ch21-p033

# ch21-p033 - F x = 3 F . Now, bead 3 exerts a leftward force...

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33. (a) We note that tan(30 ° ) = 1/ 3 . In the initial (highly symmetrical) configuration, the net force on the central bead is in the – y direction and has magnitude 3 F where F is the Coulomb’s law force of one bead on another at distance d = 10 cm. This is due to the fact that the forces exerted on the central bead (in the initial situation) by the beads on the x axis cancel each other; also, the force exerted “downward” by bead 4 on the central bead is four times larger than the “upward” force exerted by bead 2. This net force along the y axis does not change as bead 1 is now moved, though there is now a nonzero x - component F x . The components are now related by tan(30 ° ) = F x F y 1 3 = F x 3 F which implies
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Unformatted text preview: F x = 3 F . Now, bead 3 exerts a leftward force of magnitude F on the central bead, while bead 1 exerts a rightward force of magnitude F . Therefore, F F = 3 F . F = ( 3 + 1) F . The fact that Coulombs law depends inversely on distance-squared then implies r 2 = d 2 3 + 1 r = d 3 + 1 = 10 cm 10 cm 1.65 3 1 = = + 6.05 cm where r is the distance between bead 1 and the central bead. This corresponds to 6.05 cm . x = (b) To regain the condition of high symmetry (in particular, the cancellation of x-components) bead 3 must be moved closer to the central bead so that it, too, is the distance r (as calculated in part(a)) away from it....
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