Problem 10 - University Physics with Modern Physics

Question

The point masses \(m\) and 2\(m\) lie along the \(x\) -axis, with \(m\) at the origin and 2\(m\) at \(x=L .\) A third point mass \(M\) is moved along the \(x\) -axis. (a) At what point is the net gravitational force on \(M\) due to the other two masses equal to zero? (b) Sketch the \(x\) -component of the net force on \(M\) due to \(m\) and \(2 m,\) taking quantities to the right as positive. Include the regions \(x < 0,0< x < L,\) and \(x>L .\) Be especially careful to show the behavior of the graph on either side of \(x=0\) and \(x=L\) .

Step-by-Step Solution

Answer

(a) Mass \( M \) at \( x = L(\sqrt{2} - 1) \) has zero net force. (b) Sketch shows zero force at this point, changing signs across \( x=0 \) and \( x=L \).

Step 1: Understanding Gravitational Force

The gravitational force between two masses is given by the formula \( F = \frac{Gm_1m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the two masses, and \( r \) is the distance between them. In this problem, we have three masses: \( m \) at the origin, \( 2m \) at \( x = L \), and \( M \) located at \( x \) along the \( x \)-axis. We need to find where the net gravitational force on \( M \) is zero due to the masses \( m \) and \( 2m \).

Step 2: Equating Gravitational Forces

For the net gravitational force on \( M \) to be zero, the force exerted by mass \( m \) on \( M \) must be equal in magnitude and opposite in direction to the force exerted by mass \( 2m \) on \( M \). Let the position of mass \( M \) be denoted by \( x \). The force due to \( m \) is \( F_1 = \frac{GMm}{x^2} \), directed towards the origin. The force due to \( 2m \) is \( F_2 = \frac{G(2m)M}{(L-x)^2} \), directed towards \( x = L \). Set \( F_1 = F_2 \).

Step 3: Solve for Position x

Setting \( F_1 = F_2 \), we get:\[ \frac{GmM}{x^2} = \frac{G(2m)M}{(L-x)^2} \]Cancel \( GM \) from both sides:\[ \frac{m}{x^2} = \frac{2m}{(L-x)^2} \]Cancel \( m \) from both sides:\[ \frac{1}{x^2} = \frac{2}{(L-x)^2} \]Take the square root of both sides:\[ \frac{1}{x} = \frac{\sqrt{2}}{L-x} \]Rearranging gives:\[ x(\sqrt{2}) = L-x \]\[ x(\sqrt{2} + 1) = L \]\[ x = \frac{L}{\sqrt{2} + 1} \]

Step 4: Simplify the Expression

To simplify \( x = \frac{L}{\sqrt{2} + 1} \):Multiply the numerator and the denominator by the conjugate of the denominator \( \sqrt{2} - 1 \):\[ x = \frac{L(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} \]\[ x = \frac{L(\sqrt{2} - 1)}{2 - 1} \]\[ x = L(\sqrt{2} - 1) \]

Step 5: Sketch the Force Diagram

For part (b), sketch the force versus position diagram for mass \( M \). For \( x < 0 \), the force due to both \( m \) and \( 2m \) is directed towards the respective masses. For \( 0 < x < L \), the force starts positive near the origin and becomes zero at \( x = L(\sqrt{2} - 1) \), then turns negative approaching \( x = L \). For \( x > L \), force due to \( 2m \) dominates, turning the net force positive again. Ensure that the sketch captures these transitions near \( x = 0 \) and \( x = L \).

Key Concepts

Understanding Point Masses

In physics, the concept of point masses simplifies complex systems by assuming that all the mass of an object is concentrated at a single point. This abstraction makes calculations like gravitational forces easier to handle.

In the given exercise, we have three point masses along the x-axis: a mass, \(m\), at the origin, a mass \(2m\) at \(x = L\), and a third mass \(M\) which we want to place at a specific position \(x\) so that it experiences no net gravitational force. This setup helps in analyzing interactions cleanly without considering the objects' dimensions or shapes.

The simplification into point masses is useful in theoretical models, especially when the distances between objects are large compared to their sizes. This approach lets students focus on learning gravitational interactions without extra complexities.

Finding Net Gravitational Force

To determine where the net gravitational force on mass \(M\) is zero, we need to find the point where the forces acting on \(M\) cancel out each other. The gravitational force between two point masses is described by Newton's law of universal gravitation: \( F = \frac{Gm_1m_2}{r^2} \).

In this problem, mass \(M\) is influenced by mass \(m\) at the origin and mass \(2m\) at \(x = L\).

• Force \(F_1\) exerted by \(m\) is \(\frac{GmM}{x^2}\), pulling \(M\) towards the origin.
• Force \(F_2\) from \(2m\) is \(\frac{G(2m)M}{(L-x)^2}\), pulling \(M\) towards \(x = L\).

By setting these forces equal—\(F_1 = F_2\)—and solving the resulting equation, students find the point on the x-axis where \(M\) is in gravitational equilibrium. This exercise reinforces understanding of directional forces and balance.

Interpreting the Force Diagram

A force diagram, also known as a free-body diagram, represents the forces acting on an object, allowing students to visualize how various forces interact. For mass \(M\), the diagram helps to illustrate how the net force changes across different regions on the x-axis.

In the region \(x < 0\), both \(m\) and \(2m\) exert forces in the positive direction (towards themselves). In the region \(0 < x < L\), the force due to \(m\) remains positive as \(M\) moves away from the origin, but decreases in magnitude, eventually reaching zero when \(M\) is at the calculated equilibrium point. Past that, it becomes negative as \(2m\) exerts a stronger influence until \(x=L\). For \(x > L\), \(2m\)'s influence continues to dominate, pulling \(M\) to the right, making the net force positive again.

The force diagram aids in understanding transitions and reinforcing the impact of distances in gravitational interactions.

Approach to Physics Problem Solving

Solving a physics problem involves a structured approach: understanding the scenario, identifying the known and unknown variables, and applying relevant formulas and principles. In this problem, the key is to understand gravitational forces between point masses and translate that into finding equilibrium positions.

Students should follow these steps:

• Recognize the forces involved and their directions.
• Use fundamental equations, like Newton's law of universal gravitation.
• Set up equations to find conditions of equilibrium.
• Solve the equations, keeping track of units and simplifying when possible.

This method helps students not just in this exercise but also in tackling more complex problems. It builds critical thinking and analytical skills necessary for advanced studies in physics.