If the output piston (the one the car is sitting on) has an area of 70 square inches, how much force must be applied to the 0.5 square inch input piston to lift the car?
The lifting need is 3000 pounds. Since the piston has an area of 70 square inches, the required pressure is 3000 divided by 70 or 42.86 psi. To get that much pressure you would require 1/2 that amount of pressure on a 0.5 square inch input piston or 21.43 pounds of pressure.
The hydraulics give you a mechanical advantage of 70/0.5 140 so you have to supply 3000/140 21.4 pounds Didn't I answer this already? .
Pressure force/area Force pressure * area The pressures in the two cylinders will be the same. So F1/A1 F2/A2 3*10^4 /[ pi * 22^2] X / [pi * 6.38^2] X 6.38^2 * 3*10^4 / 22^2 2523 N
