### Before going ahead we need to know *Pascal’s Law*

*Pascal’s Law*

Pascal’s lawstate that the change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all point in the fluid.

In a hydraulic jack oil there is enclose hydraulic oil, hence Pascal’s law will apply. Now suppose we have to lift a load off 1000 kg and the bigger piston area is 100 cm^{2} then Pressure,

P=F/A

P=1000/100

p=10 kg/ cm^{2}

If the area of smaller piston is 10 cm^{2} and according to pascal’s law pressure will be same so, P= 10kg/ cm^{2}

required force,

F=P x A

F= 10 x 10

F = 100 kg

**Note: Unit of force is N (newton), therefore 1kg force = 1 kg x g (acceleration due to gravity) = 1 x 9.8 = 9.8 N. **

In example, we see with the help of 100 kg force we can lift 1000 kg of load. This is called **Force Multiplication. **

See the both hydraulic jack, if same amount of muscle force if applied on the lever then in which pressure will be more. Remember the pressure formula P=F/A. Because pressure has inversely relation with area, Greater the area lower the pressure. Hence small jack will create more pressure than bigger jack.

We know work= force X displacement.

In the above example let’s consider small piston displaced 10 cm.

Therefore work done= 100 kg.f X 10 cm= 1000 kg.f/cm (kg.f is also a unit of force)

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