384 Universal GravitationP13.3(a)At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objectsare oppositely directed,and from FGm mrg=122we have FG∑=−=×−50 05002000200250 1025...kgkgkgmNbgbgaftoward the 500-kg object.(b)At a point between the two objects at a distance dfrom the 500-kg objects, the net force onthe 50.0-kg object will be zero whenGdGd50 0200040050 050022...kgkgmkgkgbgafbg−=ord=0245. mP13.4mm500+=kg21=−kgFGr=⇒×=×⋅−−=×=−−−−281111228111 00 106 67 10100 1000400667 10600.......N NmkgkgmkgmNm kgkg22ejejThus, 1210−+=kgkgor3002000=kgkggiving ==kg, so kg . The answer m1=.kg and m2=.kg is physicallyequivalent.P13.5The force exerted on the 4.00-kg mass by the 2.00-kg mass isdirected upward and given byFjjj244224
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .