4.2.2 Moments

The moment ( $\displaystyle \tau$ ) of a force F about a pivot is given by

$\displaystyle \tau =F\times {{d}_{\bot }}$

where $\displaystyle {{d}_{\bot }}$ is the perpendicular distance between the (line of action of the) force and the pivot.

Most students have no problem calculating moments produced by vertical forces on horizontal beams. But some find it challenging if the forces or/and beams are “slanted”, like the one shown below.

If the angles at which F and L are inclined to the horizontal are provided, then a good approach is to resolve everything vertically and horizontally.

$\displaystyle \tau =F\sin \alpha \times L\cos \beta +F\cos \alpha \times L\sin \beta$

Notice that we have to sum up the moments produced by both components of F. Notice also that a vertical force has a horizontal perpendicular distance, whereas a horizontal force has a vertical perpendicular distance.

If the angle between the (line of action of the) force and the beam is given, then a better approach would be to resolve the force into components perpendicular to the beam $\displaystyle (F\sin \theta )$ and parallel to the beam $\displaystyle (F\cos \theta )$ .

$\displaystyle \tau =F\sin \theta \times L$

Notice that $\displaystyle F\cos \theta$ can be ignored because it produces zero moment. Its line of action passes right through the pivot.

Alternatively, we can work out the (perpendicular) distance between the force and the pivot to be $\displaystyle L\sin \theta$ . Obviously, this leads us to the same answer.