04prob_11

# 04prob_11 - F T free-body for m 1 m 1 g F N1 x y F T...

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Unformatted text preview: F T free-body for m 1 m 1 g F N1 x y F T free-body for m 2 m 2 g F N2 x y oe 1 oe 2 oe 2 oe 1 m 1 m 2 a a T 1 w forces on w T 2 forces on lower pulley T 2 T 1 T 2 forces on upper pulley T 2 T 3 F N f=mg mg T Mg N' f' N=mg F mg f m M T ae r oe oe F T F N mg F T free-body for m 1 m 1 g F N1 x y F T free-body for m 2 m 2 g F N2 x y oe 1 oe 2 oe 2 oe 1 m 1 m 2 a a T 1 w forces on w T 2 forces on lower pulley T 2 T 1 forces upper T 2 T F N f=mg mg T Mg N' N=mg F mg m M ae r oe oe F N mg Physics 9A-C Assignment #4 Cole [5.13]a) sin = r r + = sin 1 16 cm 46 cm =20.35, F x =F T sinoe -F N =0, F y =F T cosoe- mg =0 F T = mg cos = 471 N b)Subbinginto F x : F N =F T sinoe=164NByNewtons3rdlaw-tothe right. [5.55]a)The&quot;artificialgravity&quot;isnotgravityatall. Theeffectiscausedbythenormalforcerequiredto causethecentripetalacceleration. F y =N=ma c . To simulategravity,wewant N=mg mg=mr 2 .Wewantthefrequency: =2fg=r4 2 f Solving: f = 1 2 g r = 1 2 9.8 m/s 2 800 m (60s/min)=1.5rev/ min.b)Theloweraccelerationdecreasesthenormalforce,andhence,thefrequency ofrotation: f = 1 2 3.7 m/s 2 800 m (60s/min)=0.92rev/min. [5.62]Summingforcesonthefirstsegmentofropegives F=T-F=0T=F. Theropeisideal,negligible weight,sothetensionintheropeisthesameeverywhere. Second,summingtheforcesontheweight F=T 1 -w=0 T 1 =w. Third,summingtheforcesonthelowerpulley, F=2T 2 -T 1 =0 T 2 = 1 2 T 1 = 1 2 w. Fourth,summingtheforcesontheupper pulley, F=T 3 -2T 2 =0T 3 =2 1 2 w=w....
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## This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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04prob_11 - F T free-body for m 1 m 1 g F N1 x y F T...

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