ESTRUCTURA DIAGRAMA DE CUERPO LIBRE
Enviado por Carol Durán • 15 de Junio de 2022 • Informe • 666 Palabras (3 Páginas) • 140 Visitas
ESTRUCTURA[pic 1] DIAGRAMA DE CUERPO LIBRE [pic 2]
SUMATORIA DE FUERZAS
∑MC = 0
(8N × 6N) + (4N × 3N) - (Rye × 1.5M) = 0
48N/M + 12N/M – Rye ×1.5M = 0
60N/M – Rey × 1.5M = 0
= Rye[pic 3]
40N = Rye
↑+ ∑Fy = 0
-8N – 4N – Ryc + 40N = 0
28N = Ryc
↑+ ∑Fx = 0
Rxc = 0
NUDO A[pic 4]
β = = 36,86°[pic 5][pic 6]
α = 90° - 36,86° = 53,14°
[pic 7]
[pic 8]
Fab = 6N (T)
[pic 9]
Fad = 10N (C)
↑+ ∑Fy = 0
-8N + Fad × cos36,86° = 0
[pic 10]
Fad = 9,99 ≈ 10N (C)
∑Fx = 0[pic 11]
Fab – Fad × sen36,86° = 0
Fab = 5,99 ≈ 6N (T)
NUDO D
α = = 53,13°[pic 13][pic 12]
β = = 53,13°[pic 14]
Fab = Fdb = 10N (T)
[pic 15]
Fae = 12N (C)
↑+ ∑Fy= 0
10 × cos53,13° + Fdb × sen53,13° = 0
Fdb = [pic 16]
Fdb = 10N (T)
↑+ ∑Fx = 0
-Fde + 10 × cos53,13° + 10 × cos53,13° = 0
12N = Fde (C)
NUDO B
[pic 17]
α = = 53,13°[pic 18]
β = = 53,13°[pic 19]
↑+ ∑Fy= 0
-4 – 10 × () + Fbe () = 0[pic 20][pic 21]
-4 -8 + Fbe × 0,8 = 0
Fbe = [pic 22]
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