Two dimensions.In the figure, three point particles are fixed in place in anx-yplane. ParticleAhas mass${\mathit{m}}_{\mathbf{A}}$ , particleBhas mass$\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{m}}_{\mathbf{A}}$, and particleChas mass$\mathbf{3}\mathbf{.}\mathbf{00}{\mathit{m}}_{\mathbf{A}}$. A fourth particleD, with mass$\mathbf{4}\mathbf{.}\mathbf{00}{\mathit{m}}_{\mathbf{A}}$, is to be placed near the other three particles. In terms of distanced, at what (a)xcoordinate and (b)ycoordinateshould particle Dbe placed so that the net gravitational force on particle Afrom particles B, C, and Dis zero?

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

$\stackrel{\rightharpoonup }{F}=F\cos \left(\theta \right)\hat{i}+F\sin \left(\theta \right)\hat{j}$

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

.

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

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" $$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{AB}=\frac{G\times m_A\times m_B}{r_{AB}^2} \\ {\stackrel{\rightharpoonup }{F}}_{AB}=\frac{G\times m_A\times 2m_A}{d^2} \\ =\frac{2G{m}_A^2}{d^2}\hat{j} \end{gathered}$$

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:

${\stackrel{\rightharpoonup }{F}}_{AD}=F_{AD}\cos \left(\theta \right)\hat{I}+F_{AD}\sin \left(\theta \right)\hat{J}$ (iii)

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

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

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

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

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

Dividing equation (2) by (1)

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

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

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

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

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

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