Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 218: 59


a) $F=\frac{GM_em}{R_e^2}(1.229); 21.3^{\circ}$ b) $\tau = \frac{GM_em}{R_e}(-.0356)$

Work Step by Step

a) We know the gravitational force is given by: $F_g=\frac{GM_em}{R_e^2}$ (Recall, in this case, $R_e$ will be equal to multiplies of the radius of the earth, for it depends on how far from the center of the earth each mass is.) Using the above equation for the magnitude of the gravitational force, we turn the two forces into vectors and add them. After adding the vectors, we find that its magnitude is $\frac{GM_em}{R_e^2}(1.229)$, and its angle is $ 21.3^{\circ}$. b) We know that the torque is equal to $rFsin\theta$. Calling the left-hand mass the axis of rotation, we multiply our equation for the right-hand force by $2R_e$ and by the angle between the force and the position vector. Doing this, we find that $\tau = \frac{GM_em}{R_e}(-.0356)$. c) We know that the mass moves down $\sqrt5-2 $, for the mass will stop rotating when it is the same distance from the earth as the right-hand mass. (After all, this is when the torques will cancel.) Thus, we use trionometry to find: $d= \sqrt5-2 (sin(8.67) = .0356$ Hence, the proof is shown.
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