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hw4s - P53 P5.10 —(1.50)z 2F=fi 2 mg=miz Along x 0— f...

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Unformatted text preview: P53 P5.10 — (1.50)z 2F=fi+2+mg=miz Along x: 0— f + mgsin30.0°= mu f = m( g sin 30.0°—a) Along y: n + 0 - mg c0530.0°= 0 n = mgcos 30.0° f m(gsin30.0°—a) o a = — = __—————, = tan30.0 ————-—= 0.368 (b) pk 11 mg cos 300° [1" gcos 300° - (c) f = m(gsin30.0°—a), f = 3.00(9.80 sin 30.0°—1.78) = 9.37 N (d) v} =v,.2+2a(xf—-x,.) where xf —-x,- =2.00 m a? = 0 + 2(1.78)(2.00) = 7.11 mz/sz v, =1/7.11 mz/s2 = 2.67 m/s T —— fk = 5.00:1 (for 5.00 kg mass) 9.00 g — T = 9.00u (for 9.00 kg mass) Adding these two equations gives: 9.00(9.80) — 0.200(5.00)(9.80) = 140a a = 5.60 m/s2 T = 5.00(5.60) + 0.200(5.00)(9.80) = FIG. P5.9 FIG. P5.10 P5.15 P5.17 P538 m = 3.00 kg , r = 0.800 m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so Tmax = Mg = 25.0(9.80) = 245 N. When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, mvz _ (3.00)?)2 so T: r 0.800 2 _ fl _ (0.800)T < (0.800)Tmax z 0.800(245) = 65 3 m2 52 The“ U " m ' 3.00 “ 3.00 3.00 ' / and 0305 65.3 FIG. P5.15 or 051238.08 m/s . n=mg since all =0 The force causing the centripetal acceleration is the frictional force f. "102 From Newton’s second law f = maC = But the friction condition is f S #311 FIG. P517 . "1'02 < ~ 1.9:; -—-—r -flsmg v s ,[ysrg = J0.600(35.0 m)(9.80 m/sz) For the system to start to move when released, the force tending to move m2 down the incline, ngsin 19, must exceed the maximum friction force which can retard the motion: fmax = f1,max + f2,max : #5,1n1 +fls,2n2 fmax = #sflmlg +fls,2ngcosg From Table 5.1, ,usll = 0.610 (aluminum on steel) and flslz = 0.530 (copper on steel). With FIG. P538 m1 = 2.00 kg, m2 = 6.00 kg, t9=30.0°, the maximum friction force is found to be fmax = 38.9 N. This exceeds the force tending to cause the system to move, ng sin 6: 6.00 kg(9.80 m/sz)sin 30°: 29.4 N. Hence, the system will not start to move when released . The friction forces increase in magnitude until the total friction force retarding the motion, f = f1 + f2 , equals the force tending to set the system in motion. That is, until f=ngsin0=29.4N . P5.41 P5.43 2131 =m1a: —m1gsi1135.0°—fk,1 +T=m1a -—(3.50)(9.80) sin 35.0°—;15 (3.50)(9.80) cos 35.0°+T = 3.50(1.50) (1) >21:2 = mza: +m2gsin35.0°—fklz —T = mza +(8.00)(9.80) sin 35.0°—,us (8.00)(9.80) cos 35.0°—T = 8.00(1.50) (2) Solving equations (1) and (2) simultaneously gives <a> 0» FIG. P5.41 (a) First, draw a free-body diagram, (top figure) of the top block. Since (1,, = 0, n1 = 19.6 N. And fk = ,uknl =0300(19.6 N) = 5.88 N. 211-“" =maT 10.0 N — 5.88 N = (2.00 kg)aT or a7 = 2.06 m/s2 (for top block). Now draw a free- body diagram (middle figure) of the bottom block and observe that 213x = Mal, gives f = 5.88 N = (8.00 kg)aB or a3 = 0.735 m/s2 (for the bottom block). In time t, the distance each block moves (starting from rest) is dT =%¢1Tt2 and d3 =§a3t2 . For the top block to reach the right edge of the bottom block, (see bottom figure) it is necessary that dT = (1,; +L or I Top Block 4— dB —>l L 2.00 ; Bottom Block | 1 1 | E(2.06 m/sz)t2 =—2-(0.735 m/sz)t2 +3.00 m I l which gives: t = . | From above, dB = %(0.735 m/s2 )(213 s)2 =. NWAW/AS Initial position of left ‘ (b) ' edges of both blocks FIG. P5.43 ...
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