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Chapter 4: Problem 37

Two particles are located on the \(x\) axis. Particle 1 has a mass \(m\) and is at the origin. Particle 2 has a mass \(2 m\) and is at \(x=+L\). A third particle is placed between particles 1 and \(2 .\) Where on the \(x\) axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? Express your answer in terms of \(L\).

Short Answer

Expert verified
Place the third particle at \( x = \frac{L}{\sqrt{2}} \).

Step by step solution

01

Identify existing gravitational forces

First, calculate the initial gravitational force between Particle 1 and Particle 2. The gravitational force is given by Newton's law of universal gravitation: \[F = G \frac{m_1 m_2}{r^2}\]where \(G\) is the gravitational constant, \(m_1 = m\), \(m_2 = 2m\), and \(r = L\). Substituting these values, we get:\[F_{12} = G \frac{m (2m)}{L^2} = \frac{2G m^2}{L^2}\].
02

Determine the new gravitational forces

Now place a third particle with a mass \( m_3 \) at a distance \( x \) from the origin (Particle 1) so that the force of attraction between Particle 1 and this third particle plus the force between Particle 1 and Particle 2 equals twice the initial force. Similarly, do the same for Particle 2. The problem requires that these forces double, meaning:1. For Particle 1:\[ F_{13} + F_{12} = 2F_{12} \ F_{13} = 2F_{12} - F_{12} = F_{12} = \frac{2G m^2}{L^2} \]2. For Particle 2:\[ F_{23} + F_{12} = 2F_{12} \ F_{23} = 2F_{12} - F_{12} = F_{12} = \frac{2G m^2}{L^2} \]
03

Set up equations for the third particle

Now find \( F_{13} \) and \( F_{23} \) in terms of \( x \) and \( m_3 \). Let the mass of the third particle be \( m_3 = m \) (for simplicity), and its location be \( x \). The gravitational force the third particle exerts on Particle 1 is:\[F_{13} = G \frac{m_1 m_3}{x^2} = G \frac{m^2}{x^2}\]And the force it exerts on Particle 2 is:\[F_{23} = G \frac{m_2 m_3}{(L-x)^2} = G \frac{2m^2}{(L-x)^2}\]
04

Solve for the location of the third particle

From the conditions:1. \( F_{13} = \frac{2G m^2}{L^2} \) implies:\[\frac{G m^2}{x^2} = \frac{2G m^2}{L^2} \Rightarrow \frac{1}{x^2} = \frac{2}{L^2} \Rightarrow x^2 = \frac{L^2}{2} \Rightarrow x = \frac{L}{\sqrt{2}}\]2. \( F_{23} = \frac{2G m^2}{L^2} \) gives:\[\frac{2G m^2}{(L-x)^2} = \frac{2G m^2}{L^2} \Rightarrow \frac{1}{(L-x)^2} = \frac{1}{L^2} \Rightarrow L-x = L \Rightarrow x=0\] (not possible).So the third particle should only be placed at \( x = \frac{L}{\sqrt{2}} \) based on the condition for doubling the force on Particle 1.

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Key Concepts

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

Gravitational Force
Newton's Law of Universal Gravitation describes the gravitational force as an attractive force between any two masses in the universe. It can be expressed as: \[F = G \frac{m_1 m_2}{r^2}\] where:
  • \( G \) is the gravitational constant, which has a value of approximately \( 6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}} \)
  • \( m_1 \) and \( m_2 \) are the masses of the two objects.
  • \( r \) is the distance between the centers of the two masses.
In our problem, the gravitational force between any two particles depends on their mass and how far apart they are. For instance, the initial gravitational force between Particle 1 and Particle 2 is defined by their masses \( m \) and \( 2m \), and the distance \( L \) along the x-axis. This force is calculated as: \[F_{12} = G \frac{m (2m)}{L^2} = \frac{2G m^2}{L^2}\] It's important to note that gravitational forces are always attractive, drawing objects together rather than pushing them apart.
Particle Mass
The mass of a particle is a measure of its matter. In physics, mass plays a critical role in calculating forces, including gravitational ones. In the given problem, we have three particles.
  • Particle 1 has a mass \( m \), located at the origin \(x = 0\).
  • Particle 2 has a mass \(2m\), positioned at \(x = L\).
  • A third particle, with mass chosen as \( m \) (for simplicity), is introduced between the two, our task is to find its most effective position based on balancing gravitational forces.

Understanding these mass values helps in determining the gravitational interaction between particles. The force each particle feels is directly proportional to its mass, meaning that a more massive particle will experience a stronger gravitational pull if all other factors remain constant. This property of mass indicates why it鈥檚 pivotal to calculations involving Newton's Law of Universal Gravitation.
x-axis Location
The position of a particle along the x-axis, or its x-axis location, specifies where a particle sits in a defined space, crucial in this problem. The given challenge is to locate the third particle such that the gravitational forces on the existing particles double.
Initially, Particle 1 is located at the origin \(x = 0\) while Particle 2 is at \(x = L\). The key is to insert a third particle between these points in a manner that evenly influences gravitational force.
The solution involves determining a spot on this axis, represented as \( x = \frac{L}{\sqrt{2}} \), ensuring that the gravitational impact the third particle adds, aligns correctly with the goal of doubling force magnitude. This value of \( x \) optimizes the forces between particles, as derived from solving equations for the new forces introduced by the third particle. It signifies the importance of precise positioning in gravitational scenarios, highlighting the balance needed for specified force changes.

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Most popular questions from this chapter

Scientists are experimenting with a kind of gun that may eventually be used to fi re payloads directly into orbit. In one test, this gun accelerates a 5.0-kg projectile from rest to a speed of \(4.0 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The net force accelerating the projectile is \(4.9 \times 10^{5} \mathrm{N} .\) How much time is required for the projectile to come up to speed?

In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to \(45 \mathrm{m} / \mathrm{s}\) (about \(100 \mathrm{mi} / \mathrm{h}\) ) in a time of 7.0 s. The combined mass of the car and riders is \(5.5 \times 10^{3} \mathrm{kg}\). Find the average net force exerted on the car and riders by the magnets.

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 N. The coefficient of static friction between his hands and the book is 0.40. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

The speed of a bobsled is increasing because it has an acceleration of \(2.4 \mathrm{m} / \mathrm{s}^{2} .\) At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 450N. The combined mass of the bobsled and its riders is \(270 \mathrm{kg}\) (a) What is the magnitude of the force propelling the bobsled forward? (b) What is the magnitude of the net force that acts on the bobsled?

Two horizontal forces, \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2},\) are acting on a box, but only \(\overrightarrow{\mathbf{F}}_{1}\) is shown in the drawing. \(\overrightarrow{\mathbf{F}}_{2}\) can point either to the right or to the left. The box moves only along the \(x\) axis. There is no friction between the box and the surface. Suppose that \(\overrightarrow{\mathbf{F}}_{1}=+9.0 \mathrm{N}\) and the mass of the box is \(3.0 \mathrm{kg}\). Find the magnitude and direction of \(\overrightarrow{\mathbf{F}}_{2}\) when the acceleration of the box is (a) \(+5.0 \mathrm{m} / \mathrm{s}^{2},\) (b) \(-5.0 \mathrm{m} / \mathrm{s}^{2},\) and (c) \(0 \mathrm{m} / \mathrm{s}^{2}.\)
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