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Prob06132
Course: EGM 2511, Spring 2010School: University of Florida
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Complete COSMOS: Online Solutions Manual Organization System Chapter 6, Problem 132. Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the c ouple M A required to hold the system in equilibrium. Vector for Mechanics Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J....
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Complete COSMOS: Online Solutions Manual Organization System
Chapter 6, Problem 132.
Two rods are connected by a slider block as shown. Neglecting the effect
of friction, determine the c ouple M A required to hold the system in
equilibrium.
Vector for Mechanics 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 Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 133.Rod CD is attached to the collar D and passes through a collar welded toend B of lever AB. Neglecting the effect of friction, determine the coupleM required to hold the
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 134.Rod CD is attached to the collar D and passes through a collar welded toend B of lever AB. Neglecting the effect of friction, determine the coupleM required to hold the
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 135.A small barrel having a weight of 250 lb is lifted by a pair of tongs asshown. Knowing that a = 5 in., determine the forces exerted at B and Don tong ABD.Vector Mechan
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 136.The tongs shown are used to apply a total upward force of 45 kNon a pipe cap. Determine the forces exerted at D and F on tongADF.Vector Mechanics for Engineers: Static
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 137.The pallet puller shown is used to pull a loaded pallet to the rearof a truck. Knowing that P = 2.1 kN, determine the forcesexerted at G and H on tong FGH.Vector Mecha
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 138.The drum lifter shown is used to lift a steel drum. Knowing thatthe weight of the drum and its contents is 110 lb, determine theforces exerted at F and H on member DFH.
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 139.A hand-operated hydraulic cylinder has been designed for use wherespace is severely limited. Determine the magnitude of the force exertedon the piston at D when two 90-
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 140.The tool shown is used to crimp terminals onto electric wires. Knowingthat a worker will apply forces of magnitude P = 135 N to the handles,determine the magnitude of t
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 141.The compound-lever pruning shears shown can be adjusted byplacing pin A at various ratchet positions on blade ACE.Knowing that 1.5-kN vertical forces are required to co
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 142.1-in. steel plate.4Determine the magnitude of the gripping forces produced when two 30-lbforces are applied as shown.A locking C-clamp is used to clamp two pieces of
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 143.Determine the force P which must be applied to the toggle BCD tomaintain equilibrium in the position shown.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Fe
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 144.In the locked position shown, the toggle clamp exerts at A avertical 270-lb force on the wooden block, and handle CF restsagainst the stop at G. Determine the force P r
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 145.The bone rongeur shown is used in surgical procedures to cutsmall bones. Determine the magnitude of the forces exerted on thebone at E when two 100-N forces are applied
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 146.A lopping pruner is used to cut a small branch at F. Handles ADand BE pivot about bolt B and blade AC pivots about fixed boltA while bolt C can slide freely in the slot
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 147.The telescoping arm ABC is used to raise a worker to theelevation of overhead electric and telephone wires. For theextension shown, the center of gravity of the 1400-lb
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 148.The position of the spindle DE of a lift truck is partiallycontrolled by two identical linkage-and-hydraulic-cylindersystems, only one of which is shown. A 3000-lb spoo
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 149.The telescoping arm ABC is used to provide an elevated platformfor construction workers. The workers and the platform togetherhave a mass of 240 kg and have a combined
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 150.A 500-kg concrete slab is supported by a chain and slingattached to the bucket of the front-end loader shown. The actionof the bucket is controlled by two identical mec
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 151.In the planetary gear system shown, the radius of the central gear A isa = 24 mm, the radius of the planetary gears is b, and the radius of the outergear E is ( a + 2b
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 152.Gears A and D are rigidly attached to horizontal shafts that are held byfrictionless bearings. Determine (a) the couple M 0 that must be applied toshaft DEF to maintain
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 153.Two shafts AC and CF, which lie in the vertical xy plane, are connected by auniversal joint at C. The bearings at B and D do not exert any axial force. Acouple of magni
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 154.Solve Prob. 6.153 assuming that the arm of the crosspiece attached to shaftCF is vertical.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer,
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 155.The large mechanical tongs shown are used to grab and lift a thick 1500-lbsteel slab HJ. Knowing that slipping does not occur between the tong gripsand the slab at H an
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 156.Using the method of joints, determine the force in each member of the trussshown. State whether each member is in tension or compression.Vector Mechanics for Engineers:
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 157.Determine the force in member FG and in each of the members located tothe right of member FG for the scissor roof truss shown. State whethereach member is in tension or
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 158.Determine the force in each member of the truss shown. State whether eachmember is in tension or compression.Vector Mechanics for Engineers: Statics and Dynamics, 8/e,
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 159.A pitched flat roof truss is loaded as shown. Determine the forcein members CE, DE, and DF.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer,
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 160.A stadium roof truss is loaded as shown. Determine the force in membersAB, AG, and FG.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. R
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 161.Determine the force in members AF and EJ of the truss shown whenP = Q = 2 kips. (Hint: Use section aa.)Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdin
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 162.For the frame and loading shown, determine the components of all forcesacting on member ABC.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 163.Knowing that the pulley has a radius of 1.25 in., determine thecomponents of the reactions at B and E.Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdina
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 164.Knowing that P = 15 N and Q = 65 N, determine the components ofthe forces exerted (a) on member BCDF at C and D, (b) on memberACEG at E.Vector Mechanics for Engineers:
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 165.For the system and loading shown, determine (a) the force P required forequilibrium, (b) the corresponding force in member BD, (c) thecorresponding reaction at C.Vecto
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 166.Determine the magnitude of the gripping forces exerted along line aa onthe nut when two 240-N forces are applied to the handles as shown.Assume that pins A and D slide
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Problem 167.The garden shears shown consist of two blades and two handles. The twohandles are connected by pin C and the two blades are connected by pinD. The left blade and the ri
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 1.\Joint FBDs:Joint B:FAB 800 lb FBC = = 15 8 17 soFAB = 1500 lb TFBC = 1700 lb CJoint C:FAC Cx 1700 lb = = 8 15 17 FAC = 800 lb TVector Mechanics for Engineers: Sta
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 2.Joint FBDs: Joint B:Fx = 0: Fy = 0: 1 4 FAB - FBC = 0 5 2 1 3 FAB + FBC - 4.2 kN = 0 5 2so7 FBC = 4.2 kN 5FBC = 3.00 kN C !FAB = 3.39 kN C !Joint C:FAB =12 2 kN 5
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 3.Joint FBDs:Joint B:FAB FBC 450 lb = = 12 13 5 so FAB = 1080 lb T FBC = 1170 lb CJoint C:Fx = 0:3 12 FAC - (1170 lb ) = 0 5 13 FAC = 1800 lb CVector Mechanics for Eng
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 4.Joint FBDs:Joint D:FCD FAD 500 lb = = 8.4 11.6 8 FAD = 725 lb T FCD = 525 lb CJoint C:Fx = 0:FBC - 525 lb = 0 FBC = 525 lb CThis is apparent by inspection, as is FAC
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 6.FBD Truss:M A = 0:(10.2 m ) C y + ( 2.4 m )(15 kN ) - ( 3.2 m )( 49.5 kN ) = 0C y = 12.0 kNJoint FBDs: Joint FBDs: Joint C:FBC F 12 kN = CD = 7.4 7.4 8 FBC = 18.50 kN
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 7.Joint FBDs:Joint E:FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T FDE = 4.00 kN C Fx = 0: - FAB + 4 (5 kN) = 0 5 FAB = 4.00 kN T Fy = 0: FBD - 6 kN - 3 (5 kN) = 0 5 FBD = 9.00 k
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 8.Joint FBDs:Joint B:By inspection: FAB = 12.00 kips C FBD = 0Joint A:FAC FAD 12 kips = = 5 13 12 FAC = 5.00 kips C FAD = 13.00 kips TJoint D:Fx = 0: FCD -12 (13 kips
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 9.First note that, by inspection of joint H:FCH = 0 ! andFDH = FGH FCG = 0 ! and FBC = FCD FBG = 0 ! and FFG = FGH FBF = 0 ! and FAB = FBCthen, by inspection of joint C:
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 11.FBD Truss:Fx = 0: M G = 0:Ax = 03a Ay - 2a (3 kN) - a (6 kN) = 0A y = 4 kNby inspection of joint C, by inspection of joint D, FAC = FCE and FBC = 0 FBD = FDF and FDE
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 12.FBD Truss:Fx = 0:Ax = 0By symmetry: A y = B y = 4.90 kNandFAB = FEG , FAC = FFG , FBC = FEF FBD = FDE , FCD = FDFJoint FBDs: Joint A:Fx = 0: Fy = 0:5 4 FAC - FAB
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 13.FBD Truss:Fx = 0: A x = 0M A = 0: (8 m) Gy - (4 m)(4.2 kN) - (2m)(2.8 kN) = 0G y = 2.80 kNFy = 0:Ay - 2.8 kN - 4.2 kN + 2.8 kN = 0A y = 4.2 kNJoint FBDs: Joint A:
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 14.FBD Truss:Fx = 0: A x = 0M A = 0:4a H y - 3a (1.5 kN) - 2a (2 kN)- a (2 kN) = 0H y = 3.625 kNFy = 0:Ay - 1 kN - 2 kN - 2 kN - 1.5 kN - 1 kN+ 3.625 kN = 0A y = 3.
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 15.FBD Truss:Fx = 0: A x = 0M A = 0: 8a ( J y - 1 kN) - 7a (1 kN) - 6a (2.8 kN) - 4a (4.5 kN) - 2a (4 kN) - a(1 kN) = 0J y = 6.7 kNFy = 0: Ay - 1 kN - 1 kN - 1.4 kN - 4.
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 16.\FBD Truss:Fx = 0:M A = 0:Ax = 0 8a( J y - 1 kN) - 7a(1 kN) - 6a(2.8 kN)- 4a(4.5 kN) - 2a(4 kN) - a(1 kN) = 0J y = 6.7 kNFy = 0:Ay - 1 kN - 1 kN - 1.4 kN - 4.5 kN
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 17.FBD Truss: By load symmetry, A y = H y = 1600 lbFx = 0:Ax = 0 = tan -16.72 ft = 16.2602 23.04 ft (1600 lb - 400 lb) cos - FAC sin = 0Joint FBDs: Joint A:Fy = 0:FAC
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 18.FBD Truss:By load symmetry, Fx = 0:Ax = 0A y = H y = 1600 lb = tan -16.72 ft = 16.2602 23.04 ftJoint FBDs: Joint H:Fy = 0: 1600 lb - 400 lb - FFH sin = 0 FFH = 428
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 19.FBD Truss:Fx = 0:M L = 0:Ax = 0(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN)+ (5.5 m)(6 kN) - (11 m) Ay = 0A y = 4.5 kNJoint FBDs: Joint A:Fy = 0:4.5 kN - 6 kN - 2.2 kN -
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 20.FBD Truss:Fx = 0:M L = 0:Ax = 0(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN)- (11 m) Ay = 0A y = 4.5 kNJoint FBDs: Joint L:Fy = 0:4.5 kN - 6 kN - 2.2 kN -
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 21. Joint FBDs: Joint F:FDF F 500 lb = EF = 916 916 480FDF = FEF = 954.17 lb or FDF = 954 lb T ! FEF = 954 lb C !Joint D:FBD FDE 954.17 lb = = 884 240 916 FBD = 920.84 lb
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 22. Joint FBDs: Joint L:FJL F 500 lb = KL = 1212 1212 480FJL = FKL = 1262.50 lb FJL = 1263 lb T !FKL = 1263 lb C ! Comparing joint G to joint L:FGH = 1263 lb T ! FGI = 12
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 24.FBD Truss:Fx = 0:M A = 0:Ax = 0(12 m ) (M y - 1 kN)- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0M y = 5.05 kNFy = 0: Ay - 2(1 kN) - 5(1.5 kN) + M y = 0Joi
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 25.FBD Truss:Fx = 0: 180 lb - Ax = 0A x = 180 lbM A = (12 ft ) G y - ( 6 ft )( 480 lb ) - ( 2 ft )(180 lb )- (8 ft)(120 lb) = 0, Fy = 0: Ay - 480 lb - 120 lb + 350 lb =
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 26. Joint FBDs: Joint A:Fy = 0:Fx = 0:9 FAC - 1.8 kips = 0, 4140 FAB - 8.20 kips = 0, 41FAC = 8.20 kips T !FAB = 8.00 kips C ! FBD = 8.00 kips C ! FBC = 0.600 kips C !
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 27. P6.14Starting with ABC, add, in order, joints E, D, F, G, H simple trussP6.15Starting withDEF, add, in order, G, C, B, A, I, H, J simple trussP6.23Starting withABC
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 28. P6.21Starting with ABC, add, in order, joints E, D, F, H, J, K, L, I, G, N, P, Q, R, O, M, S, T simple trussP6.25Starting with ABC, add, in order, joints D, E, F, G si
University of Florida - EGM - 2511
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 29.Fx = 0:Fx = 0FFG = 0 FGH = 0 FIJ = 0 FHI = 0 FEI = 0Then, by inspection of joint F, Then, by inspection of joint G, By inspection of joint J, Then, by inspection of jo