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### Soln08117

Course: CIVE 1150, Spring 2008
School: Toledo
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Word Count: 145

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 117. Geometry and force rotation: Let EBC = = cos -1 50 mm = 60 = 100 mm s DBE , FAE , GAE Then contact angles are B = 360 - 120 = 240 = upper cylinder, and A = 30 = contact on lower cylinder. Let the force in section FC = TF Let the force in section DG = TG With A fixed and the cord moving, 0.25 6 6 4 rad for cord on 3 rad...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 8, Solution 117. Geometry and force rotation: Let EBC = = cos -1 50 mm = 60 = 100 mm s DBE , FAE , GAE Then contact angles are B = 360 - 120 = 240 = upper cylinder, and A = 30 = contact on lower cylinder. Let the force in section FC = TF Let the force in section DG = TG With A fixed and the cord moving, 0.25 6 6 4 rad for cord on 3 rad for each cord TG = We k A = We ( ) = 1.13985W For W, maximum slip impends on drum B, so TB = TF e s B or TF = TG e- s B TF = 1.13985We For slip at F -0.30 43 ( ) = 0.32441W W1 = TF e k A = 0.32441We 0.25 6 ( ) = 0.36978W so W = 2.7043W1 and m = 2.7043 in. = 2.7043 ( 75 kg ) m = 203 kg 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 9, Solution 70.First note: At x = a : b = ke a or k = Nowab b x ; y = ea e eI xy = xydA= =x b a a ee 0 0xydydx1 b 2 a 2ax xe dx 2 e2 0a 2x 2 a 1b e 2 = x
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 116.(a) For minimum mC with blocks at rest, impending slip of A is down/left. Note: s = tan -1 s = tan -1 0.30 = 16.7 &lt; 30, so mC min &gt; 0 FBD A:WA = ( 6 kg ) 9.81 m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 69.HaveI xy = xydA= - b h x xydydxb00=1 0 h2 2 x - x dx 2 -b b2 =- =1 h2 4 x 8 b 2 -b01 2 2 b h 8 I xy = 1 2 2 b h 8Vector Mechanics
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 67.First notey = a 1- =x2 4a 21 4a 2 - x 2 2Have wheredI xy = dI xy + xEL yEL dA dI xy = 0( symmetry )1 1 y = 4a 2 - x 2 2 4xEL = xyEL =dA = ydx =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 68.First note: At x = a : or k = Nowb = ka3b b ; y = 3 x3 3 a aI xy = xydA= 0 a b b 3 x a3xydydx=1 a 2 b2 6 x b - a 6 x dx 2 0 ab2 1 2 1 8 =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 66.The pressure p at an arbitrary depth ( y sin ) isp = ( y sin )so that the hydrostatic force dP exerted on an infinitesimal area dA isdP = ( y sin ) dAThe
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 115.FBD Lever:M B = 0:( 40 mm ) TC - (100 mm ) TA = 0,TC = 2.5 TAFBD Drum: (a) For impending slip ccw: TC = Tmax = 4.5 kN soTA =TC = 1.8 kN 2.5M D = 0:
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 112.FBD Lever:M E = 0:( 60 mm )( 240 N ) - ( 40 mm ) FBD cos 30 = 0FBD = 415.69 NFBD Drum: Belt slip:T2 = T1 e k 0.25( 5.5851)= ( 415.69 N ) e= 1679.44 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 113.FBD Drum: (a) With M E = 125 lb ftM E = 0:( 7 in.) (TA - TC ) - (125 lb ft ) = 0 0.30 76TA - TC = 214.29 lbBelt slip: TA = TC e k = TC e so 2.0028 TC =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 110.FBD Flywheel:M C = 0:( 0.225 m )(TB - TA ) - 12.60 N m = 0TB - TA = 56 N, TB = TA + 56 NAlso, since the belt doesn't change length, the additional stretch
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 111.FBD Flywheel: Slip of belt: TB = TA e k = TA e0.20 Also, since the belt doesn't change length, the increase in stretch of spring B equals the decrease in stretch
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 109.FBD lower portion of belt:Fy = 0:48 N - N D = 0,N D = 48 NSlip on both platen and woodFD = kD N D = 0.10 ( 48 N ) = 4.8 N FE = kE N E = ( 48 N ) kEFBD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 114.FBD Lever: If brake is self-locking, no force P is requiredM B = 0:( 2 in.) TC - ( 7.5 in.) TA = 0TC = 3.75 TAFor impending slip on drum: TC = TA e s e s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 105.Horizontal pipe: Vertical pipeContact angles H =2Contact angle V = sH = 0.25For P to impend downward, sV = 0.2 P = e sH 2 Q = e sH 2 e s
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 106.Horizontal pipe Vertical pipeContact angles H =2Contact angle V = sH = 0.30For Pmin , the 100 lb force impends downward, and sV = ? 0.30 + sV )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 107.FBD motor and mount: Impending belt slip: cw rotationT2 = T1e s = T1e0.40 = 3.5136 T1 M D = 0:(12 in.)(175 lb ) - ( 7 in.) T2 - (13 in.) T1 = 02100 lb = ( 7
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 108.FBD motor and mount: Impending belt slip: ccw rotationT1 = T2e s = T2e0.40 = 3.5136 T2 M D = 0:(12 in.)(175 lb ) - (13 in.) T1 - ( 7 in.) T2T1 = 3.5136 T2 =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 64.xv of the volume is defined byxvV = xEL dVSelecting the element of volume showndV = ydA = kxdAV = k xdA = kx A AWhere x A = coordinate of the centroid of
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 63.Have where Nowx dV x = EL dVdV = ydA and xEL = xy =60 1 x= x 300 5Then1 x = 1 5 x dA =2 x 5 x dA x dA = ( I z ) A xdA ( xA) Awhere
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 60.From Sec. 9.2:R = ydAM AA = y 2dALet yP = Distance of center of pressure from AA. We must have2 M AA y dA I AA = = R yA ydARyP = M AA ' :whereyP
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 59.From Sec. 9.2:R = ydAM AA = y 2dALet yP = Distance of center of pressure from AA : We must haveRyP = M AA ' :yP =2 M AA y dA I AA = = R yA ydAD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 65.The pressure p at an arbitrary depth ( y sin ) isp = ( y sin )so that the hydrostatic force dF exerted on an infinitesimal area dA isdF = ( y sin ) dAEqui
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 61.Using equation developed on page 491 of text:yP =ThenIx I = AA yA yAR = yA = gyAR = 920kg m 9.81 2 3 m ( 0.55 0.25 ) m 2 3 m s= 3722.9 NandI
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 62.Using equations developed on page 491 of text:yP = Ix I = ss yA yAR = yANowyA = yA1 = ( h + 17 in.) 120 in. 51 in. 2 1 + ( h + 34 in.) 84 in. 51
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 58.From Sec. 9.2:R = ydAM AA = y 2dALet yP = Distance of center of pressure from AA. We must haveRyP = M AA ' :yP =I AA = y =2 M AA y dA I AA = = R y
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 54.Angle:L3 3 1 : 4A = 1.44 in 2L6 4 1 : 2I x = I y = 1.24 in 4A = 4.75 in 2Plate:I x = 6.27 in 4I y = 17.4 in 4A = ( 27 in.)( 0.8 in.) = 21.6 in
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 55.Angle:A = 2420 mm 2 I x = 3.93 106 mm 4I y = 1.06 106 mm 4Plate:A = ( 200 mm )(10 mm ) = 2000 mm 2Ix = Iy =Centroid1 ( 200 mm )(10 mm )3 = 0.01667 1
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 103.FBD A:Fx = 0:TA - WA sin 30 = 0,TA =WA 2FBD B:Fx = 0:WB sin 30 - TB = 0,TB =WB 2For mB min , motion of B impends up incline And But0.50 TA = e
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 104. = 1.5 turns = 3 radFor impending motion of W upP = We s = (1177.2 N ) e(= 4839.7 N0.15 )3For impending motion of W down- 0.15 3 P = We- s = (1177.2 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 101.Two full turns of rope (a) = 4 rad s = lnT2 T1 ors =1lnT2 T1s =1 20 000 N = 0.329066 ln 4 320 N s = 0.329(b) ==1slnT2 T11 80 00
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 100.FBD wheel: For equilibrium (constant speed), R and and tangent to the friction circle as shownW are equal and opposite 2rf = raxle sin tan -1 k = (12.5 mm ) si
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 97.FBD disk:tan = slope = 0.02b = r tan = ( 60 mm )( 0.02 ) b = 1.200 mmVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 96.FBD pipe: = sin -10.025 in. + 0.0625 in. = 1.00257 5 in.P = W tan for each pipe, so also for total P = ( 2000 lb ) tan (1.00257 ) P = 35.0 lbVector Mechani
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 93.Let normal force on A be N , and As in the text The total normal force P isN k = A rF = N , M = r F2 R k P = lim N = 0 R 2 rdr d 1 r A 0 P = 2 R 2
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