$\sum_{}^{}{Mo:\ S_{2}r - M_{1} - S_{1}r = 0}$

S₂>S₁ , więc S₂ = S₁e^αμ

$\sum_{}^{}{M_{A}:\ S_{1} \times 2r} - P \times 3r = 0$

$\sum_{}^{}\text{Px}$: −T₁ +  S₁cosα = 0,

$\sum_{}^{}{\text{Py}:\ N_{1} - Q + S_{1}\text{sinα} = 0}$

$\sum_{}^{}{Px:T_{1} + T_{2} = 0}$

$\sum_{}^{}{Py:\ N_{2} - N_{1} - G = 0}$

T₁ = N₁ × μ

T₂ = N₂ × μ

ΣP_x: T₂ + T₁sinα – N₂cosα = 0

ΣP_y: N₂ – G – T₁cosα – N₁sinα = 0