Σ F N = 0 : E1(1-L1W/l1)Sin δ 1 - T ′N1Sin(α+θ1) - W 1Cosθ1 + C′l3Cosθ1 + R 31Sin(Φ′-θ1) + R 1CosΦ′ = 0 ----------(6) Σ FT = 0 : - E1Cosδ 1 + T ′N1Cos(α+θ1) - W 1Sinθ1 + C′l1 + C′l3Sinθ1 + R 31Cos(Φ′-θ1) + R 1SinΦ′ = 0
----------(7)
From : R 1 = [(W1-C′l3)Cosθ1/CosΦ′] + [(B1-A1)/CosΦ′] - [R 31Sin(Φ′-θ1)/CosΦ′]
----------(8)
R 1 = [(W1-C′l3)Sinθ1/SinΦ′] - [C′l1/SinΦ′] + [(A11-B11)/SinΦ′] - [R 31Cos(Φ′-θ1)/SinΦ′]--(9) (8) = (9) => R 31[(Cos(Φ′-θ1)) - (Sin(Φ′-θ1)Tan Φ′)] = (A11-B11) + (W1-C′l3)Sinθ1 - (C′l1) [(B1-A1) + (W1-C′l3)Cosθ1]Tan Φ′ Rearranging and reducing to a simpler form ; R 31 = {(A11-B11) + (W1-C′l3)Sinθ1 - (C′l1) - [(B1-A1) + (W1-C′l3)Cosθ1]Tan Φ′}/(Cos(Φ′ - θ1) - Sin(Φ′-θ1)TanΦ′) ----------(10) Set 5) = 10) & remember that : 1 l A1 = E1(1-L1W/l1)Sin δ 1
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