Question

Two bodies of masses 3kg and and 12 kg are placed at a distance 12 m. A third body of mass 0.5kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point.

Answer 1

Answer:

$\boxed{\mathfrak{Position \ of \ point \ where \ gravitational \ force \ is \ zero = 4 \ m \ from \ 3 kg \ mass}}$

Explanation:

Gravitational force of attraction due to both 3 kg and 12 kg body should be equal and opposite at a point between the two bodies so that gravitational force acting on the body of mass 0.5 kg is zero.

Let the position of that point be x from 3 kg mass.

From the figure attached:

$\rm \implies \dfrac{ \cancel{G} \times 3 \times \cancel{0.5}}{x^2} = \dfrac{ \cancel{G} \times 12 \times \cancel{0.5}}{(12-x)^2} \\ \\ \rm \implies \dfrac{ 3 }{x^2} = \dfrac{ 12 }{(12-x)^2} \\ \\ \rm \implies \dfrac{ (12-x)^2 }{x^2} = \dfrac{ 12 }{3} \\ \\ \rm \implies \dfrac{ (12-x)^2 }{x^2} =4 \\ \\ \rm \implies( \dfrac{ 12-x }{x} )^{ \cancel{2}} = {2}^{ \cancel{2}} \\ \\ \rm \implies \dfrac{ 12-x }{x} =2 \\ \\ \rm \implies 12 - x = 2x \\ \\ \rm \implies 3x = 12 \\ \\ \rm \implies x = \frac{12}{3} \\ \\ \rm \implies x = 4 \: m$

$\therefore$ 0.5 kg body will be at 4 m from 3 kg body and 8 m from 12 kg body so that gravitational force acting on body is zero.

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

Two bodies of masses 3 kg and 12 kg .

Two bodies are placed at a distance 12 m .

A third body of mass 0.5 kg is to be placed at such a point that the force acting on this body is zero .

$\bf{\gray{\underbrace{\blue{TO\: FIND:-}}}}$

ᴛʜᴇ ᴘᴏsɪᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴘᴏɪɴᴛ ᴡʜᴇʀᴇ ᴛʜᴇ ᴛʜɪʀᴅ ʙᴏᴅʏ ᴡᴀs ᴘʟᴀᴄᴇᴅ .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

Let,

Mass of first body = ($\bf\pink{m_1}$) = 3 kg .

Mass of second body = ($\bf\pink{m_2}$) = 12 kg .

Distance between first body and second body = ($\bf\pink{r}$) = 12 m .

Mass of third body ($\bf\pink{m_3}$) = 0.5 kg .

Let,

Distance between first body and third body = $\bf\pink{X\:m}$

Distance between second body and third body = $\bf\pink{(12\:-\:X)\:m}$

We have know that,

$\orange\bigstar\:\bf{\red{\overbrace{\underbrace{\purple{Gravitational\:force\:of\:attraction\:=\:\dfrac{G\:M\:m}{r^2}\:}}}}}$

✨ Hence, force of attraction on $\bf\pink{m_3}$ by $\bf\pink{m_1}$ is,

$\green\bigstar\:\bf{\red{\overbrace{\underbrace{\purple{F_{31}\:=\:\dfrac{G\:m_1\:m_3}{X^2}\:}}}}}$

✨ And force of attraction on $\bf\pink{m_3}$ by $\bf\pink{m_2}$ is,

$\green\bigstar\:\bf{\red{\overbrace{\underbrace{\purple{F_{32}\:=\:\dfrac{G\:m_2\:m_3}{(12\:-\:X)^2}\:}}}}}$

✍️ According to the question, the net force on the third body is zero .

$\bf\orange{\implies\:F_{31}\:=\:F_{32}\:}</p><p>$

$\rm{\implies\:\dfrac{G\:m_1\:m_3}{X^2}\:=\:\dfrac{G\:m_2\:m_3}{(12\:-\:X)^2}\:}$

$\rm{\implies\:\dfrac{m_1}{X^2}\:=\:\dfrac{m_2}{(12\:-\:X)^2}\:}$

$\rm{\implies\:\dfrac{3}{X^2}\:=\:\dfrac{12}{(12\:-\:X)^2}\:}$

$\rm{\implies\:3\times{(12\:-\:X)^2}\:=\:12\times{X^2}\:}$

$\rm{\implies\:144\:+\:X^2\:-\:24X\:=\:4\times{X^2}\:}$

$\rm{\implies\:4X^2\:-\:X^2\:+\:24X\:-\:144\:=\:0\:}$

$\rm{\implies\:3X^2\:+\:24X\:-\:144\:=\:0\:}$

$\rm{\implies\:3X^2\:+\:36X\:-\:12X\:-\:144\:=\:0\:}$

$\rm{\implies\:3X\:(X\:+\:12)\:-\:12\:(X\:+\:12)\:=\:0\:}$

$\rm{\implies\:(X\:+\:12)\:(3X\:-\:12)\:=\:0\:}$

$\rm{\implies\:X\:+\:12\:=\:0\:\:\:\:or\:\:\implies\:3X\:-\:12\:=\:0\:}$

$\rm{\implies\:X\:=\:-12\:\:\:\:or\:\:\implies\:X\:=\:\dfrac{12}{3}\:}$

[NOTE :- Distance never be negative .]

$\bf\green{\implies\:X\:=\:4\:m\:}$ [ from 1st body ]

$\rm{\implies\:12\:-\:X}$

$\rm{\implies\:12\:-\:4}$

$\bf\green{\implies\:8\:m}$ [ from second body ]

$\rm\red{\therefore}$ The position of the point is " 4 m " from the first body and " 8 m " from the second body .