University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 13 - Gravitation - Problems - Exercises - Page 425: 13.2

Answer

(a) Zero, (b) $3.81 \times 10^{-8} N.m$, (c) No. The torque is too small. It can be increased by either increasing the masses of the spheres or by decreasing the distance between their centres.

Work Step by Step

For (a): From Newton's law of universal gravitation we know that the magnitude of the forces for each pair is $F = G\frac{m_1 m_2}{r^2}$. From Figure 13.4, we observe that both forces point in opposite directions. Since they are equal and opposite, they cancel out and $F_{net} = 0$. For (b): The net torque $\tau_{net} = F (2l) $, where $l$ is the length of each moment arm. Calculating directly . . . $\tau_{net} = 2Fl = 2G\frac{m_1 m_2}{r^2} l = 2(6.67 \times 10^{-11})\frac{(1.1)(25.0)}{(0.12)^2}(0.15) = 3.81 \times 10^{-8} \ N$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.