\({F_{\text{x}}}\) is the \(x\)-coordinate of the force (\({\text{N}}\))
\({F_{\text{y}}}\) is the \(y\)-coordinate of the force (\({\text{N}}\))
\(M\) is the moment (\(\text{N}\text{m}\))
Notes
For a body in static equilibrium, the net forces and moments acting on the body will cancel out. Assuming a 2D problem (A:Effective-Norm-Stress-Large), the \(x\)-coordinate of the force \({F_{\text{x}}}\) and \(y\)-coordinate of the force \({F_{\text{y}}}\) will be equal to \(0\). All forces and their distance from the chosen point of rotation will create a net moment equal to \(0\).
\({τ^{\text{f}}}\) is the shear strength (\({\text{Pa}}\))
\({σ_{N}}‘\) is the effective normal stress (\({\text{Pa}}\))
\(φ’\) is the effective angle of friction (\({{}^{\circ}}\))
\(c’\) is the effective cohesion (\({\text{Pa}}\))
Notes
In this model the shear strength \({τ^{\text{f}}}\) is proportional to the product of the effective normal stress \({σ_{N}}‘\) on the plane with its static friction in the angular form \(\tan\left(φ’\right)\). The \({τ^{\text{f}}}\) versus \({σ_{N}}‘\) relationship is not truly linear, but assuming the effective normal forces is strong enough, it can be approximated with a linear fit (A:Surface-Base-Slice-between-Interslice-Straight-Lines) where the effective cohesion \(c’\) represents the \({τ^{\text{f}}}\) intercept of the fitted line.
\(\boldsymbol{a}\text{(}t\text{)}\) is the acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
Notes
The net force \(\boldsymbol{F}\) on a body is proportional to the acceleration \(\boldsymbol{a}\text{(}t\text{)}\) of the body, where \(m\) denotes the mass of the body as the constant of proportionality.