Graphical equilibrium of concurrent force systems: which condition must be satisfied by the force polygon and funicular polygon for equilibrium?

Introduction / Context:
Graphical statics provides visual conditions for equilibrium. For a system of concurrent forces (all lines of action intersect at a point), the force polygon (vector polygon) represents the head-to-tail addition of all forces.

Given Data / Assumptions:

• Forces are coplanar and concurrent.
• No couples are present.
• Standard graphical construction of the force polygon applies.

Concept / Approach:
Equilibrium requires the vector sum of forces to be zero. Graphically, this is achieved when the force polygon closes (the final vector head meets the initial tail). The funicular (link) polygon is primarily used for non-concurrent systems to locate resultants and reactions; for purely concurrent forces, a funicular polygon is not necessary to demonstrate equilibrium.

Step-by-Step Solution:
Construct the force polygon by placing vectors head-to-tail.If the polygon closes, ΣF = 0 → equilibrium condition satisfied for concurrent forces.Recognize that the funicular polygon is not required for concurrency.

Verification / Alternative check:
Analytical check: resolving forces into components gives ΣF_x = 0 and ΣF_y = 0, equivalent to a closed polygon in the plane.

Why Other Options Are Wrong:

• Requiring a closed funicular polygon applies to non-concurrent systems with moments.
• Allowing an open force polygon contradicts ΣF = 0.

Common Pitfalls:
Confusing conditions for beam/load systems (where funicular polygons help find bending/shear) with concurrent force systems.

Final Answer:
Force polygon should be a closed figure