ch13-p012 - 12 All the forces are being evaluated at the...

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30.0º = 150º (measured ccw from the + x axis). The component along, say, the x axis of one of the force-vectors F is simply Fx/r in this situation (where F is the magnitude of F ). Since the force itself (see Eq. 13-1) is inversely proportional to r 2 then the aforementioned x component would have the form GmMx/r 3 ; similarly for the other components. With m A = 0.0060 kg, m B = 0.0120 kg, and m C = 0.0080 kg, we therefore have F net x = Gm A m B x B r B 3 + Gm A m C x C r C 3 = (2.77 × 10 14 N)cos( 163.8º) and F net y = Gm A m B y B r B 3 + Gm A m C y C r C 3 = (2.77 × 10 14 N)sin( 163.8º) where r B = d AB = 0.50 m, and ( x B , y B ) = ( r B cos(150º), r B sin(150º)) (with SI units understood). A fairly quick way to solve for r C is to consider the vector difference
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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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