CivilSolve

Statics & Mechanics of Materials

Moment of Forces About a Point

The moment (torque) of one or more 2D forces about any point, by the cross-product form M = rx·Fy − ry·Fx (equivalent to Varignon's theorem), plus applied couples — reporting each force's contribution, the total moment (CCW positive), and the resultant force of the system. Use for: 'find the moment of the force about point O', wrench/lever-arm questions, reducing a force system to a force and couple. Not for: 3D moments or distributed loads (beam solvers).

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
forcesThe applied forces with their application points
forces 1
couplesPure couples acting on the body (location-independent)
couples 1

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

upward 100 N at (2, 0) about the origin

Given

forces
  1. 1.magnitude 100 Nangle deg 90x 2 my 0 m

Assumptions

  • 2D (coplanar) system; counterclockwise moments are positive; the moment center may be any point — couples are the same about every point.

Solution steps

  1. Moment of each force

    Each force's moment is its lever components crossed with its force components: M = rx·Fy − ry·Fx.

    M_O = rx·Fy − ry·Fx (CCW positive)

    Contributions: 0.2 kN·m

  2. Total moment and resultant

    The sum is the single moment of the whole system about the chosen point; together with the resultant force it forms the equivalent force-couple system.

    M_total = ΣMi; R = ΣFi

    M_total = 0.2 kN·m (counterclockwise); R = 0.1 kN

    = 0.2 kN·m

Results

Total moment about the point (CCW positive)

0.2kN·m

Resultant force magnitude of the system

0.1kN

Resultant x-component

6.123e-18kN

Resultant y-component

0.1kN

Where this answer was checked
source
M = rx·Fy − ry·Fx — single force, CCW moment
verified by
hand-recomputed
derivation
F = (0, 100) at r = (2, 0): M = 2·100 − 0 = +200 N·m (CCW).

Example 2

horizontal 50 N at height 3 m gives a clockwise moment

Given

forces
  1. 1.magnitude 50 Nangle deg 0x 0 my 3 m

Assumptions

  • 2D (coplanar) system; counterclockwise moments are positive; the moment center may be any point — couples are the same about every point.

Solution steps

  1. Moment of each force

    Each force's moment is its lever components crossed with its force components: M = rx·Fy − ry·Fx.

    M_O = rx·Fy − ry·Fx (CCW positive)

    Contributions: -0.15 kN·m

  2. Total moment and resultant

    The sum is the single moment of the whole system about the chosen point; together with the resultant force it forms the equivalent force-couple system.

    M_total = ΣMi; R = ΣFi

    M_total = -0.15 kN·m (clockwise); R = 0.05 kN

    = -0.15 kN·m

Results

Total moment about the point (CCW positive)

-0.15kN·m

Resultant force magnitude of the system

0.05kN

Resultant x-component

0.05kN

Resultant y-component

0kN

Where this answer was checked
source
M = rx·Fy − ry·Fx with ry ≠ 0 — sign convention check
verified by
hand-recomputed
derivation
F = (50, 0) at r = (0, 3): M = 0 − 3·50 = −150 N·m (clockwise).

Example 3

US: two forces + a couple about a shifted point

Given

forces
  1. 1.magnitude 20 lbfangle deg 270x 4 fty 0 ft
  2. 2.magnitude 10 lbfangle deg 90x 0 fty 0 ft
about x
1 ft
about y
0 ft
couples
  1. 1.magnitude 30 lbf*ftsense ccw

Assumptions

  • 2D (coplanar) system; counterclockwise moments are positive; the moment center may be any point — couples are the same about every point.

Solution steps

  1. Moment of each force

    Each force's moment is its lever components crossed with its force components: M = rx·Fy − ry·Fx.

    M_O = rx·Fy − ry·Fx (CCW positive)

    Contributions: -60 lbf·ft; -10 lbf·ft

  2. Add the couples

    A couple contributes the same moment about every point — no lever arm needed.

    Total after couples: -40 lbf·ft

  3. Total moment and resultant

    The sum is the single moment of the whole system about the chosen point; together with the resultant force it forms the equivalent force-couple system.

    M_total = ΣMi; R = ΣFi

    M_total = -40 lbf·ft (clockwise); R = 10 lbf

    = -40 lbf·ft

Results

Total moment about the point (CCW positive)

-0.04kip·ft

Resultant force magnitude of the system

0.01kip

Resultant x-component

-3.062e-18kip

Resultant y-component

-0.01kip

Where this answer was checked
source
Varignon superposition with couples — combined system about (1, 0) ft
verified by
hand-recomputed
derivation
20 lbf down at (4,0) about (1,0): r = (3,0), F = (0,−20) → M = 3·(−20) = −60. 10 lbf up at (0,0): r = (−1,0), F = (0,10) → M = −10. Couple 30 lbf·ft ccw → +30. Total = −60 − 10 + 30 = −40 lbf·ft.