Saved Bookmarks
| 1. |
Applying to the system two force couples having torques equal in magnitude and opposite in sign, |
|
Answer» Solution :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. 
|
|