DIFFERENTIAL AREAS.— Consider thespecial situation shown in figure 2-12. Here, asingle piston (1) in a cylinder (2) has a piston rod(3) attached to one of its sides. The piston rodextends out of one end of the cylinder. Fluid underpressure is admitted equally to both ends of thecylinder. The opposed faces of the piston (1)behave like two pistons acting against each other.The area of one face is the full cross-sectional areaof the cylinder, say 6 square inches, while the areaof the other face is the area of the cylinder minusthe area of the piston rod, which is 2 squareinches. This leaves an effective area of 4 squareinches on the right face of the piston. The pressureon both faces is the same, in this case, 20 psi.Applying the rule just stated, the force pushingthe piston to the right is its area times the pressure,or 120 pounds (20 x 6). Likewise, the forcepushing the piston to the left is its area times thepressure, or 80 pounds (20 x 4). Therefore, thereis a net unbalanced force of 40 pounds acting tothe right, and the piston will move in thatdirection. The net effect is the same as if the pistonand the cylinder had the same cross-sectional areaas the piston rod.VOLUME AND DISTANCE FACTORS.—You have learned that if a force is applied to asystem and the cross-sectional areas of the inputand output pistons are equal, as in figures 2-9 and2-10, the force on the input piston will supportan equal resistant force on the output piston. Thepressure of the liquid at this point is equal to theforce applied to the input piston divided by thepiston’s area. Let us now look at what happenswhen a force greater than the resistance is appliedto the input piston.In the system illustrated in figure 2-9, assumethat the resistance force on the output piston is100 psi. If a force slightly greater than 100 poundsis applied to the input piston, the pressure in thesystem will be slightly greater than 10 psi. Thisincrease in pressure will overcome the resistanceforce on the output piston. Assume that the inputpiston is forced downward 1 inch. The movementdisplaces 10 cubic inches of fluid. The fluid mustgo somewhere. Since the system is closed and thefluid is practically incompressible, the fluid willmove to the right side of the system. Because theoutput piston also has a cross-sectional area of10 square inches, it will move 1 inch upward toaccommodate the 10 cubic inches of fluid. Youmay generalize this by saying that if two pistonsin a closed system have equal cross-sectional areasand one piston is pushed and moved, the otherpiston will move the same distance, though in theopposite direction. This is because a decrease involume in one part of the system is balanced byone equal increase in volume in another part ofthe system.Apply this reasoning to the system in figure2-11. If the input piston is pushed down a distanceFigure 2-12.—Differential areas on a piston.2-8

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