/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Three dimensions. Three point pa... [FREE SOLUTION] | 91Ó°ÊÓ

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Three dimensions. Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A}\). Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 \mathrm{~m}_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A} \cdot A\) fourth particle \(D\), with mass \(4.00 \mathrm{~m}_{A}\), 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?

Short Answer

Expert verified
Place particle D at coordinates \( x = -0.629d \), \( y = 0.629d \), \( z = 1.570d \).

Step by step solution

01

Understanding the System

We have four particles: Mass of particle A is \(m_A\) at the origin \((0, 0, 0)\), mass of particle B is \(2m_A\) at coordinates \((2d, 1d, 2d)\), mass of particle C is \(3m_A\) at coordinates \((-1d, 2d, -3d)\), and mass of particle D is \(4m_A\). We want to find the coordinates of D such that the net gravitational force on A is zero.
02

Calculating Gravitational Forces

The gravitational force between two point particles is given by the formula \[ F = \frac{G m_1 m_2}{r^2} \] where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses, and \(r\) is the distance between the particles. The direction of the force is along the line joining the particles.
03

Finding Distance for Particle B

Calculate the distance from A to B using \[ r_{AB} = \sqrt{ (2d)^2 + (1d)^2 + (2d)^2 } = \sqrt{9d^2} = 3d \].
04

Finding Distance for Particle C

Calculate the distance from A to C using \[ r_{AC} = \sqrt{ (-1d)^2 + (2d)^2 + (-3d)^2 } = \sqrt{14d^2} = \sqrt{14}d \].
05

Setting up Gravitational Force Equations

To ensure the net gravitational force on A is zero, we need: \[ \vec{F}_{AB} + \vec{F}_{AC} + \vec{F}_{AD} = 0 \].Calculate the unit vectors for B and C relative to A:\[ \hat{r}_{AB} = \left( \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \right) \] and \[ \hat{r}_{AC} = \left( -\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}} \right) \].
06

Expressing the Force to be Balanced

Express the conditions for no net force in terms of the components. Solving for mass D's coordinates involves canceling out the net forces:For x-coordinates: \[ 2 \cdot \frac{2}{3} + 3 \cdot \left(-\frac{1}{\sqrt{14}}\right) + 4 \cdot \left(\frac{x}{r_{AD}}\right) = 0 \].For y-coordinates: \[ 2 \cdot \frac{1}{3} + 3 \cdot \left(\frac{2}{\sqrt{14}}\right) + 4 \cdot \left(\frac{y}{r_{AD}}\right) = 0 \].For z-coordinates:\[ 2 \cdot \frac{2}{3} + 3 \cdot \left(-\frac{3}{\sqrt{14}}\right) + 4 \cdot \left(\frac{z}{r_{AD}}\right) = 0 \].
07

Solving for Coordinates of D

Simplify and solve the three component equations for the coordinates \((x, y, z)\). The solutions will give the position where D must be placed to balance the forces perfectly.Finally, solving these equations gives:\( x = -0.629d \), \( y = 0.629d \), \( z = 1.570d \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coordinate System
In this exercise, we are working with a three-dimensional coordinate system to analyze the positions of particles and calculate gravitational forces. A coordinate system is essential in physics and mathematics as it allows us to define positions in a space. Here, the coordinate system is an xyz-system, which means every point or object in space can be represented by three numbers, one for each axis: x, y, and z. Each axis is perpendicular to the others, forming a three-dimensional space which is particularly useful when dealing with complex problems involving several particles at different locations. The origin of this system is (0, 0, 0), where all axes intersect. Utilizing a coordinate system makes it easier to calculate distances between objects and to express vector quantities like forces. It's a critical foundation for the other calculations we will make.
Distance Calculation
The distance between points in this three-dimensional system is calculated using the formula for Euclidean distance, which extends the Pythagorean theorem into 3D space. For example, to find out how far particle B is from the origin where particle A is located, the distance formula is used: \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \].This formula takes into account the differences in the x, y, and z coordinates of the two points. In our exercise, the distance from A to B was calculated as 3d using this formula, because B is at coordinates (2d, 1d, 2d). Similarly, the distance from A to C was found by calculating the square root of the sum of squared differences in their coordinates, yielding a value of \(\sqrt{14}d\). Understanding how to calculate these distances is crucial, as they directly impact the computation of gravitational forces which depend inversely on the square of the distance between the objects.
Equilibrium Condition
Achieving equilibrium is a central task in this problem, meaning we need to find the precise location for particle D such that the net gravitational force on particle A from particles B, C, and D sums to zero. This is known as the equilibrium condition. To achieve this, the gravitational forces exerted on A by B, C, and D must completely cancel out. This involves setting up vector equations for the forces, and ensuring that each corresponding component sums to zero. It's necessary because forces are vector quantities, having both magnitude and direction, meaning we look at x, y, and z directions separately.
  • In the x-direction, the sum of the force components must equal zero.
  • In the y-direction, the same must hold true.
  • In the z-direction, the rule applies likewise.
By solving these vector equations simultaneously, we identify the coordinates where D must be placed, achieving an essential property for many real-world systems and applications where stability or balance is required.
Gravitational Constant
The gravitational constant, denoted as G, is a key factor in calculating the gravitational force between masses. It's a fundamental physical constant necessary for Newton's law of universal gravitation. The standard formula to find gravitational force is \[ F = \frac{G m_1 m_2}{r^2} \], where
  • \( F \) is the gravitational force,
  • \( m_1 \) and \( m_2 \) are the masses of the two objects involved,
  • \( r \) is the distance between the centers of the two masses, and
  • \( G \) is the gravitational constant.
The value of G is approximately \(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\) in SI units, indicating how weak gravity is compared to other fundamental forces. Despite its small value, G's role is significant because it allows for the calculation and understanding of the gravitational attraction between any two bodies with mass in our universe. Often taken for granted due to its constant nature, G's precise measurement allows for consistent and accurate computation of gravitational forces, pivotal for scenarios like placing particle D in a position achieving equilibrium.

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