The window is held open by chain AB 2


The window is held open by chain AB. Determine the length of the chain, and express the 50-lb force acting at A along the chain as a Cartesian vector and determine its coordinate direction angles.

The window is held open by chain AB

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 figure out the locations of points A and B and write it in Cartesian form. To figure out the location of point A, we will need to use trigonometry.

The window is held open by chain AB

Notice how we can construct a right angle triangle to figure out the location of point A. Using the diagram, the locations of points A and B are:

A:(5\cos40^0i+8j+5\sin40^0k)=(3.83i+8j+3.21k) ft

B:(0i+5j+12k) ft

 

We can now write our position vector from point A to B.

r_{AB}\,=\,\left\{(0-3.83)i+(5-8)j+(12-3.21)k\right\}=\left\{-3.83i-3j+8.79k\right\}

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 also the length of the chain.

magnitude of r_{AB}\,=\,\sqrt{(-3.83)^2+(-3)^2+(8.79)^2}=10 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 find the unit vector.

u_{AB}\,=\,\left(-\dfrac{3.83}{10}i-\dfrac{3}{10}j+\dfrac{8.79}{10}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)}}

 

We can now express the force in Cartesian vector form:

F_{AB}=50\left(-\dfrac{3.83}{10}i-\dfrac{3}{10}j+\dfrac{8.79}{10}k\right)

F_{AB}=\left\{-19.15i-15j+43.9k\right\} lb

 

To find the coordinate direction angles, we need to figure out the magnitude of the resultant force.

magnitude of F_R\,=\,\sqrt{(-19.15)^2+(-15)^2+(43.9)^2}=50.2

 

Now, we can figure out the coordinate direction angles by taking the cosine inverse of each component of the resultant force divided by the magnitude.

\alpha=\cos^{-1}\left(\dfrac{-19.15}{50.2}\right)=112^0

\beta=\cos^{-1}\left(\dfrac{-15}{50.2}\right)=107^0

\gamma=\cos^{-1}\left(\dfrac{43.9}{50.2}\right)=29^0

 

Final Answers:

length of chain = 10 ft

F_{AB}=\left\{-19.15i-15j+43.9k\right\} lb

\alpha=112^0

\beta=107^0

\gamma=29^0

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-111.

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