Question

A particle of mass 3\(m\) is located 1.00 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\(m\)?

Short answer

(a) 0.79 meters from 3m, on the line. (b) (i) Unstable; (ii) Neutral.

Step-by-step solution

Understand the Problem

We need to determine the position where a third mass, \(M\), experiences zero net gravitational force due to two other masses: one with mass \(3m\) and the other with mass \(m\), separated by 1 meter.

Set Up the Equation for Gravitational Forces

The force exerted on \(M\) by \(3m\) is \(F_1 = \frac{G \cdot 3m \cdot M}{x^2}\) and the force exerted by \(m\) is \(F_2 = \frac{G \cdot m \cdot M}{(1-x)^2}\), where \(x\) is the distance from the mass \(3m\) to \(M\).

Equate the Gravitational Forces to Find Position

Set the magnitudes of \(F_1\) and \(F_2\) equal: \(\frac{G \cdot 3m \cdot M}{x^2} = \frac{G \cdot m \cdot M}{(1-x)^2}\). Canceling out common terms and solving for \(x\), we obtain: \(3(1-x)^2 = x^2\).

Solve the Equation

Expand \(3(1-x)^2 = x^2\) to get \(3 - 6x + 3x^2 = x^2\). Simplify to \(2x^2 - 6x + 3 = 0\) and solve the quadratic equation, giving \(x = 0.79\, \text{m}\) and \(x = 1.21\, \text{m}\). Only \(x = 0.79\, \text{m}\) is within 0 to 1 meter.

Determine Stability Along the Line

For points along the line connecting \(m\) and \(3m\), if \(M\) is moved slightly, forces will not restore \(M\) back to the equilibrium position (unstable equilibrium).

Determine Stability Perpendicular to the Line

For points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and \(3m\), any movement does not change the gravitational forces from \(m\) and \(3m\) (neutral equilibrium).

Key concepts

Gravitational Force

Gravitational force is a fundamental concept in physics describing the attraction between objects with mass. According to Newton's law of universal gravitation, the force between two masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be expressed mathematically as:

• \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

where:

• \( F \) is the gravitational force,
• \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \ \text{Nm}^2/\text{kg}^2 \),
• \( m_1 \) and \( m_2 \) are the masses,
• \( r \) is the distance between the centers of the two masses.

Gravitational force is always attractive, aiming to bring masses closer. In our problem, we have to balance the forces exerted on a third mass \( M \) by two other masses, \( m \) and \( 3m \), by precisely positioning \( M \) such that the net force becomes zero. This is achieved by setting the gravitational force from \( m \) equal to that from \( 3m \), which involves using this fundamental force equation.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation typically of the form:

• \( ax^2 + bx + c = 0 \)

where:

• \( a \), \( b \), and \( c \) are constants with \( a eq 0 \),
• \( x \) is the variable or unknown.

In the context of the gravitational equilibrium problem, once the gravitational forces are equated, manipulating and simplifying the expressions leads to a quadratic equation, \( 2x^2 - 6x + 3 = 0 \). Quadratic equations can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Solving the quadratic equation helps find the specific position \( x \) where the third mass \( M \) experiences zero net gravitational force. In this scenario, the solution of the quadratic equation gives two results: \( x = 0.79 \ ext{m} \) and \( x = 1.21 \ ext{m} \). Only \( x = 0.79 \ ext{m} \) is a valid solution within the 0 to 1-meter limit, placing \( M \) in equilibrium.

Stability Analysis

Stability analysis is crucial to determine whether an equilibrium position is stable, unstable, or neutrally stable. In mechanical systems, stability often refers to the tendency of a system to return to its equilibrium position after a disturbance.In the given exercise, the equilibrium position of the third mass \( M \) is evaluated for stability:

• **Along the connecting line:** the equilibrium is unstable. If \( M \) is slightly shifted from its equilibrium point along the line connecting \( m \) and \( 3m \), the gravitational forces contribute to further displacing \( M \) rather than bringing it back. Thus, any disturbance will lead to an increased displacement from the equilibrium position.


• **Perpendicular to the connecting line:** the analysis shows neutral equilibrium. Here, if \( M \) is displaced sideways (perpendicular to the connecting line), the net gravitational forces from both masses \( m \) and \( 3m \) remain unchanged, as they act along the line connecting the masses. Hence, \( M \) does not experience any restorative or displacing forces, maintaining neutral stability.

Stability analysis provides insights into how systems respond to various disturbances, highlighting potential vulnerabilities and assisting in predicting system behaviors under different scenarios.