Pascal Laws

Overview

Engineering reference data for Pascal Laws in miscellaneous.

Key Formulas

Unit Conversion

$y = x \cdot k$

Multiply by conversion factor.

Linear Interpolation

$y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}$

Estimate between two known points.

Percentage

$p = \frac{\text{part}}{\text{whole}} \times 100\%$

Part as fraction of whole.

Variables

Pascal's Laws: Key Principles

• If the weight of a fluid is neglected, the pressure throughout an enclosed volume will be the same.
• The static pressure in a fluid acts equally in all directions.
• The static pressure acts at right angles to any surface in contact with the fluid.

Practical Examples

Example 1: Pressure in a Hydraulic Cylinder

The pressure of 2000 Pa in a hydraulic cylinder acts equally on all surfaces. The force on a piston with area 0.1 m² can be calculated using the formula:

$F = p \cdot A$

where:

• $F$ is the force (N)
• $p$ is the pressure (Pa, N/m²)
• $A$ is the area (m²)

Plugging in the values:

$F = (2000 \text{ Pa}) \cdot (0.1 \text{ m}^2) = 200 \text{ N}$

Example 2: Force in a Hydraulic Jack

The pressure acting on both pistons in a hydraulic jack is equal. The force equations for the small and large cylinders are:

$F_s = p \cdot A_s$

$F_l = p \cdot A_l$

where:

• $F_s$ is the force on the small piston (N)
• $F_l$ is the force on the large piston (N)
• $A_s$ is the area of the small cylinder (m²)
• $A_l$ is the area of the large cylinder (m²)
• $p$ is the pressure (Pa, N/m²)

From these, we can derive the relationship:

$\frac{F_s}{A_s} = \frac{F_l}{A_l}$

or

$F_s = F_l \cdot \frac{A_s}{A_l}$

This indicates that the effort force required in the small cylinder to lift a load on the large cylinder depends on the area ratio between the two cylinders.

Example 3: Hydraulic Jack Lifting a Car

Consider a hydraulic jack lifting the back end (half the weight) of a car with a mass of 2000 kg. The area ratio $A_s / A_l$ is 0.1 (the large cylinder area is 10 times the small cylinder area). The force (weight) acting on the large cylinder is calculated using Newton's second law:

$F_l = m \cdot a$

where:

• $m = \frac{1}{2} \cdot 2000 \text{ kg} = 1000 \text{ kg}$ (since only half the weight is lifted)
• $a = 9.81 \text{ m/s}^2$

Thus:

$F_l = (1000 \text{ kg}) \cdot (9.81 \text{ m/s}^2) = 9810 \text{ N}$

Then the force on the small cylinder is:

$F_s = F_l \cdot \frac{A_s}{A_l} = (9810 \text{ N}) \cdot 0.1 = 981 \text{ N}$

References

• Original Source