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.
3 Pages
Soln08070
Course: CIVE 1150, Spring 2008School: Toledo
Rating:
Word Count: 150
Document Preview
Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 70. FBD joint D: By symmetry: FAD = FCD Fy = 0: 2FAD sin 25 - 4 kN = 0 FAD = FCD = 4.7324 kN FBD joint A: By symmetry: FAE = FAD Fx = 0: FAC - 2 ( 4.7324 kN ) cos 25 = 0 FAC = 8.5780 kN Block and incline A: = tan -1 2 mm = 4.8518 ( 7.5 mm ) s = tan -1 s = tan -1 0.15 = 8.5308 Vector Mechanics for Engineers: Statics and...
Unformatted Document Excerpt
Coursehero >> Ohio >> Toledo >> CIVE 1150Course 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 8, Solution 70.
FBD joint D:
By symmetry:
FAD = FCD
Fy = 0: 2FAD sin 25 - 4 kN = 0
FAD = FCD = 4.7324 kN
FBD joint A:
By symmetry:
FAE = FAD
Fx = 0: FAC - 2 ( 4.7324 kN ) cos 25 = 0
FAC = 8.5780 kN
Block and incline A:
= tan -1
2 mm = 4.8518 ( 7.5 mm )
s = tan -1 s = tan -1 0.15 = 8.5308
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 The 2007 McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
PROBLEM 8.70 CONTINUED
Q = ( 8.578 kN ) tan (13.3826 )
= 2.0408 kN Couple at A:
M A = rQ
7.5 = mm ( 2.0408 kN ) 2 = 7.653 N m
By symmetry: Couple at C:
M C = 7.653 N m
Total couple M = 2 ( 7.653 N m )
M = 15.31 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.
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:
Below is a small sample set of documents:
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 68.FBD large gear:M C = 0:(12 in.)W- 7.2 kip in. = 0W = 0.600 kips = 600 lbBlock on incline: = tan -10.375 in. = 2.2785 2 (1.5 in.) s = tan -1 s = ta
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 65.WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N Slip impends at all surfaces simultaneouslyFBD I: A + B()()Fx = 0: Fy = 0:N A -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 55.Assume no impending motion of board on ground. Then there will be impending slip at all wood/wood interfaces.FBD Top wedge:Wedge is a two-force member so R 2 = -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 58.As the plates are moved, the angle will decrease. (a) s = tan -1 s = tan -1 0.2 = 11.31. As decreases, the minimum angle at the contact approaches 12.5 > s =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 60.FBD Cylinder:W = ( 80 kg ) 9.81 m/s 2 = 784.8 N()M G = 0:M D = 0:FA - FB = 0,FA = FBN A = NB + W 3(1) (2)dN B - dN A + rW = 0,soN A > NB,FA m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 88.FBD Each wheel:rf = raxle sin = raxle sin tan -1 ()Fx = 0: Fy = 0:P - R sin = 0 4 R cos - W =0 4 P Wrf rw tan =sin =or =P = W tanbut (a) For
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 38.FBD A:Note: Rod is a two force member. For impending slip the reactions are at angle s = tan -1 s = tan -1 ( 0.40 ) = 21.801Consider first impending slip to
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 84.FBD gate:W1 = 66 kg 9.81 m/s 2 = 647.46 N W2 = 24 kg 9.81 m/s 2 = 235.44 N rf = rs sin s = rs sin tan -1 s()()()= ( 0.012 m ) sin tan -1 0.2 = 0.00
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 81.Pulley FBD's: Left:rp = 30 mmrf = raxle sin k = raxle sin tan -1 k()*= ( 5 mm ) sin tan -1 0.2 = 0.98058 mmLeft: Right: or M C = 0:()( rp - rf )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 78.FBD Windlass:rf = r sin s = r sin tan -1 s()= (1.5 in.) sin tan -1 0.50 = 0.67082 in.()M A = 0:( 8 + 0.67082 ) in. P - ( 5 - 0.67082 ) in. (160 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 75.FBD Bucket:rf = r sin s = r sin tan -1 s= ( 0.18 m ) sin tan -1M A = 0:(() 0.30 ) = 0.05172 m=0(1.6 m + 0.05172 m ) T - ( 0.05172 m )WT = 0.031314W
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 72.FBD lower jaw:By symmetry B = 540 N Fy = 0: - 540 N + A - 540 N = 0,A = 1080 N(a) since A > B when finished, adjust A first when there will be no forceBlock/
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 36.FBD Collar:Note: BC is a two-force member, and for M max , slip will impend to the right. Fy = 0: Impending slip: Fx = 0:FBD AB:FBC cos - N = 0, F = s N = s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 69.Block/incline analysis: = tan -10.125 in. = 2.4238 2.9531 in. s = tan -1 ( 0.35 ) = 19.2900Q = 47250 tan ( 21.714 ) = 18.816 lbCouple =d 0.94 in. (18.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 66.FBD jack handle:See Section 8.6 M C = 0:FBD block on incline:aP - rQ = 0 or P =r Q a(a)Raising loadQ = W tan ( + s )P=r W tan ( + s ) acontinued
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 63.FBD plank + wedge:M A = 0:(8 ft ) N B - (1.5 ft )( 48 lb/ft )( 3 ft )- ( 2 ft )1 ( 48 lb/ft )( 3 ft ) 2 5 - 3 + ft ( 96 lb/ft )( 5 ft ) = 0 3 NW = 185
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 56.FBD Cylinder:Slip impends at B SC = tan -1 ( 0.35 ) = 19.2900M A = 0:r RC cos (12 + 19.29 ) -4r W =0 3RC = 0.49665, W = 124.163 lbFBD Wedge: SF = tan -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 35.Geometry:1 m - ( 0.5 m ) cos tan = ( 0.5 m ) sin tan ( 2 - cos ) = sin = 30 = 60then LAB = (1 m ) cos 30 =FBD B:3 m 21 kN 3 m - m Fs = k (
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 34.FBD Collar:Impending motion down:Stretch of spring x = AB - a =a -a cos a 1 - a = (1.5 kN/m )( 0.5 m ) - 1 Fs = kx = k cos cos 1 = ( 0.75 kN )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 33.FBD Plate:Assuming reactions as shown, at ends of sleeves, For impending slip Fx = 0: Fy = 0: Solving:FA = s N A , FB = s FB N A + N B = 2.5 P sin N A - N B -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 30.Geometry of four-bar:Considering the geometry when = 0,2 2 LCD = ( 60 mm - 52 mm ) + ( 36 mm + 22 mm ) 1/ 2= 58.549 mmIn general, 52 mm - ( 36 mm ) sin
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 31.FBD ABD: D = 0:Impending slip: So(15 mm ) N A - (110 mm ) FA = 0FA = SA N A 15 - 110 SA = 0 SA = 0.136364 SA = 0.1364FBD Pipe:Fx = 0:FA - Dx = 0, Dx
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 32.FBD Plate:Assume reactions as shown, at ends of sleeves, For impending slipFx = 0:FA = s N A ,FB = s N BP sin - s N A - s N B = 0 N A + N B = 2.5 P si
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 29.FBD table + child:WC = 18 kg 9.81 m/s 2 = 176.58 N WT2( ) = 16 kg ( 9.81 m/s ) = 156.96 N(a) Impending tipping about E , N F = FF = 0, andM E = 0:( 0.05
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 23. FBD rod:(a) Geometry:BE =L cos 2L DE = cos tan 2 EF = L sin So or Also, orDF =L cos 2 tan s1 L cos L cos tan + sin = 2 2 tan stan + 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 24. FBD:Assume the weight of the slender rod is negligible compared to P. First consider impending slip upward at B. The friction forces will be directed as shown and
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 25.FBD ABC:M C = 0:0.045 m + ( 0.30 m ) sin 30 ( 400 N ) sin 30 + 0.030 m + ( 0.30 m ) cos 30 ( 400 N ) cos 30 12 5 - ( 0.03 m ) FBD - ( 0.045 m ) FBD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 26.FBD CD: Note: The plate is a 3-force member, and for minimum s , slip impends at C and D, so the reactions there are at angle s from the normal. From the FBD, and
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 27.FBD pin A: From FBD Whole the force at A = 750 lbFx = 0: Fy = 0:FBD Casting:4 ( FAB - FAB ) = 0, FAB = FAB 5 3 750 lb - 2 FAB = 0, FAB = 625 lb 5Fx = 0:Impe
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 28.From FBD Whole, and neglecting weight of clamp compared to 550 lb plate, P = - W Since AB is a two-force member, B is vertical and B = W .FBD BCD:M C = 0:(1.85
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 75.HaveySP = r ( - 0 ) , yB = l AB cos ,r = 4 in., l AB = 18 in. 0 = 20 =9radPotential Energy:V = =1 2 kySP + WyB 2 1 2 2 kr ( - 0 ) + Wl AB cos 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 21.2.4 kN Force:Fx = ( 2.4 kN ) cos 50Fx = 1.543 kNFy = ( 2.4 kN ) sin 50Fy = 1.839 kN1.85 kN Force:Fx = (1.85 kN ) cos 20Fx = 1.738 kNFy = (1.85 kN ) sin
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 22.FBD ladder:Slip impends at both A and B, FA = s N A , FB = s N B Fx = 0: Fy = 0:FA - N B = 0,N B = FA = s N A N A + FB = WN A - W + FB = 0, N A + s N B =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 21.FBD ladder: Note: slope of ladder =4.5 m 12 13 = , so AC = ( 4.5 m ) = 4.875 1.875 m 5 12 AD = 1 L 2L = 6.5 m, so AC =4.875 m 3 = L, 6.5 m 4and DC = BD =1
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 19.FBD's: For slipping, F = k N = 0.30 N (a) For cw rotation of drum, the friction force F is as shown. From FBD arm:M A = 0:( 6 in.)( 600 lb ) + ( 6 in.) F - (18
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 18.FBD's: (a) FBD Drum:M D = 0: 10 ft F - 50 lb ft = 0 12 F = 60 lbImpending slip: N = FBD arm:Fs=60 lb = 150 lb 0.40M A = 0:( 6 in.) C + ( 6 i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 15.FBD:For impending tip the floor reaction is at C.W = ( 40 kg ) 9.81 m/s 2 = 392.4 NFor impending slip = s = tan -1 s = tan -1 ( 0.35 )() = 19.2900tan
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 13.FBD A:Note that slip must impend at both surfaces simultaneously. Fy = 0: N1 + T sin - 16 lb = 0 N1 = 16 lb - T sin Impending slip: F1 = s N1 = ( 0.20 )(16 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 12.FBD top block:Note that, since s = tan -1 s = tan -1 ( 0.40 ) = 21.8 > 15, no motion will impend if P = 0, with or without cable AB. (a) With cable, impending
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 10.FBD A with pulley:Tension in cord is T throughout from pulley FBD'sFy = 0:2T - 20 lb = 0,T = 10 lbFBD E with pulley:For max , motion impends to right, an
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 9.FBD Block:For impending motion. s = tan -1 s = tan -1 ( 0.40 ) s = 21.801Note 1,2 = 1,2 - s From force triangle:10 lb 15 lb = sins sin1,21,2 = sin -1
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 7. FBD Block:For Pmin motion will impend down the incline, and the reaction force R will make the angle s = tan -1 s = tan -1 ( 0.35 ) = 19.2900with the normal, as
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 16.First assume slip impends without tipping, so F = s N FBDFy = 0:N + P sin 40 - W = 0,N = W - P sin 40F = s N = 0.35 (W - P sin 40 ) Fx = 0:F - P cos 40 =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 20.FBD: (a)M C = 0:r ( F - T ) = 0,T = FImpending slip: F = s N or N =Fs=TsFx = 0:F + T cos ( 25 + ) - W sin 25 = 0T 1 + cos ( 25 + ) = W s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 17. FBD Cylinder:For maximum M, motion impends at both A and B FA = A N A; Fx = 0: N A - FB = 0 FB = B N B N A = FB = B N BFA = A N A = A B N BFy = 0: or and
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 14.FBD's: A:Fy = 0:Note: Slip must impend at both surfaces simultaneously.N1 - 20 lb = 0,N1 = 20 lbImpending slip: Fx = 0: Fy = 0:F1 = s N1 = ( 0.25 )( 20
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 11.FBD top block:Fy = 0:N1 - 196.2 N = 0N1 = 196.2 N(a) With cable in place, impending motion of bottom block requires impending slip between blocks, so F1 = s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 8.FBD block (impending motion downward) s = tan -1 s = tan -1 ( 0.25 ) = 14.036(a) Note: For minimum P, So andPR = = 90 - ( 30 + 14.036 ) = 45.964P = ( 30 l
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 6.FBD Block: W = ( 20 kg ) 9.81 m/s 2 = 196.2 N()For min motion will impend up the incline, so F is downward and F = s N Fy = 0: N - ( 220 N ) sin - (196.2 N )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 2.FBD Block B:Tension in cord is equal to WA = 40 lb from FBD's of block A and pulley. (a)Fy = 0: N - ( 52 lb ) cos 25 = 0,N = 47.128 lbFmax = s N = 0.35 ( 47.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 1.FBD Block B: Tension in cord is equal to W A = 25 lb from FBD's of block A and pulley. Fy = 0: N - WB cos 30 = 0,N = WB cos 30(a) For smallest WB , slip impends
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 152.For equivalence F: FA + FB + FC + FD = R C R C = - ( 5 lb ) j - ( 3 lb ) j - ( 4 lb ) k - ( 7 lb ) i R C = ( -7 lb ) i - ( 8 lb ) j - ( 4 lb ) k Also for equiva
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 151.For equivalence Fx : - ( 90 N ) sin 30 + (125 N ) cos 40 = Rx or Rx = 50.756 N Fy : - ( 90 N ) cos30 - 200 N - (125 N ) sin 40 = Ry or Ry = -358.29 N Then and tan
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 150.Have F : PAB = FC where PAB = AB PAB =( 2.0 in.) i + ( 38 in.) j - ( 24 in.) k44.989 in.( 45 lb )or FC = ( 2.00 lb ) i + ( 38.0 lb ) j - ( 24.0 lb ) k Hav
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 149.Require the equivalent forces acting at A and C be parallel and at an angle of with the vertical. Then for equivalence, Fx :( 250 lb ) sin 30 =FA sin + FB s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 148.(a) Based onF : FA = T = 1000 N or FA = 1000 N M A : M A = (T sin 50 )( dA ) = (1000 N ) sin 50 ( 2.25 m ) = 1723.60 N m or M A = 1724 N m 20(b) Based onF
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 146.Based on whereM O = rA/O TBAM O = M xi + M y j + M zk = M xi + (100 lb ft ) j - ( 400 lb ft ) k rA/O = ( 6 ft ) i + ( 4 ft ) j TBA = BATBA=( 6 ft ) i -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 145.First noteAC = rC/ A =( -2.4 )2 + (1.8)2m = 3mAD = rD/ A =and(1.2 )2 + ( -2.4 )2 + ( 0.3)2m = 2.7 mrC/ A = - ( 2.4 m ) j + (1.8 m ) k rD/ A = (1.2 m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 147.Have where DB =M DB = DB rC/D TCF()( 48 in.) i - (14 in.) j50 in.= 0.96i - 0.28 jrC/D = ( 8 in.) j - (16 in.) kTCF = CF TCF =( 24 in.) i - (
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 143.HaveM C = rB/C FBNoting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx where x = 144 mm - 78 mm = 66 mm y = 86 m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 141.First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action ( AA