91Ó°ÊÓ

Equilibrium is a state in which all forces acting on an object are balanced, so the object remains at rest or moves with a constant velocity. For an object in equilibrium under gravitational forces, the sums of gravitational attractions in all directions must cancel out. Here, we explore a situation where we need to place a third mass, labeled as \(M\), in such a position that the net gravitational force on it is zero when influenced by two other masses, \(m\) and \(3m\).

This involves calculating where to position \(M\) along the line between the two masses such that the attractions pulling from each mass are equal. This leads to a balanced state or equilibrium. By setting the gravitational forces equal and solving for \(x\), the position of \(M\), we can find points that provide the desired zero net force.

Stable and unstable equilibrium are conditions based on how an object reacts to small displacements.

• **Stable Equilibrium**: This occurs when an object returns to its initial position after being slightly displaced. For mass \(M\) closer to the larger mass \(3m\), this situation is present because any small shift is corrected by a stronger gravitational pull that returns \(M\) to the equilibrium point.


• **Unstable Equilibrium**: This arises when a slight displacement causes the object to move further away from its initial position. In the case of \(M\) placed closer to the smaller mass \(m\), a small displacement increases the imbalance in forces, pushing \(M\) away from equilibrium.

For lines perpendicular to \(m\) and \(3m\), equilibrium is generally unstable since any displacement causes \(M\) to shift towards either \(m\) or \(3m\), disrupting balance.

A quadratic equation takes the form \(ax^2 + bx + c = 0\) and is used to find values that satisfy a specific condition. In the context of gravitational forces, a quadratic equation helps locate the balance points where net forces are zero. By substituting our situation into \( x^2 = 3(1.00-x)^2 \), we derive a quadratic that needs solving.

To find \( x \), we expand and rearrange the equation to get it in standard quadratic form, and then solve by factoring or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This provides us with the required positions along the line, helping assess equilibrium points for mass \(M\) with respect to \(m\) and \(3m\).

Newton's Law of Universal Gravitation is fundamental in understanding the attraction between two masses. It states that every point mass attracts every other point mass by a force acting along the line intersecting both points.

The equation representing this law is given by \(F = G \frac{m_1 m_2}{r^2}\), where:

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

This law helps calculate the specific distances where the gravitational forces exerted on a third mass \(M\) by two other masses cancel out, leading to equilibrium. Employing this concept is essential for determining the zero-net force position of \(M\), thereby deriving equilibrium solutions.