# Made between the axes OA of the flag pole

Determine the angles θ and Φ made between the axes OA of the flag pole and AB and AC, respectively, of each cable.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first find three position vectors from points A to B,  A to C and A to O. To do so, we will first write the locations of points A, B, C, and O in vector Cartesian form.

From the diagram, the locations of the points are:

$A:(0i+4j+3k)$ m

$B:(1.5i+0i+6k)$ m

$C:(-2i+0i+4k)$ m

$O:(0i+0j+0k)$ m

Let us now find the position vectors.

$r_{AB}\,=\,\left\{(1.5-0)i+(0-4)j+(6-3)k\right\}$

$r_{AB}\,=\,\left\{1.5i-4j+3k\right\}$ m

$r_{AC}\,=\,\left\{(-2-0)i+(0-4)j+(4-3)k\right\}$

$r_{AC}\,=\,\left\{-2i-4j+1k\right\}$ m

$r_{AO}\,=\,\left\{(0-0)i+(0-4)j+(0-3)k\right\}$

$r_{AO}\,=\,\left\{0i-4j-3k\right\}$ 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 will now find the magnitude of each position vector.

magnitude of $r_{AB}\,=\,\sqrt{(1.5)^2+(-4)^2+(3)^2}\,=\,5.22$ m

magnitude of $r_{AC}\,=\,\sqrt{(-2)^2+(-4)^2+(1)^2}\,=\,4.58$ m

magnitude of $r_{AO}\,=\,\sqrt{(0)^2+(-4)^2+(-3)^2}\,=\,5$ 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.

Next, we will find the dot product between $r_{AB}$ and $r_{AO}$.

$r_{AB}\cdot r_{AO}\,=\,\left[(1.5)(0)+(-4)(-4)+(3)(-3)\right]$

$r_{AB}\cdot r_{AO}\,=\,7$

The dot product is the product of the magnitudes of two vectors and the cosine angle $\theta$ between them. For example, if we had two vectors, $\mathbf{A}$ and $\mathbf{B}$, the dot product of the two is: $A\cdot B\,=\,AB\cos \theta$. In cartesian vector form, $A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z$. The angle formed between two vectors can be found by: $\theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)$

Now, we will find the dot product between $r_{AC}$ and $r_{AO}$.

$r_{AC}\cdot r_{AO}\,=\,\left[(-2)(0)+(-4)(-4)+(1)(-3)\right]$

$r_{AC}\cdot r_{AO}\,=\,13$

We can now figure out the two angles required using the dot product.

$\theta\,=\,\cos^{-1}\left(\dfrac{\text{dot product of the two vectors}}{(\text{magnitude of 1st vector})(\text{magnitude of 2nd vector})}\right)$

$\theta\,=\,\cos^{-1}\left(\dfrac{7}{(5.22)(5)}\right)$

$\theta\,=\,74.4^0$

$\phi\,=\,\cos^{-1}\left(\dfrac{13}{(4.58)(5)}\right)$

$\phi\,=\,55.4^0$

$\theta\,=\,74.4^0$
$\phi\,=\,55.4^0$