Three-force equilibrium on a rigid body – necessary geometric condition Three non-parallel forces acting on a rigid body keep it in equilibrium. Besides being coplanar, what must be true of their lines of action?

Introduction / Context:
The three-force rule is central in statics for frames, arches, and linkages. When exactly three forces act on a rigid body in equilibrium, their geometric relation must satisfy both force and moment balance simultaneously.

Given Data / Assumptions:

• Exactly three forces act; the body is rigid.
• Forces are coplanar and not all parallel.
• No couple moments applied externally.

Concept / Approach:

Equilibrium requires ΣF = 0 and ΣM = 0. With three non-parallel forces, if their lines of action are not concurrent, a net moment remains. The only way for both force and moment sums to vanish is for the three lines of action to intersect at a common point (or be parallel with magnitudes balancing, the special case excluded here).

Step-by-Step Solution:

Verification / Alternative check:

Graphical statics confirms: draw the force polygon closed; concurrency of lines of action follows from the funicular polygon closing to a point pole.

Why Other Options Are Wrong:

(b) All-parallel is a different special case. (c) Concurrent and parallel together is contradictory unless all forces are the same line. (d) Skew lines violate coplanarity. (e) Magnitude triangle alone does not ensure zero moment unless concurrency holds.

Common Pitfalls:

Forgetting the moment equilibrium; assuming a force triangle is sufficient without checking concurrency of lines of action.

Final Answer:

They must be concurrent (intersect at a single point)