382_Physics ProblemsTechnical Physics

382_Physics - 384 P13.3 Universal Gravitation(a At the midpoint between the two objects the forces exerted by the 200-kg and 500-kg objects are

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384 Universal Gravitation P13.3 (a) At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objects are oppositely directed, and from F Gm m r g = 12 2 we have F G = 50 0 500 200 0200 250 10 2 5 . . . kg kg kg m N bg b g af toward the 500-kg object. (b) At a point between the two objects at a distance d from the 500-kg objects, the net force on the 50.0-kg object will be zero when G d G d 50 0 200 0400 50 0 500 22 . . . kg kg m kg kg b g a f b g = or d = 0245 . m P13.4 mm 500 += k g 21 =− k g FG r =⇒ × = ×⋅ −= × = −− 2 81 1 11 2 2 8 11 1 00 10 6 67 10 100 10 00400 667 10 600 .. . . . . . N N m k g kg m kg m Nm kg kg 2 2 ej e j Thus, 1 2 1 0 −+ = kg kg or 300 200 0 = kg kg giving == kg, so kg . The answer m 1 = . kg and m 2 = . kg is physically equivalent. P13.5 The force exerted on the 4.00-kg mass by the 2.00-kg mass is directed upward and given by Fj j j 24 42 24
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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 .

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