Four identical balls of mass m each are placed as shown. The three balls that lie on the ground are prevented from moving by applying horizontal forces ( not shown). The reaction force between the fourth ball (placed above) and any of the other three ball is

Solutions:

1. Symmetry:

Due to symmetry, the vertical downward weight of the fourth ball (mg) is equally shared among the three ground balls. Each of the ground balls provides a normal reaction force R to the fourth ball at an angle θ with the horizontal.

2. Forces Acting on the Fourth Ball:

• Gravitational Force: mg, acting vertically downward.
• Reaction Forces: R (from each ground ball), acting along the lines joining the centers of the fourth ball and the respective ground ball.

3. Geometry of the Tetrahedron:

The arrangement forms a regular tetrahedron. The angle θ between the vertical line and the line joining the center of the fourth ball to any ground ball is given by:

cosθ = 1/√3
sinθ = √(2/3)

4. Force Balance (Vertical Direction):

The vertical components of the reaction forces from the three ground balls must balance the weight of the fourth ball:

3R cosθ = mg

Substituting cosθ = 1/√3:

3R (1/√3) = mg
R = (√3 mg) / 3

5. Horizontal Component of Reaction Force:

Each reaction force has a horizontal component Rx = R sinθ. Substituting R = (√3 mg)/3 and sinθ = √(2/3):

Rx = (√3 mg)/3 * √(2/3)
Rx = (√6 mg) / (3√3)

6. Net Reaction Force:

The net reaction force Rnet between the fourth ball and any ground ball is obtained by combining the vertical and horizontal components:

Rnet = √(Rx2 + Ry2)

Here:

Ry = R cosθ = (√3 mg)/3 * (1/√3) = mg/3

Substituting Rx and Ry:

Rnet = √[(√6 mg / 3√3)2 + (mg / 3)2]
Rnet = √[6m2g2/27 + m2g2/9]
Rnet = √[9m2g2/27]
Rnet = mg/√6

Final Answer:

The reaction force between the fourth ball and any one of the three ground balls is:

R = mg/√6