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### Prob02073

Course: MAE 271, Spring 2009
School: University of Alabama -...
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Complete COSMOS: Online Solutions Manual Organization System Chapter 2, Problem 73. To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AB is 950 lb, determine (a) the components of the force exerted this by cable on the tree, (b) the angles x , y , and z...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 2, Problem 73. To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AB is 950 lb, determine (a) the components of the force exerted this by cable on the tree, (b) the angles x , y , and z that the force forms with axes at A which are parallel to the coordinate axes. 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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University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 74.To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground.
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 75.Determine (a) the x, y, and z components of the 900-N force, (b) the angles x , y , and z that the force forms with the coordinate axes.Vector Mechanics for Engineers: St
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 76.Determine (a) the x, y, and z components of the 1900-N force, (b) the angles x , y , and z that the force forms with the coordinate axes.Vector Mechanics for Engineers: S
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 77.A gun is aimed at a point A located 20 west of north. Knowing that the barrel of the gun forms an angle of 35 with the horizontal and that the maximum recoil force is 180
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 78.Solve Prob. 2.77 assuming that point A is located 25 north of west and that the barrel of the gun forms an angle of 30 with the horizontal. Problem 2.77: A gun is aimed at
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 79.The angle between the spring AB and the post DA is 30. Knowing that the tension in the spring is 220 N, determine (a) the x, y, and z components of the force exerted by th
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 80.The angle between the spring AC and the post DA is 30. Knowing that the x component of the force exerted by spring AC on the plate is 180 N, determine (a) the tension in s
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 81.Determine the magnitude and direction of the force F = ( 65 N ) i (80 N ) j ( 200 N ) k.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E.
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 82.Determine the magnitude and direction of the force F = ( 450 N ) i +( 600 N ) j (1800 N ) k.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer,
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 83.A force acts at the origin of a coordinate system in a direction defined by the angles x = 43.2 and z = 83.8. Knowing that the y component of the force is 50 lb , determin
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 84.A force acts at the origin of a coordinate system in a direction defined by the angles x = 113.2 and y = 78.4. Knowing that the z component of the force is 35 lb, determin
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 85.A force F of magnitude 250 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, y = 72.4, and Fz &gt; 0, determine (a) the components Fy and Fz , (b) the angl
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 86.A force F of magnitude 320 N acts at the origin of a coordinate system. Knowing that x = 104.5, Fz = 120 N, and Fy &lt; 0, determine (a) the components Fx and Fy , (b) the an
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 87.A steel rod is bent into a semicircular ring of radius 36 in. and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 88.A steel rod is bent into a semicircular ring of radius 36 in. and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 89.A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 2100 N, determine the components of the force exerted by the
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 90.A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 91.Two cables BG and BH are attached to the frame ACD as shown. Knowing that the tension in cable BG is 450 N, determine the components of the force exerted by cable BG on th
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 92.Two cables BG and BH are attached to the frame ACD as shown. Knowing that the tension in cable BH is 600 N, determine the components of the force exerted by cable BH on th
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 93.Determine the magnitude and direction of the resultant of the two forces shown knowing that P = 4 kips and Q = 8 kips.Vector Mechanics for Engineers: Statics and Dynamics
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 94.Determine the magnitude and direction of the resultant of the two forces shown knowing that P = 6 kips and Q = 7 kips.Vector Mechanics for Engineers: Statics and Dynamics
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 95.The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension is 510 N in cable AB and 765 N in cable AC, determine the magnitude and dire
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 96.Assuming that in Prob. 2.95 the tension is 765 N in cable AB and 510 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by th
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 97.For the tree of Prob. 2.73, knowing that the tension in cable AB is 760 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, dete
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 98.For the tree of Prob. 2.73, knowing that the tension in cable AC is 980 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, dete
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Problem 99.For the boom of Prob. 2.95, knowing that = 0, the tension in cable AB is 600 N, and the resultant of the load P and the force exerted at A by the two cables is directed al
University of Alabama - Huntsville - MAE - 271
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 20Fy = 0.
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 22.5 kips:Fx = ( 5 kips ) cos 40or Fx = 3.83 kipsFy = ( 5 kips ) sin 40or Fy = 3.21 kips7 kips:Fx = - ( 7 kips ) cos 70or Fx = -2.39 kipsFy = ( 7 kips ) sin 70or Fy
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 23.Determine the following distances:dOA = dOB = dOC = 680 N Force:( -160 mm )2 + ( 300 mm )2 ( 600 mm )2 + ( 250 mm )2( -160 mm )340 mm= 340 mm= 650 mm = 610 mm( 600
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 24.We compute the following distances:OA = OB = OC =Then: 204 lb Force:( 48)2 + ( 90 )2 ( 56 )2 + ( 90 )2 (80 )2 + ( 60 )2= 102 in. = 106 in. = 100 in.Fx = - ( 204 lb )
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 25.(a)P=Py sin 35=960 N sin 35or P = 1674 N(b)Px =Py tan 35=960 N tan 35or Px = 1371 NVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Bee
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 26.(a)P= P=Px cos 40 30 lb cos 40 or P = 39.2 lb !(b)Py = Px tan 40 Py = ( 30 lb ) tan 40 or Py = 25.2 lb !Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Fe
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 27.(a)Py = 100 N P= P= Py sin 75 100 N sin 75or P = 103.5 N &quot;(b)Px = Px =Py tan 75 100 N tan 75 or Px = 26.8 N &quot;Vector Mechanics for Engineers: Statics and Dynamics, 8
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 28.We note: CB exerts force P on B along CB, and the horizontal component of P is Px = 260 lb. Then: (a)Px = P sin 50P=Px sin 50=260 lb sin 50 P = 339 lb != 339.40 lb
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 29.(a)P=45 N cos 20 or P = 47.9 N !(b)Px = ( 47.9 N ) sin 20 or Px = 16.38 N !Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell J
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 30.(a)P=18 N sin 20 or P = 52.6 N !(b)Py =18 N tan 20 or Py = 49.5 N !Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 31.From the solution to Problem 2.21:F2.4 = (1.543 kN ) i + (1.839 kN ) j F1.85 = (1.738 kN ) i + ( 0.633 kN ) j F1.40 = (1.147 kN ) i - ( 0.803 kN ) j R = F = ( 4.428 kN )
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 32.From the solution to Problem 2.22:F5 = ( 3.83 kips ) i + ( 3.21 kips ) j F7 = - ( 2.39 kips ) i + ( 6.58 kips ) j F9 = - ( 8.46 kips ) i + ( 3.08 kips ) j R = F = - ( 7.
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 33.From the solution to Problem 2.24:FOA = - ( 48.0 lb ) i + ( 90.0 lb ) j FOB = (112.0 lb ) i + (180.0 lb ) j FOC = - ( 320 lb ) i - ( 240 lb ) j R = F = - ( 256 lb ) i +
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 34.From Problem 2.23:FOA = - ( 320 N ) i + ( 600 N ) j FOB = ( 360 N ) i + (150 N ) j FOC = ( 600 N ) i - (110 N ) j R = F = ( 640 N ) i + ( 640 N ) jR=( 640 N )2 + ( 640
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 35.Cable BC Force:Fx = - (145 lb ) Fy = (145 lb ) 84 = -105 lb 11680 = 100 lb 116 3 = -60 lb 5 4 = -80 lb 5100-lb Force:Fx = - (100 lb ) Fy = - (100 lb )156-lb Force:F
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 36.(a)Since R is to be horizontal, Ry = 0 Then, Ry = Fy = 090 lb + ( 70 lb ) sin - (130 lb ) cos = 0(13) cos = ( 7 ) sin + 913 1 - sin 2 = ( 7 ) sin + 9169 1 - sin 2 =
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 37.300-N Force:Fx = ( 300 N ) cos 20 = 281.91 N Fy = ( 300 N ) sin 20 = 102.61 N400-N Force:Fx = ( 400 N ) cos85 = 34.862 N Fy = ( 400 N ) sin 85 = 398.48 N600-N Force:
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 38.Fx :Rx = Fx Rx = ( 600 N ) cos 50 + ( 300 N ) cos85 - ( 700 N ) cos 50 Rx = - 38.132 NFy :Ry = Fy Ry = ( 600 N ) sin 50 + ( 300 N ) sin 85 + ( 700 N ) sin 50 Ry = 1294
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 39.We have:Rx = Fx = - 84 12 3 TBC + (156 lb ) - (100 lb ) 116 13 5or andRx = -0.72414TBC + 84 lbR y = Fy =80 5 4 TBC - (156 lb ) - (100 lb ) 116 13 5Ry = 0.68966TBC -
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 40.(a)Since R is to be vertical, Rx = 0 Then, Rx = Fx = 0( 600 N ) cos + ( 300 N ) cos ( + 35) - ( 700 N ) cos = 0Expanding: 3 ( cos cos 35 - sin sin 35 ) - cos = 01 cos
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 41.Selecting the x axis along aa, we writeRx = Fx = 300 N + ( 400 N ) cos + ( 600 N ) sin R y = Fy = ( 400 N ) sin - ( 600 N ) cos (1) (2)(a) Setting R y = 0 in Equation
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 42.(a)Require Ry = Fy = 0:( 900 lb ) cos 25 + (1200 lb ) sin 35 - TAE sin 65 = 0or TAE = 1659.45 lbTAE = 1659 lb !(b)R = Fx R = - ( 900 lb ) sin 25 - (1200 lb ) cos 35
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 43.Free-Body DiagramForce TriangleLaw of Sines:FAC TBC 400 lb = = sin 25 sin 60 sin 95(a)FAC = TBC =400 lb sin 25 = 169.691 lb sin 95 400 sin 60 = 347.73 lb sin 95FAC
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 44.Free-Body Diagram:Fx = 0:4 21 - TCA + TCB = 0 5 29or 29 4 TCB = TCA 21 5 Fy = 0:3 20 TCA + TCB - ( 3 kN ) = 0 5 29 3 20 29 4 TCA + 21 5 TCA - ( 3 kN ) = 0 5 29 The
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 45.Free-Body Diagram:Fy = 0:- FB sin 50 + FC sin 70 = 0FC = Fx = 0:sin 50 ( FB ) sin 70- FB cos 50 - FC cos 70 + 940 N = 0 sin 50 FB cos 50 + cos 70 = 940 sin 70 FB =
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 46.Free-Body Diagram:Fx = 0: Fy = 0:- TAB cos 25 - TAC cos 40 + ( 70 lb ) cos10 = 0 TAB sin 25 - TAC sin 40 + ( 70 lb ) sin10 = 0(1) (2)Solving Equations (1) and (2) sim
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 47.Free-Body Diagram:(a)Fx = 0:- TAB cos 30 + R cos 65 = 0 R= cos 30 TAB cos 65Fy = 0:- TAB sin 30 + R sin 65 - ( 550 N ) = 0 cos 30 TAB - sin 30 + sin 65 - 550 = 0 cos
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 48.Free-Body Diagram At B:Fx = 0:-12 17 TBA + TBC = 0 13 293 TBA = 1.07591 TBCorFy = 0:5 TBA + 132 TBC - 300 N = 0 2935 293 TBC = 300 - TBA 13 2 TBC = 2567.6 - 3.291
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 49.Free-Body Diagram:Fx = 0: - 8 kips + 15 kips - TD cos 40 = 0 TD = 9.1378 kips TD = 9.14 kips !Fy = 0:( 9.1378 kips ) sin 40 - TC=0 TC = 5.87 kips !Vector Mechanics f
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 50.Free-Body Diagram:Fy = 0:- 9 kips + TD sin 40 = 0 TD = 14.0015 kips TD = 14.00 kipsFx = 0:- 6 kips + TB - (14.0015 kips ) cos 40 = 0 TB = 16.73 kipsTB = 16.73 kipsV
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 51.Free-Body Diagram:Fx = 0:FC + ( 2.3 kN ) sin15 - ( 2.1 kN ) cos15 = 0orFC = 1.433 kNFy = 0:FD - ( 2.3 kN ) cos15 + ( 2.1 kN ) sin15 = 0orFD = 1.678 kNVector Mech
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 52.Free-Body Diagram:Fx = 0:- FB cos15 + 2.4 kN + (1.9 kN ) sin15 = 0orFB = 2.9938 kN FB = 2.99 kNFy = 0:FD - (1.9 kN ) cos15 + ( 2.9938 kN ) sin15 = 0FD = 1.060 kNV
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 53.From Similar Triangles we have:L2 - ( 2.5 m ) = ( 8 - L ) - ( 5.45 m )2 2 2- 6.25 = 64 - 16 L - 29.7025or Andcos =L = 2.5342 m5.45 m 8 m - 2.5342 mor Then = 4.35
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 54.From Similar Triangles we have:L2 - ( 3 m ) = ( 8 - L ) - ( 4.95 m )2 2 2- 9 = 64 - 16 L - 24.5025or Then or And or Free-Body Diagram At B: (a)L = 3.0311 m 4.95 m 8
University of Alabama - Huntsville - MAE - 271
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 55.Free-Body Diagram At C:3 15 15 Fx = 0: - TAC + TBC - (150 lb ) = 0 5 17 17orFy = 0:-17 TAC + 5 TBC = 750 5(1)4 8 8 TAC + TBC - (150 lb ) - 190 lb = 0 5 17 17or Th