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Textbook › Engineering Mechanics › Chapter 1 › Lesson 1.4

1.4 Moment of a force and Varignon's theorem

Engineering Mechanics Chapter 1: Force systems and equilibrium Reviewed 25 Oct 2026 Author: Indiacon

Learning objectives

After this lesson you should be able to:

  • Define the moment of a force about a point and about an axis, with its sign convention.
  • Calculate moments using perpendicular distance and using components.
  • State, prove and use Varignon's theorem.

Theory

A force tends to translate a body along its line of action and to rotate it about any point not on that line. The turning effect is the moment of the force. In a plane, the moment about a point O is the magnitude of the force times the perpendicular distance from O to its line of action: M_O = F d. The unit is the newton-metre (N·m) or, in structures, the kilonewton-metre (kN·m).

Moments have a sense. In this textbook anticlockwise moments are positive and clockwise negative, matching the right-hand rule with the z-axis out of the page. Any consistent convention works, but it must be declared and kept throughout a calculation.

OFdAFxFyM_O = F d = x Fy − y FxIndiacon original, not to scale
Figure 1.4: Moment of F about O by perpendicular distance d, or by its components at A (Varignon). Indiacon original, not to scale.

Finding the perpendicular distance can be awkward when a force is inclined. Varignon's theorem removes the difficulty: the moment of a force about a point equals the sum of the moments of its components about the same point. So we resolve the force at any convenient point on its line of action into x and y components and add their moments, each of which has an easy lever arm.

In vector form the moment is M_O = r × F, where r is a position vector from O to any point on the line of action. For plane problems this reduces to M_O = x Fy − y Fx, a formula worth memorising because it handles signs automatically.

Derivation

  1. Let r run from O to point A on the line of action of F. In vector form, M_O = r × F; its magnitude is r F sin φ = F (r sin φ) = F d, where φ is the angle between r and F and d is the perpendicular distance.
  2. Choosing a different point A′ on the same line changes r by a vector parallel to F, whose cross product with F is zero, so the moment does not depend on which point of the line is used.
  3. Varignon: if F = F₁ + F₂, then r × F = r × (F₁ + F₂) = r × F₁ + r × F₂, because the cross product distributes over addition. The moment of the resultant equals the sum of the moments of the components.
  4. In 2D with r = (x, y) and F = (Fx, Fy): r × F = (x Fy − y Fx) k, giving M_O = x Fy − y Fx.
Formulas, symbols and SI units
  • M_O = F d (N·m or kN·m)
  • M_O = x Fy − y Fx (anticlockwise positive)
  • Varignon: M_O(R) = Σ M_O(Fi)
  • Moment of an inclined force at a lever arm L: M = F L sin β, β the angle between force and arm
Assumptions and limits
  • The body is rigid, so the point of application can move along the line of action.
  • Forces and the moment centre lie in one plane (for 3D moments see lesson 1.10).
  • Anticlockwise positive.

Codes and standards

Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.

  • No design code governs moments as such. Design codes such as IS 456:2000 and IS 800:2007 use bending moments computed from these principles; the codes then govern how sections resist them.

Worked examples

Example 1 (exam style)

Problem. A 400 N force is applied at the end of a 250 mm spanner, at 60° to the handle. Find the moment about the bolt centre, first with the perpendicular distance and then by Varignon's theorem.

  1. Perpendicular distance d = 0.250 × sin 60° = 0.2165 m; M = 400 × 0.2165 = 86.60 N·m.
  2. By components: along the handle 400 cos 60° = 200 N (passes through the bolt, no moment); perpendicular 400 sin 60° = 346.4 N at 0.250 m.
  3. M = 346.4 × 0.250 = 86.60 N·m, the same answer.

Answer. 86.60 N·m

Example 2 (site style)

Problem. A 5 kN force acts through point A(3 m, 2 m) on a bracket. Its direction is given by the slope 4 horizontal to 3 vertical (both positive). Find its moment about the origin O and the perpendicular distance from O to its line of action.

  1. Unit vector (4, 3)/5, so Fx = 4 kN and Fy = 3 kN.
  2. M_O = x Fy − y Fx = 3 × 3 − 2 × 4 = 9 − 8 = 1 kN·m (anticlockwise).
  3. d = M_O / F = 1 / 5 = 0.2 m.

Answer. 1 kN·m anticlockwise; d = 0.2 m

Step-by-step procedure

  1. Choose the moment centre, often a support or an unknown force's point, to remove unknowns.
  2. Fix the sign convention.
  3. Resolve inclined forces at a convenient point on their lines of action.
  4. Use M = x Fy − y Fx or the component moments, checking signs.
  5. Sum the moments and state the sense of the result.

Site notes

  • The tightening torque on bolts and anchor rods is a moment; torque wrenches are calibrated in N·m.
  • A horizontal wind or impact force at the top of a pole or hoarding produces a base moment equal to force times height; foundations and base plates are sized for it.
  • Lifting a long element off-centre creates a moment about the hook; rigging layouts place the hook directly above the centre of gravity for this reason.
Common mistakes
  • Multiplying the force by the distance to its point of application instead of the perpendicular distance to its line of action.
  • Mixing clockwise and anticlockwise signs within one sum.
  • Forgetting that a force through the moment centre has zero moment.
  • Using y Fx − x Fy, which flips every sign.
  • Writing moments in kN instead of kN·m.
Quick check

The moment of a force about any point cannot exceed F times the distance from that point to the force's point of application. If your moment is larger, the lever arm is wrong.

Summary

  • M_O = F d, with d the perpendicular distance to the line of action.
  • Anticlockwise positive; M_O = x Fy − y Fx in 2D.
  • Varignon: moment of a resultant equals the sum of moments of its components.
  • Moments are independent of where along its line of action the force is drawn.

Practice questions

Tap a question to see the answer.

1A 50 N force acts at 0.3 m perpendicular distance from a hinge. Find its moment.

M = 50 × 0.3 = 15 N·m.

2A force F = (2, −5) kN acts through (1 m, 4 m). Find its moment about the origin.

M = 1 × (−5) − 4 × 2 = −13 kN·m, i.e. 13 kN·m clockwise.

3A horizontal force of 2 kN acts at the top of a 6 m pole. Find the moment at the base.

M = 2 × 6 = 12 kN·m.

4What is the moment of a force about a point lying on its line of action?

Zero, because the perpendicular distance is zero.

5A 10 kN force at 30° above the horizontal (pointing right) acts at the top of a 3 m vertical post whose base is at the origin. Find the moment about the base.

Fx = 8.660 kN, Fy = 5 kN at (0, 3): M = 0 × 5 − 3 × 8.660 = −25.98 kN·m, i.e. 25.98 kN·m clockwise.

References

Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.

  • F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
  • R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
  • J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
  • S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.