Question 2.6: The force F of Sample Problem 2.5 is shown again in Fig. (a)...

Part 1

Referring to Fig. (a), we see that it is not practical to calculate the moment about the axis CE by the scalar method. Because the axis CE is not parallel to a coordinate axis, the task of determining the perpendicular distance between F and CE would be tedious. However, if the vector method is used, the calculation of the moment is straightforward.To employ the vector method we first express the force F in vector form. This was already done in the solution to Sample Problem 2.5:

F=45i + 180j − 60k kN

Next we calculate the moment of F about any convenient point on the axis CE. Inspection of Fig. (a) reveals that there are only two convenient points from which to choose—points C and E. Let us choose point C. Because we will use the cross product r × F to compute the moment about C, our next step is to choose the vector r and to write it in vector form (remember that r must be a vector from point C to any point on the line of action of F). From Fig. (a) we see that there are two convenient choices for r: either the vector from C to A or the vector from C
to B. Choosing the latter, we have

r=r_{CB}= −4k m

The moment of F about point C then becomes

Note that the z-component ofM_{C} is zero. To understand this result, recall that the z-component of M_{C} equals the moment of F about the axis BC (the line parallel to the z-axis passing through C). Because F intersects BC, its moment about BC is expected to be zero. Next, we calculate the unit vector λ_{CE} directed from point C toward point E:

The moment of MC about the axis CE can now be obtained from Eq. (2.11) M_{AB}=M_{O}\cdot \lambda =r\times F\cdot \lambda:

M_{CE} = M_{C} ·λ_{CE}=(720i − 180j) · (−0.5571i + 0.3714j − 0.7428k)=−468 kN · m

The negative sign indicates that the sense of the moment is as shown in Fig. (b)—that is, opposite to the sense associated with λ_{CE}.We could also compute M_{CE} without first determining M_{C} by using the scalar triple product:

M_{CE}=r_{BC}\times F\cdot \lambda _{CE}=\left|\begin{matrix} 0 & 0 & -4 \\ 45 & 180 & -60 \\ −0.5571 & 0.3714 & −0.7428 \end{matrix} \right|=−468 kN · m

This agrees, of course, with the result determined previously.

Part 2

To express the moment of F about the axis CE in vector form, we multiply M_{CE} by the unit vector λ_{CE}, which gives

M_{CE} =M_{CE}λ_{CE}=−468(−0.5571i + 0.3714j − 0.7428k) = 261i − 174j + 348k kN · m

There is no doubt that using the vector method is convenient when one wishes to calculate the moment about an axis such as CE, which is skewed relative to the coordinate system. However, there is a drawback to vector formalism: You can easily lose appreciation for the physical nature of the problem.