Tarcie w cięgnach

From the equilibrium condition

\begin{aligned} &S+dS=S+\mu dN\\ &dS=\mu dN\\ \end{aligned}

At the same time

\begin{aligned} &dN=Sd\varphi\\ &dS=\mu Sd\varphi\\ &\frac{dS}{S}=\mu d\varphi\\ \end{aligned}

By integrating both sides, we obtain

\begin{aligned} &\ln{S}=\mu \varphi+C\\ &S=e^C e^{\mu \varphi}\\ \end{aligned}

Substituting the boundary condition

\begin{aligned} &\varphi=0, S=P\\ &P=e^C e^{0}=e^C\\ &S=Pe^{\mu \varphi}\\ \end{aligned}

Substituting \(\varphi\) with the wrapping angle \(\alpha\)

\begin{aligned} &S=Pe^{\mu \alpha}\\ \end{aligned}