Pascal's Law
1.

In a hydraulic machine, the two pistons are of area of cross–section in the ratio 1 : 10. What force is needed on the narrow piston to overcome a force of 120 N on the wider piston?

Sol:
A1 : A2 = 1 : 10
F1 = ?
F2 = 120 N
By the principle of hydraulic machine
F1/A1 = F2/A2
∴ F1 = (F2 × A1) / A2
= (120 × 1) / 10
= 12 N
2.

The press plunger of a hydraulic press is 4 m2 in cross–section and is required to overcome a resistive load of 400 kgf.
(i) Calculate the force required on the pump plunger if the area of cross–section of the pump plunger is 0.01 m2 ?
(ii) If the mechanical advantage of the handle is 10, calculate the force which should be applied to operate the press ?

Sol:
(i) Let F be the force required on the pump plunger.
By Pascal's law, Pressure on pump plunger = pressure on press plunger.
F/0.01 = 400/4
∴F = 100 × 0.01 = 1 kgf.
(ii) Mechanical advantage = Load/effort
∴10 = 1 kgf / effort
Hence effort = 1/10
= 0.1 kgf.