problem 3.6 - determine the moment about the line ab due to force f.

Answer

Explanation:

Step1: Define position vectors

Let $\vec{r}{AD}$ be the position - vector from point $A$ to point $D$. If we assume $A=(0,0,0)$ (for simplicity of calculation), then $\vec{r}{AD}=(18\vec{i}+ 12\vec{j}-20\vec{k})\text{ m}$. The force vector $\vec{F}$ needs to be resolved. Given the magnitude $F = 6\text{ kN}$, assume $\vec{F}$ has components based on the right - angled triangle formed by its direction. If we consider the right - angled triangle with sides $15\text{ m}$ and $12\text{ m}$, the direction cosines can be used to find $\vec{F}$. The magnitude of the hypotenuse of the triangle in the $x - y$ plane is $\sqrt{15^{2}+12^{2}}=\sqrt{225 + 144}=\sqrt{369}=3\sqrt{41}\text{ m}$. The unit vector in the direction of $\vec{F}$ in the $x - y$ plane is $\frac{15\vec{i}+12\vec{j}}{3\sqrt{41}}$. Let's assume $\vec{F}=6\left(\frac{15\vec{i}+12\vec{j}}{3\sqrt{41}}\right)\text{ kN}$ (assuming no $z$ - component for simplicity as not enough information about the $z$ - direction of $\vec{F}$ is given from the right - angled triangle shown). So $\vec{F}=\frac{6\times15}{3\sqrt{41}}\vec{i}+\frac{6\times12}{3\sqrt{41}}\vec{j}=\frac{30}{\sqrt{41}}\vec{i}+\frac{24}{\sqrt{41}}\vec{j}\text{ kN}$.

Step2: Calculate the moment about point $A$

The moment of the force $\vec{F}$ about point $A$, $\vec{M}{A}=\vec{r}{AD}\times\vec{F}$. [ \begin{align*} \vec{M}_{A}&=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\ 18&12& - 20\ \frac{30}{\sqrt{41}}&\frac{24}{\sqrt{41}}&0 \end{vmatrix}\ &=\vec{i}(0+\frac{480}{\sqrt{41}})-\vec{j}(0 + \frac{600}{\sqrt{41}})+\vec{k}(\frac{432}{\sqrt{41}}-\frac{360}{\sqrt{41}})\ &=\frac{480}{\sqrt{41}}\vec{i}-\frac{600}{\sqrt{41}}\vec{j}+\frac{72}{\sqrt{41}}\vec{k}\text{ kN}\cdot\text{m} \end{align*} ]

Step3: Find the unit vector along $AB$

The vector $\vec{r}{AB}=(0\vec{i}+0\vec{j}- 20\vec{k})\text{ m}$, and the unit vector $\hat{u}{AB}=\frac{\vec{r}{AB}}{\vert\vec{r}{AB}\vert}=-\vec{k}$.

Step4: Calculate the moment about the line $AB$

The moment about the line $AB$, $M_{AB}=\vec{M}{A}\cdot\hat{u}{AB}$. [ \begin{align*} M_{AB}&=\left(\frac{480}{\sqrt{41}}\vec{i}-\frac{600}{\sqrt{41}}\vec{j}+\frac{72}{\sqrt{41}}\vec{k}\right)\cdot(-\vec{k})\ &=-\frac{72}{\sqrt{41}}\text{ kN}\cdot\text{m}\approx - 11.1\text{ kN}\cdot\text{m} \end{align*} ]

Answer:

$-\frac{72}{\sqrt{41}}\text{ kN}\cdot\text{m}\approx - 11.1\text{ kN}\cdot\text{m}$