Explanation - The figure displays two joined cylindrical chambers. They both have different diameters and including the connecting tube, are filled with liquid.
Since the second piston is larger, it produces a larger force than the force applied to the smaller piston:
F2 = F1(A2/A1)
These ideas benefit us every day.
The hydraulic car lift relates to Pascal's Principle. It’s used in garages to lift a car off the ground for repairs. A small force by a small-area piston can be converted to a large force at a large area piston. It works relatively close to a lever, where a small force passes through a great distance to transport a heavy object a short distance. The work is the same when lifting the heavy object or when applying a small force. In the case of a lever a bar and a fulcrum convert the work, in the hydraulic lift the fluid performs the work.
Question – A hydraulic car lift has a pump piston with radius r1 = 0.0120 m. The resultant
piston has a radius of r2 = 0.150 m. The total weight of the car and plunger is F2 = 2500 m. If
the bottom ends of the piston and plunger are at the same height, what input force is
required to stabilize the car and output plunger?
Answer – We need to use the area for circular objects, A=3.14r2 for both the piston and
plunger. Apply Pascal's Principle:
F1= F2(A1/A2) = F2(3.14r2/3.14r2) = (20 500N)[(0.01202)/(0.1502)] = 131 N