Ch13_Sec1-4_Ex-5-6-18-22-33-34

Ch13_Sec1-4_Ex-5-6-18-22-33-34 - 13-5 A block having a mass...

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Unformatted text preview: 13-5. A block having a mass of 2 kg is placed on a spring scale located in an elevator that is moving downward. If the scale re ding, which measures the force in the spring, is 20 N, dltermine the acceleration of the elevator. Neglect the mass of the scale. 20 I0 Equation of Motion : +T25l=ma,; 20-—l9.62=2a a=0.l90m/szT Ans 2148!) =/ 5761/ *13-6. As a point of historical interest, the acceleration of gravity g used to be determined using an “Atwood Machine,” which is simply the arrangement of the blocks and smooth pulley shown. ln an experiment of this type, consider block A to have a mass of 2 kg and block 3 a mass of 2.6 kg. If it is observed that B descends 3 "l in 2.2 5 starting from rest, determine the value of g. Di r-cuss what effects must be considered in order to increas- the accuracy of this experiment. 1 Kinematics : Applying equation s = so + on t+ 54,12, we have (+t) ‘3=o+o+—;-a,(2.2‘) a,,=1.240m/s2 Equation of Motion : Here. «A = a, = 1.240 m/sz. Site: .56 cord passes over a smooth pulley, the tension developed in the cord is the 5:5" zhroughout the entire cord. Applying Eq.13—7 to FBD(a) , we have +TZI';=ma,: r—2g=2(1.2403 [ll Applying [5413—7 to FBD(a) . we have +1Tzr. =ma,; T—2.6g= 2.5(— vJ)‘ [2] Solving Eqs.[l] a‘tid t? . yields ‘ g = 9.50 m/s2 Ans T= 21.49 N ‘ The following effects list be considered in order :- use the accuracy of the experiment: ‘ Reduce the dis or fall and thus the final spea- % the blocks so that air resistance can It: .zeed. ' kednee the Ms of pulley and cord or account for these mass in the alculation (as in ct. m 17). ~eduoe friction“ the , allcy's axle. 128 a. l ##240/0/5” (a) ”i (b) de‘r'i'wm/S ll 13-18. The man pushes on the 60-lb crate with a force F. Thelforjrce is always directed down at 30° from the horizo taléas shown, and its magnitude is increased until the cr te begins to slide. Determine the crate‘s initial acceler tion if the static coefficient of friction is m. = 0.6 '5 g and m kinetic coefficient of friction is ”=03. Force to produce motion : .3}; =b; Fcos30°-0.6N=0 +T2fi=l0; N-60-Fsin30"=0 N=9l.801b F=63,601b smoc~=91.al)ib. Jam". ‘ so = ,; 53. so°-o. 91. — ma 6060: 3( 80)=(32'2)a , l 1 i l l a = 14.8 ml Am 13-22.; The 1le block 'A is traveling to the right at 21,1 = Zlft/s at the instant shown. lf the coefficient of kinetic frictiorLis pk == 0.2 between the surface and A, determine the veil city of A when it has moved 4 ft. Block B has a weight lot 20 ll). BlockAj: 9-4 Io/L l . l ; 10 (—zE‘Fmbx. -T+2=(m)a4 (I) T. ; ‘ egg/apes Weights: N=/0/£ ‘ i r 7' +~L2F,¢=nta,: 2047:6293)“ (2) g Kinematicss ‘ J“! = 34+2.r.i=l? aA='2d‘l (3} 20/6 Solving slur-(3): 54 a4é-l7‘l73fi/sz a.=l3.5237n/s2 T=7.331b I V=Vé+ cts-So) v1 = (2)’+2(l7.173)(4—0) v=ll.9fL/s Am _——"?—"’— —>a &. V .—-—__w l_.. .—_~._.—__.. 13in. The IO-lb block A and the 20-lb block B are initially at rest. If a force of P = 20 lb is applied to B as sh wh, determine the acceleration of each block. The coiffibient of kinetic friction between any two surfaces is pk‘= 0.2, and the coefficient of static friction is u, = 0.3. NafimA slips on B as B slips. Blodt A: 2 Q- 10 —9 EF, l= 1pm,; 2=(§TZ)“‘ 04 = 6.441’t/s2 Bl B : . 20 —) ‘5} =ma,; 20-6—2=(fi)a, T a. = 19.323131 NotetlmtforA. F", = 0.3M = 3 lbmd ifA was none slip onB then. fotB. c ‘ 2° — all}; =ma,; 20-6—3=(§Z§)a, a, = 17.71 fils’ Thus. ForA. a,‘ = 6.44 ft/s2 -.+Zlv‘,=ma,; 3=(‘° )a. =9.cs:sn/s’<17.7m/s2 OK AllipaonB. m a, = 19.3 ft/sz Also. 1“ the ground. F... = 01300) = 9 lb < 20 lb. so indeed motion occurs. 143 \ i i i i \ 13-34. If 5 Horizontal force of P = 10 lb is applied to block/1, etermine the acceleration of block B. Neglect friction. int: Show that a” = a,‘ tan 15°. Equation of Motion : Applying Eq. 13~7 to FBD(I) , we have ¢ 8 = ; lO—N ' ° —— —) Elf, ma, ”sin 15 = (32.2)aA l3-r7 to FBD(b), we have 15 Fi-TEF;==ma,; N.COS 150—15=(-3—23)fl. K incmatic: : me the geometry of Fig. (c). .r, = a" tan 15° Taking time derivative twice to the above expression yids a, =aAlan 15° (Q.E.D.) [1]. 1212mm yidds a, = 5.68 ids’ «A = 2122 {115’ N, = 13.27 lb ...
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