# The door is held opened by means of two chains

The door is held opened by means of two chains. If the tension in AB and CD is $F_A = 300$ N and $F_C = 250$ N, respectively, express each of these forces in Cartesian vector form.

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

#### Solution:

We will first write the locations of points A, B, C and D in Cartesian vector form.

Notice how we can figure our the z-coordinates of point A and C using the triangle we formed. Using sine function (opposite over hypotenuse), we can figure out $h$ which is the height (z-component). The height of points A and C is:

$\sin30^0\,=\,\dfrac{h}{1.5}$

$h\,=\,0.75$ m

We can also figure out the $l$ portion as well, by using cosine (adjacent over hypotenuse). This will allow us to figure out the total y-lengths of points A and C.

$\cos30^0\,=\,\dfrac{l}{1.5}$

$l\,=\,1.3$ m

Let us now write the coordinates of each point.

Using the diagram and the lengths we found, we can now write the location of each point.

$A:(0i-2.3j+0.75k)$ m

$B:(0i+0j+0k)$ m

$C:(-2.5i-2.3j+0.75k)$ m

$D:(-0.5i+0j+0k)$ m

Let us now write the position vectors for points from A to B and C to D.

$r_{AB}\,=\,\left\{(0-0)i+(0-(-2.3))j+(0-0.75)k\right\}$

$r_{AB}\,=\,\left\{0i+2.3j-0.75k\right\}$

$r_{CD}\,=\,\left\{(-0.5-(-2.5))i+(0-(-2.3))j+(0-0.75)k\right\}$

$r_{CD}\,=\,\left\{2i+2.3j-0.75k\right\}$

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Let us now find the magnitude of each position vector.

magnitude of $r_{AB}\,=\,\sqrt{(0)^2+(2.3)^2+(-0.75)^2}$

magnitude of $r_{AB}\,=\,2.42$ m

magnitude of $r_{CD}\,=\,\sqrt{(2)^2+(2.3)^2+(-0.75)^2}$

magnitude of $r_{CD}\,=\,3.14$ 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.

We can now write our unit vectors for $r_{AB}$ and $r_{CD}$.

$u_{AB}\,=\,\left(\dfrac{0}{2.42}i\,+\,\dfrac{2.3}{2.42}j\,-\,\dfrac{.75}{2.42}k\right)$

$u_{CD}\,=\,\left(\dfrac{2}{3.14}i\,+\,\dfrac{2.3}{3.14}j\,-\,\dfrac{.75}{3.14}k\right)$

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 write each force in Cartesian vector form by multiplying the force by the corresponding unit vector.

$F_A\,=\,300\left(\dfrac{0}{2.42}i\,+\,\dfrac{2.3}{2.42}j\,-\,\dfrac{.75}{2.42}k\right)$

$F_A\,=\,\left\{0i+285j-93k\right\}$ N

$F_C\,=\,250\left(\dfrac{2}{3.14}i\,+\,\dfrac{2.3}{3.14}j\,-\,\dfrac{.75}{3.14}k\right)$

$F_C\,=\,\left\{159i+183j-60k\right\}$ N

$F_A\,=\,\left\{0i+285j-93k\right\}$ N
$F_C\,=\,\left\{159i+183j-60k\right\}$ N