Varignon's Theorem: Moment Shortcut and Resultant Position

Varignon's theorem states that the moment of a resultant force about any point equals the algebraic sum of the moments of the forces it replaces, taken about the same point. In symbols, M₀(R) = ΣM₀(Fᵢ). A moment is the turning effect of a force about a point: force × perpendicular distance, measured in N·m.

Do not confuse moment with momentum. Momentum is mass × velocity and has nothing to do with this theorem.

Work out what the question is asking before you compute anything:

Everything on this page is two-dimensional. That matches the syllabus course outcome, which asks you to understand "forces, moments and couples in two-dimension force system" (SPPU 2024 pattern syllabus, course CVL107).

Resolve the force into components, multiply each by the distance measured at right angles to it, and add with signs. Each component has an easy moment arm: the vertical component's arm is the horizontal distance, and the horizontal component's arm is the vertical distance. Pairing a force with the wrong distance is the usual error.

The four-step routine:

Worked example. A 100 N force acts at 30° above the horizontal at point P, 4 m to the right of and 3 m above the pivot O.

• Fx = 100 cos 30° = 86.60 N. Fy = 100 sin 30° = 50.00 N.
• Fy, upward at 4 m to the right, turns anticlockwise: +4 × 50 = +200 N·m.
• Fx, rightward at 3 m above O, turns clockwise: −3 × 86.60 = −259.81 N·m.
• M₀ = 200 − 259.81 = −59.81 N·m, so 59.8 N·m clockwise.

The long method agrees. OP is 5 m long, at about 36.87° above the horizontal. The force line is at 30°, so the angle between them is about 6.87°. The perpendicular distance is 5 sin 6.87° ≈ 0.598 m, and 100 × 0.598 ≈ 59.8 N·m. That distance is hard to see on a sketch, which is why splitting the force is quicker.

Add the forces to get R, add their moments about a convenient point, and divide: x̄ = ΣM ÷ R. The result is the distance from that point to the resultant's line of action. This is the second use of the principle of moments, and exams often ask for it.

Worked example. Four vertical forces act on a beam. Take upward as positive, distances x from the left end, and moments about the left end.

Here R means the signed sum ΣFy (upward positive), so R = −1,000 N in this example, a 1,000 N force pointing downward. R × x̄ = ΣM gives (−1,000) × x̄ = −2,800, so x̄ = 2.8 m from the left end.

Second-point check. Take moments about x = 3 m, using (x − 3) × Fy for each force: +900, +500, +400 and −1,600, a total of +200 N·m. The resultant gives (2.8 − 3) × (−1,000) = +200 N·m. The two agree.

Be clear about what this check proves. It passes whenever your addition is right, so it catches arithmetic slips. If you got one force's direction wrong in both the R sum and the moment sum, it still passes, so compare the answer with a sketch as well.

Couple case. If ΣFy is zero but ΣM is not, no single resultant exists. A 100 N downward force at x = 0 and a 100 N upward force at x = 2 m give R = 0 and a net moment of +200 N·m about x = 0. That is a couple. The same moment-sum idea also underlies the centroid formulas, because for a uniform body the weight acts at its centroid (its centre of gravity).

No. Varignon's theorem holds for forces that do not meet at one point (non-concurrent forces), but the one-line proof assumes concurrent forces, meaning forces whose lines of action pass through a single point. For non-concurrent forces, the theorem fixes where the resultant acts (our inference from the moment equation).

Proof to write in an exam. Let F₁ and F₂ meet at point A, let r be the vector from O to A, and let R = F₁ + F₂. Then:

M₀(R) = r × (F₁ + F₂) = r × F₁ + r × F₂ = M₀(F₁) + M₀(F₂)

The cross product distributes over addition, so the same line extends to any number of forces meeting at A (Wikipedia writes it out for N forces).

Link to the Fx, Fy method. Fx and Fy both act at the point of application and add up to F. So M₀(F) = M₀(Fx) + M₀(Fy), which is exactly x·Fy − y·Fx from step 3.

Geometric version. Let θ be the angle between F and the line OA, with OA of length r. Resolve F into a part along OA and a part F⊥ at right angles to it. The part along OA passes through O, so its moment is zero. The perpendicular part has arm r, so the moment is F⊥ × r = F sin θ × r. That equals F × (r sin θ), which is force × perpendicular distance. LibreTexts shows the same equivalence, M = F⊥ d = F d⊥ = F d sin θ.

Why the second-point check always works. Moving the reference point from O to O′ changes the moment of each force Fᵢ by −(OO′ × Fᵢ). The total change is −(OO′ × ΣFᵢ) = −(OO′ × R), which is exactly the change in the moment of R. So the two sides stay equal.

Four slips are common in principle-of-moments numericals (our recommendation from the working above, not exam-marking data). Each can be tested before you submit. Each of these four slips is easy to catch by running the test beside it.

• Pairing a component with its own axis. Fx must be multiplied by y, and Fy by x. Pass if every product uses a distance measured at right angles to that component; fail if any product uses the matching axis.
• Inconsistent signs. Choose anticlockwise positive once. Pass if every product's sign matches the turning direction on your sketch; fail if any one does not.
• Using different points. Every moment, including the resultant's, must be about the same point. Pass if the point is named once at the top of your working and used throughout; fail if any line switches points without saying so.
• Dividing by zero or ignoring a couple. Pass if ΣF is written down and non-zero before you divide; if ΣF = 0, stop and treat it as a couple.

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