Question

A weight of 50N is hung from one end of a uniform beam 12 m long. If the beam weighs 25 N, where and  with what forces should the beam picked up so that it remains horizontal ?

Answer 1

$$R_1+R_2=50+25=75\ N\\ 25(3-x)=50(x)+R_2(6-2x)\\25(3-x)+R_1(6-2x)=50(12-x)$$

$$x=3\ m\\R_1=50\ N\\R_2=25\ N$$