# The tower is held in place by three cables 1

The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles $\alpha , \beta , \gamma$ of the resultant force. Take x = 15 m, y = 20 m. 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 determine where the locations of point A, B, C, and D is and write it in Cartesian vector form. (note we set x = 15 m, and y = 20 m as stated by the question)

The points are at the following coordinates:

$A:(15i+20j+0k)$

$B:(-6i+4j+0k)$

$C:(16i-18j+0k)$

$D:(0i+0j+24k)$

We will now write the position vectors for cables DA, DB, and DC.

$r_{DA}\,=\,\left\{(15-0)i+(20-0)j+(0-24)k\right\}$

$r_{DA}\,=\,\left\{15i+20j-24k\right\}$ m

$r_{DB}\,=\,\left\{(-6-0)i+(4-0)j+(0-24)k\right\}$

$r_{DB}\,=\,\left\{-6i+4j-24k\right\}$ m

$r_{DC}\,=\,\left\{(16-0)i+(-18-0)j+(0-24)k\right\}$

$r_{DC}\,=\,\left\{16i-18j-24k\right\}$ 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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Next, we will calculate the magnitude of each position vector.

magnitude of $r_{DA}\,=\,\sqrt{(15)^2+(20)^2+(-24)^2}$

magnitude of $r_{DA}\,=\,34.65$ m

magnitude of $r_{DB}\,=\,\sqrt{(-6)^2+(4)^2+(-24)^2}$

magnitude of $r_{DB}\,=\,25.05$ m

magnitude of $r_{DC}\,=\,\sqrt{(16)^2+(-18)^2+(-24)^2}$

magnitude of $r_{DC}\,=\,34$ 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 take the square root of that value.

Using the magnitude, we can find the unit vector.

$u_{DA}\,=\,\dfrac{15}{34.65}i\,+\,\dfrac{20}{34.65}j\,-\,\dfrac{24}{34.65}k$

$u_{DB}\,=\,-\dfrac{6}{25.05}i\,+\,\dfrac{4}{25.05}j\,-\,\dfrac{24}{25.05}k$

$u_{DC}\,=\,\dfrac{16}{34}i\,-\,\dfrac{18}{34}j\,-\,\dfrac{24}{34}k$

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 write each force in Cartesian vector form.

$F_{DA}\,=\,400\left(\dfrac{15}{34.65}i\,+\,\dfrac{20}{34.65}j\,-\,\dfrac{24}{34.65}k\right)$

$F_{DA}\,=\,\left\{173.2i+230.9j-277k\right\}$ N

$F_{DB}\,=\,800\left(-\dfrac{6}{25.05}i\,+\,\dfrac{4}{25.05}j\,-\,\dfrac{24}{25.05}k\right)$

$F_{DB}\,=\,\left\{-191.6i+127.7j-766.5k\right\}$ N

$F_{DC}\,=\,600\left(\dfrac{16}{34}i\,-\,\dfrac{18}{34}j\,-\,\dfrac{24}{34}k\right)$

$F_{DC}\,=\,\left\{282.3i-317.6j-423.5k\right\}$ N

The resultant force is equal to:

$F_R\,=\,F_{DA}\,+\,F_{DB}\,+\,F_{DC}$

$F_R\,=\,263.9i+41j-1467k$ N

(Remember, we simply add each corresponding coordinate of the vectors together)

We can now find the magnitude of the resultant force like so:

magnitude of $F_R\,=\,\sqrt{(263.9)^2+(41)^2+(-1467)^2}$

magnitude of $F_R\,=\,1491.1$ N

Finally, we can calculate the coordinate direction angles.

$\alpha\,=\,\cos^{-1}\left(\dfrac{263.9}{1491.1}\right)\,=\,79.8^0$

$\beta\,=\,\cos^{-1}\left(\dfrac{41}{1491.1}\right)\,=\,88.4^0$

$\gamma\,=\,\cos^{-1}\left(\dfrac{-1467}{1491.1}\right)\,=\,169.7^0$

magnitude of $F_R\,=\,1491.1$ N

$\alpha\,=\,79.8^0$

$\beta\,=\,88.4^0$

$\gamma\,=\,169.7^0$

## One thought on “The tower is held in place by three cables”

• Ankomah Desmond Marfo

Good ?