Question

If initially the force on \(m_0\) is \(F_0\). When the positions of \(4m\) and \(3m\) are interchanged, the force becomes \(F'\). If \[ \frac{F_0}{F'}=\frac{\alpha}{\sqrt{5}}, \] find \(\alpha\).

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For force problems involving symmetry:
Resolve forces into perpendicular components
Use proportionality to avoid unnecessary constants
Only differences in opposite forces contribute to the net force

Answer 1

The Correct Option is B

Concept: 
Gravitational force is proportional to the product of masses and inversely proportional to the square of the distance between them. The total gravitational force on a mass due to multiple surrounding masses is the vector sum of individual forces. The formula for gravitational force is: \[ F = G \frac{m_0 m}{r^2} \] where: - \( F \) is the gravitational force, - \( G \) is the gravitational constant, - \( m_0 \) and \( m \) are the masses, - \( r \) is the distance between the masses. 
Step 1: Identify forces acting on \( m_0 \).
Given that four masses \( 4m, 3m, m, 2m \) are placed at the corners of a square with side length \( L \), the distance from the center mass \( m_0 \) to each corner is: \[ r = \frac{L}{\sqrt{2}} \] Using the formula for gravitational force, the force due to each mass \( km \) (where \( k = 4, 3, 2, 1 \)) is: \[ F_k = G \frac{m_0 (km)}{(L/\sqrt{2})^2} = \frac{2 G m_0 m}{L^2} k \] 
Step 2: Compute the initial force \( F_0 \).
In the initial configuration: - The vertical pair of masses are \( 4m \) and \( 2m \), - The horizontal pair are \( 3m \) and \( m \). Now, resolve the forces along the \( x \)- and \( y \)-axes: - Along the \( x \)-axis: \( F_{x0} \propto (3m - m) = 2m \), - Along the \( y \)-axis: \( F_{y0} \propto (4m - 2m) = 2m \). The resultant force \( F_0 \) is the vector sum of these components: \[ F_0 \propto \sqrt{(2m)^2 + (2m)^2} = 2m \sqrt{2} \] 
Step 3: Compute the final force \( F' \) after swapping \( 4m \) and \( 3m \).
After the swap: - The vertical pair consists of \( 3m \) and \( 2m \), - The horizontal pair consists of \( 4m \) and \( m \). Resolve the forces along the \( x \)- and \( y \)-axes: - Along the \( x \)-axis: \( F_{x'} \propto (4m - m) = 3m \), - Along the \( y \)-axis: \( F_{y'} \propto (3m - 2m) = m \). The resultant force \( F' \) is: \[ F' \propto \sqrt{(3m)^2 + (m)^2} = m \sqrt{10} \] 
Step 4: Compute the ratio of forces \( \frac{F_0}{F'} \).
Now, compute the ratio of the initial force \( F_0 \) to the final force \( F' \): \[ \frac{F_0}{F'} = \frac{2m \sqrt{2}}{m \sqrt{10}} = \frac{2 \sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}} \] This matches the form: \[ \frac{F_0}{F'} = \frac{\alpha}{\sqrt{5}} \] Thus, \( \alpha = 2 \). 
Final Answer: \[ \boxed{\alpha = 2}