[tex]\Large\boxed{\begin{array}{l}\underline{\rm Observe\,a\,figura\,que\,eu\,anexei.}\\\sf Note\,que\,a\,componente\,da\,forc_{\!\!,}a\,resultante\\\sf no\,eixo\,y(F_Ry)~representa\,o\,cateto\,oposto\\\sf ao\,\hat angulo\,de\,60^\circ.Perceba\,que\,a\,forc_{\!\!,}a\,resultante\\\sf \acute e\,a\,hipotenusa.\\\sf A\,relac_{\!\!,}\tilde ao\,trigonom\acute etrica\,que\\\sf envolve\,cateto\,oposto\,e\,hipotenusa\,\acute e\,o\,\boldsymbol{S\!E\!N\!O}.\end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\sf DA\acute I\\\sf sen(60^\circ)=\dfrac{F_Ry}{F_R}\implies F_Ry=F_Rsen(60^\circ)\\\\\sf F_Ry=5\cdot\dfrac{\sqrt{3}}{2}\\\\\sf F_Ry=2,5\sqrt{3}\\\sf adotando\,\sqrt{3}=1,73\,temos:\\\sf F_Ry=2,5\cdot1,73\\\sf F_Ry=4,32N\end{array}}[/tex]
