Exam (elaborations)
Ejercicios de cargas aplicadas a un sistema
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Tres ejercicios, donde se aplican cargas a un sistema, temas: tensiones, cargas, equilibrio, reacciones.
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Recuperación parcial 1Nombre: Gizell Camila Gordillo Ramirez Documento: 1001275565
1. Una barra AC está soportando las dos cargas P como se muestra en la figura. Los rodillos A
y C descansan en superficies sin fricción y un cable BD esta unido a B. Determinar:
a) La máxima fuera P que puede aplicarse en la barra AC si la máxima tensión a
resistir por el cable BD es 80kN.
Como primer ítem las alturas de P, para ello, aunque la magnitud será la misma, la
posición de estos son distintas, por lo tanto, serán P1 y P2. Para encontrar las alturas
se hará por relacion de triángulos, teniendo de referencia ∆𝐴𝐶𝐷. Las longitudes se
tomarán en metros.
Altura P1:
0.5𝑚 → 0.25𝑚
0.1𝑚 → 𝑃1ℎ
Teniendo una altura de 0.05m.
Altura P2:
0.5𝑚 → 0.25𝑚
0.4𝑚 → 𝑃2ℎ
Teniendo una altura de 0.2m.
Altura B:
0.5𝑚 → 0.25𝑚
0.15𝑚 → 𝐵ℎ
Teniendo una altura de 0.075m.
Ahora sacaremos los vectores fuerza.
⃗⃗⃗⃗⃗
𝐴𝑦 = 𝐴𝑦 𝑗̂ ; ⃗⃗⃗⃗
𝐶𝑥 = −𝐶𝑥 𝑖̂ ; ⃗⃗⃗1 = −𝑃𝑗̂
𝑃 ; ⃗⃗⃗⃗
𝑃2 = −𝑃𝑗̂
, 0.35𝑚𝑖̂ − 0.075𝑚𝑗̂
⃗⃗⃗⃗⃗⃗⃗
𝑇 𝐵𝐷 = 80𝑘𝑁 ( )
√205
40 𝑚
⃗⃗⃗⃗⃗⃗⃗
𝑇 𝐵𝐷 = 78.22𝑘𝑁𝑖̂ − 16.76𝑘𝑁𝑗̂
Se encontrarán los vectores posición.
𝑟𝐴𝑃1 = 0.1𝑚𝑖̂ + 0.05𝑚𝑗̂ → 𝐹(𝑃
⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗1 )
𝑟𝐴𝑃 = 0.4𝑚𝑖̂ + 0.2𝑚𝑗̂ → 𝐹(𝑃
⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗2 )
2
𝑟𝐴𝑇𝐵𝐷 = 0.15𝑚𝑖̂ + 0.075𝑚𝑗̂ → 𝐹(𝑇
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗
𝐵𝐷 )
𝑟𝐴𝐶 = 0.5𝑚𝑖̂ + 0.25𝑚𝑗̂ → 𝐹(𝐶
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗𝑥 )
Teniendo estos datos, haremos la sumatoria de momentos alrededor del punto A.
𝑖̂ 𝑗̂ 𝑘̂
𝑃
𝑀𝐴 1 = ⃗⃗⃗⃗⃗⃗⃗
𝑟𝐴𝑃1 × 𝑃⃗⃗⃗1 = |0.1𝑚 0.05𝑚 0𝑚 | = −0.1𝑚𝑃𝑘̂
0𝑘𝑁 −𝑃 0𝑘𝑁
𝑖̂ 𝑗̂ 𝑘̂
𝑃
𝑀𝐴 2 = ⃗⃗⃗⃗⃗⃗⃗
𝑟𝐴𝑃2 × 𝑃⃗⃗⃗⃗2 = |0.4𝑚 0.2𝑚 0𝑚 | = −0.4𝑚𝑃𝑘̂
0𝑘𝑁 −𝑃 0𝑘𝑁
𝑖̂ 𝑗̂ 𝑘̂
𝑇
𝑀𝐴 𝐵𝐷 = 𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗
𝐴𝑇𝐵𝐷 × 𝑇𝐵𝐷 = | 0.15𝑚 0.075𝑚 0𝑚 | = −8.38𝑘𝑁𝑚𝑘̂
78.22𝑘𝑁 −16.76𝑘𝑁 0𝑘𝑁
𝑖̂ 𝑗̂ 𝑘̂
𝐶 ⃗⃗⃗⃗𝑥 = |0.5𝑚 0.25𝑚 0𝑚 | = 0.25𝑚𝐶𝑥 𝑘̂
𝑀𝐴 𝑥 = ⃗⃗⃗⃗⃗
𝑟𝐴𝐶 × 𝐶
−𝐶𝑥 0𝑘𝑁 0𝑘𝑁
∑ 𝑀𝐴 = −0.1𝑚𝑃 − 0.4𝑚𝑃 − 8.38𝑘𝑁𝑚 + 0.25𝑚𝐶𝑥 = 0
−0.5𝑚𝑃 = −0.25𝑚𝐶𝑥 + 8.38𝑘𝑁𝑚
𝑃 = 0.5𝐶𝑥 − 16.76𝑘𝑁
𝑃 = 0.5(78.22𝑘𝑁) − 16.76𝑘𝑁
𝑃 = 22.35𝑘𝑁
b) Las reacciones en los apoyos A y C.
Al haber encontrado la máxima fuerza en P, se encontrarán las reacciones por
sumatoria de fuerzas.
∑ 𝐹𝑥 = −𝐶𝑥 + 𝑇𝐵𝐷 = 0 → −𝐶𝑥 + 78.22𝑘𝑁 = 0 → 𝐶𝑥 = 78.22𝑘𝑁
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