Editing Resultant force (section)

==Illustration==
The diagram illustrates simple graphical methods for finding the [[line of application]] of the resultant force of simple planar systems.
#Lines of application of the actual forces <math> {\scriptstyle \vec{F}_{1}}</math> and <math>\scriptstyle \vec{F}_{2}</math> in the leftmost illustration intersect. After [[vector addition]] is performed "at the location of <math>\scriptstyle \vec{F}_{1}</math>", the [[net force]] obtained is translated so that its line of application passes through the common intersection point. With respect to that point all torques are zero, so the torque of the resultant force <math>\scriptstyle \vec{F}_{R}</math> is equal to the sum of the torques of the actual forces.
#Illustration in the middle of the diagram shows two parallel actual forces. After vector addition "at the location of <math>\scriptstyle\vec{F}_{2}</math>", the net force is translated to the appropriate line of application, whereof it becomes the resultant force <math>\scriptstyle \vec{F}_{R}</math>. The procedure is based on a decomposition of all forces into components for which the lines of application (pale dotted lines) intersect at one point (the so-called pole, arbitrarily set at the right side of the illustration). Then the arguments from the previous case are applied to the forces and their components to demonstrate the torque relationships.
#The rightmost illustration shows a [[couple (mechanics)|couple]], two equal but opposite forces for which the amount of the net force is zero, but they produce the net torque <math>  \scriptstyle\tau = Fd </math>&nbsp;&nbsp;&nbsp;where <math>\scriptstyle  d </math>&nbsp; is the distance between their lines of application. This is "pure" torque, since there is no resultant force.