Question

4-28. determine the moment of the force f about point p. express the result as a cartesian vector. problems 4-27/28 f = {-6i + 4j + 8k} kn

Explanation:

Step1: Find the position vector $\vec{r}$ from point $P$ to point of application of force $A$

The coordinates of point $P$ are $(3,0,4)$ and of point $A$ are $(6,2,1)$. So, $\vec{r}=(6 - 3)\vec{i}+(2 - 0)\vec{j}+(1 - 4)\vec{k}=3\vec{i}+2\vec{j}- 3\vec{k}$ m.

Step2: Recall the formula for moment $\vec{M}=\vec{r}\times\vec{F}$

We know $\vec{F}=-6\vec{i}+4\vec{j}+8\vec{k}$ kN and $\vec{r}=3\vec{i}+2\vec{j}-3\vec{k}$ m. The cross - product formula is $\vec{M}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\3&2&- 3\\-6&4&8\end{vmatrix}$$

$.

Step3: Calculate the determinant

$\vec{M}=\vec{i}(2\times8-(4\times(-3)))-\vec{j}(3\times8-(-6)\times(-3))+\vec{k}(3\times4-(-6)\times2)$
$=\vec{i}(16 + 12)-\vec{j}(24 - 18)+\vec{k}(12 + 12)$
$=28\vec{i}-6\vec{j}+24\vec{k}$ kN$\cdot$m

Answer:

$\vec{M}=28\vec{i}-6\vec{j}+24\vec{k}$ kN$\cdot$m