Made between the axes OA of the flag pole


Determine the angles θ and Φ made between the axes OA of the flag pole and AB and AC, respectively, of each cable.

Made between the axes OA of 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:

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We will first find three position vectors from points A to B,  A to C and A to O. To do so, we will first write the locations of points A, B, C, and O in vector Cartesian form.

Made between the axes OA of the flag pole

From the diagram, the locations of the points are:

A:(0i+4j+3k) m

B:(1.5i+0i+6k) m

C:(-2i+0i+4k) m

O:(0i+0j+0k) m

 

Let us now find the position vectors.

r_{AB}\,=\,\left\{(1.5-0)i+(0-4)j+(6-3)k\right\}

r_{AB}\,=\,\left\{1.5i-4j+3k\right\} m

 

r_{AC}\,=\,\left\{(-2-0)i+(0-4)j+(4-3)k\right\}

r_{AC}\,=\,\left\{-2i-4j+1k\right\} m

 

r_{AO}\,=\,\left\{(0-0)i+(0-4)j+(0-3)k\right\}

r_{AO}\,=\,\left\{0i-4j-3k\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

 

We will now find the magnitude of each position vector.

magnitude of r_{AB}\,=\,\sqrt{(1.5)^2+(-4)^2+(3)^2}\,=\,5.22 m

magnitude of r_{AC}\,=\,\sqrt{(-2)^2+(-4)^2+(1)^2}\,=\,4.58 m

magnitude of r_{AO}\,=\,\sqrt{(0)^2+(-4)^2+(-3)^2}\,=\,5 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.

 

Next, we will find the dot product between r_{AB} and r_{AO}.

r_{AB}\cdot r_{AO}\,=\,\left[(1.5)(0)+(-4)(-4)+(3)(-3)\right]

r_{AB}\cdot r_{AO}\,=\,7

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)

 

Now, we will find the dot product between r_{AC} and r_{AO}.

r_{AC}\cdot r_{AO}\,=\,\left[(-2)(0)+(-4)(-4)+(1)(-3)\right]

r_{AC}\cdot r_{AO}\,=\,13

 

We can now figure out the two angles required using the dot product.

\theta\,=\,\cos^{-1}\left(\dfrac{\text{dot product of the two vectors}}{(\text{magnitude of 1st vector})(\text{magnitude of 2nd vector})}\right)

\theta\,=\,\cos^{-1}\left(\dfrac{7}{(5.22)(5)}\right)

\theta\,=\,74.4^0

 

\phi\,=\,\cos^{-1}\left(\dfrac{13}{(4.58)(5)}\right)

\phi\,=\,55.4^0

 

Final Answers:

\theta\,=\,74.4^0

\phi\,=\,55.4^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-131.

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