Final answer:
To transition from a distributed force to point forces at the vertices of a square, you divide the total force by the number of vertices. For non-planar surfaces, normal force varies with surface inclination and weight is resolved into perpendicular and parallel components. The weight of an object can be simplified as a single point force acting at the object's center of gravity.
Step-by-step explanation:
When a flat square with a unit area is positioned at sea level, and experiences a downward force of approximately 105N due to the weight of the air above it (often referred to as atmospheric pressure), this force is distributed evenly across the surface. If you want to equate this to weight loads at its vertices, you would divide the total force by the number of vertices, assuming an even distribution. However, real-world scenarios often involve more complexities such as non-planar surfaces, where the force distribution can change significantly.
If the surface were non-planar, the normal force exerted by the surface would still be perpendicular to the surface at each point, but the value of the normal force might vary according to the local inclination of the surface. When dealing with non-planar surfaces such as inclined planes, the weight of an object can be resolved into two components; one that is perpendicular to the plane (normal force) and one that is parallel to the plane (causing the object to potentially slide down the plane).
The concept of distributed force and point force is illustrated by how we treat the weight of an object as a single point force acting through the center of gravity, regardless of how it is distributed over a contact area. When simplifying distributed forces to point forces for calculations, we assume the distributed force to act at a single point which typically corresponds to the center of gravity of the system.