So let us consider next an example of a simple two-dimensional problem, where we will simplify the system of forces applied to this block consisting of three forces F1, F2, and F3, and a couple M. We will simplify this system of forces with a single force and a single couple acting at A. The force F_A that we need to apply it to this equivalent system should be equal to the sum of all the forces F_i in the original system. That is, F1 plus F2 plus F3, where F1, F2, F3 are given in vector form in this equation. So the sum clearly becomes 100i and 80j minus 20j, x plus 60j. So we replace all these forces with a single force where this component is 100 and this component is 60. Okay. So this is a single force F_A. In addition, we must apply a couple MA. Notice that in the case of two-dimensional problems, the applied couples can be represented by curly arrow since we have the understanding that all the moments are in the direction perpendicular planes or in the z direction. So the only thing we need to show is the orientation whether the couple is a clockwise or counterclockwise. So in this case, the external couple MA which of course in the z direction should be such that it is equal to the sum of all the moments about point A. That means the total rotational effect about this point of F_A is zero. So we have only MA. So this MA should be equal to the total rotational effect about point A in the original system. The moment of F1 is zero, the moment of F2 about A is zero since they pass through A. The moment of C, the two components of C. This is 60 and 80. The two components. So the moment of 60 is minus 60 times 2 in the clockwise direction plus 80 times the distance of 5. So this is the moment of the force F3 minus this one effect of the couple which is minus 100. So this gives us minus 120 plus 400 gives us plus 180. All these of course indicate direction. So plus 180K Newton meter. So the applied moment, I should not show vector here, I could say the magnitude of the moment MA in this case is equal to 180. Now, let us repeat this and replace all the forces with a single force and couple applied at point B. So to replace the system of forces with a single force and couple applied at B, the single force F_B, we'd have to be again equal to the sum of the original forces. That means it should be equal to F_A equal to 100i plus 60j. So the force would look exactly the same as before. This force F_B is equal to F_A. At the same time, we would have to apply a moment, a couple MB. The couple MB in the K direction should be equal to the sum of the moments of the original forces about point B or to the sum of the forces of this system about point B because if these two are equivalent and this two are equivalent it means these two are also equivalent. So the sum of the moments about point B in this case would be, decomposing this into two components, would be minus 60 times 5. That's the position of the effect of this component. This component has no additional effect plus 180 which is the applied couple. So many K units. So this gives us the value of minus 120K Newton meter. So what we notice. So this value is minus 120. So what we see is that when we replace this system of forces with a single force and a couple applied at A, and when we later replacing a single force and couple applied at B, the single force we have to apply are identical. That means these are invariant of the point we choose to reduce our system. However, the couple changes and in this case from a positive value it changes to becoming a negative value. Since we anticipate that picking different points would continuously affect the value of the applied moment, it is clear that at some point from positive values, we should be turning into negative values. So there exist some point, a point E, at which we should be able to replace the whole system of forces with a single force because the force as we said we cannot affect. It has to remain unchanged, but we would have only single force without a couple. So let us assume that there exist a point E such that we can replace the system of forces with a single force same as before. It would have to be the same as before. That means with component of 160. The force F_E. So F_E would still be the same force as before 100i plus 60j. But we would like now in this case that we do not need to apply any moment. So let's add a couple. So let us assume that distance of point is at x. If we wanted to make this system equivalent to this or this system equivalent to this for example, then the sum of the moments about A in all cases should be the same. The sum of the moments about point A in this case we know, we already calculated it was 180 in the K direction. The sum of the moments about A in this case is, this component does not create any moment. So it is 60 times the distance of x also in the K direction. So for this moment about A and for this moment about A to be equal, we conclude that x must be equal to three meters. So this is an example where we clearly demonstrate that we can simplify system with a single force and a couple. The value of the single force is not affected by the choice of the point. However, the value of the moment depends on the choice of point, and since the applied couple is perpendicular to the force, we're able, by proper sifting off this force, to completely eliminate the couple for an appropriate choice of sifting of the line of action of the force. In this case, three meters.
