Let me set up the problem:
- Three identical balls of mass m on a horizontal frictionless surface, touching each other
- A fourth identical ball of mass m rests on top in the void created by the three balls
- All balls have the same radius R
- The system is in equilibrium
Geometry Analysis:
When three identical spheres touch each other on a plane, their centers form an equilateral triangle. The distance between any two centers is 2R.
For the fourth ball resting on top, I need to find the geometry. The center of the top ball will be directly above the centroid of the triangle formed by the three bottom balls.
In an equilateral triangle with side length 2R:
- Distance from center to vertex = (2R)/√3 × (2/3) = (4R)/(3√3) = (4R√3)/9
Wait, let me recalculate this more carefully.
For an equilateral triangle with side length 2R:
- Distance from centroid to vertex = (2R)/√3
The height of the top ball's center above the plane of the bottom balls' centers:
- The top ball touches all three bottom balls
- Distance from top ball's center to any bottom ball's center = 2R
- Using the right triangle: horizontal distance = (2R)/√3, hypotenuse = 2R
- Vertical height h = √[(2R)² - ((2R)/√3)²] = √[4R² - 4R²/3] = √(8R²/3) = (2R√6)/3
Force Analysis:
Let me define:
- F = magnitude of force between adjacent balls on the floor
- N = magnitude of normal force from each floor ball on the top ball
For the top ball (equilibrium in vertical direction):
3N cos α = mg
where α is the angle each contact force makes with the vertical.
From geometry: cos α = h/(2R) = (2R√6/3)/(2R) = √6/3
Therefore: 3N × (√6/3) = mg
N√6 = mg
N = mg/√6
For each floor ball (equilibrium in horizontal direction):
Each floor ball experiences:
- Weight mg (downward)
- Normal reaction from floor (upward) = mg + N cos α
- Contact force N from top ball (at angle α from vertical)
- Two contact forces F from adjacent floor balls
The horizontal component of force from the top ball: N sin α
From geometry: sin α = √(1 - cos²α) = √(1 - 6/9) = √(3/9) = √3/3
Horizontal component from top ball = N sin α = (mg/√6) × (√3/3) = mg√3/(3√6) = mg√3/(3√6) = mg/(3√2)
For horizontal equilibrium of each floor ball:
The net horizontal force from the two adjacent floor balls must balance the horizontal force from the top ball.
Due to symmetry, if F is the magnitude of force between adjacent floor balls, then each ball experiences two forces of magnitude F at 60° to each other. The resultant of these two forces is:
Resultant = 2F cos(30°) = 2F × (√3/2) = F√3
This resultant must balance the horizontal force from the top ball:
F√3 = mg/(3√2)
Therefore: F = mg/(3√6) = mg√6/18
The forces applied by the balls kept on the floor to each other have magnitude F = mg√6/18.
