2) Calcular el centro de gravedad de \(\vec{x}\) de una barra homogénea en forma de un arco semicircular.
x=r cos\(\theta\) \(\frac{w}{\pi}=0.5\ \frac{lb}{ft}\)
\(\vec{x}=\frac{\int_L^{ }\vec{xdl}}{\int_L^{ }dl}=\frac{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}r\ \cos\theta\ r\ d\theta}{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}r\ d\theta}\) dl=r d\(\theta\) \(1\Sigma fx=0\)
\(2\ \Sigma fy=0\)
=\(\frac{r\ sen\theta\left|_{\frac{\pi}{-2}}^{\frac{\pi}{2}}\right|}{\theta\left|_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\right|}=\frac{r\left[1+1\right]}{\pi}=\frac{2r}{\pi}=\frac{4}{\pi}=1.25\) \(3\ \Sigma MA=0\)
\(1\ Bx-Ax=0\) \(4ft\ Bx=2r^2\left(\frac{0.5lb}{ft}\right)\) \(Bx=\ 1\ lb\)
Ax=Bx
\(2\ Ay-w=0\) \(Bx=\frac{2r^2}{4ft}\left(\frac{0.5lb}{ft}\right)\) \(Ax=\ 1lb\)
Ay=w
\(3\ -\vec{x}w+Bx\left(4ft\right)=0\) \(=\frac{2\left(4ft^2\right)}{4ft^2}\left(0.5lb\right)\) \(Ay=\pi\ lb\)
\(\frac{-2r}{\pi}\left(\frac{0.5lb}{ft}\right)r\pi+Bx\left(4ft\right)=0\) \(=1\ lb\)
\(-2r^2\left(\frac{0.5lb}{ft}\right)+4ft\ Bx=0\)