The hinged plate is supported by


The hinged plate is supported by the cord AB. If the force in the cord is F = 340 lb, express this force, directed from A toward B, as a Cartesian vector. What is the length of the cord?

The hinged plate is supported by

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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Let us first determine the locations of points A and B.

The hinged plate is supported by

Using the diagram, the locations of the points are:

A:(8i+9j+0k) ft

B:(0i+0j+12k) ft

 

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

r_{AB}\,=\,\left\{(0-8)i+(0-9)j+(12-0)k\right\}=\left\{-8i-9j+12k\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 the position vector is also the length of the cord.

magnitude of r_{AB}\,=\,\sqrt{(-8)^2+(-9)^2+(12)^2}=17 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 write the unit vector for this position vector.

u_{AB}\,=\,\left(-\dfrac{8}{17}i-\dfrac{9}{17}j+\dfrac{12}{17}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)}}

 

Finally, we can express the force in the cord in Cartesian vector form. To do so, we multiply the magnitude of the force by the unit vector.

F_{AB}=340\left(-\dfrac{8}{17}i-\dfrac{9}{17}j+\dfrac{12}{17}k\right)

F_{AB}=\left\{-160i-180j+240k\right\} lb

 

Final Answer:

Length of cord = 17 ft

F_{AB}=\left\{-160i-180j+240k\right\} lb

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-149.

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