# 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.

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

#### Solution:

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.

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.

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

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

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

length of chain = 10 ft

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

$\alpha=112^0$

$\beta=107^0$

$\gamma=29^0$