hw06 - 6—2 Draw a flee-bed}F diagram of the diving beard...

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Unformatted text preview: 6—2 Draw a flee-bed}F diagram of the diving beard Shem-‘11 in Fig. P6-2. The surface at B is smooth. Neglect the weight of the diving beard. SOLUTION 6—5 Draw a free-body diagram of the bracket shown in Fig. Po-fi. The eoirraet surfaces between The cylinders and bracket are smooth. SOLUTION 6—6 Forces P are applied To the handles of the pipe pliers shown in Fig. PMS-6. Draw flee-body diagrams of each handle. SOLUTION 6—8 Forces P are applied to the handles of the bolt cutter shown in Fig. PG-S. Draw a free-body diagram of (a) The lower handle. (13) The lower cutter jaw. SOLUTION 6—11 Draw a free-body diagram of the bar shown in Fig. P6-11. The supme atA is ajoumal bearing and the supports at B and C are ball bearings. SOLUTION 6—12 Draw a flee-body diagram of the shaft shown in Fig. PIS-12. The hearing at A is a thrust bearing. and the hearing at D is a ball bearing. Neglect the weights of the shaft and the lEYEl‘S. SOLUTION 6715 A 30-11) force P is applied to the brake pedal of an automobile as shoer in Fig. P6-15. Determine the force Q applied to the brake cylinder and the reaction at support A. SOLUTION From a free-body diagram of the brake pedal. the equilibrium equations are solved to get the forces OEMA =0: 5.59—(30cos30°)(11)—(305i11300)(4)= 0 Q=62.8711b aEfi20: Ax—Q+30COS3OO:0 AI :36.8901b T 2F). : 0: A’saosnmoo : 0 A}_:151b A=39.81b 4 22.130 Q=6291b e 154 6716 A beam is loaded and supported as shovm in Fig. Fifi-16. Determine the reactions at supports A and B. SOLUTION From a free—body diagram of the beam. the equilibriinn equations are solved to get the forces +21%:0: Ax:0 own. :0: 4.53{(300)(1., —[l(4oo)(1.s) 2 [(300)(1.5)]—E A}. : 441.667 N B : 308 N T 6719 Three pipes are supported in a pipe rack as shown in Fig. P6-19. Each pipe weighs 100 1b. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the pipes and the rack. the equilibrium equations are solved to get the forces OXMA =0: lag—(looms300)(9)—(100c05300)(15) 7(10000530°)(21)73(100si1130°)(4) O 3:249.840 lb aZF‘ =0: AI+Bsin30°=0 AX=—124.920 lb T2550: AA—300+Bcos30°=0 4:83.632113 A=150.3 lb It 33 800 B : 250 lb C 60° 6-22 A pipe strut BC is loaded and supported as shown in Fig. PG-ll. Determine (a) The reactions at supports A and C'. (b) The shearing stress on a cross section of the 10-1mii—dianieter pin at C. which is in double shear. (c) The elongation of cable .43 if it is made of aluminum alloy (E = 73 GPa) and has a diameter of 6 mm. SOLUTION (a) From a free-body diagram of the pipe shut. the equilibrium equations are solved to get the forces 021-116 : 0: 1000133 7800(750) : 0 affizfl (TI—114320 CY : 600 N 157 6725 A lever is loaded and supported as shown in Fig. P6-25. Determine (a) The reactions at A and C'. (b) The normal stress in the 1frills-diameter rod CD. (c) The shearing stress in the lit-inrdiaineter pin atA. which is in double shear. (d) The change in length ofi‘od CD. which is made ofa material with a modulus of elasticity of 30(106) psi. S OLIITION 9 :tah’l (6/12): 265650 LCD : J62 +122 : 13.416 in. (a) From a free-body diagram of the lever. the equilibrium equations are solved to get the forces OEMA =0: (rm cos6)(6)—40(4)—60(s)+80(12) =0 TC :759628 lb a21220: AI+125+TCDc05320 =—71667 lb T2550: A’.740760+807TCDsi116:0 A) = —6.666 lb A:72.01b :7 531° ................................ ..Aus. To = 59.6 lb (C) (b) Dividing the normal force in the rod CD by its cross-sectional area gives the normal stress 0a, = Jug—D = —304 psi = 304 psi (C) tr(0.5)‘,:’4 (c) Dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress VJ 72.0 1. : . A 214s : 2[;t(0.5)3f4] (:1) Finally. the change in length of the rod CD is given by 159 PL E4 (i30><10"’")[;r(0.5)3j4] :7l.358><104i11.:1.358><104i11.(shrink).. 6-26 The wood plane shown in Fig. P646 moves with a constant velocity when subjected to the forces shown. Determine (a) The shearing force of the wood on the plane. (b) The noimal force. and its location. of the wood 01] the plane. SOLUTION From a free-body diagram of the plane. the equilibrium equations are solved to get the forces aZR=0: F740c05700770003160=0 T 21F“: 0: N—40sin'i'0"—'."Osi1116O = 0 OZMA=0: (70c08160)(75)7(703i11160)(220) +(40cos70°)(60)+Nd :0 (1727285 1.11111 31.5 1111.11 (from front ofplane) 6732 Pulleys A and B ofthe chain hoist shown in Fig. P6—32 are connected and rotate as a unit. The chain is continuous. and each of the pulleys contains slots that prevent the chain from slipping. Determine the force F required to hold a 450-kg block W in equilibrium if the radii of pulleys A and B are 90 111111 and 100 111111. respectively. SOLUTION A From a fi'eefbod-y diagram of the lower pulley. T T vertical equilibrium equation gives the tensron t25,:0: 2r7450(9.31):0 T: 2207.25 N Then. from a free-body diagram of the lower pulley. nioinent equilibrium equation gives the force F Hi 01 0 C35.) 162 ND MECHANICS OF MATERIAL TURG S AND MORRIS o = 0 : 100T — 90r—100F = 0 6736 The le\-'er shown in Fig. P6-36 is formed in a quarter circular arc of radius 450 mm. Determine the angle 9 if neither of the reactions atA or B can exceed 200 N. SOLUTION From a free-body diagram of the lever. the equilibrium equations are HER=02 AI—Bsit16=0 T2550: A,—125+Bcosa=0 OZMD = 0: —A) (450): 0 The fourth needed equation is either A : . + A? : 200 N or B : 200 N . In either case. the moment equation gives A}. : 0 N Now. guessing that B = 200 N gives 9 = 51 .320 A], : 15.61 N *A‘ :156.1N<200N Therefore. the guess was correct. 6-37 A man is slowly raising a 20-ft-long homogeneous pole weighing 1501b as sliovm in Fig. PG-ST’. The lower end of the pole is kept in place by smooth surfaces. Determine the force exerted by the man to hold the pole in the position shown. SOLUTION From a free-body diagram] of the pole. the equations of equilibrium give the tension OEM’A :0: (Tsi1130°)(20)7150(10cos600) 0 ...
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hw06 - 6—2 Draw a flee-bed}F diagram of the diving beard...

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