Solución:
Datos:
0.5 Ib / ft
\(\left[^{-\frac{\pi}{2},\frac{\pi}{2}}\right]\)
x= r cos \(\theta\)
y= r sen \(\theta\)
dl= rd \(\theta\)
p=\(\frac{w}{l}\) = 0.5 \(\frac{lb}{ft}\)
w= \(\left(0.5\frac{lb}{ft}\right)\pi\ 2ft\) = \(\pi\ lb\)
\(-\ w\ xc\ \) + \(\left(4\ ft\right)Bx=0\)
Resultados:
\(Ay-w=0\)
\(Ay=\pi\ lb\)
\(-\left(\frac{2r}{\pi}\right)\left(\pi\ lb\right)+4Bx=0\)
\(\left(4ft\right)Bx=\frac{2r}{\pi}\left(\pi\ lb\right)\)
\(Bx\ =\ 1\ lb\)
\(Bx=\ Ax=1lb\)