Решение.

M_(I)(F_i^E) + Σ M_(I)(R_i^E) = 0 = F₂·b – S₇·b·tg(α) = 0; (1)

S₇ = = = 5,000 кН.

Σ M_(II)(F_i^E) + Σ M_(II)(R_i^E) = 0 =

= F₂·b + F₃·b·tg(α) + S₅·b·tg(α) = 0; (2)

S₅ = = = – 15,000 кН.

Σ+ Σ= 0 = F₂ + S₆·sin(α) = 0; (3)

S₆ = – F₂/sin(α) = – 5/0,707 = – 7,072 кН.