sambil nunggu komentar dari agan yang lain, iseng-iseng ngerjain soal ini...
mencari reaksi gaya-gaya perletakan,

Misalkan titik A di sendi dan titik B di rol,

ΣM di sendi = 0

-0.5(q1)(a')^2 + (P1)(a) + (q2)(c)(a + b + c/2) - (Vb)(a + b + c) + (P2 sin 30°)(a + b + c + d) + (P3)(a + b + c + d + e) = 0

-0.5(2)(2)^2 + (2)(1) + (2)(2)(1 + 1 + 2/2) - (Vb)(1 + 1 + 2) + (2 sin 30°)(1 + 1 + 2 + 2) + (1)(1 + 1 + 2 + 2 + 3) = 0

Vb = 6.25 ton




ΣFx = 0

Ha + P2 cos 30° = 0

Ha + 2 cos 30° = 0

Ha = -√3 ton



ΣFy = 0

-(q1)(a') + Va - P1 - (q2)(c) + Vb - P2 sin 30° - P3 = 0

-(2)(2) + Va - 2 - (2)(2) + 6.25 - 2 sin 30° - 1 = 0

Va = 5.75 ton




periksa, apakah gaya reaksi di atas sudah benar...

ΣM di rol = 0

-(q1)(a')(a'/2 + a + b + c) + (Va)(a + b + c) - (P1)(b + c) - 0.5(q2)(c)^2 + (P2 sin 30°)(d) + (P3)(d + e) = 0

-(2)(2)(2/2 + 1 + 1 + 2) + (5.75)(1 + 1 + 2) - (2)(1 + 2) - 0.5(2)(2)^2 + (2 sin 30°)(2) + (1)(2 + 3) = 0

0 = 0

ok semua sudah cocok