A 500 N force acts at a 35° angle at the end of a 2 m structural member. The perpendicular distance from point O to the force's line of action is geometrically complex to compute. Applying Varignon's theorem, you decompose the force into Fx and Fy at the end of the member. Where must these component moments be calculated?
AFx about one convenient point and Fy about a different convenient point, to simplify each calculation independently
BBoth Fx and Fy must be calculated about the same reference point O
CFx and Fy about the midpoint of the member, then doubled to account for the full length
DAbout any two points, then averaged — the average equals the moment about O
Varignon's theorem states: M_O = sum of moments of components ABOUT THE SAME POINT O. Using different reference points for different components is the most common error in applying the theorem and produces a meaningless result. The theorem follows from the distributive property r × (Fx + Fy) = r × Fx + r × Fy — the same position vector r (pointing from O to the point of force application) appears in every term. The reference point must be fixed throughout the calculation.
Question 2 Multiple Choice
Varignon's theorem follows from which mathematical property of the cross product?
AThe commutative property: A × B = B × A
BThe distributive property: r × (F1 + F2) = r × F1 + r × F2
CThe associative property: (r × F) × G = r × (F × G)
DThe magnitude property: |r × F| = |r||F|sin θ
Varignon's theorem is a direct consequence of the distributive property of the cross product over vector addition. If a force F is decomposed into components F1 + F2, then M_O = r × F = r × (F1 + F2) = r × F1 + r × F2 = M1 + M2. The total moment equals the sum of the component moments — the entire theorem is just this one algebraic step. The commutative, associative, and magnitude properties are all true but don't yield Varignon's theorem.
Question 3 True / False
When applying Varignon's theorem, all component moments must be computed about the same reference point O.
TTrue
FFalse
Answer: True
This is the non-negotiable constraint of Varignon's theorem. The theorem states that M_O (moment about a specific point O) equals the sum of the component moments — all about that same point O. The derivation shows this explicitly: r × (F1 + F2) distributes over addition using the same r vector throughout, so all moments share the same reference point. Computing one component's moment about a different point corrupts the calculation entirely, producing a result that is neither M_O nor any other physically meaningful moment.
Question 4 True / False
To find the moment of a force using Varignon's theorem, you add the magnitudes of the force components and multiply by the perpendicular distance from O to the force's line of action.
TTrue
FFalse
Answer: False
This confuses the original moment formula (Fd⊥) with how Varignon's theorem actually works. The theorem does NOT combine force magnitudes — it computes each component's moment separately using its own lever arm (the perpendicular distance from O to that component's line of action) and then sums those moments with correct signs. Horizontal component Fx has lever arm equal to the y-coordinate of the point of application; vertical component Fy has lever arm equal to the x-coordinate. Adding force magnitudes and using a single distance is not part of the procedure.
Question 5 Short Answer
Explain why Varignon's theorem is useful in practice — what computational difficulty does it eliminate, and what constraint must be maintained?
Think about your answer, then reveal below.
Model answer: In direct moment calculation, you need the perpendicular distance d⊥ from the reference point O to the force's line of action — which often requires trigonometry that is awkward when the force acts at an angle to irregular geometry. Varignon's theorem eliminates this by decomposing the force into horizontal and vertical components; each component's perpendicular distance to O is simply a coordinate (the y-coordinate for a horizontal force, the x-coordinate for a vertical force). These distances are easy to read off from the geometry. The constraint: all component moments must be computed about the same reference point O, and correct signs (counterclockwise positive, for example) must be applied to each component's moment contribution.
The practical payoff is significant in truss and frame problems. If you choose O at a pin joint through which the force passes, the component's lever arm for that force vanishes — reducing the calculation to a single multiplication. This 'point at the pin' strategy appears constantly in statics and is only possible because Varignon's theorem lets you freely decompose the force and choose a convenient evaluation point, as long as the same O is used throughout.