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Multiplication of Forces

Pascal's Law states that the pressures in both cylinders are the same (p1=p2).   Thus, given a force, F1, of 10 pounds (lbs) in the right cylinder acting on a piston area, A1, of 2 square inches (sq. in.) a pressure in the right cylinder, p1, of 5 pounds per square inch (lbs/sq. in. = psi) is produced.  Now if A2 is given as 5 sq. in., then the force developed in the left cylinder is F2 = p2xA2, or 25 lbs.  This is due to the fact that p1=p2.  Thus Pascal's Law shows the way in which one can increase the output force for a given input force...regulate the areas of the pistons!

 The only disadvantage is the size of the piston stroke involved.  Let's say, piston 2 moves (up) 10 inches.  For the previous problem the work done by piston 2 is F2 times the stroke of piston 2 (10 in. x 25 lbs).  If no losses exist in the system due to friction, then work is conserved and piston 1 must do 250 in-lb of work. Therefore, the F1 must move down 25 inches (250 in-lbs/10 lb force)! To move piston 2 up, a volume of 50 cubic inches (cu. in.) of incompressible oil must be pumped in at 5 psi (since pressure times Volume is also another way to find work). The movements of the pistons are measured relative to the bottom of the cylinder with all the measurements computed to produce 100 % efficiency.

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Updated: February 23, 1999