Notes@HKU by Jax

Moment of forces

Definition of moment

The moment of a force is a measure of its tendency to cause a body to rotate about a specific point.

The moment about a point OO, when FF is applied a distance dd from the point is:

MO=F×dM_O=F\times d

Keep in mind that positive moment is anti-clockwise.

Coplanar / 2D moment

Resultant moments

The resultant moment is the sum of all moments present on the point, given by:

MR=∑MOM_R=\sum M_O

The moment of a non-linearly attached force

One simple way is to find the components of the force, and sum their individual moments together. The following is a simple example:

After finding the component forces of FF, we can deduce the resultant moment to be:

MO=Fx×0.8sin⁡45deg⁡+Fy×(1+0.8cos⁡45deg⁡)=−150.7kNm\begin{aligned} M_O &= F_x\times 0.8\sin 45\deg+F_y\times(1+0.8\cos 45\deg) \\ & =-150.7kNm \end{aligned}

Non-coplanar / 3D moment

Moments in a 3D system

Consider position vector r⃗\vec{r} drawn from OO to any point on the line of action of FF. The moment can hence be given by:

MO=r×FM_O=r\times F

Finding the moment via cross products Cartesian vectors

The cross product CC given by AA and BB is:

The cross-product for vectors going in the same direction is 0. (i.e. nk×mk=0n\boldsymbol{k}\times m\boldsymbol{k}=0)

Resultant moments

The resultant moment is simply the sum of couple moments and moments of forces:

(MR)O=∑MO+∑M(M_R)_O=\sum M_O + \sum M

You can interpret (MR)O(M_R)_O as the resultant moment about point OO.

Finding moments by force perpendicular to plane

(Exam question) Given a force FF acting perpendicular to a plane ABOABO at OO, determine the moment about a point AA.

  1. Find the position vector {r=OD}\{r=OD\} of the force. (AC×BC=ODAC\times BC=OD)
  2. Convert force to Cartesian form. (F=∥F∥×r∥r∥F=\|F\|\times\frac{r}{\|r\|})
  3. Find the cross product of the vector from point to the force. (MA=OA×FM_A=OA\times F)

Couple moments

Couples are two parallel forces that have the same magnitude but have opposite directions, separated by a perpendicular distance dd. The magnitude of the moment is given by:

M=FdM=Fd

Notice that there's no point mentioned so far. For couple moment, it is always the same about any point. Let's assume for any point OO (refer to graph), the moment is:

MO=rB×F+rA×(−F)=(rB−rA)×F=r×Fwhich is independent of O\begin{aligned} M_O &= r_B \times F + r_A \times (-F) \\ &= (r_B - r_A) \times F \\ &= r \times F \quad \text{which is independent of } O \end{aligned}

Hence, we can say that couple moments are free vectors.

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