From 4bd0306ed67e97809e12c498bc5f522c52303c3d Mon Sep 17 00:00:00 2001 From: DanSim Date: Wed, 20 Dec 2023 19:42:29 -0500 Subject: [PATCH] Update tutorial-static_equilibrium.md --- content/tutorial-static_equilibrium.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/content/tutorial-static_equilibrium.md b/content/tutorial-static_equilibrium.md index b4a9d75b..007d9495 100644 --- a/content/tutorial-static_equilibrium.md +++ b/content/tutorial-static_equilibrium.md @@ -319,7 +319,7 @@ $\sum M_{z} = 0 = (-)T_{BD}-T_{BE}+\frac{18}{7}T_{CF}$ 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. -$$ \begin{bmatrix} +$$ \begin{bmatrix} 1/3 & 1/3 & 1 & 0 & 0 \\ -2/3 & -2/3 & 0 & 1 & 0 \\ -2/3 & 2/3 & 0 & 0 & 1 \\ @@ -340,7 +340,7 @@ R_{z} \\ -130 \\ -780 \\ -1170 \\ -\end{bmatrix}$$ +\end{bmatrix} $$ ```{code-cell} # sum forces