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15° -X 1350 RBC 30° 200N 10g ΣΕ, = 0 b) Determine the tension required by cable AC to hold up the structure in a static
Posted: Sat Feb 19, 2022 3:33 pm
by answerhappygod
- 1 (28.06 KiB) Viewed 94 times
15° -X 1350 RBC 30° 200N 10g ΣΕ, = 0 b) Determine the tension required by cable AC to hold up the structure in a static condition. +1 T cos 15° - 10(9.81) – 200 cos 10º – RBC Cos 30º = 0 0.966T - 295.06 -0.866Rgc = 0 T -0.896 RBC = 305.45 – (1) + - Fx = 0 -T sin 15° - Roc sin 30° +200 sin 10° = 0 -0.259T - 0.5R2c +34.73 = 0 T +1.93 RBC = 134.09 - (2) = f = Eq (1)-(2) = 171.36 Rgc = x 2.826Rec Substitute back into (1) T = 305.45 + 0.896(60.64) = x c) Resultant forces at Cis same as reaction of Rec The resultant in x-axis = Recsin 3060.64 sin 300 The resultant in y-axis =RBC COS 30 = 60.64 cos 30º = x