Chapter 7, Problem 41PQ

Three billiard balls, the two-ball, the four-ball, and the eight-ball, are arranged on a pool table as shown in Figure P7.26. Given the coordinate system shown and that the mass of each ball is 0.150 kg, determine the gravitational field at $x=2\text{m}$, $y=0$.

To determine

The gravitational field at $x=2\text{cm}$, $y=0$.

Answer to Problem 41PQ

The gravitational field at $x=2\text{cm}$, $y=0$ is $\left(-4\times {10}^{-12}\widehat{\text{i}}+1\times {10}^{-11}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ .

Explanation of Solution

The following figure gives the direction of field vectors.

The point at which gravitational field is to be calculated is identified as $\left(2\text{m,0}\text{m}\right)$.

Use figure1, calculate the distance between two-ball and point at which field has to be calculated.

  $\begin{array}{c}{r}_2=\sqrt{{\left(1\text{m}\right)}^2+{\left(2\text{m}\right)}^2}\\ =\sqrt{5}\text{m}\end{array}$

Here, $r_2$ is the distance between two-ball and point at which field is to be calculated.

Use figure1, to calculate angle $\theta$.

  $\begin{array}{c}\tan \theta =\frac{1\text{m}}{2\text{m}}\\ \theta ={\tan }^{-1}\left(0.5\right)\\ =27°\end{array}$

Here, $\theta$ is the angle that gravitational field due to two-ball at $x=2\text{cm}$, $y=0$ makes with $x$ axis.

Write the expression for the gravitational field at a point due to a mass $m$.

  $\overrightarrow{\text{g}}=G\frac{m}{r^2}\widehat{\text{r}}$ (I)

Here, $g$ is the gravitational field at a point, $G$ is the universal gravitational constant, $m$ is the mass of the object which produce the field, $\widehat{\text{r}}$ is the unit vector along the direction of field and $r$ is the distance between the object and the point at which field is to be calculated.

Use equation (I) to write expression for the gravitational field at $\left(2\text{m,0m}\right)$ by eight-ball.

    ${\overrightarrow{\text{g}}}_8=G\frac{m_8}{r_8^2}{\widehat{\text{r}}}_8$ (II)

Use equation (I) to write expression for the gravitational field at $\left(2\text{m,0m}\right)$ by two-ball.

  ${\overrightarrow{\text{g}}}_2=G\frac{m_2}{r_2^2}{\widehat{\text{r}}}_2$ (III)

Use equation (I) to write expression for the gravitational field at $\left(2\text{m,0m}\right)$ by four-ball.

  ${\overrightarrow{\text{g}}}_4=G\frac{m_4}{r_4^2}{\widehat{\text{r}}}_4$ (IV)

Use figure1, to calculate direction of field vector due to eight-ball.

  ${\widehat{\text{r}}}_8=-\widehat{\text{i}}$

Use figure1, to calculate direction of field vector due to two-ball.

  ${\widehat{\text{r}}}_2=-\left(\cos \theta \right)\widehat{\text{i}}+\sin \theta \widehat{\text{j}}$

Use figure1, to calculate direction of field vector due to four-ball.

  ${\widehat{\text{r}}}_4=\widehat{\text{j}}$

Substitute $-\left(\cos \theta \right)\widehat{\text{i}}+\sin \theta \widehat{\text{j}}$ for ${\widehat{\text{r}}}_2$ in equation (III) to get ${\overrightarrow{\text{g}}}_2$.

  ${\overrightarrow{\text{g}}}_2=G\frac{m_2}{r_2^2}\left(-\left(\cos \theta \right)\widehat{\text{i}}+\sin \theta \widehat{\text{j}}\right)$ (V)

Write the expression for the total field at $\left(2\text{m,0}\text{m}\right)$.

  ${\overrightarrow{\text{g}}}_{\text{net}}={\overrightarrow{\text{g}}}_2+{\overrightarrow{\text{g}}}_8+{\overrightarrow{\text{g}}}_4$ (VI)

Conclusion:

Substitute $-\widehat{\text{i}}$ for ${\widehat{\text{r}}}_8$, $2\text{m}$ for $r_8$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ and $0.150\text{kg}$ for in equation(II) to get ${\overrightarrow{\text{g}}}_8$.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_8=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\frac{\left(0.150\text{kg}\right)}{{\left(2\text{m}\right)}^2}\left(-\widehat{\text{i}}\right)\\ =-2.5\times {10}^{-12}\text{m}/{\text{s}}^2\widehat{\text{i}}\end{array}$

Substitute $-\left(\cos \theta \right)\widehat{\text{i}}+\sin \theta \widehat{\text{j}}$ for ${\widehat{\text{r}}}_2$, $\sqrt{5}\text{m}$ for $r_2$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$, $27°$ for $\theta$ and $0.150\text{kg}$ for in equation(V) to get ${\overrightarrow{\text{g}}}_2$.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_2=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\frac{\left(0.150\text{kg}\right)}{{\left(2\text{m}\right)}^2}\left(-\left(\cos 27°\right)\widehat{\text{i}}+\sin 27°\widehat{\text{j}}\right)\\ =2.0\times {10}^{-12}\left(-\left(\cos 27°\right)\widehat{\text{i}}+\sin 27°\widehat{\text{j}}\right)\\ =\left(-1.8\times {10}^{-12}\widehat{\text{i}}+9.1\times {10}^{-13}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\end{array}$

Substitute $\widehat{\text{j}}$ for ${\widehat{\text{r}}}_4$, $1\text{m}$ for $r_4$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ and $0.150\text{kg}$ for in equation(IV) to get ${\overrightarrow{\text{g}}}_4$.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_4=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\frac{\left(0.150\text{kg}\right)}{{\left(1\text{m}\right)}^2}\left(\widehat{\text{j}}\right)\\ =1.0\times {10}^{-11}\text{m}/{\text{s}}^2\widehat{\text{j}}\end{array}$

Substitute $-2.5\times {10}^{-12}\text{m}/{\text{s}}^2\widehat{\text{i}}$ for ${\overrightarrow{\text{g}}}_8$, $\left(-1.8\times {10}^{-12}\widehat{\text{i}}+9.1\times {10}^{-13}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ for ${\overrightarrow{\text{g}}}_2$ and $1.0\times {10}^{-11}\text{m}/{\text{s}}^2\widehat{\text{j}}$ for ${\overrightarrow{\text{g}}}_4$ in equation (VI) to get ${\overrightarrow{\text{g}}}_{\text{net}}$.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_{\text{net}}=\left(-1.8\times {10}^{-12}\widehat{\text{i}}+9.1\times {10}^{-13}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2-2.5\times {10}^{-12}\text{m}/{\text{s}}^2\widehat{\text{i}}+1.0\times {10}^{-11}\text{m}/{\text{s}}^2\widehat{\text{j}}\\ =\left(-4\times {10}^{-12}\widehat{\text{i}}+1\times {10}^{-11}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\end{array}$

Therefore, the gravitational field at $x=2\text{cm}$, $y=0$ is $\left(-4\times {10}^{-12}\widehat{\text{i}}+1\times {10}^{-11}\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ .