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.

Chapter 2 Q. 2.5

Engineering Mechanics Statics [EXP-90698]

The force $F$ of magnitude $195\: kN$ acts along the line $AB$ . (1) Determine the moments $M_{x}$ , $M_{y}$ , and $M_{z}$ 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 $M_{O} = M_{x}i + M_{y}j + M_{z}k$ .

Step-by-Step

Report Solution

Verified Solution

We start by computing the rectangular components of $F$ :
$F=Fλ_{AB}=F\frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}=195\left(\frac{3i+12j-4k}{\sqrt{3^2+12^2+(-4)^2}}\right)=45i+180j−60k\:kN$

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 $180(4)=720\:kN\cdot m$ , clockwise. According to the right-hand rule, the positive sense of $M_{x}$ is counterclockwise, which means that $M_{x}$ is negative; that is,

M_{x}=−720\:kN\cdot 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 $45(4)=180\:kN\cdot m$ , counterclockwise. Therefore, we have
$M_{y}=45(4)=180\:kN\cdot m$
The sign of the moment is positive, because the right-hand rule determines positive $M_{y}$ to be counterclockwise.
Moment about the $z$ -Axis The moment of $F$ about the $z$ -axis is zero, because $F$ intersects that axis. Hence
$M_{z}= 0$

Part 2
Recognizing that the vector from $O$ to $A$ in Fig. (a) is $r_{OA}=4k\:m$ , the moment of $F$ about point $O$ can be computed as follows.
$M_{O}=r_{OA}×F=\left|\begin{matrix}i&j&k\\0&0&4\\45&180&-60\end{matrix} \right|=−i(4)(180)+j(4)(45)=−720i+180j\:kN\cdot m$

Comparing with $M_{O}=M_{x}i+M_{y}j+M_{z}k$ , we see that
$M_{x}=-720\:kN\cdot m$ $M_{y}=180\:kN\cdot m$ $M_{z}=0$
which agree with the results obtained in Part 1.