Statics · Rigid Body Mechanics

Moment of a Force — the r × F cross product

How a force applied away from a support twists it. One operation, the vector cross product, tells you the size and the sense of that twist.

MO = r × F   |   |MO| = |r||F|sin θ   |   direction by the right-hand rule

Why a cross product? Concept

A force doesn't just push or pull — if it's applied away from a support or pivot, it also twists that point. That twisting effect is called the moment of the force about that point.

Two things decide how hard it twists:

  • How big the force is, and how it's angled
  • How far away — and in what direction — it's applied, measured by the position vector r from the support to the point of application

The cross product packages both of these into a single vector operation:

moment about O = MO = r × F

r and F are both vectors. Their cross product gives a third vector — the moment — that's perpendicular to both, with a magnitude and a direction that tell you exactly how the support is being twisted.

x y O (support) r F M_O

r runs from the support O to where F is applied. The moment MO curls around O, out of the page.

Computing it: the determinant method Mechanics

Write r = (rx, ry, rz) and F = (Fx, Fy, Fz). The cross product is the determinant of this matrix:

| i   j   k |
| rx   ry   rz |
| Fx   Fy   Fz |

Expanding it gives three components:

Mx = ryFz − rzFy
My = rzFx − rxFz
Mz = rxFy − ryFx

For a purely planar problem — r and F both lying in the xy-plane, which is the usual beam/bracket case — the first two components vanish and only Mz survives:

Mz = rxFy − ryFx
Sign convention: a positive Mz means counter-clockwise when viewed from the +z axis (standard right-hand rule). Negative means clockwise.

Worked example Walkthrough

A bracket is bolted to a wall at O. A cable pulls with force F = (40, −30, 0) N at a point located at r = (0.5, 0.2, 0) m from the bolt.

Mz = rxFy − ryFx
Mz = (0.5)(−30) − (0.2)(40)
Mz = −15 − 8 = −23 N·m

The negative sign means the bolt feels a clockwise moment of 23 N·m. That's the number an engineer checks against the bolt's rated torque capacity.

Worked example 2: cantilever beam Walkthrough

A beam is welded into a wall at O and sticks straight out along x. A downward load is hung at its free end, 2 m out from the wall.

r = (2, 0, 0) m
F = (0, −500, 0) N
Mz = rxFy − ryFx
Mz = (2)(−500) − (0)(0)
Mz = −1000 N·m

Negative again: the weld feels a clockwise moment of 1000 N·m. This is exactly the bending moment you'd check against the weld's strength — the further out the load sits, the bigger r, the bigger the twist, even though the force never changed.

O r = 2 m F M_O

Fixed beam, load at the tip. The clockwise curl shows the beam wants to droop and twist the wall joint clockwise.

Worked example 3: pushing a door open Walkthrough

A door hinge sits at O. You push perpendicular to the door, 0.8 m from the hinge, with 60 N, aimed to swing the door open.

r = (0, 0.8, 0) m
F = (60, 0, 0) N
Mz = rxFy − ryFx
Mz = (0)(0) − (0.8)(60)
Mz = −48 N·m

The hinge takes a 48 N·m clockwise moment — that's the swing torque. Push at the same spot but angle your hand 45° off perpendicular, and only the perpendicular component of F still does the twisting; the part of F aimed straight at the hinge contributes nothing, because it's parallel to r.

O (hinge) r = 0.8 m F = 60 N M_O

Door swinging open about its hinge. r runs up the door to your hand; F is your push, perpendicular to the door for maximum effect.

Build your own r × F Interactive

Enter a position vector r (in metres, from the support) and a force vector F (in newtons). We'll compute the full moment vector and walk through the determinant.

Position vector r (m)
Force vector F (N)
Mx = ryFz − rzFy =
My = rzFx − rxFz =
Mz = rxFy − ryFx =
|MO| =

Quick checks Self-test

Five short ones. Three come with a picture of the actual object — work the vectors out from the diagram before you calculate.

Check 1

r = (0.3, 0, 0) m, F = (0, 50, 0) N. What is Mz?

Check 2 — torque wrench

You pull straight up on a wrench handle, 0.2 m from the bolt, with 100 N applied perpendicular to the handle.

r = (0.2, 0, 0) m
F = (0, 100, 0) N
bolt O r = 0.2 m F = 100 N
Check 3 — cantilever beam

A beam is fixed into a wall at O and a load hangs straight down at its tip, 1.5 m out.

r = (1.5, 0, 0) m
F = (0, −200, 0) N
O r = 1.5 m F = 200 N
Check 4 — pushing a door

Hinge at O. You push perpendicular to the door, 0.9 m from the hinge, with 40 N.

r = (0, 0.9, 0) m
F = (40, 0, 0) N
O (hinge) r = 0.9 m F = 40 N
Check 5

r = (0.6, −0.2, 0) m, F = (10, 30, 0) N. What is Mz?

Score: 0 / 12
Question 1 — Concept

What does the magnitude of r × F physically represent?

Question 2 — Calculation

r = (0.4, 0, 0) m, F = (0, 25, 0) N, both in the xy-plane. Find Mz in N·m.

Question 3 — Calculation

r = (0.2, 0.6, 0) m, F = (10, −20, 0) N. Find Mz in N·m.

Question 4 — Sign convention

A calculated Mz comes out negative. What does that mean about the support?

Question 5 — Calculation, cantilever beam

A beam fixed at O carries a downward load at its tip, 3 m out.

r = (3, 0, 0) m
F = (0, −150, 0) N

Find Mz in N·m.

O r = 3 m F = 150 N
Question 6 — Sense of rotation, wrench

You pull straight up on a wrench handle, 0.2 m from the bolt, with 100 N perpendicular to the handle (same setup as Practice, Check 2). Which way does the bolt get twisted?

bolt O r = 0.2 m F = 100 N
Question 7 — Concept, order of the cross product

How does F × r compare to r × F?

Question 8 — Calculation

r = (1, 1, 0) m, F = (5, −5, 0) N. Find Mz in N·m.

Question 9 — Calculation, door

Hinge at O. Push perpendicular to the door, 0.9 m out, with 40 N (same setup as Practice, Check 4).

r = (0, 0.9, 0) m
F = (40, 0, 0) N

Find Mz in N·m.

O (hinge) r = 0.9 m F = 40 N
Question 10 — Concept, line of action

The line of action of F passes directly through the support O. What is the moment about O?

Question 11 — Full 3D moment vector

A signpost is bolted to the ground at O. Wind load hits the sign plate at a point offset in all three directions.

r = (0.5, 0.3, 0.2) m
F = (10, −20, 15) N

Find Mx, My, and Mz in N·m.

x y z O r F

r and F both have x, y, and z components — none of M_x, M_y, M_z drops out this time.

Question 12 — Full 3D moment vector

An L-shaped bracket is fixed to a wall at O. It extends out along x then up along z, and carries a load there.

r = (1, 0, 0.5) m
F = (0, 40, −10) N

Find Mx, My, and Mz in N·m.

O 1 m (x) 0.5 m (z) F (in y & z)

r bends through x and z; F sits in the y–z plane. Compute the determinant carefully, term by term.