Chapter 7, Problem 28PQ

To determine

Find the net gravitational force on the sphere at the origin.

Answer to Problem 28PQ

The net gravitational force on the sphere at the origin is $\underset{\_}{\left(-1.04\times {10}^{-10}\widehat{i}+2.67\times {10}^{-10}\widehat{j}\right)\text{N}}$.

Explanation of Solution

Force on mass $m_1$ due to mass $m_2$ is along the x axis and is directed along the negative x direction.

Write the equation to find the gravitational force on mass $m_1$ due to $m_2$.

    $\overrightarrow{F_{21}}=G\frac{m_1{m}_2}{r_{12}{}^2}\left(-\widehat{i}\right)$                                                                                                  (I)

Here, $\overrightarrow{F_{21}}$ is the force on mass $m_1$ due to $m_2$, $G$ is gravitational constant, $m_1$ is the mass of first sphere, $m_2$ is the mass of second sphere, $r_{12}$ is the distance between first and second sphere

Gravitational force on mass $m_1$ due to $m_3$ is along the y axis and is directed in the positive y direction.

Write the equation to find the gravitational force on mass $m_1$ due to $m_3$.

    $\overrightarrow{F_{31}}=G\frac{m_1{m}_3}{r_{13}{}^2}\left(\widehat{j}\right)$                                                                                                 (II)

Here, $\overrightarrow{F_{31}}$ is the force on mass $m_1$ due to $m_3$, $G$ is gravitational constant, $m_3$ is the mass of third sphere, $r_{13}$ is the distance between first and third sphere

Write the equation to find the net force on $m_1$ due to $m_2$ and $m_3$.

    $\overrightarrow{F_{net}}=\overrightarrow{F_{21}}+\overrightarrow{F_{31}}$                                                                                                      (III)

Here, $\overrightarrow{F_{net}}$ is the net force

Conclusion

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $12.5\text{kg}$ for $m_1$, $4.50\text{kg}$ for $m_2$ and $6.00\text{m}$ for $r_{12}$ in equation (I) to get $\overrightarrow{F_{21}}$

    $\begin{array}{c}\overrightarrow{F_{21}}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\frac{\left(12.5\text{kg}\right)\left(4.50\text{kg}\right)}{\left(6.00{\text{m}}^2\right)}\left(-\widehat{i}\right)\\ =\left(-1.04\times {10}^{-10}\text{N}\right)\left(\widehat{i}\right)\end{array}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $12.5\text{kg}$ for $m_1$, $8.00\text{kg}$ for $m_2$ and $5.00\text{m}$ for $r_{13}$ in equation (II) to get $\overrightarrow{F_{31}}$

      $\begin{array}{c}\overrightarrow{F_{31}}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\frac{\left(12.5\text{kg}\right)\left(8.00\text{kg}\right)}{\left(5.00{\text{m}}^2\right)}\left(\widehat{j}\right)\\ =\left(2.67\times {10}^{-10}\text{N}\right)\left(\widehat{j}\right)\end{array}$

Substitute $\left(-1.04\times {10}^{-10}\text{N}\right)\left(\widehat{i}\right)$ for $\overrightarrow{F_{21}}$ and $\left(2.67\times {10}^{-10}\text{N}\right)\left(\widehat{j}\right)$ for $\overrightarrow{F_{31}}$ in equation (III) to get $\overrightarrow{F_{net}}$

    $\overrightarrow{F_{net}}=\left(\left(-1.04\times {10}^{-10}\text{N}\right)\widehat{i}+\left(2.67\times {10}^{-10}\text{N}\right)\widehat{j}\right)\text{N}$

Therefore, the net gravitational force on the sphere at the origin is $\underset{\_}{\left(-1.04\times {10}^{-10}\widehat{i}+2.67\times {10}^{-10}\widehat{j}\right)\text{N}}$.