# Determine the angle between the edges

Determine the angle between the edges of the sheet-metal bracket.

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 two position vectors along the top edge and right edge of the hinge (shown in the diagram below).

The position vectors are as follows:

$r_1\,=\,\left\{400i+0j+250k\right\}$ mm

$r_2\,=\,\left\{50i+300j+0k\right\}$ mm

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

Next, we will find the magnitude of each position vector.

magnitude of $r_1\,=\,\sqrt{(400)^2+(0)^2+(250)^2}\,=\,471.7$ mm

magnitude of $r_2\,=\,\sqrt{(50)^2+(300)^2+(0)^2}\,=\,304.1$ mm

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 will now find the dot product between the position vectors $r_1$ and $r_2$.

$r_1\cdot r_2\,=\,\left[(400)(50)+(0)(300)+(250)(0)\right]$

$r_1\cdot r_2\,=\,20000$

The dot product is the product of the magnitudes of two vectors and the cosine angle $\theta$ between them. For example, if we had two vectors, $\mathbf{A}$ and $\mathbf{B}$, the dot product of the two is: $A\cdot B\,=\,AB\cos \theta$. In cartesian vector form, $A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z$. The angle formed between two vectors can be found by: $\theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)$

Now, we can find the angle between the two vectors like so:

$\theta\,=\,\cos^{-1}\left(\dfrac{20000}{(471.7)(304.1)}\right)$

$\theta\,=\,82^0$

$\theta\,=\,82^0$