domingo, 17 de abril de 2011

Ejercicios Resueltos de Física - Estática

Problemas que se resuelven usando la primera y segunda condición de equilibrio.
Institución: I.E. San Agustín 
                   Chiclayo - Perú
                   4to Año de secundaria
Nº de Ejercicios:  1 de 10 (9 en proceso de subida)

ESTÁTICA – EJERCICIOS RESUELTOS

1. La barra mostrada pesa 20N y está en reposo. Calcular la longitud de la barra, si además se sabe que la reacción en el apoyo B es 5N.

Solución
Sea \(L\) la longitud de la barra, entonces su diagrama de cuerpo libre (DCL) para la barra es:
De la segunda condición de equilibrio “La suma de momentos con respecto al punto O es nula”, se tiene:
\[\sum {{M_0}}  = 0\; \ldots \;(1.1)\]
reemplazando el lado izquierdo de (1.1):
\[M_0^{{F_1}} + M_0^W + M_0^{{F_2}} = 0\] calculando los momentos de \({F_1}\), \(W\) y \({F_2}\) respectivamente: \[\boxed{2 \cdot {F_1} - \frac{L}{2} \cdot W + L \cdot {F_2} = 0}\; \ldots \;(1.2)\] por dato del problema: \[\begin{gathered}
{F_2} = 5\,\,{\text{N}} \\
W = 20\,\,{\text{N}} \\
\end{gathered} \] entonces: \[\begin{gathered}
2 \cdot {F_1} - \frac{L}{2} \cdot 20 + L \cdot 5 = 0 \\
2{F_1} - 10L + 5L = 0 \\
\end{gathered} \] \[\boxed{2{F_1} - 5L = 0}\; \ldots \;(1.3)\] Por otro lado de la primera condición de equilibrio “la suma de todas las fuerzas en y es igual a cero” entonces:
\[\sum {{F_y}}  = 0\]\[\begin{gathered}
{F_1} + {F_2} - W = 0 \\
{F_1} + 5 - 20 = 0 \\
{F_1} - 15 = 0 \\
\end{gathered} \] \[\left. {\underline {\,
{{F_1} = 15} \,}}\! \right| \] Reemplazando \({F_1}\)  en (1.3): \[\begin{gathered}
2(15) - 5L = 0 \\
2(15) = 5L \\
2(3) = L \\
\left. {\underline {\,
{L = 6\,\,{\text{m}}} \,}}\! \right| \\
\end{gathered} \]
Rpta: a)

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