solución:
\(x=2\cos\theta\)
\(y=2\sin\theta\)
\(dL=2d\theta\)
\(x=\frac{\int_a^bxdL}{\int_a^bdL}=\frac{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2\cos\theta2d\theta}{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2d\theta}\)
                       \(=\frac{4\left[\sin\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}{\left[2\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}\)
                            =\(\frac{4}{\pi}\) ft
\(Arc\ length=\pi r=2\pi\)
\(W=2\pi\left(0.5\right)lb\)
\(\Sigma M_A=0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\pi\left(0.5\right)\left(\frac{4}{\pi}\right)+B_A\left(4\right)=0\)
                                                         \(B_1=1lb\)
\(\Sigma F_x=0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_x=1lb\)
\(\Sigma F_y=0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_y=3.14lb\)