The Hydraulic Jack simulation shows how when the piston of the master cylinder is pumped, oil flows past non-return valves and pushes the piston of the slave cylinder up. Pressure gauges show how the pressures in both cylinders are the same, and yet the force multiplication is achieved.
The master cylinder has a small diameter d and hence a small cross-sectional area a. The force f that is applied to the master piston produces a pressure p which is equal to f/a.
The slave cylinder has a large diameter D and hence a large cross-sectional area A. The load on the slave piston is F and it exerts pressure P on the oil beneath it which is equal to F/A.
Since the pressures of the oil in the two arms are equal, p = P.
f /a = F / A so F = f (A/a)
Now
A/a = (D/d)2 so F = f (D/d)2
This means that even if the diameter D is 5 times larger than d, F can be 25 times larger than f. That is why a weak man can lift a two tonne truck with a jack.
Pressure 22.3
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