Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 37

Answer

Particle 3 should be placed at the point $x=\frac{L}{1+\sqrt2}$

Work Step by Step

Take particle 3 with mass $M$, its distance to particle 1 to be $a$ and its distance to particle 2 to be $b$ (thus, $a+b=L$) 1) After particle 3 is introduced, there are 2 gravitational forces exerted on particle 1 - The gravitational force exerted by particle 2 $F_{21}$ - The gravitational force exerted by particle 3 $F_{31}$ So the magnitude of the gravitational force on particle 1 doubles when $$F_{31}+F_{21}=2F_{21}$$ $$F_{31}=F_{21}$$ 2) Similarly, there are 2 gravitational forces exerted on particle 2 - The gravitational force exerted by particle 2 $F_{12}$ - The gravitational force exerted by particle 3 $F_{32}$ The magnitude of the gravitational force on particle 2 doubles when $$F_{32}+F_{12}=2F_{12}$$ $$F_{32}=F_{12}$$ We know that $F_{21}=F_{12}$. Therefore, $F_{31}=F_{32}$ As particle 1 has mass $m$ and particle 2 has mass $2m$, $$G\frac{Mm}{a^2}=G\frac{M2m}{b^2}$$ $$\frac{1}{a^2}=\frac{2}{b^2}$$ $$\frac{b}{a}=\sqrt2$$ As $a+b=L$, we can figure out that $a=\frac{L}{1+\sqrt2}$, which is the point on the x-axis particle 3 should be placed.
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