Question

Three dimensions. Three point particles are fixed in place in an xyz coordinate system. Particle A, at the origin, has mass mA. Particle B, at xyz coordinates (1.00d, 3.00d, 3.00d), has mass 2.00mA, and particle C, at coordinates (–3.00d, 3.00d, –4.00d), has mass 4.00mA. A fourth particle D, with mass 3.00mA, is to be placed near the other particles. If distance d = 3.60 m, at what (a) x, (b) y, and (c) z coordinate should D be placed so that the net gravitational force on A from B, C, and D is zero?

Answer 1

here we have to request that the vector sum of the forces acting on mass A has to be equal to zero vector, namely:
\[\Large {{\mathbf{F}}_{{\mathbf{A}}{\mathbf{,D}}}} + {{\mathbf{F}}_{{\mathbf{A}}{\mathbf{,B}}}} + {{\mathbf{F}}_{{\mathbf{A}}{\mathbf{,C}}}} = {\mathbf{0}}\]
where:
F_A,D is the force exerted by the particle D on particle A, similarly for F_A,B, and F_A,C

Answer 2

that equation can be rewritten as below:
\[\large - G\frac{{{m_A}{m_D}}}{{r_{A,D}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} - G\frac{{{m_A}{m_B}}}{{r_{A,B}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} - G\frac{{{m_A}{m_C}}}{{r_{A,C}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {\mathbf{0}}\]

Answer 3

where:
\[\Large {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}}\]
is the radius from point A to point D, similarly for r_A,B, and r_A,C
and after a simplification, we get:
\[\Large \frac{{3{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}}}}{{r_{A,D}^3}} + \frac{{2{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}}}}{{r_{A,B}^3}} + \frac{{4{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}}}}{{r_{A,C}^3}} = {\mathbf{0}}\]

Answer 4

now, we have:
\[\Large \begin{gathered}
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {{\mathbf{r}}_{\mathbf{C}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{C}}} \hfill \\
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} = {{\mathbf{r}}_{\mathbf{B}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{B}}} \hfill \\
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} = {{\mathbf{r}}_{\mathbf{D}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{D}}} \hfill \\
\end{gathered} \]

Answer 5

next, from your data, we can write:
\[\Large \begin{gathered}
r_{A,D}^3 = {\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)^3} \hfill \\
r_{A,B}^3 = {d^3}{\left( {\sqrt {19} } \right)^3} \hfill \\
r_{A,C}^3 = {d^3}{\left( {\sqrt {34} } \right)^3} \hfill \\
\end{gathered} \]

Answer 6

and:
\[\Large \begin{gathered}
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {{\mathbf{r}}_{\mathbf{C}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{C}}} = \left( { - 3d,3d, - 4d} \right) \hfill \\
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} = {{\mathbf{r}}_{\mathbf{B}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{B}}} = \left( {d,3d,3d} \right) \hfill \\
{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} = {{\mathbf{r}}_{\mathbf{D}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{D}}} = \left( {x,y,z} \right) \hfill \\
\end{gathered} \]

Answer 7

now you have to substitute those quantities into this equation. After that you should get three equations for x, y and z:
\[\Large \frac{{3{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}}}}{{r_{A,D}^3}} + \frac{{2{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}}}}{{r_{A,B}^3}} + \frac{{4{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}}}}{{r_{A,C}^3}} = {\mathbf{0}}\]

Answer 8

Thank you @Michele_Laino :D

Answer 9

:)