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## Q. 7.9

BILLIARDS, ANYONE?

GOAL Use vectors to find the net gravitational force on an object.

PROBLEM (a) Three 0.300-kg billiard balls are placed on a table at the corners of a right triangle, as shown from overhead in Figure 7.17. Find the net gravitational force on the cue ball (designated as $m_1$ ) resulting from the forces exerted by the other two balls. (b) Find the components of the gravitational force of $m_2$ on $m_3$.

STRATEGY (a) To find the net gravitational force on the cue ball of mass $m_1$, we first calculate the force $\overrightarrow{\mathbf{F}}_{21}$ exerted by $m_2$ on $m_1$. This force is the $y$-component of the net force acting on $m_1$. Then we find the force $\overrightarrow{\mathbf{F}}_{31}$ exerted by $m_3$ on $m_1$, which is the $x$ component of the net force acting on $m_1$. With these two components, we can find the magnitude and direction of the net force on the cue ball. (b) In this case, we must use trigonometry to find the components of the force $\overrightarrow{\mathbf{F}}_{23}$.

## Verified Solution

(a) Find the net gravitational force on the cue ball.

Find the magnitude of the force $\overrightarrow{\mathbf{F}}_{21}$ exerted by $m_2$ on $m_1$ using the law of gravitation, Equation 7.21:

$F=G \frac{m_1 m_2}{r^2}$      [7.21]

\begin{aligned}&F_{21}=G \frac{m_2 m_1}{r_{21}^2}=\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right) \frac{(0.300 \mathrm{~kg})(0.300 \mathrm{~kg})}{(0.400 \mathrm{~m})^2} \\&F_{21}=3.75 \times 10^{-11} \mathrm{~N}\end{aligned}

Find the magnitude of the force $\overrightarrow{\mathbf{F}}_{31}$ exerted by $m_3$ on $m_1$, again using Newton’s law of gravity:

$F_{31}=G \frac{m_3 m_1}{r_{31}^2}=\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right) \frac{(0.300 \mathrm{~kg})(0.300 \mathrm{~kg})}{(0.300 \mathrm{~m})^2}$

$F_{31}=6.67 \times 10^{-11} \mathrm{~N}$

The net force has components $F_x=F_{31}$ and $F_y=F_{21}$. Compute the magnitude of this net force:

\begin{aligned}F &=\sqrt{F_x^2+F_y^2}=\sqrt{(6.67)^2+(3.75)^2} \times 10{ }^{-11} \mathrm{~N} \\&=7.65 \times 10^{-11} \mathrm{~N}\end{aligned}

Use the inverse tangent to obtain the direction of $\overrightarrow{\mathbf{F}}$ :

$\theta=\tan ^{-1}\left(\frac{F_y}{F_x}\right)=\tan ^{-1}\left(\frac{3.75 \times 10^{-11} \mathrm{~N}}{6.67 \times 10^{-11} \mathrm{~N}}\right)=29.3^{\circ}$

(b) Find the components of the force of $m_2$ on $m_3$.

First, compute the magnitude of $\overrightarrow{\mathbf{F}}_{23}$ :

$F_{23}=G \frac{m_2 m_1}{r_{23}^2}$

$=\left(6.67 \times 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^3 \mathrm{s}^{-2} \right) \frac{(0.300 \mathrm{~kg})(0.300 \mathrm{~kg})}{(0.500 \mathrm{~m})^2}$

$=2.40 \times 10^{-11} \mathrm{~N}$

To obtain the $x$-and $y$-components of $F_{23}$, we need $\cos \phi$ and $\sin \phi$. Use the sides of the large triangle in Figure 7.17:

\begin{aligned}&\cos \phi=\frac{\text { adj }}{\text { hyp }}=\frac{0.300 \mathrm{~m}}{0.500 \mathrm{~m}}=0.600 \\&\sin \phi=\frac{\text { opp }}{\text { hyp }}=\frac{0.400 \mathrm{~m}}{0.500 \mathrm{~m}}=0.800\end{aligned}

Compute the components of $\overrightarrow{\mathbf{F}}_{23}$. A minus sign must be supplied for the $x$-component because it’s in the negative $x$-direction.

\begin{aligned}F_{23 x} &=-F_{23} \cos \phi=-\left(2.40 \times 10^{-11} \mathrm{~N}\right)(0.600) \\&=-1.44 \times 10^{-11} \mathrm{~N} \\F_{23 y} &=F_{23} \sin \phi=\left(2.40 \times 10^{-11} \mathrm{~N}\right)(0.800)=1.92 \times 10^{-11} \mathrm{~N}\end{aligned}

REMARKS Notice how small the gravity forces are between everyday objects. Nonetheless, such forces can be measured directly with torsion balances.