Pascal's Principle

Hydrostatic pressure at depth $h$ in an incompressible fluid depends on surface pressure $p_0$ and fluid density $\rho$ via $p = p_0 + \rho g h$ . Evaluating how a change in $p_0$ propagates through the medium yields the foundation of hydraulic systems:

Pascal's Principle

Pascal's Principle states that when a change in pressure is applied to an enclosed fluid, it is transmitted undiminished to all portions of the fluid and to the walls of its container.

To see why this holds mathematically from hydrostatic equilibrium:

Derivation of Pascal's Principle from Hydrostatic Pressure

Let the original pressure at depth $h$ be:

p_{\text{old}} = p_0 + \rho g h

Increase the surface pressure $p_0$ by an applied change $\Delta p_0$ . The new total pressure $p_{\text{new}}$ at depth $h$ becomes:

p_{\text{new}} = (p_0 + \Delta p_0) + \rho g h

Regrouping the terms gives:

p_{\text{new}} = \underbrace{(p_0 + \rho g h)}_{p_{\text{old}}} + \Delta p_0 = p_{\text{old}} + \Delta p_0

Subtracting $p_{\text{old}}$ yields the pressure change at depth $h$ :

\Delta p = p_{\text{new}} - p_{\text{old}} = \Delta p_0

Because depth $h$ is arbitrary, any applied pressure change $\Delta p_0$ transmits undiminished to every point in the fluid.

Hydraulic Systems and Mechanical Advantage

A primary application of Pascal's Principle is the hydraulic lift. Consider two fluid-filled cylinders connected by a pipe, fitted with airtight pistons of cross-sectional areas $A_1$ and $A_2$ .

Applying a downward force $\vec{F}_1$ to the smaller piston creates a pressure change:

p_1 = \frac{F_1}{A_1}

By Pascal's Principle, this pressure change is transmitted undiminished throughout the fluid, so $p_1 = p_2$ :

\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \left(\frac{A_2}{A_1}\right)

Because $A_2 > A_1$ , the output force $F_2$ is greater than the input force $F_1$ by the ratio of the piston areas.

Work and Conservation of Energy

Although the force increases, total energy is conserved. For an incompressible fluid, the volume of fluid displaced by piston 1 ( $\Delta V_1 = A_1 d_1$ ) must equal the volume displaced at piston 2 ( $\Delta V_2 = A_2 d_2$ ):

A_1 d_1 = A_2 d_2 \implies d_2 = d_1 \left(\frac{A_1}{A_2}\right)

Evaluating the work done on the system:

W_2 = F_2 d_2 = \left( F_1 \frac{A_2}{A_1} \right) \left( d_1 \frac{A_1}{A_2} \right) = F_1 d_1 = W_1

Thus, the work output equals the work input; a smaller force applied over a larger distance produces a larger force over a smaller distance.

Example 1

A hydraulic car lift has a small input piston with a cross-sectional area of $0.05\text{ m}^2$ and a large output piston with an area of $2.5\text{ m}^2$ . Calculate the input force $F_1$ required to lift a $1,500\text{ kg}$ vehicle.

Hint Click to unblur

Determine the force needed to support the car's weight, then equate the pressures on both pistons using $p_1 = p_2$ .

Solution Click to unblur

First, calculate the required output force $F_2$ equal to the vehicle's weight:

F_2 = m g = (1500\text{ kg})(9.8\text{ m/s}^2) = 14,700\text{ N}

By Pascal's Principle, pressure is transmitted undiminished throughout the fluid:

\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_1 = F_2 \left(\frac{A_1}{A_2}\right)

Substitute the given areas and force:

F_1 = (14,700\text{ N})\left(\frac{0.05\text{ m}^2}{2.5\text{ m}^2}\right) = (14,700\text{ N})(0.02) = 294\text{ N}

Example 2

A technician applies a force of $120\text{ N}$ to the small circular piston of a hydraulic jack. The small piston has a radius of $1.5\text{ cm}$ , while the large circular piston has a radius of $12\text{ cm}$ . What maximum force can the large piston exert?

Solution

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Remember that area scales quadratically with radius ( $A = \pi r^2$ ), causing the constant $\pi$ to cancel out.

Solution Click to unblur

Express the surface areas in terms of their radii $r_1$ and $r_2$ :

A_1 = \pi r_1^2 \quad \text{and} \quad A_2 = \pi r_2^2

Substitute these into the pressure balance relation:

F_2 = F_1 \left(\frac{A_2}{A_1}\right) = F_1 \left(\frac{\pi r_2^2}{\pi r_1^2}\right) = F_1 \left(\frac{r_2}{r_1}\right)^2

Evaluate using the given values:

F_2 = (120\text{ N}) \left(\frac{12\text{ cm}}{1.5\text{ cm}}\right)^2 = (120\text{ N}) (8)^2 = (120\text{ N})(64) = 7,680\text{ N}

Example 3

Using the hydraulic system from Example 2, if the small piston is depressed downward by a distance of $d_1 = 20\text{ cm}$ , how far up is the load lifted ( $d_2$ )? Verify that total work done on the system is conserved.

Hint Click to unblur

Equate the displaced fluid volumes $V_1 = V_2$ to find $d_2$ , then check $W = F d$ for both sides.

Solution Click to unblur

Equating the displaced fluid volumes ( $A_1 d_1 = A_2 d_2$ ):

d_2 = d_1 \left(\frac{A_1}{A_2}\right) = d_1 \left(\frac{r_1}{r_2}\right)^2 = (20\text{ cm}) \left(\frac{1.5\text{ cm}}{12\text{ cm}}\right)^2 = \frac{20}{64}\text{ cm} = 0.3125\text{ cm}

Next, verify work conservation by evaluating $W_1$ and $W_2$ (converting displacements to meters):

W_1 = F_1 d_1 = (120\text{ N})(0.20\text{ m}) = 24\text{ J}

W_2 = F_2 d_2 = (7,680\text{ N})(0.003125\text{ m}) = 24\text{ J}

Because $W_1 = W_2$ , energy is conserved.