Moment about an axis

Why a new quantity is needed. The moment of a force about a point, $\vec M_O = \vec r \times \vec F$, is a vector that can point anywhere in 3-D space. A door hinge, a bolted shaft, or any rigid pivot line only lets the body rotate about one fixed direction, and only the component of $\vec M_O$ along that direction produces rotation — the rest is absorbed by the supports holding the axis fixed. So the quantity worth computing is a projection, not the full vector.

The definition. The moment of a force $\vec F$ about an axis $aa'$ with unit direction $\hat u$ is the axis-component of $\vec M_O$, taken about any point $O$ lying on that axis:

where: $\hat u$ is the unit vector along the axis, $\vec r$ is the position vector from any point $O$ on the axis to any point on the force's line of action, and $\vec F$ is the force. The result $M_{axis}$ is a signed scalar, not a vector.

Computing it as a determinant. The scalar triple product expands directly as a $3\times3$ determinant, the practical way to evaluate it:

where: the rows are the components of $\hat u$, $\vec r$, and $\vec F$ in that order. Expanding along the first row gives the same number as computing $\vec r \times \vec F$ first and then dotting with $\hat u$ — the determinant is just a faster route to the same scalar.

Sign and physical reading. The sign follows the right-hand rule relative to the chosen positive $\hat u$: curl the fingers in the sense of rotation the moment tends to produce, and if the thumb points along $+\hat u$, the sign is positive. A negative $M_{axis}$ means the tendency to rotate is opposite to the chosen positive sense — not that the force is weaker, only that it turns the body the other way.

Independence of the point on the axis. $M_{axis}$ does not depend on which point $O$ on the axis is used to build $\vec r$ — any point on the axis gives the same scalar, mirroring the property that a couple's moment does not depend on the reference point. Pick whichever axis point makes the arithmetic easiest.

Special cases where $M_{axis} = 0$. If the force's line of action intersects the axis, or is parallel to it, then $\hat u$, $\vec r$, and $\vec F$ become coplanar, the determinant's rows become linearly dependent, and $M_{axis} = 0$. A fast sanity check: a force passing through the hinge line, or running alongside it, cannot turn the body about that axis at all.

Extend the same door problem to check that the answer does not depend on which point on the hinge axis is used, and see the same tool applied to a second force.

Step 1. Take a fresh force, $\vec F = 60\hat i - 20\hat k\ \text{N}$, applied at $A = (2, 3, 0)\ \text{m}$, and find its moment about the $z$-axis using $O = (0,0,0)$. Here $\vec r_{OA} = 2\hat i + 3\hat j$, so

Step 2. Repeat with a different point on the same axis, $O' = (0,0,5)$, to confirm the axis-point choice does not matter. Here $\vec r_{O'A} = 2\hat i + 3\hat j - 5\hat k$, so

The extra $-5\hat k$ term sits in the row-3, column-3 slot, which the $\hat u = (0,0,1)$ top row picks out through a cofactor that never involves that column — so it cannot change the answer.

Step 3. Compare and conclude. Both choices of $O$ give $M_{axis} = -180\ \text{N}\,\cdot\,\text{m}$, confirming the moment about an axis is a property of the axis itself, not of an arbitrarily chosen point on it. The moment about the $z$-axis is $-180\ \text{N}\,\cdot\,\text{m}$, magnitude $180\ \text{N}\,\cdot\,\text{m}$.

Competitive stretch

If the same force instead acted at a point on the $z$-axis itself, say $(0,0,4)\ \text{m}$, then $\vec r$ would be parallel to $\hat u = \hat k$, the determinant's rows would become linearly dependent, and $M_{axis}$ would come out zero — the special case in action, worth checking before a full expansion.

• Taking $\vec r$ from a point that is not on the axis — any point on the axis works, but off-axis points give a wrong answer.
• Using a non-unit $\hat u$ — a non-normalized axis direction scales the determinant incorrectly.
• Confusing $M_{axis}$ with $|\vec M_O|$ — $M_{axis}$ is only the along-axis projection; the full moment about the point is generally larger.
• Skipping the intersects-or-parallel check — a force meeting or running alongside the axis gives $M_{axis} = 0$ immediately, saving a full determinant expansion.

Ask the tutor: if the choice of axis point, the sign convention, or expanding the determinant still feels shaky, ask the AI tutor to walk through one more axis-moment example with you, step by step.