Chapter 6, Problem 11P

To determine

To Find: The resultant gravitational force on each particle.

Explanation of Solution

Given:

Mass of each particle, $m_1=m_2=m_3=m=2.00\text{kg}$ . $d=1.00\text{m}$

Distance between particles, $d=1.0\text{m}$

Formula used:

The gravitational force between two bodies having mass m and M , d distance away is:

  $F=\text{G}\frac{mM}{d^2}$

G is the gravitational constant.

Calculation:

Free body diagram:

  

Gravitational force on $m_1$ due to $m_2$ and $m_3$ :

  $\begin{array}{l}{F}_{13}=G\frac{m_1{m}_3}{d^2}\\ \Rightarrow F_{13}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\begin{array}{l}{F}_{12}=G\frac{m_1{m}_2}{d^2}\\ \Rightarrow F_{12}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\Rightarrow F_{12}=F_{13}$

  $\begin{array}{l}{\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{12}=\left(F_{12}\sin 30°-F_{13}\sin 30°\right)\stackrel{⌢}{i}+\left(F_{13}+F_{12}\right)\cos 30°\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{12}=0+2\times 26.68\times {10}^{-11}\times \frac{\sqrt{3}}{2}\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{12}=46.2\times {10}^{-11}\text{N}\stackrel{⌢}{j}\end{array}$

The gravitational force on

  $m_2$ due to $m_1$ and $m_3$ :

  $\begin{array}{l}{F}_{21}=G\frac{m_1{m}_2}{d^2}\\ \Rightarrow F_{21}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\begin{array}{l}{F}_{23}=G\frac{m_2{m}_3}{d^2}\\ \Rightarrow F_{23}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\Rightarrow F_{21}=F_{23}$

  $\begin{array}{l}{\overrightarrow{F}}_{21}+{\overrightarrow{F}}_{23}=\left(F_{23}+F_{21}\sin 30°\right)\left(-\stackrel{⌢}{i}\right)+F_{21}\cos 30°\left(-\stackrel{⌢}{j}\right)\\ \Rightarrow {\overrightarrow{F}}_{21}+{\overrightarrow{F}}_{23}=-\left(26.68\times {10}^{-11}+26.68\times {10}^{-11}\times \frac{1}{2}\right)\stackrel{⌢}{i}-26.68\times {10}^{-11}\times \frac{\sqrt{3}}{2}\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{12}=-40.02\times {10}^{-11}\text{N}\stackrel{⌢}{i}-23.11\times {10}^{-11}\stackrel{⌢}{j}\end{array}$

Gravitational force on $m_3$ due to $m_1$ and $m_2$ :

  $\begin{array}{l}{F}_{31}=G\frac{m_1{m}_3}{d^2}\\ \Rightarrow F_{31}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\begin{array}{l}{F}_{32}=G\frac{m_2{m}_3}{d^2}\\ \Rightarrow F_{32}=\left(6.67\times {10}^{-11}\right)\frac{{2.0}^2}{{1.0}^2}=26.68\times {10}^{-11}\text{N}\end{array}$

  $\begin{array}{l}{\overrightarrow{F}}_{31}+{\overrightarrow{F}}_{32}=\left(F_{32}+F_{31}\sin 30°\right)\stackrel{⌢}{i}+F_{31}\cos 30°\left(\stackrel{⌢}{j}\right)\\ \Rightarrow {\overrightarrow{F}}_{31}+{\overrightarrow{F}}_{32}=\left(26.68\times {10}^{-11}+26.68\times {10}^{-11}\times \frac{1}{2}\right)\stackrel{⌢}{i}-26.68\times {10}^{-11}\times \frac{\sqrt{3}}{2}\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{31}+{\overrightarrow{F}}_{32}=40.02\times {10}^{-11}\text{N}\stackrel{⌢}{i}-23.11\times {10}^{-11}\stackrel{⌢}{j}\end{array}$

Conclusion:

Thus, resultant gravitational force on $m_1$ is $46.2\times {10}^{-11}\text{N}\stackrel{⌢}{j}$ , on $m_2$ is $-40.02\times {10}^{-11}\text{N}\stackrel{⌢}{i}-23.11\times {10}^{-11}\stackrel{⌢}{j}$ and on $m_3$ is $40.02\times {10}^{-11}\text{N}\stackrel{⌢}{i}-23.11\times {10}^{-11}\stackrel{⌢}{j}$ .