# The cable attached to the tractor at B

The cable attached to the tractor at B exerts a force of 350 lb on the framework. Express this force as a Cartesian vector. 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 a position vector from A to B. To do so, let us first write the locations of points A and B in Cartesian vector. The locations of the points can be calculated using the diagram. Also, note that we will use trigonometry to figure out the location of point B.

$A:(0i+0j+35k)$ ft

$B:(50\sin20^0i+50\cos20^0j+0k)=(17.1i+47j+0k)$ ft

Let us now write a position vector from A to B.

$r_{AB}\,=\,\left\{(17.1-0)i+(47-0)j+(0-35)k\right\}=\left\{-17.1i-47j+35k\right\}$ ft

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

The magnitude of this position vector is:

magnitude of $r_{AB}\,=\,\sqrt{(17.1)^2+(47)^2+(-35)^2}=61$ ft

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 our unit vector:

$u_{AB}\,=\,\left(\dfrac{17.1}{61}i+\dfrac{47}{61}j-\dfrac{35}{61}k\right)$

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

The force expressed in Cartesian vector form is:

$F_{BA}=350\left(\dfrac{17.1}{61}i+\dfrac{47}{61}j-\dfrac{35}{61}k\right)$

$F_{BA}=\left\{98.1i+270j-201k\right\}$ lb

$F_{BA}=\left\{98.1i+270j-201k\right\}$ lb