8.2 Pressure

Pressure is a fundamental concept in physics, measuring how concentrated a force is on a surface. It's calculated by dividing force by area and plays a crucial role in understanding fluid behavior and engineering applications.

Fluid pressure arises from particle interactions and varies with depth. It's essential in designing structures like dams and submarines. Understanding absolute and gauge pressure helps in practical applications, from weather forecasting to tire inflation.

Pressure on surfaces

Force per unit area

Pressure represents how concentrated a force is when distributed across a surface area. When we push on something, the pressure depends not just on how hard we push, but also on how much area we contact.

• Pressure $P$ equals the magnitude of the perpendicular force component $F$ exerted over a given surface area $A$, expressed by the equation $P = \frac{F}{A}$
• Increasing the force while keeping the area constant results in higher pressure
• Decreasing the area while maintaining the same force also increases pressure
• Units of pressure include pascals (Pa), which equal newtons per square meter ($\text{N}/\text{m}^2$)
• This explains why snow shoes have large surface areas (to reduce pressure on snow) and why knife edges are sharp (to increase pressure for cutting)

Scalar nature of pressure

Pressure is uniquely different from many quantities we study in physics because it lacks directionality. It pushes equally in all directions within a fluid.

• Pressure lacks directionality, having only magnitude without a specific orientation
• Differs from force, which is a vector quantity possessing both magnitude and direction
• Allows pressure to be evenly distributed throughout a fluid, regardless of the direction of the force causing it
• This scalar nature explains why a water balloon expands equally in all directions and why deep-sea creatures experience pressure from all sides

Incompressible fluid properties

Incompressible fluids maintain their volume regardless of the pressure applied to them, which is why hydraulic systems can be so effective at transmitting force.

• Incompressible fluids (liquids) maintain a constant volume and density under varying pressures
• Applying pressure to an incompressible fluid does not significantly alter its volume or density
• Examples of incompressible fluids include water, oil, and honey
• In contrast, compressible fluids (gases) experience changes in volume and density when subjected to pressure variations
• This property is the foundation for hydraulic systems like car brakes and construction equipment

Fluid pressure

Particle interactions in fluids

Fluid pressure emerges from countless microscopic collisions between fluid particles and surfaces. Understanding this helps explain how fluid pressure works at the molecular level.

• Fluid pressure arises from the collective interactions between the fluid's particles and the surface they contact
• Each particle exerts a minuscule force on the surface, contributing to the overall pressure
• The greater the number of particle collisions with the surface, the higher the pressure exerted by the fluid
• Factors influencing particle interactions include fluid density, temperature, and velocity
• This explains why gases are more compressible than liquids—their particles have more space between them and fewer collisions per unit volume

Absolute vs gauge pressure

We often need to distinguish between the total pressure (absolute) and the pressure relative to atmospheric conditions (gauge). This distinction is crucial in many practical applications.

• Absolute pressure $P$ at a point equals the sum of a reference pressure $P_0$ (e.g., atmospheric pressure $P_{atm}$) and the gauge pressure $P_{gauge}$
• Relevant equation: $P = P_0 + \rho gh$
• Gauge pressure measures the pressure relative to the surrounding atmospheric pressure
  • Positive gauge pressure indicates pressure above atmospheric pressure
  • Negative gauge pressure (vacuum) signifies pressure below atmospheric pressure
• Absolute pressure provides a standardized measurement independent of atmospheric conditions
• Examples include tire pressure gauges (measuring gauge pressure) and barometers (measuring absolute pressure)

Vertical fluid pressure

The pressure in a fluid increases with depth due to the weight of the fluid above. This vertical pressure variation explains many phenomena we observe in lakes, oceans, and even the atmosphere.

• Gauge pressure $P_{gauge}$ in a vertical column of fluid depends on fluid density $\rho$, gravitational acceleration $g$, and height $h$ of the fluid column
• Described by the equation $P_{gauge} = \rho gh$
• Pressure increases linearly with depth in a static fluid due to the weight of the fluid above
• At the surface of the fluid (h = 0), gauge pressure equals zero (atmospheric pressure)
• Doubling the depth in a fluid results in a doubling of the gauge pressure
• Applications include understanding pressure in swimming pools, aquariums, and underwater environments
• This explains why your ears "pop" when diving underwater or changing elevation rapidly

Practice Problem 1: Force per Unit Area

Solution: First, we need to find the force exerted by the student on the floor: $F = mg = 75 \text{ kg} \times 9.8 \text{ m/s}^2 = 735 \text{ N}$

a) To find the pressure, we use $P = \frac{F}{A}$: $P = \frac{735 \text{ N}}{180 \text{ cm}^2} = \frac{735 \text{ N}}{0.018 \text{ m}^2} = 40,833 \text{ Pa}$

b) When standing on one foot, the contact area is halved to 90 cm² or 0.009 m²: $P = \frac{735 \text{ N}}{0.009 \text{ m}^2} = 81,667 \text{ Pa}$

The pressure doubles when standing on one foot because the same force is distributed over half the area.

Practice Problem 2: Incompressible Fluid Properties

Solution: a) The pressure in the hydraulic fluid is: $P = \frac{F}{A} = \frac{50 \text{ N}}{10 \text{ cm}^2} = \frac{50 \text{ N}}{0.001 \text{ m}^2} = 50,000 \text{ Pa}$

b) Since pressure is the same throughout an incompressible fluid, we can calculate the force on the large piston: $F_2 = P \times A_2 = 50,000 \text{ Pa} \times 500 \text{ cm}^2 = 50,000 \text{ Pa} \times 0.05 \text{ m}^2 = 2,500 \text{ N}$

This demonstrates how hydraulic systems can multiply force—in this case, by a factor of 50.

Practice Problem 3: Vertical Fluid Pressure

Solution: a) The gauge pressure at the bottom is: $P_{gauge} = \rho gh = 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 2.5 \text{ m} = 24,500 \text{ Pa}$

b) The absolute pressure at the bottom is: $P_{absolute} = P_{atm} + P_{gauge} = 101,300 \text{ Pa} + 24,500 \text{ Pa} = 125,800 \text{ Pa}$

c) The force on the bottom of the tank is: $F = P_{gauge} \times A = 24,500 \text{ Pa} \times (3 \text{ m} \times 4 \text{ m}) = 24,500 \text{ Pa} \times 12 \text{ m}^2 = 294,000 \text{ N}$

This shows why dam walls are built thicker at the bottom—they must withstand significantly greater forces due to increasing water pressure with depth.