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