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3 Pages

### Soln08032

Course: CIVE 1150, Spring 2008
School: Toledo
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Word Count: 188

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 32. FBD Plate: Assume reactions as shown, at ends of sleeves, For impending slip Fx = 0: FA = s N A , FB = s N B P sin - s N A - s N B = 0 N A + N B = 2.5 P sin Fy = 0: N A - N B - P cos = 0, P ( 2.5sin + cos ) , 2 N A - N B = P cos NB = P ( 2.5sin - cos ) (1) 2 Solving: N A = M B = 0: ( 23.5 in.) P sin - (16 in.) N...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 32. FBD Plate: Assume reactions as shown, at ends of sleeves, For impending slip Fx = 0: FA = s N A , FB = s N B P sin - s N A - s N B = 0 N A + N B = 2.5 P sin Fy = 0: N A - N B - P cos = 0, P ( 2.5sin + cos ) , 2 N A - N B = P cos NB = P ( 2.5sin - cos ) (1) 2 Solving: N A = M B = 0: ( 23.5 in.) P sin - (16 in.) N A + (1 in.) FA = 0 P - 16 in. - 0.4 (1 in.) ( 2.5sin + cos ) = 0 2 (2) ( 23.5 in.) P sin 4sin - 7.8cos = 0, = 62.9 For > 62.9, panel the will be self locking, motion for 62.9. As decreases, N B will reverse direction at 2.5sin - cos = 0, (see equ. 1) or at = 21.8. So for 21.8 Fx = 0 : P sin - s ( N A + N B ) = 0 N A + N B = 2.5 P sin Fy = 0: N A + N B - P cos = 0, N A + N B = P cos 2.5sin = cos , = 21.8 So impending motion for 21.8 62.9 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 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 &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
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
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 142.(a) HaveM B = rC/B FN = ( 0.1 m )( 800 N ) = 80.0 N m or M B = 80.0 N m(b) By definition M B = rA/B P sin where = 90 - ( 90 - 70 ) - = 90 - 20 - 10 = 60
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 144.(a) Have whereM A = rE/ A TDE rE/ A = ( 92 in.) j TDE = DETDE =( 24 in.) i + (132 in.) j - (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.= ( 48 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 140.First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus,
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 138.DefineFA = ( FA ) x i + ( FA ) y j FB = ( FB ) x i + ( FB ) y jThenFx :( FA ) x + ( FB ) x = 0( FA ) x = - ( FB ) x( FA ) y + ( FB ) y = R ( FA ) y= R
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 139.First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wren
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 134.First reduce the given force system to a force-couple at the origin at B. (a)15 8 Have F : - ( 79.2 lb ) k - ( 51 lb ) i + j = R 17 17 R = - ( 24.0 lb ) i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 135.(a)First reduce the given force system to a force-couple at the origin. Have F : P BA + P DC + P DE = R 4 12 3 3 4 -9 4 R = P j - k + i - j + i - j+ k
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 133.B First replace the given couples with an equivalent force-couple system R, M O at the origin.()F :M O :R = - ( 35 lb ) i - (12 lb ) kR M O = - ( 200 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 137.First, reduce the given force system to a force-couple at the origin. Have F : FA + FG = R ( 4 in.) i + ( 6 in.) j - (12 in.) k R = (10 lb ) k + 14 lb = ( 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 136.First reduce the given force-couple system to an equivalent force-couple system ( R, M B ) at point B. d BD =( - 480 mm )2 + ( 560 mm )2 + ( - 480 mm )2= 880
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 132.(a) HaveFirst, reduce the given force system to a force-couple system. F : - ( 50 N ) k - ( 50 N ) j + ( 50 N ) k = R R = - ( 50 N ) j;R = - ( 50.0 N ) jR =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 131.First, reduce the given force system to a force-couple at the origin. Have F : - (10 N ) j - (11 N ) j = R R = - ( 21 N ) jR Have M O : ( rO F ) + M C = M O
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 130.First, reduce the given force system to a force-couple at the origin. HaveF : Pi - Pi - Pk = R R = - PkR M O : - P ( 3a ) k - P ( 3a ) j + P ( -ai + 3aj) = MO
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 129.First, reduce the given force system to a force-couple system at the origin. Have F :( 2P ) i - ( P ) j + ( P ) j = R R = ( 2P ) iHaveR M O : ( rO F ) =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 127.For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer,
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 122.From - ( 0.6 ft ) i + ( 4.2 ft ) j - (1.5 ft ) k R C = R = ( 60 lb ) AB = 60 lb 2 2 2 ( - 0.6 ft ) + ( 4.2 ft ) + ( -1.5 ft ) R C = - ( 8.00 lb ) i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 126.HaveF : FB + FC + FD + FE = R - ( 470 kips ) j - ( 66 kips ) j - ( 28 kips ) j - (116 kips ) j = R R = - ( 680 kips ) jHaveM x : FB ( z B ) + FC ( zC ) + F
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 128.For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of the trailer, the box must be as close as possib
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 120.(a) Have whereR = F = FF + FEFF = 50 N ( sin 60 ) j + ( cos 60 ) k = ( 43.301 N ) j + ( 25 N ) k FE = - ( 25 N ) k R = ( 43.301 N ) j or R = ( 43.3 N ) j
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 121.Have whereF : R = R A = R BC BC =( 42 in.) i - ( 96 in.) j - (16 in.) k106 in. RA =21.2 lb ( 42i - 96j - 16k ) 106or R A = ( 8.40 lb ) i - (19.20 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 124.Have:M x :FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) + FF ( z F ) = R ( zG )(80 kN )( 0 ) + FB ( 3 m ) sin 60 + ( 40 kN ) ( 3 m ) sin 60
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 125.HaveF : FA + FB + FC + FD = R - (116 kips ) j - ( 470 kips ) j - ( 66 kips ) j - ( 28 kips ) j = R R = - ( 680 kips ) j R = 680 kipsHaveM x : FA ( z A ) +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 119.(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:R R M By = j M B = j rF /B FF + rE/B FE = 0()whererF /B = (1.125 m ) i -