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## Q. 2.5

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 $M_{O} = M_{x}i + M_{y}j + M_{z}k$. ## 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.  