Problem 9 - University Physics with Modern Physics

Question

A particle of mass 3\(m\) is located 1.00 \(\mathrm{m}\) from a particle of mass \(m\) (a) Where should you put a third mass \(M\) so that the net gravitational force on \(M\) due to the two masses is exactly zero? (b) Is the equilibrium of \(M\) at this point stable or unstable (i) for points along the line connecting \(m\) and \(3 m,\) and (ii) for points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and 3\(m ?\)

Step-by-Step Solution

Answer

Place mass M at x = \(\frac{-1 + \sqrt{3}}{2}\) m from m. It's stable along the line and neutral perpendicularly.

Step 1: Understand the Problem

We need to find a point on the line between the two masses, 3m and m, where a third mass M will feel zero net gravitational force. Additionally, we need to determine the stability of this equilibrium point.

Step 2: Set Up Gravitational Force Equation

Let the distance from mass m to the point where mass M is placed be x meters. Then, the distance from mass 3m to mass M is (1 - x) meters. The gravitational force by m on M is \[ F_1 = \frac{G m M}{x^2} \] and the gravitational force by 3m on M is \[ F_2 = \frac{G (3m) M}{(1-x)^2} \]. Set these two forces equal to balance them: \[ \frac{G m M}{x^2} = \frac{G (3m) M}{(1-x)^2} \].

Step 3: Simplify the Equation

Cancel out the common factors: G and M. You are left with \[ \frac{1}{x^2} = \frac{3}{(1-x)^2} \]. Simplify this equation to find x.

Step 4: Solve for x

Cross-multiply to eliminate fractions:\[ (1-x)^2 = 3x^2 \].Expand and solve the quadratic equation: \[ 1 - 2x + x^2 = 3x^2 \]\[ 1 - 2x = 2x^2 \]\[ 2x^2 + 2x - 1 = 0 \].Use the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where a=2, b=2, and c=-1. Solve for x.

Step 5: Calculate x using the Quadratic Formula

For \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \], calculate inside the square root: \[ 2^2 - 4 \cdot 2 \cdot (-1) = 4 + 8 = 12 \].So, \[ x = \frac{-2 \pm \sqrt{12}}{4} \].Simplify: \[ x = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2} \]. Since x must be between 0 and 1, choose \[ x = \frac{-1 + \sqrt{3}}{2} \].

Step 6: Stability Analysis (i) Along Connecting Line

To test stability along the connecting line, consider moving M slightly from x = \(\frac{-1 + \sqrt{3}}{2}\). A small displacement results in a net force opposite to the displacement, confirming stable equilibrium.

Step 7: Stability Analysis (ii) Perpendicular Line

If M is displaced perpendicularly from the equilibrium position, there is no restoring force since gravitational forces act along the line connecting m and 3m. Hence, M's equilibrium is neutral (neither stable nor unstable) in this direction.

Key Concepts

Gravitational Force

Gravitational force is the attractive force that exists between any two masses. It is calculated using Newton's Universal Law of Gravitation. The formula for gravitational force between two masses, say mass 1 (\(m_1\)) and mass 2 (\(m_2\)) separated by distance (\(r\)) is:

• \(F = \frac{G m_1 m_2}{r^2}\)

where \(G\) is the gravitational constant. The force acts along the line joining the centers of these masses.

Consider a particle of mass \(3m\) and another particle of mass \(m\). If we want to find a point along the line connecting these two particles where a third mass \(M\) experiences zero net gravitational force, we balance the gravitational forces exerted by the two masses on \(M\). This involves setting the gravitational pull due to \(m\) equal to that due to \(3m\) to ensure \(M\) remains stationary. This balance is a practical application of gravitational force in achieving equilibrium positions between bodies.

Stability Analysis

Stability analysis in physics determines whether an object will return to equilibrium after a small displacement. For a third mass \(M\) positioned between two masses \(m\) and \(3m\) at a certain distance, stability is considered along different directions. Let's break this down:

If displaced slightly along the line connecting \(m\) and \(3m\), the gravitational forces exerted by these masses tend to restore \(M\) to its original position. This phenomenon is indicative of stable equilibrium. The restoring force increases the stability of \(M\) in its equilibrium position.

Conversely, if \(M\) is displaced perpendicularly to the line joining \(m\) and \(3m\), it experiences no net force to bring it back. Thus, in this direction, the forces do not act to restore equilibrium, indicating a neutrally stable equilibrium. M simply remains where it is placed once disturbed, avoiding any further motion back or forth.

Quadratic Equation in Physics

Quadratic equations commonly appear in physics to solve problems involving equilibrium, forces, and motion. For this exercise, the equilibrium condition required solving a quadratic equation. Given the gravitational forces, simplifying leads us to a form:

• \( 2x^2 + 2x - 1 = 0 \)

This step involves using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 2 \), and \( c = -1 \).

By computing the discriminant \( b^2 - 4ac \), we find the roots for \(x\). These roots help determine potential positions for \(M\) where the gravitational forces balance. The calculation reflects the power of quadratic equations to define physical scenarios, showing how algebra helps solve real-world physics problems.

Thus, achieving equilibrium requires not only understanding forces but also handling mathematical tools like quadratic equations efficiently.