Question

A $200-kg$object and a localid="1663652122187" $500-kg$object are separated by $4.00 m.$(a) Find the net gravitational force exerted by these object on a $50.0-kg$object placed midway between them.(b) At what position (other than an infinitely remote one) can the $50.0-kg$object be placed so as to experience a net force of zero from the other two objects.

Short answer

a) $\left(2.5\times {10}^{-7}N\right)\hat{i}$$\left(2.5\times {10}^{-7}N\right)\hat{i}$

b) The $50-kg$object should be at a distance 1.55 m from the$200-kg$in order to experience a zero net force.

Step-by-step solution

Given data

$$\begin{gathered} m_1=200 kg \\ {m}_2=500 kg \\ {m}_3=50 kg \end{gathered}$$

Definition and concept of Newton’s Law of Gravitation

According to Newton's Law of Universal Gravitation, every particle in the universe is drawn to every other particle with a force that is directly proportional to the product of their masses and inversely proportional to their distance from one another.

The two particles have masses $m_1 and m_2$are separated by a distance r, the force exerted by particle 1 on particle 2 is:

${\overrightarrow{F}}_{12}=G\frac{m_1{m}_2 }{r^2}\widehat{r_{12}}$

G is a constant, called the universal gravitational constant, and has a value of

$6.674\times {10}^{-11}N{m}^2/K{g}^2$

(a) Determining the net gravitational force exerted by these objects on an object placed midway between them

Force$\left(\overrightarrow{F_{13}}\right)$ exerted by$\left(m_1\right)on\left(m_3\right)$ is found by equation (i)

$\overrightarrow{F_{13}}=\frac{G{m}_1{m}_3}{{r^2}_{13}}\left(-\hat{i}\right)$

The $\left(r_{13}\right)$is the distance between $\left(m_1\right) and \left(m_3\right)$which is equal to$\left(2 m\right)$ as we see in the figure. Substitute numerical values:

$$\begin{gathered} \overrightarrow{F_{13}}=-\frac{\left({\displaystyle 6.67\times {10}^{-11}N{m}^2/K{g}^2}\right){\displaystyle \left(200Kg\right)}{\displaystyle \left(50Kg\right)}}{{\displaystyle {\left(2m\right)}^2}}\left(\hat{\mathrm{i}}\right) \\ =-1.67\times {10}^{-7} \mathrm{N} \hat{\mathrm{i}} \end{gathered}$$

Force $\left(\overrightarrow{F_{23}}\right)$exerted by$\left(m_2\right)on\left(m_3\right)$ is found by equation (i)

$\overrightarrow{F_{23}}=\frac{G{m}_2{m}_3}{{r^2}_{23}}\left(\hat{i}\right)$

The$\left(r_{23}\right)$ is the distance between $\left(m_2\right)on\left(m_3\right)$, which is equal to $\left(2 m\right)$as we see in the figure. Substitute numerical values:

$$\begin{gathered} \overrightarrow{F_{13}}=-\frac{\left({\displaystyle 6.67\times {10}^{-11}N{m}^2/K{g}^2}\right){\displaystyle \left(500Kg\right)}{\displaystyle \left(50Kg\right)}}{{\displaystyle {\left(2m\right)}^2}}\left(\hat{\mathrm{i}}\right) \\ =4.17\times {10}^{-7} \hat{\mathrm{i}} \end{gathered}$$

Hence, the net force exerted on $\left(m_3\right)$is:

$\left(\overrightarrow{F_{net}}\right)={\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{23}$

$$\begin{gathered} \overrightarrow{F_{net}}=-\left(1.67\times {10}^{-7}\right) \hat{i} +\left(4.17\times {10}^{-7} \hat{i} \right) \\ =\left(2.5\times {10}^{-7}N\right)\hat{i} \end{gathered}$$

(b) Determining the position (other than an infinitely remote one) ofobject be placed so as to experience a net force of zero from the other two objects.

Assume is placed at a distance $r_{13}=x$ from $\left(m_1\right)$Hence, $m_3$is at a distance $r_{23}=\left(4-x\right)$ from $m_2$

First:

The force $\left({\overrightarrow{F}}_{13}\right)$exerted by $\left(m_1\right) on \left(m_3\right)$is found by Equation (i)

$\overrightarrow{F_{13}}=\frac{G{m}_1{m}_3}{{r^2}_{13}}\left(-\hat{i}\right)$

Substitute numerical values:

$$\begin{gathered} \overrightarrow{F_{13}}=-\frac{\left({\displaystyle 1.67\times {10}^{-7}N{m}^2/K{g}^2}\right){\displaystyle \left(200Kg\right)}{\displaystyle \left(50Kg\right)}}{{\displaystyle \left(x^2\right) m^2}}\left(\hat{\mathrm{i}}\right) \\ =-\frac{6.67\times {10}^{-7}N}{x^2}\hat{i} \end{gathered}$$

Second:

The force $\left({\overrightarrow{F}}_{23}\right)$ exerted by $m_{2 }on m_3$is found by equation (i)

$\overrightarrow{F_{23}}=\frac{G{m}_2{m}_3}{{r^2}_{23}}\left(\hat{i}\right)$

$$\begin{gathered} \overrightarrow{F_{13}}=-\frac{\left({\displaystyle 1.67\times {10}^{-7}N{m}^2/K{g}^2}\right){\displaystyle \left(500Kg\right)}{\displaystyle \left(50Kg\right)}}{{\displaystyle \left(4-x^2\right)m^2}}\left(\hat{\mathrm{i}}\right) \\ =\frac{4.17\times {10}^{-7} N}{\left(4-x^2\right)} \hat{i} \end{gathered}$$

Hence, the net force exerted on $\left(m_3\right)$is:

$\overrightarrow{F_{net}}={\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{23}$

Substitute$\left(F_{net}=0\right)$as required in the problem

$0=-\frac{{\displaystyle 6.67\times {10}^{-7} N}}{{\displaystyle x^2}}\hat{i}+\frac{4.17\times {10}^{-7} N}{{\displaystyle \left(4-x^2\right)}}\hat{i}$

Rearrange:

$$\begin{gathered} \frac{6.67\times {10}^{-7}}{x^2}=\frac{4.17\times {10}^{-7}}{\left(4-x^2\right)} \\ \frac{\left(4-x^2\right)}{x^2}=\frac{4.17\times {10}^{-7}}{6.67\times {10}^{-7}} \end{gathered}$$

$$\begin{gathered} \frac{4}{x^2}-1=\frac{4.17}{6.67} \\ =0.625 \\ \frac{4}{x^2}=1.625 \end{gathered}$$

$$\begin{gathered} x^2=\frac{4}{1.625} \\ =2.46 \\ x=1.55 m \end{gathered}$$

The $50-kg$ object should be at a distance 1.55 m from 200 Kg in order to experience a zero net force.