Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 13 - Gravitation - Problems - Page 379: 6

Answer

$F = (1.18\times 10^{-14}~N)~\hat{i}+ (1.18\times 10^{-14}~N)~\hat{j}$

Work Step by Step

By symmetry, the gravitational forces on the central sphere from $m_1$ and $m_4$ cancel with each other. $m_2 = 3.00~g$ and $m_3 = 1.00~g$ and the gravitational forces on the central sphere from these two masses are directed in opposite directions. We can replace these two masses with a single mass of $2.00~g$ at the location of $m_2$ We can find the distance between $m_2$ and $m_5$: $r = \sqrt{(10.0~cm)^2+(10.0~cm)^2}$ $r = 14.14~cm$ We can find the magnitude of the net gravitational force on the central sphere: $F = \frac{G~(2.00~g)(m_5)}{r^2}$ $F = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)~(0.0020~kg)(0.0025~kg)}{(0.1414~m)^2}$ $F = 1.668\times 10^{-14}~N$ Note that this force is directed from the central sphere toward $m_2$ We can find the x-component of this force: $F_x = (1.668\times 10^{-14}~N)~cos~45^{\circ}$ $F_x = 1.18\times 10^{-14}~N$ We can find the y-component of this force: $F_x = (1.668\times 10^{-14}~N)~sin~45^{\circ}$ $F_x = 1.18\times 10^{-14}~N$ We can express the net gravitational force in unit-vector notation: $F = (1.18\times 10^{-14}~N)~\hat{i}+ (1.18\times 10^{-14}~N)~\hat{j}$
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