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Soln09083
Course: CIVE 1150, Spring 2008School: Toledo
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Word Count: 182
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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...
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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.
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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 > T2 = T1 > 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 > T1 = T2 > 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 < 30, so mC min > 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