Applying to the system two force couples having torques equal in magnitude and opposite in sign,

Explanation

The solid arrows in Fig. show the forces `F_1` and `F_2` applied to a rigid body at points `A_1 and A_2` Apply the forces `T_1 and T_2` at the same points and the forces `T_3 and T_4` at the point C so that is `T_1= T_3= F_1 and T_2 = T_4 = F_2` . The point C is chosen so that `F_1a_1 = F_ya_y` . Show that the now nystem of six forces is equivalent to the old one, i.e. that
`(F_1,F_2,T_1,T_2,T_3,T_4)~(F_1,F_2)`
Indeed, the system of these forces reduces to the former forces `F_1 and F_2` and to two force couples `(T_1, T_3) and (T_2, T_4)` whose moments are
`M_1=T_1a_1 sin alpha and M_2=-T_2a_2sin alpha`
But point was chosen go that `|M_1| = |M_2| ` Therefore these moments are compensated and do not act on the rigid body.
On the other hand, the system of six forces is equivalent to the system of forces `T_3 and T_4` , i.e.
`(F_1, F_2, T_1, T_2, T_3, T_4) ~ (T_3, T_4)`
Indeed, the forces `F_1` and `T_1` , as well as `F_1 and T_2` are compensated, only the forces `T_3 and T_4` remaining uncompensated.
Hence, the system of forces `(F_1, F_2)` is equivalent to the system `( T_3 T_4 )` , and this gives the solution to the problem.