The view from the y axis reveals that is perpendicular to the x axis and that its line of action does not intersect the x axis. Therefore, the moment of about the x axis isfound as where is the momentarm of the force with respect to the x axis. In this case, themoment axis is pointing in the positive x direction as shown.Similarly, the view from the x axisreveals that isperpendicular to the y axis as well. Hence, the moment of about the y axis isfound as | |

where is the momentarm of the force with respect to the y axis. In this case, themoment axis is pointing in the negative y direction. Also as to beexpected, since is parallel to the zaxis. You are watching: Moment of a force about a line |

We can now expand this discussion to the case of calculating themoment about an arbitrary line aa. Twocommonly encountered cases are described next.

Case 1: The line of action of the forceis perpendicular to aa, and that the twolines do not intersect each other.In this case, the moment about aa isfound as (6)
An example of this case is shown in the figure where the line ofaction of the force is in z direction and line aa is in xy plane; clearly, they areperpendicular to each other. | |

Case 2: The line of action of the forceis NOT perpendicular to aa, and that thetwo lines do not intersect each other.In this case, the moment about aa isfound in two steps using the vector approach. First, the momentabout a point lying on line aa iscalculated as Then, the projection of along line aa is found usingthe dot product(Magnitude of the component) | |

The previous two equations can be combined into a triple scalarproduct as(7) | |

If comes out negative, it simply means that its direction is oppositeto that defined by This projection can also be put in vector form as |