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 \(x = L\).

Short answer

(a) At \(x = \frac{L}{1 + \sqrt{2}}\), the net gravitational force on \(M\) is zero. (b) Graph force as described.

Step-by-step solution

Define the Problem

We have three point masses: \(m\) at the origin \((x = 0)\), \(2m\) at \(x = L\), and a third mass \(M\) that can be moved along the x-axis. We need to find the position \(x\), where the net gravitational force on \(M\) by the other two masses is zero.

Write the Gravitational Force Equations

The gravitational force exerted by a mass \(m_1\) on mass \(M\) is given by \[ F = G \frac{m_1 M}{r^2} \]where \(G\) is the gravitational constant and \(r\) is the distance between the masses.

Expression for Force from m

The force \(F_m\) from mass \(m\) at the origin on \(M\) at position \(x\) is \[ F_m = G \frac{mM}{x^2} \] It is directed towards the origin.

Expression for Force from 2m

The force \(F_{2m}\) from mass \(2m\) at \(x = L\) on \(M\) at position \(x\) is \[ F_{2m} = G \frac{2mM}{(L-x)^2} \] It is directed towards \(L\).

Set Net Force to Zero

For the net gravitational force on \(M\) to be zero:\[ F_m = F_{2m} \]This becomes:\[ G \frac{mM}{x^2} = G \frac{2mM}{(L-x)^2} \]

Simplify and Solve for x

Cancel \(G\), \(m\), and \(M\) from both sides:\[ \frac{1}{x^2} = \frac{2}{(L-x)^2} \] Taking the square root and solving gives:\[ (L-x)^2 = 2x^2 \] \[ L-x = \sqrt{2}x \] \[ L = x(1 + \sqrt{2}) \] Finally, solving for \(x\) results in:\[ x = \frac{L}{1 + \sqrt{2}} \]

Analyze Regions for Force Behavior

For \(x < 0\), the forces from both masses point to the right (positive), so net force is positive.For \(0 < x < L\), compute as above, with regions closer to a weight exerting stronger force.For \(x > L\), both forces point to the left (negative), causing net force to be negative.

Sketch the x-Component of the Net Force

The net force is positive for \(x < 0\), decreases to zero at \(x = \frac{L}{1 + \sqrt{2}}\), and switches from positive to negative as \(x\) passes this point. It remains negative for \(x > L\). Near \(x=0\), the force is steeply positive; near \(x=L\), it is steeply negative.

Key concepts

Point Masses

In physics, point masses are idealized objects that have mass but occupy no volume. They are treated as if all their mass is concentrated at a single point. This allows for simpler calculations of gravitational forces since the only factor to consider is the distance between these point masses. In our exercise, we are dealing with three point masses: a mass \(m\) at the origin \((x = 0)\), a mass \(2m\) at position \(x = L\), and a third mass \(M\) that can be positioned anywhere along the x-axis. The simplicity of point masses makes them a fundamental premise in many physics problems, particularly when dealing with gravitational force calculations.

X-Axis Analysis

X-axis analysis is crucial when dealing with problems involving movement and forces along a straight line, like in our scenario. Here, the three masses are aligned along the x-axis at given points. The goal is to analyze the impact of gravitational forces as mass \(M\) moves along this axis. The position of mass \(M\) is variable, and our task is to figure out where the forces exerted by the other two masses on \(M\) balance out exactly. This requires examining different positions along the x-axis and calculating the gravitational interactions to find the point of equilibrium. Such analysis often involves identifying regions and points of interest, like where the forces balance to zero. Specifically in this problem, detailed x-axis analysis helps determine the conditions under which the gravitational forces cancel each other out.

Net Force

Net force is the overall force acting on an object when all individual forces are combined. For mass \(M\) in this particular exercise, the net force is the sum of gravitational forces exerted by the point masses \(m\) and \(2m\). To find when the net force on \(M\) is zero, we must find a position where these forces are equal and opposite in direction. This is done by calculating the magnitude and direction of each gravitational force acting on \(M\). If these forces sum to zero, \(M\) will not accelerate, reaching an equilibrium state. Understanding net force concepts aids in predicting the resulting motion or lack of motion of the system in question.

Force Equations

Gravitational forces are calculated using force equations that stem from Newton's law of universal gravitation. The force between two masses is given by the equation \[ F = G \frac{m_1 M}{r^2} \] where \(G\) is the gravitational constant, \(m_1\) and \(M\) are the two masses, and \(r\) is the distance between them. In our task, we use this equation to find the force from both \(m\) and \(2m\) upon \(M\). By writing down these equations, we can set the net force to zero because we seek a balance between \(F_m\) and \(F_{2m}\), resulting in simplified algebraic equations that we solve to find the critical position \(x = \frac{L}{1 + \sqrt{2}}\). These force equations are essential tools that allow us to progress from theoretical physics into practical solutions.