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:
Show me the final answer↓
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
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
Final answer:
\theta\,=\,82^0