Chapter 6, Problem 40P

To determine

To Find: The vector sum of all gravitational forces acting on the given system.

Explanation of Solution

Given:

Mass of A, $m_A=1000\text{kg}$

Mass of B, $m_B=2000\text{kg}$

Mass of C, $m_C=3000\text{kg}$

Mass of D, $m_D=4000\text{kg}$

Distance between A and B, $d_{AB}=2.0\text{m}$

Distance between B and C, $d_{BC}=2.0\text{m}$

Distance between A and C, $d_{AC}=4.0\text{m}$

Distance between C and D, $d_{CD}=2.0\text{m}$

Distance between B and D, $d_{BD}=\sqrt{{2.0}^2+{2.0}^2}=2.82\text{m}$

Formula used:

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:

  

Distance between A and D,

  $\begin{array}{l}{d}_{AD}=\sqrt{{\left(4.0\right)}^2+{\left(2.0\right)}^2}\\ d_{AD}=4.47\text{m}\end{array}$

Gravitational force on A due to B:

  $\begin{array}{l}{\overrightarrow{F}}_{AB}\text{= G}\frac{m_A{m}_B}{d_{AB}{}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{AB}=\left(6.67\times {10}^{-11}\right)\frac{\left(1000\text{kg}\right)\left(2000\text{kg}\right)}{{\left(2.0\text{m}\right)}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{AB}=3.34\times {10}^{-5}\text{N}\stackrel{⌢}{i}\end{array}$

Gravitational force on A due to C:

  $\begin{array}{l}{\overrightarrow{F}}_{AC}\text{= G}\frac{m_A{m}_C}{d_{AC}{}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{AC}=\left(6.67\times {10}^{-11}\right)\frac{\left(1000\text{kg}\right)\left(3000\text{kg}\right)}{{\left(4.0\text{m}\right)}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{AC}=1.25\times {10}^{-5}\text{N}\stackrel{⌢}{i}\end{array}$

Gravitational force on A due to D:

  $\begin{array}{l}\tan \theta =\frac{4}{2}\\ \Rightarrow \theta ={\tan }^{-1}\left(2\right)\\ \Rightarrow \theta =63.43°\end{array}$

  $\begin{array}{l}{\overrightarrow{F}}_{AD}\text{= G}\frac{m_A{m}_D}{d_{AD}{}^2}\sin \theta \stackrel{⌢}{i}+\text{G}\frac{m_A{m}_D}{d_{AD}{}^2}\cos \theta \stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{AD}=\left(6.67\times {10}^{-11}\right)\frac{\left(1000\text{kg}\right)\left(4000\text{kg}\right)}{{\left(4.47\text{m}\right)}^2}\sin \left(63.43°\right)\stackrel{⌢}{i}+\left(6.67\times {10}^{-11}\right)\frac{\left(1000\text{kg}\right)\left(4000\text{kg}\right)}{{\left(4.47\text{m}\right)}^2}\cos \left(63.43°\right)\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{AD}=1.19\times {10}^{-5}\text{N}\stackrel{⌢}{i}+0.59\times {10}^{-5}\text{N}\stackrel{⌢}{j}\end{array}$

Gravitational force on B due to C:

  $\begin{array}{l}{\overrightarrow{F}}_{BC}\text{= G}\frac{m_A{m}_C}{d_{AC}{}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{BC}=\left(6.67\times {10}^{-11}\right)\frac{\left(2000\text{kg}\right)\left(3000\text{kg}\right)}{{\left(2.0\text{m}\right)}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{BC}=1.00\times {10}^{-4}\text{N}\stackrel{⌢}{i}\end{array}$

Gravitational force on B due to D:

  $\begin{array}{l}\tan \theta \text{'}=\frac{2}{2}\\ \Rightarrow \theta \text{'}={\tan }^{-1}1\\ \Rightarrow \theta \text{'}=45°\end{array}$

  $\begin{array}{l}{\overrightarrow{F}}_{BD}\text{= G}\frac{m_B{m}_D}{d_{BD}{}^2}\sin \theta \stackrel{⌢}{i}+\text{G}\frac{m_B{m}_D}{d_{BD}{}^2}\cos \theta \stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{BD}=\left(6.67\times {10}^{-11}\right)\frac{\left(2000\text{kg}\right)\left(4000\text{kg}\right)}{{\left(2.82\text{m}\right)}^2}\sin \left(45°\right)\stackrel{⌢}{i}+\left(6.67\times {10}^{-11}\right)\frac{\left(2000\text{kg}\right)\left(4000\text{kg}\right)}{{\left(2.82\text{m}\right)}^2}\cos \left(45°\right)\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{BD}=4.74\times {10}^{-5}\text{N}\stackrel{⌢}{i}+4.74\times {10}^{-5}\text{N}\stackrel{⌢}{j}\end{array}$

Gravitational force on B due to D:

  $\begin{array}{l}{\overrightarrow{F}}_{CD}\text{= G}\frac{m_C{m}_D}{d_{CD}{}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{CD}=\left(6.67\times {10}^{-11}\right)\frac{\left(3000\text{kg}\right)\left(4000\text{kg}\right)}{{\left(2.0\text{m}\right)}^2}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{CD}=2.00\times {10}^{-4}\text{N}\stackrel{⌢}{i}\end{array}$

Thus, the vector sum of all the gravitational forces acting on the system is:

  $\begin{array}{l}{\overrightarrow{F}}_{net}={\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{AD}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{BD}+{\overrightarrow{F}}_{CD}\\ \Rightarrow {\overrightarrow{F}}_{net}=3.34\times {10}^{-5}\text{N}\stackrel{⌢}{i}+1.25\times {10}^{-5}\text{N}\stackrel{⌢}{i}+1.19\times {10}^{-5}\text{N}\stackrel{⌢}{i}\\ +0.59\times {10}^{-5}\text{N}\stackrel{⌢}{j}+1.00\times {10}^{-4}\text{N}\stackrel{⌢}{i}+4.74\times {10}^{-5}\text{N}\stackrel{⌢}{i}\\ +4.74\times {10}^{-5}\text{N}\stackrel{⌢}{j}+2.00\times {10}^{-4}\text{N}\stackrel{⌢}{i}\\ \Rightarrow {\overrightarrow{F}}_{net}=\left(3.34\times {10}^{-5}+1.25\times {10}^{-5}+1.19\times {10}^{-5}+1.00\times {10}^{-4}+4.74\times {10}^{-5}\text{+}2.00\times {10}^{-4}\right)\text{N}\stackrel{⌢}{i}\\ +\left(0.59\times {10}^{-5}+4.74\times {10}^{-5}\right)\text{N}\stackrel{⌢}{j}\\ \Rightarrow {\overrightarrow{F}}_{net}=40.52\times {10}^{-5}\text{N}\stackrel{⌢}{i}+5.33\times {10}^{-5}\text{N}\stackrel{⌢}{j}\end{array}$

Conclusion:

Thus, the vector sum of all gravitational forces acting on the given system is: $40.52\times {10}^{-5}\text{N}\stackrel{⌢}{i}+5.33\times {10}^{-5}\text{N}\stackrel{⌢}{j}$ .