386 Universal GravitationP13.8Let θrepresent the angle each cable makes with the vertical, Lthecable length, xthe distance each ball scrunches in, and d=1 m theoriginal distance between them. Then rd x=−2 is the separation ofthe balls. We haveFy∑=0:Tmgcos−=0Fx∑=TGmmrsin20FIG. P13.8Thentan=Gmmrmg2xLxGmgdx2222−=−afxdxGmg−22af.The factor Gmgis numerically small. There are two possibilities: either xis small or else dx−2 issmall.Possibility one: We can ignore xin comparison to dand L, obtainingx1667 101009845211mNm kgkg±msm2afejbg=×⋅−..x=×−306 108. m.The separation distance is r×=−−123061010006138m mmnm....Possibility two: If −2 is small, x≈05. m and the equation becomes450 5211....m±Nkgmafafafr=−
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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 .