Figure 9-2.—Balances.Unfortunately, the more springs are used, the more theylose their ability to snap back to their original position.Hence, an old spring or an overloaded spring will giveinaccurate readings.Balanced ScaleThe problem with the spring-type scale eventuallyled to the invention of the balanced scale, shown infigure 9-2. This type of scale is an application of first-class levers. The one shown in figure 9-2, A, is thesimplest type. Since the distance from the fulcrum to thecenter of each platform is equal, the scales balance whenequal weights are placed on the platforms. With yourknowledge of levers, you can figure outyard shown in figure 9-2, B, operates.PRESSUREhow the steelPressure is the amount of force within a specificarea. You measure air, steam, and gas pressure and thefluid pressure in hydraulic systems in pounds per squareinch (psi). However, you measure water pressure inpounds per square foot. You’ll find more about pressuremeasurements in chapter 10. To help you betterunderstand pressure, let’s look at how pressure affectsyour ability to walk across snow.Have you ever tried to walk on freshly fallen snowto have your feet break through the crust when you putyour weight on it? If you had worn snowshoes, youcould have walked across the snow without sinking; butdo you know why? Snowshoes do not reduce yourweight, or the amount of force, exerted on the snow; theymerely distribute it over a larger area. In doing that, thesnowshoes reduce the pressure per square inch of theforce you exert.Let’s figure out how that works. If a man weighs160 pounds, that weight, or force, is more or less evenlydistributed by the soles of his shoes. The area of the solesof an average man’s shoes is roughly 60 square inches.Each of those square inches has to carry 160 ÷ 60= 2.6pounds of that man’s weight. Since 2 to 6 pounds persquare inch is too much weight for the snow crest tosupport, his feet break through.When the man puts on snowshoes, he distributes hisweight over an area of about 900 square inches,depending on the size of the snowshoes. The forceon each of those square inches is equal to only160 ÷ 900 = 0.18 pounds. Therefore, with snowshoeson, he exerts a pressure of 0.18 psi. With this decreasedpressure, the snow can easily support him.9-2
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