Chapter 6, Problem 10P

(a)

To determine

To Find: The resultant force on $0.100\text{kg}$ mass.

Explanation of Solution

Given:

Mass 1, $m_1=0.400\text{kg}$

Mass 2, $m_2=0.200\text{kg}$

Mass 3, $m_3=0.100\text{kg}$

Distance between mass 1 and 2, $d_{12}=10.0\text{cm}$ .

Distance between mass 2 and 3, $d_{23}=10.0\text{cm}$ .

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:

Resultant force on $0.100\text{kg}$ mass can be calculated as follows:

  $\begin{array}{l}F=\text{G}\frac{mM}{d^2}\\ \Rightarrow F_{net}=F_{13}+F_{23}\\ \Rightarrow F_{net}=G\frac{m_1{m}_3}{d_{13}{}^2}+G\frac{m_2{m}_3}{d_{23}{}^2}\\ \Rightarrow F_{net}=G{m}_3\left(\frac{m_1}{d_{13}{}^2}+\frac{m_2}{d_{23}{}^2}\right)\\ \Rightarrow F_{net}=\left(6.67\times {10}^{-11}\right)\left(0.100\right)\left(\frac{0.400}{{0.10}^2}+\frac{0.200}{{0.10}^2}\right)\\ \Rightarrow F_{net}=40.02\times {10}^{-11}\text{N }\left(\text{towards}\text{left}\right)\end{array}$

Conclusion:

Thus, the net force on $0.100\text{kg}$ is $40.02\times {10}^{-11}\text{N}$ towards left.

(b)

To determine

To Find: The resultant force on $0.200\text{kg}$ mass.

Explanation of Solution

Given:

Mass 1, $m_1=0.400\text{kg}$

Mass 2, $m_2=0.200\text{kg}$

Mass 3, $m_3=0.100\text{kg}$

Distance between mass 1 and 2, $d_{12}=10.0\text{cm}$ .

Distance between mass 2 and 3, $d_{23}=10.0\text{cm}$ .

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:

Resultant force on $0.200\text{kg}$ mass can be calculated as follows:

  $\begin{array}{l}F=\text{G}\frac{mM}{d^2}\\ \Rightarrow F_{net}=F_{12}+F_{32}\\ \Rightarrow F_{net}=G\frac{m_1{m}_2}{d_{12}{}^2}-G\frac{m_2{m}_3}{d_{23}{}^2}\\ \Rightarrow F_{net}=G{m}_2\left(\frac{m_1}{d_{12}{}^2}-\frac{m_2}{d_{23}{}^2}\right)\\ \Rightarrow F_{net}=\left(6.67\times {10}^{-11}\right)\left(0.200\right)\left(\frac{0.400}{{0.10}^2}-\frac{0.100}{{0.10}^2}\right)\\ \Rightarrow F_{net}=40.02\times {10}^{-11}\text{N }\left(\text{towards left}\right)\end{array}$

Conclusion:

Thus, the net force on $0.200\text{kg}$ is $40.02\times {10}^{-11}\text{N}$ towards left.