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Q13 P

Expert-verifiedFound in: Page 410

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A $200-kg$ ****object and a localid="1663652122187" $500-kg$****object are separated by $4.00\u200a\u200am.$ ****(a) Find the net gravitational force exerted by these object on a $50.0-kg\u200a\u200a$****object placed midway between them.(b) At what position (other than an infinitely remote one) can the $50.0-kg\u200a\u200a$****object be placed so as to experience a net force of zero from the other two objects.**

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.

${m}_{1}=200\u200a\u200akg\phantom{\rule{0ex}{0ex}}{m}_{2}=500\u200akg\phantom{\rule{0ex}{0ex}}{m}_{3}=50\u200akg$

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 $\u200a{m}_{1}\u200a\u200aand\u200a\u200a\u200a{m}_{2}$ are separated by a distance r, the force exerted by particle 1 on particle 2 is:

${\overrightarrow{F}}_{12}=G\frac{{m}_{1}{m}_{2}\u200a\u200a}{{r}^{2}}\hat{{r}_{12}}$

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

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

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 $\left({m}_{1}\right)\u200a\u200aand\u200a\u200a\left({m}_{3}\right)$which is equal to $\left(2\u200am\right)$ as we see in the figure. Substitute numerical values:

$\overrightarrow{{F}_{13}}=-\frac{\left(6.67\times {10}^{-11}N{m}^{2}/K{g}^{2}\right)\left(200Kg\right)\left(50Kg\right)}{{\left(2m\right)}^{2}}\left(\hat{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}=-1.67\times {10}^{-7}\u200a\mathrm{N}\u200a\u200a\hat{\mathrm{i}}\u200a$

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 $\left(2\u200am\right)$as we see in the figure. Substitute numerical values:

$\overrightarrow{{F}_{13}}=-\frac{\left(6.67\times {10}^{-11}N{m}^{2}/K{g}^{2}\right)\left(500Kg\right)\left(50Kg\right)}{{\left(2m\right)}^{2}}\left(\hat{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}=4.17\times {10}^{-7}\u200a\u200a\u200a\hat{\mathrm{i}}\u200a$

Hence, the net force exerted on $\left({m}_{3}\right)$ is:

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

$\overrightarrow{{F}_{net}}=-\left(1.67\times {10}^{-7}\right)\u200a\hat{i}\u200a+\left(4.17\times {10}^{-7}\u200a\hat{i}\u200a\right)\phantom{\rule{0ex}{0ex}}=\left(2.5\times {10}^{-7}N\right)\hat{i}\u200a$

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 $\left({m}_{1}\right)\u200a\u200aon\u200a\u200a\left({m}_{3}\right)$ is found by Equation (i)

$\overrightarrow{{F}_{13}}=\frac{G{m}_{1}{m}_{3}}{{{r}^{2}}_{13}}\left(-\hat{i}\right)$

Substitute numerical values:

$\overrightarrow{{F}_{13}}=-\frac{\left(1.67\times {10}^{-7}N{m}^{2}/K{g}^{2}\right)\left(200Kg\right)\left(50Kg\right)}{\left({x}^{2}\right)\u200a{m}^{2}}\left(\hat{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}=-\frac{6.67\times {10}^{-7}N}{{x}^{2}}\hat{i}\u200a\u200a$

Second:

The force $\left({\overrightarrow{F}}_{23}\right)$ exerted by ${m}_{2\u200a\u200a}on\u200a\u200a{m}_{3}$ is found by equation (i)

$\overrightarrow{{F}_{23}}=\frac{G{m}_{2}{m}_{3}}{{{r}^{2}}_{23}}\left(\hat{i}\right)$

$\overrightarrow{{F}_{13}}=-\frac{\left(1.67\times {10}^{-7}N{m}^{2}/K{g}^{2}\right)\left(500Kg\right)\left(50Kg\right)}{\left(4-{x}^{2}\right){m}^{2}}\left(\hat{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}=\frac{4.17\times {10}^{-7}\u200aN}{\left(4-{x}^{2}\right)}\u200a\hat{i}\u200a\u200a$

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

$0=-\frac{6.67\times {10}^{-7}\u200a\u200aN}{{x}^{2}}\hat{i}+\frac{4.17\times {10}^{-7}\u200a\u200aN}{\left(4-{x}^{2}\right)}\hat{i}$

Rearrange:

$\frac{6.67\times {10}^{-7}}{{x}^{2}}=\frac{4.17\times {10}^{-7}}{\left(4-{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}\frac{\left(4-{x}^{2}\right)}{{x}^{2}}=\frac{4.17\times {10}^{-7}}{6.67\times {10}^{-7}}$

$\frac{4}{{x}^{2}}-1=\frac{4.17}{6.67}\phantom{\rule{0ex}{0ex}}=0.625\phantom{\rule{0ex}{0ex}}\frac{4}{{x}^{2}}=1.625$

${x}^{2}=\frac{4}{1.625}\phantom{\rule{0ex}{0ex}}=2.46\phantom{\rule{0ex}{0ex}}x=1.55\u200am$

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

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