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Expert-verified Found in: Page 410 ### Physics For Scientists & Engineers

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 $4.00 m.$ (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.

a) $\left(2.5×{10}^{-7}N\right)\stackrel{^}{i}$$\left(2.5×{10}^{-7}N\right)\stackrel{^}{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.

See the step by step solution

## Step 1: Given data

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

## Step 2: 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:

${\stackrel{\to }{F}}_{12}=G\frac{{m}_{1}{m}_{2} }{{r}^{2}}\stackrel{^}{{r}_{12}}$

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

$6.674×{10}^{-11}N{m}^{2}/K{g}^{2}$ ## Step 3: (a) Determining the net gravitational force exerted by these objects on an object placed midway between them

Force $\left(\stackrel{\to }{{F}_{13}}\right)$ exerted by $\left({m}_{1}\right)on\left({m}_{3}\right)$ is found by equation (i)

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

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:

Force $\left(\stackrel{\to }{{F}_{23}}\right)$exerted by $\left({m}_{2}\right)on\left({m}_{3}\right)$ is found by equation (i)

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

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:

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

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

$\stackrel{\to }{{F}_{net}}=-\left(1.67×{10}^{-7}\right) \stackrel{^}{i} +\left(4.17×{10}^{-7} \stackrel{^}{i} \right)\phantom{\rule{0ex}{0ex}}=\left(2.5×{10}^{-7}N\right)\stackrel{^}{i}$

## Step 4: (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 $\left({\stackrel{\to }{F}}_{13}\right)$ exerted by $\left({m}_{1}\right) on \left({m}_{3}\right)$ is found by Equation (i)

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

Substitute numerical values:

Second:

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

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

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

$\stackrel{\to }{{F}_{net}}={\stackrel{\to }{F}}_{13}+{\stackrel{\to }{F}}_{23}$

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

Rearrange:

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

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

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

The $50-kg$ object should be at a distance 1.55 m from 200 Kg in order to experience a zero net force. ### Want to see more solutions like these? 