# Along the diagonal of the parallelepiped

A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment of F about point A, using $M_A=r_B\times F$ and $M_A=r_C\times F$. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let us first figure out the locations of the points, A, B and C and write them in Cartesian vector form. Using the image, the locations of the points are:

$A:(0.4i+0.6j+0k)$ m

$B:(0.4i+0j+0k)$ m

$C:(0i+0.6j+0.2k)$ m

Let us now write position vectors for $r_B$ and $r_C$. It’s especially important to note that $r_B$ is actually a position vector from A to B and $r_C$ is a position vector from A to C.

$r_{AB}\,=\,\left\{(0.4-0.4)i+(0-0.6)j+(0-0)k\right\}=\left\{0i-0.6j+0k\right\}$ m

$r_{AC}\,=\,\left\{(0-0.4)i+(0.6-0.6)j+(0.2-0)k\right\}=\left\{-0.4i+0j+0.2k\right\}$ m

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

We now need to express the force being applied in Cartesian vector form. To do so, we need to look at the path the force is going though. That path can be shown by a position vector from B to C. Thus, let us now figure out the position vector $r_{BC}.$

$r_{BC}\,=\,\left\{(0-0.4)i+(0.6-0)j+(0.2-0)k\right\}=\left\{-0.4i+0.6j+0.2k\right\}$ m

The magnitude of this position vector is:

magnitude of $r_{BC}\,=\,\sqrt{(-0.4)^2+(0.6)^2+(0.2)^2}=0.748$ m

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

The unit vector for this position vector is:

$u_{BC}\,=\,\left(-\dfrac{0.4}{0.748}i+\dfrac{0.6}{0.748}j+\dfrac{0.2}{0.748}k\right)$

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

We can now express the force in Cartesian vector form by multipling the magnitude of the force by the unit vector representing the path it takes:

$F=100\left(-\dfrac{0.4}{0.748}i+\dfrac{0.6}{0.748}j+\dfrac{0.2}{0.748}k\right)$

$F=\left\{-53.5i+80.2j+26.7k\right\}$

The moment at A using $r_B$ is:

$M_A=r_B\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\0&-0.6&0\\-53.5&80.2&26.7\end{bmatrix}$

$M_A=\left\{-16i-32.1k\right\}\,N\cdot m$

The moment at A using $r_C$ is:

$M_A=r_C\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\-0.4&0&0.2\\-53.5&80.2&26.7\end{bmatrix}$

$M_A=\left\{-16i-32.1k\right\}\,N\cdot m$

$M_A=\left\{-16i-32.1k\right\}\,N\cdot m$