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### Soln02052

Course: CIVE 1150, Spring 2008
School: Toledo
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Complete COSMOS: Online Solutions Manual Organization System Chapter 2, Solution 52. Free-Body Diagram: Fx = 0: - FB cos15 + 2.4 kN + (1.9 kN ) sin15 = 0 or FB = 2.9938 kN FB = 2.99 kN Fy = 0: FD - (1.9 kN ) cos15 + ( 2.9938 kN ) sin15 0 FD = = 1.060 kN Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 2, Solution 52. Free-Body Diagram: Fx = 0: - FB cos15 + 2.4 kN + (1.9 kN ) sin15 = 0 or FB = 2.9938 kN FB = 2.99 kN Fy = 0: FD - (1.9 kN ) cos15 + ( 2.9938 kN ) sin15 0 FD = = 1.060 kN 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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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
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 69.(a)M = Fd12 N m = F ( 0.45 m ) or F = 26.7 N (b)M = Fd12 N m = F ( 0.24 m ) or F = 50.0 N (c)M = FdWhere d =( 0.45 m )2 + ( 0.24 m )2= 0.51 m12 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 97.(a) FBD Entire Machine:M A = 0:(145 in.) (2B) - ( 35 in.)(18 kips )- (170 in.)( 50 kips ) = 0 2B = 62.966 kips, B = 31.5 kips!Fy = 0:2 A + 62.966 kips - 1
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 50.For steel/steel contact, s1 = tan -1 s1 = tan -1 ( 0.3) = 16.6992For steel/concrete interface, s2 = tan -1 s2 = tan -1 ( 0.6 ) = 30.964FBD Plate CD:Fy = 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 131.l = total length yB = c cosh or xB , c L + c = c cosh 1 = cosh L L - 2c c L = 4.933 c L/2 cSolving numerically, sB = c sinh So xB cl L = c sinh 2 2cl = 10 ft
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 62.FBD Section above a-a:M G = 0:( 5.9 m ) 27 FIK - ( 2.7 m )( 40 kN ) - ( 5.4 m )( 40 kN ) = 0 793 FIK = 57.275 kN,Fy = 0: 27 ( 57.275 kN - FGJ ) = 0 793
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 46.First note that by symmetry y = 0. Using the area element shown in the figure,xEL =2 2 r cos = R cos 2 cos 3 3dA =1 2 1 r d = R 2 cos 2 2 d 2 2 1 1 4 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 131.FBD BD: = 90 - 25 = 65M B = 0: 2 ( 0.2 m ) cos 25 D sin 65 -15 N m = 0 D = 45.654 NFBD AC:M A = 0:M A = ( 0.2 m )( 45.654 N ) = 0M A = 9.13 N m!
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 61.\FBD Truss:Fx = 0: A x = 0By symmetry Ay = Ny = 1.6 kipsFBD Joint J:Fx = 0: - FGJ - FHJ = 0, FHJ = - FGJBy inspection of joint I, FIJ = 0.6 kip CFBD Sec
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 130.Free-Body Diagram:Express forces and moments in terms of rectangular components:FDE = FDE- 40 i - 70 j + 40k( - 40 )2+ ( - 70 ) + ( 40 )22=FDE
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 34.By inspection of joint A: By inspection of joint C : By inspection of joint E : By inspection of joint L : By inspection of joint N :FAF = 0 FCH = 0 FDE = FEI =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 137.FBD Joint A:TB TC 2.1 kN = = 17 17 30FBD BF and CF w/cables to A:TB = TC = 1.19 kN Ex = Fx = 1.05 kN and E y = FyBy symmetryFBD Machine:By symmetryH x
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 67.Free-Body Diagram At C: Since TAB = TBC = 140 lb, Force triangle is isosceles:With2 + 75 = 180 = 52.5Then = 90 - 52.5 - 30 = 7.50P = (140 lb ) cos 52.5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 22.Free-Body Diagram:(a) x = 0:or (b) 2.4 in. ( 3.6 lb ) - ( 0.9 in.)(12 lb ) = 0 cos cos = 0.80000, or = 36.870 = 36.9Fx = 0:( 3 lb ) sin 36.870 +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 23.Free-Body Diagram:From free-body diagram for (a): A = 0: Fx = 0: Fy = 0:- B ( 0.2 m ) - (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) - 10 N m = 0 = -187.50 or B
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 85.(a)Based on Fz : - 200 N + 200 N + 240 N = FA FA = 240 N or FA = ( 240 N ) k Based on M A :(b)( 200 N )( 0.7 m ) - ( 200 N )( 0.2 m ) = M AM A = 100 N m or
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 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 mM B = 13.02 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 19.(a)HaveMO = r Fijk= - 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
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 79.(a)We note that in ABC, 1 A = B = 180 - ( 90 + ) 2 = 45 -2 Thus AB = 2a cos 45 - 2 = 2a cos 45 cos + sin 45 sin 2 2 = 2a 2 cos + sin 2 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 120.Free-Body Diagram:Geometry: Using triangle ACD and the law of sinessin sin 50 or = 20.946 = 15 in. 7 in. = 50 + 20.946 = 70.946Expressing FCD in terms of
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 59.FBD Truss: Notes: = 20, = 40, = 60, = 80, = 22.5, = 45, = 60 outer members AC, CE, etc. are each 1.0 m radial members AB, CD, etc. are each 0.4 m By symmetry
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 38.Free-Body Diagram:Note thatWC = - ( 80 kg ) 9.81 m/s 2 = 784.80 N WD2( ) = - ( 52 kg ) ( 9.81 m/s ) = 510.12 N(a)M A = 0:- ( 784.80 N )( 0.8 m ) cos 30
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 124.Free-Body Diagram:Express tension, weight in terms of rectangular components: uuu r EF = ( 300 mm ) i + (1350 mm ) j - ( 700 mm ) k uuu r EF 300 i + 1350 j - 700
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 49.FBD Truss:Fx = 0: A x = 0By symmetry: A y = L y = 6 kNFBD Section:Notes:yI = yH2 3 2 = 3m 5 5 = m 2 3soyH - yI = 1 mcontinuedVector Mechanics
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 65.At Collar A .Fy = 0:- 9 lb +h 12 + h 22Fs = 0or NowhFs = 9 144 + h 2Fs = k ( L - LAB ) ABWhere the stretched lengthL = AB(12 in.)2 + h2LAB =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 10.Slope of line EC = Then and Have35 in. 7 = 112 in. + 8 in. 24 TABx = TABy = 24 TAB 25 7 TAB 25M D = TABx ( y ) + TABy ( x ) 7840 lb in. =24 7 TAB ( 0 ) + T
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 79.FBD Frame:M A = 0:( 0.25 m ) Dx - 400 N m = 0D x = 1600 NFBD member DF: Note BE is a two-force member, so Ex = EyFx = 0: 1600 N - E x = 0, E x = 1600 N E
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 154.Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke can exert a couple along the arm of the crosspiece it c
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 89.(a) FBD ACF:Note: BC is a two-force memberM F = 0:( 0.1 m )(120 N ) - ( 0.3 m ) 1 FBC = 0, 26 FBC = 40 26 N,B x = 200 N B y = 40.0 N!! ! !Fx =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 19.FBD Frame and pipe:W = (10 ft )(18.5 lb/ft ) = 185 lb M A = ( 24 in.) Cx - (12.6 in.)185 lb = 0C x = 97.125 lbFBD pipe: By symmetry N E = N D = NFy = 0:Also
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 125.Since the spherical cup is uniform, the center of gravity will coincide with the centroid. Also, because the cup is thin, it can be treated like an area in finding
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 96.! &quot; AB = - ( 600 mm ) i + ( 360 mm ) j + ( 270 mm ) kAB =( - 600 mm )2 + ( 360 mm)2 + ( 270 mm) 2= 750 mmAB = 750 mm !&quot; AC = - ( 600 mm ) i + ( 320 mm ) j
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 122.First note that by symmetry:y =0! z = 0!Choose as a volume element a disk of radius y and thickness dx. Then:xEL = x, and dV = y 2dx, ordV = h 2 cos 2 Usi
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 109.Apply vertical load P at D.M H = 0: - P (12 m ) + E ( 36 m ) = 0E= Fy = 0: P 33 P FBF - =0 5 3 FBF = 5 P 9Virtual Work: We remove member BF and replace it
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 68.Free-Body Diagram of Pulley (a)Fy = 0: 2T - ( 280 kg ) 9.81 m/s 2 = 0()T =1 ( 2746.8 N ) 2T = 1373 N(b)Fy = 0: 2T - ( 280 kg ) 9.81 m/s 2 = 0 T = 1
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 55.Free-Body Diagram At C:3 15 15 Fx = 0: - TAC + TBC - (150 lb ) = 0 5 17 17orFy = 0:-17 TAC + 5 TBC = 750 5(1)4 8 8 TAC + TBC - (150 lb ) - 190 lb = 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 114.d DA = d DB = d DC =( 400 mm )2 + ( - 600 mm )2= 721.11 mm = 663.32 mm = 663.32 mm( - 200 mm )2 + ( - 600 mm )2 + ( 200 mm )2( - 200 mm )2 + ( - 600 mm )2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 48.FBD Truss:Fx = 0: A x = 0By symmetry: A y = L y = 6 kNFBD Section:Notes:15 m 6 2 5 5 yD = = m 3 2 3 2 2 yE = 1 = m 3 3 5 yF - yD = m 6 yG = 1 m yF = yD -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 76.HaveM = M1 + M 2 + M PFrom Problem 3.73 solution: M1 = ( 400 lb in.) i + ( 960 lb in.) j M 2 = ( 220 lb in.) i + ( 528 lb in.) j + (192 lb in.) k Now M P
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 73.HaveM = M1 + M 2 rC/B = ( 38.4 in.) i - (16 in.) j PIC = - ( 25 lb ) kWhere M1 = rC/B PICi j k M1 = 38.4 -16 0 lb in. 0 0 -25 = ( 400 lb in.) i + ( 960 l
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 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 = 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 53.Replacing the 1500 N force with equivalent force and couple at D,M C = 0:(1.25 m ) P - (1.2 m )(1500 N ) - 1800 N m+ ( 2.4 m ) E y - ( 3.65 m ) 2P = 0assum
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 43. FBD Truss:Fx = 0: M A = 0:Kx = 06a K y - 125 lb - 5a ( 250 lb ) - 4a ( 250 lb ) - 3a ( 375 lb ) - 2a ( 500 lb ) - a ( 500 lb ) = 0K y = 937.5 lb()Fy = 0:
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 90. (a) FBD AF: Note: BC is a two-force memberM F = 0:( 0.3 m ) 1 FBC - 48 N m = 0, 26 FBC = 160 26 N soB x = 800 N B y = 160.0 N!! ! !Fx = 0:(b &amp; c
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 40.(a) To check for simple truss, start with ABDE and add three members at a time which meet at a single new joint, successively adding joints G, F, H and C. This is
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 39.Free-Body Diagram:Equilibrium for rod: (a)M E = 0:( 6 lb ) cos 60 ( dOE ) - (T cos 45) ( dOE ) = 0T = 4.2426 lbT = 4.24 lb(b)Fx = 0:( 4.2426 lb ) cos
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 69.Free-Body Diagram:From the free-body diagram: = tan -1 (900 mm)sin 50 = 82.726 88 mm From the force triangle:FN = (130 N ) tan = (130 N ) tan 82.726
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 65.(a)Shear Diag: dV V is zero at A with constant slope = - w = -1 kip/ft decreasing to dx - 3.6 kips at C. V then jumps 9 kips to 5.4 kips and is constant t
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 10.Using the Law of Sines:60 N 80 N = sin sin10 or = 7.4832 = 180 - (10 + 7.4832 )= 162.517Then:R 80 N = sin162.517 sin10 or R = 138.405 N(a) (b) = 7.48
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 7.Using the triangle rule and the Law of Cosines, Have: = 180 - 45 = 135Then:R 2 = ( 900 ) + ( 600 ) - 2 ( 900 )( 600 ) cos 1352 2or R = 1390.57 NUsing the
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 143.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
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 132.FBD BD:M B = 0: 2 ( 0.2 m ) cos 25 C -15 N m = 0 C = 41.377 NFBD AC: M A = 0: M A - ( 0.2 m )( 41.377 N ) sin 65 = 0M A = 7.50 N m!Vector Mechanics