# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

2 Pages

### Soln09083

Course: CIVE 1150, Spring 2008
School: Toledo
Rating:

Word Count: 182

#### Document Preview

Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 83. From Problem 9.74 I xy = -0.1596 106 mm 4 From Figure 9.13 I x = 0.166 106 mm 4 I y = 0.453 106 mm 4 Now 1 I x + I y = 0.3095 106 mm 4 2 ( ) 1 I x - I y = -0.1435 106 mm 4 2 ( ) Using Equations (9.18), (9.19), and (9.20) Equation (9.18): I x = Ix + I y 2 + Ix - I y 2 cos 2 - I xy sin 2 = 0.3095 106 + -0.1435...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Ohio >> Toledo >> CIVE 1150

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.

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.
Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 83. From Problem 9.74 I xy = -0.1596 106 mm 4 From Figure 9.13 I x = 0.166 106 mm 4 I y = 0.453 106 mm 4 Now 1 I x + I y = 0.3095 106 mm 4 2 ( ) 1 I x - I y = -0.1435 106 mm 4 2 ( ) Using Equations (9.18), (9.19), and (9.20) Equation (9.18): I x = Ix + I y 2 + Ix - I y 2 cos 2 - I xy sin 2 = 0.3095 106 + -0.1435 106 cos ( -90 ) - -0.1596 106 sin ( -90 ) mm 4 = 0.1499 106 mm 4 or I x = 0.1499 106 mm 4 ( ) ( ) 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 Cornwell J. 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Equation (9.19): I y = Ix + I y 2 - Ix - I y 2 cos 2 + I xy sin 2 = 0.3095 106 - -0.1435 106 cos ( -90 ) + -0.1596 106 sin ( -90 ) mm 4 ( ) ( ) = 0.4691 106 mm 4 or I y = 0.469 106 mm 4 Equation (9.20): I xy = Ix - I y 2 sin 2 + I xy cos 2 = -0.1435 106 sin ( -90 ) + 0.1596 106 cos ( -90 ) mm 4 = 0.1435 106 mm 4 or I xy = 0.1435 106 mm 4 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.
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 9, Solution 82.From Problem 9.75 Now where andI xy = 1.5732 106 mm 4I x = ( I x )1 + ( I x )2 + ( I x )3( I x )1 = ( I x )2 = ( I x )3=1 (150 mm )(12 mm )3 = 21 600 mm 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 79.From Figure 9.12:Ix ==16( 2a )( a )38a4Iy ==From Problem 9.67: First note16( 2a )3 ( a )2a4 1 4 a 2I xy =1 1 5 I x + I y = a4 +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 78.Have For each rectangleI xy = I xy( )1 + ( I xy )2I xy = 0 (symmetry)I xy = I xy + x yAThenand I xy = x yA = ( -0.75 in.)( -1.5 in.) ( 3 in.)( 0.5 in
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 75.Have Now symmetry implies and for the other rectangles ThusI xy = I xy( )1 + ( I xy )2 + ( I xy )3( I xy )1 = 0I xy = I xy + x y A where I xy = 0 (symmetry)
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 74.Have For each rectangle ThusI xy = I xy( )1 + ( I xy )2and I xy = 0 (symmetry)I xy = I xy + Ax y I xy = x y AA, mm 21 2x , mmy , mmAx y , mm 476
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 72.Note: Orientation of A3 corresponding to a 180 rotation of the axes. Equation 9.20 then yieldsI xy = I xySymmetry implies Using Sample Problem 9.6 and Similarl
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 80.From the solution to Problem 9.72I xy = 501.1875 in 4A2 = A3 = 20.25 in 2First compute the moment of inertiaI x = ( I x )1 + ( I x )2 + ( I x )3with( I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 76.Symmetry implies( I xy )1 = 0Using Sample Problem 9.6 and Equation 9.20, note that the orientation of A2 corresponds to a 1 2 2 90 rotation of the axes; thus I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 71.Have Where and NowI xy = I xy( )1 - ( I xy )2 - ( I xy )30and( I xy )1 = 0I xy = I xy + A x y for areas A2 = ( 50 mm )( 20 mm )= 1000 mm 2x2 = -15 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 81.From Problem 9.73,I xy = 138.24 106 mm 4I x = ( I x )1 + ( I x )24 = 2 (120 mm ) 8 ( I x )1 = ( I x )2= 51.84 106 mm 4Iy = Iy( )1 + ( I y )2( I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 77.Have Where Then Section :I xy = I xy + x yAI xy = 0 for each rectangleI xy = I xy( )1 + ( I xy )2 + ( I xy )3 = x yAx1 = - ( 8.92 in. - 8 in.) = - 0.92 i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 73.Have For each semicircle ThusI xy = I xy( )1 + ( I xy )2andI xy = I xy + x y A I xy = x y AI xy = 0 (symmetry)A, mm 21x , mmy , mmAx y , mm 469.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 128.Fn = 0:or N - T + (T + T ) sin =0 2N = ( 2T + T ) sin Ft = 0: 2 (T + T ) - T cos - F = 0 2 F = T cosF = s Nor Impending slipping: So 2T cos
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 129.Small belt section: Side view: End view:Fy = 0: Fx = 0: Impending slipping:2N sin - T + (T + T ) sin =0 2 2 2 (T + T ) - T cos - F = 0 2F = s N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 130.FBD motor and mount: Impending belt slip, cw rotations T2 = T1esin 2T2 = T1e sin18 = 58.356T1( 0.40 )M D = 0:(12 in.)(175 lb ) - (13 in.) T1 - ( 7 in
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 127.FBD wrenchNote: EC =( 0.20 m ) ,sin 75EA = EC - 0.03 m = 75so = 285 = 4.9742 radM E = 0: 0.20 m 0.20 m - 0.03 m F - cos 75 - 0.03 m T = 0 sin 7
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 126.FBD wrench:Note: EC =( 0.2 m ) ,sin 65EA = EC - 0.03 m = 65so = 295 = 5.1487 rad 0.20 m 0.20 m - 0.03 m F - cos 65 - 0.03 m T = 0 sin 65 sin
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 125. FBD pin B:(a) By symmetry: 2 Fy = 0: B - 2 2 T1 = 0 T1 = T2or B= 2T1(1)FBD Drum:For impending rotation : T4 &gt; T2 = T1 &gt; T3 , so T4 = Tmax = 5.6 kN
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 121.Geometry and force notation: = sin -15 in. = 30 = rad, so contact angles are: 10 in. 6C = -6=5 2 = , D = + , 6 2 6 3E =2(a) D and E fixed,
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 122.FBD drum B:M B = 0:( 0.02 m )(TA - T ) - 0.30 N m = 0TA - T = 0.30 N m = 15 N 0.02 mImpending slip: TA = Te s = Te0.40 Solving; T e0.40 - 1 = 15 N()
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 123.FBD drum B:M B = 0:( 0.02 m )(TA - T ) - 0.3 N m = 0TA - T = 15 NImpending slip:TA = Te s B = Te0.40Solving, T e0.40 - 1 = 15 N()T = 5.9676 NIf
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 119.Geometry and force notation: = sin -1r = 30 = , so contact angles are: 2r 6C = D =2+6=2 , 3E = (a) D and E fixed, so slip on these surfaces
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 120.Geometry and force notation: = sin -15 in. = 30 = rad, so contact angles are: 10 in. 6C = -6=5 2 = , D = + , 6 2 6 3E =2(a) All pulleys loc
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 124. FBD pin B:(a) By symmetry: 2 Fy = 0: B - 2 T =0 2 1 T1 = T2or B= 2T1 = 2T2(1)Drum:For impending rotation : T3 &gt; T1 = T2 &gt; T4 , so T3 = Tmax = 5.6 k
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 118.Geometry and force notation:Note: = sin -1 r = 30 = rad, so contact angles are: 2r 6C = D =2+6=2 , 3E = (a) For all pulleys locked, slip im
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 117.Geometry and force rotation: LetEBC = = cos -1 50 mm = 60 = 100 mm s DBE , FAE , GAEThen contact angles are B = 360 - 120 = 240 = upper cylinder, and A = 30
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
Toledo - CIVE - 1150
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