Ejercicio 2.137 - Beer and Johnston PDF

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Ejercicio #2.137 Beer and Johnston Estática...

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Ejercicio 2.137 Ingeniería mecatrónica

Mecánica Vectorial Grupo B

Datos: a) b)

𝑃 = 314 𝑁 Alambre = 525𝑚𝑚

c)

Coordenadas 𝐴 = (0, 155, 0) 𝑚𝑚 𝐵 = (200, 0, 𝑧) 𝑚𝑚

𝑦 = 155 𝑚𝑚

Fuerzas 𝑃 = < 0, 341, 0 > 𝐹 = < 𝐹𝑥 , 0, 𝐹𝑧 >  𝑇 = 𝑇 ⋅ 𝐴𝐵

𝑄 = < 0, 0, 𝑄 > 𝐺 = < 𝐺𝑥 , 𝐺𝑦 , 0 >

 = < 200, −155, 𝑧 > 𝐴𝐵 | = 525 𝑚𝑚 |𝐴𝐵 5252 = 2002 + 1552 + 𝑧 2

𝒛 = 460

 = < 200, −155, 460 > 𝐴𝐵  𝐴𝐵 8 31 92  = > 𝐴𝐵 | = < 21 , − 105 , 105 |𝐴𝐵 𝑇 = <

8 31 92 𝑇> 𝑇, − 𝑇, 21 105 105

→Para A: 8 ∑𝐹𝑥 = 𝐹𝑥 + 𝑇=0 21 31 𝑇=0 ∑𝐹𝑦 = 341 − 105 92 ∑𝐹𝑧 = 𝐹𝑧 + 𝑇=0 105 𝑇 = 1155 𝑁

→Para B: 8 𝑇=0 21 31 𝑇=0 ∑𝐹𝑦 = 𝐺𝑦 + 105 92 ∑𝐹𝑧 = 𝑄 − 𝑇=0 105 𝑄 = 1012 𝑁 ∑𝐹𝑥 = 𝐺𝑥 −

• En función de 𝒚 | = 525 𝑚𝑚 |𝐴𝐵  = < 200, −𝑦, 𝑧 > 𝐴𝐵 525 = √2002 + 𝑦2 + 𝑧 2 𝑧 = √235625 − 𝑦 2 −485.41 ≤ 𝑦 ≤ 485.41

 = < 200, −𝑦, √235625 − 𝑦 2 > 𝐴𝐵  = 𝐴𝐵

 8 −𝑦 √235625 − 𝑦2 𝐴𝐵 =< , , >  | |𝐴𝐵 21 525 525

→Para A: 8 ∑𝐹𝑥 = 𝐹𝑥 + 𝑇=0 21 𝑦 ∑𝐹𝑦 = 341 − 𝑇=0 525 √235625 − 𝑦 2 𝑇=0 ∑𝐹𝑧 = 𝐹𝑧 + 525 179025 𝑇= 𝑦 Gráficas 𝑻=

179025 𝑦

𝑸=

341√235625 − 𝑦 2 𝑦

→Para B: 8 𝑇=0 21 𝑦 ∑𝐹𝑦 = 𝐺𝑦 + 𝑇=0 525 √235625 − 𝑦2 𝑇=0 ∑𝐹𝑧 = 𝑄 − 525 341√235625 − 𝑦2 𝑄= 𝑦 ∑𝐹𝑥 = 𝐺𝑥 −...