Mechanical Properties of Fluids • Topic 1 of 3

Pressure & Pascal's Law

A fluid is anything that can flow — both liquids and gases. Unlike a solid, a fluid cannot sustain a shearing stress at rest, so it takes the shape of its container. The single most useful idea for fluids at rest is pressure.

Pressure is the normal (perpendicular) force exerted by a fluid per unit area of the surface in contact with it:

  • $P=\frac{F}{A}$, where $F$ is the force acting perpendicular to a surface of area $A$.
  • The SI unit of pressure is the pascal, $1\ \text{Pa}=1\ \text{N/m}^2$. Pressure is a scalar — it has no direction, even though force does.
  • At a point inside a fluid the pressure acts equally in all directions; this is why a balloon inflates evenly.

Pressure due to a liquid column (pressure at depth). Consider a point at depth $h$ below the free surface of a liquid of density $\rho$. The weight of the liquid column above it pushes down, so the pressure increases with depth:

  • $P=P_0+\rho gh$, where $P_0$ is the atmospheric (or surface) pressure and $\rho gh$ is the gauge pressure due to the liquid.
  • Pressure depends only on the depth, density and $g$ — not on the shape or cross-sectional area of the container. This is the hydrostatic paradox: liquid stands at the same level in connected vessels of different shapes.

Pascal's law states that a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and to the walls of the container. This principle is the heart of all hydraulic machines.

Hydraulic lift and brakes. In a hydraulic lift two pistons of areas $A_1$ (small) and $A_2$ (large) are connected by fluid. A force $F_1$ on the small piston creates a pressure $P=\frac{F_1}{A_1}$, transmitted to the large piston, which then experiences $F_2=P\,A_2=F_1\frac{A_2}{A_1}$. Because $A_2>A_1$, a small input force lifts a large load — a force multiplier. Hydraulic brakes use the same idea to apply equal braking force at all wheels.

Atmospheric pressure and the barometer. The atmosphere exerts pressure because of the weight of air above us; at sea level $P_0\approx1.013\times10^5\ \text{Pa}=1\ \text{atm}$. Torricelli's mercury barometer measures it: atmospheric pressure supports a column of mercury about $76\ \text{cm}$ high, so $P_0=\rho_{Hg}\,g\,h$. A height of $76\ \text{cm}$ of mercury equals 1 atm.

Hydraulic lift using Pascal's lawF1 (small)F2 (large)A1A2F2 = F1 (A2 / A1)
Solved Examples
1
Worked Example
A force of 200 N acts perpendicular to a surface of area $0.5\ \text{m}^2$. Find the pressure exerted.
Solution
  1. Step 1: Use $P=\frac{F}{A}$.
  2. Step 2: Substitute: $P=\frac{200}{0.5}$.
  3. Step 3: Compute: $P=400\ \text{Pa}$.

Answer: $P=400\ \text{Pa}$.

2
Worked Example
Find the total pressure at a depth of 10 m in a lake. (Take $\rho=1000\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$, $P_0=1.013\times10^5\ \text{Pa}$.)
Solution
  1. Step 1: Use $P=P_0+\rho gh$.
  2. Step 2: Gauge pressure $\rho gh=1000\times9.8\times10=9.8\times10^4\ \text{Pa}$.
  3. Step 3: Total $P=1.013\times10^5+0.98\times10^5=1.993\times10^5\ \text{Pa}$.

Answer: $P\approx1.99\times10^5\ \text{Pa}$ (about 2 atm).

3
Worked Example
In a hydraulic lift the small piston has area $0.01\ \text{m}^2$ and the large piston has area $0.5\ \text{m}^2$. A force of 100 N is applied on the small piston. What is the force on the large piston?
Solution
  1. Step 1: By Pascal's law the pressure is the same: $\frac{F_1}{A_1}=\frac{F_2}{A_2}$.
  2. Step 2: So $F_2=F_1\frac{A_2}{A_1}=100\times\frac{0.5}{0.01}$.
  3. Step 3: Compute: $F_2=100\times50=5000\ \text{N}$.

Answer: $F_2=5000\ \text{N}$.

4
Worked Example
What is the gauge pressure at the bottom of a 5 m tall column of oil of density $800\ \text{kg/m}^3$? (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: Gauge pressure $=\rho gh$.
  2. Step 2: Substitute: $P=800\times10\times5$.
  3. Step 3: Compute: $P=40000\ \text{Pa}=4\times10^4\ \text{Pa}$.

Answer: Gauge pressure $=4\times10^4\ \text{Pa}$.

5
Worked Example
A mercury barometer reads a height of 76 cm. Calculate the atmospheric pressure. (Take $\rho_{Hg}=13600\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Atmospheric pressure $P_0=\rho_{Hg}\,g\,h$ with $h=0.76\ \text{m}$.
  2. Step 2: Substitute: $P_0=13600\times9.8\times0.76$.
  3. Step 3: Compute: $P_0\approx1.013\times10^5\ \text{Pa}$.

Answer: $P_0\approx1.013\times10^5\ \text{Pa}=1\ \text{atm}$.

6
Worked Example
At what depth in sea water (density $1030\ \text{kg/m}^3$) is the gauge pressure equal to one atmosphere ($1.013\times10^5\ \text{Pa}$)? (Take $g=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Set gauge pressure $\rho gh=1.013\times10^5$, so $h=\frac{1.013\times10^5}{\rho g}$.
  2. Step 2: Substitute: $h=\frac{1.013\times10^5}{1030\times9.8}=\frac{1.013\times10^5}{1.0094\times10^4}$.
  3. Step 3: Compute: $h\approx10.0\ \text{m}$.

Answer: $h\approx10\ \text{m}$ below the surface.

Key Points

  • Pressure is normal force per unit area: $P=\frac{F}{A}$, a scalar measured in pascals ($1\ \text{Pa}=1\ \text{N/m}^2$).
  • Pressure at depth $h$ in a liquid: $P=P_0+\rho gh$; it depends on depth, density and $g$, not on container shape.
  • Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.
  • Hydraulic machines multiply force: $F_2=F_1\frac{A_2}{A_1}$, the basis of lifts and brakes.
  • Atmospheric pressure $\approx1.013\times10^5\ \text{Pa}=76\ \text{cm}$ of mercury, measured by a barometer.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The SI unit of pressure is the:
Explanation: Pressure $P=\frac{F}{A}$ is measured in $\text{N/m}^2=$ pascal (Pa).
Q2.The pressure at a depth $h$ inside a liquid of density $\rho$ (open to atmosphere $P_0$) is:
Explanation: The weight of the liquid column adds $\rho gh$ to the surface pressure, giving $P=P_0+\rho gh$.
Q3.Pascal's law is the working principle of a:
Explanation: A hydraulic lift transmits pressure undiminished through an enclosed fluid, as Pascal's law states.
Q4.The fact that liquid pressure at a point depends only on depth (not container shape) is called the:
Explanation: Since $P=P_0+\rho gh$ has no area term, liquid rises to the same level in differently shaped connected vessels — the hydrostatic paradox.
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