Determine the angle between the edges


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

Determine the angle between the edges of the sheet-metal

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

Solution:

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

Determine the angle between the edges

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

 

Final answer:

\theta\,=\,82^0
This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-113.

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