91Ó°ÊÓ

i) The mass of particle A is $m_A$, and its coordinates are (0,0,0)

ii) The mass of particle B is $2m_A$, and its coordinates are (2d,1d,2d)

iii) The mass of particle C is$3m_A$, and its coordinates are (-1d,2d,-3d)

iv) The mass of particle D is$4m_A$

v) The net force acting on A is$F_{net}=0$

Using the formula for gravitational force, we can find all components of forces acting on a particle A due to particles B and C. Then, equating the net force acting on particle A to zero, we can find the force acting on A due to particle D. From this, we can its position vector. Then we can write the formulae for x,y, and z components of force acting on D due to A. From this, we will get x, y, and z-coordinates of particle D.

Formula:

The gravitational force of attraction between two bodies of masses M and m separated by distance d is, $F=\frac{GMm}{r^2}$ (i)

Using equation (i), the z-component of the force between A & B

$$\begin{gathered} F_{AB,z}=\frac{G{m}_A{m}_B{z}_B}{{r_{AB}^{}}^2} \\ =\frac{G{m}_A\left(2m_A\right)\left(2d\right)}{{\left(\sqrt{{\left(2d\right)}^2+d^2+{\left(2d\right)}^2}\right)}^3} \\ =\frac{4G{m}_A^2}{27d^2} \end{gathered}$$

Similarly, we can write,

The z-component of the force between A & C,$F_{AC,z}=\frac{-9\sqrt{14}G{m}_A^2}{196d^2}$

The x-component of the force between A & B,role="math" localid="1657267419100" $F_{AB,x}=\frac{4G{m}_A^2}{27d^2}$

The x-component of the force between A & C,role="math" localid="1657267428282" $F_{AC,x}=\frac{-3\sqrt{14}G{m}_A^2}{196d^2}$

The y-component of the force between A & B,$F_{AB,y}=\frac{2G{m}_A^2}{27d^2}$

The y-component of the force between A & C,$F_{AC,y}=\frac{-3\sqrt{14}G{m}_A^2}{98d^2}$

The net force acting on A is

role="math" localid="1657268886350" $$\begin{gathered} F_{net}=\sqrt{F{x}^2+F_y^2+F_z^2} \\ 0=\left[{\left(F_{AB.x}-F_{AC.x}\right)}^2+{\left(F_{AB.y}-F_{AC.y}\right)}^2+{\left(F_{AB.z}-F_{AC.z}\right)}^2+F_{AD}^2\right] \\ {F}_{AD}^2=\left[{\left(\frac{4}{27}-\frac{3\sqrt{14}}{196}\right)}^2+{\left(\frac{2}{27}-\frac{3\sqrt{14}}{98}\right)}^2+{\left(\frac{4}{27}-\frac{9\sqrt{14}}{196}\right)}^2\right]\left(\frac{G{m}_A^2}{d^2}\right) \\ {\left(\frac{G{m}_A^{}{m}_D}{r^2}\right)}^2=0.4439\left(\frac{G{m}_A^2}{d^2}\right) \\ {\left(\frac{G{m}_A^{}\left(4m_A\right)}{r^2}\right)}^2=0.4439\left(\frac{G{m}_A^2}{d^2}\right) \\ {\left(\frac{4}{r^2}\right)}^2=\frac{0.4439}{d^4} \\ r=d{\left(\frac{16}{0.4439}\right)}^{1/4} \\ =4.357d \\ \mathrm{Then},\mathrm{the}\mathrm{x}\mathrm{component}\mathrm{of}\stackrel{â‡\P Ä}{\mathrm{r}}\mathrm{can}\mathrm{be}\mathrm{calculated}\mathrm{as} \\ \frac{G{m}_A^{}{m}_{D}X}{r^3}=-{\left(\frac{4}{27}-\frac{3\sqrt{14}}{196}\right)}^2\frac{G{m}_A^2}{d^2} \\ \frac{G{m}_A^{}{m}_{D}X}{r^3}=-0.0909\frac{G{m}_A^2}{d^2} \\ x=-\frac{0.0909m_A{r}^3}{m_D{d}^2} \\ =-\frac{0.0909m_{A_{}}{\left(4.357d\right)}^3}{4m_A{d}^2} \\ =-1.88d \end{gathered}$$

The x-component of particle D is -1.88d

Similarly, we can find

$$\begin{gathered} \frac{G{m}_A{m}_{D}y}{r^3}=-{\left(\frac{2}{27}-\frac{3\sqrt{14}}{98}\right)}^2\frac{G{m}^2}{d^2} \\ \frac{G{m}_A{m}_{D}y}{r^3}=-0.1886\frac{G{m}^2}{d^2} \\ y=-\frac{0.1886m_A{r}^3}{m_D{d}^2} \\ =-\frac{0.1886m_A{\left(4.357d\right)}^3}{4m_A{d}^2} \\ =-3.90d \\ They-componentofparticleDis-3.90d \end{gathered}$$

$$\begin{gathered} \frac{G{m}_A{m}_{D}z}{r^3}=-{\left(\frac{4}{27}-\frac{9\sqrt{14}}{196}\right)}^2\frac{G{m}_A^2}{d^2} \\ \frac{G{m}_A{m}_{D}y}{r^3}=-0.0236\frac{G{m}^2}{d^2} \\ z=-\frac{0.236m_A{r}^3}{m_D{d}^2} \\ =-\frac{0.236m_A{\left(4.357d\right)}^3}{4m_A{d}^2} \\ =-0.489d \\ They-componentofparticleDis-0.489d \end{gathered}$$