The pipe is supported at its end


The pipe is supported at its end by a cord AB. If the cord exerts a force of F = 12 lb on the pipe at A, express this force as a Cartesian vector.

The pipe is supported at its end

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 write the locations of points A and B in Cartesian vector form. To do so, we will construct a right angle triangle to figure out the location of point A using trigonometry.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Using the diagram, the locations of the points are:

A:(5i+3\cos20^0j-3\sin20^0k)=(5i+2.82j-1.03k) ft

B:(0i+0j+6k) ft

 

We can now write a position vector from A to B.

r_{AB}\,=\,\left\{(0-5)i+(0-2.82)j+(6-(-1.03))k\right\}=\left\{-5i-2.82j+7.03k\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{(-5)^2+(-2.82)^2+(7.03)^2}=9.07 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.

 

Let us now write the unit vector for this position vector:

u_{AB}\,=\,\left(-\dfrac{5}{9.07}i-\dfrac{2.82}{9.07}j+\dfrac{7.03}{9.07}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)}}

 

Now, we can express this force as a Cartesian vector.

F_{AB}=12\left(-\dfrac{5}{9.07}i-\dfrac{2.82}{9.07}j+\dfrac{7.03}{9.07}k\right)

F_{AB}=\left\{-6.61i-3.73j+9.3k\right\} lb

 

Final Answer:

F_{AB}=\left\{-6.61i-3.73j+9.3k\right\} lb

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-105.

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