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:
This question involves expressing forces in Cartesian vector form. If you are unsure on how to do this, read the detailed guide on expressing forces in Cartesian notation.
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.
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)}}
