Solución
\(\vec{f_1}=mg\ \cos\ 15^oi+mg\ \sin\ 15^oj\)
\(\vec{f_2}=2mg\ \cos\ 120^oi+2mg\ \sin\ 120^oj\)
\(\vec{f_T}=mg\ \left(\cos\ 15^o+2\ \cos\ 120\right)i+mg\ \left(\sin\ 15^o\ +2\ \sin\ 120^o\right)j\)
\(\left|\vec{Ft}\right|=\sqrt{m^2g^2\cos15^{\ }+2\cos120^2\ +m^2g^2\ \left(\sin15+2\sin\ 120^2\ \right)\ }\)
\(=mg\sqrt{\left(\cos15+2\cos120\right)^2+\left(\sin15+2\sin120\right)^2}\) =1.9911 mg