Determine the projection of the force F along the pole.
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 find a position vector from point O to the bottom of the pole.
Let us write a position vector from point O to A.
r_{OA}\,=\,\left\{(2-0)i+(2-0)j+(-1-0)k\right\}
r_{OA}\,=\,\left\{2i+2j-1k\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
Now, we will find the magnitude of this position vector.
magnitude of r_{OA}\,=\,\sqrt{(2)^2+(2)^2+(-1)^2}\,=\,3 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 then take the square root of that value.
We can now write a unit vector for our position vector.
u_{OA}\,=\,\left(\dfrac{2}{3}i+\dfrac{2}{3}j-\dfrac{1}{3}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)}}
To find the projection, we have to remember the following equation. This is derived from the dot product.
\text{proj}\,F\,=\,F\cdot u_{OA}
\text{proj}\,F\,=\,\left\{2i+4j+10k\right\}\cdot\left(\dfrac{2}{3}i+\dfrac{2}{3}j-\dfrac{1}{3}k\right)
\text{proj}\,F\,=\,\left[(2)(\dfrac{2}{3})+(4)(\dfrac{2}{3})+(10)(-\dfrac{1}{3})\right]
\text{proj}\,F\,=\,0.67 kN
\text{proj}\,F\,=\,\left\{2i+4j+10k\right\}\cdot\left(\dfrac{2}{3}i+\dfrac{2}{3}j-\dfrac{1}{3}k\right)
\text{proj}\,F\,=\,\left[(2)(\dfrac{2}{3})+(4)(\dfrac{2}{3})+(10)(-\dfrac{1}{3})\right]
\text{proj}\,F\,=\,0.67 kN
The dot product is the product of the magnitudes of two vectors and the cosine angle \theta between them. For example, if we had two vectors, \mathbf{A} and \mathbf{B}, the dot product of the two is: A\cdot B\,=\,AB\cos \theta. In cartesian vector form, A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z. The angle formed between two vectors can be found by: \theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)
Final Answer:
Projection of the force F along the pole is 0.67 kN.