Question: This, follow this format solution 9:00 PM O _ _ ... all all 2 21 Hy= 44.58 The moment about H=0 - 5 x 3

This, follow this format solution

9:00 PM O _ _ ... all all 2 21 Hy= 44.58 The moment about H=0 - 5 x 3 + 10 x 4 + 20 x 6 - Fix4=0 5 x 3 + 10 x 4 + 20 x 6 ->Fi(kN) 4 = 43.75 Sum of vertical forces=0 # HY - 5 - 20 + F2 x sin (53.13) = 0 44.58 - 5 - 20 =>F2(kN) = sin (53.13 x 180 = -24.48 Sum of horizontal forces=0 -F2 x cos (0) - F3 - F1 + 10 =0 7 => F3(KN) - -(-24.48) x cos ( 53.13 x 180 - 43.75 + 10 = -19.06 Explanation: Using method of section, calculations have been done. Answer A 1. The force in marked members F1 - 43.75 kN.... T F2 = 24.48 kN . C F3 = 19.06 kN . ... C Was this solution helpful? More matches

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