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3 Pages

### Soln08070

Course: CIVE 1150, Spring 2008
School: Toledo
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Word Count: 150

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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...

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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.
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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
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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 = -
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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 &gt; s =
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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 &gt; NB,FA m
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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
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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
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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
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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 )
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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 &gt; B when finished, adjust A first when there will be no forceBlock/
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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
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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 )
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 &gt; 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
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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 =
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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
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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
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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
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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 )
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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.
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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
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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
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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
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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
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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
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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
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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 -
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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
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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 - (
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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
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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