A hydraulic jack has two pistons. The diameters of the larger piston and smaller piston are 15 cm and 1.25 cm respectively.It is necessary to lift a 2000 pound car. How large a force must be applied to the small piston? Now assume that a small piston moves through a distance of 10 cm. By what distance will the large piston move?
The hydraulic jack works on the principle of Pascal's law:
pressure (p1) = pressure (p2)
=>F1/A1 = F2/A2
where F1 and F2 are the forces and A1 and A2 are the corresponding sides cross-sectional ...
The solution shows the formulas and calculations with explanations to arrive at the answer.