# The tower is held in place by three cables

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:

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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
$\alpha\,=\,79.8^0$
$\beta\,=\,88.4^0$
$\gamma\,=\,169.7^0$