To solve this problem, we need to analyze the forces acting on the system of hemispheres and the sphere placed on top. The setup consists of three solid hemispheres arranged in a triangular formation, with their flat faces resting on a rough horizontal surface. A sphere of mass 'm' and radius 'R' is positioned symmetrically on top of these hemispheres. Our goal is to determine the normal reaction force between the top sphere and any one of the hemispheres while the system is in equilibrium.

Understanding the Forces at Play

In this scenario, we have several forces acting on the top sphere:

• Weight of the Sphere: The weight of the sphere (W) is given by the formula W = mg, where 'g' is the acceleration due to gravity.
• Normal Reaction Force: Each hemisphere exerts a normal reaction force (N) on the sphere. Since the sphere is symmetrically placed, the normal forces from the three hemispheres will be equal.

Equilibrium Condition

For the system to be in equilibrium, the sum of the vertical forces acting on the sphere must equal zero. This can be expressed mathematically as:

Sum of upward forces = Sum of downward forces

In our case, the only downward force is the weight of the sphere (mg), while the upward forces are the normal reactions from the three hemispheres. Since the sphere is symmetrically placed, we can denote the normal reaction from each hemisphere as N.

Calculating the Normal Reaction Force

Since there are three hemispheres, the total upward force can be expressed as:

3N = mg

From this equation, we can solve for the normal reaction force (N):

N = mg / 3

Final Thoughts

Thus, the normal reaction between the top sphere and any one of the hemispheres, assuming the system is in equilibrium, is:

N = mg / 3

This result shows that the weight of the sphere is evenly distributed among the three hemispheres, leading to each hemisphere supporting one-third of the total weight. This principle of equilibrium and distribution of forces is fundamental in understanding how objects interact in a static system.