Q15P Three dimensions.聽Three point p... [FREE SOLUTION]

91影视

Chapter 13: Q15P (page 379)

Three dimensions.Three point particles are fixed in place in axyzcoordinate system. ParticleA, at the origin, has mass $m_{A}$ . ParticleB, atxyzcoordinates $(2.00 d, 1.00 d, 2.00 d)$ , has mass $2.00 m_{A}$ , and particleC, at coordinates $(- 1.00 d, 2.00 d, - 3.00 d)$ , has mass $3.00 m_{A}$ . A fourth particleD, with mass $4.00 m_{A}$ , is to be placed near the other particles. In terms of distanced, at what (a)x, (b)y, and (c)zcoordinate shouldDbe placed so that the net gravitational force onAfromB,C, andDis zero?

Short Answer

Expert verified

a) The x-coordinate of particle D is -1.88 d

b) The y-coordinate of particle D is -3.90d

c) The z-coordinate of particle D is 0.489d

Step by step solution

The given data

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

ii) The mass of particle B is $2 m_{A}$ , and its coordinates are (2d,1d,2d)

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

iv) The mass of particle D is $4 m_{A}$

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

Understanding the concept of Newton’s gravitational law

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, $F = \frac{G M m}{r^{2}}$ (i)

a) Calculation of the x-coordinate of Particle D

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

$F_{A B, z} = \frac{G m_{A} m_{B} z_{B}}{{r_{A B}^{}}^{2}} = \frac{G m_{A} (2 m_{A}) (2 d)}{{(\sqrt{{(2 d)}^{2} + d^{2} + {(2 d)}^{2}})}^{3}} = \frac{4 G m_{A}^{2}}{27 d^{2}}$

Similarly, we can write,

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

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

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

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

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

The net force acting on A is

role="math" localid="1657268886350" $F_{n e t} = \sqrt{F x^{2} + F_{y}^{2} + F_{z}^{2}} 0 = [{(F_{A B . x} - F_{A C . x})}^{2} + {(F_{A B . y} - F_{A C . y})}^{2} + {(F_{A B . z} - F_{A C . z})}^{2} + F_{A D}^{2}] F_{A D}^{2} = [{(\frac{4}{27} - \frac{3 \sqrt{14}}{196})}^{2} + {(\frac{2}{27} - \frac{3 \sqrt{14}}{98})}^{2} + {(\frac{4}{27} - \frac{9 \sqrt{14}}{196})}^{2}] (\frac{G m_{A}^{2}}{d^{2}}) {(\frac{G m_{A}^{} m_{D}}{r^{2}})}^{2} = 0.4439 (\frac{G m_{A}^{2}}{d^{2}}) {(\frac{G m_{A}^{} (4 m_{A})}{r^{2}})}^{2} = 0.4439 (\frac{G m_{A}^{2}}{d^{2}}) {(\frac{4}{r^{2}})}^{2} = \frac{0.4439}{d^{4}} r = d {(\frac{16}{0.4439})}^{1 / 4} = 4.357 d Then, the x component of \overset{鈬赌}{r} can be calculated as \frac{G m_{A}^{} m_{D} X}{r^{3}} = - {(\frac{4}{27} - \frac{3 \sqrt{14}}{196})}^{2} \frac{G m_{A}^{2}}{d^{2}} \frac{G m_{A}^{} m_{D} X}{r^{3}} = - 0.0909 \frac{G m_{A}^{2}}{d^{2}} x = - \frac{0.0909 m_{A} r^{3}}{m_{D} d^{2}} = - \frac{0.0909 m_{A_{}} {(4.357 d)}^{3}}{4 m_{A} d^{2}} = - 1.88 d$

The x-component of particle D is -1.88d

b) Calculating the y-component of particle D

Similarly, we can find

$\frac{G m_{A} m_{D} y}{r^{3}} = - {(\frac{2}{27} - \frac{3 \sqrt{14}}{98})}^{2} \frac{G m^{2}}{d^{2}} \frac{G m_{A} m_{D} y}{r^{3}} = - 0.1886 \frac{G m^{2}}{d^{2}} y = - \frac{0.1886 m_{A} r^{3}}{m_{D} d^{2}} = - \frac{0.1886 m_{A} {(4.357 d)}^{3}}{4 m_{A} d^{2}} = - 3.90 d T h e y - c o m p o n e n t o f p a r t i c l e D i s - 3.90 d$

c) Calculating the z-component of the particle D

$\frac{G m_{A} m_{D} z}{r^{3}} = - {(\frac{4}{27} - \frac{9 \sqrt{14}}{196})}^{2} \frac{G m_{A}^{2}}{d^{2}} \frac{G m_{A} m_{D} y}{r^{3}} = - 0.0236 \frac{G m^{2}}{d^{2}} z = - \frac{0.236 m_{A} r^{3}}{m_{D} d^{2}} = - \frac{0.236 m_{A} {(4.357 d)}^{3}}{4 m_{A} d^{2}} = - 0.489 d T h e y - c o m p o n e n t o f p a r t i c l e D i s - 0.489 d$