Chapter 5, Problem 37P

To determine

The gravitational force exerted by balls 1 and 2 on ball 3.

Answer to Problem 37P

The gravitational force exerted by balls 1 and 2 on ball 3 is $\underset{\_}{4.0\times {10}^{-9}\text{N}}$, $34°$ below the negative $x$ axis.

Explanation of Solution

Write the expression for the force between ball 1 and 3.

  $F_{13}=\frac{G{m}_1{m}_3}{r_{13}^2}$        (I)

Here, $G$ is the gravitational constant, $m_1$ is the mass of ball 1, $m_3$ is the mass of ball 3, $r_{13}$ is the distance between ball 1 and 3.

Write the expression for the force between ball 1 and 3.

  $F_{13}=\frac{G{m}_1{m}_3}{r_{13}^2}$        (II)

Here, $G$ is the gravitational constant, $m_2$ is the mass of ball 2, $r_{23}$ is the distance between ball 2 and 3.

Write the component of force along $x$ axis.

  $F_{x,grav}=-F_{13}\cos \theta -F_{23}\cos \varphi$        (III)

Write the component of force along $y$ axis.

  $F_{y,grav}=-F_{13}\sin \theta -F_{23}\sin \varphi$        (IV)

Write the expression for the magnitude of net force.

  $\left|F\right|=\sqrt{F_{x,grav}^2+F_{y,grav}^2}$        (V)

Write the expression for the direction of net force.

  $\theta ={\tan }^{-1}\left[\frac{F_{y,grav}}{F_{x,grav}}\right]$        (VI)

Conclusion:

Substitute, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $15\text{kg}$ for $m_1$, $9\text{kg}$ for $m_3$, $\sqrt{{\left(0.5\text{m}\right)}^2}+{\left(3\text{m}\right)}^2$ for $r_{13}$ in equation (I).

  $\begin{array}{c}{F}_{13}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(15\text{kg}\right)\left(9\text{kg}\right)}{\left(\sqrt{{\left(0.5\text{m}\right)}^2}+{\left(3\text{m}\right)}^2\right)}\\ =9.7\times {10}^{-10}\text{N}\end{array}$

Substitute, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $25\text{kg}$ for $m_1$, $9\text{kg}$ for $m_3$, $\sqrt{{\left(1.5\text{m}\right)}^2}+{\left(1.5\text{m}\right)}^2$ for $r_{23}$ in equation (II).

  $\begin{array}{c}{F}_{13}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(25\text{kg}\right)\left(9\text{kg}\right)}{\left(\sqrt{{\left(1.5\text{m}\right)}^2}+{\left(1.5\text{m}\right)}^2\right)}\\ =3.3\times {10}^{-10}\text{N}\end{array}$

From the graph the values of trigonometric expressions are.

  $\begin{array}{c}\cos \theta =\frac{\text{adj}}{\text{hyp}}\\ =\frac{3\text{m}}{\sqrt{{\left(3\text{m}\right)}^2+{\left(0.5\text{m}\right)}^2}}\\ =0.986\\ \sin \theta =\frac{\text{oppo}}{\text{hyp}}\\ =\frac{0.5\text{m}}{\sqrt{{\left(3\text{m}\right)}^2+{\left(0.5\text{m}\right)}^2}}\\ =0.164\\ \cos \varphi =\frac{1.5\text{m}}{\sqrt{{\left(1.5\text{m}\right)}^2+{\left(1.5\text{m}\right)}^2}}\\ =0.707\\ \sin \varphi =\frac{1.5\text{m}}{\sqrt{{\left(1.5\text{m}\right)}^2+{\left(1.5\text{m}\right)}^2}}\\ =0.707\end{array}$

Substitute, $9.7\times {10}^{-10}\text{N}$ for $F_{13}$, $3.3\times {10}^{-10}\text{N}$ for $F_{23}$, $0.986$ for $\cos \theta$, $0.707$ for $\cos \varphi$ in equation (III).

  $\begin{array}{c}{F}_{x,grav}=-\left(9.7\times {10}^{-10}\text{N}\right)\cos 0.986-\left(3.3\times {10}^{-10}\text{N}\right)\cos 0.707\\ =-3.3\times {10}^{-9}\text{N}\end{array}$

Substitute, $9.7\times {10}^{-10}\text{N}$ for $F_{13}$, $3.3\times {10}^{-10}\text{N}$ for $F_{23}$, $0.164$ for $\sin \theta$, $0.707$ for $\sin \varphi$ in equation (IV).

  $\begin{array}{c}{F}_{y,grav}=-\left(9.7\times {10}^{-10}\text{N}\right)\cos 0.164-\left(3.3\times {10}^{-10}\text{N}\right)\cos 0.707\\ =-2.2\times {10}^{-9}\text{N}\end{array}$

Substitute, $-3.3\times {10}^{-9}\text{N}$ for $F_{x,grav}$, $-2.2\times {10}^{-9}\text{N}$ for $F_{y,grav}$ in equation (V) to find the magnitude of force.

  $\begin{array}{c}\left|F\right|=\sqrt{{\left(-3.3\times {10}^{-9}\text{N}\right)}^2+{\left(-2.2\times {10}^{-9}\text{N}\right)}^2}\\ =-4.0\times {10}^{-9}\text{N}\end{array}$

Substitute, $-3.3\times {10}^{-9}\text{N}$ for $F_{x,grav}$, $-2.2\times {10}^{-9}\text{N}$ for $F_{y,grav}$ in equation (VI) to find the direction of force.

  $\begin{array}{c}\theta ={\tan }^{-1}\left[\frac{-2.2\times {10}^{-9}\text{N}}{-3.3\times {10}^{-9}\text{N}}\right]\\ =34°\text{ below the negative }x\text{ axis}\end{array}$

Therefore, the gravitational force exerted by balls 1 and 2 on ball 3 is $\underset{\_}{4.0\times {10}^{-9}\text{N}}$, $34°$ below the negative $x$ axis.