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BALANCING MOMENTS You know that the sailor in figure 3-4 would land flat on his face if the anchor hawser snapped. As long as nothing breaks, he must continue to push on the capstan bar. He is working against a clockwise moment of force that is equal in magnitude, but opposite in direction, to his counterclockwise moment of force. The resisting moment, like the effort moment, depends on two factors. In the case of resisting moment, these factors are the force (R2) with which the anchor pulls on the hawser and the distance (L2) from the center of the capstan to its rim. The existence of this resisting force would be clear if the sailor let go of the capstan bar. The weight of the anchor pulling on the capstan would cause the whole works to spin rapidly in a clockwise directionand good-bye anchor! The principle involved here is that whenever the counterclockwise and the clockwise moments of force are in balance, the machine either moves at a steady speed or remains at rest. This idea of the balance of moments of force can be summed up by the expression
Since a moment of force is the product of the amount of the force times the distance the force acts from the center of rotation, this expression of equality may be written El x L1 = E2 x L2,in that EI = force of effort,L1 = distance from fulcrum or axle to point where you apply force,E2 = force of resistance, and L2= distance from fulcrum or center axle to the point where you apply resistance. EXAMPLE 1 Put this formula to work on a capstan problem. You grip a single capstan bar 5 feet from the center of a capstan head with a radius of 1 foot. You have to lift a 1/2-ton anchor. How big of a push does the sailor have to exert? First, write down the formula El x L1 = E2 x L2, Here L1=5 E2 = 1,000 pounds, and L2=l. Substitute these values in the formula, and it becomes: E1 X 5 = 1,000 x 1 and
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 Sams shoulder.Heres how they made out. Use Slims shoulder as a fulcrum (Fl). Look at the clockwise movement caused by the 300-pound load. That load is 5 feet away from Slims shoulder. If RI is the load, and Ll the distance from Slims shoulder to the load, the clockwise moment (MA) isMA = R1 x Ll = 300 x 5 = 1,500 ft-lb.With Slims shoulder still acting as the fulcrum, the resistance of Sams effort causes a counterclockwise moment (MB) acting against the load moment. This counterclockwise moment is equal to Sams effort (E2) times the distance (L3) from his shoulder to the fulcrum (F1) at Slims shoulder. Since L2 = 8 ft, the formula isMB = E2 x L3 = E2 X 8 = 8E2There is no rotation, so the clockwise moment and the counterclockwise moment are equal. MA = MB.Hence 1,500 = 8E2 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 Sams shoulder as the fulcrum (F2). The counterclockwise moment (Mc) is equal to the 300-pound load (Rl) times the distance (b = 3 feet) from Sams shoulder. Mc 300 x 3 = 900 foot-pounds. The clockwise moment (mD) is the result of Slims lift (EI) acting at a distance (L3) from the fulcrum. L3 = 8 feet. Again, since counterclockwise moment equals clock-wise moment, you have900 = E1 X 8
Figure 3-6.-A couple.
Figure 3-7.-Valves. and
Slim, the smart sailor, has to lift only 112.5 pounds. Theres a sailor who really puts his knowledge to work. |
 
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= 187.5 pounds.
