In the given situation, the force at the center on 1 kg mass is \( F_1 \). Now if \( 4m \) and \( 3m \) are interchanged, the force is \( F_2 \). Given \( \frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}} \), find \( \alpha \).
Show Hint
When forces depend on distance, changing the distance between masses can significantly affect the force, which is inversely proportional to the square of the distance.
To find the value of α, we are given the relation:
F₁ / F₂ = 2 / √α
A 1 kg mass is placed at the center of a square, and masses are placed at the four
corners of the square. The gravitational force acting on the central mass due to
each corner mass acts along the line joining that corner to the center.
Step 1: Force due to a single corner mass
The gravitational force between two masses separated by a distance r is:
F = G × m₁ × m₂ / r²
Here, the distance r from each corner to the center is the same for all masses.
Thus, the magnitude of force due to each corner mass is directly proportional
to the mass placed at that corner.
Step 2: Calculation of F₁
In the first arrangement, the masses at the corners are:
4m, 3m, m, and 2m
The forces due to these masses act along different directions, and the net force
on the central mass is obtained by vector addition of these individual forces.
Step 3: Calculation of F₂
In the second arrangement, the masses 4m and 3m are interchanged.
Since the distances remain unchanged, only the directions of the forces associated
with these two masses are swapped, which alters the resultant force on the central mass.
Step 4: Ratio of forces
Using symmetry and vector addition of forces for both configurations, the ratio
of the magnitudes of the resultant forces is given as:
F₁ / F₂ = 2 / √α
Comparing the numerical values obtained from the vector sums, this condition is
satisfied when:
