The force F of magnitude 195 kN acts along the line AB. (1) Determine the moments Mx , My , and Mz of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that MO = Mxi + Myj + Mzk.

Question

The force F of magnitude 195 kN acts along the line AB. (1) Determine the moments [latex]M_{x}[/latex] , [latex]M_{y}[/latex] , and [latex]M_{z}[/latex] of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that [latex]M_{O}[/latex] = [latex]M_{x}i + M_{y}j + M_{z}k[/latex].

Accepted Answer

We start by computing the rectangular components of F:

[latex]F=F\lambda _{AB}=F\frac{\overrightarrow{AB} }{\left|\overrightarrow{AB} ight| }=195\left(\frac{3i + 12j − 4k}{\sqrt{3^{2}+12^{2}+(-4)^{2}} } ight)=45i + 180j − 60k kN[/latex]

When calculating the moment of a force, the force may be placed at any point on its line of action. As shown in Fig. (a), we chose to have the force acting at point A.

Part 1

The moment of F about a coordinate axis can be computed by summing the moments of the components of F about that axis (the principle of moments).

Moment about the x-Axis Figure (b) represents a two-dimensional version of Fig. (a), showing the yz-plane. We see that the 45-kN and the 60-kN components of the force contribute nothing to the moment about the x-axis (the former is parallel to the axis, and the latter intersects the axis). The perpendicular distance (moment arm) between the 180-kN component and the x-axis is 4 m. Therefore, the moment of this component about the x-axis (which is also the moment of F) is 180(4)=720 kN · m, clockwise. According to the right-hand rule, the positive sense of [latex]M_{x}[/latex] is counterclockwise, which means that [latex]M_{x}[/latex] is negative; that is,

[latex]M_{x}[/latex]=−720 kN · m

Moment about the y-Axis To compute the moment about the y-axis, we refer to Fig. (c), which represents the xz-plane. We note that only the 45-kN force component has a moment about the y-axis, because the 180-kN component is parallel to the y-axis and the 60-kN component intersects the y-axis. Because the moment arm of the 45-kN component is 4 m, the moment of F about the y-axis is 45(4)=180 kN · m, counterclockwise. Therefore, we have

[latex]M_{y}[/latex] =45(4)=180 kN · m

The sign of the moment is positive, because the right-hand rule determines positive [latex]M_{y}[/latex] to be counterclockwise.

Moment about the z-Axis The moment of F about the z-axis is zero, because F intersects that axis. Hence

[latex]M_{z}[/latex] =0

Part 2

Recognizing that the vector from O to A in Fig. (a) is [latex]r_{OA}[/latex] =4k m, the moment of F about point O can be computed as follows.

[latex]M_{O}=r_{OA} imes F=\left|\begin{matrix} i & j & k \ 0 & 0 & 4 \ 45 & 180 & -60 \end{matrix} ight|=-i(4)(180)+j(4)(45)=−720i + 180j kN · m[/latex]

Comparing with [latex]M_{O} = M_{x}i + M_{y}j + M_{z}k[/latex], we see that

[latex]M_{x}[/latex]= −720 kN · m [latex]M_{y}[/latex] =180 kN · m [latex]M_{z}[/latex] =0

which agree with the results obtained in Part 1.

Question 2.5: The force F of magnitude 195 kN acts along the line AB. (1) ...

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