Pressure in Fluids and Atmospheric Pressure

A. Thrust vs Pressure:

• Thrust ($T$): Total normal compressive force acting perpendicular to a surface. SI unit: $\text{N}$.
• Pressure ($P$): Thrust acting perpendicularly per unit surface area: $$\mathbf{P = \frac{\text{Thrust}}{\text{Area}} = \frac{F}{A}}$$ SI unit: Pascal ($1\text{ Pa} = 1\text{ N/m}^2$). Common units: $1\text{ bar} = 10^5\text{ Pa}$, $1\text{ millibar} = 100\text{ Pa}$, $1\text{ atmosphere (atm)} = 1.013 \times 10^5\text{ Pa} = 760\text{ mm of Hg} = 760\text{ torr}$.

B. Derivation of Hydrostatic Pressure ($P = h\rho g$):

Consider an imaginary cylindrical column of liquid of cross-sectional area $A$ and height $h$ in a liquid of density $\rho$:

1. $\text{Volume of column } V = A \times h$
2. $\text{Mass of liquid column } m = V \times \rho = A h \rho$
3. $\text{Thrust (Weight) on the base } W = mg = A h \rho g$
4. $$\mathbf{\text{Liquid Pressure } P = \frac{\text{Thrust}}{\text{Area}} = \frac{A h \rho g}{A} = h\rho g}$$

Total (Absolute) Pressure at Depth $h$: If atmospheric pressure $P_0$ acts on the open surface:

where $h\rho g$ is called the gauge pressure.

Statement of Pascal's Law:

"Pressure exerted at any point in an enclosed, incompressible liquid is transmitted equally and undiminished in all directions throughout the liquid."

Principle of the Hydraulic Press / Jack:

Consists of two cylinders fitted with airtight, frictionless pistons of unequal cross-sectional areas $A_1$ (narrow piston) and $A_2$ (wide piston), connected by a liquid pipe:

• A small input force $F_1$ applied to the narrow piston creates a pressure: $$P_1 = \frac{F_1}{A_1}$$
• By Pascal's law, this exact pressure $P_1$ is transmitted to the wide piston: $P_2 = P_1$.
• The upward load force exerted on the wide piston is: $$F_2 = P_2 \times A_2 = \left(\frac{F_1}{A_1}\right) A_2 = F_1 \left(\frac{A_2}{A_1}\right)$$
• Mechanical Advantage (MA): $$\mathbf{\text{MA} = \frac{\text{Load } (F_2)}{\text{Effort } (F_1)} = \frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{D_2}{D_1}\right)^2}$$
• Since $A_2 \gg A_1$, a tiny effort $F_1$ lifts a massive load $F_2$! (Energy is strictly conserved: the small piston must move down through a much greater distance $d_1$ than the large piston lifts $d_2$: $F_1 d_1 = F_2 d_2$).

A. Torricellian Mercury Barometer:

Invented by Evangelista Torricelli in 1643. A $1\text{ meter}$ long glass tube sealed at one end is filled with pure mercury and inverted into a trough of mercury:

• The mercury level drops until the vertical height of the mercury column is exactly $76\text{ cm}$ ($760\text{ mm}$) above the trough level at sea level.
• The empty space above the mercury column is called the Torricellian Vacuum (contains only a trace of invisible mercury vapor).
• Standard Atmospheric Pressure: $$P_0 = h\rho g = 0.76\text{ m} \times 13,600\text{ kg/m}^3 \times 9.8\text{ m/s}^2 = \mathbf{1.013 \times 10^5\text{ Pa}} \approx 1.013\text{ bar}$$

B. Why Mercury is Chosen Over Water:

1. Ultra-High Density: Mercury density is $13.6\text{ g/cm}^3$ ($13.6$ times denser than water). A mercury barometer requires a column of only $0.76\text{ m}$. If water were used, the column height would be $h = 0.76 \times 13.6 = \mathbf{10.34\text{ meters}}$—requiring a four-story building!
2. Does Not Wet Glass: Mercury has high surface tension and does not stick to glass, forming a crisp convex meniscus.
3. Opaque & Shining: Easily visible through the glass tube.
4. Low Vapor Pressure: Mercury vapor exerts negligible pressure in the vacuum space.

Problem 1: Calculate the pressure exerted by a column of water of height $20\text{ meters}$ at the bottom of a lake. Take density of water $\rho = 1000\text{ kg/m}^3$ and $g = 9.8\text{ m/s}^2$. If atmospheric pressure is $1.01 \times 10^5\text{ Pa}$, find the total absolute pressure.

Solution:

1. Gauge pressure due to water column:

2. Total Absolute Pressure:

Problem 2: In a hydraulic lift, the diameters of the smaller and larger pistons are $4\text{ cm}$ and $20\text{ cm}$ respectively. What force must be applied to the smaller piston to lift an automobile of mass $1500\text{ kg}$ on the larger piston? (Take $g = 10\text{ m/s}^2$).

Solution:

Load on larger piston $F_2 = mg = 1500 \times 10 = 15,000\text{ N}$.

Ratio of piston diameters: $\frac{D_2}{D_1} = \frac{20}{4} = 5$.

Using the hydraulic mechanical advantage relation:

Notice: A modest effort of only $600\text{ N}$ (approx. $60\text{ kgf}$) effortlessly lifts a massive $1500\text{ kg}$ automobile!