Determine the angle ϴ between the pipe segments 2


Determine the angle ϴ between the pipe segments BA and BC.

Determine the angle ϴ between the pipe segments

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 write two position vectors, from points B to A and B to C. To do so, we must first write the locations of points A, B and C in Cartesian vector form.

Determine the angle ϴ between the pipe segments

Using the diagram, the locations of the points are:

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

B:(1.5i+2j+0k) m

C:(3.5i+3j-2k) m

 

We can now write our position vectors.

r_{BA}\,=\,\left\{(0-1.5)i+(0-2)j+(0-0)k\right\}

r_{BA}\,=\,\left\{-1.5i-2j+0k\right\} m

 

r_{BC}\,=\,\left\{(3.5-1.5)i+(3-2)j+(-2-0)k\right\}

r_{BC}\,=\,\left\{2i+1j-2k\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

 

Let us now find the magnitude of our position vectors.

magnitude of r_{BA}\,=\,\sqrt{(-1.5)^2+(-2)^2+(0)^2}\,=\,2.5 m

magnitude of r_{BC}\,=\,\sqrt{(2)^2+(1)^2+(-2)^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.

 

Next, we will find the dot product between the two position vectors.

r_{BA}\cdot r_{BC}\,=\,\left[(-1.5)(2)+(-2)(1)+(0)(-2)\right]

r_{BA}\cdot r_{BC}\,=\,-5

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 that we know the magnitudes of each position vector, and the dot product value of our position vectors, we can find the angle between the two pipe segments.

\theta\,=\,\cos^{-1}\left(\dfrac{-5}{(2.5)(3)}\right)

\theta\,=\,131.8^0

 

Final Answer:

\theta\,=\,131.8^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-130.

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