anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
that equation can be rewritten as below: \[\large - G\frac{{{m_A}{m_D}}}{{r_{A,D}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} - G\frac{{{m_A}{m_B}}}{{r_{A,B}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} - G\frac{{{m_A}{m_C}}}{{r_{A,C}^3}}{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {\mathbf{0}}\]
Michele_Laino
  • Michele_Laino
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: \[\Large \frac{{3{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}}}}{{r_{A,D}^3}} + \frac{{2{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}}}}{{r_{A,B}^3}} + \frac{{4{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}}}}{{r_{A,C}^3}} = {\mathbf{0}}\]

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Michele_Laino
  • Michele_Laino
now, we have: \[\Large \begin{gathered} {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {{\mathbf{r}}_{\mathbf{C}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{C}}} \hfill \\ {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} = {{\mathbf{r}}_{\mathbf{B}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{B}}} \hfill \\ {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} = {{\mathbf{r}}_{\mathbf{D}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{D}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
next, from your data, we can write: \[\Large \begin{gathered} r_{A,D}^3 = {\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)^3} \hfill \\ r_{A,B}^3 = {d^3}{\left( {\sqrt {19} } \right)^3} \hfill \\ r_{A,C}^3 = {d^3}{\left( {\sqrt {34} } \right)^3} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and: \[\Large \begin{gathered} {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}} = {{\mathbf{r}}_{\mathbf{C}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{C}}} = \left( { - 3d,3d, - 4d} \right) \hfill \\ {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}} = {{\mathbf{r}}_{\mathbf{B}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{B}}} = \left( {d,3d,3d} \right) \hfill \\ {{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}} = {{\mathbf{r}}_{\mathbf{D}}} - {{\mathbf{r}}_{\mathbf{A}}} = {{\mathbf{r}}_{\mathbf{D}}} = \left( {x,y,z} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now you have to substitute those quantities into this equation. After that you should get three equations for x, y and z: \[\Large \frac{{3{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,D}}}}}}{{r_{A,D}^3}} + \frac{{2{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,B}}}}}}{{r_{A,B}^3}} + \frac{{4{{\mathbf{r}}_{{\mathbf{A}}{\mathbf{,C}}}}}}{{r_{A,C}^3}} = {\mathbf{0}}\]
anonymous
  • anonymous
Thank you @Michele_Laino :D
Michele_Laino
  • Michele_Laino
:)

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