# If FB = 560 N and FC = 700 N determine 3

If $F_B$ = 560 N and $F_C$ = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag 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 points A, B and C are in Cartesian vector form. Looking at the diagram, we can write the following:

$A:(0i+0j+6k)$

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

$C:(3i+2j+0k)$

We can now find the position vectors from A to B and A to C.

$r_{AB}\,=\,\left\{(2-0)i+(-3-0)j+(0-6)k\right\}$

$r_{AB}\,=\,\left\{2i-3j-6k\right\}$ m

$r_{AC}\,=\,\left\{(3-0)i+(2-0)j+(0-6)k\right\}$

$r_{AC}\,=\,\left\{3i+2j-6k\right\}$ m

We now need to find the magnitude of each position vector so that we can use it to find the unit vector in the next step.

Magnitude of $r_{AB}\,=\,\sqrt{(2^2)+(-3)^2+(-6)^2}$

Magnitude of $r_{AB}\,=\,7$ m

Magnitude of $r_{AC}\,=\,\sqrt{(3^2)+(2)^2+(-6)^2}$

Magnitude of $r_{AC}\,=\,7$ m

Let us now find the unit vector for each force. Remember, this is found by dividing each component of the position vector by it’s magnitude.

$u_{AB}\,=\,\left(\dfrac{2}{7}i\,-\,\dfrac{3}{7}j\,-\,\dfrac{6}{7}k\right)$

$u_{AC}\,=\,\left(\dfrac{3}{7}i\,+\,\dfrac{2}{7}j\,-\,\dfrac{6}{7}k\right)$

We can now write force $F_B$ and $F_C$ in Cartesian form.

$F_B\,=\,560\left(\dfrac{2}{7}i\,-\,\dfrac{3}{7}j\,-\,\dfrac{6}{7}k\right)$

$F_B\,=\,\left\{160i\,-\,240j\,-\,480k\right\}$ N

$F_C\,=\,700\left(\dfrac{3}{7}i\,+\,\dfrac{2}{7}j\,-\,\dfrac{6}{7}k\right)$

$F_C\,=\,\left\{300i\,+\,200j\,-\,600k\right\}$ N

(Here we multiplied each individual component by the force. In other words, expand the brackets using FOIL)

We can find the resultant force by adding both forces together like so:

$F_R\,=\,F_B\,+\,F_C$

$F_R\,=\,\left\{160i\,-\,240j\,-\,480k\right\}\,+\,\left\{300i\,+\,200j\,-\,600k\right\}$

$F_R\,=\,\left\{460i-40j-1080k\right\}$ N

To figure out the coordinate direction angles, we need to find the magnitude of the resultant force we just found.

magnitude of $F_R\,=\,\sqrt{(460^2)+(-40)^2+(-1080)^2}$

magnitude of $F_R\,=\,1174.56$ N

Coordinate direction angles can be found taking the cosine inverse of each component of the resultant divided by the magnitude:

$\alpha\,=\,\cos^{-1}\left(\dfrac{460}{1174.56}\right)\,=\,67^0$

$\beta\,=\,\cos^{-1}\left(\dfrac{-40}{1174.56}\right)\,=\,92^0$

$\gamma\,=\,\cos^{-1}\left(\dfrac{-1080}{1174.56}\right)\,=\,157^0$

magnitude of $F_R\,=\,1174.56$ N

$\alpha\,=\,67^0$

$\beta\,=\,92^0$

$\gamma\,=\,157^0$

## 3 thoughts on “If FB = 560 N and FC = 700 N determine”

• Mark

• • 