= − −i j k Also ( ) ( )/ 3.1 m 1.2 mA C = +r i k Have /C A C A= ×M r R 0 3.1 1.2 N m 216 1017 108 = ⋅ − − i j k ( ) ( ) ( )885.6 N m 259.2 N m 669.6 N m= ⋅ + ⋅ − ⋅i j k ( ) ( ) ( )886 N m 259 N m 670 N mC = ⋅ + ⋅ − ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 22. Have /A C A= ×M r F where ( ) ( ) ( )/ 215 mm 50 mm 140 mmC A = − +r i j k ( )36 N cos 45 sin12xF = − ° ° ( )36 N sin 45yF = − ° ( )36 N cos 45 cos12zF = − ° ° ( ) ( ) ( )5.2926 N 25.456 N 24.900 N∴ = − − −F i j k and 0.215 0.050 0.140 N m 5.2926 25.456 24.900 A = − ⋅ − − − i j k M ( ) ( ) ( )4.8088 N m 4.6125 N m 5.7377 N m= ⋅ + ⋅ − ⋅i j k ( ) ( ) ( )4.81 N m 4.61 N m 5.74 N mA = ⋅ + ⋅ − ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. (b) Have /O B O BE= ×M r T where ( ) ( )2.5 m 2 mB/O = +r i j BE BE BET BE =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0.5 m 2 m 4 m 675 N 0.5 m 2 m 4 m  − − − = + − + − i j k ( ) ( ) ( )75 N 300 N 600 N= − − −i j k Then 2.5 2 0 N m 75 300 600 O = ⋅ − − − i j k M ( ) ( ) ( )1200 N m 1500 N m 600 N mO = − ⋅ + ⋅ − ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 25. Have /C A C= ×M r P where / / /A C B C A B= +r r r ( )( )= 16 in. cos80 cos15 sin80 cos80 sin15− ° ° − ° − ° °i j k ( )( )15.2 in. sin 20 cos15 cos 20 sin 20 sin15+ − ° ° + ° − ° °i j k ( ) ( ) ( )7.7053 in. 1.47360 in. 2.0646 in.= − − −i j k and ( )( )150 lb cos5 cos70 sin 5 cos5 sin 70= ° ° + ° − ° °P i j k ( ) ( ) ( )51.108 lb 13.0734 lb 140.418 lb= + −i j k Then 7.7053 1.47360 2.0646 lb in. 51.108 13.0734 140.418 C = − − − ⋅ − i j k M ( ) ( ) ( )233.91 lb in. 1187.48 lb in. 25.422 lb in.= ⋅ − ⋅ − ⋅i j k or ( ) ( ) ( )19.49 lb ft 99.0 lb ft 2.12 lb ftC = ⋅ − ⋅ − ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 26. Have /C A C BA=M r F× where ( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mA C = − +r i j k and BA BA BAF=F λ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 0.1 m 1.8 m 0.6 m 228 N 0.1 1.8 0.6 m  − + − =   + +   i j k ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 0.96 0.12 0.72 N m 12.0 216 72 C∴ = − ⋅ − − i j k M ( ) ( ) ( )146.88 N m 60.480 N m 205.92 N m= − ⋅ + ⋅ + ⋅i j k or ( ) ( ) ( )146.9 N m 60.5 N m 206 N mC = − ⋅ + ⋅ + ⋅M i j k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 29. Have O ABT d=M where d = Perpendicular distance from O to rope AB with /O A O AB= ×M r T and ( ) ( )/ 30 ft 3 ftA O = +r j k AB AB ABT AB =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 5 ft 30 ft 6 ft 62 lb 5 ft 30 ft 6 ft  − + = + − + i j k ( ) ( ) ( )10 lb 60 lb 12 lb= − +i j k Then 0 30 3 lb ft 10 60 12 O = ⋅ − i j k M ( ) ( ) ( )540 lb ft 30 lb ft 300 lb ft= ⋅ + ⋅ − ⋅i j k and ( ) ( ) ( )2 2 2540 lb ft 30 lb ft 300 lb ftO = ⋅ + ⋅ + − ⋅M 618.47 lb ft= ⋅ ( )618.47 lb ft = 62 lb d∴ ⋅ or 9.98 ftd = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 30. Have C BDT d=M where d = Perpendicular distance from C to cable BD with / /C B C B D= ×M r T and ( )/ 2 mB C =r j BD BD BDT BD =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 1 m 2 m 2 m 900 N 1 m 2 m 2 m  − − + = − + − + i j k ( ) ( ) ( )300 N 600 N 600 N= − − +i j k Then 0 2 0 N m 300 600 600 C = ⋅ − − i j k M ( ) ( )1200 N m 600 N m= ⋅ + ⋅i k and ( ) ( )2 21200 N m 600 N mC = ⋅ + ⋅M 1341.64 N m= ⋅ ( )1341.64 = 900 N d∴ or 1.491 md = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 31. Have CM Pd= From the solution of problem 3.25 ( ) ( ) ( )233.91 lb in. 1187.48 lb in. 25.422 lb in.C = ⋅ − ⋅ − ⋅M i j k Then ( ) ( ) ( ) 2 2 2233.91 1187.48 25.422CM = + − + − 1210.57 lb in.= ⋅ and 1210.57 lb.in. 150 lb CMd P = = or 8.07 in.d = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 34. (a) Have / /sinC A AB C Ad r θ= = × rλ where d = Perpendicular distance from C to pipe AB with ( ) ( ) ( )2 2 2 7 4 32 7 4 32 AB AB + − = = + + − AB i j k λ ( )1 7 4 32 33 = + −i j k and ( ) ( ) ( )/ 14 ft 5 ft 22 ftC A L = − + + − r i j k Then / 1 7 4 32 ft 33 14 5 22 AB C A L × = − − − i j k rλ ( ) ( ) ( ) ( ) ( ) ( ){ }1 4 22 32 5 32 14 7 22 7 5 4 14 ft33 L L     = − + + − − + +     i j k ( ) ( )1 4 72 7 602 91 ft 33 L L = + + − + + i j k and ( ) ( ) ( )2 2 21 4 72 7 602 91 33 d L L= + + − + + For ( )( ) ( )( ) 2 min 2 1( ) , 2 4 4 72 2 7 7 602 0 d 33 ddd L L L  = + + − − + =  or 65 3926 0L − = or 60.400 ftL = But greenhouseL L> so 30.0 ftL = (b) with ( ) ( ) ( )2 2 2130 ft, 4 30 72 7 30 602 91 33 L d= = × + + − × + + or 13.51 ftd = Note: with 60.4 ft,L = ( ) ( ) ( )2 2 21 4 60.4 72 7 60.4 602 91 11.29 ft 33 d = × + + − × + + = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 35. ( ) ( )4 8 3 9 7⋅ = − + − ⋅ − −P Q i j k i j k ( )( ) ( )( ) ( )( )4 9 8 1 3 7= − + − + − − 23= − or 23⋅ = −P Q ( ) ( )4 8 3 5 6 2⋅ = − + − ⋅ − +P S i j k i j k ( )( ) ( )( ) ( )( )4 5 8 6 3 2= − + − + − 74= − or 74⋅ = −P S ( ) ( )9 7 5 6 2⋅ = − − ⋅ − +Q S i j k i j k ( )( ) ( )( ) ( )( )9 5 1 6 7 2= + − − + − 37= or 37⋅ =Q S COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 36. By definition ( )cosBC α β⋅ = −B C where ( ) ( )cos sinB β β = + B i j ( ) ( )cos sinC α α = + C i j ( )( ) ( )( ) ( )cos cos sin sin cosB C B C BCβ α β α α β∴ + = − or ( )cos cos sin sin cosβ α β α α β+ = − (1) By definition ( )cosBC α β′⋅ = +B C where ( ) ( )cos sinβ β′  = − B i j ( )( ) ( )( ) ( )cos cos sin sin cosB C B C BCβ α β α α β∴ + − = + or ( )cos cos sin sin cosβ α β α α β− = + (2) Adding Equations (1) and (2), ( ) ( )2 cos cos cos cosβ α α β α β= − + + or ( ) ( )1 1cos cos cos cos 2 2 α β α β α β= + + − COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 39. (a) By definition (1) (1)cosBC EF θ+ =λ λ where ( ) ( ) ( ) ( ) ( ) ( )2 2 2 32 ft 9 ft 24 ft 32 9 24 ft BC − − = + − + − i j k λ ( )1 32 9 24 41 = − −i j k ( ) ( ) ( ) ( ) ( ) ( )2 2 2 14 ft 12 ft 12 ft 14 12 12 ft EF − − + = − + − + i j k λ ( )1 7 6 6 11 = − − +i j k Therefore ( ) ( )32 9 24 7 6 6 cos 41 11 θ − − − − + ⋅ = i j k i j k ( )( ) ( )( ) ( )( ) ( )( )32 7 9 6 24 6 41 11 cosθ− + − − + − = cos 0.69623θ = − or 134.1θ = ° (b) By definition ( ) ( )cosEG EFBCT T θ= ( )( )110 lb 0.69623= − 76.585 lb= − or ( ) 76.6 lbEF BCT = − COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 40. (a) By definition (1) (1)cosBC EG θ⋅ =λ λ where ( ) ( ) ( ) ( ) ( ) ( )2 2 2 32 ft 9 ft 24 ft 32 9 24 ft BC − − = + − + − i j k λ ( )1 32 9 24 41 = − −i j k ( ) ( ) ( ) ( ) ( ) ( )2 2 2 16 ft 12 ft 9.75 16 12 9.75 ft EG − + = + − + i j k λ ( )1 16 12 9.75 22.25 = − +i j k Therefore ( ) ( )32 9 24 16 12 9.75 cos 41 22.25 θ − − − + ⋅ = i j k i j k ( )( ) ( )( ) ( )( ) ( )( )32 16 9 12 24 9.75 41 22.25 cosθ+ − − + − = cos 0.42313θ = or 65.0θ = ° (b) By definition ( ) ( )cosEG EGBCT T θ= ( )( )178 lb 0.42313= 75.317 lb= or ( ) 75.3 lbEG BCT = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 41. First locate point B: 3.5 22 14 d = or 5.5 md = (a) ( ) ( ) ( ) 2 2 25.5 0.5 22 3 23 mBAd = + + − + − = 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 ( ) ( ) ( )2 2 28.6226 5.5 19.3033 22 1.37260 0 mBDd = − + + − + − 4.3482 m= and ( ) ( ) ( )( ) 6 22 3 3.1226 2.6967 1.37260 cos 23 4.3482 BA BD ABD BA BD d d d d θ − − ⋅ − − +⋅ = = i j k i j k 0.36471= or 68.6ABDθ = ° (b) ( ) cosBA BA ABDBDT T θ= ( )( )230 N 0.36471= or ( ) 83.9 NBA BDT = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 44. For the vectors to all be in the same plane, the mixed triple product is zero. ( ) 0⋅ × =P Q S 3 7 5 0 2 1 4 8 6yS − − ∴ = − − − 10 12 84 40y yS S0 = 18 + 224 − − + − So that 22 286yS = 13yS = or 13.00yS = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 45. Have ( )2.25 mC =r k CE CE CET CE =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0.90 m 1.50 m 2.25 m 1349 N 0.90 1.50 2.25 m CE  + − = + + − i j k T ( ) ( ) ( )426 N 710 N 1065 N= + −i j k Now O C CE= ×M r T 0 0 2.25 N m 426 710 1065 = ⋅ − i j k ( ) ( )1597.5 N m 958.5 N m= − ⋅ + ⋅i j 1598 N m, 959 N m, 0x y zM M M∴ = − ⋅ = ⋅ = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 46. Have ( ) ( )0.90 m 1.50 mE = +r i j DE DE DET DE =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2.30 m 1.50 m 2.25 m 1349 N 2.30 1.50 2.25 m  − + − = − + + − i j k ( ) ( ) ( )874 N 570 N 855 N= − + −i j k Now O E DE= ×M r T 0.90 1.50 0 N m 874 570 855 = ⋅ − − i j k ( ) ( ) ( )1282.5 N m 769.5 N m 1824 N m= − ⋅ + ⋅ + ⋅i j k 1283 N m, 770 N m, 1824 N mx y zM M M∴ = − ⋅ = ⋅ = ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 49. Based on ( ) ( ) ( ) ( )cos 0.225 m sin sin 0.225 m cosxM P Pφ θ φ θ   = −    (1) ( )( )cos 0.125 myM P φ= − (2) ( )( )sin 0.125 mzM P φ= − (3) By ( ) ( ) ( )( ) ( )( ) Equation 3 sin 0.125 : Equation 2 cos 0.125 z y PM M P φ φ − = − or 4 tan 9.8658 23 φ φ− = ∴ = ° − or 9.87φ = ° From Equation (2) ( )( )23 N m cos9.8658 0.125 mP− ⋅ = − ° 186.762 NP = or 186.8 NP = From Equation (1) ( ) ( )26 N m 186.726 N cos9.8658 0.225 m sinθ   ⋅ = °    ( ) ( )186.726 N sin 9.8658 0.225 m cosθ   − °    or 0.98521sin 0.171341cos 0.61885θ θ− = Solving numerically, 48.1θ = ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 50. Based on ( ) ( ) ( ) ( )cos 0.225 m sin sin 0.225 m cosxM P Pφ θ φ θ   = −    (1) ( )( )cos 0.125 myM P φ= − (2) ( )( )sin 0.125 mzM P φ= − By ( ) ( ) ( )( ) ( )( ) Equation 3 sin 0.125 : Equation 2 cos 0.125 z y PM M P φ φ − = − or 3.5 tan ; 9.9262 20 φ φ− = = ° − From Equation (3): ( )( )3.5 N m sin 9.9262 0.125 mP− ⋅ = − ° 162.432 NP = From Equation (1): ( )( )( )162.432 N 0.225 m cos9.9262 sin 60 sin 9.9262 cos60xM = ° ° − ° ° 28.027 N m= ⋅ or 28.0 N mxM = ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 51. First note: BA BA BAT BA =T ( ) ( ) ( ) ( ) ( ) ( ) ( )22 2 4 1.5 1 6 70 lb 4 1.5 1 6 BC BC L L  + − + + − =  + − + + −  i j k ( ) ( ) ( )2 4 0.5 6 70 lb 52 0.5 BC BC L L + − − = + − i j k ( ) ( ) ( )4 ft 1.5 ft 12 ftA = + −r i j k Have O A BA= ×M r T ( ) ( ) 2 70 lb 4 ft 1.5 ft 12 ft 52 + 0.5 4 0.5 6BC BCL L = − − − − i j k For the i components: ( ) ( ) ( ) 2 70763 lb ft 1.5 6 12 0.5 lb ft 52 + 0.5 BC BC L L  − ⋅ = − + − ⋅  − or ( )210.9 52 0.5 3 12BC BCL L+ − = + or ( ) ( )22 210.9 52 0.5 9 72 144BC BC BCL L L + − = + +   or 225.19 190.81 6198.8225 0BC BCL L+ − = Then ( ) ( )( ) ( ) 2190.81 190.81 4 25.19 6198.8225 2 25.19BC L − ± − − = Taking the positive root 12.35 ftBCL = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 54. Have ( )/AD AD B A BGM = ⋅ ×λ r T where ( ) ( )0.8 m 0.6 mAD = −λ i k ( )/ 0.4 mB A =r i ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0.4 m 0.74 0.32 m 1125 N 0.4 m 0.74 m 0.32 m BG BG BGT BG  − + − = = − + + − i j k T ( ) ( ) ( )500 N 925 N 400 N= − + −i j k Then 0.8 0 0.6 0.4 0 0 500 925 400 ADM − = − − or 222 N mADM = − ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 55. Have ( )/AD AD E A EFM = ⋅ ×λ r F where AD AD AD =λ ( ) ( ) ( ) ( )2 2 7.2 m 0.9 m 7.2 m 0.9 m AD + = + i j λ 0.99228 0.124035= +i j ( ) ( )/ 2.1 m 0.9 mE A = −r i j ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0.3 m 1.2 m 2.4 m 24.3 kN 0.3 m 1.2 m 2.4 m EF EF EFF EF  + + = = + + i j k F ( ) ( ) ( )2.7 kN 10.8 kN 21.6 kN= + +i j k Then 0.99228 0.124035 0 2.1 0.9 0 kN m 2.7 10.8 21.6 ADM = − ⋅ 19.2899 5.6262= − − 24.916 kN m= − ⋅ or 24.9 kN mADM = − ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 56. Have ( )/AD AD G A EFM = ⋅ ×λ r E Where ( ) ( ) ( ) ( )2 2 7.2 m 0.9 m 7.2 m 0.9 m AD + = + i j λ 0.99228 0.124035= +i j ( ) ( )/ 6 m 1.8 mG A = −r i j ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 1.2 m 2.4 m 2.4 m 21.3 kN 1.2 m 2.4 m 2.4 m GH GH GHF GH  − + + = = − + + i j k F ( ) ( ) ( )7.1 kN 14.2 kN 14.2 kN= − + +i j k Then 0.99228 0.124035 0 6 1.8 0 kN m 7.1 14.2 14.2 ADM = − ⋅ − 25.363 10.5678= − − 35.931 kN m= − ⋅ or 35.9 kN mADM = − ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 59. Have ( )/DI DI F I EFM = ⋅ ×λ r T where ( ) ( ) ( ) ( )2 2 4.8 ft 1.2 ft 4.8 ft 1.2 ft DI DI DI − = = + − i j λ 0.97014 0.24254= −i j ( )/ 16.2 ftF I =r k EF EF EFT EF =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 3.6 ft 10.8 ft 16.2 ft 29.7 lb 3.6 ft 10.8 ft 16.2 ft  − + = + − + i j k ( ) ( ) ( )5.4 lb 16.2 lb 24.3 lb= − +i j k Then 0.97014 0.24254 0 0 0 16.2 lb ft 5.4 16.2 24.3 DIM − = ⋅ − 21.217 254.60= − + 233.39 lb ft= ⋅ or 233 lb ftDIM = ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 60. Have ( )/DI DI G I EGM = ⋅ ×λ r T where ( ) ( ) ( ) ( )2 2 4.8 ft 1.2 ft 4.8 ft 1.2 ft DI DI DI − = = + − i j λ 0.97014 0.24254= −i j ( )/ 35.1 ftG I = −r k EG EG EGT EG =T ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 3.6 ft 10.8 ft 35.1 ft 24.6 lb 3.6 ft 10.8 ft 35.1 ft  − − = + − + − i j k ( ) ( ) ( )2.4 lb 7.2 lb 23.4 lb= − −i j k Then 0.97014 0.24254 0 0 0 35.1 lb ft 2.4 7.2 23.4 DIM − = − ⋅ − − 20.432 245.17= − 224.74 lb ft= − ⋅ or 225 lb ftDIM = − ⋅ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 61. First note that 1 1 1F=F λ and 2 2 2F=F λ Let 1 2 moment of M = F about the line of action of 1M and 2 1 moment of M = F about the line of action of 2M Now, by definition ( ) ( )1 1 / 2 1 / 2 2B A B AM F= ⋅ = ⋅r F rλ × λ × λ ( ) ( )2 2 / 1 2 / 1 1A B A BM F= ⋅ = ⋅r F rλ × λ × λ Since 1 2F F F= = and / /A B B A= −r r ( )1 1 / 2B AM F= ⋅ rλ × λ ( )2 2 / 1B AM F= ⋅ −rλ × λ Using Equation (3.39) ( ) ( )1 / 2 2 / 1B A B A⋅ = ⋅ −r rλ × λ λ × λ so that ( )2 1 / 2B AM F= ⋅ rλ × λ 12 21 M M∴ = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 64. From the solution of Problem 3.59: 0.97014 0.24254DI = −i jλ ( ) ( ) ( )5.4 lb 16.2 lb 24.3 lb 29.7 lb EF EFT = − + = T i j k 233.39 lb ftDIM = ⋅ Only the perpendicular component of EFT contributes to the moment of EFT about line DI. The parallel component of EFT will be used to find the perpendicular component. Have ( )ParallelEF DI EF= ⋅T Tλ [ ] ( ) ( ) ( )0.97014 0.24254 5.4 lb 16.2 lb 24.3 lb = − ⋅ − + i j i j k ( )5.2388 3.9291= + 9.1679 lb= Since ( ) ( )Perpendicular ParallelEF EF EF= +T T T Then ( ) ( ) ( )2 2Perpendicular ParallelEF EF EFT = −T T ( ) ( )2 229.7 9.1679= − 28.250 lb= and ( )PerpendicularDI EF d=M T ( )233.39 lb ft 28.250 lb d⋅ = 8.2616 ftd = or 8.26 ftd = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 65. From the solution of Problem 3.60: 0.97014 0.24254DI = −i jλ ( ) ( ) ( )2.4 lb 7.2 lb 23.4 lb 24.6 lb EG EGT = − − = T i j k 224.74 lb ftDIM = − ⋅ Only the perpendicular component of EGT contributes to the moment of EGT about line DI. The parallel component of EGT will be used to find the perpendicular component. Have ( )ParallelEG DI EG= ⋅T Tλ [ ] ( ) ( ) ( )0.97014 0.24254 2.4 lb 7.2 lb 23.4 lb = − ⋅ − − i j i j k ( )2.3283 1.74629= + 4.0746 lb= Since ( ) ( )Perpendicular ParallelEG EG EGT= +T T Then ( ) ( ) ( )2 2Perpendicular ParallelEG EG EGT T T= − ( ) ( )2 224.6 4.0746= − 24.260 lb= and ( )PerpendicularDI EGM T d= ( )224.74 lb ft 24.260 lb d⋅ = 9.2638 ftd = or 9.26 ftd = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 66. From the solution of Prob. 3.55: 0.99228 0.124035AD = +i jλ ( ) ( ) ( )2.7 kN 10.8 kN 21.6 kNEF = + +F i j k 24.3 kNEFF = 24.916 kN mADM = − ⋅ Only the perpendicular component of EFF contributes to the moment of EFF about edge AD. The parallel component of EFF will be used to find the perpendicular component. Have ( )ParallelEF AD EF= ⋅F Fλ [ ] ( ) ( ) ( )0.99228 0.124035 2.7 kN 10.8 kN 21.6 kN = + ⋅ + + i j i j k 4.0187 kN= Since ( ) ( )Perpendicular ParallelEF EF EF= +F F F Then ( ) ( ) ( )2 2Perpendicular ParallelEF EF EFF F F= − ( ) ( )2 224.3 4.0187= − 23.965 kN= and ( )PerpendicularAD EFM F d= ( )24.916 kN m 23.965 kN d⋅ = 1.039683md = or 1.040 md = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 69. (a) M Fd= ( )12 N m 0.45 mF⋅ = or 26.7 NF = (b) M Fd= ( )12 N m 0.24 mF⋅ = or 50.0 NF = (c) M Fd= ( ) ( )2 2Where 0.45 m 0.24md = + 0.51 m= ( )12 N m 0.51 m= F⋅ or 23.5 NF = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 70. (a) Note when 8 in.,a = /C Fr is perpendicular to the inclined 10 lb forces. Have dM F= Σ ( ) ( ) ( ) ( ) ( )10 lb 8 in. + 2 1 in. 10 lb 2 2 2 1 in.a a  = − + − +    For 8 in.,a = ( )( )10 lb 18 in. 24.627 in.M = − + 426.27 lb in.= − ⋅ or 426 lb in.M = ⋅ (b) Have 480 lb in.= ⋅M Also ( )dM M F= Σ + ( ) = Moment of couple due to horizontal forces at A and D + Moment of force-couple systems at C and F about C. Then ( ) ( ) ( )480 lb in. 10 lb 8 in. 2 1 in. 8 in. 2C F X ya M M F a F a  − ⋅ = − + + + + + + +    Where ( )( )10 lb 1 in. 10 lb in.CM = − = − ⋅ 10 lb in.F CM M= = − ⋅ 10 lb 2x F −= 10 lb 2y F −= ( )480 lb 10 in. 10 lb in. 10 lb in.in. 10 lb a∴ − + − ⋅ − ⋅⋅ = − ( ) ( )10 lb 10 lb8 in. 2 2 2 a a− + − 303.43 = 31.213 a or 9.72in.a = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 71. (a) Have dM F= Σ ( ) ( )( ) ( )( )9 lb 13.8 in. 2.5 lb 15.2 in.= − ( )86.2 lb in.= ⋅ 86.2 lb in.= ⋅M (b) Have 86.2 lb in.M Td= = ⋅ For T to be a minimum, d must be maximum. minT∴ must be perpendicular to line AC. 15.2 in.tan 11.4 in. θ = 53.130θ = ° or 53.1θ = ° (c) Have min maxM T d= Where 86.2 lb in.M = ⋅ ( ) ( ) ( )2 2max 15.2 in. 11.4 in. 2 1.2 in.d = + + 21.4 in.= ( )min86.2 lb in. 21.4 in.T∴ ⋅ = min 4.0280 lbT = or min 4.03 lbT = COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/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 2 2620 1488 192 lb in.M  = + + ⋅   1623.39 lb in.= ⋅ or 1.623 kip in.M = ⋅ ( ) ( ) ( )620 lb in. 1488 lb in. 192 lb in. 1623.39 lb in.M ⋅ + ⋅ + ⋅ = = ⋅ i j kM λ 0.38192 0.91660 0.118271= + +i j k cos 0.38192xθ = or 67.5xθ = ° cos 0.91660yθ = or 23.6yθ = ° cos 0.118271zθ = or z 83.2θ = ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 74. Have 1 2= +M M M Where 1 /E D D= ×M r F ( ) ( )0.7 m 80 N= − ×k j ( )56.0 N m= ⋅ i And 2 /G F B= ×M r F Now ( ) ( ) ( )2 2 20.300 m 0.540 m 0.350 mBFd = − + + 0.710= m Then B BF BFλ=F ( ) ( ) ( ) ( )0.300 m 0.540 m 0.350 m 71 N 0.710 m − + + = i j k ( ) ( ) ( )30 N 54 N 35 N= − + +i j k ∴ ( ) ( ) ( ) ( )2 0.54 m 30 N 54 N 35 N = × − + + M j i j k ( ) ( )18.90 N m 16.20 N m= ⋅ + ⋅i k Finally ( ) ( ) ( )56.0 N m 18.90 N m 16.20 N m = ⋅ + ⋅ + ⋅ M i i k ( ) ( )74.9 N m 16.20 N m= ⋅ + ⋅i k and ( ) ( )2 274.9 N m 16.20 N mM = ⋅ + ⋅ 76.632 N m= ⋅ or 76.6 N m= ⋅M 74.9 0 16.20cos cos cos 76.632 76.632 76.632x y z θ θ θ= = = or 12.20 90.0 77.8x y zθ θ θ= ° = ° = ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 75. Have ( )1 2 P= + +M M M M From the solution to Problem 3.74 ( ) ( ) ( )1 2 74.9 N m 16.20 N m+ = ⋅ + ⋅M M i k Now /P D E E= ×M r P ( ) ( ) ( )0.54 m 0.70 m 90 = + × j k N i ( ) ( )63.0 N m 48.6 N m= ⋅ − ⋅j k ∴ ( ) ( )74.9 16.20 63.0 48.6= + + −M i k j k ( ) ( ) ( )74.9 N m 63.0 N m 32.4 N m= ⋅ + ⋅ − ⋅i j k and ( ) ( ) ( )2 2 274.9 N m 63.0 N m 32.4 N mM = ⋅ + ⋅ + − ⋅ 103.096 N m= ⋅ or 103.1 N mM = ⋅ and 74.9 63.0 32.4cos cos cos 103.096 103.096 103.096x y z θ θ θ −= = = or 43.4 52.3 108.3x y zθ θ θ= ° = ° = ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 77. Have 1 2 3= + +M M M M Where ( ) ( )1 1.2 lb ft cos 25 1.2 lb ft sin 25= − ⋅ ° − ⋅ °M j k ( )2 1.3 lb ft= − ⋅M j ( ) ( )3 1.4 lb ft cos 20 1.4 lb ft sin 20= − ⋅ ° + ⋅ °M j k ∴ ( ) ( )1.08757 1.3 1.31557 0.507142 0.478828= − − − + − +M j k ( ) ( )3.7031 lb ft 0.028314 lb ft= − ⋅ − ⋅j k and ( ) ( )2 23.7031 0.028314 3.7032 lb ft= − + − = ⋅M or 3.70 lb ft= ⋅M 3.7031 0.028314 3.7032 − − = = M j k M λ 0.99997 0.0076458= − −j k cos 0xθ = or 90xθ = ° cos 0.99997yθ = − or 179.6yθ = ° cos 0.0076458zθ = − or 90.4zθ = ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 78. (a) :B =F P 160.0 NBF∴ = 50.0° cos10B BAM r P= − ° ( )( )0.355 m 160 N cos10= − ° 55.937 N m= − ⋅ or 55.9 N mBM = ⋅ (b) :C =F P 160.0 NCF∴ = 50.0° ( )C B CB BM M r F ⊥= − sin 55B CB BM r F= − ° ( )( )55.937 N m m 160 N sin 55= − ⋅ − 0.305 ° or 95.9 N mC = ⋅M COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 79. (a) : 135 NBΣ =F F or 135 NB =F : B BM P dΣ =M ( )( )135 N 0.125 m= 16.875 N m= ⋅ or 16.88 N mB = ⋅M (b) :B B CM F dΣ =M ( )16.875 N m = 0.075 mCF⋅ 225 NCF = or 225 NC =F : 0 B CF FΣ = − +F 225 NB CF F= = or 225 NB =F COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 81. : 2.8cos65 cos cosx A CF F Fθ θΣ ° = + ( )cosA CF F θ= + (1) : 2.8sin 65 sin siny A CF F Fθ θΣ ° = + ( )sinA CF F θ= + (2) Then (2) tan 65 tan (1) θ⇒ ° = or 65.0θ = ° ( )( ) ( )( ): 27 m 2.8 kN sin 65 72 m sin 65A CM FΣ ° = ° or 1.050 kNCF = From Equation (1): 2.8 kN 1.050 kNAF= + or 1.750 kNAF = 1.750 kNA∴ =F 65.0° 1.050 kNC =F 65.0° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 82. Based on ( ): 54 lb cos30 cos cosx B CF F Fα αΣ − ° = − − ( ) ( )cos 54 lb cos30B CF F α+ = ° (1) ( ): 54 lb sin 30 sin siny B CF F Fα αΣ ° = + or ( ) ( )sin 54 lb sin 30B CF F α+ = ° (2) From ( ) ( ) 2 : tan tan 30 1 Eq Eq α = ° 30α∴ = ° Based on ( ) ( ) ( ) ( )( ): 54 lb cos 30 20 10 in. cos10 24 in.C BM F Σ ° − ° = °  22.5 lbBF∴ = or 22.5 lbB =F 30° From Eq. (1), ( ) ( )22.5 cos30 54 cos30CF+ ° = ° 31.5 lbCF = or 31.5 lbC =F 30° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 83. (a) Based on ( ): 54 lb cos30 cos30x CF FΣ − ° = − ° 54 lbCF∴ = or 54.0 lbC =F 30° ( ) ( ): 54 lb cos10 10 in.C CM M Σ ° =  531.80 lb in.CM∴ = ⋅ or 532 lb in.C = ⋅M (b) Based on ( ): 54 lb sin 30 siny BF F αΣ ° = or sin 27BF α = (1) ( ) ( ): 531.80 lb in. lb cos10 24 in.BM  Σ ⋅ − 54 °  ( )24 in. cos 20CF  = − °  33.012 lbCF = or 33.0 lbC =F And ( ): 54 lb cos30 33.012 lb cosx BF F αΣ − ° = − − cos 13.7534BF α = (2) From ( ) ( ) 1 27: tan 63.006 2 13.7534 Eq Eq α α= ∴ = ° From Eq. (1), ( ) 27 30.301 lb sin 63.006B F = = ° or 30.3 lbB =F 63.0° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 86. Let R be the single equivalent force... : A CΣ = +F R F F ( )( ) ( )( )260 N cos10 sin10 320 N cos8 sin8= ° − ° + − ° − °i k i k ( ) ( )60.836 N 89.684 N= − −i k or ( ) ( )60.8 N 89.7 N= − −R i k : cos8A AD x AC CM r R r FΣ = ° ( ) ( )( )60.836 N 0.690 m 320 N cos8ADr = ° 3.5941 mADr = ∴R Would have to be applied 3.59 m to the right of A on an extension of handle ABC. COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 87. (a) Have : B C D AΣ + + =F F F F F Since B D= −F F 22 lbA C∴ = =F F 20° or 22.0 lbA =F 20° Have ( ) ( ) ( ):A BT CT DT AM F r F r F r MΣ − − + = ( ) ( ) ( ) ( )28 lb sin15 8 in. 22 lb sin 25 8 in.   − ° − °    ( ) ( )28 lb sin 45 8 in. AM + ° =  26.036 lb in.AM = ⋅ or 26.0 lb in.A = ⋅M (b) Have : A EΣ =F F F or 22.0 lbE =F 20° [ ]( ): cos 20A EM F aΣ = °M ( ) ( )26.036 lb in. 22 lb cos 20 a ∴ ⋅ = °  1.25941 in.a = or 1.259 in. Belowa A= COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 88. (a) Let R be the single equivalent force. Then ( )120 N=R k 120 NR = ( ) ( )( ) ( )( ): 120 N 0.165 m 90 N cos15 0.201 m 90 N sin15BM aΣ − = − ° + ° 0.080516 ma = ∴The line of action is 201 mm 80.516 mm 19.984 mm 2 y = − = or 19.98 mmy = (b) ( ) ( ) ( )( ) ( )( ): 0.201 0.040 m 120 N 0.165 m 90 N cos 0.201 m 90 N sinBM θ θ Σ − − = − +  or cos 1.21818sin 1.30101θ θ− = or ( )22cos 1.30101 1.21818sinθ θ= + or 2 21 sin 1.69263 3.1697sin 1.48396sinθ θ θ− = + + or 22.48396sin 3.1697sin 0.69263 0θ θ+ + = Then ( ) ( )( ) ( ) 23.1697 3.1697 4 2.48396 0.69263 sin 2 2.48396 θ − ± − = or 16.26 and 85.0θ θ= − ° = − ° COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 91. Have : ABΣ =F T F where AB AB ABT AB =T ( ) ( ) ( ) ( )2 2 2 2.25 18 954 lb 2.25 18 9 − + = + − + i j k ( ) ( ) ( )6 lb 48 lb 24 lb= − +i j k So that ( ) ( ) ( )6.00 lb 48.0 lb 24.0 lb= − +F i j k Have /:E A E ABΣ × =M r T M 0 22.5 0 lb ft 6 48 24 ⋅ = − i j k M ( ) ( )540 lb ft 135 lb ft∴ = ⋅ − ⋅M i k or ( ) ( )540 lb ft 135.0 lb ft= ⋅ − ⋅M i k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 92. Have : CDΣ =F T F where CD CD CDT CD =T ( ) ( ) ( ) ( )2 2 2 0.9 16.8 7.261 lb 0.9 16.8 7.2 − − + = − + − + i j k ( ) ( ) ( )3 lb 56 lb 24 lb= − − +i j k So that ( ) ( ) ( )3.00 lb 56.0 lb 24.0 lb= − − +F i j k Have /O C D CDΣ = × =M r T M 0 22.5 0 lb ft 3 56 24 ⋅ = − − i j k M ( ) ( )540 lb ft 67.5 lb ft∴ = ⋅ + ⋅M i k ( ) ( )540 lb ft 67.5 lb ft= ⋅ + ⋅M i k COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 3, Solution 93. Have : ABΣ =F T F where AB AB ABT AB =T ( ) ( ) ( ) ( )2 2 2 4.75 210.5 kN 1 4.75 2 − − + = − + − + i j k ( ) ( ) ( )2 kN 9.5 kN 4 kN= − − +i j k So that ( ) ( ) ( )2.00 kN 9.50 kN 4.00 kN= − − +F i j k Have :O A ABΣ × =M r T M 3 4.75 0 kN m 2 9.5 4 ⋅ = − − i j k M ( ) ( ) ( )19 kN m 12 kN m 19 kN m∴ = ⋅ − ⋅ − ⋅M i j k ( ) ( ) ( )19.00 kN m 12.00 kN m 19.00 kN m= ⋅ − ⋅ − ⋅M i j k