Basic Angle Explanation for Statics

  • #1
ekpm
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My textbook introduces this angle concept really early on and I still don't understand it. It just shows that a normal to a line and some other random angle shown is the same. I don't see any transversal angles or anything. Where did they get the secondary line to form theta for the normal line?
 

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  • #2
I interpeted which is which as I draw on your sketch below. Does it make sense in the story of the textbook ? What is the title of the textbook ?

1677071394922.png
 
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  • #3
ekpm said:
My textbook introduces this angle concept really early on and I still don't understand it. It just shows that a normal to a line and some other random angle shown is the same. I don't see any transversal angles or anything. Where did they get the secondary line to form theta for the normal line?
Are you familiar with the geometrical theorem, "Two lines that have sides mutually perpendicular are equal"? Also, in this case, the figure on the right is the figure on the left rotated clockwise by 90°.
 
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  • #4
ekpm said:
My textbook introduces this angle concept really early on and I still don't understand it. It just shows that a normal to a line and some other random angle shown is the same. I don't see any transversal angles or anything. Where did they get the secondary line to form theta for the normal line?
That is a useful concept for studying friction, for which “normal forces” are important.
Those are forces that are oriented in a normal or perpendicular direction respect to the surface where friction is happening.

In the diagram posted by @anuttarasammyak in post #2, lines in red represent a horizontal and a vertical surface.
Lines in blue represent a normal force to each of those surfaces.

If any of those horizontal and vertical surfaces is rotated certain angle (for any reason), its normal force acting on it must rotate exactly the same angle in order to remain being considered a “normal force”.
 
  • #5
Once more an example, where vectors help a lot. Take a Cartesian coordinate system such that
$$\vec{g}=-g \vec{e}_3.$$
Now introduce the new basis vector
$$\vec{e}_1'=\begin{pmatrix} \cos \alpha \\ 0 \\ \sin \alpha \end{pmatrix}$$
and the normal vector
$$\vec{e}_3'=\begin{pmatrix}-\sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}.$$
The plane is then described as the ##1'2##-plane. For a particle on the plane there's the gravitational force
$$\vec{F}_g=m \vec{g} =\begin{pmatrix}0 \\ 0 \\ -g \end{pmatrix}.$$
Its components in the new frame are
$$\vec{F}_g'=\begin{pmatrix} \vec{e}_1' \cdot \vec{F}_g \\ \vec{e}_2 \cdot \vec{F}_g \\ \vec{e}_3' \cdot \vec{F}_g \end{pmatrix}= \begin{pmatrix}-m g \sin \alpha \\ 0 \\ -m g \cos \alpha \end{pmatrix}.$$
Then there's a contact force from the surface acting normally to the surface such that it compensates the corresponding 3'-component of ##\vec{F}_g##, i.e.,
$$\vec{F}_{\text{N}}'=\begin{pmatrix} 0 \\ 0 \\ mg \cos \alpha \end{pmatrix}.$$
The equation of motion thus reads
$$m \ddot{\vec{x}}'=\vec{F}_{g}' + \vec{F}_{\text{N}}'=\begin{pmatrix}-m g \sin \alpha \\ 0 \\ 0 \end{pmatrix}.$$
 
  • #6
Use geometry to fill in the missing angles of the various triangles formed by the intersecting lines in terms of ##\theta##.

1677080953309.png
 
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  • #7
anuttarasammyak said:
I interpeted which is which as I draw on your sketch below. Does it make sense in the story of the textbook ? What is the title of the textbook ?

View attachment 322705
I actually interpreted the line as the blue one. The angle the line makes with respect to the horizontal is equal to the angle between the normal and the vertical, and vice versa.
 
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