ch13-p014

# ch13-p014 - 2 d 2 = Gm A m D r 2 . This yields r = 1.29 d ....

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14. Using Eq. 13-1, we find F AB = 2 Gm A 2 d 2 j ^ and F AC = 4 Gm A 2 3 d 2 i ^ . Since the vector sum of all three forces must be zero, we find the third force (using magnitude-angle notation) is F AD = Gm A 2 d 2 (2.404 –56.3º) . This tells us immediately the direction of the vector r (pointing from the origin to particle D ), but to find its magnitude we must solve (with m D = 4 m A ) the following equation: 2.404 © ¨ § ¹ ¸ · Gm A
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Unformatted text preview: 2 d 2 = Gm A m D r 2 . This yields r = 1.29 d . In magnitude-angle notation, then, r = (1.29 56.3) , with SI units understood. The exact answer without regard to significant figure considerations is r = ( 2 6 13 13 , 3 6 13 13 ) . (a) In ( x, y ) notation, the x coordinate is x =0.716 d . (b) Similarly, the y coordinate is y = 1.07 d ....
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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