# Pascal law , Applications of Pascal Law , Hydraulic Lift

### What is Pascal’s law ?

When external pressure is applied to a confined fluid, the fluid transmits this pressure to every point within, undiminished.
At every point on the wall of the container this transmitted pressure acts normal to the wall of the container.

Let us understand this through an example. Consider a U-tube with each limb having different cross-sections as shown in the figure.

An incompressible fluid fills the U-tube. Suppose that the area of cross-section of the right limb is A1and that of the left limb is A2. Both the limbs are fitted with frictionless pistons. Suppose that a force of magnitude F1 is applied to the right piston downward.

The pressure developed on the right piston is $\displaystyle P = \frac{F_1}{A_1}$
which acts normally at every point within the fluid and also at every point on the wall as well as on the pistons.

Suppose that the force experienced by the left piston due to this pressure P action on it is F2.

According to Pascal’s law,

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2} = P$     ……(1)

Therefore ,

$\displaystyle F_2 = \frac{A_2}{A_1}F_1$    …….. (2)

Since A2 > A1

F2 > F1      ……..    (3)

Thus, we obtain a larger force F2 by applying a smaller force F1. Now as F2 > F1, does it violate the principle of conservation of energy? Does it mean that greater work is done by the left piston than the right piston?

Suppose that the right (smaller) piston comes is displaced downward through a distance x1 and the left (larger) piston is consequently displaced upward through a distance x2 . As the fluid is incompressible,

x1A1 = x2A2 = volume of fluid displaced     ……(4)

where x1A1 = volume of fluid displaced by right piston and x2A2 = volume of the fluid displaced by left piston.

Therefore, $\displaystyle x_2 = \frac{A_1}{A_2}x_1$  …..(5)

The work done by the force F1 on the piston on the right is

W1 = F1x1        ………(6)

The work done by the force F2 on the piston on the left is

W2 = F2x2          ……….(7)

From equations (2) , (5) and (7), we get,

$\displaystyle W_2 = \frac{A_2}{A_1}.F_1 \frac{A_1}{A_2}x_1$

$\displaystyle W_2 = F_1 .x_1 = W_1$     ……(8)

Therefore, W2 = W1    ……….(9)

The work done by the pistons are identical. Therefore, energy is conserved.

Solved Example : In a hydraulic press the narrower limb has an area of cross-section 0.25 m2 and while that of the broader limb is 1 m2. How much force should be applied on to the liquid in the narrower limb to lift a vehicle of 1000 kg kept on top of the fluid in the larger limb ?

Solution: Let the required force be F1

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2} \, = Pressure$

$\displaystyle F_1 = \frac{A_1}{A_2}.F_2$

$\displaystyle F_1 = \frac{0.25}{1} (1000\times 9.8)$

$\displaystyle F_1 = \frac{9800}{4}$ = 2450 N

Solved Example: A hydraulic brake has brake lever piston of area 0.01 m2. The wheel drum piston has area of 0.25 m2. If the driver exerts a force on the lever of 10 N, find the force on wheel drum.

Solution : According to Pascal’s law,

$\large \frac{F_1}{A_1} = \frac{F_2}{A_2}$

Therefore , $\large F_2 = \frac{A_2}{A_1} \times F_1$

$\large F_2 = \frac{0.25}{0.01} \times 10 = 250 N$

Therefore, force on wheel drum is F2 = 250 N