Calculate the moment of 2D forces and applied couples about a point. See each contribution, the total moment, and the resultant force.
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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.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
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
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.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
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
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.magnitude 20 lbfangle deg 270x 4 fty 0 ft
2.magnitude 10 lbfangle deg 90x 0 fty 0 ft
about x
1 ft
about y
0 ft
couples
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
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
Add the couples
A couple contributes the same moment about every point — no lever arm needed.
Total after couples: -40 lbf·ft
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.