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.
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 8, Solution 102.FBD A:WA = (10 kg ) 9.81 m/s 2 = 98.1 N()Fx = 0:TA - WA sin 30 = 0,TA =WA 2FBD B:Fx = 0:WB sin 30 - TB = 0,TB =WB 2(a) Motion of B impend
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 98.FBD wheel:r = 230 mm b = 1 mm = sin -1b rb P = W tan = W tan sin -1 for each wheel, so for total r 1 P = (1000 kg ) 9.81 m/s 2 tan sin -1 230 P
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 94.Let normal force on A be N , and N = k AN = k, A s = r sin soA = r swhere is the azimuthal angle around the symmetry axis of rotation Fy = N sin = kr r
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 92.Let the normal force on A be N , and As in the text The total normal forceN k = r AF = N , M = r F2 R k P = lim N = 0 0 rdr d A 0 r P = 2 0 kdr =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 99.FBD wheel:rf = raxle sin = raxle sin tan -1 ,rw = rf sin + b tan ()s or kFor small , sin tan , so tan rf + b rwFy = 0: Fx = 0:R cos -W =0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 95.If normal force per unit area (pressure) of the center is PO , then as a functionr of r, P = PO 1 - R r 2 R FN = W = PdA = 0 0 PO 1 - rdrd R 2 R3 R2 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 91.Eqn. 8.8 gives M =3 3 2 R3 - R1 1 D3 - D1 s P 2 = s P 2 2 2 3 3 R2 - R12 D2 - D12so1 M = ( 0.15 )( 80 kg ) 9.81 m/s 2 3(( 0.030 m )3 - ( 0.024 m )3 ) ( 0.0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 98.From the analysis of Problem 10.97 withl = 400 mm and k = 1.25 kN/mP < 0.382kl = 0.382 (1250 N/m )( 0.4 m ) = 191 N0 P < 191.0 NVector Mechanics for Engine
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 81.Note:xSPA = rA sin xSPB = xSPA xSP = xSPA + xSPB= 2rA sin ,rA = 150 mm r = 200 mmyBLOCK = r ,Potential Energy:V ==1 2 kxSP - mgyBLOCK 2 1 2 k ( 2r
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 82.(a)sSP = l cos 0 - l cos= l ( cos 0 - cos ) Potential Energy:V ==1 2 ksSP - mg ( 2l sin ) 2 1 2 2 kl ( cos 0 - cos ) - 2mgl sin 2 (1)dV = kl 2 ( cos 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 85.First note, by Law of Cosines: 2 d 2 = (16 ) + 16 sin - 2 (16 ) 16 sin cos 2 2 2 d = 16 1 + sin 2Also note Potential Energy:22- sin in.xSP =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 93.From geometry:xC = - a sin = - 2a sin For small values of , = 2or = 1 2y A = a cos + 3 a cos = a cos + 3cos 2 For spring:s = xC = - a sin P
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 43.FBD A:For impending motion A must start up and C down the incline. Since the normal force between A and B is less than that between B and C, and the friction coeff
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 44.FBD rod:M A = 0: or Impending motion: 3 in. N B - ( 4.5 in.) cos W = 0 cosN B = (1.5cos 2 )WFB = s N B = (1.5 s cos 2 )W= ( 0.3cos 2 )W Fx = 0: or Impe
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 46. s = tan -1 s = tan -1 ( 0.4 ) = 21.801, slip impends at wedge/block wedge/wedge and block/inclineFBD Block:R2 530 lb = sin 41.801 sin 46.398R2 = 487.84 lbF
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 45.FBD pin A:Fx = 0: Fy = 0: Solving:FBD B:12 3 FAB - FAC = 0 13 5 5 4 FAB + FAC - P = 0 13 5 FAB = 13 P, 21 FAC = 20 P 21Fx = 0:NB -12 13 P = 0, 13 21NB
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 42.FBD pulley: Note that SA = tan -1 SA = tan -1 ( 0.5 ) = 26.565 < 30, Cable is needed to keep A from sliding downward.Fy = 0:2T - WB = 0,T =WB , 2WB = 2T
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 40.FBD yoke:Fx = 0:P - N = 0,N = P = 8 lb F = s N = 125 ( 8 lb ) F = 2 lbFor impending slip,For M max , F on yoke is down as shownFBD wheel and slider:Fo
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 37.Geometry: = cos -1L L L + - 2 4 2 = 60 L 2For minFBD AB: s = tan -1 s = tan -1 ( 0.35 ) = 19.2900 from normal.Note: AB is a three-force membera Lsli
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 48.WD = (18 kg ) 9.81 m/s 2 = 176.58 N Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N() s = tan -1 s = tan -1 ( 0.25 ) = 14.0362FBD Lever:M C = 0:( 0.3
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 39.FBD AB:M A = 0:8 in 2 + 4 in 2 ( N ) - M A = 0N =Impending motion:(12 lb ft )(12 in./ft )8.9443 in.= 16.100 lbF = s N = 0.3 (16.100 lb ) = 4.83 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 83.(1)Law of Sines:yA l = sin ( 90 + - ) sin ( 90 - ) yA l = cos ( - ) cos yA = l cos ( - ) cos From Eq. (1): Potential Energy:yB = lcos ( - ) - l c
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 90. FBDM O = 0:(8 in.) Q - M= 0,Q=M 8 in.but, from equ. 8.9,M =2 2 7 in. k WR = ( 0.60 )(10.1 lb ) 3 3 2 = 14.14 lb so, Q= 14.14 , 8 Q = 1.768 lb
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 89.FBD Each wheel:For equilibrium (constant speed) the two forces R and and opposite, tangent to the friction circle, so rf sin = where = tan -1 ( slope ) rwW mu
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 87.It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single weight of( 90 kg )( 9.81 N/kg ) = 882.9 N,right of B. FBD pulley + gate:locat
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 86.FBD gate:W1 = 66 kg 9.81 m/s 2 = 647.46 NW22( ) = 24 kg ( 9.81 m/s ) = 235.44 N( )*rf = rs sin s = rs sin tan -1 s= ( 0.012 m ) sin tan -1 0.2 = 0.0023
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 85.It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single combined weight of( 90 kg ) ( 9.81 m/s2 ) = 882.9 N,right of B.located at a d
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 82.Pulley FBDs: Left:rp = 30 mmrf = raxle sin k = raxle sin tan -1 k()*= ( 5 mm ) sin tan -1 0.2 = 0.98058 mmM C = 0: or()( rp + rf ) ( 600 N ) - 2r
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 79.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 76.FBD Windlass:rf = rb sin s = rb sin tan -1 s()= (1.5 in.) sin tan -1 0.5 = 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 73.FBD lower jaw:By symmetry B = 540 N Fy = 0: - 540 N + A - 540 N = 0,A = 1080 Nsince A > B, A should be adjusted first when no force is required. If instead, B
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 67.FBD large gear:M C = 0:(12 in.)W- 7.2 kip in. = 0,W = 0.600 kips= 600 lbBlock on incline: = tan -10.375 in. = 2.2785 2 (1.5 in.) s = tan -1 s = t
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 64.WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N Slip must impend at all surfaces simultaneously, F = s NFBD I: A + B()()Fy = 0:
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 62.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 1 - 3 + ft ( 96 lb/ft )( 5 ft ) = 0 3 2 N B
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 61. FBD Cylinder:W = ( 80 kg ) 9.81 m/s 2 = 784.8 NFor impending slip at B, M A = 0:()FB = sB N B = 0.30 N B( r cos 30 ) N B - r (1 + sin 30 )( 0.30 N B )-
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 59.FBD Wedge: s = tan -1 s = tan -1 ( 0.35 ) = 19.2900by symmetry R1 = R2 Fy = 0: 2 R1 sin 22.29 - 60 lb = 0R2 = 79.094 lbWhen P is removed, the vertical compon
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 57.FBD tip of screwdriver:s = tan -1 s = tan -1 ( 0.12 ) = 6.8428by symmetry R1 = R2Fy = 0:2R1 sin ( 6.8428 + 8 ) - 3.5 N = 0R1 = R2 = 6.8315 NIf P is remov
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 54.Since vertical forces are equal and s ground > s wood, assume no impending motion of board. Then there will be impending slip at all wood/wood contacts, s = tan
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 52.FBD Wedge: s = tan -1 s = tan -1 ( 0.4 ) = 21.801By symmetry RB = RC Fy = 0: 2RC sin ( 29.801 ) - P = 0,P = 0.9940 RCFBD Block C:RC 175 lb , = sin 41.801 si
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 53.FBD Block C: s = tan -1 s = tan -1 ( 0.4 ) = 21.8014Fx = 0: Fy = 0: soRCFy RCFx RACx - RCFx = 0 RCFy - RACy - 175 lb = 0-RACy RACx=175 lb RACxcot ( 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 83.FBD link AB: Note: That AB is a two-force member. For impending motion, the pin forces are tangent to the friction circles. r = sin -1 f 25 in. whererf = rp sin
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 80.(a) FBD lever (Impending CW rotation): M C = 0:( 0.2 m + rf ) ( 75 N ) - ( 0.12 m - rf ) (130 N ) = 0rf = 0.0029268 m = 2.9268 mmsin s = rf rs* -1 rf -1 2.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 77.FBD Windlass:rf = r sin s = r sin tan -1 s()= (1.5 in.) sin tan -1 0.5 = 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 74.Block/incline: = tan -10.25 in. = 2.4302 1.875 in. s = tan -1 s = tan -1 ( 0.10 ) = 5.7106Q = (1000 lb ) tan ( 8.1408 ) = 143.048 lbCouple = rQ = ( 0.9375
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 71.FBD joint D:By symmetry:FAD = FCDFy = 0: 2FAD sin 25 - 4 kN = 0FAD = FCD = 4.7324 kNFBD joint A:By symmetry:FAE = FADFx = 0: FAC - 2 ( 4.7324 kN ) cos
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 70.FBD joint D:By symmetry:FAD = FCDFy = 0: 2FAD sin 25 - 4 kN = 0FAD = FCD = 4.7324 kNFBD joint A:By symmetry:FAE = FADFx = 0: FAC - 2 ( 4.7324 kN ) cos
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