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Unformatted text preview: (Hint: sketch FBD of bar BC, and then write equilibrium equations. Neglect masses of bars AB and BC.) g 9.807 m s 2 = W 20kg g ⋅ := W 196N = Fab Fab A B FBD: Geometry: The perpendicular distance from ab to C is: Ldc 4m cos 45° ( ) ⋅ := Ldc 2.828m = Σ Mc= 0 Fab − 2.83 ⋅ m 196N 2.00 ⋅ m + = Fab 138.5 N ⋅ := Why Mc? Fab passes through A and B, so we can't sum moments about them and get an answer. Summing forces doesn't help, because these equations contain no distance information....
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- Fall '07
- red line, old test problem, two-force member AB