Book edition 9th Edition

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

Pages 1624 pages

ISBN 9781133947271

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Short Answer

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

a) $(2.5 \times 10^{- 7} N) \hat{i}$ $(2.5 \times 10^{- 7} N) \hat{i}$

b) The $50 - k g$ object should be at a distance 1.55 m from the $200 - k g$ in order to experience a zero net force.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

$m_{1} = 200 k g m_{2} = 500 k g m_{3} = 50 k g$

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 $m_{1} a n d m_{2}$ are separated by a distance r, the force exerted by particle 1 on particle 2 is:

${\vec{F}}_{12} = G \frac{m_{1} m_{2}}{r^{2}} \hat{r_{12}}$

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

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

Step 3: (a) Determining the net gravitational force exerted by these objects on an object placed midway between them

Force $(\vec{F_{13}})$ exerted by $(m_{1}) o n (m_{3})$ is found by equation (i)

$\vec{F_{13}} = \frac{G m_{1} m_{3}}{{r^{2}}_{13}} (- \hat{i})$

The $(r_{13})$ is the distance between $(m_{1}) a n d (m_{3})$ which is equal to $(2 m)$ as we see in the figure. Substitute numerical values:

$\vec{F_{13}} = - \frac{(6.67 \times 10^{- 11} N m^{2} / K g^{2}) (200 K g) (50 K g)}{{(2 m)}^{2}} (\hat{i}) = - 1.67 \times 10^{- 7} N \hat{i}$

Force $(\vec{F_{23}})$ exerted by $(m_{2}) o n (m_{3})$ is found by equation (i)

$\vec{F_{23}} = \frac{G m_{2} m_{3}}{{r^{2}}_{23}} (\hat{i})$

The $(r_{23})$ is the distance between $(m_{2}) o n (m_{3})$ , which is equal to $(2 m)$ as we see in the figure. Substitute numerical values:

$\vec{F_{13}} = - \frac{(6.67 \times 10^{- 11} N m^{2} / K g^{2}) (500 K g) (50 K g)}{{(2 m)}^{2}} (\hat{i}) = 4.17 \times 10^{- 7} \hat{i}$

Hence, the net force exerted on $(m_{3})$ is:

$(\vec{F_{n e t}}) = {\vec{F}}_{13} + {\vec{F}}_{23}$

$\vec{F_{n e t}} = - (1.67 \times 10^{- 7}) \hat{i} + (4.17 \times 10^{- 7} \hat{i}) = (2.5 \times 10^{- 7} N) \hat{i}$

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 $(m_{1})$ Hence, $m_{3}$ is at a distance $r_{23} = (4 - x)$ from $m_{2}$

First:

The force $({\vec{F}}_{13})$ exerted by $(m_{1}) o n (m_{3})$ is found by Equation (i)

$\vec{F_{13}} = \frac{G m_{1} m_{3}}{{r^{2}}_{13}} (- \hat{i})$

Substitute numerical values:

$\vec{F_{13}} = - \frac{(1.67 \times 10^{- 7} N m^{2} / K g^{2}) (200 K g) (50 K g)}{(x^{2}) m^{2}} (\hat{i}) = - \frac{6.67 \times 10^{- 7} N}{x^{2}} \hat{i}$

Second:

The force $({\vec{F}}_{23})$ exerted by $m_{2} o n m_{3}$ is found by equation (i)

$\vec{F_{23}} = \frac{G m_{2} m_{3}}{{r^{2}}_{23}} (\hat{i})$

$\vec{F_{13}} = - \frac{(1.67 \times 10^{- 7} N m^{2} / K g^{2}) (500 K g) (50 K g)}{(4 - x^{2}) m^{2}} (\hat{i}) = \frac{4.17 \times 10^{- 7} N}{(4 - x^{2})} \hat{i}$

Hence, the net force exerted on $(m_{3})$ is:

$\vec{F_{n e t}} = {\vec{F}}_{13} + {\vec{F}}_{23}$

Substitute $(F_{n e t} = 0)$ as required in the problem

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

Rearrange:

$\frac{6.67 \times 10^{- 7}}{x^{2}} = \frac{4.17 \times 10^{- 7}}{(4 - x^{2})} \frac{(4 - x^{2})}{x^{2}} = \frac{4.17 \times 10^{- 7}}{6.67 \times 10^{- 7}}$

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

$x^{2} = \frac{4}{1.625} = 2.46 x = 1.55 m$

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

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Definition and concept of Newton’s Law of Gravitation

Step 3: (a) Determining the net gravitational force exerted by these objects on an object placed midway between them

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.

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Definition and concept of Newton’s Law of Gravitation

Step 3: (a) Determining the net gravitational force exerted by these objects on an object placed midway between them

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.

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