# The guy wires are used to support the telephone pole

The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole. 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 write the locations of point A, B, C, and D in Cartesian vector form. Using the diagram, the locations of the points are:

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

$B:(0i+0j+5.5k)$ m

$C:(-1i+4j+0k)$ m

$D:(2i-3j+0k)$ m

We will now write position vectors for points from A to C and B to D.

$r_{AC}\,=\,\left\{(-1-0)i+(4-0)j+(0-4)k\right\}$

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

$r_{BD}\,=\,\left\{(2-0)i+(-3-0)j+(0-5.5)k\right\}$

$r_{BD}\,=\,\left\{2i-3j-5.5k\right\}$ 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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Let us now find the magnitude of each position vector.

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

magnitude of $r_{AC}\,=\,5.74$ m

magnitude of $r_{BD}\,=\,\sqrt{(2)^2+(-3)^2+(-5.5)^2}$

magnitude of $r_{BD}\,=\,6.58$ 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.

We can now write the unit vectors for $r_{AC}$ and $r_{BD}$.

$u_{AB}\,=\,\left(-\dfrac{1}{5.74}i\,+\,\dfrac{4}{5.74}j\,-\,\dfrac{4}{5.74}k\right)$

$u_{BD}\,=\,\left(\dfrac{2}{6.58}i\,-\,\dfrac{3}{6.58}j\,-\,\dfrac{5.5}{6.58}k\right)$

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, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

Now, we will write each force in Cartesian vector form by multiplying the force by each corresponding unit vector.

$F_A\,=\,250\left(-\dfrac{1}{5.74}i\,+\,\dfrac{4}{5.74}j\,-\,\dfrac{4}{5.74}k\right)$

$F_A\,=\,\left\{-43.5i+174j-174k\right\}$ N

$F_B\,=\,175\left(\dfrac{2}{6.58}i\,-\,\dfrac{3}{6.58}j\,-\,\dfrac{5.5}{6.58}k\right)$

$F_B\,=\,\left\{53.2i-79.8j-146k\right\}$ N

$F_A\,=\,\left\{-43.5i+174j-174k\right\}$ N
$F_B\,=\,\left\{53.2i-79.8j-146k\right\}$ N