In one line
Varignon's theorem says the moment of the resultant about any point equals the algebraic sum of the moments of its component forces about the same point.
Explained simply
If you split a pull into two parts, the total turning effect of the two parts must equal the turning effect of the original pull. Nothing is gained or lost just by splitting it. Varignon's theorem is this common-sense idea written precisely.
Statement: for concurrent forces \( F_1, F_2, \ldots \) with resultant R, \( M_O(R) = \Sigma M_O(F_i) \) about any point O in the plane. It is also called the principle of moments.
How it works
- Proof (two forces): let P and Q meet at A with resultant R, and let O be any point in their plane. Take the x-axis along AO, and let P, Q, R make angles \( \alpha, \beta, \theta \) with it.
- Only the component perpendicular to AO creates a moment about O; with \( AO = d \): \( M_O(P) = P\sin\alpha \cdot d \), \( M_O(Q) = Q\sin\beta \cdot d \), \( M_O(R) = R\sin\theta \cdot d \).
- The y-component of R equals the sum of the y-components: \( R\sin\theta = P\sin\alpha + Q\sin\beta \).
- Multiplying by d gives \( M_O(R) = M_O(P) + M_O(Q) \). The argument extends to any number of forces.
- Use: find the moment of an inclined force by adding moments of its x and y components, and locate the resultant: \( R \cdot d = \Sigma M_O \).
Key formula
\( R \cdot d = \Sigma F_i d_i \), where R is the resultant (N), d its perpendicular distance from O (m), and \( F_i d_i \) each force's moment about O (N·m), with signs.
Quick example
A 100 N force at 30° to the horizontal acts at point (2, 1) m from O. Components: \( F_x = 86.6 \) N, \( F_y = 50 \) N. Moment about O (anticlockwise positive) \( = xF_y - yF_x = 2(50) - 1(86.6) = 13.4 \) N·m anticlockwise, found without ever measuring the perpendicular distance.
Common mistakes
- Stating it only for parallel forces; it holds for concurrent forces and, through the resultant, for any coplanar system.
- Writing the proof without a diagram or without explaining why only perpendicular components give moments.
- Losing signs of individual moments when adding.
How MU asks it
A classic 4 to 5-mark theory part: 'State and prove Varignon's theorem'. Write the statement, draw P, Q, R meeting at A with point O and the axes, derive in 3 to 4 lines as above, and end with one application. It is also used in nearly every resultant-location numerical.