Static Equilibrium Extension

content/tutorial-static_equilibrium.md

@@ -39,8 +39,9 @@ import matplotlib.pyplot as plt
 
 In this tutorial you will use the following NumPy tools:
 
-* [`np.linalg.norm`](https://numpy.org/doc/stable/reference/generated/numpy.linalg.norm.html) : this function determines the measure of vector magnitude
-* [`np.cross`](https://numpy.org/doc/stable/reference/generated/numpy.cross.html) : this function takes two matrices and produces the cross product
+* [`np.linalg.norm`](https://numpy.org/doc/stable/reference/generated/numpy.linalg.norm.html) : This function determines the measure of vector magnitude
+* [`np.cross`](https://numpy.org/doc/stable/reference/generated/numpy.cross.html) : This function takes two matrices and produces the cross product
+* [`np.linalg.solve`](https://numpy.org/doc/stable/reference/generated/numpy.linalg.solve.html) : This function produces the solution (x) to a linear system of equations in the form of A * x = B
 
 +++
 
@@ -245,10 +246,10 @@ forceCord = cordUnit * cordTension
 print("Force from the cord =", forceCord)
 ```
 
-In order to find the moment you need the cross product of the force vector and the radius.
+In order to find the moment you need the cross product of the distance and the force vector.
 
 ```{code-cell}
-momentCord = np.cross(forceCord, poleDirection)
+momentCord = np.cross(poleDirection, forceCord)
 print("Moment from the cord =", momentCord)
 ```
 
@@ -263,107 +264,105 @@ print("Reaction moment =", M)
 ```
 
 ### Another Example
-Let's look at a slightly more complicated model.  In this example you will be observing a beam with two cables and an applied force.  This time you need to find both the tension in the cords and the reaction forces of the beam. *(Source: [Vector Mechanics for Engineers: Statics](https://www.mheducation.com/highered/product/vector-mechanics-engineers-statics-beer-johnston/M9780077687304.html), Problem 4.106)*
+Let's look at a slightly more complicated model.  In this example you will be observing a beam with two cables and an applied force.  This time you need to find both the tension in the cords and the reaction forces of the beam. *(Source: [Vector Mechanics for Engineers: Statics, 12th Edition](https://www.mheducation.com/highered/product/vector-mechanics-engineers-statics-beer-johnston/M9781259977268.html), Problem 4.106. ISBN13: 9781259977268)*
 
 
 ![image.png](_static/problem4.png)
 
-Define distance *a* as 3 meters
+Define distance *a* as 3 meters. The ball joint at A can apply reaction forces, but no reation torques.
 
 
-As before, start by defining the location of each relevant point as an array.
+As before, start by defining the location of each relevant point as an array. For this problem vertical arrays are more convenient.
 
 ```{code-cell}
-A = np.array([0, 0, 0])
-B = np.array([0, 3, 0])
-C = np.array([0, 6, 0])
-D = np.array([1.5, 0, -3])
-E = np.array([1.5, 0, 3])
-F = np.array([-3, 0, 2])
+A = np.array([[0], [0], [0]])
+B = np.array([[0], [3], [0]])
+C = np.array([[0], [6], [0]])
+D = np.array([[1.5], [0], [-3]])
+E = np.array([[1.5], [0], [3]])
+F = np.array([[-3], [0], [2]])
 ```
 
 From these equations, you start by determining vector directions with unit vectors.
 
 ```{code-cell}
-AB = B - C
+AB = B - A
 AC = C - A
 BD = D - B
 BE = E - B
 CF = F - C
 
-UnitBD = BD / np.linalg.norm(BD)
-UnitBE = BE / np.linalg.norm(BE)
-UnitCF = CF / np.linalg.norm(CF)
+Unit_BD = BD / np.linalg.norm(BD)
+Unit_BE = BE / np.linalg.norm(BE)
+Unit_CF = CF / np.linalg.norm(CF)
 
-RadBD = np.cross(AB, UnitBD)
-RadBE = np.cross(AB, UnitBE)
-RadCF = np.cross(AC, UnitCF)
+Rad_BD = np.cross(AB, Unit_BD, axis=0)
+Rad_BE = np.cross(AB, Unit_BE, axis=0)
+Rad_CF = np.cross(AC, Unit_CF, axis=0)
 ```
 
 This lets you represent the tension (T) and reaction (R) forces acting on the system as
 
-$$\left[
-\begin{array}
-~1/3 & 1/3 & 1 & 0 & 0\\
--2/3 & -2/3 & 0 & 1 & 0\\
--2/3 & 2/3 & 0 & 0 & 1\\
-\end{array}
-\right]
-\left[
-\begin{array}
-~T_{BD}\\
-T_{BE}\\
-R_{x}\\
-R_{y}\\
-R_{z}\\
-\end{array}
-\right]
-=
-\left[
-\begin{array}
-~195\\
-390\\
--130\\
-\end{array}
-\right]$$
-
-and the moments as
-
-$$\left[
-\begin{array}
-~2 & -2\\
-1 & 1\\
-\end{array}
-\right]
-\left[
-\begin{array}
-~T_{BD}\\
-T_{BE}\\
-\end{array}
-\right]
-=
-\left[
-\begin{array}
-~780\\
-1170\\
-\end{array}
-\right]$$
+$\sum F_{x} = 0 = \frac{1}{3}T_{BD}+\frac{1}{3}T_{BE}-\frac{3}{7}T_{CF}+R_{x}$
 
-Where $T$ is the tension in the respective cord and $R$ is the reaction force in a respective direction. Then you just have six equations:
+$\sum F_{y} = 0 = (-\frac{2}{3})T_{BD}-\frac{2}{3}T_{BE}-\frac{6}{7}T_{CF}+R_{y}$
 
+$\sum F_{z} = 0 = (-\frac{2}{3})T_{BD}+\frac{2}{3}T_{BE}+\frac{2}{7}T_{CF}+R_{z}$
 
-$\sum F_{x} = 0 = T_{BE}/3+T_{BD}/3-195+R_{x}$
+and the moments as
 
-$\sum F_{y} = 0 = (-\frac{2}{3})T_{BE}-\frac{2}{3}T_{BD}-390+R_{y}$
+$\sum M_{x} = 0 = (-2)T_{BD}+2T_{BE}+\frac{12}{7}T_{CF}$
 
-$\sum F_{z} = 0 = (-\frac{2}{3})T_{BE}+\frac{2}{3}T_{BD}+130+R_{z}$
+$\sum M_{y} = 0 = (0)T_{BD}-(0)T_{BE}+(0)T_{CF}$
 
-$\sum M_{x} = 0 = 780+2T_{BE}-2T_{BD}$
+$\sum M_{z} = 0 = (-)T_{BD}-T_{BE}+\frac{18}{7}T_{CF}$
 
-$\sum M_{z} = 0 = 1170-T_{BE}-T_{BD}$
+Where $T$ is the tension in the respective cord and $R$ is the reaction force in a respective direction. $M_{y}$ contains no information and can be discarded. $T_{CF}$ is known to be 455N and can be moved to the opposite side of the equation. You now have five unknowns with five equations that can be represented by a linear system. Stacking the vectors solved above produces a matrix, a 2D array. With the matrix of coefficients for each of the unkown variables on the left hand side of the equation and all of the known values on the right hand side, we can use NumPy's linear solver to obtain the solution.
 
+$$ \begin{bmatrix}
+1/3 & 1/3 & 1 & 0 & 0 \\
+-2/3 & -2/3 & 0 & 1 & 0 \\
+-2/3 & 2/3 & 0 & 0 & 1 \\
+-2 & 2 & 0 & 0 & 0 \\
+-1 & -1 & 0 & 0 & 0 \\
+\end{bmatrix}
+\begin{bmatrix}
+T_{BD} \\
+T_{BE} \\
+R_{x} \\
+R_{y} \\
+R_{z} \\
+\end{bmatrix}
+=
+\begin{bmatrix}
+195 \\
+390 \\
+-130 \\
+-780 \\
+-1170 \\
+\end{bmatrix} $$
 
-You now have five unknowns with five equations, and can solve for:
+```{code-cell}
+# sum forces
+unknown_Forces = np.hstack((Unit_BD, Unit_BE, np.eye(3)))
+# sum torques
+unknown_Torques = np.hstack((Rad_BD, Rad_BE, np.zeros((3,3))))
+# -1 due to being moved to the RHS
+T_CF = 455
+known_Forces = -1 * T_CF * Unit_CF
+known_Torques = -1 * T_CF * Rad_CF
+
+# remove M_y
+unknown_Torques = np.delete(unknown_Torques, 1, 0)
+known_Torques = np.delete(known_Torques, 1, 0)
+
+# combine into a single system
+LHS = np.vstack((unknown_Forces, unknown_Torques))
+RHS = np.vstack((known_Forces, known_Torques))
+
+solution = np.linalg.solve(LHS, RHS)
+print(solution)
+```
 
 $\ T_{BD} = 780N$
 
@@ -387,5 +386,5 @@ This same process can be applied to kinetic problems or in any number of dimensi
 
 ### References
 
-1. [Vector Mechanics for Engineers: Statics (Beer & Johnston & Mazurek)](https://www.mheducation.com/highered/product/vector-mechanics-engineers-statics-beer-johnston/M9780077687304.html)
+1. [Vector Mechanics for Engineers: Statics (Beer & Johnston & Mazurek)](https://www.mheducation.com/highered/product/vector-mechanics-engineers-statics-beer-johnston/M9781259977268.html)
 2. [NumPy Reference](https://numpy.org/doc/stable/reference/)