Figure 3-5.-A practical application.Example 2Consider now the sad case of Slim and Sam, asillustrated in figure 3-5. Slim has suggested that theycarry the 300-pound crate slung on a handy 10-foot pole.He was smart enough to slide the load up 3 feet fromSam’s shoulder.Here’s how they made out. Use Slim’s shoulder asa fulcrum (Fl). Look at the clockwise movement causedby the 300-pound load. That load is 5 feet away fromSlim’s shoulder. If RI is the load, and Ll the distancefrom Slim’s shoulder to the load, the clockwise moment(MA) isA4~=R1xLl= 300 x 5 = 1,500 ft-lb.With Slim’s shoulder still acting as the fulcrum, theresistance of Sam’s effort causes a counterclockwisemoment (MB) acting against the load moment. Thiscounterclockwise moment is equal to Sam’s effort (Ez)times the distance (LJ) from his shoulder to the fulcrum(F,,at Slim’s shoulder. Since L~= 8 ft, the formula isMB= Ezx L3 =E2X8=8EZThere is no rotation, so the clockwise moment andthe counterclockwise moment are equal. MA = MB.Hence1,500 = 8Ez= 187.5 pounds.So poor Sam is carrying 187.5 pounds of the330-pound load.What is Slim carrying? The difference between 300and 187.5 = 112.5 pounds, of course! You can checkyour answer by the following procedure.This time, use Sam’s shoulder as the fulcrum (FI).The counterclockwise moment (MJ is equal to the300-pound load (Rl) times the distance (b = 3 feet) fromSam’s shoulder. Mc 300 x 3 = 900 foot-pounds. Theclockwise moment (m~, is the result of Slim’s lift (EI)acting at a distance (LJ from the fulcrum. L? = 8 feet.Again, since counterclockwise moment equals clock-wise moment, you have900 = E1X8Figure 3-6.-A couple.3-4