Moment of a force about a point

A rigid body resists rotation about a point the same way it resists translation under a net force — the measure of "how hard a given force is trying to spin it about $O$" is the moment, and geometry alone decides the answer: same force, same body, but push near the axis and the moment is tiny; push far from it and the moment is large.

The vector definition. Pick any point $A$ on the force's line of action and draw the position vector $\vec r$ from $O$ to $A$. The moment is the cross product

where: $\vec r$ is the position vector from $O$ to any point on the line of action of $\vec F$, and $\vec F$ is the force vector. Because of transmissibility (a force can slide along its own line of action without changing its external effect), the answer does not depend on which point along the line you pick for $A$ — a convenient freedom worth using to choose the simplest $\vec r$ available.

Magnitude and the scalar shortcut. Expanding the cross product's magnitude gives

where: $\theta$ is the angle between $\vec r$ and $\vec F$, $r=|\vec r|$, $F=|\vec F|$, and $d = r\sin\theta$ is the perpendicular (moment-arm) distance from $O$ to the line of action. In 2-D problems this scalar version is usually all you need: find $d$ by inspection or trigonometry, multiply by $F$, and fix the sign by the right-hand rule — curl your fingers from $\vec r$ toward $\vec F$ and your thumb gives the direction of $\vec M_O$, which in the plane reduces to counterclockwise-positive by the usual sign convention.

Component (determinant) form. Writing $\vec r = x\hat i + y\hat j + z\hat k$ and $\vec F = F_x\hat i + F_y\hat j + F_z\hat k$, the cross product expands to

where: each bracketed term is a $2\times2$ determinant obtained by crossing out the row and column of that unit vector in the $3\times3$ array $\begin{vmatrix}\hat i & \hat j & \hat k\\ x & y & z\\ F_x & F_y & F_z\end{vmatrix}$. For a purely 2-D problem ($z=0$, $F_z=0$) only the $\hat k$ term survives, giving the handy scalar $M_O = xF_y - yF_x$.

Varignon's theorem. The moment of a force about a point equals the sum of the moments of its components about that same point. Instead of hunting for one awkward perpendicular distance to a tilted line of action, resolve $\vec F$ into $F_x$ and $F_y$, find each component's (usually trivial) moment arm, and add the two scalar moments algebraically. This is the single biggest time-saver in moment problems.

A force $\vec F = 80\ \text{N}$ acts at point $A=(4,0)\ \text{m}$, directed $60°$ above the horizontal. Find its moment about the origin $O$, then use the result to also find the moment of the same force about the point $B=(4,-3)\ \text{m}$.

Step 1. Resolve the force into components: $F_x = 80\cos60° = 40\ \text{N}$ and $F_y = 80\sin60° = 69.3\ \text{N}$.

Step 2. Moment about $O$ using the 2-D component formula with $\vec r_{OA} = 4\hat i\ \text{m}$: $M_O = xF_y - yF_x = (4)(69.3) - (0)(40) = 277.1\ \text{N}\,\cdot\,\text{m}$, positive so counterclockwise.

Step 3. Cross-check with Varignon's theorem: $F_x = 40\ \text{N}$ lies along $y=0$, so its moment arm about $O$ is zero, contributing $0$; $F_y = 69.3\ \text{N}$ acts at $x=4\ \text{m}$, giving moment $69.3 \times 4 = 277.1\ \text{N}\,\cdot\,\text{m}$ CCW. The two components sum to the same $277.1\ \text{N}\,\cdot\,\text{m}$, confirming Step 2.

Step 4. Now shift the reference point to $B=(4,-3)\ \text{m}$. The new position vector is $\vec r_{BA} = 0\hat i + 3\hat j\ \text{m}$. Using the same 2-D formula, $M_B = xF_y - yF_x = (0)(69.3) - (3)(40) = -120\ \text{N}\,\cdot\,\text{m}$, negative, so clockwise — sharply different from $M_O$, confirming a single force's moment is reference-point dependent. Final answers: $M_O = 277.1\ \text{N}\,\cdot\,\text{m}$ counterclockwise; $M_B = 120\ \text{N}\,\cdot\,\text{m}$ clockwise.

Competitive stretch

For a 3-D check, $\vec F = 10\hat i - 24\hat j\ \text{N}$ at $A=(2,1,3)\ \text{m}$ has $\vec r_{OA} = 2\hat i + 1\hat j + 3\hat k$, giving $\vec M_O = 72\hat i + 30\hat j - 58\hat k\ \text{N}\,\cdot\,\text{m}$ about the origin. Recomputing with $\vec r_{BA} = 1\hat i+1\hat j+3\hat k$ for $B=(1,0,0)\ \text{m}$ gives $\vec M_B = 72\hat i + 30\hat j - 34\hat k\ \text{N}\,\cdot\,\text{m}$ — only the $\hat k$ component changes, again confirming the moment tracks the reference point.

• Treating a single force's moment as reference-point-independent — that property belongs only to a couple; always recompute $\vec r$ freshly when the reference point changes, as the worked example above shows.
• Measuring $\vec r$ to a point that is not on the force's line of action — any point on the line works by transmissibility, but a point off the line gives a wrong moment.
• Losing track of the sign convention in the scalar method — fix counterclockwise-positive (or whichever convention the problem states) before assigning $+$ or $-$ to $Fd$, and apply it consistently across every force in the problem.
• Searching for one awkward perpendicular distance by trigonometry when Varignon's theorem would let you resolve the force into simple components first.

Ask the tutor: If setting up the cross-product determinant or tracking the sign when the reference point changes feels error-prone, ask the AI tutor to walk through a moment calculation with you step by step.