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.

#### Solution:

Show me the final answer↓

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:

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\{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

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

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

Using the magnitude, we can find the unit vector.

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

We can now write each force in Cartesian vector form.

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\,=\,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\,=\,1491.1 N

Finally, we can calculate the coordinate direction angles.

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

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

#### Final Answers:

\alpha\,=\,79.8^0

\beta\,=\,88.4^0

\gamma\,=\,169.7^0

Good ?