# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

2 Pages

### Soln06117

Course: CIVE 1150, Spring 2008
School: Toledo
Rating:

Word Count: 236

#### Document Preview

Complete COSMOS: Online Solutions Manual Organization System Chapter 6, Solution 117. (a) Member FBDs: FBD I: FBD II: M A = 0: aF1 - 2aP = 0 M B = 0: - aF2 = 0 F1 = 2P; F2 = 0 Fy = 0: Ay - P = 0 Ay = P Fx = 0: Bx + F1 = 0, Bx = - F1 = -2P B x = 2P so B = 2P ! Fy = 0: By = 0 FBD I: Fx = 0: - Ax - F1 + F2 = 0 Ax = F2 - F1 = 0 - 2P A x = 2P so A = 2.24P 26.6 ! Frame is rigid ! (b) FBD left: FBD whole:...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Ohio >> Toledo >> CIVE 1150

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Complete COSMOS: Online Solutions Manual Organization System Chapter 6, Solution 117. (a) Member FBDs: FBD I: FBD II: M A = 0: aF1 - 2aP = 0 M B = 0: - aF2 = 0 F1 = 2P; F2 = 0 Fy = 0: Ay - P = 0 Ay = P Fx = 0: Bx + F1 = 0, Bx = - F1 = -2P B x = 2P so B = 2P ! Fy = 0: By = 0 FBD I: Fx = 0: - Ax - F1 + F2 = 0 Ax = F2 - F1 = 0 - 2P A x = 2P so A = 2.24P 26.6 ! Frame is rigid ! (b) FBD left: FBD whole: FBD I: M E = 0: a a 5a P + Ax - Ay = 0 2 2 2 Ax - 5 Ay = - P Ax - 5 Ay = -3P This is impossible unless P = 0 not rigid ! FBD II: M B = 0: 3aP + aAx - 5aAy = 0 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Jr., Johnston, Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (c) Member FBDs: FBD I: Fy = 0: A - P = 0 M D = 0: aF1 - 2aA = 0 Fx = 0: F2 - F1 = 0 FBD II: M B = 0: 2aC - aF1 = 0 F1 = 2 P F2 = 2 P A= P ! C = F1 = P 2 Bx = P - P = 0 By = -C = - P C= P ! Fx = 0: F1 - F2 + Bx = 0 Fx = 0: By + C = 0 B= P ! Frame is rigid ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 152. Free-Body Diagram:From free-body diagram of beamFx = 0: Bx = 0so thatB = ByFy = 0: A + B - (100 + 200 + 300 ) N = 0orA + B = 600 Nthe other support re
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 14.First, by symmetryX = 90 mmA, mm 21 2 3y , mmyA, mm31 296 000(180 )(120 ) = 21 600- -604( 90 )(120 ) = - 8482.3 ( 90 )(120 ) = - 8482.34635.4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 116.Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 4.FBD Frame:Note: AC is a two-force member, resolve FAC at C:M E = 0: ( 0.25 m + 0.25 m )1 FAC - ( 0.15 m )( 320 N ) = 0 2FAC = 96 2 NFBD sect. AB:Fx = 0:
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 158.Free-Body Diagram: Since Slope of ED is45 45yED = xED = a, slope of HC isAlso andDE =2aa 1 DH = HE = DE = 2 2 a/ 2 = R a 2RFor triangles DHC and EHC
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 159. Free-Body Diagram:First noteW = mg = (17 kg ) 9.81 m/s 2 = 166.77 N()h=From free-body diagram of plywood sheet(1.2 )2 - (1.125)2= 0.41758 m (1.125
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 150.Free-Body Diagram:Express forces in terms of their rectangular components: uuur FG 18 i - 6 j - 9 k T TFG = TFG = TFG = FG ( 6 i - 2 j - 3 k ) 2 2 2 FG 7 (18) +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 101. FBD Frame:M E = 0:( 6 in.)(84 lb ) - ( 42 in.) Ay = 0Ay = 12.00 lbFBD CF: (I)FBD ABC: (II)I: II:M D = 0: M B = 0:( 3.5 in.) Cx + ( 7 in.) C y - ( 5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 75.Structure (a): Rigid truss with r = 3, m = 14, n = 8sor + m = 17 &gt; 2n = 16 so completely constrained but indeterminate !Structure (b):Simple truss (start wi
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 136.Free-Body Diagram:Fx = 0: Fy = 0: Fy = 0:M O = 0:Bx + C x = 0, or Bx = - C x(1) (2) (3)Ay + By + C y = 0 Az - P = 0, or Az = P = 40.0 lbrOA A + rOB B +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 137.Free-Body Diagram:Fx = 0: Fy = 0: Fz = 0:M O = 0:Bx + C x = 0, or Bx = - C x(1) (2) (3)Ay + By + C y = 0Az - P = 0, or Az = P = 60.0 NrOA A + rOB B +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 128.Free-Body Diagram:Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net moment about this point is zero. (a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 129.Free-Body Diagram:Express the tension in terms of its rectangular components: uuuu r DG = - ( 4.8 in.) j - ( 9 in.) kTDC = TDGuuuu r DG = TDG DG- 4.8 j - 9
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 24.L, mm1 2 3 4 5 75 150x , mmy , mmxL, mm 22812.5 0 45 000 0 5625.0 53 437yL, mm 20 11 250 037.5 0 95.492 0 47.7460 75 0(150 )= 471.2475-112.5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 94. FBD AC:M c = 0:( 0.6 m ) Ax - ( 0.06 m )(170 N ) = 0,A x = 17.00 N-17 N + 15 (170 N ) - Cx = 0, 17 Cx = 133 NFx = 0:FBD CE:Fx = 0:133 N -15 (170 N )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 3.A, mm 21x , mmxA, mm3 729 000 2 460 375 3 189 3751 90 270 = 12 150 2 1 135 270 = 18 225 22 ( 90 ) = 60 3 90 + 1 (135) = 135 3230375xA 3189375 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 88.(a)Let R be the single equivalent force. ThenR = (120 N ) kM B : - a (120 N ) = - ( 0.165 m )( 90 N ) cos15 + ( 0.201 m )( 90 N ) sin15 a = 0.080516 mR = 12
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 82.F= F =Fx2 + Fy2 + Fz2( 450 N )2 + ( 600 N )2 + ( -1800 N )2F = 1950 N !cos x =Fx 450 N = F 1950 N x = 76.7 !cos y =Fy F=600 N 1950 N y = 72.1 !
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 144.Free-Body Diagram:Express forces in terms of their rectangular components: uuur BG - 40 i + 74 j - 32 k 37 16 20 TBG = TBG j- k = TBG = TBG - i + 2 2 2 BG 45
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 145.Free-Body Diagram:Express forces in terms of their rectangular components: uuur BG - 40 i + 74 j - 32 k 37 16 20 = TBG = TBG - i + TBG = TBG j- k 2 2 2 BG 45
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 146.Free-Body Diagram:Express forces in terms of their rectangular components:W1 = W2 = - ( 30 lb ) j uuuu r ( x - 6) i + y j - 6 k BH T=T =T BH ( x - 6 )2 + y 2 +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 70.Structure (a): Non-simple truss with r = 3, m = 16, N = 10, so 2N &gt; m + r partially constrained ! Structure (b): Non-simple truss with r = 3, m = 15, N = 10, so 2N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 59.The mass of the lamp shade is given bym = V = Atwhere A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 87.(a) FBD AC:Note: CE is a two-force memberM A = 0: 2 FCE = 0, 120 N m - ( 0.4 m ) 2 FCE = 150 2 NE x = 150.0 N E y = 150.0 N Fx = 0:(b) FBD CE:!! !
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 31.\By inspection of joint C:FBC = 0 FBE = 0Then, by inspection of joint B: Then, by inspection of joint E:FDE = 0 FFH = 0 and FHI = 0By inspection of joint
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 71.The distributed load can be represented in terms of resultants: R1 = ( 8 m )( 300 N/m ) = 2400 NR2 =1 ( 8 - a ) m ( 2400 N/m ) = 1200 ( 8 - a ) N 21 - 8 Ay
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 142.Free-Body Diagram:There is only one unknown of interest and, therefore only one equation is needed:M AB = 0Geometry: = tan -1 1.05 m = 16.2602 3.5 m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 49.Based on M x = ( P cos ) ( 0.225 m ) sin - ( P sin ) ( 0.225 m ) cos (1) (2) (3)M y = - ( P cos )( 0.125 m ) M z = - ( P sin )( 0.125 m )ByEquation
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 118.From the solutions to Problems 2.111 and 2.112, haveTBE + TCF + TDG = 0.2 65-TBE sin 45 + TCF sin 30 - TDG sin15 = 0(2)(3)TBE cos 45 + TCF cos 30 - TDG co
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 119.d AB = d AC =( - 30 ft )2 + ( 24 ft )2 + ( 32 ft )2= 50 ft = 38 ft( - 30 ft )2 + ( 20 ft )2 + ( -12 ft )2TAB = TAB AB =TAB - ( 30 ft ) i + ( 24 ft ) j
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 12.FBD Truss:Fx = 0:Ax = 0By symmetry: A y = B y = 4.90 kNandFAB = FEG , FAC = FFG , FBC = FEF FBD = FDE , FCD = FDFJoint FBDs: Joint A:Fx = 0: Fy = 0:5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 82.FBD Frame:M A = 0:( 0.2 m ) Bx - ( 0.36 m ) 2 180 N = 0 2 Bx = 162 2 N, Fx = 0: -162 2 N + Ax - Ax = 252 2 N, Fy = 0: By -B y = 90 2 N,B x = 229 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 97.Free-Body Diagram:Fx = 0:M D = 0:Dx = 0( - 7 in.) i C + ( 2 in.) i + ( 3 in.) k ( 530 lb ) j + ( -192 lb ) k + ( -3 in.) i + ( 6 in.) j ( -96 lb )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 43.Free-Body Diagram:Geometry: Equation of the slot: y =x2 4 2x dy = 1.20000 = slope of slot at C = 4 dx C ( x = 2.4 in.) It follows for the angles tha
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 81.FBD Ring:(a)M A = 0: FBC =(8 in.) ( FBC cos 20 - 6 lb ) = 0FBC = 6.39 lb C 6 lb cos 20 = 0, cos 20 Ax = 6 lb6 lb , cos 20 Ax -(b)Fx = 0:Fy = 0:Ay -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 41.For y2 ThenNow and foratx = a, y = b : a = kb2y2 = b 1/2 x aork =a b2xEL = x 0 x a y b x1/2 x1/2 : yEL = 2 = , dA = y2dx = b dx 2 2 2 a aFora 1 b
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 77.(a)Fx = (180 lb ) cos 35 sin 20 = 50.430 lbFx = 50.4 lb !Fy = - (180 lb ) sin 35 = -103.244 lb Fy = -103.2 lb ! Fz = (180 lb ) cos 35 cos 20 = 138.555 lb Fz =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 125.FBD half-span:sB = 65 m,hB = 30 mw = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m2 2 yB = c 2 + sB()( c + hB )2c=2 = c 2 + sB2 22 2 ( 65 m ) - ( 30 m )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 92.FBD Pipe 2:Fx = 0:Fy = 0:NG - ND -3 ( 220 lb ) = 0, 5 4 ( 220 lb ) = 0, 5NG = 132 lb N D = 176 lbFBD Pipe 1: = 90 - 2 tan -1Fx = 0:3 = 16.2602 43 (
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 88.! &quot; EB = ( 36 in.) i - ( 45 in.) j + ( 48 in.) kEB =( 36 in.)2 + ( - 45 in.)2 + ( 48 in.)2! &quot; EB EB= 75 in.TEB = TEBEB = TEB TEB =60 lb ( 36 in.) i - ( 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 89.! &quot; BA = ( 4 m ) i + ( 20 m ) j - ( 5 m ) kBA =F = F BA( 4 m )2 + ( 20 m )2 + ( - 5 m )2= 21 m! &quot; BA 2100 N ( 4 m ) i + ( 20 m ) j - ( 5 m ) k = F = 21
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 90.! &quot; DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) kDA =F = F DA( 4 m )2 + ( 20 m )2 + (14.8 m )2= 25.2 m! &quot; DA 1260 N ( 4 m ) i + ( 20 m ) j + (14.8 m ) k = F
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 138.The forces applied at A are:TAB , TAC , TAD and Pwhere P = Pj . To express the other forces in terms of the unit vectors i, j, k, we write uuu r AB = - ( 0.72
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 72.Structure (a): Simple truss (start with ABC and add joints alphabetical to complete truss), withr = 4, m = 13, n = 8sor + m = 17 &gt; 2n = 16Structure is comple
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 140.Free-Body Diagram:Express the forces in terms of rectangular components:W = - ( mg ) j = - ( 3 kg ) 9.81 m/s 2 j = - ( 29.43 N ) j()N B = N B ( 0.8j + 0.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 96.First note For small values of and :A = a sin = b sin a = b= V = P ( a + b ) cos - 2Q ( a + b ) cos a = ( a + b ) P cos - 2Q cos b a dV a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 136.The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Fig. 5.8A
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 26.Have where andM C = rA/C FBA rA/C = ( 0.96 m ) i - ( 0.12 m ) j + ( 0.72 m ) k FBA = BA FBA - ( 0.1 m ) i + (1.8 m ) j - ( 0.6 m ) k ( 228 N ) = ( 0.1)2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 119. FBD Blade:Fy = 0:3 kN -12 FBD = 0: FBD = 3.25 kN C 13FBD ABC:M C = 0: 0.045 m + ( 0.3 m ) sin 30 P sin 30 + 0.030 m + ( 0.3 m ) cos 30 P cos 30 -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 62.(a) and (b)M B = 0:( 0.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) - ( 9.6 ft ) AyA y = 12.625 kips=0Shear Diag: dV V is piecewise co
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 52.Free-Body Diagram:Fx = 0:- FB cos15 + 2.4 kN + (1.9 kN ) sin15 = 0orFB = 2.9938 kN FB = 2.99 kNFy = 0:FD - (1.9 kN ) cos15 + ( 2.9938 kN ) sin15 = 0FD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 78.(a)Fx = (180 lb ) cos 30 cos 25 = 141.279 lbFx = 141.3 lb !Fy = - (180 lb ) sin 30 = - 90.000 lb Fy = - 90.0 lb ! Fz = (180 lb ) cos 30 sin 25 = 65.880 lb Fz
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 103.Given:l AB = 3.6 in. lBC = 1.6 in. lCD = 1.2 in.lDE = 1.6 in.lEF = 1.6 in. lFG = 4.8 in.Assume y A : yC =(1.6 in.) y = 4 y ( 3.6 in.) A 9 A4 9 yD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 33.For the element (EL) shownAtx = a, y = h: h = k1a3ork1 =h a3ora = k 2 h3Hence, on line 1k2 =a h3y =and on line 2h 3 x a3 h 1/3 x a1/3y =Th
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 10.Free-Body Diagram:For Qmin , TD = 0 M B = 0:( 7.5 kN )( 0.5 m ) - Qmin ( 3 m ) = 0Qmin = 1.250 kNFor Qmax , TB = 0 M D = 0:( 7.5 kN )( 2.75 m ) - Qmax ( 0.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 148.FBD Whole:M A = 0: ( 5.4 ft ) Bx - ( 7.8 ft )( 90 lb ) = 0B x = 130 lbFBD BE with pulleys and cord:M E = 0: ( 5.4 ft )(130 lb ) - ( 7.2 ft ) By+ ( 4.8 ft
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 65.From the solution of Problem 3.60: DI = 0.97014 i - 0.24254 jTEG = ( 2.4 lb ) i - ( 7.2 lb ) j - ( 23.4 lb ) k TEG = 24.6 lbM DI = - 224.74 lb ftOnly the pe
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 68FBD Section ABDC: Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be slack, FCF = 0Fx = 0:M D = 0:20
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 64.At Collar A .Have For stretched lengthL = ABFs = k ( L - LAB ) AB(12 in.)2 + (16 in.)2L = 20 in. ABFor unstretched lengthLAB = 12 2 in.ThenFs = 4 l
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 63.TBC must be perpendicular to FAC to be as small as possible.Free-Body Diagram: CForce Triangle is a Right Triangle(a) We observe: (b)TBC = ( 400 lb ) sin 60
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 74.(a)Fx = - ( 810 lb ) cos 45 sin 25 = - 242.06 lbFx = -242 lb !Fy = - ( 810 lb ) sin 45 = - 572.76 lb Fy = - 573 lb ! Fz = (810 lb ) cos 45 cos 25 = 519.09 lb