# Determine the lengths of wires AD, BD, and CD

Determine the lengths of wires AD, BD, and CD. The ring at D is midway between A and B.

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 determine where points A, B, C and D are with respect to the origin (0i+0j+0k).

From the diagram, we can determine the location of each point and write it in Cartesian vector form.

$A:(2i+0j+1.5k)$ m

$B:(0i+2j+0.5k)$ m

$C:(0i+oj+2k)$ m

$D:\left(\dfrac{2}{2}i+\dfrac{2}{2}j+\dfrac{1.5+0.5}{2}k\right)\,=\,(1i+1j+1k)$ m

(Remember D is midway between A and B. That means it is half way from each direction, including a total height of 2 m)

The next step is to figure out the position vector for each segment. The segments are DA, DC and DB.

$r_{DA}\,=\,\left\{(2-1)i+(0-1)j+(1.5-1)k\right\}\,=\,\left\{1i-1j+0.5k\right\}$ m

$r_{DC}\,=\,\left\{(0-1)i+(0-1)j+(2-1)k\right\}\,=\,\left\{-1i-1j+1k\right\}$ m

$r_{DB}\,=\,\left\{(0-1)i+(2-1)j+(0.5-1)k\right\}\,=\,\left\{-1i+1j-0.5k\right\}$ m

Here, we subtract each corresponding component from each point. For example, when we found $r_{DA}$ we subtracted the corresponding components of D from A.

We can now find the magnitude of each segment, which is also the length of each segment.

Magnitude of $r_{DA}\,=\,\sqrt{(1^2)+(-1)^2+(0.5)^2}\,=\,1.5$ m

Magnitude of $r_{DC}\,=\,\sqrt{(-1^2)+(-1)^2+(1)^2}\,=\,1.73$ m

Magnitude of $r_{DB}\,=\,\sqrt{(-1^2)+(1)^2+(-0.5)^2}\,=\,1.5$ m

To find the magnitude, we take each term of the position vector and we find the square root value of each term squared and added together.