= Solution
Use the same conditional <optimal normal-fault dip relation> as in question 1, with $\delta=55^\circ$:
$$
\varphi=2(55^\circ)-90^\circ=20^\circ,\qquad
\boxed{\mu_f=\tan20^\circ=0.364.}
$$
The second panel of the original <Mohr circle> figure has $\sigma_3'=1$ and $\sigma_1'=2.03961$; its tangent from the origin has slope $0.364$. Resolving the stresses onto a $55^\circ$ plane places its stress point at that tangency. The diagram assumes negligible <cohesion>, a vertical maximum <principal stress> and an optimally oriented new or equivalently oriented sliding plane. Without those assumptions, a dip and a statement that sliding began do not uniquely determine the <coefficient of friction>.
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