1.0 Introduction to Fluids and Pressure
A substance that can flow is called a fluid. Both liquids and gases are fluids. Unlike solids, fluids do not have a fixed shape and exert pressure in all directions. In this section, we differentiate between the total force applied and the pressure felt by a surface.
Thrust vs. Pressure
- Thrust: The total force acting normally (perpendicularly) on a surface. It is a vector quantity and its S.I. unit is the Newton (N).
- Pressure: The thrust acting per unit area of a surface. It is a scalar quantity.
Mathematical Definition
$$P = \frac{F}{A}$$
Where: $P$ = Pressure, $F$ = Normal Force (Thrust), $A$ = Area.
S.I. Unit: Pascal (Pa) or $N/m^2$
1.1 Factors Affecting Pressure
From the formula, we can conclude that for a constant thrust, the pressure is inversely proportional to the area. This explains many everyday phenomena:
- Wide Straps on Bags: School bags have wide straps to increase the area, thereby reducing the pressure on the shoulders.
- Sharp Knives: A sharp knife has a very small area at the edge, producing high pressure to cut through objects easily.
- Foundations of Buildings: They are made wide to reduce the pressure exerted on the ground.
In numericals, area is often given in $cm^2$. You MUST convert it to $m^2$ to get the answer in Pascals.
Remember: $1\,cm^2 = 10^{-4}\,m^2$.
Do not simply divide by 100!
A block of weight $500\,N$ is placed on a table. If the area of the base of the block is $0.25\,m^2$, calculate the pressure exerted on the table.
Solution:
1. Thrust ($F$): $500\,N$ (Weight acts as normal force).
2. Area ($A$): $0.25\,m^2$.
3. Calculation: $P = F/A = 500 / 0.25 = \mathbf{2000\,Pa}$.
Final Answer: The pressure exerted is $2000\,Pa$ (or $2\,kPa$).
Camels can walk easily on soft sand because they have very large, broad feet. This increases the surface area in contact with the sand, reducing the pressure and preventing them from sinking!
2.0 Pressure in Liquids (Hydrostatic Pressure)
A liquid contained in a vessel exerts pressure at all points and in all directions. This pressure is due to the weight of the liquid column above that point. This is why deep-sea divers experience immense pressure as they go deeper into the ocean.
Pressure in a Liquid Column
$$P = h \times \rho \times g$$
Where: $h$ = depth of the point from the free surface, $\rho$ (rho) = density of liquid, $g$ = acceleration due to gravity.
2.1 Laws of Liquid Pressure
Based on the formula $P = h\rho g$, we can observe several key behaviors of liquids at rest:
- Depth Dependency: Pressure increases with the depth below the free surface.
- Horizontal Level: In a stationary liquid, pressure is the same at all points on the same horizontal plane.
- Independent of Shape: The pressure at a certain depth does not depend on the shape or size of the container (Hydrostatic Paradox).
- All Directions: A liquid exerts pressure in all directions—upwards, downwards, and laterally.
Pascal's Law
This law states that if pressure is applied to any part of a confined, incompressible fluid, it is transmitted undiminished to every other part of the fluid and the walls of the container.
Applications: Hydraulic Brakes, Hydraulic Jacks, and Hydraulic Presses.
The formula $h\rho g$ gives the pressure due to the liquid only. If a vessel is open to the atmosphere, the Total Pressure at depth $h$ is:
Total Pressure = Atmospheric Pressure ($P_o$) + $h\rho g$.
Read the numerical carefully to see if "total" pressure is asked!
Find the pressure at the bottom of a water tank $5\,m$ deep. (Density of water = $1000\,kg/m^3$, $g = 9.8\,m/s^2$)
Solution:
1. Depth ($h$): $5\,m$.
2. Density ($\rho$): $1000\,kg/m^3$.
3. Gravity ($g$): $9.8\,m/s^2$.
4. Calculation: $P = h\rho g = 5 \times 1000 \times 9.8 = \mathbf{49,000\,Pa}$.
Final Answer: The pressure at the bottom is $49,000\,Pa$ or $49\,kPa$.
Dam walls are always made thicker at the bottom than at the top. This is because the pressure of the water increases with depth, and the thick base is needed to withstand the massive lateral pressure at the bottom.
3.0 Atmospheric Pressure
Our Earth is surrounded by a thick layer of air called the Atmosphere, extending up to nearly 300 km. Just like a column of liquid, this massive column of air exerts a thrust on the surface of the Earth. The thrust exerted per unit area of the Earth's surface by the air column is called Atmospheric Pressure.
Standard Atmospheric Pressure
At sea level, the atmospheric pressure is equivalent to the pressure exerted by a 76 cm (or 760 mm) high column of mercury.
- 1 atm: $1.013 \times 10^5\,Pa$ (approx. $10^5\,Pa$).
- Torr: A unit equal to 1 mm of Hg. ($1\,atm = 760\,Torr$).
- Bar: Another common unit. ($1\,bar = 10^5\,Pa$).
3.1 Measurement: The Simple Barometer
A barometer is an instrument used to measure atmospheric pressure. The first barometer was designed by Torricelli. It consists of a long glass tube filled with mercury and inverted into a trough of mercury.
Why Mercury?
- Mercury is very dense ($\rho = 13600\,kg/m^3$), so the tube height is manageable ($76\,cm$). Water would require a tube over $10\,m$ high!
- It does not wet the glass.
- Its vapor pressure is negligible, ensuring a true Torricellian Vacuum at the top.
As we go higher (e.g., climbing a mountain), the density of air decreases and the height of the air column above us also decreases. Therefore, atmospheric pressure decreases with altitude.
Effects: Breathing becomes difficult, and fountain pens may leak due to higher pressure inside the pen than outside.
Pressure to Mercury Height Conversion
$$P = h \cdot \rho_{Hg} \cdot g$$
Using $h = 0.76\,m, \rho = 13600\,kg/m^3, g = 9.8\,m/s^2$ results in $\approx 101,293\,Pa$.
The atmospheric pressure at a place is $75\,cm$ of Hg. Calculate this pressure in S.I. units. (Take $\rho_{Hg} = 13600\,kg/m^3$ and $g = 10\,m/s^2$)
Solution:
1. Height ($h$): $75\,cm = 0.75\,m$.
2. Density ($\rho$): $13600\,kg/m^3$.
3. Gravity ($g$): $10\,m/s^2$.
4. Calculation: $P = h\rho g = 0.75 \times 13600 \times 10 = \mathbf{1,02,000\,Pa}$.
Final Answer: The pressure is $1.02 \times 10^5\,Pa$.
A rapid fall in barometric height is a warning of an approaching storm or cyclone. It indicates that the air pressure in that region has dropped significantly, and high-pressure air will soon rush in to fill the gap!
4.0 Upthrust and Archimedes' Principle
When an object is partially or completely immersed in a fluid, it experiences an upward force that opposes its weight. This is why you feel lighter while swimming or why a heavy stone is easier to lift while it is underwater. This upward force is called Upthrust or Buoyant Force.
Archimedes' Principle
The principle states: "When a body is immersed partially or completely in a fluid, it experiences an upthrust which is equal to the weight of the fluid displaced by it."
Upthrust ($F_B$) = Weight of displaced fluid = $V \times \rho \times g$
- $V$: Volume of the submerged part of the body.
- $\rho$: Density of the fluid.
- $g$: Acceleration due to gravity.
4.1 Factors Affecting Upthrust
Upthrust depends on two main factors:
- Volume of submerged body: Larger the volume of the body inside the fluid, greater is the upthrust.
- Density of the fluid: Denser the fluid, greater is the upthrust it exerts. This is why it is easier to float in salt water (higher density) than in fresh water.
Apparent Weight Formula
The weight you feel inside a fluid:
$$W_{apparent} = W_{true} - \text{Upthrust}$$
The "loss in weight" of a body is not a real loss of mass. It is simply the upward support provided by the fluid.
Weight of displaced fluid = Upthrust = Loss in weight.
A body of volume $100\,cm^3$ weighs $5\,N$ in air. It is completely immersed in water. Find the upthrust and its apparent weight. (Density of water = $1000\,kg/m^3$, $g = 10\,m/s^2$)
Solution:
1. Volume ($V$): $100\,cm^3 = 100 \times 10^{-6}\,m^3 = 10^{-4}\,m^3$.
2. Upthrust ($F_B$): $V\rho g = 10^{-4} \times 1000 \times 10 = \mathbf{1\,N}$.
3. Apparent Weight: $W_{true} - F_B = 5\,N - 1\,N = \mathbf{4\,N}$.
Final Answer: Upthrust is $1\,N$ and apparent weight is $4\,N$.
Legend says Archimedes discovered this principle while stepping into his bathtub! He was so excited that he forgot to dress and ran through the streets of Syracuse shouting "Eureka!" (I have found it!).
5.0 Relative Density and Floatation
To understand why a heavy iron ship floats while a small iron nail sinks, we must look at the density of the object compared to the fluid. This comparison is quantified as Relative Density (R.D.), also known as Specific Gravity.
Relative Density (R.D.)
Relative Density is the ratio of the density of a substance to the density of water at $4^\circ C$.
R.D. = $\frac{\text{Density of Substance}}{\text{Density of Water at } 4^\circ C}$
- No Units: Since it is a ratio of similar quantities, R.D. has no unit.
- Pure Water Density: $1000\,kg/m^3$ or $1\,g/cm^3$.
5.1 Law of Floatation
The Law of Floatation is a special case of Archimedes' Principle. It states that a floating body displaces an amount of fluid whose weight is exactly equal to the weight of the whole body.
| Case | Condition | Observation |
|---|---|---|
| Sink | $\text{Weight} > \text{Max Upthrust}$ | Body sinks to the bottom. |
| Float (Just) | $\text{Weight} = \text{Max Upthrust}$ | Body floats completely immersed. |
| Float (Partial) | $\text{Weight} < \text{Max Upthrust}$ | Body floats with a part above surface. |
Fraction Submerged Formula
For a body floating in a liquid:
$$\frac{V_{in}}{V_{total}} = \frac{\rho_{body}}{\rho_{liquid}}$$
Density of ice is approx. $0.9\,g/cm^3$ and sea water is $1.03\,g/cm^3$. This means about 9/10ths of an iceberg remains hidden underwater! This makes icebergs extremely dangerous for ships like the Titanic.
A wooden block of density $600\,kg/m^3$ floats in water ($1000\,kg/m^3$). What fraction of the block's volume is above the water surface?
Solution:
1. Fraction Submerged: $\frac{\rho_{body}}{\rho_{liquid}} = \frac{600}{1000} = 0.6$.
2. Volume Inside ($V_{in}$): $60\%$ of total volume.
3. Volume Above: $Total - V_{in} = 1 - 0.6 = 0.4$.
Final Answer: $2/5$ or $40\%$ of the volume is above the water surface.
A Submarine uses ballast tanks to control its floatation. To dive, it fills these tanks with water (increasing its average density). To surface, it pumps compressed air into the tanks to blow the water out, making it lighter than the displaced water!