Solución:
a)
para Q1;
Ecuaciones de equilibrio:
\(\Sigma_{Fx}=0\)
\(\Sigma_{Fy}=0\)
\(r=l\ sen\ \theta1+\ l\ sen\ \theta2\)
\(=l\ \left(sen\ \theta1\ +\ sen\ \theta2\right)\)
\(=l\left(\theta1+\theta2\right)\)
\(\Sigma_{Fx}:\)
\(Tx-Fc=0\)
\(T\ \cos\ \left(\frac{\Pi}{2}-\theta1\right)=Fc\)
\(T\ sen\ \theta1\ =\frac{2k\ Q^2}{l^2\left(Q1+Q2\right)}\)
\(TQ1=\frac{2K\ Q^2}{l^2Q1\left(\theta1+\theta2\right)^2}\)
\(\Sigma_{Fy}:\)
\(Ty-mg=0\)
\(T\ sen\ \left(\frac{\Pi}{2}-\theta1\right)=mg\)
\(T\ \cos\ \theta1=mg\)
\(T=mg\)
Para     Q2:
\(\Sigma_{Fx}:\)
\(Fc=T\ \cos\ \left(\frac{\Pi}{2}-Q2\right)\)
\(T\ sen\ \theta2=\frac{2KQ^2}{l^2\left(Q1+Q2\right)^2}\)
\(T\ Q2=\frac{2K\ \theta^2}{l^2\left(Q1+Q2\right)}\)
\(\frac{T\theta1}{T\theta2}=1\)
\(\frac{\theta_1}{\theta_2}=1\)
b)
\(\Sigma_{Fy}:\)
\(T_2\ sen\left(\frac{\Pi}{2}-\theta_2\right)=2\ mg\)
\(T_2\ \cos\ \theta_2=2\ mg\)
\(T_2=2mg\)
\(\frac{T_1\theta_1}{T_2\theta_2}=1\)
\(\frac{mg\ \theta_1}{2\ mg\ \theta_2}=1\)
\(\frac{\theta_1}{\theta_2}=2\)
c)
\(r=\ l\ sen\ \theta_1+l\ sen\ \theta_2\)
\(=l\ \left(sen\ \theta_1+sen\ \theta_2\right)\)
\(=l\left(\theta_1+\theta_2\right)\)