1. The Wrench: Let’s suppose that we have already reduced a complicated force system to single (resultant ) force,, and a single couple with the moment,, as shown in the figure: And, we want to reduce this force system to the simplest one possible, a wrench. Step 1: The unit vector parallel to : Step 2: The component of parallel to : Step 3: The component of perpendicular to : Continued on the next page We note that Hence, and are perpendicular.

2. Continued on the next page The moment of the couple has been resolved into a perpendicular component,, and a parallel component,. Theses vector could shown anywhere in space because a couple produces the same moment about all points in space. Here they are shown at the origin to make it easier to see that they are perpendicular and parallel to.

3. couple P ( x,y,z ) Step 4: in a new position, denoted by P, such that the new resultant moment of all the couples,, is given by: Because is perpendicular to both and, there is no way that can be ; we can only eliminate. We put Continued on the next page B A

4. We note that which indicates that there are not three independent equations. In fact, we see that the sum of the first two equations in B is equal to the third. As in the two- dimensional case, we have the equation of a straight line. To see which line, we arbitrarily put x = 0 and then find y = - 250 and z = 200. This is one point on the line. Then we arbitrarily put y = 0 and find x = 125 and z = - 50 to find another. The vector to the first point from the second is given by To show that this vector is parallel to, we note that. Now we have reduced the force system to a single force acting along the line defined by B (point P is on this line) and a couple with a moment parallel to the force. Such a force system is called a wrench. Because the moment is in the direction opposite to that of the force, the wrench is said to be negative. An Alternative Approach: We know that the final moment in the simplest force system is parallel to the resultant force, so we can go back to A and set the resultant moment equal to a vector parallel to the force: where M is the magnitude of the moment parallel to the force (unknown at this point) and u is the unit vector parallel to R. Recalling that we can arbitrarily assign a value to one of the coordinates, we can put x = 0, y = 0, z = 0 in the three equations above and obtain the following three sets of equations with three equations in each set:

5. We can solve these three sets of equations using, for example, MATLAB or MATHEMATICA: We note that. This solution agrees with the one obtained earlier. The solution for M is the same for all three arbitrary choices of the coordinates (indeed for all choices).