ch13-p015 - 2 , which yields x = 1.88 d . (b) Similarly, y...

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Gm A m B z B r B 3 = Gm A (2 m A )(2 d ) ((2 d ) 2 + d 2 + (2 d ) 2 ) 3 = 4 Gm A 2 27 d 2 . In this way, each component can be written as some multiple of Gm A 2 /d 2 . For the z component of the force exerted on particle A by particle C , that multiple is –9 14 /196. For the x components of the forces exerted on particle A by particles B and C , those multiples are 4/27 and –3 14 /196, respectively. And for the y components of the forces exerted on particle A by particles B and C , those multiples are 2/27 and 3 14 /98, respectively. To find the distance r to particle D one method is to solve (using the fact that the vector add to zero) © ¨ § ¹ ¸ · Gm A m D r 2 2 = [(4/27 –3 14 /196) 2 + (2/27 +3 14 /98) 2 + (4/27 –9 14 /196) 2 ] © ¨ § ¹ ¸ · Gm A 2 d 2 2 (where m D = 4 m A ) for r . This gives r = 4.357 d . The individual values of x , y and z (locating the particle D ) can then be found by considering each component of the Gm A m D / r 2 force separately. (a) The x component of r G would be Gm A m D x/r 3 = –(4/27 –3 14 /196) Gm A 2 /d
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Unformatted text preview: 2 , which yields x = 1.88 d . (b) Similarly, y = 3.90 d , (c) and z = 0.489 d . In this way we are able to deduce that ( x, y, z ) = (1.88 d , 3.90 d , 0.49 d ). 15. All the forces are being evaluated at the origin (since particle A is there), and all forces are along the location-vectors r which point to particles B, C and D. In three dimensions, the Pythagorean theorem becomes r = x 2 + y 2 + z 2 . 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. For example, the z component of the force exerted on particle A by particle B is...
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