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3 Pages

Soln08093

Course: CIVE 1150, Spring 2008
School: Toledo
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Word Count: 127

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 93. Let normal force on A be N , and As in the text The total normal force P is N k = A r F = N , M = r F 2 R k P = lim N = 0 R 2 rdr d 1 r A 0 P = 2 R 2 kdr = 2 k ( R2 - R1 ) R 1 or k = P 2 ( R2 - R1 ) The total couple is k 2 R M worn = lim M = 0 R 2 rdr r d A 0 r 1 R2 R1 M worn = 2 k ( rdr ) = k ( 2...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 93. Let normal force on A be N , and As in the text The total normal force P is N k = A r F = N , M = r F 2 R k P = lim N = 0 R 2 rdr d 1 r A 0 P = 2 R 2 kdr = 2 k ( R2 - R1 ) R 1 or k = P 2 ( R2 - R1 ) The total couple is k 2 R M worn = lim M = 0 R 2 rdr r d A 0 r 1 R2 R1 M worn = 2 k ( rdr ) = k ( 2 R2 - R12 )= 2 P R2 - R12 2 ( R2 - R1 ) ( ) M worn = 1 P ( R2 + R1 ) 2 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 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 &lt; 0.382kl = 0.382 (1250 N/m )( 0.4 m ) = 191 N0 P &lt; 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 &lt; 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 &gt; 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 &gt; 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 &gt; 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 &gt; 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 &gt; 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