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.


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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\{-1.5i-2j+0k\right\} m



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.




Final Answer:



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

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