UCSD Phys 4A Intro Mechanics Winter 2016 Ch 5 Solutions

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Henry Young
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1 UCSD Phs 4 Into Mechanics Winte 016 Ch 5 Solutions 0. Since the uppe bloc has a highe coefficient of iction, that bloc will dag behind the lowe bloc. Thus thee will be tension in the cod, and the blocs will hae the same acceleation. Fom the ee-bod diagams fo each bloc, we wite ewton s second law fo both the and diections fo each bloc, and then combine those equations to find the acceleation and tension. (a) loc : F F m gcos 0 F m gcos F m gsin F F m a T ma mgsin F F mgsin mgcos F T T loc : F F m gcos 0 F m gcos F m gsin F F m a T ma mgsin F F mgsin mgcos F T T dd the final equations togethe om both analses and sole fo the acceleation. ma mgsin mgcos F ; ma mgsin mgcos F T T mama mgsin mgcos Fmgsin mgcos F T T m a g 9.80 m s sin cosm sin cos m m m s 3.1m s 5.0 gsin 30.0cos 35.0 gsin 30.30cos g (b) Sole one of the equations fo the tension foce. ma mgsin mgcos F T F m gsin gcos a T 5.0 g 9.80 m s sin 3 0.0cos m s.1 F mg F T F F F T mg F 18. (a) Conside the ee-bod diagam fo the cate on the suface. Thee is no motion in the diection and thus no acceleation in the diection. Wite ewton s second law fo both diections. F F mgcos 0 F mgcos F mgsin F ma ma mg sin F mg sin mg cos a gsin cos 9.80 m s sin cos m s.5m s (b) ow use Eq. -1c, with an initial elocit of 0, to find the final elocit. a a.454m s 8.15m 6.3m s F F mg

2 3. (a) Fo m to not moe, the tension must be equal to mg, and so mg F. Fo m to not T moe, the tension must be equal to the foce of static iction, and so F F. ote that the S T nomal foce on m is equal to its weight. Use these elationships to sole fo m. m.0 g mg F F mg m 5.0 g m 5.0 g T s s 0.40 s (b) Fo m to moe with constant elocit, the tension must be equal to mg. Fo m to moe with constant elocit, the tension must be equal to the foce of inetic iction. ote that the nomal foce on m is equal to its weight. Use these elationships to sole fo m. mg F mg m 4. We define f to be the action of the cod that is handing down, between m and the pulle. m.0 g g Thus the mass of that piece of cod is fm. C F F T We assume that the sstem is moing to the ight as well. We tae the tension in the cod mg to be F at the pulle. We teat the hanging T 1 f mcg mass and hanging action of the cod as one mass, and the sliding mass and hoizontal pat of the cod as anothe mass. See the ee-bod diagams. We wite ewton s second law fo each object. F F m 1 f m g 0 1 C F F F F F m f m a T T C F m fm g F m fm a C T C Combine the elationships to sole fo the acceleation. In paticula, add the two equations fo the -diection, and then substitute the nomal foce. a m fm m 1 f m m m m C C C g F F T m g fm g 5. (a) Conside the ee-bod diagam fo the bloc on the suface. Thee is no motion in the diection and thus no acceleation in the diection. F Wite ewton s second law fo both diections, and find the acceleation. F F mgcos 0 F mgcos F F mgsin F ma mg ma mg sin F mg sin mg cos a g sin cos C

3 ow use Eq. -1c, with an initial elocit of, a final elocit of 0, and a displacement of 0 d to find the coefficient of inetic iction. a 0 g sin cos d tan gd cos (b) ow conside the ee-bod diagam fo the bloc at the top of its motion. We use a simila foce analsis, but now the magnitude of the iction foce is gien b F F, and the acceleation is 0. s F F mgcos 0 F mgcos F mgsin F ma 0 F mgsin F F mgsin mgcos tan s s s 31. Daw a ee-bod diagam fo each bloc. F F mg F T F mg loc (top) F F T F F F F mg loc (bottom) F

4 is the foce of iction between the two blocs, F F two blocs, F is the nomal foce of contact between the is the foce of iction between the bottom bloc and the floo, and F nomal foce of contact between the bottom bloc and the floo. eithe bloc is acceleating eticall, and so the net etical foce on each bloc is zeo. top: F m g 0 F m g bottom: F F m g 0 F F m g m m g is the Tae the positie hoizontal diection to be the diection of motion of each bloc. Thus fo the bottom bloc, positie is to the ight, and fo the top bloc, positie is to the left. Then, since the blocs ae constained to moe togethe b the connecting sting, both blocs will hae the same acceleation. Wite ewton s second law fo the hoizontal diection fo each bloc. top: F F m a bottom: F F F F m a T T (a) If the two blocs ae just to moe, then the foce of static iction will be at its maimum, and so the ictions foces ae as follows. F F m g ; F F m m g s s s s Substitute into ewton s second law fo the hoizontal diection with a 0 and sole fo F. top: F m g 0 F m g T s T s bottom: F F m g m m g 0 T s s F F m g m m g m g m g m m g T s s s s s m m g g 9.80 m s s (b) Multipl the foce b 1.1 so that F gain use ewton s second law fo the hoizontal diection, but with a 0 and using the coefficient of inetic iction. top: F m g m a bottom: sum: T F F m g m m g m a T F m g m g m m g m m a F mg mg m m g F 3m m g a m m m m g 9.80 m s m s 4.5m s 8.0 g 37. We assume the wate is otating in a etical cicle of adius. When the bucet is at the top of its motion, thee would be two foces on the wate (consideing the wate as a single mass). The weight of the wate would be diected down, and the nomal foce of the bottom of the bucet pushing on the wate would also be down. See the ee-bod diagam. If the wate is moing in a cicle, then the net downwad foce would be a centipetal foce. F F mg ma m F m g The limiting condition of the wate falling out of the bucet means that the wate loses contact with the bucet, and so the nomal foce becomes 0. F m g m g 0 g citical citical Fom this, we see that es, it is possible to whil the bucet of wate fast enough. The minimum speed is g. F mg

5 43. The obit adius will be the sum of the Eath s adius plus the 400 m obit height. The obital peiod is about 90 minutes. Find the centipetal acceleation om these data sec 6380 m 400 m 6780 m m T 90 min 5400 sec 1 min a m 1 g R 9.18 m s g's T 5400 sec 9.80 m s otice how close this is to g, because the shuttle is not e fa aboe the suface of the Eath, elatie to the adius of the Eath. 47. (a) See the ee-bod diagam fo the pilot in the jet at the bottom of the loop. We hae a g. R 6 1m s 100 m h 3.6 m h m 6.0g m s g (b) The net foce must be centipetal, to mae the pilot go in a cicle. Wite ewton s second law fo the etical diection, with up as positie. The nomal foce is the appaent weight. F F mg m R The centipetal acceleation is to be 6.0 g. F mg m mg g 9.80m s (c) See the ee-bod diagam fo the pilot at the top of the loop. otice that the nomal foce is down, because the pilot is upside down. Wite ewton s second law in the etical diection, with down as positie. F F mg m 6 mg F 5mg 3800 R 60. (a) The object has a unifoml inceasing speed, which means the tangential acceleation is constant, and so constant acceleation elationships can be used fo the tangential motion. The object is moing in a cicle of adius.0 metes. tan 0 1 tan 4 tan.0 m t m s tan tan 0 t tan t.0s (b) The initial location of the object is at.0 mj ˆ, and the final location is.0 m ˆi..0 mˆi.0 mˆj m s ag ˆi ˆj t.0s (c) The elocit at the end of the.0 seconds is pointing in the j ˆ diection. 0 msˆj a ag m s ˆj t.0s F F mg mg

6 80. Since mass m is dangling, the tension in the cod must be equal to the weight of mass m, and so F mg That same tension is in the othe end of the cod, maintaining the cicula motion of mass T. M, and so F F Ma M Equate the epessions fo tension and sole fo the elocit.. T R R M mg mgr M 94. n object at the Eath s equato is otating in a cicle with a adius equal to the adius of the Eath, and a peiod equal to one da. Use that data to find the centipetal acceleation and then compae it to g m T 4 ar 86, 400s 3 a R T g 9.80 m s 1000 So, fo eample, if we wee to calculate the nomal foce on an object at the Eath s equato, we could not sa F F mg 0. Instead, we would hae the following. F F mg m F mg m If we then assumed that F mg mg m, then we see that the effectie alue of g is eff g g g 0.003g g. eff

Physics 111 Lecture 5 Circular Motion

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