Three dimensions. Three point particles are fixed in place | StudySoup

Textbook Solutions for Fundamentals of Physics

Chapter 13 Problem 15

Question

Three dimensions. Three point particles are fixed in place in an xyz coordinate system. Particle A, at the origin, has mass \(m_A\). Particle B, at xyz coordinates (2.00d, 1.00d, 2.00d), has mass \(2.00 m_{A}\), and particle C, at coordinates (-1.00d, 2.00d, -3.00d), has mass \(3.00 m_{A}\). A fourth particle D, with mass \(4.00 m_{A}\), is to be placed near the other particles. In terms of distance d, 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?

Solution

Step 1 of 5

The figure below shows the arrangement of the three particles in the xyz coordinate system.

Here, A, B, and C are the three particles in the xyz coordinate system.

Particle  A is located at the origin of the xyz coordinate system. Using the Pythagorean Theorem, the position of each particle from the origin or point A is as follows.

\(r = \sqrt {{x^2} + {y^2} + {z^2}}\)  

Here, x, y, and z are the components of each particle.

The distance of the particle B from the particle A is,

\({r_B} = \sqrt {{{\left( {2d} \right)}^2} + {{\left( {1d} \right)}^2} + {{\left( {2d} \right)}^2}} \)

\({r_B} = 3d\)

The distance of the particle C from the particle A is,

\({r_C} = \sqrt {{{\left( { - d} \right)}^2} + {{\left( {2d} \right)}^2} + {{\left( { - 3d} \right)}^2}} \)

\({r_C} = \sqrt {14} d\)

The  x component of the force exerted on particle A by the particle B is,

\({\left( {{F_{AB}}} \right)_x} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{x_B}\)

Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A,\( {m_B}\) is the mass of the particle B, and \({x_B}\) is the x component of the position of the particle B.

Substitute 2.00\({m_A}\) for \({m_B}\), 3d  for \({r_B}\), and 2.00d for \({x_B}\) in the above expression, and we get,

\({\left( {{F_{AB}}} \right)_x} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {2.00d} \right)\)

\({\left( {{F_{AB}}} \right)_x} = \frac{{4Gm_A^2}}{{27{d^2}}}\)

The y component of the force exerted on particle A by the particle B is,

\({\left( {{F_{AB}}} \right)_y} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{y_B}\)

Here, \( {y_B}\) is the y component of the position of the particle B.

Substitute 2.00\({m_A}\)  for \({m_B}\), 3d  for \({r_B}\), and 1.00d for \({y_B}\) in the above expression, and we get,

\({\left( {{F_{AB}}} \right)_y} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {1.00d} \right)\)

\({\left( {{F_{AB}}} \right)_y} = \frac{{2Gm_A^2}}{{27{d^2}}}\)

The z component of the force exerted on particle A by the particle B is,

\({\left( {{F_{AB}}} \right)_z} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{z_B}\)

Here, \({z_B}\) is the z component of the position of the particle B.

Substitute 2.00\({m_A}\) for \({m_B}\), 3d  for \({r_B}\), and 2.00d for \({z_B}\) in the above expression, and we get,

\({\left( {{F_{AB}}} \right)_z} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {2.00d} \right)\)

\({\left( {{F_{AB}}} \right)_z} = \frac{{4Gm_A^2}}{{27{d^2}}}\)

 The  x component of the force exerted on particle A by the particle C is,

\({\left( {{F_{AC}}} \right)_x} = \frac{{G{m_A}{m_C}}}{{r_C^3}}{x_C} \)

Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A, \({m_C}\) is the mass of the particle C, and \({x_C}\) is the x component of the position of the particle C.

Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d  for \({r_C}\), and -1.00d for \({x_C}\) in the above expression, and we get,

\({\left( {{F_{AC}}} \right)_x} = \frac{{G{m_A}3.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( { - 1.00d} \right)\)

\({\left( {{F_{AC}}} \right)_x} = \frac{{ - 3\sqrt {14} Gm_A^2}}{{196{d^2}}}\)

The y component of the force exerted on particle A by the particle C is,

\({\left( {{F_{AC}}} \right)_y} = \frac{{G{m_A}{m_C}}}{{r_C^3}}{y_C}\)

Here,  \({y_C}\) is the y component of the position of the particle C.

Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d  for \({r_C}\), and -1.00d for \({y_C}\) in the above expression, and we get,

\({\left( {{F_{AC}}} \right)_y} = \frac{{G{m_A}3.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( {2.00d} \right)\)

\({\left( {{F_{AC}}} \right)_y} = \frac{{3\sqrt {14} Gm_A^2}}{{98{d^2}}}\)

The z component of the force exerted on particle A by the particle C is,

\({\left( {{F_{AC}}} \right)_z} = \frac{{G{m_A}{m_B}}}{{r_C^3}}{z_C}\)

Here, \({z_C}\) is the z component of the position of the particle B.

Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d  for \({r_C}\), and -1.00d for \({z_C}\) in the above expression, and we get,

\({\left( {{F_{AC}}} \right)_z} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( { - 3.00d} \right)\)

\({\left( {{F_{AC}}} \right)_z} = \frac{{ - 9\sqrt {14} Gm_A^2}}{{196{d^2}}}\)

The  x component of the force exerted on particle A by the particle D is,

\(\left( {{F_{AD}}} \right) = \frac{{G{m_A}{m_D}}}{{r_D^2}} \)

Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A,\( {m_D}\) is the mass of the particle D, and \({r_D}\) is the distance between points A and D.

Substitute 4.00\({m_A}\) for \({m_D}\) in the above expression, and we get,

\(\left( {{F_{AD}}} \right) = \frac{{G{m_A}4.00{m_A}}}{{r_D^2}}\)

\(\left( {{F_{AD}}} \right) = \frac{{4Gm_A^2}}{{r_D^2}}\)

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full solution

Title Fundamentals of Physics 10 
Author David Halliday; Robert Resnick; Jearl Walker
ISBN 9781118230725

Three dimensions. Three point particles are fixed in place

Chapter 13 textbook questions

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