91Ó°ÊÓ

a) Mass of particle A $=m_A$

b) Mass of particle B, $m_B=2m_A$

c) Mass of particle C, $m_C=3m_A$

d) Mass of particle D, $m_D=4m_A$

e) Distance between particle A and particle C, $r_{AC}=1.5d$

f) Distance between particle A and particle B, $r_{AB}=d$

This problem is based on Newton’s law of gravitation. We can find the force along the x-axis due to particle C and force along the y-axis due to particle B. Using these forces, we can find the x and y coordinates of particle D by equating the x and y component of the force due to particle D.

Formula:

Gravitational force of attraction,$F=\frac{GMm}{r^2}$ (i)

Force between two particles in vector form can be written as

$\stackrel{â‡\P Ä}{F}=F\cos \left(\mathrm{θ}\right)\hat{i}+F\sin \left(\mathrm{θ}\right)\hat{j}$

Using equation (i), Force between A & C can be written as:

.

$$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_{AC}=\frac{G×m_A×3m_A}{{\left(1.5d\right)}^2} \\ =\frac{3G{m}_A^2}{{\left(1.5d\right)}^2}\hat{i} \end{gathered}$$

Negative sign indicates that force is directed along negative x axis.

Using equation (i), Force between A & B can be written as:

role="math" localid="1657257917304" $$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_{AB}=\frac{G×m_A×m_B}{r_{AB}^2} \\ {\stackrel{â‡\P Ä}{F}}_{AB}=\frac{G×m_A×2m_A}{d^2} \\ =\frac{2G{m}_A^2}{d^2}\hat{j} \end{gathered}$$

Similarly using equation (i), the force between particle A and D can be written as: $F_{AD}=\frac{4G{m}_A^2}{r^2}$ (ii)

Force between A and D in vector form can be written as:

${\stackrel{â‡\P Ä}{F}}_{AD}=F_{AD}\cos \left(\mathrm{θ}\right)\hat{I}+F_{AD}\sin \left(\mathrm{θ}\right)\hat{J}$ (iii)

Net force on the particle A should be zero, so we have,

$$\begin{gathered} F_{netA}=0 \\ {F}_{AB}+F_{AC}+F_{AD}=0 \\ \frac{3G{m}_A^2}{{\left(1.5d\right)}^2}\hat{I}+\frac{2G{m}_A^2}{d^2}\hat{j}+{\stackrel{â‡\P Ä}{F}}_{AD}=0 \end{gathered}$$

Hence, ${\stackrel{â‡\P Ä}{F}}_{AD}=\frac{3G{m}_A^2}{{\left(1.5d\right)}^2}\hat{I}+\frac{-2G{m}_A^2}{d^2}\hat{j}$ (iv)

Now equating the components by comparing equation (iii) and (iv), we get

$$\begin{gathered} F_{AD}\cos \left(\mathrm{θ}\right)=\frac{3G{m}_A^2}{{\left(1.5d\right)}^2}..............\left(1\right) \\ {F}_{AD}\sin \left(\mathrm{θ}\right)=\frac{-2G{m}_A^2}{d^2}..............\left(2\right) \end{gathered}$$

Dividing equation (2) by (1)

$$\begin{gathered} \tan \left(\mathrm{θ}\right)=-1.5 \\ \mathrm{θ}={\tan }^{-1}\left(-105\right) \\ =-56.3° \end{gathered}$$ (v)

Now, substituting the values ofandfrom equation (ii) and (v) in equation (1), we get

$\frac{4G{m}_A^2}{r^2}\cos \left(-56.3\right)=\frac{2G{m}_A^2}{1.5d^2}$

By simplifying above:

$r=1.29d$

Hence, x-coordinate of the fourth particle D,

$$\begin{gathered} r_x=r\cos \left(\mathrm{θ}\right) \\ =1.29×d×\cos \left(-56.3\right) \\ =0.716d \end{gathered}$$

The x-coordinate of particle D is $0.716d$

Similarly the y-coordinate of the fourth particle can be written as:

$$\begin{gathered} r_y=r\sin \left(\mathrm{θ}\right) \\ =1.29×d×\sin \left(-56.3\right) \\ =-1.07d \end{gathered}$$

The y-coordinate of particle D is$-1.07d$