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Soln08048
Course: CIVE 1150, Spring 2008School: Toledo
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Word Count: 106
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Complete COSMOS: Online Solutions Manual Organization System Chapter 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 ) = 14.0362 FBD Lever: M C = 0: ( 0.3 m )( 350 N ) - ( 0.4 m )(176.58 N ) - ( 0.525 m ) RA cos 4.0362 + ( 0.05 m ) RA sin 4.0362 = 0 RA = 66.070 N Fx = 0: ( 66.07 N ) sin 4.0362 + Cx 0, Cx = = - 4.65 N...
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Complete COSMOS: Online Solutions Manual Organization System
Chapter 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 ) = 14.0362
FBD Lever:
M C = 0:
( 0.3 m )( 350 N ) - ( 0.4 m )(176.58 N )
- ( 0.525 m ) RA cos 4.0362 + ( 0.05 m ) RA sin 4.0362 = 0 RA = 66.070 N
Fx = 0:
( 66.07 N ) sin 4.0362 + Cx 0,
Cx = = - 4.65 N
Fy = 0:
FBD Wedge:
( 66.07 N ) cos 4.0362 - 350 N - 176.58 N = 0
P 66.070 N = sin18.072 sin 75.964 P = 21.1 lb (a) (b) P = 21.1 lb
C x = 4.65 N C y = 461 N
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 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
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