Como ya conocemos "\(x\)" y \(dl\) sustituimos en las integrales para de esta manera poder obtener "\(x\)" barra. Sustituimos para obtener "\(x\)" barra, evaluar de \(\frac{\pi}{2}\ \ \ a\ \ \ -\frac{\pi}{2}\)
\(x=\frac{\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}r\cos\theta rd\theta}{\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}rd\theta}=\frac{\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}r\cos\theta d\theta}{\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta}=\frac{r\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\theta rd\theta}{\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta}=\frac{r\left[sen\theta\right]}{\theta}\)
\(x=\frac{r\left[sen\frac{\pi}{2}-sen\left(-\frac{\pi}{2}\right)\right]}{\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)}=\frac{2r}{\pi}=\frac{2ft\left(2\right)}{\pi}=1.273ft\)
Componentes
\(\Sigma M_A=0\)
\(-r\gamma x+B_x\cdot2r=0\)
\(B_x=\frac{\pi\cdot r\cdot\gamma\cdot x}{2r}=\frac{\pi\left(2\right)\left(.5\frac{lb}{ft}\right)\left(1.273\right)}{2\left(2\right)}=1lb\)
\(\Sigma F_{x=0}\)
\(-Ax+Bx=0\)
\(Ax=Bx\)
\(Ax=1lb\)
\(\Sigma F_{y=0}\ \)
\(\ A_y=\pi rγ=\pi\left(2ft\right)\left(.5\frac{lb}{ft}\right)\)=\(\pi lb\)
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