So let us consider a system of forces comprised of N forces and M couples acting on this body. We will show that there is always a much simpler system comprised only of a single force and a single couple that is equivalent to the system. For that, we pick up a point A, and we claim that there exist a single force F prime_ A, and a single couple, let's call this M_A prime. So a single force and a single couple that is equivalent to the system. To find F_A prime and M_A prime based on their previous definition of equivalent systems, the force F_A here should be equal to the sum of the forces F_i for the original system. The single couple moment, M_A prime, should be equal to the sum of all the total rotational effect about point A, that is, to the sum of all the moments of the forces F_i. That means sum of r_i cross F_i plus the rotational effect of all the couples plus the sum of all M_j's. So we can always replace a very complicated system of force and couples with a single force and a single couple acting at a pre-specified point A. If we would have picked up a different point, say point B, we could also do the same and replace everything with a single force F_B and a couple, let's call this maybe M_B prime, where similar conditions should hold. That is, F_B should be still the sum of all these F_i's and therefore the single force would be equal to the force F_A prime. That means we can always replace the system forces with a single force acting at different points, and this single force is always the same. That means the sum of all the forces which governs the translational behavior of this body is represented by this single force which is equivalent to the sum of all these forces. Now, the moment about B, the couple M_B prime that we need to apply here, however, varies, it depends on point B because it is easy to show that in this case, for this two systems to be equivalent, the moment M_B prime, the moment about this point and about this point should be equal, and therefore, we would have that BA cross with this force F_A prime, which is the rotational effect of F_A prime, plus M_A prime should be the total rotational effect about B, and in this case, the total rotational effect is M_B prime. So the relationship between M_B prime and M_A prime is given by this expression. So we can replace a complicated system of forces with always with a system of force and couple, force and couple. The force is invariant of the point we choose, but the couple itself could be varying. Depending on the choice of point B, we can show that under certain conditions, we can simplify the expression for this moment because as we move around from point to point, we affect the vector M_B prime, and we can show that we could affect it. We could effect as a matter of fact only that component of M_A prime which is perpendicular to F_A prime, that means this couple has two components. As we move around from point to point, this then affects the magnitude and the direction of M_A prime. It affects is perpendicular component, so there exist a particular point C such that you could have replaced everything with a single force F_C prime, which is equal to F_B prime, equal to F_A prime, that means the force is invariant, and we would have only one couple which is along the direction of F_C prime, and this is what we call [inaudible] system. That means a system where the force and the torque are in the same direction, the same as when you use a screwdriver and you're pushing and screwing in the same direction. Let's look at an example where we are going to calculate a simplified equivalent system for a 2D problem.