Triangle of forces for a rigid body — equivalent couple: Three non-concurrent forces act on a rigid body and are represented (in magnitude, direction, and line of action) by the three sides of a triangle taken in order. These forces are not in simple translational equilibrium and reduce to a pure couple whose moment is equal to:

Introduction / Context:
This question tests the triangle (polygon) law of forces for a rigid body when lines of action are non-concurrent. When three forces form a closed triangle taken in order, their vector sum is zero; however, if the forces do not meet at a single point, the system is not a null-resultant but a pure couple. The magnitude of this couple connects directly to the geometry (area) of the force triangle.

Given Data / Assumptions:

• Three coplanar forces with specified magnitudes, directions, and lines of action.
• The forces correspond to the three sides of a triangle taken sequentially (head-to-tail), so the polygon closes.
• Forces do not share a common point of concurrency on the rigid body.

Concept / Approach:
For a particle, a closed force polygon implies equilibrium (net force = 0). For a rigid body, additionally all moments must cancel. If lines of action are not concurrent, the translational resultant may be zero while a non-zero free couple remains. Classical statics shows that the moment of this couple equals twice the area of the triangle of forces (with an appropriate force–length scale).

Step-by-Step Solution:
Construct the triangle of forces by placing the three vectors head-to-tail in the given order; the triangle closes, indicating ΣF = 0.Because lines of action are distinct (non-concurrent), translate forces to a common reference and evaluate net moment; a residual couple remains.By geometric/statics proof, the moment of this couple M equals 2 * Δ, where Δ is the signed area of the triangle of forces in the chosen scale.Therefore, the equivalent system is a pure couple of moment equal to twice the triangle’s area.

Verification / Alternative check:
Take any two sides of the triangle as equal-and-opposite forces separated by a perpendicular distance. Their moment equals force * arm. A symmetric construction around the third side recovers M = 2Δ consistently, independent of the chosen pair, confirming uniqueness of the couple.

Why Other Options Are Wrong:

• area of the triangle / half the area: Underestimates the standard result; derivation yields a factor of 2.
• none of these: Incorrect, because a well-known closed-form exists.

Common Pitfalls:

• Confusing particle equilibrium (ΣF = 0) with rigid-body equilibrium (ΣF = 0 and ΣM = 0).
• Assuming concurrency just because the force polygon closes. Concurrency concerns lines of action on the body, not the drawn polygon.

Final Answer:
twice the area of the triangle