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ch5 - P5.11(a See Figure to the right T mz—r(b...

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Unformatted text preview: P5.11 (a) See Figure to the right. T mz—r. (b) 68.0—T—m2g=mza (Block#2} Lymlgzmla (Block #1) 1’“ i i T": s MI —I» T1— 171: —'f . ‘— Adding, 1 = #1:": I 2 = {-59:52 I sac—m +msg=Im1+sza “=11” “28:1?“ :1 : fiflug :— FIG. P5.11 1 Z P513 (Case I, impending upward motion) fl. Setting Pegs 50° |:| ‘— i=5’A 2P3 =0: Pc0550.0°—n:0 ; I $1;me fa m = flan: f8,“ = #31390550-00 _5 mg‘ = U.250{0.643)P = 0.161P 1’ ' P‘sIIn 50° Setting 1 f5, max = JIL- 1'! I ii 2 Fy : 0: P sin 50.00—0.161P— 3.00900) : 0 pm 50» I— Eb Pm = 48.6 N :/ (Case 2, impending downward motion) P l _ _ __ Mg As in Case 1, Pam 50° fa. m = “-161? FIG. P5.13 Setting Z I; = 0: P sin 50.U°+U.161P— 3.00[9.00) : 0 Pmin : 31.7 N P5-19 TcosBflU": mg: [00.0 kg)[9.00 mfg!) I | | (a) T = 787 N: T = (68.6 N)i + (734 N)j F: :* 5-00” | (b) Tsin5.0°: mac: .0: : 0.85? mills2 toward the center of _ _ _ 51 the circle. ME The length of the wire is urmecessary information. We could, on the other hand, use :it to find the radius of the circle, the speed of the 13013, and the period of the motion. FIG. P5.19 P539 P550 (a) The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be :2 and the friction force, f8. Resolving vertically: n : F: + Paine. Horizontally: Pcosn? : f8. But, f5 31 pan i.e., PEOSQEpJFx. +Psint9) or P{cosQ—,us sinug) filial-"g. Divide by cosé': Pfl —p.s tan9)gpsl:z secE'. Then HP _. = graFrsecs' . m 1 — its tan 6' 0.400 100 N secsJ (b) P:# 1 — 0.400 tans? 9(deg) | 0.00 15.0 30.0 45.0 60.0 P(N) | 40.0 46.4 60.1 94.3 2.60 It the angle were 68.2D or more, the expression for P would go to infinity and motion would become impossible. (it) Since the object of mass m2 isinequilibrium, 21:3: =T—ng=0 or T=-- (b) The tension in the string provides the required centripetal acceleration of the puck. Thus, FC=T= mag . m '02. (c) From F: = 1 R we have 1): RF” 2 flJSR m1 m1 P555 (21) HIT f—ngU ZJER f—F‘s” 3-? T: 431313;!!! 3 (b) T: 2545 FIG. P5.55 ...
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