HW8_Sol_F10 - Oct 18 2010 PHYS 2101 HW#8 WileyPlus Problem...

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Oct. 18, 2010 PHYS 2101 HW#8 WileyPlus Problem Solutions 1 HW #8 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 2) [13-6] The gravitational forces on m 5 from the two 5.00 g masses m 1 and m 4 cancel each other. Contributions to the net force on m 5 come from the remaining two masses: The force is directed along the diagonal between m 2 and m 3 , toward m 2 . In unit-vector notation, we have . 3) [13-9] Both the Sun and the Earth exert a gravitational pull on the space probe. The net force can be calculated by using superposition principle. At the point where the two forces balance, we have , where M e is the mass of Earth, M s is the mass of the Sun, m is the mass of the space probe, r 1 is the distance from the center of Earth to the probe, and r 2 is the distance from the center of the Sun to the probe. We substitute r 2 = d ! r 1 , where d is the distance from the center of Earth to the center of the Sun, to find Using the values for M e , M s , and d given in Appendix C, we take the positive square root of both sides to solve for r 1 . A little algebra yields Note: The fact that indicates that the probe is much closer to the Earth than the Sun. 4) [13-10] 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 $ % & ' ( ) Gm A 2 d 2 = Gm A m D r 2 . This yields r = 1.29 d . In magnitude-angle notation, then, r " = (1.29 # –56.3º) , with SI units understood. The “exact” answer without regard to significant figure considerations is (a) In ( x, y ) notation, the x
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