Figure 3-5.-A practical application.

Figure 3-5.-A practical application. Example 2 Consider now the sad case of Slim and Sam, as illustrated in figure 3-5. Slim has suggested that they carry the 300-pound crate slung on a handy 10-foot pole. He was smart enough to slide the load up 3 feet from Sam’s shoulder. Here’s how they made out. Use Slim’s shoulder as a fulcrum (Fl). Look at the clockwise movement caused by the 300-pound load. That load is 5 feet away from Slim’s shoulder. If RI is the load, and Ll the distance from Slim’s shoulder to the load, the clockwise moment (MA) is A4~=R1xLl= 300 x 5 = 1,500 ft-lb. With Slim’s shoulder still acting as the fulcrum, the resistance of Sam’s effort causes a counterclockwise moment (MB) acting against the load moment. This counterclockwise 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 is MB = Ez x L3 =E2X8=8EZ There is no rotation, so the clockwise moment and the counterclockwise moment are equal. MA = MB. Hence 1,500 = 8Ez = 187.5 pounds. So poor Sam is carrying 187.5 pounds of the 330-pound load. What is Slim carrying? The difference between 300 and 187.5 = 112.5 pounds, of course! You can check your answer by the following procedure. This time, use Sam’s shoulder as the fulcrum (FI). The counterclockwise moment (MJ is equal to the 300-pound load (Rl) times the distance (b = 3 feet) from Sam’s shoulder. Mc 300 x 3 = 900 foot-pounds. The clockwise 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 have 900 = E1X8 Figure 3-6.-A couple. 3-4